This document contains the table of contents for a chemistry textbook, outlining topics such as gases, liquids, atomic structure, chemical bonding, and organic chemistry. It provides an overview of the fundamental concepts covered in each chapter from introduction to chemistry through macromolecules and chemical formulas. The table of contents serves as a high-level outline of the essential information presented in the textbook.

1. Introduction to fundamental concepts of
chemistry. 2
Gases. 23
Liquids. 60
Atomic Structure. 88
Chemical Bonding. 114
Thermochemistry. 154
Chemical equilibrium. 183
Electrochemistry. 220
Reaction kinetics. 258
S and P Block elements. 286
Transition metals. 321
Fundamental principles of
organic chemistry. 340
Chemistry of hydrocarbon. 365
Alkyl halides. 401
Alcohol and phenol. 421
Aldehyde and ketones. 447
TABLE OF CONTENTS
Book Title

2. Carboxylic acid. 475
Macromolecules. 495
Chemical formula. 520
TABLE OF CONTENTS
Book Title

3. 1
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4. 2
Introduction of Fundamental Concepts of
Chemistry
 Atomic mass
 Empirical formula
 Molecular formula
 Concept of mole
 Construction of mole ratios as conversion factors in
Stoichiometry calculations
 Avogadro’s number
 Important assumptions of stoichiometric
calculations
 Stoichiometry
 Limiting reactant
 Percentage yield

5. 3
Relative Atomic Mass
The relative atomic mass is the mass of one atom of an element compared with the mass of one atom of
carbon taken as 12.
Why Carbon is taken as a Standard?
(i) It is a stable element.
(ii) Its isotope 12
6C can be found in the purest form.
(iii) It exists abundantly.
Units of Atomic Mass
The unit used for the atomic mass is Atomic mass unit (a.m.u.).
The mass of 1/12th of an atom of carbon 12 is called atomic mass unit."
Molecular and Empirical Formula
Molecular Formula
Actual number of atoms of different elements present in a molecule is called molecular formula.
Empirical Formula
The simplest whole number ratio of atoms of different elements in a compound is called empirical
formula.
Examples
Compouds Molecular Formula Empirical Formula
Glucose C6H12O6
CH
2
O
Benzene C6H6
CH
Methane CH4
CH
4
Water H2O H
2
O
Did you know?
The Empirical Formula
of sand is SiO2

6. 4
Determination of Empirical Formula:
Empirical formulas can be calculated from the given
1. Percentage composition of the compounds.
% of element =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑎 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
x 100
2. Number of gram atoms
No of gram atoms =
𝐌𝐚𝐬𝐬
𝐀𝐭𝐨𝐦𝐢𝐜 𝐌𝐚𝐬𝐬
x NA
3. Atomic ratio
Divide each number of moles by smallest number of moles to get the mole ratio of elements.
4. Whole number ratio
If ratio is simple whole number, then it gives empirical formula, otherwise multiply with a suitable digit to
get the whole number ratio.
Determination of Molecular Formula:
It can be determined by the following expression:
Molecular Formula = n x Empirical Formula
Here
n =
𝐌𝐨𝐥𝐚𝐫 𝐌𝐚𝐬𝐬
𝐄𝐦𝐩𝐢𝐫𝐢𝐜𝐚𝐥 𝐅𝐨𝐫𝐦𝐮𝐥𝐚 𝐦𝐚𝐬𝐬
Combustion Analysis
It is a quantitative analysis that is used to determine empirical formula of those compounds that contain
carbon, hydrogen and oxygen.
Combustion
It is the burning of a compound in excess of oxygen to form carbon dioxide and water Examples:
(1) Methane combusts in the presence of oxygen
CH
4
+ 2O
2
→CO
2
+ 2H
2
O

7. 5
Figure 1.1: Combustion Analysis
Process:
 Take pre weighed sample in combustion tube fitted in furnace.
 Supply oxygen to burn the compound.
 Hydrogen is converted to H
2
O and is absorbed in Magnesium per chlorate, Mg(ClO
4
)
2
.
 Carbon is converted to CO
2
and is absorbed in 50% KOH.
 The difference in the masses of these absorbers gives us the amounts of H
2
O and CO
2
.
Finding the Percentages of Elements
Following formulae are used to get the percentages of carbon hydrogen and oxygen.
% of carbon =
Mass of CO
2
Mass of Compound
x
12
44
x 100
% of Hydrogen =
Mass of H
2
O
Mass of Compound
x
2.016
18
x 100
The percentage of oxygen is determined by the method of difference.
% of Oxygen = 100 – (% of C + % of H)

8. 6
Concept of Mole
Substance No of particles Quantity Special term
H 6.02 x 1023 atoms 1 mole Gram atom
Cl- 6.02 x 1023 ions 1 mole Gram ion
H2O 6.02 x 1023 molecules 1 mole Gram molecule
NaCl 6.02 x 1023 formula units 1 mole Gram formula
Avogadro’s Number
It is the number of atoms, molecules and ions in one gram atom of an element, one gram molecule of a
compound and one gram ion of a substance, respectively.
Examples
1.008g of H = 1 mole of H = 6.02x10
23
atoms of H
23 g of Na = 1 mole of Na = 6.02x10
23
atoms of Na
18 g of H
2
O = 1 mole of H
2
O = 6.02x10
23
molecules of H
2
O
63 g of NO
3
-1
= 1 mole of NO
3
-1
= 6.02x10
23
ions of NO
3
-1
Relationships
No of atoms of an element =
Mass of element
Atomic mass
x NA
No of molecules of a compound =
Mass of 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
Molecular mass
x NA
No of ions of an ionic specie =
Mass of ion
Ionic mass
x NA

9. 7
Stoichiometry
The branch of chemistry which deals with the quantitative relationship between reactants and products
in a balanced chemical equation is called Stoichiometry.
Conditions to study Stoichiometry
1. All reactants are completely converted into the products.
2. No side reactions occurs.
3. While doing calculations, law of conservation of mass and law of definite proportions are obeyed.
Stoichiometric Amounts:
The amount of reactants and products in a balanced chemical equations are called Stoichiometric
amounts.
e.g. 2H2 + O2 →
2H2O
In above equation 4 g of H2, 32 g of O2 and 36 g of H2O are called Stoichiometric amounts.
Relationships to study stoichiometry
The following type of relationship can be studied with the help of a balanced chemical equation.
(1) Mass - mass relationship:
If we are given the mass of the one substance, we can calculate the mass of the other substance.
(2) Mass - mole relationship or mole - mass relationship:
If we are given the mass of one substance, we can to calculate the moles of others Substance and vice -
versa.
(3) Mass - volume relationship:
If we are given the mass of one substance, we can calculate the volume of the other substance and vice -
versa.
Limiting Reactant
 It is a reactant that is
 Taken in small amount.
 Consumed earlier in a reaction
 Controls (limits) the amount of product.
 Gives minimum amount of product.

10. 8
Example of Sandwiches
1. If we have 30 shami kababs and five breads “having 58 slices”, then we can only prepare 29 sandwiches.
One kabab will be extra. So slices will be the limiting reactant.
2.When 4 g of H2 reacts with 32 g of O2, 36 g of water is produced.
In this reaction there is no limiting reactant because both reactants are in stoichiometric ratio
3.When 6 g of H2 reacts with 32 g of O2, 36 g of water is produced.
In this reaction, formation of H2O is limited by O2 reactant. 2g H2 remains un-reactive therefore in excess
or non-limiting reactant.
Identification of Limiting Reactant
There are three steps to identify the limiting reactant:
 Calculate the number of moles from the given amount of reactants.
 Find out the number of moles of product with the help of balanced chemical equation.
 Identify the reactant which produces the least amount of product as a limiting reactant.
Yield
The amount of product obtained as a result of chemical reaction is called yield.
Theoretical Yield
It is the maximum amount of product that can be produced from given amount of reactants.
A chemical reaction rarely produces the theoretical yield ofproduct.
A chemist determines the actual yield of a reaction through a careful experiment in which the mass of the
product is measured.

11. 9
Actual Yield
It is the amount of product produced when the chemical reaction is carried out in an experiment.
Actual Yield Is Less than Theoretical Yield
Reaction may be reversible.
Side reaction may occur .
Byproduct formation.
Personal error due to inexperienced worker.
Instrumental error due to faulty apparatus.
Mechanical loss of product during separation by filtrations, separation by distillationseparation by
a separating funnel, washing, drying and crystallization is not properlycarried out decreases the
yield.
Efficiency of Reaction
One way of measuring efficiency is by means of percent yield. So efficiency of reaction is expressed in
terms of percentage yield.
Percent yield
Percent yield of product is the ratio of the actual yield to the theoretical yield expressed as a percent.
Percentage yield =
Actual Yield
Theoratical Yield
x 100

12. 10
Assessment 1
1. Mass of 1.5 mole electron is
a. 0.525 mg
b. 0.625 mg
c. 0.725 mg
d. 0.825 mg
2. Glucose and acetic acid have same
a. Molecular mass
b. Molecular formula
c. Empirical formula
d. No of atoms
3. A compound having empirical formula CH2O gives information about
a. Total no of atoms
b. Total number of hydrogen atoms
c. Simple ratio between atoms
d. All of these
4. Which contain Avogadro’s number of particles
a. 23 g of sodium
b. 22 g of carbon dioxide
c. 1 g of hydrogen gas
d. 23 g of calcium
5. A compound having n= 6 and molecular formula mass 78 g/mol, its empirical
formula may be
a. CH2O
b. CHO
c. CH
d. C2H2O
6. No of formula units in 29.25 g of sodium chloride is
a. 6.02x1023
formula units
b. 6.02x1022
formula units
c. 3.01x1023
formula units
d. 3.01x1022
formula units
7. Simple sugars (monosaccharides) generally have the empirical formula
a. CH
b. CH2O
c. C2H6O
d. CHO

13. 11
8. Which has the more number of molecules
a. 1g of hydrogen
b. 1g of carbon dioxide
c. 1g of methane
d. 1g of ammonia
9. The efficiency of the reaction can be checked by percentage yied. Percentage
yield is always found as
a. Theoretical yield/actual yield
b. Actual yield/theoretical yield
c. Actual yield/theoretical yield x 100
d. Calculated from balanced chemical equation
10. Consider the following reaction:
2Mg + O2 →2MgO
Using 0.05 mole of magnesium, how many no of molecules of magnesium oxide
are produced?
a. 3.01x1023
molecules
b. 3.01x1022
molecules
c. 3.01x1021
molecules
d. 3.01x1024
molecules

14. 12
Assessment 2
1. Which of the following is present in 1 mol of CO2:
a. 6.02x1022 molecules of Carbon dioxide.
b. 6.02x1023 atoms of oxygen
c. 1.2x1024 atoms of carbon
d. 3 moles of atoms.
2. Total number of atoms present in one mole of CuSO4.5H2O is:
a. 1 mole atoms
b. 12 mole atoms
c. 21 mole atoms
d. 6 mole atoms
3. The mass of one molecule of water is:
a. 18 g
b. 3x10-26 g
c. 3x10-26 kg
d. Both a and c
4. The no. of covalent bonds present in 9 gm of water are
a. 6.070 x 1022
b. 6.02 x 1023
c. 3.01 x 1024
d. 3.01 x 1024
5. If two compounds have the same empirical formula but different molecular formula,
they may have:
a. Different kind of atoms
b. Different molecular masses
c. Different simple whole number ratio of atoms
d. All of these
6. In which of the following pairs of compounds the ratio of C, Hand O is same:
a. Acetic acid and methyl alcohol
b. Glucose and acetic acid
c. Fructose and benzene
d. All of these
7. Which of the following is the limitation of writing simple chemical equations?
a. Rate of reaction
b. Mass of product produced
c. Prediction of Intermediate
d. Both a and c

15. 13
8. While dealing with the stoichiometry, the following law does not necessarily to be
obeyed?
a. Law of conservation of mass
b. Law of definite proportion
c. Graham’s law of diffusion
d. None of these
9. In a given reaction; 2KOH(aq) + H2SO4(aq) →
K2SO4(aq) + 2H2O(l), which of the
following is not true?
a. 2 mole of KOH produces 2 mole of water
b. In a reaction 2NA of total molecules are produced.
c. In a reaction, 2 moles of water molecules are produced.
d. All of these
10. The percentage composition of carbon in 180 g of glucose is:
a. 40%
b. 6.66%
c. 60%
d. 46.6%

16. 14
Assessment 3
1. 10 moles of H2 reacts with 7 moles of O2 to produce water in the following reaction:
2H2 + O2 →
2H2O
Which of the following is limiting reactant?
a. Only H2
b. Only O2
c. H2O
d. Both H2 and O2
2. Which of the following statement is not true about the yield?
a. Theoretical yield is always greater than actual yield.
b. Theoretical yield is the yield calculated by experiment.
c. Practically inexperience worker may get the lower actual yield.
d. None of these
3. Efficiency of chemical reaction can be checked by:
a. Limiting reactant
b. Theoretical yield
c. Percentage yield
d. All of these
4. For a given reaction, CaCO (s) →
CaO(s) + CO (s); 100g of limestone is decomposed to
give 28g of lime. Percentage yield of the given reaction is:
a. 100%
b. 75%
c. 50%
d. 25%
5. For the given reaction;
Mg (s) + 2HCl (aq)→
MgCl2(aq) + H2(g)
12g of magnesium is made to react with excess of HCl. The volume of hydrogen gas
produced at STP is:
a. 22.414 dm3
b. 2.2414 dm3
c. 11.2 dm3
d. 1.12 dm3
6. A compound (X) having empirical formula C3H3O; molar mass equal to 110 g/mole.
Its molecular formula is:
a. C3H3O
b. C6H6O
c. C6H6O2
d. C6H6
7. No of gram atoms contained in 66 g of C is:
a. 1.82
b. 5.50
c. 11
d. 12

17. 15
8. Hydrogen burns in chlorine to produce hydrogen chloride. The ratio of masses of
reactants in this chemical reaction is:
a. 2:35.5
b. 1:35.5
c. 1:71
d. 2:71
9. Calculate the no of mole of oxygen atoms in 10.6 g of sodium carbonate:
a. 0.1
b. 0.2
c. 0.3
d. 0.4
10. Actual yield is always less than theoretical yield because of the following reason:
a. Production of side product
b. Reversibility of reaction
c. Mechanical loss
d. All of these

18. 16
Assessment 4
1. A polymer of empirical formula CH2 has a molar mass of 14000 g/mol. Its
molecular formula will be:
a. 100 times that of its empirical formula mass
b. 1000 times that of its empirical formula mass
c. 200 times that of its empirical formula mass
d. 2000 times that of its empirical formula mass.
2. In combustion analysis, which of the following can be measured by method of
differences?
a. Percentage of Oxygen
b. Mass of Carbon dioxide
c. Percentage of hydrogen
d. Both a and b
3. In order to find out the molecular formula, n is the ratio of:
a.
Molecular mass
Atomic mass
b.
Molecular mass
Molar mass
c.
Molecular mass
Empirical formula mass
d. All of these
4. 58.5g of the compound (MCl) contains 35.5 g of chlorine. The metal is:
a. Li
b. Na
c. K
d. Rb
5. Which of the following is true about the combustion analysis?
a. The amount of oxygen is determined by the method of difference.
b. 50% KOH is used to absorb CO2
c. Mg(ClO4)2 is used to absorb H2O
d. All of these
6. During combustion analysis, 6g of organic compound is burnt to produce 4g of
carbon dioxide. %age of carbon in it is calculated as:
a. 10%
b. 18%
c. 20%
d. 50%

19. 17
7. 𝑛 =
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠
is unity for which of the following
a. Benzene
b. Glucose
c. Ammonia
d. Hydrogen peroxide
8. The molecular mass of glucose is 180 g/mol. Its empirical formula is CH2O. The value of
“n” is:
a. 1
b. 2
c. 6
d. 3
9. One mole of ethanol and one mole of ethane have equal
a. Mass
b. No of atoms
c. No of electrons
d. No of molecules
10. In a given reaction, SiO2 + 3C →
SiC + 2CO; The amount of silicon carbide produced
when 60 g of sand is heated with excess of carbon:
a. 20 g
b. 40 g
c. 10 g
d. 60 g

20. 18
Assessment 2
Key
Assessment 1
1. D
2. C
3. C
4. A
5. C
6. C
7. B
8. A
9. C
10. B
Assessment 2
1. D
2. C
3. C
4. B
5. B
6. B
7. D
8. C
9. B
10. A

21. 19
Assessment 4
1. A
2. B
3. C
4. C
5. C
6. C
7. B
8. B
9. C
10. D
Assessment 3

22. 20

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24. Learning Objectives:
Properties of gases
Gas laws
Boyle’s law
Charles’s law
General gas equation
Kinetic molecular theory of gases
Ideal gas equation
Gases

25. States of Matter
Matter exists in the following four states:
➢ Gas
➢ Liquid
➢ Solid
➢ Plasma
Note:
➢ The simplest form of matter is the gaseous state.
➢ The most of matter around us (on earth) is in the solid state.
➢ Liquid are less common than solids, gases and plasmas because they can exist only
within a relatively narrow range of temperature and pressure.
Properties of Solids, Liquids and Gases
Sr# Property Gases
1 Volume No definite volume
2 Shape No definite shape
3 Density Very low density
4 Diffusion and effusion Rate of diffusion an diffusion is very high
5 Compressibility Much compressible
6
Expansion
Rate of expansion is very high and sudden expansion is
called Joule Thomson Effect
7 Motion of particles Random motion with high energy
8 Intermolecular forces Very weak intermolecular forces of attraction
Pressure
“The force per unit area is called pressure.”
P = F/A
Units of Pressure:
1. Atmosphere
“The pressure of air that can support 760 mm or 76 cm Hg column at sea level at 0o C is
called an atmosphere.”
OR
“The force exerted by 760 mm or 76 cm Hg column on an area of 1 cm2 at sea level at 0oC is
called one atmosphere.”
1 atm = 760 torr = 760 mm of Hg = 101325 Nm–2 =101325 Pa = 101.325 kPa = 14.7 Psi.

26. Gas law
“The relationships between volume of a given amount of gas and the prevailing
conditions of temperature and pressure are called the gas law.”
Boyle’s law
Statement:
“The volume of a given mass of a gas at constant temperature is inversely proportional
to the pressure applied to the gas.”
Mathematically:
1
V
p
 (at constant temperature and number of moles)
k
V
p
 Here ‘k’ is proportionality constant
PV k
= (When ‘T’ and ‘n’ are constant)
The value of k is different for the different amounts of the same gas.
Another Statement:
“The product of pressure and volume of a fixed amount of a gas at constant
temperature is a constant quantity.”
So, 1 1 2 2
PV k and P V k
= =
Hence 1 1 2 2
PV P V
=
1 1
PV are the initial values of pressure and volume while 2 2
P V are the final values of
pressure and volume.
EXPERIMENTAL VARIFICATION
➢ Take a gas cylinder with a moveable piston.
➢ This cylinder is also attached with a manometer to read the pressure of the gas directly.

27. ➢ Let the initial volume of gas is 1 dm3and its pressure is 2 atmospheres when the piston
has one weight on it.
➢ When the piston is pressed twice with the help of two equal weights, the pressure now
becomes 4 atmospheres.
➢ Similarly, when the piston is pressed with a mass three times greater, then the pressure
becomes 6 atmospheres.
➢ The initial volume of the gas at 2 atmospheres in 1 dm3. It is reduced to 3
1
dm
2
and then
3
1
dm
3
with increase of weights, respectively.
In short,
3 3
1 1
PV 2atm 1dm 2dm atm k
= = =
3 3
2 2
1
P V 4atm dm 2dm atm k
2
=  = =
3 3
3 3
1
P V 6atm dm 2dm atm k
3
=  = =
It is observed that the value of k will remain the same for the same quantity of a gas at the
same temperature. Hence, Boyle’s law is verified.
Graphical explanation of boyle’s law
➢ If we plot a graph between pressure on the x–axis (abscissa) and volume
on the y–axis (ordinate) at a constant temperature, then a curve is
obtained.
➢ This curve is called isotherm (‘Iso’ means same, ‘therm’ mean heat).
➢ On increasing temperature from 0 0C to 25 0C, isotherm moves away
from both axis.
Reason:
➢ At high temperature the volume of the gas has increased.
➢ Similarly, if we increase the temperature further, make it constant and plot another
isotherm, it goes further away from the axes.

28. Graph between
➢
1
V
(inverse of volume) on x–axis and the pressure P on the y–axis then a straight line is
obtained.
➢ It means the pressure and inverse of volume are directly proportional to each other. By
increasing the temperature of the same gas from T1 to T2and keeping it constant, one
can vary pressure and volume.
Graph between P and PV
➢ If a graph is plotted between pressure on x-axis and the product PV on y-axis, a straight
line parallel to the pressure axis is obtained. This straight line indicates that ‘k’ is a
constant quantity.
➢ This type of straight line will help us to understand the non-ideal behavior of gases.
Boyle’s law is applicable only to ideal gases.

29. Charles Law
It is a quantitative relationship between temperature and volume of a gas and was given by
French scientist J. Charles in 1787.
Statement:
“The volume of the given mass of gas is directly proportional to the absolute
temperature at constant pressure.”
Mathematically:
V T
 (at constant pressure and number of moles)
Or V kT
=
Or
V
k
T
=
If temperature is changed from T1 to T2,and volume changes from V1 to V2then
1 2
1 2
V V
k and k
T T
= =
1 2
1 2
V V
constant
T T
= =
Another Statement:
“The ratio of volume to temperature remains constant form same amount of gas at
same pressure.”
Experimental Verification
➢ Consider a certain amount of a gas enclosed in a cylinder fitted with a moveable piston.
➢ The volume of the gas is V1 and its temperature T1.
➢ When the gas in the cylinder is heated, both volume and temperature of the gas
increase.
➢ The new values of volume and temperature are V2 and T2 respectively.
Result:
The experiment shows that:
1 2
1 2
V V
k and k
T T
= =

30. 1 2
1 2
V V
constant
T T
= =
Hence, Charles’s law is verified.
Derivation of Absolute Zero
Absolute Zero:
“The hypothetical temperature at which the volume of a gas is supposed to become
zero if the gas remains in gaseous state.”
0K 237.16 C 459 F
= −  = 
Quantitative Definition of Charles’s Law
“At constant pressure, the volume of the given mass of a gas increases or decreases by
1/273 of its original volume at 0o C for every 1o C rises or fall in temperature respectively.”
General Equation for Volume of Gas:
The equation to know the volumes of the gas at various temperatures is:
t o
t
V V 1
273
 
= +
 
 
Where t
V = volume of gas at temperature T.
o
V = volume of gas at 0o C.
t = temperature of centigrade or Celsius scale.
Explanation:
➢ If a gas is warmed by 1o C, it expands by 1/273 of its original volume at 0o C. since,
original volume is 546 cm3.
➢ So for 1o C rise in temperature, 2 cm3increase the volume will take place.
➢ 2 cm3is the 1/273 of 546 cm3.
➢ Similarly for 100o C rise in temperature, a change of 200 cm3will take place.
➢

31. Volume-Temperature Data for a Given Amount of a Gas at Constant Pressure
Volumes
(cm3)
Celsius Temperature
(oC)
Temperature
(K)
𝐕
𝐓
= K cm3 K–1
1092 273 546 2
846 150 423 2
746 100 373 2
646 50 323 2
566 10 283 2
548 0 274 2
546 0 273 2
544 –1 272 2
526 –10 263 2
400 –73 200 2
346 –100 173 2
146 –200 73 2
0 –273 0 2
Calculation of Volume at 0oC and -273oC:
at
546 1
V 546
273

= +
3
546 2 248 cm
= + =
at 273 C
546 273
V 546
273
− 
−
= +
546 546
= −
3
0 cm
=
Conclusion:
➢ It is observed from above table shows that at 0o C, the volume of the gas taken is 546
cm3.
➢ This is twice of 273 cm3, and is being supposed for the sake of convinced of
understanding.
➢ At 273o C, the volume of the gas has doubled (1092 cm3) and it should become
practically zero at –273o C.
➢ The volume does not increase corresponding to increase in temperature on Celsius
scale. e.g. the increase in temperature from 1 2
T 10 C to T 100 C
=  =  increase the
volume from 3 3
1 2
v 566 cm to v 746cm .
= =

32. Application of Charles’s Law
By Using Celsius Scale:
Applying Charles’s law:
1 2
1 2
V V
T T
=
➢ It is observed that, two sides of equation
are not equal.
➢ So it is concluded that, Charles’s law is
not being obeyed when temperature is
measured on the Celsius scale.
By using Kelvin Scale:
➢ Charles’s law is obeyed when the
temperature is taken on the Kelvin scale.
➢ For example, at T1 = 283 K (10o C), the
volume v1 = 566 cm3, while at T2 = K
(100oC), the volume is v2 = 746 cm3.
According to Charles’s law:
1 2
1 2
V V
k
T T
= =
566 746
2 k
283 373
= = =
So Kelvin scale was developed to verify Charles’s law.
Graphical Explanation of Charles Law
If we plot a graph between temperature on x-axis and the volume of one mole of an ideal gas
on y-axis, we get a straight line in which cuts the temperature axis at –273.16oC.
Lowest possible temperature:
➢ This can be possible only if we extrapolate the graph upto -273.16oC.
➢ This temperature is the lowest possible temperature, which would have been achieved if
the substance remains in the gaseous state.
➢ Actually, all the gases are converted into liquids above this temperature.

33. Effect of mass on slope:
➢ Greater the mass of gas taken, greater will be the slop of straight line.
➢ The reason is that greater the number of moles, greater the volume occupied. All these
straight lines when extrapolated meet at a single point of –273.16oC (0K).
Behavior of a Real Gas:
➢ It is apparent that this temperature of –273.16oC will be attained when the volume
becomes zero.
➢ But for a real gas, the zero volume is impossible which shows that this temperature
cannot be attained for a real gas.
➢ This is how we recognize that –273.16oC must represent the coldest temperature.
Scales of Thermometry
The following three scales are used for temperature measurements.
i. Centigrade Scale or Celsius Scale:
➢ It has a zero mark (0oC) for the temperature of ice at one atmospheric pressure.
➢ The mark 100oC indicates temperature of boiling water at one atmospheric pressure.
➢ The space between these temperature marks is divided into 100 equal parts and each
part is 1oC.
ii. Fahrenheit Scale:
➢ The melting point of ice at 1 atmospheric pressure has a mark 32oF and that of
boiling water is 212oF.
➢ The space between these temperature marks is divided into 180 equal parts and
each part is 1oF.
iii. Kelvin Scale or Absolute Scale:
➢ The melting point of ice at 1 at 1 atmospheric pressure 273 K.
➢ Water boils at 373 K or more precisely at 373.16K.

34. Interconversion of Temperature Scales:
The following relationships help us to understand the interconversion of various scales of
temperature.
➢ oC to K: K = oC + 273.16
➢ oFtooC:oC = 5/9 (oF–32)
➢ oC to oF: oF = 9/5 (oC) + 32
General Equation
According to Boyle’s law:
1
V
P
 (When n and T are constant)
According to Charles’s law:
V T
 (When n and P are constant)
According to Avogadro’s law:
V n
 (When P and T are constant)
By combining all the gas laws:
nT
V
P

nT
V constant
P
=

35. General Gas Constant:
The constant suggested is R which is called general gas constant.
nT
V R
P
=
PV nRT
=
➢ The general gas equation shows that if we have any quantity of an ideal gas then the
product of its pressure and volume is equal to the product of number of moles, general
gas constant and absolute temperature.
For one mole of a gas:
➢ For one mole of a gas, the general gas equation is:
n 1
=
PV
PV RT or R
T
= =
It means that ratio of PV to T is a constant quantity (molar gas constant).
Hence,
1 1 2 2
1 2
PV P V
R R
T T
= =
Therefore,
1 1 2 2
1 2
PV P V
T T
=
Applications of The General Gas Equation
i. Calculation of the Density of a Gas:
The general gas equation is
PV nRT
=
n = Number of moles of the gas
n =
mass of the gas in g m
molar massofthegas M
=
Hence,
m
PV RT
M
=
m
PM RT
V
=

36. mass
PM dRT d
Volume
PM
d
RT
 
= =
 
 
=
ii. Calculation of the Mass of a Gas:
PM
d
RT
=
m PM
V RT
=
PMV
m
RT
=
Ideal Gas Constant (R)
Avogadro’s principle is used to calculate values and units of R. According to this principle
“The volume of one mole of an ideal gas at STP (one atmospheric pressure an d273.16K) is
22.414 dm3.”
➢ Its value depends upon the units chose for pressure, volume and temperature.
➢ General gas equation is used to calculate the value of R.
➢ To evaluate ‘R’, the general gas equation can be written as
PV nRT
=
PV
R
nT
=
a. When P in atm and V in dm3:
P = 1 atm n = 1 mole
V = 22.414 dm3 T = 273.16 K
3
PV 1atm 22.414dm
R
nT 1mol 273.16K

= =

3 -1 -1
R = 0.0821 atmdm mol K
Physical meanings of value of R:
➢ If we have one mole of an ideal gas at 273.16 K and one atmospheric pressure and its
temperature is increased by 1K, then it will absorb 0.0821 dm3 atm is the unit of energy
in this situation.

37. ➢ Hence, the value of R is a universal parameter for all the gases.
➢ It tells us that the Avogadro’s number of molecules of all the ideal gases have the same
demand of energy.
b. When P in torr or mmHg and V in dm3:
P = 760 torr n = 1 mole
V = 22.414 dm3 T = 273.16 K
3
PV 1atm 22.414dm
R
nT 1mol 273.16K

= =

s
1 1
K
− −
3
R = 62.4 dm torr or mmHg mol
c. When P in torr or mmHg and V in cm3:
P = 760 torr n = 1 mole
V = 22.414 dm3 T = 273.16 K
3
PV 760torr 22414cm
R
nT 1mol 273.16K

= =

1 1
K
− −
3
R = 62400 cm torr or mmHg mol
d. Value and Units of R in SI:
The SI units of pressure are Nm–2 and of volume are m3. By using Avogadro’s principle.
n = 1mol T = 273.16K
V = 22.414 dm3 = 0.022414 m3
3
PV 1atm 22.414dm
R
nT 1mol 273.16K

= =

-1 -1 -1
R = 8.3143 NmK =8.3143JK mol (1Nm = 1J)
Keep in mind that, wherever the pressure is given in Nm–2 and the volume in m3, then the
value of R used must be 8.3143 JK–1mol–1.
e. Since 1cal. = 4.18 J:
So, 1 1
8.3143
R 1.987 calK mol
4.18
− −
= =
f. When energy is expressed in ergs:
Since Ij 10erg
=
So, 1 1
R 8.3143 J.mol K
− −
=
So, 7 -1 -1
R = 8.3143×10 ergsmol K

38. Avogadro’s Law
Statement:
“Equal volumes of all the ideal gases at the same temperature and pressure contain
equal number of molecules.”
Explanation:
One mole of an ideal gas at 273.16 K and one atmospheric pressure (STP) has a volume of
22.414 dm3 and one mole of a gas has Avogadro’s number of molecules.
So, 22.414 dm3 of ideal gas at STP will have Avogadro’s number of molecules i.e., 23
6.02 10

molecules.
Examples:
3 23
2
H 2g 1 mole 22.414dm at STP 6.02 10 molecules
= = = = 
3 23
2
O 32g 1 mole 22.414dm at STP 6.02 10 molecules
= = = = 
3 23
2
N 28g 1 mole 22.414dm at STP 6.02 10 molecules
= = = = 
3 23
4
CH 16g 1 mole 22.414dm at STP 6.02 10 molecules
= = = = 
➢ So, one mole of all gases at STP will have same volume of 22.414 dm3 and same number
of molecules i.e., 23
6.02 10
 but their masses are not equal.
➢ Similarly if we have one dm3 of H2, O2, N2 and CH4in separate vessels at STP, then they
have number of molecules i.e., 22
2.68 10
 molecules.
➢ Although, oxygen molecule is 16 times heavier than hydrogen, but this does not disturb
the volume occupied, because molecules of the gases are widely separated from each
other at STP.
➢ One molecule is approximately at a distance of 300 times its own diameter from its
neighbor at room temperature. Analytical chemistry is the science of chemical
characterization.
Kinetic Molecular Theory Of Gases (Kmt)
“A set of postulates that describes the nature and behavior of an ideal gas is called
kinetic molecular theory of gases.”
Fundamental postulates:
➢ Every gas consists of a large number of very small particles called molecules. Gases like
He, Ne, Ar have mono-atomic molecules.
➢ The molecules of a gas move haphazardly (randomly), colliding among themselves and
with the walls of the container and change their directions.

39. ➢ The pressure exerted by a gas is due to collisions of its molecules with the walls of a
container. The collisions among the molecules are perfectly elastic.
➢ The molecules of a gas are widely separated from one another and there are sufficient
empty spaces among them.
➢ The molecules of a gas have no forces of attraction for each other.
➢ The actual volume of molecules of a gas is negligible as compared to the volume of gas.
➢ The motion imparted to the molecules by gravity is negligible as compared to the effect
of the continued collisions between them.
➢ The average kinetic energy of the gas molecules varies directly as the absolute
temperature of the gas.
Clausius’ Kinetic Equation
On the basic of kinetic molecular theory, R.J. Clausius deduced an expression for the pressure of
an ideal gas as:
2
1
PV mNc
3
=
Where P = Pressure, V = Volume, m = Mass of one molecule of the gas
N = Number of molecules of gas in the vessel
2
c = Mean square velocity
Mean Square Velocity:
➢ All the molecules of a gas under the given conditions don’t have the same velocities.
Rather different velocities are distributed among the molecules.
➢ It is explained in Maxwell’s law of distribution of velocities. If there are n1 molecules
with velocity c1, n2 molecules with velocity c2 and so on then:
2 2 2
2 1 2 3
1 2 3
c c c .....
c
n n n .....
+ + +
=
+ + +
In this reference,
1 2 3
n n n ..... N
+ + + =
Root Mean Square Velocity:
➢ 2
c is the average of the squares of all the possible velocities.
➢ When we take the square root of this 2
c , then it is called root mean square velocity
(Crms). So,
2
rms
C c
=
The expression for the root mean square velocity deduced from the kinetic equation is written
as follows:
rms
3RT
C
M
=

40. Where,
rms
C = Root mean square velocity
M = Molar mass of the gas
T = Temperature
Conclusion:
The above equation is a quantitative relationship between absolute temperature and the
velocities of the gas molecules.
According to this equation, higher the temperature of a gas, greater the velocities.
Explanation of Gas Laws From Kinetic Theory Of Gases
i. Boyle’s Law:
According to one of the postulates of kinetic molecular theory of gases,
“The kinetic energy is directly proportional to the absolute temperature of the gas.”
The kinetic energy of ‘N” molecules is 2
1
mNc
2
So, 2
1
mNc T
2

2
1
mNc kT .....(i)
2
=
Where ‘k’ is the proportional constant.
According to the kinetic equation of gases.
2
1
PV mNC
3
=
Multiplying and dividing by 2 on right hand side
2
2 1
PV mNc
2 3
 
=  
 
2
2 1
PV mNc ........(ii)
3 3
 
=  
 
Putting eq. (i) into eq. (ii)
2
PV kT ........(iii)
3
=
If the temperature (T) is constant then right hand side of eq. (iii)
2
kT
3
is constant.
Let that constant be ‘k’.
PV = k’ (Boyle’s law)
Conclusion:
At the constant temperature and number of moles, the product PV is a constant quantity.

41. ii. Charles’s Law:
According to one of the postulates of kinetic molecular theory of gases,
“The kinetic energy is directly proportional to the absolute temperature of the gas.”
The kinetic energy of ‘N’ molecules is 2 1
c mN
2
So, 2
1
mNc T
2

2
1
mNc kT .....(i)
2
=
Where ‘k’ is the proportional constant.
According to the kinetic equation of gases
2
1
PV mNc
3
=
Multiply and dividing by 2 on right hand side
2
2 1
PV mNc
2 3
 
=  
 
2
2 1
PV mNc ......(ii)
3 2
 
=  
 
Putting eq. (i) into eq. (ii)
2
PV kT ......(iii)
3
=
2k
V T
3P
 
=  
 
At constant pressure,
2k
k
3P

= (a new constant)
Therefore, V k T

=
Or
V
k (Charles's law)
T

=
Conclusion:
At a constant pressure and number of moles, the ration of V/T is a constant quantity.
iii. Avogadro’s Law
Consider two gases 1 and 2 at same pressure P and having same volume V. Their number of
molecules are N1 and N2, masses of molecules are m1 and m2 and mean square velocities
are 2
1
c and 2
2
c respectively. Their kinetic equation cam be written as follows:
For gas 1: 2
1 1 1
1
PV m N c
3
=

42. For gas 2: 2
2 2 2
1
PV m N c
3
=
Equalizing:
2 2
1 1 1 2 2 2
1
m N c m N c
3
=
Hence 2 2
1 1 1 2 2 2
m N c m N c ........(i)
=
When the temperature of both gases in the same, their mean kinetic energies per molecule
will also be same, so
2 2
1 1 2 2
1
m c m c
2
=
2 2
1 1 2 2
m c m c ........(ii)
=
Dividing eq. (i) by eq. (ii)
2 2
1 1 1 2 2 2
2 2
1 1 2 2
m N c m N c
m c m c
=
1 2
N = N
Conclusion:
Equal volumes of all the gases at the same temperature and pressure contain equal number
of molecules, which is Avogadro’s law.
iv. Graham’s Law of Diffusion
According to kinetic equation
2
1
PV mNc .....(i)
3
=
If we take one mole of a gas having Avogadro’s number of molecules (N=NA) then the
equation (i) can be written as:
2
A
1
PV mN c
3
=
Or 2
A
1
PV Mc .....(ii) (M mN )
3
= =
Where M is the molecular mass of the gas.
Or 2 3PV
c
M
=
Taking square root.
2 3PV
c
M
=

43. 2 3P
c
M / V
=
2 3P M
c d
d V
 
= =
 
 
‘V’ is the molar volume of the gas at given conditions. Since the root mean square velocity
of the gas is proportional to the rate of diffusion of he gas.
2
c r

So,
3P
r
d
=
At constant pressure,
1
r
d

This Graham’s law of diffusion.
Kinetic Interpretation of Temperature
The kinetic gas equation is givens as:
2
1
PV mNc ......(a)
3
=
Where P = Pressure
V = Pressure
m = Mass of one molecule of the gas
N = Number of molecules of the gas
2
c = Mean square velocity
The average kinetic energy associated with one molecule of a gas due to its translational
motion is given below:
2
K
1
E mc ......(b)
2
=
In the above equation EK represents the average translational kinetic energy of gas molecules.
2
2 1
PV mNc
2 3
= 
2
2 1
PV N mc .....(c)
3 2
 
=  
 
Putting equation (b) into equation (c), we get
K
2
PV NE .....(d)
3
=
If we use one mole of a gas, then N = NA
Now put the value of N in the equation (d)

44. A K
2
PV N E .....(e)
3
=
According to the general gas equation for one mole of a gas
A K
2
RT N E
3
=
K
A
3R
E
2N T
=
But
A
3R
2N
is a constant quantity. So the above equation can be written as:
K
E Constant T
= 
K
E T

➢ The above equation shows that absolute temperature or Kelvin temperature of a gas is
directly proportional to the average translational kinetic energy its molecules.
➢ This suggests that a change in temperature mean change in the intensity of molecular
motion
Flow of Heat:
➢ When het flows from one body to another body, the molecules in the hotter body give
up some of their kinetic energy through collisions to the molecules in the colder body.
➢ This process of flow of heat continues until the average translational kinetic energies of
all the molecules in both bodies become equal.
Interpretation of temperature of gases, liquids and solids
➢ In gases and liquids, temperature is the measure of average translational kinetic
energies of molecules.
➢ In solids,Where molecules cannot move freely temperature becomes a measure of
vibration kinetic energy.
Interpretation of Absolute Zero:
➢ In the light of kinetic molecular interpretation, Absolute temperature is the temperature
at which molecular motion ceases.
➢ The absolute zero is unattainable.
➢ However current attempts have resulted in a temperature as low as 10–5K.
Non-Ideal Behavior Of Gases

45. The gases that obey gas laws (Boyle’s law and Charles’s law) and resulting general gas
equation under all conditions of temperature and pressure are called ideal gases.
The gases that donot obey gas laws (Boyle’s law and Charles’s law) and resulting general
gas equation under all conditions of temperature and pressure are called non-ideal gases.
Explanation:
Compressibility Factor:
➢A graph is plotted between pressure on x-axis and PV/RT on y-axis for ideal gas.
The factor PV/RT is called compressibility factor.
➢Its value is unity for 1mole of an ideal gas.
Behavior of Ideal Gas:
For an ideal gas, increase of pressure
decreases the volume such that PV/RT remains
constant at a constant temperature and a straight
line is obtained parallel to x-axis.
Behavior of Real Gases at 0o
C:
All the real gases have been found to
show marked deviations from this behaviour as
discussed below.
He Gas:
Graph for He gas goes along with expected
horizontal line to some extent but goes above this
line at very high pressures. It means that at very
high pressure, the decrease in
volume is not according to general gas equation and the value of PV/RT has increased from the
expected values. With this type of behavior, we would say that the gas is non-ideal.
H2 Gas:
In the case of H2, the deviation starts even at low pressure in comparison to He.
N2 Gas:
N2 shows a decrease in PV/RT value at the beginning and shows marked deviation even
at low pressure than H2.
CO2 Gas:
CO2 gas has a very strange behaviour as it is
evident from the graph.
Limitations For Gases
2.00
1.50
1.00
0.5
PV
RT
0 200 400 600 800 1000
CO
N
2
2
H
He
2
Ideal
gas
T = 0
o
C or 273K

46. The extent of deviation of these four gases
shows that these gases have their own limitations
for obeying general gas equation. It depends upon
the nature of the gas that at which value of
pressure, it will start disobeying.
Behavior of real gases at 100 o
C
If the behaviour of all these four gases at
elevated temperature (100oC) is studied, then the
graph comes closer to the expected straight line
and the deviations are shifted towards higher
pressure. This means that the increase in
temperature makes the gases ideal.
Conclusion:
This discussion on the basis of experimental observations convinces us that:
1. Gases are ideal at low pressure and non-ideal at high pressure.
2. Gases are ideal at high temperature and non-ideal at low temperature.
Note: The gases are ideal under the conditions where they have negligible forces between their
molecules and vice versa.
Causes for Deviation From Ideality
In 1873, Van der Waal’s attributed the deviation of real gases from ideal behaviour to
two of the eight postulates of kinetic molecular theory of gases.
Faulty Postulates Of Kinetic Molecular Theory:
The faulty postulates of kinetic molecular theory are as follows:
1) There are no forces of attraction among the molecules of a gas.
2) The actual molecular volume of a gas is negligible as compared to volume of the
vessel.
Forces Of Attraction:
When the pressure on a gas is high and the
temperature is low, then the attractive forces
among the molecules become significant, so the
ideal gas equation i.e. PV = nRT does not hold.
Actually, under these conditions, the gas does not
remain ideal.
`
Actual Volume Of Gas Molecules:
Volume of the
molecules of a
gaseous state
(a) (b)
2.00
1.50
1.00
0.5
0 200 400 600 800 1000
P (atm)
CO
N
2
2
H
He
2
Ideal
gas
PV
RT
T = 100 C or 373K
o

47. The actual volume of the molecules of a gas is usually very small as compared to the
volume of the vessel and hence, it can be neglected. This volume, however, does not remain
negligible when the gas is subjected to high pressure as shown in the figures.
Vander Waal’s Equation For Real Gases
Vander Waal’s pointed out that both pressure and volume factors in ideal gas equation needed
correction to make it applicable to the real gases.
1) Volume Correction:
Volume of the gas molecules cannot be ignored in the vessel. They do occupy certain
space. Because the molecules of a real gas are rigid spherical particles which possess a definite
volume. Hence, the actual volume of molecules cannot be ignored in the highly compressed
gas. This volume is called effective volume of molecules if we have taken initially one mole of
the gas, then the effective volume is represented by ‘b’.
Free Volume:
The volume available to the gas molecules is the volume of the vessel minus the volume
of gas molecules. This available volume is also called free volume. This was considered
to be the ideal volume.
Vfree = Vvessel – b ---------------------- (1)
‘b’ is also called excluded volume. It depends upon the size of the gas molecules.
Vi = V – b for 1 mole.
Vi = V – nb for total no. of moles.
Minimum volume occupied by 1mole of highly compressed gas in gaseous state is called effective
volume. It is also called excluded volume or incompressible volume. It is represented by ‘b’.
The volume occupied by the 1mole gas molecules in their closest approach is called actual volume. It is
represented by Vm.
Effective volume (b) is four times greater than actual volume (Vm).
b = 4Vm
2) Pressure Correction:
A molecule in the interior of the gas is attracted by other molecules on all sides. These
forces of attractions cancel the effect of each other. When a molecule strikes the wall of a
container, it experiences a force of attraction towards other molecules in centre of gas. This
decreases the impact of its force on the wall.
Decrease of pressure due to the attractive forces:
Consider the molecule ‘A’ which is unable to create pressure on the wall due to
presence of attractive forces of ‘B’ type molecules. This is shown in the figure below.
Let the observed pressure on the wall
of container is P. This pressure is less
than the ideal pressure Pi (calculated
A
B
B
B
B
A
B
B
B
Inward
pull on A
Molecular attractions
are balanced
Wall of
vessel
Velocity of A
reduced B
B
B
A

48. from general gas equation) by an
amount P/. So,
P = Pi – P/
Pi = P + P/-------(i)
Pi is the true kinetic pressure if forces
of attractions would have been
absent.
P is the pressure measured and called actual pressure. As the volume increases, P
decreases.
P/ is the amount of pressure lessened (decreased) due to attractive forces. P/ is expressed
in terms of a constant ‘a’ which accounts for attractive forces. That is why ‘a’ is
called Co-efficient of attraction for one mole of gas. P/ for 1mole of a gas can be
represented as.
P/ = 2
a
V
How to Prove It:
P/ is determined by the forces of attraction between molecules of type A (which are
striking the wall of the container) and molecules of type B (which are pulling them
inward). The net force of attraction is proportional to the concentrations of A type and B
type molecules.
P/  CA.CB
Let n is the number of moles of A and B separately and total volume of both types of
molecules is ‘V’. The n/V is moles dm-3 of A and B, separately.
P/  n n
.
V V
P/ 
2
2
n
V
P/
2
2
an
=
V
P/ = 2
a
V
-----------(ii) (for 1 mole n = 1)
This ‘a’ is a constant of proportionality and is called co-efficient of attraction or
attraction per unit volume. It has a constant value for a particular real gas.
The value of lessened pressure P/ is greater when
(i) Attractive forces among the gas molecules are stronger
(ii) Volume of vessel is smaller

49. Thus, effective kinetic pressure of a gas is given by Pi, which is the pressure if the gas
would have been ideal.
Put the value of the P/ from eq. (ii) into eq. (i) to get Pi
Pi = P + 2
a
V
Once the corrections for pressure and volume are made, the general gas equation for
one mole of a gas can be constructed by taking pressure as (P + 2
a
V
) and volume as (V -
b).
Now,
(P + 2
a
V
) (V - b) = RT
For ‘n’ moles of a gas
(P +
2
2
a n
V
) (V - nb) = nRT
This is called Van der Waal’s equation. ‘a’ and ‘b’ are called Van der Waal’s constants.
Units of ‘a’:
Since, P/ =
2
2
an
V
So, a =
2
2
P'V
n
i) In common units, pressure is in atm and volume is in dm3
a =
3 2
2
atm×(dm )
(mol)
a = atm.dm6.mol-2
ii) In SI units, pressure is in Nm-2 and volume in m3
a =
-2 3 2
2
Nm ×(m )
(mol)
a = Nm-2. m6.mol-2
a = Nm+4mol-2
Unit of ‘b’:
bis excluded or incompressible volume /mol-1 of gas, b = V / n. Hence, its units should be
i) Common units: b =dm3 mol-1
ii) SI units: b = m3mol-1

50. Assessment 1
1. At STP, Carbon dioxide bubbles through water and rises up because of:
a. High External pressure
b. Low Density
c. High molar mass
d. Less no of atoms
2. Which of the following has more ability to diffuse?
a. Carbon dioxide gas
b. Red ink
c. Ice vapours
d. Hydrogen gas
3. The pressure of air that can support 760 mmHg column at sea level, is called:
a. Standard pressure
b. Atmospheric pressure
c. Sea level pressure
d. All of these
4. The most widely used unit in Engineering work which is equivalent to one atm
a. 1.47 psi
b. 14.7 psi
c. 17.4 psi
d. 17psi
5. The SI unit of pressure is usually expressed in terms of
a. Nm2
b. torr
c. N/m2
d. atm
6. Intermolecular forces are the cohesive forces of attraction by which molecules cling to each
other. The strength of these forces in gases are
a. Strong
b. Stonger
c. Weak
d. Moderate
7. The human body temperature is 98.6 0F. In centigrade it is
a. 320C
b. 370C
c. 320C
d. 31 0C

51. 8. Robert Boyle investigated the behavior of gases experimentally to explain which of the
following relationships
a. Volume is proportional to pressure at constant temperature
b. Volume is directly proportional to temperature at constant pressure
c. Volume is inversely proportional to pressure at constant temperature
d. Volume is proportional to no of moles at constant P and T
9. Keeping the pressure constant, the temperature of an ideal gas is changed from 10 K to 30 K,
its volume changes from 1 dm3 to
a. 1.2 dm3
b. 2.4 dm3
c. 3.0 dm3
d. 4.8 dm3
10. The number of molecules in 0.0224m3 of oxygen gas at 25°C and 1 atm pressure
a. NA
b. NA/2
c. 2NA
d. 1.5NA

52. Assessment 2
1. Gases are compressed by applying pressure on them. Why compression is significant and
easy in case of Gases than liquids and solids
a. Stronger attractive forces
b. Low densities
c. Large empty spaces among gas molecules
d. Large available space to gas
2. Keeping the pressure constant, the temperature of gas is changed from 0 0C to 546 K, its
volume changes from 1 dm3 to
a. 2.0 dm3
b. 2.4 dm3
c. 3.6 dm3
d. 4.8 dm3
3. The following curve is obtained when we plot a graph between:
a. Pressure on x-axis and volume on y-axis keeping temperature changed
b. Pressure on x-axis and pv on y-axis keeping temperature unchanged
c. Pressure on x-axis and 1/v on y-axis keeping temperature unchanged
d. Pressure on x-axis and volume on y-axis keeping temperature unchanged
4. General gas equation is used to determine the density of gas. Density of the gas can be
measured as
a. d =PM/R
b. d = nRT
c. d = PM/RT
d. d = P/RT
5. The mass of 22.414 dm3 of ammonia at 00C and 760 torr
a. 0.907g
b. 9.07g
c. 17g
d. 1.7g
6. At constant temperature, density changes with pressure. If pressure is decreased then
density will
a. remains same
b. decreases
c. change but not significantly
d. increases three times

53. 7. Which of the following is true when a gas, enclosed in a cylinder at standard temperature and
pressure having volume 22.414 dm3, is heated to 298K?
a. The kinetic energy of gas molecules increase
b. Collisions between molecules decrease
c. Volume of gas is decreased
d. Mass of gas is increased
8. According to kinetic molecular theory of gases, gas molecules collide with each other and with
the walls of the container. The collisions among them are perfectly elastic. Which of the
following is true when the gas molecules undergo elastic collisions?
a. Energy of system is changed
b. Kinetic energy is changed
c. Potential energy is changed
d. Total energy of system remains same
9. Root mean square velocity is determined by taking square root of the mean square velocity of the
gas molecules. Its expression is:
a. √
3𝑅
𝑀
b. √
3𝑅𝑇
𝑀
c. √
3𝑅𝑇
𝜋𝑀
d. √
3𝑅𝜋
𝑀
10. Which of the following curve can elaborate the ideal behaviour of given mass of gas at
constant temperature?
a.
b.
c.
d.

54. Assessment 3
1. Kinetic interpretation of temperature was done by Clausius with the help of kinetic gas
equation. According to him, temperature is the measure of _____in case of liquids and gases
is:
a. Average vibrational kinetic energy
b. Average translational kinetic energy
c. Average rotational kinetic energy
d. Both a and b
2. According to kinetic interpretation, the temperature is the measure of average translational
kinetic energy of molecules of gas. The expression of EK is:
a.
2𝑅𝑇
2𝑁𝐴
b.
2𝑅𝑇
3𝑁𝐴
c.
3𝑅𝑇
2𝑁𝐴
d.
2𝑅𝑇
2𝑁𝐴
3. Kinetic interpretation of temperature suggests that a change in temperature has results the
change in:
a. Direction of motion of particles
b. Intensity of motion of molecules
c. Behaviour of molecules
d. Shape of molecules
4. A gas that does not obey gas law and kinetic molecular theory is called non ideal gas. At which
conditions the gas shows such non ideal behavior.
a. Low temperature
b. High pressure
c. High temperature
d. Both a and b
5. Which of the following is an ideal gas?
a. Nitrogen
b. Carbon dioxide
c. Helium
d. None of these
6. The least value of ‘a’ for H2 is due to its:
a. More polar character
b. Small size
c. Non-polar character
d. Both b and c

55. 8. 2.016 g of hydrogen gas will occupy the volume of ______ at the closest approach in the gaseous
state:
a. 133 cm3
b. 266 cm3
c. 22414 cm3
d. 24000 cm3
9. Excluded volume according to Vander Waal’s equation is equal to:
a. Vm
b.
𝑉𝑚
4
c. 4Vm
d.
4
𝑉𝑚
10. The factor
𝑝𝑣
𝑛𝑅𝑇
is known as the compressibility factor which has the value of ____ for the gas
showing ideal behaviour.
a. 1
b. 2
c. 3
d. 4
11. Excluded volume is _____ times the actual volume of molecules:
a. ½
b. Two
c. Three
d. Four

56. Assessment 4
1. General gas equation needs corrections for gas deviate from ideal behavior
a. Mass correction
b. Volume correction
c. Pressure correction
d. Both b and c
2. The following may be resulted because of non ideal behavior
a. Intermolecular attraction
b. Finite volume
c. Infinite volume
d. Both a and b
3. Which of the following temperature favour more ideal behavior of gases
a. 273K
b. 373K
c. 473K
d. 573K
4. Units of excluded volume b exhibited by non ideal gas at STP
a. volume/mol
b. dm3/mole
c. m3/mole
d. all of the above
5. Among the given non ideal gases which of the following has least value for coefficient of attraction
“a”
a. Hydrogen
b. Oxygen
c. Nitrogen
d. Carbon dioxide
6. The value of excluded volume “b” for hydrogen gas is
a. 0.0266 dm3
b. 0.0318 dm3
c. 0.0391 dm3
d. 0.0428 dm3

57. 12. In general gas equation, “R” is the gas constant whose value depends upon the:
a. composition of gas
b. nature of gas
c. units of measurements of variables
d. pressure of the gas
13. One atmospheric pressure is
a. 76 torr
b. 76 mmHg
c. 101325 pa
d. 101.523 Kpa
14. Which parameter is kept constant while studying the relationship between pressure and volume
of gas
a. Mass of gas
b. No of moles
c. Temperature
d. All of these
15. The weather balloons are filled with helium gas. If a small balloon has one dm3 of gas at rtp,
calculate mass of gas filled in the balloon
a. 1/6 g
b. 1/3 g
c. 1/2g
d. ¼ g

58. Key
Assessment 1
Assessment 2
1. b
2. d
3. d
4. b
5. c
6. c
7. b
8. c
9. c
10. a
1. c
2. a
3. d
4. c
5. c
6. b
7. a
8. d
9. b
10. c

59. Key
Assessment 3
Assessment 4
1. b
2. c
3. b
4. d
5. d
6. d
7. b
8. c
9. a
10. d
1. d
2. d
3. d
4. b
5. a
6. a
7. c
8. c
9. d
10. a

60. Oldest, Largest and Most Credible Platform
MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org

61. Learning Objectives:
Properties of liquids
Intermolecular forces
Hydrogen bonding
Vapor pressure
Boiling point and external pressure
Liquids

62. Liquids
The existence of matter in our surrounding in different states (solid, liquids and gas) depends upon the
strength of intermolecular forces between the constituent particles. Intermolecular forces are responsible
for a substance to be solid, liquid or gas.
Properties of liquids:
i) Liquids have definite volume.
ii) Liquids donot have definite shape. Instead, they adopt the shape of container because liquids
molecules slip over each other.
iii) Liquid molecules are in constant state of motion. Evaporation and diffusion of liquid
molecules is due to this motion.
iv) Diffusion operates to some extent in liquids.
v) The densities of liquids are closer to solids. But their densities are much greater than gases.
vi) The spaces amongst liquid molecules are negligible.
vii) The intermolecular forces in liquids are intermediate between gases and solids.
viii) Molecules of liquids possess kinetic energy due to their motion.
ix) Molecules of liquids exchange their kinetic energy when they collide with each other. Such
collisions are called inelastic collision.
x) By decreasing their kinetic energy, liquids are converted into solids.
xi) Evaporation in liquids causes cooling.
Intermolecular Forces
The forces of attraction present between the molecules of a substance are called Intermolecular forces.
Example:
The forces of attraction between HCl molecules are:
These forces operate between all kinds of molecules when they are sufficiently close to each other.
Intermolecular forces are called Van der Waal’s forces.
Valence electrons are not involved in these forces. That is why intermolecular forces are much
weaker than intramolecular forces.
The force of attraction that holds the two atoms together within a molecule is called intramolecular
force.
Example:
The covalent bond between Cl and H atoms in HCl molecule is intramolecular force.
Intramolecular force (covalent bond) develops due to mutual sharing of electrons between the two
atoms. This satisfies the outermost shells of both the bonded atoms.
That is why, this is their firm need to stay together. As a result, this force is very strong.
δ+ δ- δ+ δ-
H Cl H Cl
⎯ ⎯
.......

63. Types of Intermolecular Forces:
There are four main types of intermolecular forces.
(i) Dipole-dipole force
(ii) Ion-dipole force
(iii) Dipole-induced dipole force (Debye force)
(iv) Instantaneous dipole-induced dipole force (London dispersion force)
Dipole-Dipole Force
The intermolecular force of attraction between positive end of a polar molecule (a dipole) and the
negatively charged end of neighbouring polar molecule (another dipole) is called as dipole-dipole force.
Example:
Polar Molecule:
A molecule between two dissimilar atoms (having appreciable difference of electro negativities) is
called polar molecule. There exists dipole-dipole force between such molecules.
Example:
In HCl, Cl being more electronegative develops partial negative charge and hydrogen develops
partial positive charge. As a result, molecule become dipole (polar molecule).
Note: Dipole – dipole forces are 1 % as effective as a covalent bond.
Factors Affecting Dipole-Dipole Forces:
(i) Electronegativity difference
(ii) The inter-molecular distance
(i) Electronegativity difference:
Greater the electronegativity difference ( EN) between the boded atoms, greater will be the
polarity, stronger will be the intermolecular forces. Hence, greater are the values of
thermodynamic properties like:
• melting point
• boiling point
• heat of vaporization
• heat of sublimation …. etc.

64. (ii) The inter-molecular distance:
In gas phase, the intermolecular forces are weaker due to greater distance between
molecules. But in liquid phase, the forces are reasonably strong due to less distance between
the molecules.
The strength of intermolecular forces affect the physical properties like melting points,
boiling points, heat of vaporization etc. whereas intramolecular forces affect the chemical
properties mainly.
Dipole-Induced Dipole Forces (Debye Forces)
The force of attraction between positive end of the permanent dipole and the negative end of induced
dipole is called dipole-induced dipole force or Debye force.
Example:
Note: Polarity created in a non-polar molecule under the influence of a permanent dipole is called
induced dipole.
Dipole-induced dipole force exists between a polar and a non-polar molecule.
Explanation:
Sometimes, we have a mixture of substances containing polar and non–polar molecules. The positive
end of the polar molecule attracts the mobile electrons of the nearby non-polar molecule. In this way,
polarity is induced in non-polar molecule and both molecules become dipole. These forces are called dipole–
induced dipole forces or Debye forces as shown in figure below:
Instantaneous Dipole-Induced Dipole Force:
The force of attraction between of oppositely charged end of an instantaneous dipole and an induced
dipole is called instantaneous dipole-induced dipole force or London dispersion force.
This is momentary, short lived force that vanishes as quickly as it is formed. But this force extends
and holds when molecules are close to each other and are more polarizable.
Example:
The forces of attraction between chlorine molecules.
Note: • It is the only force that is present between non-polar molecules.
• These force were discovered by a German physicist, Fritz London in 1930.
Instantaneous Dipole:
A temporary dipole created in a non-polar molecule as a result
of repulsion of its electronic cloud with electronic cloud of a
neighbouring non-polar molecule is called instantaneous dipole.
dipole
…

65. Induced Dipole:
A dipole created in a non-polar molecule under the influence of
a temporary or an instantaneous dipole is called induced dipole.
Explanation:
Let us understand this concept of London force with the help of
example of helium gas.
When electrons of one atom come close to electrons of other atom, they are pushed away from each
other. At this moment, the electron density of atom no more remains symmetrical. It has more negative charge
one side than the other and helium atom becomes a momentary dipole called instantaneous dipole.
The instantaneous dipole disturbs the electronic cloud of nearby atom and induces dipole in it. This
is called induced dipole.
Note: 1. London force is a short-lived attraction because electrons keep on moving. This movement of
electrons causes the dipole to vanish as quickly as they are formed.
2. London forces are present in all the types of molecules whether polar or non-polar but are very
significant between non-polar molecules like H2, Cl2 He, Ne, Ar etc.
Factors Affecting London Forces:
The main factor that affects the strength of London force is polarizability. The quantitative measure of
extent to which electronic cloud can be polarized or distorted or dispersed is called polarizability. The
polarizability in turn depends upon the following factors.
(i) Size of electronic cloud.
(ii) Number of atoms in a molecule. (atomicity)
(1) A large sized atom has large electronic cloud. As a result, its
outermost electrons move away from nuclei and become loosely
bound. Hence, dispersion of electronic cloud becomes more and
more easy. As a result, atoms become more polar and strong forces
are created between such atoms and hence, higher will be the
boiling point.
Greater the size of electronic cloud, greater will be its polarizability
and stronger will be the London force.
Examples:
(i) The boiling points of Noble gases increase down the group
due to increasing.
• Size
• Polarizability
• Strength of London force
All halogens have non-polar diatomic molecules.

66. (ii) The physical states of halogens change from gas to solid
through liquid and boiling point changes from -188.1oC to
184.4oC. This is due to increasing size, polarizability and
strength of London force down the group.
(2) No. Of Atoms Per Molecule (Atomicity):
Greater the number of atoms, greater is the polarizability and
strength of London force. Actually, when number of atoms in a
molecule increases, molecule becomes longer, the number of polar
sites increases where molecules can attract each other. As a result,
strong forces are created.
Example:
Boiling point of C2H6 and C6H14 are -88.6oC and 68.7oC respectively.
The reason is that longer molecules have more polar sites and
hence strong forces with other molecules.
Note: In hydrocarbons, trend changes from gaseous state to liquid and
finally to solid with increasing molecular mass.
Boiling points and physical states of some hydrocarbons:
Name
B.P
o
C (1atm)
Physical state at
S.T.P
Name
B.P
o
C (1atm)
Physical state at
S.T.P
Methane -16.5 Gas Pentane 36.1 Liquid
Ethane -88.6 Gas Hexane 68.7 Liquid
Propane -42.1 Gas Decane 174.1 Liquid
Butane -0.5 Gas Isodecane 327 Solid
Hydrogen Bonding:
The electrostatic force of attraction between highly partial positive hydrogen and lone pair of highly
electronegative atom is called Hydrogen bonding.
Example: The force of attraction between the lone pair of oxygen and hydrogen in water.
Note: • It is a special type of dipole-dipole force.
• It is much stronger than simple dipole-dipole force.
• Its strength is 20 times less than that of a covalent bond.
Condition For Hydrogen Bonding:
i) Highly partial positive Hydrogen.
ii) Highly electronegative atom.

67. Explanation:
In water, oxygen being more electronegative withdraws shared
pair strongly from hydrogen. As a result, hydrogen become
highly partial positive and creates strong electric field due to its
small size. This strong electric field attracts the lone pair of
electronegative atom strongly. And resulting electrostatic force
is called hydrogen bonding.
Examples:
1) Hydrogen bonding in NH3:
There is only one hydrogen bond per NH3 molecule.
Hydrogen bonding in NH3 molecules is shown below.
2) Hydrogen bonding in HF:
There is only one hydrogen bond per HF molecule.
HF molecules join each other in Zig-Zag manner due to the
presence of hydrogen bonding.
HF shows exceptionally low acidic strength than HCl, HBr and HI
due to
i. Hydrogen bonding between its molecules.
ii. Hydrogen atom being entrapped between two highly
electronegative atoms (F). As a result, release of hydrogen
atom becomes difficult. This makes it weaker acid.
3) Hydrogen bonding in H2O:
There are two hydrogen bonds per water molecule. This is the reason, why its intermolecular
hydrogen bonding is stronger than that of HF and NH3.
4) Hydrogen bonding between acetone and chloroform molecules.
(i) In Chloroform, three Chlorine atoms and one
hydrogen atom is directly attached to one Carbon.
Chlorine atoms being highly electronegative
withdraw electrons form carbon atom which in turn
withdraws electrons from hydrogen atom. As a
result, hydrogen atom becomes highly partial
positive.
(ii) In acetone, oxygen being more electronegative
withdraws electrons from carbon atom and become
partial negative.
When acetone and Chloroform come closer, highly partial positive hydrogen of chloroform and
lone pair of partial negative oxygen forms hydrogen bond with each other. That is why chloroform
and acetone are miscible into each other.
Cl — C — H O C
Cl
Cl
 +
 +
 −
 −
 −
 −
CH3
CH3
 +

68. Note: The intermolecular force between acetone and chloroform is hydrogen bonding when they are
mixed. But intermolecular force between Chloroform molecules in pure state or between acetone
molecules in pure state is dipole-dipole force.
Hydrogen bonding is not limited to fluorine, oxygen and nitrogen. But it may be present where
hydrogen atom is bonded to some other element but is highly partial positive
Example: Chloroform.
Properties and applications of compounds Containing Hydrogen Bonding:
1. Thermodynamics properties of covalent Hydrides.
Hydrogen bonding influences the physical properties like melting points and boiling points of
covalent hydrides.
Boiling points of covalent Hydrides of IV – A group elements.
The boiling points of covalent hydrides of group IV-A elements are lower than those of V-A, IV-A
and VII-A group elements due to least electronegativities of IV-A group elements.
CH4 is the hydride of top member of IV-A group. It has lowest boiling point due to its smallest size
and least polarizability. The boiling points of hydrides of this group members increase down the group due
to increasing sizes and polarizabilities of hydrides.
Boiling points of covalent hydrides of V-A group elements.
The boiling point of NH3 is greatest than hydrides of its group members (except AsH3) due to
stronger hydrogen bonding. This in turn is due to greater electronegativity difference between Nitrogen
and Hydrogen.
There is only one hydrogen bond per NH3 molecule due to the presence of only one lone pair on
central nitrogen atom in NH3. There is less elelctronegativity difference (1.1) between N (EN = 3.2) and
H (EN = 2.1).
That is why NH3 is less polar and have weak intermolecular hydrogen bonding between its molecules. As a
result, it exists as a gas with boiling point -33oC.
Boiling points of hydrides of Group VI-A and VII- A.
The boiling points of H2O and HF are greater than those of hydrides of their respective group
members due to strong intermolecular Hydrogen bonding. This in turn is due to greater electronegativities
of oxygen and fluorine respectively.
There is only one hydrogen bond per HF molecule due to the presence of only one hydrogen atom
bonded to fluorine atom. There is greater electronegativity difference (1.9) between F (EN = 4) and
H (EN = 2.1). That is why HF is strongly polar and has strong intermolecular hydrogen bonding between its
molecules. As a result, it exists as a liquid with boiling point 19.9oC.
Comparison of Hydrogen bonding and boiling points of H2O and HF.
There are two hydrogen bonds per H2O molecule due to the presence of two lone pairs and two
hydrogen atoms bonded to oxygen.There is larger electronegativity difference (1.4) between O (EN = 3.5) and
H (EN = 2.1) (Less than HF).
That is why H2O is strongly polar (lesser than HF) and has strong intermolecular hydrogen bonding
(stronger than HF) between its molecules. As a result, it exists as a liquid with boiling point 100oC.

69. Although water is less polar than HF, still its intermolecular Hydrogen bonding is stronger than HF due
to the presence of two hydrogen bonds per H2O molecule and three dimensional hydrogen bonding.
Comparison of boiling point of HCl, HBr and HI.
HCl is considered border line case between simple
dipole-dipole force and hydrogen bonding.
Boiling point of HBr is greater than that of HCl
Due to greater polarizability of Br atom than that of chlorine
atom. On the similar basis, the boiling point of HI is greater
than HBr.
Note: The order of boiling points of HCl, HBr and HI is as
follows HI > HBr > HI
Comparison of boiling points of Hydrides of 3rd and 4th period.
The boiling points of Hydrides of fourth period (GeH3,
AsH3 , H2Se , HBr) are greater that those of third period (SiH4,
PH3, H2S, HCl) due to larger size and greater polarizabilities.
2. Solubility of Hydrogen bonded molecules:
The compounds that have hydrogen bonding between
their molecules are soluble in water due to Hydrogen bonding
between water molecules and molecules of those compounds.
Example:
i) Ethyl alcohol (C2H5OH) is soluble in water due to
Hydrogen bonding between the molecules of
water and ethyl alcohol.
ii) Small sized carboxylic acids are soluble in water
due to hydrogen bonding between the molecules
of water and carboxylic acids.
iii) Hydrocarbons being non polar are completely
insoluble in water because they will not form
hydrogen bonding with water.
3. Structure of Ice
i) The molecule of water in ice has tetrahedral
structure. Two lone pairs of electrons and two
hydrogen atoms on oxygen atom occupy four
corners of tetrahedron.
The structure of ice is like diamond because each
atom of carbon in diamond is at the center of a
tetrahedron just like oxygen of water in the
structure of ice.
ii) In liquid state, associations of water molecules break and reform because water molecules
are mobile. When temperature of water is decreased below 4oC and ice is formed at 0oC,
then molecules become more regular.

70. This regularity extends throughout the structure in geometrical patterns (Hexagonal
patterns). As a result, empty spaces are created in the structure of ice.
iii) Due to empty spaces in the structure of ice, it occupies 9% more space than liquid water.
Hence, density of ice is 9% less than liquid water. As a result, ice being lighter than water
floats on surface of water.
iv) When temperature of water reaches below 4oC by fall in temperature in atmosphere, the
water on the surface become less dense. This less denser water stays at the top of slightly
warmer water underneath. A stage reaches, when it freezes at 0oC. This layer of ice insulates
the water which is below it for further heat loss. This is the reason, why water below a layer
of ice stays at 4oC. As a result, fish and plants continue to survive under the blanket of ice.
NOTE
The density of water is maximum at 4oC. When temperature decreases from 100oC to 4oC, its
volume decreases and density increases. Whereas, when temperature decreases from 4oC to
0oC. Volume increases and density decreases by 9% due to Hydrogen bonding and
geometrical arrangement (Hexagonal) of water molecules.
The change in the density of water with a changing temperature is shown below.
Volume Decreases Volume Increases
Density Increases Density Decreases
4. Cleansing action of soaps and detergents.
Soaps and detergents perform their cleansing action due to hydrogen bonding between polar parts
of their molecules and water molecules. The polar parts of the molecules are water soluble and non-polar
parts are water insoluble. Non-polar parts of molecules are either alkyl or aryl groups which remain
outside water.
5. Hydrogen bonding in biological compounds and food materials.
i) In Proteins:
Proteins are the important part of living organisms. Fibers
like those found in the hairs, silk and muscles consist of
long chains of amino acids. These long chains are coiled
around one another into a spiral. This spiral is called a helix
which may be right handed or left handed. In case of right
handed helix, the groups like >NH and >C=O are vertically
adjacent to one another and they are linked together by
hydrogen bonds which link one spiral to the other as
shown in figure.
ii) In Deoxyribonucleic acid (DNA):
DNA has two spiral chains which are coiled about each
other on a common axis. In this way, they give a double
helix. This is 18-20 in diameter. They are linked together

71. by hydrogen bonding between their sub-units as shown in
the figure.
iii) In Food Materials:
The food materials like carbohydrates include glucose,
fructose and sucrose. They all have –OH group in them
which are responsible for hydrogen bonding in them
6. Hydrogen Bonding in Paints, Dyes and Textile materials
i) One of the most important properties of paints and dyes is
their adhesive action. This property is developed due to
hydrogen bonding.
ii) Hydrogen bonding makes glue and honey as sticky substances.
iii) We use cotton, silk or synthetic fibres for clothing. Hydrogen bonding is of vital importance in
these thread making materials. This hydrogen bonding is responsible for their rigidity and the
tensile strength.
Evaporation
The spontaneous change of high energy liquid molecules into vapours at any temperature is called
evaporation. The escape of high energy molecules from the surface of liquid takes place at all
temperatures is called Evaporation.
Evaporation:
• is surface phenomenon
• is continuous process
• continues at all temperatures.
• is an endothermic process.
Explanation:
The energy of the liquid molecules in not equally distributed. Some of the molecules have low
kinetic energy and move slowly while others have high kinetic energy and move faster. If one of the high
energy molecules reaches the surface, it may break the attractions of neighboring molecules and leaves
the bulk of liquid.
Evaporation causes cooling:
Actually high energy molecules leave the surface leaving behind the low energy molecules. As a
result, temperature of liquid falls and heat moves from surrounding to the liquid. Finally temperature of
surrounding also falls.
Rate of evaporation
The number of vapours formed per unit time is called rate of evaporation
Factors affecting the rate of evaporation:
(i) Surface area
(ii) Intermolecular forces

72. (iii) Temperature
NOTE:
Evaporation also depends upon external pressure and humidity and is directly proportional to both
these factors
(i) Surface area:
When surface area is increased, then more molecules are able to escape from the surface of
liquid. As a result, liquid evaporates quickly.
Example:
Hot tea placed in a pirch gets cool at a faster rate than in a cup due to greater surface area of
pirch.
(ii) Intermolecular forces:
The molecules of liquids having weaker intermolecular forces have greater chances to
escape from the surface of liquid. As a result, rate of evaporation is greater.
Example:
Gasoline evaporates much faster than water due to weak London forces of attraction
between its molecules.
(iii) Temperature:
The rate of evaporation increases with increasing temperature. At high temperature, greater
number of molecules have high kinetic energy to overcome the attractions of their
neighbouring molecules. As a result, rate of evaporation is greater.
Example:
Wet cloths dry up faster in hot summer in comparison to cold winter due to greater rate of
evaporation at high temperature in summer.
Vapour Pressure:
The pressure exerted by vapours on the surface of the liquid in equilibrium with the liquid at a given
temperature is called vapour pressure.
Explanation:
In a closed container, high energy molecules leave the surface of
liquid and accumulate above the surface. This is called evaporation
(liquid into vapours). The molecules (vapours) collide with the walls of
container as well as on the surface of liquid. The vapours colliding on the
surface are usually recaptured by the surface of liquid. This is called
condensation (vapours into liquid). The process of evaporation and
process of condensation continue until rate of evaporation becomes
equal to the rate of condensation. This is called state of dynamic
equilibrium.
Temperature
(oC)
Vapour
Pressure (Torr)
0
10
20
30
37
40
50
60
70
80
90
100
4.579
9.209
17.54
31.82
47.07
55.32
92.51
149.4
233.7
355.1
527.0
760.0

73. At the state of dynamic equilibrium, the number of molecules
leaving the surface is just equal to number of molecules coming back
into it at constant temperature.
Liquid Vapours
The pressure exerted by vapours on the surface of liquid at this state is called vapour pressure.
Vapour pressure does not depend upon:
(i) Amount of liquid
(ii) Volume of container
(iii) Surface area of liquid
Factors affecting the vapour pressure:
(i) Intermolecular attractions and sizes of molecules (nature of liquid).
(ii) Temperature
(i) Intermolecular forces
Vapour pressure increases with decreasing
strength of intermolecular forces and vice versa.
Strong intermolecular forces hold the
molecules tightly and chances of liquid
molecules to leave the surface decreases. As a
result, vapour pressure also decreases.
Example:
Vapour pressure of water having stronger
intermolecular forces is lesser (17.5 torr)
than ether (44.2 torr) at 20oC. The
intermolecular forces in ether are weaker.
(ii) Temperature
Vapour pressure increases with the increase in temperature and vice versa. At higher
temperature, weak forces hold molecules loosely and chances of liquid molecules to escape
from surface increases. As a result, vapour pressure also increases.
Example:
i) Vapour pressure of H2O is 92.51 torr at 50oC.
ii) Vapour pressure of H2O is 527 torr at 90oC.
Note: The increase of vapour pressure at high temperature is greater for same difference of temperature.
It is because intermolecular forces are weaker at higher temperature.
Example:
i) When temperature increases from 0oC to 10oC (10oC change of T) then increase of vapour
pressure of is from 4.579 torr to 9.209 torr.
ii) When temperature increases from 90oC to 100oC (10oC change of T), then increase in vapour
pressure is from 527.8 torr to 760 torr.
Name of compound
Vapour pressure at
20oC (torr)
Isopentane 580
Ethyl ether 442.2
Chloroform 170
Carbon Tetrachloride 87
Water 17.54
Mercury 0.012
Glycerol 0.00016

74. Measurement of Vapour Pressure:
There are many methods but one of the most important method i.e. manometric method is given below.
Manometric method: (Accurate method)
The liquid whose vapour pressure is to be determined is taken in flask placed in the thermostat.
One end of the tube from flask is connected to manometer and other end is connected to vacuum pump.
The air above the liquid is removed in the following steps.
(i) Freezing:
The liquid is frozen with the help of freezing mixture and space above the liquid is evacuated.
In this way, air is removed from the surface of liquid alongwith the vapours of liquid.
(ii) Melting:
The frozen liquid is melted to release entrapped air.
(iii) Re-freezing:
Liquid is again frozen and released air is evacuated. This process is repeated many times till
almost all the air is removed.
Measurement of vapour pressure:
The liquid in the flask is warmed in thermostat at a temperature at which vapour pressure is to be
determined. Difference in the heights of Hg column in two limbs of manometer determines the vapour
pressure of liquid.
The pressure on the surface of liquid is equal to sum of atmospheric pressure and vapour pressure
of liquid. That is the reason why column of manometer facing the liquid is more depressed than facing the
atmosphere.
The vapour pressure of liquid is given by following equation.
P = Pa + h
P = vapour pressure of liquid at one atmosphere.
Pa = atmospheric pressure
h = difference in the heights of Hg column in two limbs of manometer. It gives us the vapour
pressure of liquid.

75. Boiling Point:
The temperature at which vapour pressure (internal
pressure) of liquid becomes equal to external
pressure is called boiling point.
Examples:
i) Boiling point of H2O = 100oC
ii) Boiling point of C2H5OH = 78.26oC
iii) Boiling point of C2H5OC2H5 = 34.6oC
Explanation:
When a liquid is heated, vapour pressure of liquid
goes on increasing and ultimately becomes equal to
external pressure. On further heating at this stage,
bubbles of vapours which are formed in the interior
of liquid have greater internal pressure than external
pressure on the surface of liquid. This makes the bubble to come out of the liquid and burst on the
surface of liquid. A constant stream of bubbles come out at boiling.
Temperature remains constant at boiling point.
When liquid is heated below the boiling point, kinetic energy of its molecules and temperature
increases. At boiling point, kinetic energy of molecules become maximum and heat given will be
utilized to break the attractions and to convert liquid into the vapours. At this stage, the escaping
vapours carry the absorbed heat alongwith them. As a result, temperature remains constant.
Molar heat of vaporization:
The amount of heat required to vapourize 1mole of a liquid at its boiling point is called its molar heat of
vapourization. It is represented by Hv.
Example:
Molar heat of vapourization of water is 40.6 kJ/mole.
Comparative variation of vapour pressures of different liquids with temperature.
Following graph shows the variation of vapour pressure of
i) water ii) ethyl alcohol iii) ethylene glycol iv) diethyl ether with temperature.
• The vapour pressure-temperature curve for water starts at 4.6 torr and that of diethyl ether
starts around 200 torr. This is due to weak intermolecular forces between the molecules of
diethyl ether than water.
• Vapour pressure-temperature curve for water goes along the temperature axis to a greater
extent at the start than that for diethyl ether. It is due to the reason that it is difficult to
overcome strong intermolecular forces between water molecules at low temperature.
Note:Vapour pressure temperature curve shows that vapour pressure increases rapidly when the
liquids are closer to their boiling points.
Liquids Boiling point (oC)
Acetic Acid 118.50
carbon tetrachloride 76.50
Acetone 56.00
Ethanol 78.26
Aniline 184.4
Naphthalene 218.00
Benzene 80.15
Phenol 181.80
Carbon disulphide 46.30
Water 100.00

76. Factor affecting the boiling points:
Boiling points of liquids depend upon the following factors.
i) Intermolecular forces
ii) External pressure
(i) Intermolecular forces:
Stronger the intermolecular forces, greater will be the boiling point and vice versa. Greater amount
of heat will be required to
(i) Over come strong intermolecular attractions
(ii) Equalize vapour pressure with external pressure.
Hence, boiling point will be high.
Examples:
The boiling points of some liquid at 1atom.
i) Boiling point of C2H5OC2H5 is 34.6oC (weak dipole-dipole forces)
ii) Boiling point of C2H5OH is 78.26oC (weak hydrogen bonding)
iii) Boiling point of H2O is 100oC (strong hydrogen bonding)
(ii) External pressure:
Greater the external pressure, greater will be the boiling point
and vice versa. Liquid absorbs greater amount of heat to
equalize the vapour pressure with greater external pressure.
Hence, boiling temperature is high.
Example:
i) Boiling point of H2O is 100oC at 760 torr.
ii) Boiling point of H2O is 98oC at 700 torr.
Boiling and External Pressure
Since, a liquid boils at a temperature where vapour pressure becomes equal to external pressure,
hence boiling point varies with external pressure.
Greater the external pressure greater will be the boiling point and vice versa. When external
pressure is high, liquid requires greater amount of heat to equalize its vapour pressure to external
pressure and vice versa.
Examples:
i) Boiling point of H2O is 120oC at 1489 torr.
ii) Boiling point of H2O is 100oC at 760 torr (at sea level).
iii) Boiling point of H2O is 98oC at 700 torr (at Murree hill).
iv) Boiling point of H2O is 69oC at 323 torr (at Mount Everest)
v) Boiling point of H2O is 25oC at 24.7 torr.

77. Practical applications of effect of external pressure on boiling point:
(i) Pressure cooker (boiling of liquid under increased external pressure)
(ii) Vacuum distillation (boiling of liquid under decreased external pressure)
(i) Pressure cooker (boiling under increased pressure)
Principle: The boiling point increases with external pressure.
Working: Pressure cooker is a closed container. When liquid is heated in it, vapours are formed
which accumulate on the surface of liquid and are not allowed to escape. These vapours exert more
pressure on the surface of liquid. As a result, boiling point increases. As more heat is absorbed by
water, food is cooked quickly under increased pressure.
(ii) Vacuum distillation (boiling under reduced pressure).
Principle: heT boiling point decreases with decreases in external pressure.
Working: The liquids that decompose when they are distilled at their boiling points, their
decomposition can be avoided by distilling at lower temperature and lower external pressure.
Example:
Glycerine boils as well as decomposes at 290oC and 760 torr (1atm). Hence, it cannot be distilled at
290oC. It can be distilled without decomposition at 210oC under reduced pressure of 50 torr. As a
result, it can be purified easily.
Advantages of vacuum distillation:
Vacuum distillation has following advantages.
i) It decreases the time for distillation.
ii) It is economical as it consumes less fuel.
iii) The decomposition of many liquids can be avoided.

78. Assessment 01
1. Which of the following attractive force has nothing to do with valence electrons?
a. Intramolecular force
b. Ionic bond
c. Covalent bond
d. Intermolecular force
2. Intermolecular forces have no significance when we consider:
a. Melting point
b. Heat of vaporization
c. Oxidation
d. None of these
3. The cohesive force between the molecules of liquids due to electronegativity difference between
atoms is:
a. Dipole-dipole forces
b. London dispersion forces
c. Covalent bond
d. None of these
4. In dipole dipole interaction, the dipoles which can exist are called:
a. Temporary dipoles
b. Permanent dipoles
c. Instantaneous dipoles
d. Induced dipoles
5. The number of poles which are involved when two hydrochloric acid molecules approach each
other:
a. 1
b. 2
c. 4
d. 6
6. In which of the following dipole-dipole interaction exists:
a. HCl
b. H2O
c. CHCl3
d. All of these
7. Which of the following causes the molecules not to have a perfect alignment in dipole-dipole
interaction?
a. Low kinetic energy
b. High potential energy

79. c. Thermal energy
d. All of these
8. The strength of dipole-dipole forces depends upon the
a. EN difference between atoms
b. Distance between molecules
c. Shape of molecules
d. Both a and b
9. The strength of dipole-dipole force does not influence the following property of polar liquid?
a. Melting point
b. Heat of vaporization
c. Molar mass
d. Boiling point
10. Permanent dipole in case of polar molecules is originated due to:
a. Electronegativity difference
b. Shape of atoms
c. Molar mass of atoms
d. Thermal energy

80. Assessment 02
1. The intermolecular force that exists between permanent dipole and temporary dipole is:
a. Hydrogen bonding
b. Dipole-Dipole interaction
c. Dipole-Ion dipole force
d. Dipole-induced dipole forces
2. The force of attraction between sodium ion and water molecules is:
a. Dipole-dipole interaction
b. Hydrogen bonding
c. Ion-dipole forces
d. Debye forces
3. Debye force is the special term which is assigned to the following intermolecular force:
a. Dipole-dipole interaction
b. Ion-dipole forces
c. Dipole-induced dipole force
d. Instantaneous dipole induced dipole forces
4. The intermolecular force that exists within non-polar molecules is:
a. Dipole-Dipole interaction
b. Dipole-induced dipole force
c. Instantaneous dipole-induced dipole forces
d. Hydrogen bonding
5. Which of the following noble gas has more boiling point by keeping external pressure same?
a. Helium
b. Neon
c. Argon
d. Xenon
6. London dispersion force affects the physical state of halogens. The following halogen is solid at
room temperature:
a. F2
b. Cl2
c. Br2
d. I2
7. The following statement corresponds to the strength of London dispersion force:
a. Hexane is hydrocarbon which exists in liquid state at room temperature.
b. Helium has more boiling point among noble gases
c. The molecule with longer chain length experiences stronger attractive force.
d. Both a and c

81. 8. The correct order of strength of intermolecular force is:
a.
b.
c. London dispersion force < Debye force < Dipole-dipole < H-bonding
d.
9. Heat of vaporization of which of the following is greater due to the presence of hydrogen bonding:
a. H2S
b. H2Se
c. H2O
d. All have same
10. The decrease in volume when ice melts to form water is:
a. 1%
b. 9%
c. 6%
d. No change in volume

82. Assessment 03
1. The accurate order of boiling point for the given hydrides:
H2O, CH4, HF and NH3
a. CH4
b. CH4 <NH3 <HF < H2O
c. NH3
d. CH4
2. The compounds which are soluble in water due to establishing H-bonding with water molecules:
a. Acetic acid
b. Ethyl alcohol
c. Acetone
d. All of these
3. The adhesive property of glue is usually attributed to:
a. Hydrogen bonding
b. High molar mass
c. Large molar mass
d. Low density
4. Two strands of DNA are coiled to each other forming double helix. The attractive force present
between nitrogen bases of two strands is:
a. Ion-dipole forces
b. H-bonding
c. Dipole-Dipole interaction
d. Debye force
5. Water has maximum density at the temperature of:
a. 0 o
C
b. 4 o
C
c. 25 o
C
d. Same at all temperature
6. In extreme cold season fish and aquatic life can survive within frozen water due to:
a. Less density of ice
b. More density of water which sinks
c. Insulation of ice blanket from chilling weather
d. All of these
7. The detergent molecule interacts with grease molecule as well as get soluble in water due to
hydrogen bonding. The following part of soap/detergent interacts with grease that is stick to the
fiber:

83. a. A only
b. B only
c. Either A or B
d. Both A and B
8. The biological compound which forms helical structure due to hydrogen bonding is:
a. DNA
b. Protein
c. RNA
d. Both a and b
9. Which of the following statement is true about the boiling point of the liquid?
a. Different liquids have same boiling point on the same place.
b. Same liquid has same boiling points on different places
c. Boiling point depends upon the amount of liquid
d. A liquid X has different boiling points on different places.
10. Which of the following intermolecular force is?
a. Dipole-dipole forces
b. London dispersion forces
c. H-bonding
d. Debye forces
B
A
+ -
__
+

84. Assessment 04
1. At the particular instant, the helium atom when have distorted electron density is called:
a. Permanent dipole
b. Induced dipole
c. Instantaneous dipole
d. Non-polar
2. The intermolecular force that exist between Cl2 molecules:
a. Dipole-dipole interaction
b. Debye forces
c. Hydrogen bonding
d. London dispersion forces
3. Temperature at which vapour pressure of liquid becomes equal to external pressure is:
a. Boiling point
b. Freezing point
c. Saturation point
d. Melting point
4. In order to boil the water at standard temperature (25 o
C), external pressure should be:
a. 760 mmHg
b. 700 mmHg
c. 23.7 mmHg
d. 1520 mmHg
5. Which of the following statement is true about the boiling point of any liquid?
a. Boiling point of liquid can be changed by changing the external pressure.
b. Boiling point of liquid is high at mountains.
c. Boiling point of liquid is less at sea level and more at land area.
d. Boiling point of water at K-2 is more than at Karachi.
6. The factor which affects the evaporation of water:
a. Temperature
b. Surface area
c. Intermolecular force
d. All of these
7. Which of the following is more volatile in nature?
a. Water
b. Ethanol
c. Diethyl Ether
d. Glycerine

85. 8. The physical phenomenon that can be explained by increasing evaporation:
a. Drying of clothes in summer.
b. More volatile nature of petrol than water
c. Cooling of water in earthen ware vessel
d. All of these
9. Vapour pressure does not depend upon the following factor:
a. Amount of liquid
b. Surface area
c. Intermolecular force
d. Both a and b
10. Water boils at high temperature than ethanol and diethyl ether due to:
a. Stronger intermolecular force
b. More mass
c. Low vapour pressure
d. Both a and c

86. Key
Assessment 01
1. d
2. c
3. a
4. b
5. c
6. d
7. c
8. d
9. c
10. a
Assessment 02
1. d
2. c
3. c
4. c
5. d
6. d
7. d
8. c
9. c
10. b

87. Assessment 03
1. b
2. d
3. a
4. b
5. b
6. d
7. a
8. d
9. d
10. d
Assessment 04
1. c
2. d
3. a
4. c
5. a
6. d
7. c
8. d
9. d
10. d

88. Oldest, Largest and Most Credible Platform
MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org

89. Atomic Structure
Concept of orbital’s
Electronic configuration
Discovery and properties of proton
(positive rays)
Quantum numbers
Shapes of orbital’s

90. Atomic structure
Discovery Of Proton (Positive Rays):
In 1886, German physicist, Eugene
Goldstein discovered protons or positive
rays or canal rays.
Apparatus:
A discharge tube provided with a
cathode having extremely fine holes in it.
Procedure:
➢ When a large potential difference is applied between electrodes, it is observed that
while cathode rays are traveling away from cathode, there are other rays produced at
the same time moving towards cathode. These are called positive rays.
➢ They are also called as canal rays since they pass through the canals or holes in cathode.
Reason For Production:
When high speed cathode rays strike the residual gas molecules, they knock out
electrons from them and positive ions are produced.
M + 1e- ⎯⎯
→M+ + 2e-
Observation:
These rays after passing through the perforated cathode produce a reddish glow on the
opposite wall.
Properties Of Positive Rays:
1) These rays travel in a straight line towards the cathode.
2) They show deflection by electric field and magnetic field. It shows
that they are positively charged.
3) They cause flash upon ZnS plate.
4) Their e/m ratio varies with the residual gas.
5) Their e/m ratio is maximum in case of hydrogen.
6) Their e/m ratio is smaller than that of cathode rays.
7) These rays consist of tiny particles called protons (when hydrogen gas is used) having
mass 1836 times of electrons.
The e/m value of positive rays is
different for different gases
because the mass of every gas is
different. The positive particle
obtained from hydrogen is lightest
and have maximum e/m value.
Rutherford named this particle as
proton

91. Planck’s Quantum Theory
Max Planck proposed the quantum theory in 1900 to explain emission and absorption of
radiation.
➢ According to his revolutionary theory, energy travels in discontinuous manner.
➢ Energy is composed of large number of tiny discrete units called quanta.
Postulates of theory:
The main postulate of theory are as follows.
1. Energy is emitted or absorbed in discontinuous manner in the form of energy packets.
Each packet of energy is called quantum which have definite amount of energy. In case
of light radiations quantum is called photon.
2. Energy of quantum is directly proportional to frequency () of radiation.
Number of waves of a radiation passing through a point per second is called frequency.
Its unit is per second (s-1).
E  
E = h………..(i)
E = energy of radiation
 = frequency of radiation
h = Planck’s constant = 6.625  10-34Js.
Planck’s constant is actually ratio of energy and frequency of radiation:
3. A body emits or absorbs energy in the form of quantum of energy (h).
E = h……(i)
Frequency of radiation is inversely proportional to wavelength. Greater the wavelength,
smaller will be the frequency.
……..…… (ii)
Put equation (ii) into (i)
E = hc/
…….. (iii)

92. This equation (iii) shows that energy inversely proportional to wavelength or radiation.
Greater the wavelength, lesser will be the energy.
1
E


The distance between two adjacent crests or troughs is called wavelength or a wave of
particular radiation.
The units of wavelength are
(i) Angstrom (Ao)
(ii) Nanometer (nm)
(iii) Picometer (pm)
(iv) Meter (m)
1Ao = 10-10m
1nm = 10-9 m
1pm = 10-12 m
4. Number of waves per unit length is called wave number. It is reciprocal of wavelength.
Put equation (iv) into (iii)
E=hc ........(v)
v
−
This equation (v) shows that energy is directly proportional to wave number. Greater
the wave number, greater will be energy of radiation.
Bohr’s Atomic Model
➢ In 1913, Neil Bohr presented the model of an atom of hydrogen.
➢ It is based upon the Planck’s quantum theory.
Postulates:
The postulates of Bohr’s atomic model are as follow.
(i) Electron revolves around the nucleus in some permitted circular paths called
orbits. Each orbit has fixed energy and quantized.
(ii) Energy of an electron is fixed in one orbit. It means it neither emits nor absorbs
energy as long as it is in one orbit.
(iii) The energy of electron changes by E when electron jumps between the orbits.
This energy is equal to energy difference of energies of two orbits (E1 and E2)
between which electron jumps. This energy is given by Planck’s theory.
∆E = E2 - E1
Crest Wave length = 
 = Wave length
trough

93. Or ∆E = h………. (i)
∆E = energy difference of two orbits.
E2 = energy of higher orbit
E1 = energy of lower orbit
(iv) These electrons can revolve around the nucleus in the orbit having fixed angular
momentum. It is integral multiple of
h
2π
and is given as follows.
nh
mvr =
2π
Where, n = 1, 2, 3 …………….. (orbit number)
The permitted values of angular momentum are , ……
Electron revolves around the nucleus with any one of these values of angular momentum. So,
angular momentum is quantised.
Hydrogen Spectrum
It is an important example of atomic spectrum.
Apparatus:
➢ Hydrogen is filled in a discharge tube at a very low pressure.
➢ A bluish light is emitted from the discharge tube.
➢ This line when viewed through a spectrometer shows several isolated sharp lines.
Spectral Lines:
➢ These sharp lines are called spectral lines.
➢ The wavelengths of these lines lie in the visible, ultraviolet and infrared regions.
Spectral Series:
These sharp lines can be classified into five groups called spectral series. These are
named after their discoverers.
(i) Lyman series (ultraviolet region)
(ii) Balmer series (visible region)
(iii) Paschen series(infrared region)
(iv) Brackett series(Infrared region)
(v) Pfund series (Infrared region)
The lines of Balmer series has been given specific names as H and H……etc.

94. Origin of Hydrogen Spectrum on the Basis of Bohr’s Model
Absorption of radiations:
➢ According to Bohr, electron in hydrogen atom may revolve in any orbit depending upon its energy.
➢ When hydrogen gas is heated or subjected to an electric discharge, its electron moves from one of
the lower orbit to higher orbit, absorbing particular wavelength of energy.
Emission of radiations:
➢ When electron in higher orbit comes back to lower
orbit, the same energy is released.
➢ This energy is observed as radiation of particular
wavelengths in the form of bright lines seen in the
certain region of the emission spectrum of
hydrogen gas.
1. Lyman series:
➢ The spectral lines of Lyman series are
produced when the electron jumps from
n2=2,3,4,5……. to n1= 1.
➢ Lyman did not know the reason for the
series.
2. Balmer series:
➢ The spectral lines of Balmer series
discovered in 1887 originated when an
electron jumps from n2 = 3,4,5,6…… to n1 =
2.
In the same way, Paschen, Brackett and Pfund series of lines are produced as a result of electronic
transitions from higher orbits to 3rd, 4th and 5th orbits, respectively.
Defects Of Bohr’s Atomic Model
1- Origin of Spectrum:
➢Bohr’s theory can only successfully explain the origin of spectrum in monoelectron (one
electron system) system like H, He+,Li2+ , Be+3 etc.
➢Bohr’s theory failed to successfully explain the origin of spectrum of multi-electron or poly-
electron system like He, Li and Be.
2. Fine Structure or Multiple Structure of Spectral Line
➢ The splitting of spectral line into component lines while passing through high power
resolving spectrometer is called Multiple or fine structure.
➢ Bohr failed to explain the multiple structure of spectral line.
Example: The H-line in the Balmer series is found to consist of 5-component lines.

95. Note:Actually, appearance of several lines in single line suggest that only one quantum number is
not sufficient to explain the origin of various spectral lines.
3. Atomic Model is not Flat:
➢ Modern researches have shown that motion of electron is three-dimensional instead of a
single plane.
➢ In contrary to that, Bohr suggested circular orbits of electron and flat atomic model.
4. Zeeman Effect
The splitting of spectral line into number of closely spaced lines in the presence of magnetic field is
called Zeeman effect.
Examples:
(i) The two lines in the emission spectrum of sodium split up into component lines in magnetic
field.
(ii) The spectral line of Hydrogen spectrum splits up into component lines in the presence of
magnetic field.
5. Stark Effect:
The splitting of a spectral line, into number of closely spaced lines in the presence of electric field is
called stark effect.
Example:
The spectral line of Hydrogen spectrum splits up into component lines in the presence of electric
field.
Sommer feld Modification
In 1915, Sommer feld suggested that moving electron describe elliptical orbits in addition to
circular orbits. In these elliptical orbits, the nucleus lies in one of the foci of ellipse.
Concept of Orbital:
The volume of space in which there is 95% chance of finding an electron is called atomic orbital.
This is also called as electron could.
Wave theory of atom (wave mechanical model of atom):
➢ Schrodinger, Heisenberg and Dirac worked out wave theory of atom.
➢ Schrodinger wave equation was best, treatment.
➢ Schrodinger set up a wave equation for Hydrogen and solved it for orbital.
➢ According to schrodinger, although the position of an electron cannot be found exactly, the
probability of finding an electron at certain position at any time can be found.
o
Elliptical orbits
Foci of ellipse
Nucleus

96. ➢ The solution of the wave equation gives the probability of finding an electron in small region of
space.
➢ The maximum probability of finding the electron is at a distance of 0.053nm.It is the same radius as
calculated for Bohr’s first orbit.
Note: Electron can be closer or away by 0.53nm from nucleus.
Quantum Numbers
Quantum numbers are the four constants that describe the permissible behaviour of the electron in a
orbital.
➢ A set of numerical values which gives the acceptable solutions to Schrodinger wave equation for
Hydrogen atom.
➢ It describes the behaviour of electron around the nucleus.
Types of Quantum Numbers:
There are four quantum numbers which describe the electron completely.
(i) Principal quantum number (n)
(ii) Azimuthal quantum number ( )
(iii) Magnetic quantum number (m)
(iv) Spin quantum number (s)
(i) Principal quantum number (n):
➢ A quantum number that describes the distance with respect to the nucleus, size and the
energy of the orbital is called principal quantum number.
➢ It gives us quantitative measure of size of electronic shell.
➢ It also tells us energy of a shell.
➢ It is represented by ‘n’ and its values are non-zero, positive integers upto infinity. n=1,
2,3,4,5 ‘n’ value indicates specific shell.
Example:
n=1 K-Shell n=2 L-Shell
n=3 M-Shell n=4 N-Shell
Greater the value of ‘n’, greater is the size of shell and greater will be the distance of electron from
nucleus.
Note: Followings are the applications of ‘n’ values:
(i) The number of sub shells in a shell is equal to its ‘n’ value (principal quantum number).
shell n-value sub-shells
K 1 1
L 2 2
M 3 3

97. (ii) The no. of orbitals in a shell can be calculated by ‘n2’ formula:
shell n-value orbitals
K 1 (1)2=1
L 2 (2)2=4
M 3 (3)2=9
(iii) The no. of electrons in a shell can be calculated by the formula 2n2.
shell n-value electrons
K 1 2(1)2= 2
L 2 2(2)2= 8
M 3 2(3)2= 18
(ii) Azimuthal quantum number ( ):
➢ This quantum number that gives us the idea of energy and shape of subshells in a shell.
➢ By concept of existence of subshells in a shell given by this quantum number, we can
explain one of the defects of Bohr (the splitting of spectral line in high power resolving
spectrometer).
➢ Azimuthal quantum no. is designated by ‘ ’ and its values are 0, 1, 2, 3, 4…………. etc.
➢ Its values relate to the shape of subshell.
Azimuthal Q. number Sub-shell Shapes
= 0
= 1
= 2
= 3
s-subshell
p-subshell
d-subshell
f-subshell
Spherical
dumbell
double dumbell or sausage shape
complicated
➢ The letters s, p, d, f stands for spherical, principal, diffused and fundamental respectively.
➢ These are the spectral terms used to describe the spectral series in the atomic spectrum.

98. Relationship between Principal and Azimuthal Quantum Number
The relationship is described as follows:
Principal
Q. No.
Shell Azimuthal
Q. No.
Sub-shells Called as
n=1
n=2
n=3
n=4
K-Shell
L-Shell
M-Shell
N-Shell
l = 0
l = 0
l = 1
l = 0
l = 1
l = 2
l = 0
l = 1
l = 2
l = 3
s
s
p
s
p
d
s
p
d
f
1s
2s
3s
4s
3s
4s
4s
4p
4d
4f
Applications of Azimuthal quantum number:
(i) The number of orbitals in a sub-shell is calculated by formula (2 +1)
Sub-shell Azimuthal
Q. number
Formula No. of orbitals
s
p
d
f
0
1
2
3
2 + l =2(0) + 1
2 + l =2(1) + 1
2 + l =2(2) + 1
2 + l =2(3) + 1
1
3
5
7
(ii) The number of electrons in a sub-shell is calculated by formula 2(2 +1)
Sub-shell Azimuthal
Q. number
Formula Electrons
s
p
d
f
0
1
2
3
2 (2 + l) = 2 (0) + 1
2 (2 + l) = 2 (2(1) + 1)
2 (2 + l) = 2 (2(2) + 1)
2 (2 + l) = 2 (2(3) + 1)
2
6
10
14
3. Magnetic or Orbital Orientation Quantum Number (m):

99. ➢ This quantum number describes about the no. of orientations (orbitals) in a subshell.
➢ It actually explains one of the defects of Bohr (the splitting of spectral lines in Magnetic
field).
➢ It is designated as ‘m’ and its values are 0,+1,+2,+3 ……..etc.
➢ The number of ‘m’ values tells us the no. of orbital in a sub-shells and can be calculated by
formula (2 +1).
Sub-shell (2 +1) No. of ‘m’ values No. of orbitals
s
p
d
f
2(0)+1
2(1)+1
2(2)+1
2(3)+1
2 (2 + l) = 2 (0) + 1
2 (2 + l) = 2 (2(1) + 1)
2 (2 + l) = 2 (2(2) + 1)
2 (2 + l) = 2 (2(3) + 1)
1
3
5
7
➢ The value of ‘m’ depends upon value of for a subshell and can be calculated by formula
m=- ⎯⎯
→0 ⎯⎯
→+ . (It means value of ‘m’ varies between - and + through zero)
-values sub-shells m-values No. of orbitals
0 s
- ⎯⎯
→0 ⎯⎯
→+
=0
One s-orbital
1 p
- ⎯⎯
→0 ⎯⎯
→+
-1 ⎯⎯
→0 ⎯⎯
→+1
=-1,0,+1
p-subshell has three
degenerate orbital
2 d
- ⎯⎯
→0 ⎯⎯
→+
-2 ⎯⎯
→0, ⎯⎯
→+2
= -2,-1,0,+1,+2
d-subshell has five
degenerate orbitals
3 f
- ⎯⎯
→0 ⎯⎯
→+
-3 ⎯⎯
→0 ⎯⎯
→+3
=-3.-2,-1,0,+1,+2,+3
f-subshell has seven
degenerate orbitals
Conclusion of these ‘m’ values
(i) s-subshell:
➢For s-subshell, =0 and m=0.
➢This ‘m’ value indicates that s-subshell of any shell has one space orientation. In other words,
the probability of finding the electron in all the directions from nucleus is same.
➢It is spherically symmetrical orbital.
➢Since, it is one in number, it is called as one fold degenerate orbital.
(ii) p-subshell:

100. ➢For p-subshell, =1 and m=-1,0,+1.
➢It indicates that in any energy level, p-subshell has three orientations i.e. can be
arranged along x,y and z-axis.
➢These three orbitals are perpendicular to each other and are named as px, py and pz.
➢In the absence of magnetic field, all the three orbitals have same energy and are called
degenerate orbitals.
➢Since, they are three in number they are called three fold degenerate orbitals or tripply
degenerate.
(iii) d-subshell:
➢For d-subshell, =3 and m=-2, -1, 0,+1,+2.
➢It indicates that d-subshell have five space orientations (orbitals).
➢They are designated as
dxy(m =-2), dyz(m =-1), dzx(m =+1), 2 2
x y
d −
(m = +2) and 2
z
d (m = 0).
➢In the absence of magnetic field, all the five d-orbitals have same energy and are called
degenerate orbitals.
➢Since, they are five in numbers, they are called five-fold degenerate orbitals.
(iv) f-subshell:
➢For f-subshell, =3 and m=-3,-2,-1,0,+1,+2,+3.
➢It indicates that they have seven space orientations (orbitals).
➢They have complicated arrangements and complicated shapes.
4. Spin Quantum Number (s):
➢ A quantum number that describes the spin of electron in a orbit to satisfy the magnetic
moment so as to avoid repulsion with each other.
➢ This quantum number explains the spin of electron and direction of magnetic field as a
result of that spin.
➢ In 1925, Goudsmit and Uhlenbech suggested that electron while revolving around the
nucleus also rotates around its own axis (spin of electron) either clockwise or anti-clock-
wise direction. This is called self-rotation.
➢ The clockwise and anti-clockwise spin of electron produces opposite magnetic field. This
spin motion of electrons is responsible for doublet line structure in the spectrum of Alkali
metals.

101. ➢ When outermost excited electron of alkali metals jumps back to ground state, each emitted
line seems to consist of a pair of lines when observed in high power resolving spectrometer.
This is called doublet line structure.
Principal
Quantum
number ‘n’
Azimuthal
Quantum
number ‘ ’
Magnetic
Quantum
number ‘m’
Spin Quantum
number ‘s’
Number of
electrons
accommodated
1 K 0 s 0
1 1
,
2 2
+ − 2
2 L
0 s 0
1 1
,
2 2
+ −
2
8
1 p +1, 0, -1
1 1
,
2 2
+ − 6
3 M
0 s 0
1 1
,
2 2
+ − 2
18
1 p +1, 0, -1
1 1
,
2 2
+ − 6
2 d +2+1, 0, -1, -2
1 1
,
2 2
+ − 10
0 s 0
1 1
,
2 2
+ − 2
32
0 p +1, 0, -1
1 1
,
2 2
+ − 6
2 p +2+1, 0, -1, -2
1 1
,
2 2
+ − 10
4 N 3 f
+3, +2+1, 0, -2, -
2, -3
1 1
,
2 2
+ − 14
Shapes of Orbitals:
(i) Shapes of S-orbitals:
➢ S-orbital is spherical in shape and is represented by a
circle (cut of sphere).
➢ Higher the value of ‘n’ for s-subshell, Greater will be its
size.
Example:
➢ 2s is larger than 1s.
➢ Nodal Surface or Nodal Plane:

102. ➢ The probability of finding the electron is zero between two orbitals. This plane is called
Nodal plane or Nodal surface.
(ii) Shapes of p-orbitals:
➢ Each p-orbital has two lobes in dumbel shape.
➢ These lobes are either oriented along x-
axis(Px), y-axis(Py) or z-axis (Pz).
➢ The size of p-orbital increases with increase in
its ‘n’ value.
Example: 3Px orbital is larger in size than 2Px but
both have same shape.
(iii) Shapes of d-orbitals:
➢ Each d-orbital has four lobes oriented in space in double dumbell or sausage shape except
one i.e. 2
z
d which has two lobes oriented in space in dumbell shape.
➢ Three d-orbitals have lobes between the
axis i.e. dxy(between x and y-axis), dyz
(between y and
z-axis) and dzx (between x and z-axis).
While rest of two d-orbitals have lobes
on the axis i.e. 2 2
x y
d −
(two lobes on x-axis and two on y-axis)
and 2
z
d (has only two lobes on z-axis).
(iv) Shapes of f-orbital:
The shape of f-orbital is very complicated.
Electronic Distribution
The arrangement or filling of electrons in the orbitals around the nucleus of
an atom is called electronic configuration.
The arrangement of electrons in subshells or orbitals is according to
following rules.
(i) (n + l) rule
(ii) Auf-bau principle
(iii) Hund’s rule
(iv) Pauli’s Exclusion Principle
Before going into the details of these rules, we must remember.
(a) A subshell can accommodate electrons as described below:
(b) An orbital can accommodate two electrons only
(1) (n+ ) rule:
N n +
1s 1 0 1 + 0 = 1
2s 2 0 2 + 0 = 2
2p 2 1 2 + 1 = 3
3s 3 0 3 + 0 = 3
3p 3 1 3 + 1 = 4
3d 3 2 3 + 2 =5
4s 4 0 4 + 0 =4
4p 4 1 4 + 1 = 5
4d 4 2 4 + 2 = 6
4f 4 3 4 + 3 = 7
5s 5 0 5 + 0 = 5
5p 5 1 5 + 1 = 6
5d 5 2 5 + 2 = 7
5f 5 3 5 + 3 = 8
6s 6 0 6 + 0 = 6
6p 6 1 6 + 1 = 7
6d 6 2 6 + 2 = 8
6f 6 3 6 + 3 = 9
7s 7 0 7 + 0 = 7

103. ➢ This gives us the arrangement of electrons in subshells. It has two parts:
• The electrons are filled in subshells in increasing order of their (n+ ) values.
• If two subshells have same (n+ ) values, then the subshell with low ‘n’ value will be
filled first.
➢ This rule gives us the arrangement of subshells in increasing order of their energy.
Ascending order of energy of sub-shells on the basis of (n + l) rule
1s < 2s < 2p < 3p < 3p < 4s < 3d < 4p 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s
(2) Auf-bau Principle:
➢ According to this principle, electrons should be filled in subshell in
order of their increasing energy values. It gives the same energy
order as given by (n + l) rule.
Note: Both (n + l) rule and Auf-bau Principle explain the filling of electrons
in subshells.
(3) Hund’s Rule:
➢ It explains the filling of electrons in degenerate orbtials.
➢ According to this rule, when degenerate orbitals are available to more than one electrons,
then they are placed in separate orbitals with same spin rather than in same orbitals with
opposite spin.
Or
➢ When more than one electrons are to be placed in degenerate orbtials, then they are
placed to have maximum number of unpaired electrons possible.
Example: 2 2 1 1 0
6 x y z
C 1s , 2s , 2p , 2p , 2p
=
(4) Pauli’s Exclusion Principle:
➢ It explains the arrangement of electrons in an orbital.
➢ According to this principle, it is impossible for two electrons residing in the same orbital of
poly-electron atom to have same values of four quantum numbers.
Or
➢ No two electrons in same orbital can have same values of all the four quantum numbers.
For example two electrons in 2s.
n=2, =0, m=0, and s = +
1
2
and -
1
2
➢ Spin Quantum number has two values, means one electron have clockwise spin and other
anti-clockwise spin.
Or
➢ No two electrons in the same orbital can have same spin.
Or
sub-shell electrons
s 2
p 6
d 10
f 14

104. ➢ Two electrons in the same orbital should have opposite spin.

105. Assessment 1
1. Positive rays are also termed as canal rays which are produce by
a. By combustion of gas.
b. By cooling of the gas.
c. By the ionization of gas by cathode rays.
d. Anode electrode as cathode rays is produced.
2. Positive rays are also termed as canal rays. These rays give flash on screen coated with
a. AgCl
b. ZnO
c. AgNO3
d. ZnS
3. Positive rays are also termed as canal rays or anode rays. Which of the following is not true:
a. Their e/m ratio is constant
b. They are deflected by electrical and magnetic field
c. They are produced by ionization of molecules of the residual gas
d. Their e/m ratio depends on nature of residual gas
4. e/m for positive rays changes with the change in gas to be filled in glass discharge tube. This
value is maximum for which gas
a. Helium
b. Helium
c. Oxygen
d. Hydrogen
5. The canal ray with one proton bears the following charge:
a. -1.602x10-19C/kg
b. -1.602x10-19C
c. +1.602x10-19C/mol
d. +1.602x10-19C
6. Max planks proposed the quantum theory in 1900 to explain the emission and absorption of
radiation. Which is true among these
a. Energy travels in continuous form
b. Energy is emitted or absorbed continuously
c. Energy is not emitted or absorbed continuously
d. In case of light, energy packet is called quanta

106. ➢
7. Electromagnetic radiation travels through vacuum at a speed of --------------- m/s.
a. 1000
b. 3.0x106
c. 3.0x108
d. Change with wavelength
8. Max Planck presented the Planck’s quantum theory. Planck’s constant is represented by “h”
and its value is
a. 6.626 x 10 -31Js
b. 6.626 x 10-34 Js
c. 3.345 x 10-34 Js
d. -2.178 x 10-18 Js
9. Planck’s constant (h) is the ratio of
a. energy and wave number of photon
b. energy and frequency of photon
c. energy and wave number of photon
d. wave number and wavelength of photon
10. The relation between energy and wave number of photon is given by the equation:
a.
b.
c.
d.

107. ➢
Assessment 2
1. Who has given the relationship between energy of the photon and its frequency?
a. Bohr’s atomic model
b. de-Broglie
c. Planck’s quantum theory
d. Einstein mass energy relationship
2. According to Planck’s quantum theory, Frequency is related to the wavelength of photon
a. Frequency is equal to the wavelength of photon
b. Frequency is inversely proportional to wavelength
c. Frequency is directly proportional to wavelength
d. Both a and c
3. Which of the following is not purposed idea given by Bohr in his theory
a. Electrons revolve around in nucleus in orbits of fixed energy
b. Energy of electron is proportional to n2
c. Electron falls to lower energy on releasing electron
d. Electron radiate energy continuously while residing in an orbit
4. According to Bohr, the characteristic spectrum shown by an atom is
a. Continuous spectrum
b. Line emission spectrum
c. Line absorption spectrum
d. Both b and c
5. Bohr’s atomic model does not account for the
a. Sodium
b. Chlorine
c. Helium
d. All of the above
6. Apple green colour of flame shown by barium salt is due to:
a. Removal of its atoms in flame
b. More vibration and collision of its atoms
c. De excitation of electrons to emit light
d. Excitation of electron

108. ➢
7. On the de-excitation of electrons, light is emitted in the ultraviolet region. which of the
following statement is correct
a. red color is emitted
b. violet color is emitted
c. white color is emitted
d. no color is detected
8. According to Bohr’s atomic model, angular momentum of an electron residing in K shell is
given by the following relation:
a.
b.
c.
d.
9. The second line of the Balmer series originates in the emission spectrum of the hydrogen
atom, is due to the transition from the
a. Third Bohr orbit to the first Bohr orbit
b. Fourth Bohr orbit to the second Bohr orbit
c. Sixth Bohr orbit to the third Bohr orbit
d. Fifth Bohr orbit to the third Bohr orbit
10. No of wave which are present in unit length is called wave number.In uni electron system, the
wave number of any spectral line is directly proportional to
a. The number of electrons present in the system
b. The velocity of electron undergoing transition
c.
d. The charge on the nucleus

109. ➢ Assessment 3
1. The model which has well defined approach about the electron in hydrogen atom is Bohr’s
atomic model.Which statement is false according to Bohr’s model of the atom?
a. Electrons cannot be between energy levels
b. Electrons orbit the nucleus
c. An electron’s path is not known exactly
d. Electrons exist in energy levels
2. Every addition of electron is carried out in the orbital of lowest energy. This is the basic idea
drawn from:
a. Pauli’s exclusion principle
b. Hund’s rule
c. Aufbau principle
d. de-Broglie’s equation
3. Line spectrum is the characteristics of an element. The line spectrum of two elements is not identical
because .
a. they do not have same number of neutrons
b. they have dissimilar mass number
c. they have different energy level schemes
d. they have different number of valence electrons.
4. An electron of mass m and charge e-
moves in circular orbit of radius r around the nucleus ofcharge +
Ze in uni electron system. The electrostatic force of attraction between nucleus and electron is given as
a.
b.
c.
d.
5. Which of the following is direct contradiction of Bohr’s concept of electrons revolving around the
nucleus in circular orbits of fixed energy?
a. Line spectra of hydrogen atom
b. Pauli’s principle
c. Planck’s theory
d. Heisenberg’s principle
6. If r is the radius of first orbit, the radius of nth
orbit of the H atom will be
a. r x n2
b. r x n
c.
d. r2
x n2

110. ➢
7. Bohr calculated the radius of nth orbit.The radius of second Bohr’s orbit is
a. 0·053 nm
b. 0·053/4 nm
c. 0·059 x 4nm
d. 0·053 x 20 nm.
8. If the radius of first Bohr orbit be a0, then the radius of third Bohr orbit would be
a. 3 x a0
b. 6 x a0
c. 9 x a0
d. 1/9 x a0
9. The radius of 5th orbit which is calculated by Bohr’s atomic model is given as:
a. 25r0
b. 16r0
c. 9r0
d. 4r0
10. Which of the following is true statement representing the postulate of Bohr’s atomic model
a. electron on H atom can have only certain values of angular momentum
b. electrons have a particle as well as wave character
c. atomic spectrum of atom should contain only five lines
d. all the above statements are correct.

111. ➢ Assessment 4
1. Which of the followings are measured by Bohr except
a. radius of nth
orbit
b. angular momentum
c. energy
d. probability of finding an electron
2. Bohr’s atomic model opposes Heisenberg’s uncertainty principle in respect of
a. assign radius
b. assign energy
c. assign position and momentum
d. all of these
3. In the early 20th century, Danish physicist Niels Bohr discovered the basic atomic structure; a positively
charged nucleus surrounded by orbiting electrons. The one which is discrete in Bohr’s theory:
a. Velocity
b. Angular momentum
c. Kinetic energy
d. Potential energy
4. Bohr’s atomic model serves a prominent position in atomic theory but had limitations which are:
a. splitting of spectral line under the action of magnetic field
b. elliptical orbits
c. stark’s effect
d. all of these
5. Which of the following expression restricts the condition of quantization of energy of an electron in an
atom?
a.
b.
c. λ
d.

112. ➢
➢
➢
6. Bohr deals with uni electronic systems. Which of the following is not dealt by Bohr?
a.
b.
c.
d. helium
7. Bohr’s atomic model failed with the advent of which of the following:
a. Compton’s effect
b. Heisenberg’s uncertainty principle
c. Plank’s quantum theory
d. Rutherford’s atomic model
8. Balmer series is the first spectral series observed. It originates when the following transition
takes place:
a. higher shells to K shell
b. higher shells to L shell
c. higher shells to M shell
d. higher shells to N shell
9. When an electron is brought from infinity to distance r from the nucleus, the potential energy
of the electron
a. increases
b. decreases
c. remains the same
d. increased two times
10. Spectral line is characterized by the frequency. Which of the following are units of frequency
except
a. cycles/sec
b. hertz
c. sec-1
d. k hertz/sec

113. ➢
➢
KEY
Assessment 1
Assessment 2
1. c
2. d
3. a
4. d
5. d
6. c
7. c
8. b
9. b
10. a
1. c
2. b
3. d
4. b
5. d
6. c
7. d
8. a
9. b
10. c

114. Key
Assessment 3
Assessment 4
1. c
2. c
3. c
4. a
5. d
6. a
7. c
8. c
9. a
10. a
1. d
2. c
3. b
4. d
5. d
6. d
7. b
8. b
9. b
10 d

115. Oldest, Largest and Most Credible Platform
MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org

116. Learning Objectives:
Energetic of bond formation
Atomic sizes
Atomic radii
Ionic radii
Covalent radii
Ionization energy
Electron affinity
Electro negativity
Bond energy
Bond length
Types of bonds
Electrovalent or Ionic Bond
Covalent bond
Co-ordinate or dative covalent bond
Ionic character of covalent bond
Sigma and Pi bond
Hybridization
sp3 -Hybridization
sp2 -Hybridization
sp-hybridization
The Valence Shell Electron Pair Repulsion theory
Postulates of VSEPR theory
Applications of VSEPR theory
Chemical Bonding

117. The attractive force that binds two or more atoms or ions or molecules together in a compound is called as
chemical bond.
Cause of Chemical Combination:
The cause of Chemical Combination is the tendency of elements
(i) to attain stable electronic configuration (like noble gases)
(ii) to lower their energy by bond formation.
How bond is formed:
(i) gaining electrons
(ii) losing electrons
(iii) or sharing electrons with other elements.
Octet Rule:
The tendency of elements to attain maximum of eight electrons in their valence shell like noble gas
(except He) is called Octet rule.
(i) Some elements complete octet by losing electrons
Deviation of Octet rule:
After sharing the electrons, some elements may have either more or less than eight electrons. This
is called deviation from octet rule.
There are two types of deviations.
(i) shortened or deficient octet (ii) extended octet
(i) Shortened or deficient Octet:
When elements have less than eight electrons after sharing, then they are said to have
shortened octet.
Example: BF3, BH3, AlCl3
(ii) Expanded Octet:
When elements have more than eight valence electrons after sharing, then they are said to
have expanded octet.
Example: PCl5, SF6

118. Atomic Sizes
1. Atomic Radius:
The average distance between the nucleus and the outermost shell of an atom is called as atomic
radius. It is measured in pm or Ao.
Measurement:
Atomic radius can be determined by measuring the distances between the centres of adjacent
atoms by X-rays or spectroscopic measurements.
The radius of atom can not be measured precisely due to the following reasons.
(i) There is no sharp boundary of atom. The probability of finding the electron never becomes
exactly zero at large distances from nucleus.
(ii) The probability of electronic distribution is affected by neighbouring atoms. This may result in
change in size.
Factors affecting the atomic radius:
(1) Nuclear force (Nuclear charge):
The atomic radius decreases with increase in nuclear force and vice versa.
(2) Shielding effect or screening effect:
The decrease in force of attraction between nucleus and valence electrons due to intervening
electrons (inner shell electrons) is called shielding or screening effect. It increases with increase in number
of electrons present between nucleus and valence electrons. Also, it increases with increase in no. of
shells.
It means nuclear force decreases with increase in shielding effect and atomic radius also increases
and vice versa.
Trends of Atomic radius
1. Along the Period:
The atomic radius decreases along the period due to increase in proton number
(or nuclear charge).
2. Down the group:
The atomic radius increases down the group due to decrease in nuclear force. The decrease in
nuclear force is due to the following reasons.
(i) The successive addition of shells as we move down the group.
(ii) The increase in shielding effect is due to increase in number of shells as we move down the
group.

119. Ionic Radii:
The radius of an ion considering it to be spherical in shape is called Ionic radius.
Units:
It is measured in pm or Ao.
Trends:
1. Along the period
The ionic radius decreases along the period.
2. Down the group:
The ionic radius increases down the group.
Types of ionic radii:
1) Cationic radius:
The radius of cation is always smaller than its neutral atom due to following reasons.
(i) Increase in effective nuclear charge:
The cation is formed when an atom losses electron(s). This leads to an increase in effective
nuclear charge which attracts the valence electrons more strongly towards the centre. As a
result, size decreases.
(ii) Loss of valence shell:
The loss of electron(s) by an atom to attain the nearest noble gas configuration is usually
accompanied by loss of valence shell. As a result, size decreases.
Example:
The radius of Na-atom is 186pm and that of Na+ ion is 95 pm.
Note: Greater the positive charge on an ion, greater is decrease in the size. The order of sizes of
iso-electronic cations (cations having same number of electrons) is as follows.
Na+ > Mg+2 > Al+3
2) Anionic radius:
The radius of negatively charged ion is called anionic radius.
The radius of anion is always greater than its neutral atom due to increase in electron-electron
repulsion in the valence shell. This repulsion is actually due to addition of an extra electron in the valence
shell.
Example:
The radius of Cl-atom is 99pm and that of Cl- ion is 191 pm.
Greater the charge on anion, greater is increase in size. It is due to increasing repulsion with the
addition of every electron. The order of anionic sizes of iso-electronic anions (Anions having same number
of electrons) is as follows:
N-3 > O-2 > F-

120. Interionic Distance in Crystal Lattice:
Ionic radius is an additive property because inter-ionic distance
R(distance between two adjacent oppositely charged ions) in the crystal
lattice is equal to sum of cationic radius r+ and anionic radius r-
R = r+ + r-
Example:
Pauling determined the distance between K+ and Cl- ions in KCl crystal and found that it was equal
to sum of ionic radii of both ions.
R = r+ + r-
R = 133pm + 181pm = 314pm
This concept is extended to calculate the radius of one ion when we know radius of other ion and
inter-ionic distance.
For cation r+ = R- r-
For anion r- = R- r+

121. Covalent Radii:
The half of the bond length between two covalently bonded similar atoms in a molecule is called
covalent radius.
Example:
(i) Covalent radius of Hydrogen = 37.7pm
(ii) Covalent radius of Carbon = 77pm
(iii) Covalent radius of Chlorine = 99pm
(iv) Covalent radius of Fluorine = 64pm
Trend: The trend of covalent radius is same as that of atomic radius.
(1) Along the period:
The covalent radius decreases along the period due to increase in
nuclear force.
(2) Down the Group:
The covalent radius increases down the group due to decreasing
nuclear force.
Ionization Energy or Ionization Potential:
The minimum energy required to remove most loosely bound electron from isolated gaseous atom
to form cation is called as ionization energy.
Example:
H(g) + heat energy ⎯⎯
→ H+
(g) + 1e- ∆H = 1313.315 kJ/mole
Types of Ionization Energies:
(i) The first ionization energy:
The minimum energy required to remove first electron from valence shell of an isolated
gaseous atom (neutral) to form monopositive ion is called first ionization energy.
Example:
Mg(g)
heat
⎯⎯→ Mg+
(g) + le- ∆H1=738 kJ/mole
(ii) Second ionization energy:
The minimum energy required to remove second electron from monopositive cation to form
dipositive cation is called second ionization energy.
Example:
Mg+
(g) ⎯⎯
→Mg+2
(g) + 1e- ∆H2 = 1450 kJ/mole
The second I.E is always greater than first one because increased nuclear charge on mono-
positive ion binds valence electrons more tightly than neutral atom. Hence, greater amount
of energy is required to remove electron.

122. (iii) Third ionization energy:
The amount of energy required to remove the electron from dipositive cation to form
tripositive cation is called third ionization energy.
Example:
Mg+2
(g)
heat
⎯⎯
⎯
→Mg+3
(g) ∆H3 = 7730 kJ/mole
The third I.E. is far greater than 2nd and 1st ones because much enhanced positive charge on
dipositive cation binds the valence electrons much more tightly than neutral atom and
monopositive cation. Hence, greatest amount of energy is required to remove the electron.
The order of ionization energy is
3rd I.E. > 2nd I.E > 1st I.E.
Factors Affecting the Ionization Energy:
(i) Atomic radius (ii) Nuclear charge (iii) Shielding effect (iv) Nature of orbital
(i) Atomic Radius:
The ionization energy is inversely related with atomic radius. The valence electrons of an
atom of smaller radius are closer to the nucleus and are tightly bound. Hence, greater
amount of energy will be required to remove electron and vice versa.
(ii) Nuclear Charge:
The ionization energy is directly related to the nuclear charge. The valence electrons of an
atom of greater nuclear charge are tightly bound to the nucleus and are difficult to remove.
Hence, greater energy will be required to remove electron and vice versa.
(iii) Shielding effect:
The ionization energy is inversely related to shielding effect. The inner electrons form a shield
between nucleus and valence electron. This shield decreases the nuclear pull on valence
electrons. The shielding effect increases with the increasing number of inner electrons (or
inner shells).
The valence electrons of an atom of greater shielding effect are loosely bound and are easily
removable. Hence, less energy will be required and vice versa.
(iv) Nature of orbital:
The ionization energy for an electron in specific orbital depends upon its attraction with
Nucleus. Greater the attraction of an electron with Nucleus in a specific orbital, Greater will
be the ionization energy. The extent of attraction and ionization energy for an electron in
different orbitals is in the following order.
s > p > d > f
Trend (Variation) of Ionization energy in the periodic table
(i) Down the group:
The ionization energy decreases down the group due to decreasing Nuclear force which is
due to
(a) Successive addition of valence shells:

123. As a result of this, valence electrons are placed away from nucleus and are easily removable.
(b) Shielding effect:
It increases with increasing number of shells.
(ii) Along the Period:
The ionization energy increases along the period with increasing Nuclear force/charge which
is due to.
(a) Increasing Proton number:
The Nuclear charge on the nucleus and negative charge (electrons) in the valence shell
increases with increasing proton number along the period. As a result, attraction increases
and electrons become more tightly bound.
(b) Constant no. of shells:
2nd period
Li
520
Be
900
B
801
C
1086
N
1402
O
1314
F
1681
Ne
2081
Ionization energy (in kJmol-1)
Increases across a period
Discrepancies (Abnormalities) of ionization energy along the period:
The ionization energies of group IIIA and VIA show
abnormal trend. This can be understand from the distribution of
electrons.
Reason:
The Groups III-A and VI-A show abnormal values of
ionization energies.
(i) The ionization energy of III-A is less than II-A, which should be reverse.
In case of II-A, electrons are removed from completely filled s-orbital in which they are
attracted more strongly by nucleus. Completely filledd orbital is more stable as well. Hence,
greater amount of energy is required to remove the electron by breaking the pair.

124. Whereas in case of III-A, electron is removed from p-orbital. As p-orbital extends out more,
the electrons in it are weakly attracted by the nucleus. Partially filled p-subshell is less stable
as well. Hence, lesser energy is required to remove electron.
(ii) The ionization energy of VI-A group is lesser than V-A Group which should be reverse.
Reason: In case of V-A, electron is removed from p-sub-shell which is half filled (p3) and extra
stable. Greater amount of energy is required to remove electron from extra stable p-subshell.
Whereas, in case of VI-A, p-subshell has 4-electrons and is less stable. Lesser energy is
required to remove electron from less stable p-subshell.
Ionization energy and valency:
The big difference between successive ionization energies indicates that valency of element is
equal to lower I.E value amongst the two.
(i) If there is big difference between 1st and 2nd I.E, then valency is one eg. I-A.
(ii) If there is sufficient gap between 2nd and 3rd I.E, then valency is two e.g. II-A.
(iii) If there is big difference of energy between 3rd and 4th, then valency is three e.g. III-A and so
on.
Ionization energy and Metallic character:
Ionization energy is an index to metallic character. Rather it is inversely related to metallic
character.
(i) Elements with lowest ionization energies are metals.
(ii) Elements with Intermediate ionization energies are metalloids.
(iii) Elements with greatest ionization energies are Non-metals.
Electron Affinity:
The amount of energy released when an electron is added to the valence shell orbital of an isolated
gaseous atom to form uninegative ion is called electron affinity. It is measured in kJ/mole units.
Cl(g)+1e- ⎯⎯
→ Cl ( )
g
−
∆He = -349 kJ/mole
Since, energy is released, hence, electron affinity for uninegative ion formation is given with the
negative sign.
Note: The first electron affinity is a measure of attraction of nucleus of an atom for extra electron.
Types of electron affinity:
(i) First electron affinity:
The energy released when first electron is added to the valence shell of an isolated gaseous
atom to form uninegative ion is called first electron affinity.
Example:
O(g)+1e- ⎯⎯
→ O( )
g
−
∆He = -141kJ/mol
Note: First electron affinity is always exothermic.

125. (ii) Second electron affinity:
The amount of energy absorbed when second electron is added to uninegative ion to form
dinegative anion is called second electron affinity.
Example:
O ( )
g
−
+le- ⎯⎯
→ O-2
(g) ∆He = +780kJ/mol.
Second electron affinity is positive (endothermic). It is because in this case electron is added
to uninegative ion where it faces a repulsion with already present negative charge on ion.
Hence, energy is absorbed to overcome this repulsion.
Factors affecting the Electron Affinity:
(i) Atomic size (ii) Nuclear charge (iii) Shielding effect
(i) Atomic radius:
Electron affinity is inversely related to atomic radius. Smaller the size of an atom, Greater will
be the electron affinity and vice versa.
(ii) Nuclear Charge:
Electron affinity is directly related to nuclear charge. Greater the nuclear charge, greater is
the attraction of nucleus for extra electron and greater will be the electron affinity and vice
versa.
(iii) Shielding effect:
Electron affinity is inversely related to shielding effect greater the shielding effect, lesser will
be the attraction of nucleus for extra electron and lesser will be electron affinity and vice
versa.
Trend of Electron Affinity:
(i) Down the Group:
Electron affinity decreases down the group due to decrease in effective nuclear force and
increase in shielding effect (due to successive addition of shells down the group). When
shielding effect of inner electrons around the nucleus increases, the attraction of Nucleus for
extra electron decreases and hence, electron affinity decreases.
(ii) Along the period:
Electron affinity increases along the period due to increasing nuclear charge (force).The
nuclear charge increases along the period due to increasing proton number in the nucleus
and electrons in the valence shell. The increasing nuclear charge has greater attraction for
extra electron. Due to which electron affinity increases.
Abnormality of Electron Affinity
i) Along the period:
The Group II-A, V-A and VIII-A have abnormally low values of electron affinity in every period.
• The electron affinity of II-A is lower than I-A. Actually II-A elements have filled s-
subshell and cannot accommodate an extra electron in it. Hence, have low electron
affinity than I-A which have half filled s-subshell and can accommodate an extra
electron.

126. • The electron affinity of V-A is lower than IV-A. In V-A elements, there are three
electrons in p-subshell (half filled) and are stable with this configuration. The addition
of extra electron will destablize the p-configuration. Due to which they show less
electron affinity than (IV-A Which have np2 configuration and addition of extra electron
will stabilize their configuration to p3).
• The VIII-A Group elements show zero values of electron affinity because they have
complete octets and no space for extra electron.
ii) Down the group VII-A:
The electron affinity of Fluorine is less than that of chlorine which should be reverse.
Reason:
Fluorine has small size and seven electrons in L-shell that results in thick electronic cloud. This
thick electronic cloud repels the incoming electron. Energy is absorbed to overcome this
repulsion. Due to which electron affinity decreases.
Electron affinity is
(i) a measure of attraction of an atom for extra electron.
(ii) a measure of stability of anion.
(iii) an index to non-metallic character.
Electronegativity:
The ability of an atom to attract shared pair of electron towards itself is called Electronegativity.
Explanation:
In a covalent bond between two similar atoms, the shared pair is equally shared between the
atoms.
Examples: H2, Cl2, N2.
Whereas, in a covalent bond between two dissimilar atoms, the shared pair is not equally shared
between the two atoms.
Example:
In HF molecule, fluorine atom has greater tendency to attract shared pair than Hydrogen.

127. Measurement:
Pauling calculated the electronegativities of elements from the difference between expected bond
energies for normal covalent bond and experimentally determined values.
He devised an electronegativity scale on which Fluorine is given an arbitrary standard value of 4.0.
It is the most electronegative element and electronegativity values of other elements are compared with
it.
Trends of Electronegativity:
(i) Down the group:
The electronegativity values of elements decrease down the group due to increase in size and
decrease in nuclear force.
(ii) Along the period:
The electronegativity values of elements increase along the period due to decrease in size
and increase in nuclear force.
Electronegativity difference as an index to bond nature:
(i) The bond between two atoms is non-polar covalent if electronegativity difference between
them is zero or less than 0.5.
Examples:
H2,Cl2,N2,O2,CH4 etc.
(ii) The bond between two atoms is polar covalent if electronegativity difference between them
is 0.5 to 1.6.
Examples:
NH3, H2O, HCl etc.
(iii) The bond between two atoms is equally covalent and ionic if electronegativity difference is 1.7.
(iv) The bond between two atoms is ionic when electronegativity difference between two atoms
is greater than 1.7.
Examples:
NaCl, KCl, CsCl etc.
Ionic bond or Electrovalent bond:
The electrostatic force of attraction between oppositely charged ions is called ionic bond or
electrovalent bond.
Explanation:
According to Lewis Concept.
Ionic bond is formed by the complete transfer of electrons from more electropositive elements to
more electronegative elements. It is easy to loose electron for electropositive elements since their
ionization energy is low whereas it is easy to gain electron for electronegative elements because their
electron affinity is high. It is their opposite tendency that help them to form cation and anion which bind
through electrostatic force to form ionic bond.
Ionic bond is non-rigid and non-directional.

128. Conditions necessary for Ionic Bond Formation
(1) Electropositive elements with low ionization energy like I-A and II-A that lose electron and
form cation.
(2) Electronegative elements with high electron affinity like VI-A and VII-A that gain electron and
form anion.
Examples:
(1) Formation of KCl:
(i) Cation formation:
Potassium (K19) with atomic no. 19 have 19 electrons. It looses one electron by absorbing 419
kJ/mole energy (Ionization energy) and form K+ with 18 electrons which is noble gas
configuration (Ar18).
Example:
K(g) ⎯⎯
→ K( )
g
+
+ le-  H = 419 kJ/mole
2, 8, 8, 1 2, 8, 8
(ii) Anion formation:
Chlorine (Cl17) with atomic no. 17 have 17 electrons. It Gains one electron lost by ‘K’ and form
Cl- with 18 electrons and attains the noble gas configuration (Ar18).
Example:
Cl(g) + le- ⎯⎯
→ Cl-
(g)  H = -349 kJ/mole
2, 8, 7 2, 8, 8
Combination of cation and anion
These cations and anions then bind up through electrostatic force to form solid Ionic
compound. When cations and anions combine to form 1 mole of solid Ionic compound,
energy is released called lattice energy.
Example:
K ( )
g
+
+Cl-
(g) ⎯⎯
→ KCl(s)  Hl = -690 kJ/mole
(2) Formation of CaO:
(i) Calcium looses two electrons to form dipositive ion and attains noble gas configuration (Ar18).
Example:
Ca(g) ⎯⎯
→ Ca+2
(g) + 2e-
2,8,8,2 2, 8, 8
(ii) Oxygen gains two electrons to form dinegative ion and attains the noble gas configuration
(Ne10).
Example:
O(g) + 2e- ⎯⎯
→ O-2
(g)
2, 6 2, 8
(iii) Then Ca+2 and O-2 combine to form CaO.

129. Ca+2
(g) + O 2
( )
g
−
⎯⎯
→CaO(s)
(3) Formation of Al2O3
(i) Aluminium loses three electrons and form tripositive cation and attains noble gas
configuration (Ne10).
Example:
Al(s) ⎯⎯
→ Al+3
(g) + 3e-
2,8,3 2, 8
(ii) Oxygen gains two electrons and form dinegative ion and attains the noble gas configuration.
Example:
O(g) + 2e- ⎯⎯
→ O 2
( )
g
−
2,6 2, 8
(iii) Al+3 and O-2 combine to form Al2O3
Al +3
(g) + O 2
( )
g
−
⎯⎯
→ Al2O3(s)
Points about Ionic bond
• Ionic bond is non-directional
• Ionic bond is non-rigid
• No bond is 100% ionic in nature
• % of ionic character in NaCl is 72% and CsF is 92% (Maximum ionic character).
Covalent Bond (Lewis and Kossel explain the covalent bond):
A bond formed by mutual sharing of electron between the two atoms is called covalent bond.
Example:
When two hydrogen atoms share their valence electrons, a covalent bond is formed.
H H
 • ⎯⎯
→ H H
⎯
Types of covalent bond (On the basis of number of bonds between the two atoms):
(i) Single covalent bond:
A bond formed by mutual sharing of one electron pair between two atoms is called single
covalent bond.
or
A bond formed by the sharing of one electron from each of the bonding atoms is called single
covalent bond. It is represented by (―).

130. Examples:
1- Hydrogen H + H H H or H H
 • •
→ ⎯
2- Chlorine
3- Methanol
4- Ethane
(ii) Double covalent bond
A bond formed by mutual sharing of two electron pairs between two atoms is called as
double covalent bond.
Or
A bond formed by sharing of two electrons from each of the bonding atoms is called double
covalent bond. It is represented by two dashes (=)
Examples:
1. Oxygen
2. Ethene
3. Carbon dioxide
(iii) Triple covalent bond
A bond formed by the mutual sharing of three electron pairs between the two bonding atoms is
called triple covalent bond.
Or
A bond formed by the sharing of three electrons from each of the two bonding atoms is called as
triple covalent bond. It is represented by three dashes ().
Examples:

131. (1) Nitrogen
(2) Ethyne
Types of covalent bond (on the basis of electronegativity difference):
(i) Non-polar covalent bond
A bond in which shared pair is equally distributed between two bonded atoms is called as non-
polar covalent bond. It is usually formed when bonded atoms have electronegativity difference
zero or less than 0.5.
Examples:
1- Hydrogen
2- Chlorine
3- Tetrachloromethane
4- Disilane
(ii) Polar covalent bond:
A bond in which shared pair is unequally distributed between the two bonded atoms is called polar
covalent bond. It is usually formed when bonded atoms have electronegativity difference 0.5 to
1.7.
1. Water
2. Hydrogen fluoride
3. Methylchloride

132. Non-Polar Covalent Bond Polar Covalent Bond
A covalent bond formed between two similar
atoms is called non-polar covalent bond.
A covalent bond formed between two dissimilar
atoms having appreciable difference of
electronegativity is called polar covalent bond.
Attraction
The shared pair of electrons is equally attracted
by both of the bonded nuclei.
The shared pair of electrons is unequally attracted by
the bonded nuclei.
Distribution
The electronic cloud is symmetrically distributed
over the whole molecule.
The electronic cloud is asymmetrically distributed
over the molecule.
Charge Separation
There is no charge separation. One end of the molecule having more electronegative
element acquires partial negative charge (-) whereas
other end acquires equal partial positive charge (+).
Electronegativity Difference
There is either no electronegativity difference
between bonded atoms or negligible (0.4).
Greater is the electronegativity difference, greater is
the polarity or ionic character in the covalent bond.
Example
H2 and CH4
H=2.1 C=2.5
H=2.1 H=2.1
EN = 0.0 0.4
(negligible difference)
H2O > NH3
O=3.5 N=3.0
H=2.1 H=2.1
EN = 1.4 0.9
Co-ordinate Covalent Bond (Dative bond):
A covalent bond formed by the donation of an electron-pair by one of the bonded atoms to other is
called Co-ordinate covalent bond or dative or Donor-acceptor bond.
This bond is formed by overlaping of one completely filled orbital with one empty orbital. An atom
that donates a pair is called donor atom (Lewis base) and one that accepts a pair is called acceptor
(Lewis acid). This bond is represented by an arrow pointing from donor to acceptor.

133. Examples:
1. Nitrogen atom in 3
NH
••
have lone pair and it donates it to Boron in BF3 (which is electron
deficient of octet) to form a co-ordinate covalent bond.
Donor Acceptor (complex)
In this case donor develops a positive charge and acceptor gets negative charge.
In Co-ordinate covalent formation
(i) If both donor and acceptor are neutral, before the bond formation, then donar will get positive charge
and acceptor will get negative charge after the bond formation.
(ii) If donar is negatively charged and acceptor is positively charge before bond formation, then both will
become neutral after bond formation.
2. Hydronium ion formation:
Oxygen in water donates a lone pair to form co-ordinate covalent bond to give H3O+. After this
bond formation, distinction between covalent and Co-ordinate covalent bond vanishes i.e. all the
bonds behave alike with 33% co-ordinate covalent and 66% covalent character.
3. Oxonium ion formation:
Just like water, oxygen of alcohols and ethers donate its lone pair to H+ to form co-ordinate
covalent bond and oxonium Ion.
(i) Alcohols (ii) Ethers
4. Compounds of nitrogen and its family members.
(i) Ammonium ion formation
Nitrogen in NH3 donates its lone pair to H+ to form Co-ordinate covalent bond and to give
+NH4 (Ammonium ion)
+ -
H F
H F
F F
H H
H N B F H N B F
 
 
 
 
 
 
⎯ + ⎯ ⎯⎯
→ ⎯ → ⎯
 
 
 
:
H
O H O
•• + +
••
••

+ ⎯⎯
→
Hydronium Ion
H
H H
H
R H R H
H
O H O
•• + +
••
••

+ ⎯⎯
→
Oxonium Ion
R R R
H
O H O
•• + +
••
••

+ ⎯⎯
→
Oxonium Ion
R
+
H
H
H N + H H N H
H H


+
 
 
 
⎯ ⎯⎯
→ ⎯ →
 
 
 
:

134. Ammonium ion
Similarly, Nitrogen in primary (RNH2) secondary (R2NH) and tertiary amines (R3N) form this
bond with H+.
(ii) Phosphonium ion formation:
Phosphorous in PH3 (Phosphene) donates its lone pair to H+ and form +PH4 (Phosphonium
ion)
Phosphonium ion
5. Co-ordinate covalent bonds are also present in many oxyacids like.
(i)
O
H O N
O
⎯ ⎯ (ii) H―O―Cl→O
Nitric acid HClO2
and HClO3 (chloric acid) and HClO4 (Perchloric acid)
Points about Co-ordinate covalent bond:
1. Co-ordinate covalent bond is also rigid and directional.
2. All the co-ordinate covalent bonds are polar in nature.
3. The distinction between covalent and co-ordinate covalent bond vanishes when all the atoms
bonded to central atom are similar.
Examples:
H3O+, PH4
+, NH4
+.
Limitation of Lewis Concept / model:
The following are the points that Lewis failed to explain.
1. It did not explain, how electrons form pair despite of their repulsion.
2. It did not explain the molecular shapes.
3. It did not explain molecular Geometries (Bond angles).
4. It did not explain bond lengths.
5. It did not explain the bond polarities.
+
H
H
P
H P + H H H
H H


+
 
 
 
⎯ ⎯⎯
→ ⎯ →
 
 
 
:

135. Valence Shell Electron Pair Repulsion Theory (VSEPR Theory):
In 1940, Sidgwick and Powell said that shapes of molecules can be understood in terms of
repulsions between electron pairs in outer orbit of the central atom.
Recently, Nylholm and Gillespie explained the shapes of molecules for non-transition elements by
VSEPR Theory.
Basic assumption:
The electron pairs (lone pairs and bond pairs) arrange themselves around central atom at
maximum distance apart so that repulsion between them is minimum.
Postulates of VSEPR Theory
(i) Both Lone pair (Non-bonding pair of electron) and bond pair (Bonding pair of electron)
participate in determining the geometry of molecules.
(ii) The electron pairs are arranged around central polyvalent atom at maximum distance apart
to keep the repulsions minimum.
(iii) The lone pair occupies more space than bond pair and hence. Lone pair causes more
repulsion than bond pair.
A bonding electron pair is attracted by both the nuclei of bonded atoms and non-bonding
pair is attracted by only one nucleus. As lone pair experiences less nuclear attraction, its
electronic cloud spreads out more in space and causes more repulsion than bond pair. As a
result, lone pair tends to compress bond pairs and reduces the bond angles.
The magnitude of repulsions between the electron pairs in a given molecule decreases in the
following order.
Lone-pair-lone-pair>Lone pair-bond pair>Bond pair-bond pair
These repulsions are called as Vander Waal’s repulsions.
(iv) Three electron pairs of triple bond have higher electron density and occupy more space than
two electron pairs of double bond which in turn has higher electron density and occupy more
space than single bond.
Order of Occupied Space:
Triple bond > double bond > single bond
As double and triple bond occupy the same region between two nuclei as single bond, hence,
they behave like single bond in determining the geometry of molecule.
(v) The bond angle reduces with the increasing electronegativity of the bonded atoms.

136. Examples:
Bond angle in NF3 reduces to 102o from 107.5o which is in NH3.
Type
Electron Pairs Arrangement
of pairs
Molecular
geometry
Shape Examples
Total Bonding Lone
AB2 2 2 0 Linear Linear B–A–B
BeCl2
HgCl2
AB3 3
3 0
Trigonal planar
Trigonal
planar
BH3,BF3 AlCl3
2 1
Bent (or
angular
SnCl2,SO2
AB4 4
4 0
Tetrahedral
Tetrahedral
CH4,SiCl4CCl4
,BF4
–
NH4
+
,SO4
-2
3 1
Trigonal
pyramidal
NH3,NF3
PH3
2 2
Bent (or
angular)
H2O, H2S
Limitation of VSEPR:
VSEPR Theory does not explain the formation of bonds. It only explains the shapes and geometries
of molecules.
Valence Bond Theory (VBT):
According to this theory, atomic orbitals of valence shell of bonding atoms overlap to form bonds.
As a result, maximum density of shared pair lies between the two nuclei.
Postulates of VBT:
(i) Overlaping is a necessary condition for bond formation.
(ii) Only half filled orbitals participate in overlaping during covalent bond formation.
(iii) Greater the overlaping, stronger is the bond formed.
(iv) Electrons with opposite spin pair up to stabilize themselves during bond formation
(overlaping).
(v) Sigma( ) bond: A bond formed by head to head (end to end) overlaping where the
probability of finding the shared pair around the line joining the two nuclei is maximum is
called as sigma bond.
(vi) Pi() bond: A bond formed by the parallel overlaping where the probability of finding shared pair
above and below the line joining the two nuclei is maximum is called pi bond.
Note: -bond is a weaker bond than sigma bond.

137. Limitations of VBT:
(i) It does not justify the valency of some elements i.e. carbon, Beryllium, Boron etc.
(ii) It does not explain equivalent tetravalency of carbon.
(iii) It does not justify the bond angles and shapes of molecules like H2O, NH3, H2S etc.
(iv) It does not explain paramagnetic behaviour of oxygen.
Atomic Orbital Hybridization (Modified VBT) and Shape of Molecules:
The process of mixing of atomic orbtials of different energy and shape to form new orbitals of same
energy and same shape is called hybridization. The new orbtials formed are called hybrid orbitals (orbitals
of mixed characters).
Important points of hybridization:
(i) Atoms undergo excitation and hybridization simultaneously before the bond formation.
Excitation is the promotion of electrons from lower energy orbital to higher energy. Due to
this, number of unpaired electrons are increased and valency is justified.
(ii) Number of hybrid orbtials formed is equal to number of atomic orbitals undergoing
hybridization.
(iii) Hybrid orbitals are involved in sigma bond formation. Hence, hybrid orbitals are equal to
number of sigma bonds in the molecule. And rest of the atomic orbitals are left unhybrid
which participate in pi bond formation.
(iv) The type of hybrid orbtials and hybridization depends upon number and type of atomic
orbitals involved.
1s + 3p = 4sp3-orbitals = sp3 hybridization
1s + 2p = 3sp2-orbitals = sp2 hybridization
1s + 1p = 2sp-orbitals = sp hybridization
Hybridization gives entirely new shape and orientation to atomic orbitals. Hence, it gives a
shape to molecule.
Hybridization Shape Angle
sp3-hybridization Tetrahedral 109.5o
sp2-hybridization Triangular planar 120o
sp-hybridization Linear 180o
However, deviations to above given shapes and geometries are observed due to the presence
of Lone pairs of electrons.
Type of Hybridization
There are following types of hybridization discussed at this level.
1. sp3 – hybridization
2. sp2 – hybridization
3. sp – hybridization

138. Bond Energy:
The average amount of energy required to break all the bonds of a particular type in one mole of a
substance.
The average amount of energy required to break Avogadro’s number of bonds (6.02x1023) of
particular type is called as bond energy.
Measurement:
It is measured experimentally by measuring the heat involved in a chemical reaction i.e. it is a
measure of enthalpy change at 298K. That is why, it is also called as Bond enthalpy.
Factors Affecting the Bond Energy:
1) Sizes of the bonded atoms. 2) Bond length
3) Electro negativity difference between bonded atoms.
1- Sizes of the bonded atoms:
The bond energy is inversely related to the sizes of the bonded atoms. Smaller the size of the
bonded atoms, greater will be attraction and bond energy and vice versa.
Example: Bond energy for Br2 is greater (small sized) than that of I2 (large sized).
2- Bond length:
The bond energy is inversely related to bond length. Smaller the bond length, greater will be
bond energy and vie versa.
Example: The order of bond energy for different C to C bonds is as follows.
C  C > C = C > C – C
C  C C = C C—C
Bond energies in kJ/mole 839 614 348
Bonding lengths in pm 120 134 154

139. Bond energy and Ionic character (Electronegativity difference between the bonded atoms):
3- Electronegativity Difference:
The bond energy is directly related to electronegativity difference (ΔEN). Greater the ΔEN, greater
will be bond energy.
Example:
The ΔEN for HF is 1.9 and bond energy is 567 whereas ΔEN for HCl is 1.1 and bond energy is 431
which is less than HF because HCl is less polar (less ΔEN) than HF.
The electronegativity difference or polarity in a molecule is called Ionic character. This develops
additional force between the bonded atoms. Due to which bond energy increases for polar molecules.
Bond Length:
The distance between the nuclei of two covalently bonded atoms is called as
bond length. Half of covalent bond length is called covalent radius.
Measurements:
The bond length is experimentally determined by physical techniques. These
may be:
i) electron diffraction
ii) X-ray diffraction
iii) Spectral studies
Factors affecting the bond length:
i) Sizes of the bonded atoms:
Bond length is directly related to the sizes of bonded atoms.
ii) Nature of hybridization:
The bond length is inversely related to s- character in the hybrid orbital. Greater the s- character,
shorter will be the bond length because it means that radius of s-orbital is smaller than p-orbitals.

140. Example:
The order of C – C bond length on the basis of % age of s-character is as follows
Ethane (C—C) > Ethene (C = C) > Ethyne (C  C)
Ethane Ethane C—C Ethene C = C Ethyne C  C
Bond length in pm 154 134 120
Hybridization sp3 sp2 sp
% age of s-character 25% 33% 50%
iii) Electronegativity difference:
The bond length decreases with the increase in electronegativity difference (ΔEN) between the
bonded atoms. With increase in ΔEN, bonded atoms develops opposite charges. Due to which
bonded atoms attract each other and bond length reduces.
Dipole moment (μ):
The product of electric charge (q) and distance between positive and negative centre (r) is called
dipole moment (μ). μ = q x r
Dipole moment is a vector quantity having.
i) Magnitude (can be calculated from formula μ = q x r)
ii) direction (from less electronegative atom to more electronegative atom).
Note: Dipole moments are directed from electropositive to electronegative atom.
Units:
1- The SI units of dipole moment are mC (unit of ‘r’ is m
and that of q is C)
2- The other unit is Debye (D)
1D = 3.336  10-30 mC

141. Dipole moments of some substances in Debyes
Compound Dipole
moment(D)
H2
HCl
HBr
HI
H2O
H2S
NH3
SO2
CO2
CO
NO
H2O2
CH4
CH3F
CH3Cl
CH3Br
CH3l
C2H5OH
0.00
1.03
0.78
0.38
1.85
0.95
1.49
1.61
0.00
0.12
0.16
2.20
0.00
1.81
1.45
1.85
1.35
1.69
Dipole moment for:
a) Diatomic molecules:
In diatomic molecules (HF, HCl, HBr, CO, NO), dipole moment is calculated by formula.
μ = q  r.
b) Polyatomic molecules:
Poly atomic molecules contains more than one dipole moments. Net dipole moment is the
resultant of vector addition of individual dipole moments.
Example:
O C O
== ==
net μ=O
Resultant moment
Net  =1.85 D
(a) (b)
Vector addition of bond moments in (a) linear CO2 molecule and (b) angular H2O molecule.
H
O
H

142. Dipole Moment and Molecular Structures:
Dipole moment gives us two types of information about molecular structure.
i) Percentage of ionic character in a bond.
ii) Angle between the bond or geometry of molecules.
(i) Percentage of ionic character is the ratio of observed dipole moment to ionic (calculated) dipole
moment multiplied by 100.
% age of ionic character = obs
ionic
μ
×100p
μ
μobs is experimentally determined dipole moment also called actual dipole moment.
μ-ionic is dipole moment of 100% ionic compound (hypothetical) and is calculated by multiplying
unit charge with actual bond length of molecule. So, it also called calculated dipole moment.
(ii) Bond angles or Geometry of molecules:
Molecular Geometry of triatomic molecules and dipole moment:
If dipole moment of triatomic molecule is zero, it means it has Linear geometry. Otherwise Bent or
angular.
a) The dipole moments of CO2 and CS2 are zero indicating their Linear geometry.
δ- δ-
δ+
C
O O
μ 0
 →
=
δ-
δ- δ+
C S
S μ 0
 →
=
b) The dipole moments of H2O, H2S and SO2 are 1.85D, 0.95D and 1.61D respectively. It indicate
their bent/ Angular geometry.
Comparison of CO2 and CO
CO2 has 0.0D and CO has 0.12 D dipole moments. Actually, CO2 is linear and has two equal but
opposite dipole moment which cancel the effect of each other.
Whereas CO is also Linear, but has only one dipole moment. As a result no cancellation effect and
value is not zero.
δ+
δ+
O
O C
C
O
μ = 0D μ =0.12D
→
 →

143. Symmetrical triangular planar molecules BF3, AlCl3 and perfectly tetrahedral molecules like CH4, SiH4, CCl4
also have zero dipole moments. The reason is the cancellation of individual dipole moments.
F
μ =o
B
F
F μ =o
δ+
H
δ-
C
δ+
H
δ+
H
δ+
H

144. Assessment 1
1. The octet rule does not always hold true for the element:
a. Carbon in methane
b. Oxygen in water
c. Fluorine in Hydrogen fluoride
d. Phosphorous in phosphorous pentachloride
2. The chemical reactivity and combination during a chemical reaction largely depends upon
a. Nature of atoms
b. Masses of atoms
c. Way of combination
d. Electronic configuration
3. Which of the following release their single valence shell to achieve nearest noble gas
configuration?
a. Transition metals
b. Alkaline earth metals
c. Alkali metals
d. Halogens
4. Alkaline earth metal in order to complete near noble gas configuration requires
a. Gain of electrons
b. Loss of one electron
c. Gain of two electrons
d. Loss of two electrons
5. Sodium is the soft metal which can be cut with knife. In order to attain the noble gas
configuration, it has to
a. Releases an electron
b. Gain electron
c. Releases two electrons
d. Gains two electrons
6. Which of the following has equilateral shape (having all of its sides equal)
a. sp
b. sp2
c. sp3
d. None of these

145. 7. Which of the following covalent bond is the most polar bond?
a. H—F
b. H—H
c. H—C
d. H—N
8. Which of the following has high ionization energies among these:
a. Nonmetals
b. Metals
c. Alkali metals
d. All have same values
9. The element which is the highest electro negativity is:
a. Lithium
b. Iodine
c. Cesium
d. Oxygen
10. Electron affinity generally increases down the group. The observed electron affinity values for
the halogens show the following trend
a.
b.
c.
d.

146. Assessment 2
1. The no of lone pair of electrons on oxygen atom in acetate ion (CH3COO-) is:
a. 5 lone pairs
b. 4 lone pairs
c. 3 lone pairs
d. 2 lone pairs
2. According to VSEPR theory, the correct order of repulsion between electron pairs in a given
molecule is:
a.
b.
c.
d.
3. According to VSEPR theory, linear structure of carbon dioxide molecule ( C ) is due to
the no of bonded electron pairs:
a. Four
b. Two
c. Three
d. One
4. Which of the following has more bond energy?
a. C-O
b. C=O
c. C O
d. C N
5. The angle between two 3 sp2 hybrid orbital is:
a. 1800
b. 1200
c. 900
d. 109.50
6. In NH3 molecule, the covalent bond is formed due to the overlap of:
a. s–sp overlap
b. s–sp2 overlap
c. s–sp3 overlap
d. sp2–sp2 overlap

147. 7. In a given compound; , is
a. sp3
b. sp2
c. sp
d. sp3d
8. In NH2
- , there are two bond pair and two lone pairs. Its shape is similar to:
a. Methane
b. Ethene
c. Ammonia
d. Water
9. The type of hybridization that is exhibited by boron in BF3
a. sp3
b. sp2
c. sp
d. sp3d
10. Bond angles and geometries in methane, ammonia and water are different because of having
a. Different electron pairs
b. Different lone pairs
c. Same bone pair
d. Same lone pair

148. Assessment 3
1. The no of hybrid orbitals which are produced by the combination of (n) no of atomic orbitals:
a. n
b. 2n
c. 2(n)2
d. 3n
2. According to VSEPR Theory, geometry of the molecule of ammonia is:
a. Trigonal planar
b. Trigonal pyramidal
c. Bend or angular
d. Trigonal bipyramidal
3. Which of the following hydrogen halide experiences more %age ionic character?
a. HI
b. HCl
c. HBr
d. HF
4. The following reason truly relate to non-polar nature of CO2 :
a. Linear geometry
b. sp hybridization
c. Zero dipole moment
d. All of these
5. is formed by the overlapping of:
a. Sidewise overlap of hybrid orbitals
b. Parallel overlap of py-py orbitals
c. Head to heat approach of px-px
d. Head to head overlap of s-px orbitals
6. Geometry of molecule (AB4 type) having four bond pairs is:
a. V-shaped
b. Linear
c. Trigonal
d. Tetrahedral
7. Which of the following is true about Boron trifluoride?
a. B-F-B bond angle is 90o
b. It has Trigonal planar geometry.
c. There is one lone pair around central boron atom.
d. It exhibit sp3 hybridizaiton

149. 8. The following compound is considered to exhibit sp2 hybridization:
a. BF3
b. C2H4
c. C6H6
d. All of these
9. Which of the following compound has the covalent bond (formed by the overlap of sp and p
orbitals)
a. CH4
b. BeCl2
c. NH3
d. H2O
10. Which of the following explains the tetravalency of carbon?
a. Lewis Structures
b. Hybridization
c. Valence bond theory
d. Covalent bonds

150. Assessment 4
1. A molecule having sp3 hybridization with two lone pairs has the geometry:
a. Linear
b. Tetrahedral
c. Trigonal
d. Bent or angular
2. The following compound has no net dipole moment:
a. BF3
b. CH4
c. CO2
d. All of these
3. The following statement best explains the fact that H2S has a net dipole moment while
BeCl2 has zero dipole moment?
a. H2S molecule is linear while BeCl2 is angular
b. H2S molecule is angular, while BeCl2 molecule is linear
c. The presence of covalent bond
d. The presence of polar covalent bond
4. The dipole moment of the system is measured as 1.6022x10-29Cm. What is its value in Debye:
a. 2.8 D
b. 3.8 D
c. 4.8 D
d. 1.8 D
5. Which of the following is not true about bond energy?
a. It is the average amount of energy required to break Avogadro’s number no of bonds
b. Bond energy of multiple bonds is stronger than bond energy of single bonds.
c. Bond energy bond is more than bond.
d. Bond energy of double bond is twice stronger than single bond.
6. The correct order of Bond energy is:
a.
b.
c.
d.

151. 7. C-C bond length is 154 pm which is additive of atomic size of component atoms. What is the
atomic size of carbon atom?
a. 154 pm
b. 77 pm
c. 100 pm
d. none of these
8. The bond length of C-Cl is when it is supposed to be the additive sum of atomic radii of two
atoms.
a. 154 pm
b. 198 pm
c. 176 pm
d. 77 pm
9. Which of the following diatomic molecule has more bond energy?
a. O-O
b. N-N
c. S-S
d. Cl-Cl
10. The physical method which is used to measure the bond length is:
a. Electron diffraction
b. X-ray diffraction
c. Spectral studies
d. All of these

152. Key
Assessment 1
Assessment 2
1. d
2. d
3. c
4. b
5. a
6. b
7. a
8. a
9. d
10. a
1. a
2. b
3. b
4. c
5. b
6. c
7. b
8. d
9. b
10. b

153. Assessment 3
Assessment 4
1. a
2. b
3. d
4. c
5. b
6. d
7. b
8. d
9. b
10. b
1. b
2. d
3. b
4. c
5. c
6. c
7. b
8. c
9. c
10. d

154. MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org
Oldest,LargestandMostCrediblePlatform

155. Learning Outcomes:
 System,surroundingandstatefunctions
 Terms used in Thermodynamics
 Standard states and standard enthalpy
changes
 Energy in chemical reactions
 First Law of Thermodynamics
 Sign of∆H
 Enthalpy of a reaction
 Hess’sLawofconstantheatsummation
Thermochemistry

156. Thermochemistry
Thebranchofchemistrywhichdealswiththestudyofheatchangesduringachemicalreactionisknown
as thermochemistry.
Heat ofreaction:
The amount of energy which is absorbed or released in a chemical reaction is called heat of
reaction.
Units ofenergy:
Energy in SI-system is usually expressed in Joule (J) and Kilo Joule (kJ).
Exothermic And Endothermic Reactions
Energy absorbed or released in a chemical reaction depends upon the energy of reactants and
products.
Comparison Between Exothermic and Endothermic
Exothermic reactions Endothermic reactions
1. Reactions in which
exothermic reactions.
heat is evolved are 1. Reactions in which heat is absorbed are
called endothermic reactions.
2. Heat released during bond formation is greater
than heat absorbed during bond breakage.
2. Heat released during bond formation is lesser
than heat absorbed during bond breakage.
3. Heat is transferred
surroundings.
from system to 3. Heat is transferred from surroundings to
system.
4. System cools down during these reactions i.e.
temperature of the system falls ultimately.
4. System become hot during these reactions
i.e. temperature of the system increases
ultimately.
5. Heat change (∆H) of these reactions is
represented with –ve sign.
5. Heat change (∆H) of these reactions is
represented with +ve sign.
Examples of Exothermic Reactions:
1) Combustion of carbon in oxygen is an exothermic reaction.
C(s) + O2(g) 
 CO2(g) ∆H= -393.7 kJ/mol
393.7 kJ heat is evolved when one mole of CO2(gas) is formed.
2) Formationofwaterisalsoanexothermicprocessand-285.58kJheatisevolvedwhenone
moleofliquidwaterisformedbyreactingappropriateamountofhydrogenandoxygengas.

157. H2(g) + ½O2(g) 
 H2O(l) ∆H= -285.58 kJ/mol
3) Thereactionsinwhichenergyisreleasedintheformof heatarecalledexothermicreactions.
4) Formation of NH3 gas is also an exothermic process.
N2(g) + 3H2(g) 
 2NH3(g) ∆H= -41.6 kJ/mol
Examples of Endothermic Reactions:
(1) Decomposition of water into oxygen and hydrogen gas is an endothermic process.
2H2O(l) 
 2H2(g) + O2(g) ∆H = 285.85 kJ/mol
(2) Formation of NO gas is an endothermic process:
N2(g) O2(g) 
2NO(g) ∆H = 180.51 kJ/mol
Spontaneous and non-spontaneous reactions
(1) Spontaneous reaction:
(i) A reaction which takes place on its own without any outside assistance and moves from a
non-equilibrium statetowardsan equilibrium state iscalledspontaneous reaction.
Examples:
Spontaneous reactions or processes are unidirectional, irreversible and real process. Some
examples are:
(1) Flow of water from higher level to lower level.
(2) Flow of current from higher voltage to lower voltage.
(3) Neutralization reaction of strong base and strong acid.
NaOH(aq) +HCl(aq) NaCl(aq)+H2O(l)
(4) The blue colour of CuSO4 solution fades when Zn is added to it. It is a spontaneous redox
reaction.
CuSO4(aq) + Zn(s) 
ZnSO4(aq)+ Cu(s)
(ii) Areactionwillalsobecalledaspontaneousreaction,ifitneedsenergytostartbutproceeds
further on its own.
Examples:
(i) Burning ofcoal:
A piece of coal does not burn in air on its own rather the reaction is initiated by a spark and
once coal starts burning, then reaction goes spontaneously to completion.
(ii) Burning ofhydrocarbons:
Burning of natural gas (a hydrocarbon) is also such spontaneous reactions

158. CH4(g)+2O2(g) 
 CO2(g)+ 2H2O(g)
(iii) Burning ofpaper:
(2) Non-Spontaneous Reaction:
A reaction which does not take place on its own without any external aid is called non-
spontaneous process.
Examples:
Some examples are
(1) Pumping of water uphill.
(2) Transfer of heat from cold interior part of the refrigerator to hot surroundings.
(3) Flow of electric current from lower voltage/potential to higher voltage/potential.
(4) When nitrogen reacts with oxygen, nitric oxide is formed. This reaction takes place by
absorbing heat. Although, N2 and O2 are present in air, but they donot react chemically at
ordinaryconditions.Thereactiontakesplacewhenenergyisprovidedbylightening.
N2(g)+O2(g) ‡
ˆ ˆ
ˆ†
ˆ 2NO(g)
Spontaneous reactions proceed with decrease in Energy:
Spontaneous processes proceed with the decrease in energy.
(i) All the exothermic reactions are spontaneous.
(ii) Many endothermic reactions are spontaneous as well.
Examples:
(i) Evaporation of water is spontaneous process, but this is endothermic
H2O(l) 
 H2O(g) ∆H = + 44 kJ/mol-1
(ii) DissolutionofNH4Clinwaterisalsospontaneous,becauseitdissolvesintowateronitsown
without anyexternal aid. But during dissolution it absorbs energy from water.That iswhy it
is endothermic but spontaneous.
NH4Cl(g) ‡
ˆ ˆˆ†
ˆ
+1
4aq
1 -
aq ∆H = + 15.1 kJmol-1
 Reversible processes constitute a limiting case between spontaneous and non-
spontaneous processes.
 Onlyenthalpychangeisnotsufficientinpredictionofspontaneityofreaction.
The concept of free energy and entropy change is also required.
NH + Cl

159. System, surrounding and state function
System:
Any portion of universe which is under study (consideration) is called system.
OR
Any material which is under consideration is called system.
Surrounding:
Portion of the universe other than system is called surrounding.
Boundary:
Therealorimaginarysurfaceseparatingthesystemfromsurroundingsiscalledboundaryofthe
system.
Examples:
(i) Onemoleofoxygengasconfinedinacylinderfittedwithapiston
isasystem.Thecylinder,thepistonandallotherobjectsoutside
the cylinder are surroundings.
(ii) A cup of water is a system. The cup, the air surrounding it, the
table on which it is lying is surroundings.
(iii) Consider, the reaction between ‘Zn’ and CuSO4 solution. This can
becalledasystembecauseitisunderobservation.Theflask,the
air etc. are the surroundings.
State of system:
The condition of the system is called state of the system.
System and surroundings
State of the system is described by set of variables such as temperature, pressure, volume, internal
energy andEnthalpy.
i) Initial state of a system:
Initial condition of the system is called as initial state of the system. OR
Description of variables of the system before occurring any change is called initial state of
the system.
ii) Final state of a system:
Final condition of the system is called final state of the system. OR
Description of the variables of the system after change happens is called final state of the system.

160. Suppose we have a beaker containing water in it at temperature T1. When we heat this, its
temperature rises to T2. T1 and T2 are called initial and final states of the system respectively.
We can find the change in state by following formula.
∆T= Final temperature – Initial temperature
∆T= T2-T1
State Function (Macroscopic property):
Amacroscopicpropertyofasystemwhichhassomedefinitevaluesforinitialandfinalstatesand
is independent of the path adopted to bring about the change.
Explanation:
Let us suppose ‘V1’ is the initial volume of a gas. A change is brought about in gas and its final
volume becomes ‘V2’. The change in the volume is given by
∆V = V2-V1
This change in volume of the gas can be brought about either by changing temperature or
pressure of the gas. As ‘V’ is state function, ∆V will be independent of the path or way by which the
volumeofthegashasbeenchanged.Itwilljustdependsupontheinitial andfinalvolumeofthegas.
Statefunctions areusually written in capitalletters. Thereare many state functions which are as
follows:
(i) Change of temperature = ∆T = T2-T1
(ii) Change ofenthalpy = ∆H = H2-H1
(iii) Change ofvolume = ∆V = V2-V1
(iv) Changeofinternalenergy = ∆E = E2-E1
(v) Change of pressure = ∆P = P2-P1
Internal energy(e)
Thesumofthekineticaswellasthepotentialenergiesof alltheparticlespresentinthesystemis
called internal energy of that system.
InternalenergyisrepresentedbyE.Changeininternalenergy(∆E)isastatefunction.Itisnotpossible
to measure the absolute value of internal energy. However, the change in internal energy can be
determined.
Kinetic energies of the particles are due to
(i) Translational motion
(ii) Vibrational motion
(iii) Rotational motion

161. Potential energy accounts for all types of attractive forces present in the system. These
attractiveforcesincludealltypesofbondsandtheVanderWaal’sforcespresentbetweenparticles.
Ways of Energy Transfer
Transferofenergytoorfromthesystemtakesplace
the following two ways
(i)heat exchange (ii) workdone
Heat:
Work:
The quantity of energy that flow across the
boundary of a system during a change in its
statedue todifference in temperaturebetween
system and surroundings is called heat. It is
represented by symbol ‘q’
(i) Heat absorbed by the system is represented by +ve sign.
(ii) Heat evolved by the system is represented by -ve sign.
The dot product of force and displacement is called work done.
 
W = F.S
In SI units, it is measured in Joules.
In chemistry usually pressure volume work is taken. In this work done, force become pressure and
distance become volume change.
W = P∆V (pressure volume work)
Sign conventions for work done in chemistry:
(With reference to internal energy change)
Workdonebythesystemistakennegativeasitleadstodecreaseininternalenergyofthesystem.
Workdoneonthesystemistakenpositiveasitleadstoincreaseininternalenergyofthesystem.
Sign conventions for work done in Physics:
(With reference to volume change)
Workdonebythesystemistakenpositiveasitleadstoincreaseinvolumeofthesystem.
Workdoneonthesystemistakennegativeasitleadstodecreaseinvolumeofthesystem.
by

162. First Law of Thermodynamics
Energy can neither be created nor be destroyed, but can be changed from one form to another
form. This is called first law of thermodynamics or law of conservation of energy.
OR
System cannot destroy or create energy but it can exchange energy with its surroundings in the
form of heat and work.
Explanation:
Sincesystemcanexchangeenergywithsurroundingsintheformofheatandwork,sotheenergy
changeisthesumofbothheatandwork.Hence,totalenergyofthesystemanditssurroundingsremains
constant.
∆E = q + W
∆E=q+P∆V-------(i)
Where,
 q is the heat energy exchanged between system and surroundings.
 W is the work done on or by the system.
i) Internal energy change at Constant Volume:
Suppose a gas is filled in a container having fixed piston that keeps the volume constant i.e.
∆V=0. When some amount of heat energy is supplied to the system, then internal energy
change is
∆E = q + W
∆E = q + (P∆V)(since, W = P∆V)
∆E = qv (∆V = 0 and P∆V)
Increase in internal energy of the system is equal to heat absorbed by the system at constant
volume and vice versa.
The change in internal energy of the system is equal to heat given to the system or lost by the
system at constant volume

163. Heat and Heat Capacity:
Heat is the transfer of thermal energy between two bodies that are at different temperature.
Different forms of energy can be converted into one another. Any other form of energy can be
completely transformed into heat energy.
Butatconstanttemperatureheatcannotbecompletelyconvertedintoanotherformofenergy.
Temperaturedifferencedetermines the directionin which heatflows spontaneously
Butthequantityofheatwithinsubstanceatagiventemperatureisdirectlyproportionaltoitsmass.
The amount of heat absorbed bysubstance isproportional to the temperature change.
“The amount of heat required to raise the temperature of given amount of a substance by 1 Kelvin is
called heat capacity.”
The amount of heat required to raise the temperature of one gram of a substance by 1 Kelvin is called
specific heat capacity.
It is expressed in Joules per gram per Kelvin.
Molar specific heat is given as:
Molar heat capacity at constant pressure:
Molar heat capacity at constant volume:

164. Enthalpy
Total heat contents of a system are called Enthalpy of the system and are equal to the sum of
internal energy and pressure volume work. It is represented by H.
H = E + W
H=E+PV (Since, W = PV)
Enthalpy of a system cannot be measured in a given state. However, the change in enthalpy can be
measured in a given state as follows:
∆H=∆E+∆(PV) ------------ (i)
Enthalpy is a state function and enthalpy change is measured ifn Joules.
Internal energy change at constant pressure:
When heat (q) is given to the gaseous system at constant pressure,
then it is used in
(i) Increase in internal energy of the system.
(ii) Work done by the system.
As a result, gas expands against constant pressure. At this stage, increase in internal energy of
system is equal to enthalpy change minus the pressure volume work.
∆E=∆H-∆(PV)------------(ii)
And enthalpy change is
∆H = ∆E + ∆(PV)
∆H = ∆E + ∆PV + P∆V
At constant pressure, ∆P = 0 and ∆PV = 0. Hence, above equation is modified as:
∆H=∆E+P∆V ------------ (iii)
Liquids and solids donot undergo significant volume change.
Thatiswhy∆V=0and∆Hand∆Earenearlysame(∆H ; ∆E).
According to first law of thermodynamics,
∆E = q + W
∆H = q + P∆V
Since, work is done by the system, sign of W = P∆V will be negative.
∆H =q -P∆V ----------------(iv)
Now, put the value of ∆E from equation (iv) into equation (iii).
∆H = q - P∆V + P∆V
∆H = qp (v)
Enthalpy change is equal to heat of reaction at constant pressure.

165. S. No Enthalpy Internal Energy
1 It is represented by ‘H’. It is represented by ‘E’
2 Sumofinternalenergyandproductof
pressureandvolumeofthesystemis
termed asenthalpy.
Itisthesumofkineticenergyand
potentialenergyof the system.
3 Itisstatefunctionsoonlychangein
enthalpy can be measured.
Italsoastatefunctionsochangeinit
can bemeasured.
4 Byusingthefollowingformulaitcanbe
measured H = E + PV
It is calculated by formula
E=
1
mv2
 PE
2
5 Its units are kcal mol-1 or kJ mol -1 Its units are Joules or calories.
Enthalpy of Reaction (∆H):
The standard enthalpy of reaction (∆H) is the enthalpy change which occurs when certain number
of moles of reactants as indicated by balanced chemical equation react together completely to give the
products under standard conditions i.e. 25oC(298K) and at one atmosphere pressure.
2H2(g)+ O2(g) 
2H2(l) ∆Ho
= - 285.8 kJ.mol-1
.
Exothermic Reaction:
In an exothermic reaction, the heat contents or
enthalpy of the products (H2) is less than enthalpy of
reactants (H1).
∆H = H2 - H1
∆H = - ve (Since H2 < H1)
The enthalpy change for exothermic reaction is
negative because it is accompanied by decrease in
heat contents of the system in termsof heat loss.
Note:
(i) Enthalpychangeistheheatexchangedbetweensystemand
surroundings at constant pressure.
∆H = qp
(ii) Internal energy change is the heat exchanged between system
and surroundings at constant volume.
∆E = qv

166. f
f
f
at
Endothermic Reaction:
In an endothermic reaction, the enthalpy of products
(H2) is greater than that of the reactants (H1).
∆H = H2 - H1
∆H= + ve (Since H2 > H1)
The enthalpy change for endothermic reactions is positive
because it is accompanied by increase in heat contents
of the system in terms of gain of heat.
EnthalpyofFormation :
The standard enthalpy of formation of a compound is the amount of heat absorbed or evolved
when one mole of compound isformed from its elements under standard conditions.
It is represented by ∆H o
It units are kJ.mol-1
Example:
i) Enthalpy of formation (∆Ho
) for MgO is -692kJmol-1 when it is formed from its elements i.e.
oxygen and magnesium
Mg(s) + ½ O2(g) 
 MgO(s) ∆H o
= -692 kJ.mol-1
ii) When carbon reacts with oxygen to form CO2, 393.7 kJmol-1
of energy is released. This is
called enthalpy of formation of CO2.
C(s) + O2(g) 
 CO2(g) ∆H o
= -393.7 kJ.mol-1
EnthalpyofAtomization :
The standard enthalpy of atomization of an element is the amount of heat absorbed when one
mole of gaseous atoms are formed from the element under standard conditions i.e. 298K and one
atmosphere pressure.
Example:
The standard enthalpy of atomization of hydrogen is given below
½ H2(g) 
 H(g) ∆H o
= + 218kJmol-1
f

167. +1
c
Enthalpy of Neutralization :
The standard enthalpy of neutralization is the amount of heat evolved when one mole of H+ ions
from an acid react with one mole of OH- ions from a base to form one mole of water under standard
conditions i.e. 298K and one atmosphere pressure.
Explanation:
Enthalpy of neutralization of strong acid and strong base such as HCl and NaOH is -57.4 kJmol-1.
Strong acid (HCl) and strong base (NaOH) ionizes completely in dilute solution.
HCl(aq) ‡
ˆ ˆ
ˆ†
ˆ H+
+ Cl-1
(aq) (aq)
NaOH(aq) ‡
ˆ ˆ
ˆ†
ˆ Na(aq) + OH-
When the solution are mixed together during the process of neutralization, the only change that
actually occurs is the formation of water molecules leaving sodium ions and chlorides ions as free ions and
are called spectator ions. Enthalpyof neutralization is merelythe heat of formation of liquid waterfrom
its ioniccomponents.
H+
+ Cl-
+ Na+1
+ OH-
‡
ˆ ˆ
ˆ†
ˆ Na
+ + Cl-
+ H O
(aq) (aq) aq (aq) (aq) (aq) 2 (l)
The main reaction that takes place during neutralization is
H+
+ OH-
‡
ˆ ˆ
ˆ†
ˆ H O ΔH = -57.4 kJ.mole-1
(aq) (aq) 2 l n
Enthalpy of neutralization of any strong acid and strong base is approximately the same i.e -57 .4
kJ mole-1.
Enthalpy of neutralization of strong acid and weak base or weak acid and strong base is always less
than 57.4 kJ.mol-1. This enthlpy is less due to the fact that some of the heat is absorbed to accomplish the
ionization of weaker electrolyte (either weaker acid or weaker base).
Enthalpy of Combustion (∆H):
Thestandardenthalpyofacombustionistheamountofheatevolvedwhenonemoleofsubstance
is completely burnt in excess of oxygen under standard conditions. It isdenoted by .(H
)
Examples:
i) The standard enthalpy of combustion of ethanol is -1368 kJ.mol-1. The reaction is
represented by following equation.
C H OH + 3O 
2CO + 3H O ΔHo
= -1368 kJ.mol-1
2 5 (l) 2(g) 2g 2 l c
ii) The standard enthalpy of combustion of carbon dioxide is -393.7 kJ/mol. The reaction is
represented by following equation.
C + O 
 CO ΔHο
= -393.7 kJ.mol-1
s 2g 2g c

168. Enthalpy of Solution (∆H):
Thestandardenthalpyofasolutionistheamountofheatabsorbedorevolvedwhenonemoleofa
substanceisdissolvedinsomuchsolventthatfurtherdilutionresultsinnodetectableheatchange.
Examples:
i) The enthalpy of solution of NH4Cl is 16.2 kJ.mol-1
.
NH Cl ˆ ˆ
H
ˆ2O
ˆ†
NH+1 + Cl-1
ΔH = + 16.2 kJ.mol-1
4 (s) ‡ ˆ ˆ ˆˆ 4 aq aq s
In this case, heat is absorbed results in cooling of solvent and process of dissolution is
endothermic. Heat flows from surrounding to the system and temperature of the
surroundings fallsultimately.
ii) The enthalpy of solution of Na2CO3 is -25 kJ/mol.
Na CO ˆ ˆH2
ˆO
† Na+1 + CO-
2 ΔH = -25 kJ.mol-1
2 3(s) ‡ ˆ ˆ ˆ (aq) 3 aq s
Inthiscase,heatisreleasedresultsinriseintemperatureofsolventandprocessofdissolutionis
exothermic.Heatflowsoutfromthesystemtothesurroundingandtemperatureofthesurroundingsrises
ultimately.
Measurement of Enthalpy of a Reaction:
Exothermic and endothermic reactions can easily be
detected by observing the temperature of reaction before and
afterthereactionaslongasheatofreactionevolvedorabsorbed
is reasonable.
For measurement enthalpy of reactions we will discuss two
colorimeters.
i)Glass Calorimeter ii) BombCalorimeter
(i) Glass Calorimeter
Construction:
This is an ordinary calorimeter made up of glass. It is insulated from atmosphere by cotton wool
and is fitted with a thermometer and stirrer.
Procedure for Determining Enthalpy Of Reaction:
Reactantsinstoichiometricamountsareplacedinthecolorimeter.Whenthereactionproceeds,
theheatenergyevolvedorabsorbedwilleitherwarmorcoolthesystem.Thetemperatureofthe

169. system is noted before and after the chemical reaction. Knowing the temperature change, the
massof reactantspresentandthespecificheatof water, we can calculate thequantityof heat ‘q’
evolved or absorbed during the reaction.
q = m x s x ΔT
Where, m = mass of reactants
S = specific heat of reaction mixture.
ΔT = change in temperature.
Note:
 The amount of heat required to raise the temperature of one gram of substance by one
Kelvin is called specific heat of the substance.
 Glasscalorimeter isusedforthemeasurementof heatof neutralization and heatof solution.
(ii) Bomb Calorimeter
Bomb calorimeter consists of strong cylindrical stealvesselusually lined with enamel to prevent
corrosion.Aknownmassoftestsubstance(aboutonegram)isplacedinaplatinumcrucibleinside
the bomb.
The lid is screwed on tightly and oxygen is provided in through a valve until the pressure inside is
about 20 atm. After closing the screw valve, the bomb calorimeter is immersed in a known mass of
water in a well insulated calorimeter. Then it is allowed to attain a steady temperature. The initial
temperature is measured by using the thermometer present in the calorimeter. The test substance
is then ignited electrically bypassing the current through ignition coil. The temperature of water,
which is stirred continuously, is recorded at 30 sec intervals. From the increase of temperature,
specific heat of reaction system and mass of reacting substances, we can calculate enthalpy
change
q= m x s xΔ T -----------------(i)
Bomb calorimeter is usually used for accurate determination of the enthalpy of combustion of
food, fuel and other compounds.

170. ΔH =-393.7kJmol
O
Hess’s law of constant heat summation
Enthalpy change for a reaction by different routs is same if initial and final states are same
regardless of path of reaction.
H = H1 + H2 ----------------------(i)
Or
The sum of enthalpy changes in closed cycle is zero.
 ΔHcycle=0
Explanation
Let ‘A’ can be converted into ‘D’ directly in a single step in one route and in three steps in second
route.
When ‘A’ is directly converted into ‘D’ in a single step then heat change is ΔH. when ‘A’ is
converted into ‘D’ in three steps then heat changes during three steps are ΔH1, ΔH2 and ΔH3 respectively.
According to Hess’s law,
ΔH = ΔH1 + ΔH2 + ΔH3
ΔH + (-ΔH1) + (-ΔH2) + (-ΔH3) = 0
Mathematically,
 ΔHcycle=0
ΔH
Examples:
(i) If the enthalpy of combustion for graphite to form CO2 and
enthalpyof combustionof COtoformCO2 areknown,then
by using Hess’s law we can determine the enthalpy of
formation ofCO.
Consider the following cycle.
ΔH1
ΔH2
ΔH3
C(s) + O2g 
CO2g
-1
for CO2 from graphite
CO(g) +
1
2 2g
1

 CO2(g) ΔH2 for CO2 from (CO) = - 283 kJ mol-1
C (graphite) +
2
O2 
CO(g) ΔH1 CO=?

171. From cycle, it is clear that
ΔH = ΔH1 + ΔH2
ΔH1 = ΔH – ΔH2
= - 393. 7 – (-283)
ΔH1 = - 110.7 kJ mol -1
So, the enthalpy change for the formation of CO(g) is -110 kJmol -1
.
(ii) TheformationofNa2CO3maybestudiedinasinglestepprocessorintwostepsviasodium
hydrogen carbonate.
(a) Single StepProcess
2NaOH(aq) +CO2g 
Na2CO3aq +H2Ol
(b) Two Step Process (Via NaHCO3)
NaOHaq +CO2g 
NaHCO3aq
NaHCO3aq + NaOHaq 
Na2CO3aq +H2Ol
ΔH = - 89.08 kJ/mol
ΔH1 = 48.06 kJ
ΔH2 =- 41.02 kJ
NaOH + CO Δ

H
Na CO +H O
NaHCO3
AccordingtoHess’sLaw,
ΔH=ΔH1 +ΔH2
- 89.08 = - 48.606 – 41.02
- 89.08 = - 89.08
Application of Hess’s Law
i) Hess’s law is used for the determination of heat of formation of compound such as CCl4
because
 Its heat of formation cannot be determined directly by calorimetric method.
 CCl4 cannot be prepared directly by combining graphite and chlorine.
ii) Heat of formation of Al2O3 and B2O3 cannot bemeasured directly. It isdifficulttoburnthese
substances completely in oxygen, as protective layer of their oxides is formed on the surface.
So,for thedetermination of heat of formation of Al2O3 andB2O3 Hess’s lawis used.
ΔH2
NaOH
2 3 2
ΔH1
2

172. f
f x l
The Born–HaberCycle:
It states that energy change in a cyclic process is always zero.
Calculation of Lattice Energy by Born Haber Cycle:
The enthalpy of formation of one mole of ionic compound from gaseous ions under standard
conditions is called lattice energy.
Na+
+Cl-1

NaCl ΔHo
=-787 kJmol-1
g g s latt.
Energyrequiredtobreakonemoleofioniccompoundintoitsgaseousionsatstandardconditions
is called lattice energy of that ionic compound.
NaCl 
Na+1
+Cl-1
ΔHo
= 787 kJmol-1
Explanation:
s (g) (g) latt.
Lattice energy cannot be measured directly but value can be obtained indirectly by means of Born
– Haber cycle.
Calculation of lattice energy of NaCl:
1 ΔH +1 -1
Na(s) + Cl2(g) 
×

Na(g)
2
+ Cl(g)
-ΔHf
+ΔH
o
o
latt
o
latt
NaCl(s)
From figure it is clear that
ΔHο
= ΔH + ΔH
+ΔH -ΔH

173. g g
  iNa
(g) atCl
g g
  e
2
x
l
Since, H0 the standard enthalpy of formation of NaCl, can be measured conveniently in a calorimeter ΔHο
f latt.
can be obtained if ΔHx which is the totalenergyinvolved in changingsodiumand chlorine from their normal
physicalstatetogaseousionscanbecalculated.
Hlatt. = Hf - Hx
PreviousenergytrianglehasbeextendedtoshowthevariousstagesinvolvedinfindingΔHn.
It is clear that
i) Atomization of Na:
ΔHx =ΔHat(Na) +ΔHi(Na) +ΔHatCl +ΔHeCl
Whensolidsodiumisheated,itisconvertedintogaseoussodiumatombypassingthrough
different steps. This process is endothermic
Na(s) 
 Na(g) ΔHat (Na) = + 108 kJ mol-1
ii) Ionization Energy Of Sodium
Whenthesegaseousatomsofsodiumarefurtherheatedthenelectronpresentinoutermost
shell absorbs reasonable amount of energy and escapes from the shelland sodium ion is
formed. Ionization of Na is an endothermic process.
Na 
Na+1
+1e-
ΔH = + 496 kJmol-1
iii) Atomization OfChlorine
Atomization of chlorine isendothermic process, energy is absorbed to break bonds within
chlorine molecule.
1
Cl
2
2g 
Cl ΔH = + 121kJmol-1
iv) Electron Affinity of Chlorine
Whenelectronisaddedtotheoutermostshallofisolatedgaseouschlorineatom,energyis
released. Addition of electron in chlorine is exothermic process.
Cl +e

Cl-1
ΔHo
= -349 kJmol-1
ΔHx =ΔHatNa +ΔHiNa +ΔHatCl  +ΔHeCl
ΔHx =108 + 496 + 121 - 349
ΔH = 376 kJmol-1
The lattice energy can be calculated by following equation.
ΔHo
=ΔHo
-ΔH
latt. f x
=-411-376
ΔHo
= -787 kJ mol-1

174. Assessment 1:
1. Generally all reactions exchange energy with the surrounding. Which of the following is not true
about this energy exchange:
a. Exothermic reactions involve the rise of temperature of surrounding
b. Combustion of carbon in oxygen is exothermic process.
c. Endothermicinvolvesthe riseintemperatureofthesystemwiththedecreasein
temperature ofsurrounding
d. Haber’s process is an endothermic reaction.
2. Endothermic reactions involve the decrease of the temperature of surrounding. Which of the
following is an example of it:
a. C(s) + O2(g) →
CO2 (g)
b. H2(g) + H2O(l)
c. N2(g) + 3H2(g) 2NH3
d. N2(g) + O2(g) →
2NO(g)
3. Reversible processes constitute a limiting case between:
a. spontaneous
b. non-spontaneous processes
c. Thermal process
d. both a and b
4. If the enthalpy of (B) is greater than (A), there is probability that the reaction is:
A →B
a. Exothermic
b. Endothermic
c. Does not evolve any heat change
d. Both b and c
5. Which of the following Enthalpy of reaction is positive?
a. Enthalpy of combustion
b. Enthalpy of neutralization
c. Enthalpy ofatomization
d. All of these
6. Molarheatcapacityatconstantpressureisgivenas:
a.
b.
c.
d. None of these
Assessment 1:
1. Generally all reactions exchange energy with the surrounding. Which of the
following is not true about this energy exchange:
a. Exothermic reactions involve the rise of temperature of surrounding
b. Combustion of carbon in oxygen is exothermic process.
c. Endothermicinvolvesthe riseintemperatureofthesystemwiththe
decrease in temperature of surrounding
d. Haber’s process is an endothermic reaction.
2. Endothermic reactions involve the decrease of the temperature of surrounding.
Which of the following is an example of it:
a. C(s) + O2(g) →
CO2 (g)
b. H2(g) + H2O(l)
c. N2(g) + 3H2(g) 2NH3
d. N2(g) + O2(g) →
2NO(g)
3. Reversible processes constitute a limiting case between:
a. spontaneous
b. non-spontaneous processes
c. Thermal process
d. both a and b
4. If the enthalpy of (B) is greater than (A), there is probability that the reaction is:
A →B
a. Exothermic
b. Endothermic
c. Does not evolve any heat change
d. Both b and c
5. Which of the following Enthalpy of reaction is positive?
a. Enthalpy ofcombustion
b. Enthalpy ofneutralization
c. Enthalpy ofatomization
d. All ofthese
6. Molar heat capacity at constant pressure is given as:
a.
b.
c.
d. None of these

175. 7. For which of the following substances, ΔE = ΔH due to no pressure volume work?
a. Liquidsandsolids
b. Liquidsandgases
c. Gases and solids
d. Liquids, solids and gases
8. Thequantityofheatwhichissuppliedtothesystembykeepingthevolumeconstant,isusedto
bring about the change in?
a. Enthalpy
b. Free energy
c. Internal energy
d. All of these
9. Which of the following enthalpies could be exothermic as well as endothermic?
o
a. ΔHat
o
b. ΔHn
o
c. ΔHc
o
d. ΔHf
10. Which of the following is not a state function of thermal system?
a. Work
b. Temperature
c. Internal energy
d. Change in enthalpy
7. which of the following substances, ΔE = ΔH due to no pressure volume
work?
a. Liquidsandsolids
b. Liquids and gases
c. Gases andsolids
d. Liquids, solids and gases
8. Thequantityofheatwhichissuppliedtothesystembykeepingthe
volume constant, is used to bring about the change in?
a. Enthalpy
b. Free energy
c. Internal energy
d. All ofthese
9. Whichofthefollowingenthalpiescouldbeexothermicaswellas
endothermic?
a. ΔHat
o
b. ΔHn
o
c. ΔHc
o
d. ΔHf
o
10. Which of the following is not a state function of thermal system?
a. Work
b. Temperature
c. Internal energy
d. Change inenthalpy

176. Assessment 2:
1. Which of the following involves the absorption of energy from the surrounding?
a. Conversion of graphite to diamond
b. Evaporation of water
c. Decomposition of water
d. All of these
2. The thermal energy of a system is the fraction of which of the following:
a. Internal energy
b. Thermal energy
c. Enthalpy
d. Heat of combustion
3. Inthermochemistry,therearetwofundamentalwaysoftransferingenergyacrossthethesy.
These are:
a. Heat and work
b. Heat and volume
c. Pressure and temperature
d. Pressure and volume
4. For an endothermic reaction, the value of activation energy is?
a. Less than ΔH
b. More than ΔH
c. Equal toΔH
d. Zero
5. The nutritional caloric content mentioned in food labels can be measured by:
a. Glass calorimeter
b. Coffee cup with lid
c. Bomb calorimeter
d. All of these
6. Which of the followings statement is a statement of 1ST
law of thermodynamics?
a. The internal energy of a system is constant
b. The heat content of a system is constant
c. Energy is neither created nor destroyed
d. There is an equivalence between energy and mass
Assessment 2:
1. Whichofthefollowinginvolvestheabsorptionofenergyfromthe
surrounding?
a. Conversion of graphite to diamond
b. Evaporation ofwater
c. Decomposition ofwater
d. All ofthese
2. The thermal energy of a system is the fraction of which of the following:
a. Internal energy
b. Thermal energy
c. Enthalpy
d. Heat ofcombustion
3. In thermochemistry, there are two fundamental ways of transfering energy
across the boundary of the system. These are:
a. Heat andwork
b. Heat andvolume
c. Pressure andtemperature
d. Pressure andvolume
4. For an endothermic reaction, the value of activation energy is?
a. Less thanΔH
b. More thanΔH
c. Equal toΔH
d. Zero
5. The nutritional caloric content mentioned in food labels can be measured by:
a. Glass calorimeter
b. Coffee cup with lid
c. Bomb calorimeter
d. All ofthese
6. Whichofthefollowingsstatementisastatementof1ST
lawof
thermodynamics?
a. The internal energy of a system is constant
b. The heat content of a system is constant
c. Energy is neither created nor destroyed
d. There is an equivalence between energy and mass

177. 7. Acyclicprocesswhichisusedtocalculatethelatticeenergyofpotassiumbromideis:
a. Haber’s process
b. Hess’s law
c. Born-Haber’s cycle
d. All of these
8. For a cyclic enthalpy process, ΔH and ΔE is:
a. Greater than zero
b. Less than Zero
c. Equal to zero
d. Change in Enthalpy is zero but internal energy is negative
9. The lattice energy of binary ionic compound is:
a. Directly proportional to the sum of radii of cation and anion.
b. Inversely proportional to the sum of radii of cation and anion.
c. Inversely related to only ionic radius of cation
d. Directly related to only ionic radius of anion
10. Which of the following compounds has negative heat of solution?
a. NH4Cl
b. Na2CO3
c. NaCl
d. All of these
7. A cyclic process which is used to calculate the lattice energy of potassium
bromide is:
a. Haber’s process
b. Hess’s law
c. Born-Haber’s cycle
d. All ofthese
8. For a cyclic enthalpy process, ΔH and ΔE is:
a. Greater thanzero
b. Less thanZero
c. Equal tozero
d. Change in Enthalpy is zero but internal energy is negative
9. The lattice energy of binary ionic compound is:
a. Directly proportional to the sum of radii of cation and anion.
b. Inversely proportional to the sum of radii of cation and anion.
c. Inversely related to only ionic radius of cation
d. Directly related to only ionic radius of anion
10. Which of the following compounds has negative heat of solution?
a. NH4Cl
b. Na2CO3
c. NaCl
d. All ofthese

178. Assessment 3:
1. According to Hess’s law, the enthalpy change in a chemical reaction
a. Depends on the state of the system
b. Independent of the path
c. Does not deal with those reactions in which heat is evolved
d. both a and b
2. For which of the following, ΔH cannot be measured directly by calorimeter?
a. neutralization reaction
b. combustion reactions
c. Halogenation of methane in diffusedlight
d. None of these
3. When 1M H2SO4 is completely neutralized by sodium hydroxide, the heat liberated is 114.64 kJ.
The enthalpy of neutralization is:
a. -11.46 KJ/mol
b. -57.32 KJ
c. -114.64 KJ
d. None of these
4. The enthalpies of formation of an element in its standard state is assumed to be:
a. Zero at 298 K
b. Unity at 298 K
c. Zero at all temperatures
d. None of these
5. Which of the following is not true statement about the standard enthalpy of neutralization?
a. Itinvolvestheformationofonemoleofwaterbyreactionofstrongacidwithstrongbase.
b. Enthalpy of neutralization of sodium hydroxide by hydrochloric acid is -57.4 KJ
c. It is -57.32 KJ/mol in case of weak acid reacting with weak base.
d. None ofthese
6. One mole of hydrochloric acid reacts with one mole of sodium hydroxide to:
a. Produceonemoleofwater
b. Release -57.32 KJ of energy
c. Produce one mole of sodium ions and one mole of chloride ions
d. All of these
Assessment 3:
1. According to Hess’s law, the enthalpy change in a chemical reaction
a. Depends on the state of the system
b. Independent of the path
c. Does not deal with those reactions in which heat is evolved
d. both a and b
2. For which of the following, ΔH cannot be measured directly by calorimeter?
a. neutralization reaction
b. combustion reactions
c. Halogenation of methane in diffused light
d. None ofthese
3. When 1M H2SO4 is completely neutralized by sodium hydroxide, the heat
liberated is 114.64 kJ. The enthalpy of neutralization is:
a. -11.46 KJ/mol
b. -57.32 KJ
c. -114.64 KJ
d. None of these
4. Theenthalpiesofformationofanelementinitsstandardstateisassumedto
be:
a. Zero at 298 K
b. Unity at 298 K
c. Zero at all temperatures
d. None ofthese
5. Whichofthefollowingisnottruestatementaboutthestandardenthalpyof
neutralization?
a. Itinvolvestheformationofonemoleofwaterbyreactionofstrongacid
with strongbase.
b. Enthalpy of neutralization of sodium hydroxide by hydrochloric acid is -
57.4 KJ
c. It is -57.32 KJ/mol in case of weak acid reacting with weak base.
d. None ofthese
6. One mole of hydrochloric acid reacts with one mole of sodium hydroxide to:
a. Produce one mole of water
b. Release -57.32 KJ of energy
c. Produce one mole of sodium ions and one mole of chloride ions
d. All ofthese

179. 7. The standard enthalpy of a solution is the amount of heat absorbed or evolved when
a. Onemoleofasubstanceisdissolvedinsomuchsolventthatfurtherdilutionresultsin
detectable heatchange
b. Onemoleofsubstanceisdissolvedinsomuchsolventthatfurtherdilutionresultsinno
detectable heatchange.
c. Onemoleofsubstanceisdissolvedinonedm3 ofsolventtomakeaninfinitedilution.
d. Onemoleofsubstanceisdissolvedinonedm3ofsolution.
8. Fermentation is a biochemical reaction catalyzed by yeast. This is
a. Endothermic
b. Exothermic
c. Reversible
d. Both a and c
9. For a reaction in which heat is given out from the system to surrounding:
a. Energy of the products is less than energy of the reactants
b. Energy of the products is almost triple than energy of the reactants
c. Energy of the products is equal to energy of the reactants
d. Change in enthalpy is always positive
10. Which of the following can be assumed for enthalpy (H) for one mole of an ideal gas?
a. H =E +nRT
b. H =E +RT
c. H+E= PV
d. H= E
7. Thestandardenthalpyofasolutionistheamountofheatabsorbedorevolved
when
a. Onemoleofasubstanceisdissolvedinsomuchsolventthatfurther
dilution results in detectable heat change
b. Onemoleofsubstanceisdissolvedinsomuchsolventthatfurther
dilution results in no detectable heat change.
c. Onemoleofsubstanceisdissolvedinonedm3
ofsolventtomakean
infinite dilution.
d. One mole of substance is dissolved in one dm3
of solution.
8. Fermentation is a biochemical reaction catalyzed by yeast. This is
a. Endothermic
b. Exothermic
c. Reversible
d. Both a and c
9. For a reaction in which heat is given out from the system to surrounding:
a. Energy of the products is less than energy of the reactants
b. Energy of the products is almost triple than energy of the reactants
c. Energy of the products is equal to energy of the reactants
d. Change in enthalpy is always positive
10. Whichofthefollowingcanbeassumedforenthalpy(H)foronemoleofan
ideal gas?
a. H =E+nRT
b. H =E+RT
c. H+E= PV

180. Assessment 4:
1. Which of the following is defined by the first law of thermodynamics?
a. Change in Enthalpy
b. Heat of neutralization
c. Temperature
d. Change in Internal energy
2. Bomb calorimeter is used to measure enthalpy of combustion. It works at
a. Constant volume
b. Constant temperature
c. Constant pressure
d. None of these
3. Forglasscalorimeter,thequantityofheatqevolvedorabsorbedduringthereactionisgivenas:
a. q = mΔT
b. q= H/m
c. q= msΔT
d. q = m s/ΔT
4. Specific heat is a measurement used in thermodynamics. For water it is equal to:
a. 2.4 JKg
b. 4.2 JKg
c. 2.4 JK-1g-1
d. 4.2 JK-1g-1
5. The pressure inside the bomb vessel is kept at about:
a. 1 atm
b. 10 atm
c. 20 atm
6. Consider a balloon of given volume, V1 containing the gas at temperature T1. After some time
theTemperature T1 drops to T2 WHen it moves to colder part of the room. Its Q and W are:
a. –Q, +W
b. –Q, -W
c. +Q, +W
d. +Q, -W
7. Potential energy of a system at temperature T is due to the following:
a. Translational motion
b. Vibration motion
c. Rotational motion
d. None of these
Assessment 4:
1. Which of the following is defined by the first law of thermodynamics?
a. Change in Enthalpy
b. Heat of neutralization
c. Temperature
d. Change in Internal energy
2. Bomb calorimeter is used to measure enthalpy of combustion. It works at
a. Constant volume
b. Constant temperature
c. Constant pressure
d. None of these
3. For glass calorimeter, the quantity of heat q evolved or absorbed
during the reaction is given as:
a. q = mΔT
b. q= H/m
c. q= msΔT
d. q = m s/ΔT
4. Specific heat is a measurement used in thermodynamics. For water it is equal to:
a. 2.4 JKg
b. 4.2 JKg
c. 2.4 JK-1g-1
d. 4.2 JK-1g-1
5. The pressure inside the bomb vessel is kept at about:
a. 1 atm
b. 10 atm
c. 20 atm
6. Consider a balloon of given volume, V1 containing the gas at temperature
T1. After some time the Temperature T1 drops to T2 when it moves to colder
part of the room. Its Q and W are:
a. –Q, +W
b. –Q, -W
c. +Q, +W
d. +Q, -W
7. Potential energy of a system at temperature T is due to the following:
a. Translational motion
b. Vibration motion
c. Rotational motion
d. None of these

181. 8. Ifchangeininternalenergyofasystemisequaltoheat(q),theworkdonebythesystemis:
a. Positive
b. Negative
c. Zero
d. Uncertain
9. In case of atomization of an element (A), the heat of surrounding:
a. Decreases
b. Increases
c. Remains same
d. First decreases then increases
10. Which of the following is state function:
a. Change in temperature
b. Change in pressure
c. Change in volume
d. All of these
8. Ifchangeininternalenergyofasystemisequaltoheat(q),thework
done by the system is:
a. Positive
b. Negative
c. Zero
d. Uncertain
9. In case of atomization of an element (A), the heat of surrounding:
a. Decreases
b. Increases
c. Remains same
d. First decreases then increases
10.Which of the following is state function:
a. Change intemperature
b. Change inpressure
c. Change involume
d. All ofthese

182. Key
Assessment 1
1. d
2. d
3. d
4. b
5. c
6. b
7. a
8. c
9. d
10. a
Assessment 2
1. d
2. a
3. a
4. b
5. c
6. c
7. c
8. c
9. b
10. b
Key
Assessment 1
1. D
2. D
3. D
4. B
5. C
6. B
7. A
8. C
9. D
10. A
Assessment 2
1. D
2. A
3. A
4. B
5. C
6. C
7. C
8. C
9. B
10. . B

183. Assessment 3
1. d
2. c
3. b
4. a
5. c
6. d
7. b
8. b
9. a
10. b
Assessment 4
1. d
2. a
3. c
4. d
5. c
6. d
7. d
8. c
9. a
10. d
Key
1. D
Assessment 3
2. C
3. B
4. A
5. C
6. D
7. B
8. B
9. A
10. B
Assessment 4
1. D
2. A
3. C
4. D
5. C
6. D
7. D
8. C
9. A
10. D

184. Oldest, Largest and Most Credible Platform
MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org

185. Learning Objectives:
Reversible and irreversible reactions
State of chemical equilibrium
Equilibrium constant expression for important
reaction
Applications of equilibrium constant
Solubility product
The Le Chatelier’s principle
Synthesis of ammonia by Haber’s Process
Common ion effect
Buffer solutions
Equilibrium of slightly soluble ionic compounds
(solubility product)
Chemical Equilibrium

186. Chemical Equilibrium
Reversible and Irreversible Reactions
Irreversible Chemical Reaction Reversible Chemical Reaction
i) A reaction that proceeds only in one
direction is called as Irreversible
chemical reaction.
i) A reaction that proceeds in two
directions is called as Reversible
chemical reaction.
ii) The products are stable and do not
convert back into the reactants in
these reaction.
ii) The products are unstable and
convert back into the reactants
iii) These are represented by full headed
arrow.
iii) These are represented by two
oppositely directed half headed
arrows.
iv) Reactants may be completely
converted into products in these
reactions.
iv) Reactants are never completely
converted into products in these
reactions.
Examples:
a) 2H2 + O2 → 2H2O
b) NaOH + HCl → NaCl + H2O
Examples:
a) PCl5 ⇇ PCl3 + Cl2
b) N2 +3H2 ⇇ 2NH3.
Chemical Equilibrium is a Dynamic Equilibrium:
A type of equilibrium in which two opposing processes are proceeding at the equal rate is called
dynamic equilibrium. Since, two opposing processes i.e forward and backward reactions are not stopped
but proceeds at equal rate at equilibrium.
Characteristics Of Chemical Equilibrium:
Following are the characteristics of chemical equilibrium.
At the state of equilibrium
i) Rate of forward and backward reactions becomes equal.
ii) The concentration of reactants and products becomes constant.
iii) The progress of reaction ceases.
iv) This state is attained for the reversible reaction no matter in which direction the reaction is
proceeding.

187. v) At equilibrium, both reactants and products are present in different ratios.
Example:
Lets consider the reversible reaction between
H2 gas and Iodine vapours to form HI at 450oC. The
equilibrium is established when rising Curve of the
products [HI] and falling curve of the reactants [H2] and
[I2] become parallel to time axis (x-axis). It is shown in
the following fig.
H2(g) +I2(g)
450𝑜𝐶
⇇ 2HI(g)
Note: This reaction attains equilibrium, no matter it is
started by combination of H2 and I2 or by
decomposition of HI.
Law Of Mass Action:
This law and its mathematical relationship was given by C.M. Guldberg and P.Waage in 1864.
According to this law,
Rate of reaction is directly proportions to the product of active masses of reacting species.
Rate ∝ Concentration
Active Mass:
The concentration in moles dm-3 of reactants or products for a dilute solution is called as active
mass.
Explanation:
A general reaction between reactants A and B to form products C and D is as follows.
A + B ⇇ C + D
The active masses of A,B,C and D at equilibrium (equilibrium concentration) are represented in
square brackets, like [A], [B], [C] and [D] respectively.
i) Rate Of Forward Reaction:
The rate of forward reaction is directly proportional to molar concentration (active mass) of
reactants, A and B.
Rate of forward Reaction (Rf) 𝛼 [A] [B]
Rf = Kf [A] [B] ………….. (1)
Kf is proportionality constant and is called rate constant for forward reaction. The ratio of
rate of forward reaction to the product of molar concentration of reactants is called rate
constant for forward reaction (Kf)
Concentration
[HI]
[H ] or [I ]
2 2
Time
X
o teq

188. Kf =
𝑅𝑓
[𝐴][𝐵]
………………. (2)
ii) Rate Of Backward (Reverse) Reaction:
The rate of backward or reverse reaction is directly proportional to the product of
concentrations of products (reactants for reverse reaction).
Rate of reverse Reaction (Rr) ∝ [C] [D]
Rr = Kr [C] [D] …………..(3)
Kr is the proportionality constant and is called rate constant for reverse reaction. The ratio of
rate of reverse reaction to the product of active masses of products is called rate constant for
reverse reaction (Kr).
Kr =
𝑅𝑟
[𝐶][𝐷]
………………. (4)
iii) At Equilibrium Point:
Rf = Rr
Kf [A] [B] = Kr [C] [D]……. (5)
Rearranging the equation. (5)
𝐾𝑓
𝐾𝑟
=
[𝐶] [𝐷]
[𝐴] [𝐵]
…………… (6)
𝐾𝑓
𝐾𝑟
= K𝐶
We can write
𝐾𝐶 =
[𝐶] [𝐷]
[𝐴] [𝐵]
……………. (7)
Kc is the equilibrium constant and is defined as the ratio of rate constant of forward reaction
to the rate constant of backward reaction or the ratio of product of reactants concentration
to the product of products concentration at equilibrium point is called equilibrium constant.
Equation number (7) is called equilibrium constant expression.
Conventionally:
Equilibrium constant expression is written by taking concentration of products as numerator and
that of reactants as denominator as
𝐾𝐶 =
[Products]
[Reactants]
or KC =
𝑅𝑎𝑡𝑒 constant for forward reaction
𝑅𝑎𝑡𝑒 constant for reverse reaction
General Reaction with Co-Efficient Of Balanced Equation:
aA + bB ⇇ cC + dD
a,b,c and d are Co-efficient of balanced equation and represents the number of moles of A,B,C and
D respectively.
The letters a, b, c and d are written as exponents of respective concentration terms in equilibrium
constant expression.

189. Kc =
[𝐶]𝑐 [𝐷]𝑑
[𝐴]𝑎 [𝐵]𝑏
Units of Equilibrium Constants (Kc):
Formally, we don’t write the units of Kc or Kp. Although sometimes they have units.
i) Kc have no units:
When number of moles of reactants and that of products are equal.
Example:
CH3COOH + C2H5OH ⇇ CH3COOC2H5 + H2O
KC =
[𝐶𝐻3𝐶𝑂𝑂𝐶2𝐻5 [𝐻2𝑂]
[𝐶𝐻3𝐶𝑂𝑂𝐻] [𝐶2𝐻5OH]
=
[mol.dm-3] [mol.dm-3]
[𝑚𝑜𝑙.𝑑𝑚−3] [mol.dm-3]
= no units.
ii) KC have units when no. of moles of reactants and products are not equal:
When no. of moles of reactants and products are unequal.
Example:
a) N2 + 3H2 ⇇ 2NH3.
KC =
[𝑁𝐻3]2
[𝑁2] [𝐻2]3
=
[mol.dm-3]2
[𝑚𝑜𝑙.𝑑𝑚−3] [𝑚𝑜𝑙.𝑑𝑚−3]3
= dm+6
. 𝑚𝑜𝑙−2
or mol-2
. 𝑑𝑚+6
b) PCl5 ⇌ PCl3 + Cl2
KC =
[𝑃𝐶𝑙3][𝐶𝑙2]
[𝑃𝐶𝑙5]
=
[mol.dm-3][mol.dm3]
[𝑚𝑜𝑙.𝑑𝑚3]
= mol.𝑑𝑚−3
Equilibrium Constant Expression For Some Important Reactions:
i) Formation of an ester: (Aqueous phase reaction)
Kc =
𝑥2
(𝑎−𝑥)(𝑏−𝑥)
This expression donot involve volume term which means that equilibrium position and Kc value is
not affected by volume change at equilibrium stage.
ii) Dissociation of PCl5: (Gaseous phase reaction)
Kc =
𝑥2
𝑉(𝑎−𝑥)
Since, this expression involves the volume term. Hence, equilibrium position of this reaction will be
affected by volume change.
iii) Decomposition of N2O4 (Gaseous phase reaction):
Kc =
4𝑥2
𝑉(𝑎−𝑥)

190. The equilibrium position of this reaction will also be affected by volume change at equilibrium
stage because this expression involves the volume term.
iv) Synthesis of NH3 (Gaseous phase reaction):
Kc =
4𝑥2𝑣2
(𝑎−𝑥)(𝑏−3𝑥)3
The equilibrium position of this reaction will also be affected by volume change at equilibrium
stage since, this expression involves the volume term.
v) Synthesis of SO3 (Gaseous phase reaction):
Kc =
4x2𝑣
(a-2x)2(b-x)
The equilibrium position of this reaction will also be affected by volume change, since, this
expression involves volume term.
Relationships between equilibrium constants (Kp and Kc):
The equilibrium constant depends upon units used for concentration of reactants and products.
Let us consider the following general reaction.
aA + bB ⇇ cC + dD
The equilibrium constant expression for this reaction can be written by taking the concentration in
(i) moles/dm3
(ii) partial pressure (for gaseous system)
Relationship between Kp and Kc:
The Kp and Kc are related to each other by following equation derived from Dalton’s law of parital
pressure.
Kp = Kc (RT)n --------------- (1)
Where ∆n = no. of moles of products –no of moles of reactants.
R = general gas constant
T = absolute temperature
Comparison of Kp and Kc for different reactions:
(i) Kp is equal to Kc for all those reactions where the no. of moles of products and number of
moles of reactants are equal i.e. ∆n is zero.
OR

191. Kp is equal to Kc for all those reactions which proceeds with no change in total no. of moles.
Examples:
(1) H2 + F2 2HF
(2) N2 + O2 2NO
(ii) Kp is greater than Kc when number of moles of products is greater than that of reactants i.e.
reaction proceeds with increase in no. of moles and ∆n is positive.
Examples:
(1) PCl5 PCl3 + Cl2
(2) N2 O4 2NO2
(iii) Kp is smaller than Kc when no. of moles of products are lesser than that of reactants i.e.
reaction proceeds with decrease in no. of moles and ∆n is negative.
Examples:
(1) N2 +3H2 2NH3
(2) 2SO2+O2 2SO3
Applications of equilibrium constant:
The study of equilibrium constant provides us the following information.
(i) Direction of reaction
(ii) Extent of reaction
(iii) Effect of various factors on equilibrium constant and equilibrium position.
(i) Direction of Reaction: We know that for any reaction.
Kc =
[𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠]
[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]
We can determine the direction of reaction by taking the sample from reaction mixture and
calculating the ratio between concentration of products and reactants. There are three
possibilities.
(a) If [Products]/[Reactants] < Kc then reaction will proceed in forward direction to attain
equilibrium.
(b) If [Products]/[Reactants] > Kc then reaction will move in reverse direction to attain
equilibrium.
(c) If [Products]/[Reactants] = Kc then it means reaction is already at equilibrium.
ii) Extent of Reaction: There are three cases,
(i) If the value Kc is very large it means that reaction is almost complete i.e. almost all the
reactants are converted into products.

192. Example:
Equilibrium constant for the decomposition of Ozone to form Oxygen is 1055 at 25oC.
2O3 ⇇ 3O2 Kc =1055 at 25o C
This indicates that Ozone is unstable at room temperature and decomposes rapidly to O2. This
reaction is almost complete.
(b) If Kc value is small, it reflects that reaction does not proceed appreciably in the forward direction.
Example:
The value of equilibrium constant for the formation of NH3 at 400oC is 101.
N2(g) + 3H2(g) ⇇ 2NH3(g) Kc = 101 at 400oC
(c) If value of Kc is very small this shows a very little forward reaction.
Example:
The value of equilibrium constant for the decomposition of HF at 2000oC is 10-13.
2HF(g) ⇇ H2(g) + F2(g) Kc = 10-13 at 2000oC
It indicates high stability and negligible decomposition of HF even at 2000oC
The Le-Chatelier’s Principle:
If a stress is applied to a system at equilibrium then system acts in such a way so as to nullify the
effect of stress as far as possible.
System cannot cancel the effect of change but minimizes it.
By using this principle we can study the effect of different parameters on equilibrium position and
equilibrium constant. They are as follows.
1. Effect of change in concentration.
2. Effect of change in pressure or volume.
3. Effect of change in temperature.
4. Effect of a catalyst on equilibrium constant.
Note:
i) Change in concentration and change of pressure or volume only effects the equilibrium
position.
ii) Change in temperature effects both equilibrium position and equilibrium constant.
iii) Addition of a catalyst neither effects the equilibrium position nor equilibrium constant.
However it helps the equilibrium to establish earlier.
(A) Effect Of Change In Concentration:

193. o At equilibrium, when a substance among the reactants is added or substance among
products is removed then equilibrium position is disturbed and reaction moves in forward
direction to attain the equilibrium position again.
o At equilibrium, when a substance among products is added or substance among reactants is
removed then reaction will move in backward direction to nullify the effect of stress.
Example:
In order to understand above explanation of the effect of change in concentration on the reversible
reaction, consider the reaction in which BiCl3 reacts with water to give a white Insoluble compound
BiOCl.
BiCl3 + H2O ⇇ BiOCl + 2HCl
The equilibrium constant for above reaction can be written as.
𝐾𝑐 =
[𝐵𝑖𝑂𝐶𝑙][𝐻𝐶𝑙]2
[𝐵𝑖𝐶𝑙3 ][𝐻2 𝑂]
Aqueous solution of BiCl3 is cloudy. Because of the hydrolysis and formation of BiOCl. i) If small
amount of HCl is added at equilibrium the reaction will move in backward direction to minimize the
effect of addition of HCl on equilibrium. When reaction moves in backward direction, clear solution is
obtained.
ii) However, if H2O is added in the system at equilibrium then reaction will move in forward
direction to minimize the effect of stress and again cloudy solution is obtained.
(B) Effect Of Change In Pressure Or Volume:
Change in pressure or volume are important for those gaseous phase reversible reactions in which
number of moles of reactants and products are not equal.
(I) Effect of change in pressure or volume when reaction proceeds with decrease in number of
moles:
When gas phase reaction proceeds with decrease in number of moles then it leads to decrease in
volume at equilibrium state.
i) When pressure is increased or volume is decreased at equilibrium state then reaction will
move in forward direction to minimize the effect of stress.
ii) When pressure is decreased or volume is increased for such reaction, then reaction will move
in backward direction.
2SO3(g) + O2(g) ⇇ 2SO3(g)
(II) Effect of change in pressure or volume when reaction Proceeds with increase in number of
moles:
For such reaction volume of equilibrium mixture is greater than the initial volume of reactants.

194. i) When pressure is increased or volume is decreased at equilibrium then equilibrium will be
disturbed and reaction will move in backward direction to minimize the effect of disturbance.
ii) When pressure is decreased or volume is increased at equilibrium then reaction will proceed
in forward direction.
Example:
PCl5(g) ⇇ PCl3(g) + Cl2(g)
(C) Quantitative effect of volume or pressure change on equilibrium position:
(I) For the Reaction in which Reaction Proceed with Decrease in Number of Moles:
Consider the formation of SO3 from SO2 and O2. This reaction proceeds with the decrease in
number of moles.
2SO2(g) + O2(g) ⇇ 2SO3(g)
Expression of Kc for this reaction is.
𝐾𝑐 =
4𝑋2
𝑉
(𝑎 − 2𝑥)2 ( 𝑏 − 𝑥)
Where, ‘V’ is the volume of reaction mixture at equilibrium state, a and b are the number of moles
of SO2 and O2 taken initially respectively. Similarly, ‘x’ and 2x are the number of moles of O2 and
SO2 which has reacted at equilibrium according to above expression. i) When volume is increased
then ‘x’ has to be decreased to keep ‘Kc’ constant. It means reaction will move in backward
direction.
ii) Similarly, if volume is decreased then ‘x’ has to be increased to keep ‘Kc’ constant and
reaction will move in forward direction.
For the reaction which proceeds with decrease in number of moles, If the volume is decreased at
equilibrium then reaction will proceed in forward direction and vice versa.
(II) For Those Reaction Which Proceeds With Increase In Number Of Moles:
1. Consider the dissociation of PCl5 to form PCl3 and Cl2. This reaction proceeds with the
increase in number of moles.
PCl5(g) ⇇ PCl3(g) + Cl2(g)
Expression for Kc is:
𝐾𝑐 =
𝑋2
𝑉( 𝑎 − 𝑥)
Where, ‘V’ is the volume of reaction mixture at equilibrium state and ‘a’ is the number of
moles of PCl5 taken initially. ‘x’ is the number of moles of PCl5 which has reacted at
equilibrium.

195. i) When volume is decreased for such reaction, the above expression indicates that ‘x’
has to be decreased to keep Kc constant. So, reaction will move in backward direction.
ii) When volume is increased ‘x’ has to be increased to keep ‘Kc’ constant. So, reaction will
proceeds in forward direction.
2. N2O4(g) ⇇ 2NO2(g)
Expression for Kc is:
𝐾𝑐 =
4𝑥2
𝑉( 𝑎 − 𝑥)
When volume is decreased at equilibrium, ‘x’ has to be decreased to keep Kc constant. So,
reaction will proceeds in backward direction and vice versa.
Conclusion:
i) The increase in volume shifts the equilibrium in backward direction for those reactions
which proceeds with decrease in number of moles and vice versa.
ii) The increase in volume shifts the equilibrium in forward direction for those reactions
which proceeds with increase in number of moles and vice versa.
(D) Effect Of Change In Temperature:
Most of the chemical reactions are disturbed by changing temperature.
(I) Decrease In Temperature:
Decrease in temperature favours exothermic reaction.
Reason: is that when temperature is decreased at equilibrium state, then exothermic reaction will
proceed to release energy.
Example:
CO(g) + H2O(g) ⇇ CO2(g) + H2(g) ∆H= - 41.84 kJ mol-1
At equilibrium state, if we take out heat and keep the system at this new lower temperature, then
system will readjust itself so as to compensate the loss of heat energy. So, exothermic reaction will
proceed and this reaction releases energy. Hence, above reaction will proceed in forward direction
when temperature of the system is decreased and vice versa.
(II) Increase In Temperature:
Increase in temperature favours endothermic reaction.
Reason: is that when energy is given to the system at equilibrium then temperature increases
which disturbs the equilibrium state. To minimize the effect of this rise in temperature, the
reaction system will readjust itself in such a way that it absorbs the heat. So, endothermic reaction

196. will proceed. Above reaction will proceed in backward direction when temperature of the system is
increased at equilibrium.
(III) Effect Of Temperature On Solubility:
i) The salts whose dissolution in water is an endothermic process, their solubility increases with
the rise of temperature.
Example:
Dissolution of KI in water is an endothermic process:
KI(s) ⇇ KI(aq) ∆H = 21.4 kJ mol-1
At equilibrium, when temperature is increased then more amount of KI dissolves in water and its
solubility increases. And reverse is true when temperature is decreased.
ii) The salts whose dissolution in water is exothermic process, their solubility will decrease with
increase in temperature. And reverse is true when temperature in decreased.
Example:
Heat of solution of LiCl and Li2CO3 is negative.
Na2CO3(s) ⇇ Na2CO3(aq) ∆H = -25 kJ/mol
iii) For those salts which have heat of solution close to zero, there is no effect of temperature
change on their solubility.
Example:
Solubility of NaCl remains ineffective when temperature is increased or decreased because its
heat of solution is almost zero.
(E) Effect Of Catalyst On Equilibrium Position And Constant:
Catalyst has no effect on the position of equilibrium. It just increases the rate of both forward and
backward reactions and equilibrium is attained within a short time.
In most of the reversible reactions, equilibrium is not always reached within a suitable short time.
So, a catalyst is added which speeds up the reaction and equilibrium is attained in a reduced time.
Actually, catalyst equally lowers the energy of activation for both forward and reverse step by
giving new path to the reaction.
Synthesis of Ammonia By Haber’s Process:
Formation of ammonia gas from nitrogen and hydrogen gas is reversible process.
We apply the concept of equilibrium constant to maximize the yield of ammonia by keeping in view
the chemical reaction for the formation of ammonia.
N2(g) + 3H2(g) ⇇ 2NH3(g) ∆H = -92.46 kJ/mol
1. By continual removal of ammonia, reaction will move in forward direction.

197. 2. Increase in pressure decreases the volume that causes the reaction to proceed in forward
direction and yield of ammonia increases.
3. Decrease in temperature favours exothermic reactions. As forward reaction is exothermic, the
decrease in temperature favours the reaction in forward direction.
So, high pressure, low temperature and continual removal of ammonia increase the yield of
ammonia.
Rate of Formation and Yield:
The yield of ammonia is favoured at low temperature, but at low temperature rate of reaction is
very slow and process becomes uneconomical. Temperature is raised to a moderate value and catalyst is
added to increases the rate of reaction. Without a catalyst, faster rate is only achieved at higher
temperature where yield of ammonia decreases.
Optimum Conditions For The Reasonable Yield Of
Ammonia:
1. Temperature around
673K (400oC)
2. Pressure about 200-
300atm.
3. Pieces of iron crystals embedded in fused mixture of
MgO, Al2O3 and SiO2.
These conditions will optimize the yield of ammonia.
Table below shows the effect of rise in temperature on Kc
value. At 200Co Kc value is very high but rate is too slow to
make the process economical.
Figure shows percentage yield of NH3 versus temperature
at five different operating pressures. At very high pressure
and low temperature, the yield of ammonia is high but rate
of formation is slow. So, industrial conditions are denoted
by circle are b/w 200-300 atm pressure and 400oC.
Removal of Ammonia:
T(K) Kc
200 7.17 × 1015
300 2.69 × 108
400 3.94 × 101
500 1.72 × 102
600 4.53 × 100
700 2.96 × 10-1
800 3.69 × 10-2
Temp.
oC
K
Mole%
of SO3
200 5500 98
300 690 91
400 160 75
500 55 61
600 25 46
700 13 31

198. The equilibrium mixture has 35% by volume of NH3. The
mixture is cooled by refrigeration coils until ammonia
condenses (B.P = -33.4oC) and is removed. The boiling
points of N2(g) and H2(g) are very low and they remain in gas
phase and are recycled by pumps back into reaction
chamber.
Application Of Haber’s Process And Uses Of Ammonia:
• Nearly 13% of all nitrogen fixation on earth is accomplished through Haber’s process.
• This process produces 110 million tons of ammonia annually.
• About 80% of NH3 is used for the production of fertilizers.
• Ammonia is also used in the manufacture of explosives.
• It is also used for the production of nylon and other polymers.
Preparation Of Sulphur Trioxide:
In contact process for the preparation of H2SO4, the conversion of SO2 into SO3 is achieved by
reversible reaction.
2SO2(𝑔)
+ 𝑂2(𝑔) ⇇ 2SO3(𝑔)
ΔH= -194 kJmol-1
1. By continual removal of SO3 from the reaction mixture. The reaction will proceed in forward
direction and yield of SO3 increases.
2. By decreasing temperature, reaction will also move in forward direction.
3. By increasing pressure or decreasing volume the reaction will proceed in forward direction.
Instead of removal of SO3 and maintain high pressure, the concentration of O2(g) is increased to
increase the yield of SO3.
The given table shows the effect of temperature on the yield of SO3. To have best possible yield of
SO3 within a reasonable time, a mixture of SO2(g) and O2(g) at one atmospheric pressure is passed over a
solid catalyst (V2O5 or finely divided Platinum) at 650oC. The equilibrium mixture is recycled at low
temperature 400-500oC to increase the yield of SO3.
SO3 is dissolved in H2SO4 to get Oleum (H2S2O7) which is diluted in water to H2SO4.
H2SO4 is a king of chemicals. A country’s industrial progress is measured by the amount of H2SO4 manufactured each
year.
Ionic Product of Water
Pure water is a very poor conductor of electricity, but its conductance is measurable. Water
undergoes self-ionization which is also called auto-ionization of water
𝐻2O + H2𝑂 ⇇ 𝐻3𝑂+
+ OH−
Simply,

199. 𝐻2𝑂 ⇇ 𝐻+
+ OH−
Equilibrium constant expression for this reaction can be written as
𝐾𝑐=
[𝐻+][OH−]
[𝐻2𝑂]
= 1.8×10-16
mol/dm3
--------------(i)
Kc value for this reaction is 1.8 x 10-16 moles / dm3. Concentration of pure water [H2O] can be
calculated from its density i.e d = 1g / cm3 or 1000g / dm3
Concentration =
1000
18
moles / dm3
= 55.55 moles / dm3
------------(ii)
If we consider that [H2O] remains constant i.e 55.55 moles / dm3.
𝐾𝑐 =
[𝐻+][OH−]
[𝐻2𝑂]
𝐾𝑤= K𝑐 × [𝐻2𝑂] = [𝐻+][OH−]--------(iii)
Putting value of ‘Kc’ and [H2O] from equation (i) and (ii) into equation
(iii).
𝐾𝑤= 1.8×10-16
×55.55 = [𝐻+][OH−]
𝐾𝑤= 1.01×10-14
= [𝐻+][OH−] at 25𝑜
𝐶
[𝐻+][OH−] = 1×10-14
at250
𝐶
Kw is called ionic product or dissociation constant of water at 25oC.
The [𝐻+] = [OH−]butvaluesare10-7
moles/dm3
. At 100oC, values are greater than 10-7 moles /
dm3 while [H+] = [ OH-].
Relation Between Kw and Temperature:
The value of dissociation constant of water depends upon temperature. Greater the temperature,
greater is the value of Kw and vice versa. As the temperature increases, the decomposition of H2O into its
ions increases. So, [H+] and [OH-] concentration also increases, but remains equal. That is why water
remains neutral. The value of Kw increases almost 75times when temperature is increased from 0oC to
100oC. Anyhow, the increase in Kw is not regular.
Effect of temperature on Kw is shown in table below
In case of neutral water,
[H+] = [OH-]
As we know
[𝐻+][OH−]=10-14
moles/dm3
[𝐻+][𝐻+]=10-14
moles/dm3
( since, [H+]=[OH-])
[𝐻+]2
=10-14
moles/dm3
[𝐻+]2
= (10-7)2
moles/dm3
Taking square root on both sides, we get
T(oC) Kw
0 0.11 × 10-14
10 0.30 × 10
25 1.0 × 10
40 3.00 × 10
100 7.5 × 10-14

200. [𝐻+]=10-7
moles/dm3
at 25oC
As [𝐻+] = [OH−]
So [OH−]=10-7
moles / dm3 at 25oC.
In case of addition of small amount of an acid,
[𝐻+] > [OH−]
In case of addition small amount of a base,
[OH−] > [𝐻+]
During both of these additions, the value of ‘Kw’ remains the same i.e 10-14 at 25oC.
pH and pOH:
Concentration of ‘H+’ and OH- are too low to express easily. That is why idea of pH and pOH is
introduced.
pH is negative ‘log’ of concentration of hydrogen ions.
pH = - log [H+]
pOH is negative log of concentration of OH- ions.
pOH = - log [OH-]
For neutral water, at 25oC.
[𝐻+] = [OH−] = 10-7
moles/dm3
pH = - log[𝐻+] = - log10-7
= 7
pOH = - log[OH−]= - log10-7
=7
So,
pH = 7 solution is neutral
pH < 7 solution is acidic
pH > 7 solution is basic
pKw is negative log of Kw.
𝐾𝑤 = [𝐻+][OH−] = 10-14
at 25𝑜
𝐶
Taking negative ‘log’ on both the sides of equations:
- logK𝑤= - log[𝐻+][OH−]= - log10-14
pK𝑤= - log[𝐻+] +[- logOH−] = (-1)(-14)log10
pK𝑤= pH + pOH = 14(𝑙𝑜𝑔 1 = 0)at 25𝑜
C.
pKw value is less than ‘14’ at higher temperature.
pH values normally varies from

201. 0 → 14 at 25oC. Solution with zero pH value and negative pH value and pH values greater than 14
are known. The table represents the relationship of pH and pOH.
An Acid with greater [H+] ion concentration and low pH value is stronger and vice versa.
Table shown represents pH values of different materials. This represents acidic or basic nature of
commonly used solutions.
Material pH pOH Material pH pOH
1.0 M HCl 0.1 13.9 Bread 5.5 8.5
0.1 M HCl 1.1 12.9 Potatoes 5.8 8.2
0.1M CH3COOH 2.9 11.10 Rain water 6.2 7.8
Gastric juice 2.0 12.00 Milk 6.5 7.5
Lemons 2.3 11.7 Saliva 6.5-3.9 7.5-7.1
Vinegar 2.8 11.2 Pure water 7.0 7.00
Soft drinks 3.0 11.00 Eggs 7.8 6.2
Apples 3.1 10.9 0.1M NaHCO3 8.4 5.6
Grapefruit 3.1 10.9 Seawater 8.5 5.5
Oranges 3.5 00.5 Milk of magnesia 10.5 3.5
Tomatoes 4.2 9.8 0.1 M NH3 11.1 2.9
Cherries 3.6 104 0.05 M Na2CO3 11.6 2.4
Bananas 4.6 9.4 0.1 M NaOH 13.0 1.00
Ionization constant of acid (ka)
Ionization constant of acid or dissociation constant of acids gives us information about the strength
of acid.
Let us consider ‘HA’ is an acid when it dissolves into water is produces H+ and A-1
HA + H2O ⇌ 𝐻3𝑂+
+ A-1
Kc for reversible reaction is given by

202. 𝐾𝑐 =
[𝐻3𝑂+][𝐴-1]
[HA][𝐻2𝑂]
𝐾𝑐[𝐻2𝑂] =
[𝐻3𝑂+][𝐴-1]
[HA]
------------------(i)
At equilibrium state, the concentration of water is almost same as at initial stage because water
has been taken in large excess. A reasonable approximation is to take the concentration of water
effectively constant
𝐾𝑐[𝐻2𝑂]=K𝑎
So, equation (i) becomes,
𝐾𝑎 =
[𝐻3𝑂+][𝐴-1]
[HA]
Value of ‘Ka’ of an acid gives us information about the extent to which an acid can ionize i.e greater
the ‘Ka’ value greater will be the strength of an acid and vice versa.
Value of ‘Ka’ can be calculated by using above equation if we know the value of [𝐻+]or[𝐻3𝑂+] and
initial concentration of an acid. Similarly, we can calculate the equilibrium concentration of H3O+ and A-1
produced if we know the initial concentration of acid HA and its Ka value
When
Ka < 10-3 acid is weak
Ka = 1 to 10-3 acid is moderately strong
Ka > 1 acid is strong
pKa of acid: Larger the pKa value, weaker is the acid and vice versa.
i) If the difference of the pKa values of two acids is one then acid with smaller pKa is ten times
stronger than the other.
ii) If the difference of pKa values of two acids is two then acid with smaller pKa is 100 times
stronger than the other.
Percentage of Ionization Of Acids:
The ratio of amount of acid ionized to the amount of acid initially available multiplied by 100 is
called percentage ionization.
%ageionization =
amountofacidionized
amountofacidinitiallyavailable
×100
The percentage ionization of weak acid depends upon the extent of ionization which in turn
depends upon depends upon the extent of dilution of aqueous solution. Greater the dilution, greater will
be the percentage ionization and vice versa. This is called Ostwald’s dilution law (dilution increases the
degree of dissociation of weak acid).
Ionization Constant Of Bases (Kb)

203. The substances which have ability to accept proton are called Lowry – Bronsted base.
When such substances are dissolved into water then they take hydrogen ions and release hydroxyl
ions from water.
Example:
NH3(ag)+ H2O ⇇ NH4
+1
(aq) + OH−
(aq)
CO3
-2
(aq) + H2𝑂(aq) ⇇ HCO3
-1
(aq) + OH−
(aq)
Strengths of bases are compared by their ionization constant values.
Suppose a general reaction of ionization of a base
𝐵(aq) + H2𝑂(ℓ) ⇇ BH(aq)
+
+ OH−
(aq)
𝐾𝑐 =
[BH+][OH−]
[𝐵][𝐻2𝑂]
Since, concentration of water is considered as constant, being present in large excess
𝐾𝑐[𝐻2𝑂] =
[BH+][OH−]
[𝐵]
-------------- (i)
𝐾𝑐[𝐻2𝑂] = K𝑏
Hence, equation (i) becomes,
𝐾𝑏 =
[BH+][OH−]
[𝐵]
‘Kb’ gives us information about the strength of base. Greater the ‘Kb’ value, stronger will be the
base and vice versa.
‘Kb’ values of some bases are given below in table.
pKb of base:
Kb values for weak bases are small numbers usually expressed in exponential form. It is convenient
to convert them into whole number by taking their negative log. Then we obtain pKb.
pKb = - log Kb
Greater the pKb value, weaker is the base and vice versa.
If the difference of pKb values of two bases is 1 then base with smaller pKb is ten times stronger
than the other.
Common Ion Effect
The suppression of ionization of weak electrolyte by strong electrolyte having common ion is called
common ion effect.
Examples:

204. (1) In purification of sodium chloride by passing hydrogen chloride gas through brine, common
ions are produced which suppresses the ionization of NaCl.
Equilibrium constant expression for this process can be written as follows:
NaCl(s) ⇇ Na (aq)
+1
+ Cl (aq)
-1
Kc =
[Na+][Cl−]
[NaCl]
On passing ‘HCl’ gas it ionizes in brine as
HCl ⇇ H (aq)
+1
+ Cl (aq)
-1
On passing HCl gas, concentration of Cl-1 ions is increased. Therefore, NaCl crystallizes out of
the solution to maintain the constant value of equilibrium constant.
(2) The solubility of less soluble salts KClO3 in water is suppressed by the addition of more
soluble salt KCl by common ion effect. K+1 is a common ion. The ionization of KClO3 is
suppressed and it settles down as precipitate.
KClO3 (s) ⇇ K (aq)
+1
+ ClO 3(aq)
-1
KCl(s) ⇇ K (aq)
+1
+ Cl (aq)
-1
(3) The dissociation of weak acid ‘H2S’ in water can be suppressed by the addition of stronger
acid HCl. HCl produces common H+1 ions.
H2S ⇇ 2H (aq)
+
+ S (aq)
-2
HCl ⇇ H (aq)
+`
+ Cl (aq)
-1
During salt analysis mixture of H2S and HCl is used for the precipitation of second group basic radicals. This
mixture is a group reagent for second group basic radicals and maintains required low S-2
ion concentration.
(4) By the addition of NH4Cl, the ionization of NH4OH is suppressed because NH4OH is weak
electrolyte.
NH4OH ⇇ NH 4(aq)
+
+ OH (aq)
−
NH4Cl ⇇ NH4 (aq)
+1
+ Cl (aq)
-1
Here NH4 (aq)
+1
is the common ion.
Mixture of NH4OH and NH4Cl is used for the precipitation of third group basic radicals. This
mixture is a group reagent for third group basic radicals and maintains required low OH- ion
concentration.
Buffer Solutions
The solutions which show resistance to change in their pH when a small amount of an acid or base
is added are called Buffer solutions.

205. There are two types Buffer solutions
(i) Acidic Buffers
(ii) Basic Buffers
Preparation of Buffers Solution
Buffer solutions can be prepared by mixing two substances. There are two methods for the
preparation of Buffer solution.
(i) By mixing a weak acid and salt of it with strong base.
Examples:
(a) CH3COOH + CH3COONa
(b) H2CO3 + NaHCO3
These give acidic buffer with pH values less than 7.
(ii) By mixing weak base and salt of it with strong acid.
Examples:
(a) NH4OH + NH4Cl
(b) Ca(OH)2 + CaCl2.
These give basic buffer with pH values greater than 7.
How do Buffers Act?
Buffer action can be explained by keeping in view the concepts of common ion effect and Le-
Chatelier’s principle.
Let us take the example of an acidic buffer consisting of CH3COOH and CH3COONa CH3COOH being
a weak electrolyte undergoes very little dissociation. Whereas, CH3COONa which is a strong electrolyte,
will produce CH3COO- ions as common ions. The dissociation of CH3COOH is suppressed due to the
common ion effect of CH3COO-.
CH3COOH + H2O ⇇ CH3COO (aq)
−
+ H3O (aq)
+
CH3COONa(aq) ⇇ CH3COO (aq)
−
+ Na (aq)
+
The buffer mentioned above is the large reservoir of CH3COOH and CH3COO- components.
i) Whenever an acid is added to this buffer and ‘H+’ concentration increases, at this stage
CH3COO- reacts to form undissociated CH3COOH. So, pH of the buffer will remain almost
same because ‘H+’ ions which are added are captured by CH3COO- ion and pH will not vary
too much.
ii) If base is added in this buffer, base will produce OH- ions which which combine with H3O+ to
form neutral compound i.e H2O. So, pH will remain almost unchanged.
The buffer solution consisting of NH4Cl and NH4OH can resist the change in pH and pOH when acid
or base is added from outside.

206. Calculation of pH of a Buffer:
OR
Handerson’s Equation for the preparation of Buffer Solution of Required pH
Value:
pH = pKa – log
[acid]
[salt]
pH = pKa + log
[salt]
[acid]
Two factors govern the pH of buffer solution.
1- pKa of the acid.
2- The ratio of concentration of salt and acid.
Equal concentration of acid and its salt:
The best buffer can be prepared by taking equal concentration of acid and its salt. In this case pH of
buffer is controlled by pKa of the acid.
Handerson’s Equation for Base:
pOH = pKb+log
[salt]
[base]
Buffer Capacity:
The extent of resistance offered by the buffer to change in its pH is called buffer capacity.
The capability of buffer to resist the change in pH is called buffer capacity.
OR
The no. of moles of acid or base which are required by one dm3 of buffer solution for changing its
pH by one unit, is called buffer capacity of a solution.
Explanation:
Buffer capacity is the quantitative measure that how much extra acid or base buffer solution can
absorb before the buffer is essentially destroyed. The molarities of the two components of buffer solution
determine the buffer capacity.

207. Equilibria Of Slightly Soluble Ionic Compounds (Solubility Product = Ksp)
The product of equilibrium concentration of ions of sparingly soluble salt is called solubility product
of that salt.
Explanation:
When a soluble ionic compound like NaCl is dissolved in water, it dissociates completely into ions. But
for slightly soluble salts, the dissociation is not complete at equilibrium stage.
Example:
i) When PbCl2 is shaken with water, the solution contains undissociated PbCl2 and dissolved
ions (Pb+2 and Cl-1 ions).
PbCl2(s) ⇇ PbCl2(aq) ⇇ Pb (aq)
+2
+ 2Cl (aq)
−1
According to law of mass action
Kc =
[Pb(aq)
+2
][Cl(aq)
-1
]2
[PbCl2(𝑠)]
For sparingly soluble salts, the concentration of salt is almost constant.
Kc[PbCl(s)] = [Pb (aq)
+2
][Cl (aq)
−1
]2
Ksp = [Pb (aq)
+2
][Cl (aq)
−1
]2 (Kc[PbCl(s)] = Ksp)
ii) Lead sulphate is sparingly soluble compound and it dissociates to a very small extent.
PbSO4(s) ⇇ PbSO4(aq) ⇇ Pb (aq)
+2
+ SO (aq)
-2
Kc =
[Pb(aq)
+2
][SO4
-2
(aq)]
[PbSO4(𝑠)]
Being a sparingly soluble salt the concentration of lead sulphate PbSO4 almost remains
constant.
( ) ( ) ( )
2 2
c 4 s aq 4 aq
K PbSO Pb SO
+ −
     
=
     
Ksp = [Pb (aq)
+2
][SO 4(aq)
-2
] ([PbSOP4(s)])
The value of Ksp is temperature dependent. Usually it increases with increase in temperature.
For a general, sparingly soluble substance AxBy
AxBy ⇇ XA+y + YB-x
Ksp= [A+y]x [B-x]y
The solubility product is the product of the concentration of ions raised to an exponent equal to
the co-efficient of the balanced equation. The following table shows the solubility product of slightly
soluble ionic compounds.
Applications of Solubility Product:

208. Followings are the applications of solubility product.
(i) Determination of Ksp from solubility.
(ii) Determination of solubility from Ksp
(iii) Effect of common ion on solubility.
a) Determination of Ksp from solubility
From the solubility of compound we can calculate the solubility product of a salt. Solubility is
defined in two ways.
(i) No. of grams of solute dissolved in 100g of solvent to form saturated solution at a particular
temperature is called its solubility.
(ii) No. of moles of solute dissolved in 1kg of solvent to prepare saturated solution at a particular
temperature is called its solubility.
Since, the salt is sparingly soluble, amount of solute will be very small and amount of solvent is
considered as the amount of solution. From this, volume of solution can be calculated. The no. of
moles of solute in one dm3 of solution can be calculated by dividing given mass of solute with its
molar mass. Then by using the balanced equation, we can find the molarity of each ion and then
Ksp.
b) Determination of Solubility from Ksp
For this purpose we need the
• Formula of compound
• Ksp value of the compound
And unknown molar solubility is calculated and the concentration of ions are determined.
Effect of common ion on solubility:
The presence of common ion decreases the solubility of a slightly soluble ionic compound.
Example:
Let us consider saturated solution of PbCrO4 which is sparingly soluble ionic salt.
PbCrO4(aq) ⇇ Pb (aq)
+2
+ CrO 4(aq)
-2
Ksp = [Pb (aq)
+2
][CrO 4(aq)
-2
]
The addition of Na2CrO4 decreases the solubility of PbCrO4 due to common ions CrO 4(aq)
-2
to keep
Ksp constant.
Na2CrO4(aq) ⇇ 2Na (aq)
+1
+ CrO 4(aq)
-2
Ksp > Ionic Product ------------ Unsaturated
Ksp = Ionic Product ------------ Saturated
Ksp < Ionic Product ------------ Super Saturated

210. Assessment 1
1. Which of the following is true about the reversible reaction?
a. The products never react to form reactants again.
b. Its ultimate goal is to complete steadily.
c. Reversible reaction is continuous and never complete
d. It is unidirectional.
2. The given reaction, CaCO3 CaO + CO2 becomes unidirectional in lime kiln if:
a. CO2 is added to system.
b. Temperature is increased.
c. CO2 is decomposed
d. CO2 escapes continuously
3. In the given reaction, N2 + 3H2 2 NH3 , Chemical equilibrium is meant for:
a. Concentration of all reactants and products are changing continuously.
b. Concentration of reactants is constant where as concentration of product is changing
c. Concentration of reactants is changing where as concentration of products is constant
d. Concentration of reactants and products is constant.
4. The process which corresponds to the equilibrium state is:
a. Boiling of water in beaker over Bunsen burner, keeping the temperature of water
constant
b. Presence of few drops of water mixed with air in a balloon, keeping the temperature of
balloon constant
c. Ice-water system at 00C, keeping the temperature constant
d. Both a and b
5. For the given reaction; , the position of chemical equilibrium can be shifted to
right by:
a. Addition of HI to the reaction mixture at constant volume
b. Increasing the volume
c. Decreasing pressure
d. Addition of H2 at constant pressure
6. Which of the following is true for the rate of chemical reaction according to law of mass action?
a. Rate of a chemical reaction is directly proportional to molar concentration of product
b. Rate of a chemical reaction is inversely proportional to molar concentration of reactants
c. Rate of a chemical reaction is directly proportional to molar concentration of reactants
d. None of these

211. Assessment 1
7. The active mass of 34 g of ammonia in 2 dm3 flask would be:
a. 2 mole dm-3
b. 1 mole dm-3
c. 0.5 mole dm-3
d. 0.25 mole dm-3
8. For a given reaction, 2H2(g) + O2(g) → 2H2O(g) ; noticeable quantity of water decomposed
to give hydrogen and oxygen gas by heating at 1500 0C which means:
a. At low temperature, reverse reaction is too small to be noticed
b. It is reversible only at low temperature
c. It may be reversible by providing high temperature.
d. Both a and c
9. On a given condition, the equilibrium concentration of PCl5, PCl3 and Cl2 are 0.1
mole/dm3, 0.01 mole/dm3 and 0.1 mole/dm3 respectively for the reaction; PCl5 PCl3 +
Cl2. The value of Kc is
a. 0.1 mole dm-3
b. 0.01 mole-1dm+3
c. 0.01 mole dm-3
d. 0.1 mole-1dm+3
10. For the reaction; A + B C. Which of the following value of Kc will favour more the
completion of the reaction?
a. Kc = -1
b. Kc = 1
c. Kc = 102
d. Kc = 103

212. Assessment 2
1. 6.0 moles each of hydrogen and iodine heated in a sealed ten dm3 container. At equilibrium, 5
moles of HI were found. The equilibrium constant for H2 + I2 2HI is:
a. 1
b. 1.02
c. 2
d. 2.04
2. For a given reaction; C2H5OH + CH3COOH CH3COOC2H5 + H2O .The rate constant for forward
and backward reactions are 1.1 x 10-2 and 1.5 x 10-3 respectively. Equilibrium constant for the
reaction is:
a. 1.5
b. 2.5
c. 7.3
d. 6
3. In order to get the maximum yield for Haber’s process, which of the following condition is most
appropriate?
a. High temperature, high pressure and high concentrations of the reactants
b. High temperature, low pressure and low concentrations of the reactants
c. Low temperature and high pressure
d. low pressure and low concentration of H2
4. Kw is the ionic product of water. It is
a. Temperature independent
b. It decreases with increase in temperature
c. It increased almost 25 times when the temperature is increased from 00C to 100 0C.
d. It increases with increase in temperature.
5. Kw is thedissociation constant or Ionic product of water which increases, if
a. Pressure is decreased
b. H+ is added
c. Temperature increases
d. All of these
6. Under a given set of experimental conditions, with increase in the concentration of the
reactants, the rate of a chemical reaction
a. Decreases
b. Increases
c. Remains unaltered
d. First decreases and then increases

213. 7. The unit of Kc for the reaction; CH3COOH + C2H5OH CH3COOC2H5 + H2O is:
a. no unit
b. mole dm-3
c. mole-1 dm+3
d. mole -2 dm+6
8. A basic Buffer solution can be prepared by mixing
a. weak acid and its salt with strong base
b. strong acid and its salt with weak base
c. weak base and its salt with strong acid
d. strong base and its salt with weak acid
9. Which of the following corresponds to the precipitation reaction to occur if:
a. Ionic concentration is less than Ksp
b. Ionic concentration is more than Ksp
c. Ionic concentration is equal to Ksp
d. if it is present at any moment
10. The magnitude of equilibrium constant (Kp) of a reversible gaseous reaction is changed by:
a. Change in Pressure
b. The application of catalyst
c. Keeping the no of moles of reactants and products same.
d. none of above

214. Assessment 3
1. Which of following change will favour the formation of more product (SO3) at equilibrium in a
given reversible reaction; 2SO2 + O2 2SO3 + heat
a. By adding SO3 at equilibrium
b. By increasing temperature
c. By decreasing temperature
d. By decreasing pressure
2. By increasing the temperature of an equilibrium system will:
a. Favour the exothermic reaction only
b. Favour the endothermic reaction only
c. Favour both the exothermic and endothermic reactions
d. Favour neither the exothermic nor endothermic reactions
3. For a given reaction, Kp = Kc . Which of the following is assumed from it?
a. At constant temperature, pressure of gas is proportional to its concentration
b. At constant temperature, partial pressure is either same or changed to concentration of
gas
c. At high temperature, the pressure of gas is decreased and is proportional to
concentration.
d. All of these
4. For a given species, whose 100 molecules are ionized out of 10,000. Its %age degree of
ionization is :
a. 1
b. 0.1
c. 0.001
d. 10
5. The ability of the buffer to resist the change in pH on addition of small amount of acid or base
is termed as:
a. Buffer action
b. Buffer resistance
c. Buffer capacity
d. All of these
6. Three moles of ethanol and three moles of ethanoic acid were reacted together according to
the equation
C2H5OH + CH3COOH ⇋ CH3COOC2H5 + H2O
At equilibrium; there were 2 moles each of ethyl ethanoate (CH3COOC2H5) and water formed.
What is the equilibrium constant for this reaction?
a. 6
b. 4
c. 2
d. 1

215. 7. When temperature is increased from 0 oC to 40 oC, the Kw is increased in how many times?
a. 3 times
b. 10 times
c. 30 times
d. 75 times
8. Which of the following is the correct statement about chemical Equilibrium?
a. At point A, reactant is minimum in concenration
b. At point B, Rf becomes equal to Rr
c. At point C, Product is maximum in concentration
d. Forward reaction becomes equal to reverse reaction and stops.
9. The relationship between Kc and Kp for the following general reaction is :
A+ B C
a.
b.
c.
d.
10. For the reaction; A + 2B C + D, the equilibrium conc. each of [A], [B] ,[C] and [D] are 0.1 mole
dm-3 and value of Kc is 8.0. The position and the direction of reaction would be:
a. 1 (Reverse)
b. 10 (Reverse)
c. 1 (Forward)
d. 10 (Forward)
A
B
C

216. Assessment 4
7. According to Le-Chatelier’s principle, temperature does not have any significant effect on
the solubility of:
a. KCl
b. NaCl
c. LiCl
d. Li2CO3
8. The active mass of one dm3 of water at 25 0C is:
a. 5.55 mole dm-3
b. 55.5 mole dm-3
c. 18 mole dm-3
d. 1000 mole dm-3
9. The combination of which of the following two substances is used as a group reagent in
third group basic radicals:
a. H2S and HCl
b. NaCl and HCl
c. KCl and KClO3
d. NH4OH and NH4Cl
10. A buffer containing [HCOOH]= 0.1 mole dm-3 and [HCOONa] = 0.1 mole dm-3. What is the
pH of buffer (pKa of HCOOH is 3.78)?
a. 3.78
b. 2.78
c. 4.78
d. 5.78
11. Buffer capacity of a buffer solution is determined by the:
a. Temperature of the solution
b. Actual molarities of its components
c. pH of one of the solution
d. All of these
12. In case of sparingly soluble salts, which of the following is a measure of how far to the right
dissolution proceeds at equilibrium i.e. saturation?
a. Kw
b. Ksp
c. pH
d. pKa

217. 13. For the precipitation reaction to occur, the following is being followed:
a.
b.
c.
d.
14. For a given chemical equilibrium, A + B C + D; when one mole of each of A and B are
mixed, 0.5 mole each of C and D are formed. The Kc calculated is:
a. 1
b.
c. 2.5
d.
15. For which one of the following reactions Kp = Kc?
a. PCl5 PCl3 + Cl2
b. N2 + 3H2 2 NH3
c. H2 + Cl2 2 HCl
d. 2NO2 N2O4
16. A chemical reaction which is catalyzed by a catalyst. The catalyst
a. Increases the equilibrium constant of reaction
b. Decreases rate constant of the reaction
c. Increases activation energy of the reaction
d. Does not affect equilibrium constant of reaction

218. KEY
Assessment 1
Assessment 2
1. c
2. d
3. d
4. c
5. a
6. c
7. b
8. d
9. c
10. d
1. d
2. c
3. c
4. d
5. c
6. b
7. a
8. c
9. b
10. a

219. Assessment 3
Assessment 4
1. c
2. b
3. a
4. a
5. c
6. b
7. c
8. b
9. c
10. b
1. b
2. b
3. d
4. a
5. b
6. b
7. a
8. d
9. c
10. d

220. Electrochemistry
Chapter 10
MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org
Oldest, Largest and Most Credible Platform

221. Electrochemistry
Chapter 10
ELECTRO CHEMISTRY
Learning Objectives:
 Oxidation number or state
 Explanation of electrolysis
 Electrode Potential
 Balancing of redox reactions by
Ion-electron method
 Balancing of redox reactions by
oxidation number change
method.
Electrochemistry

222. Electrochemistry
Chapter 10
Electrochemistry
The inter-conversion of electrical and chemical energy in an electrochemical cell is called
Electrochemistry.
Oxidation reaction Reduction reaction
The reactions in which substances
i) lose electrons
ii) Gain oxygen
iii) Lose hydrogen
iv) Undergoanincreaseinoxidationstate
are called oxidation reactions.
The reactions in which substances
i) Gain electrons
ii) lose oxygen
iii) Gain hydrogen
Undergo a decrease in oxidation state
are called oxidation reactions.
Oxidation Number or State
The apparent charge on an atom of an element in a molecule or in an ion is called
oxidation state or oxidation number of that atom.
Example:
In
1 1
H Cl
 
, oxidation state of hydrogen is +1 and that of Chlorine is -1.
Rules for Assigning Oxidation Number:
(i) Oxidation number of elements which are in free state:
The oxidation number of all elements which are in free state is zero. This is often
shown as a zero written on the symbol.
Example:
o o o
2
H ,Na ,Mg etc.
(ii) Oxidation number of an ion consisting of single atom:
The oxidation number of an ion consisting of a single element is same as the charge on
the ion.
Example:
The oxidation number of K+1, Ca+2, Al+3, Br-1, S-2, are +1, +2, +3, -1, and -2 respectively.
(iii) Oxidation number of hydrogen:
The oxidation number of hydrogen in all its compounds is +1 except metal hydride where
its oxidation number is -1 e.g. Na+1H-1, Mg+2H(-1)2.

223. Electrochemistry
Chapter 10
(iv) Oxidation number of oxygen:
The oxidation number of oxygen in all its compounds is -2 except in peroxides, OF2 and
in super oxides where its oxidation number in -1 , +2 and -1/2 respectively.
(v) For neutral molecule:
In neutral molecules, the algebraic sum of the oxidation numbers of all the elements is
zero.
Example:
In
2( 1) 2
H O
 
, the total charge is zero.
(vi) Oxidation number of an ion consisting of more than one atom:
In such type of ions, the algebraic sum of oxidation number of all the atoms is equal to
the charge on the ion.
(vii) Electronegativity and oxidation number:
In any substance, the more electronegative atom has the negative oxidation number
while less electronegative element is given the positive oxidation number.
Balancing of Redox Equations by Oxidation Number
Method:
Steps to balance redox equation:
(i) Writedown the skeleton equation of theredoxreaction under consideration.
(ii) Identify the elements, which show change in their oxidation number during
reaction.
(iii) Mention the oxidation number above the symbols of those elements which
undergo change in their oxidation number.
(iv) Indicate the change in oxidation number by arrows joining the atoms on both sides
of the equation. It shows number of electrons gained or lost.
(v) Equate the increase or decrease in oxidation number i.e electrons gained or lost by
multiplying with a suitable digit.
(vi) Balance the rest of the equation by inspection method.
Example:
Balance the following equation by oxidation number method.
K2Cr2O7 + HCl KCl + CrCl3 + Cl2 + H2O
Solution:
Let us balance the equation stepwise:
1. Write the equation with the oxidation number of each element

224. Electrochemistry
Chapter 10
1. Identify, those elements whose oxidation numbers have changed.
i) Equation shows that the oxidation state of Cr changes from +6 to +3.It means that it
is reduced.
ii) The oxidation state of Cl changes from -1 to zero. It means that it is oxidized.
iii) Moreover, the oxidation number of chlorine remains the same i.e. -1 when KCl and
CrCl3 are produced.
iv) We should write HCl twice on the left hand side.
 One HCl on left side shows those Cl atoms which do not undergo a change in their
oxidation number.
 Second HCl on the left hand side shows those Cl atoms which undergo a change
in their oxidation numbers.
v) Draw the arrows between the same elements whose oxidation numbers have
changed. Also, point out the change in oxidation number.
Reduction
( 1)3
1 6 1 1 3 ο
2 3
2 7 2 2
HCl+K Cr O +HCl KCl+CrCl +Cl +H O

    


Oxidation
vi) Cr has changed its oxidation number from +6 to +3 and chlorine has changed from -1 to
zero. It means 6 electrons have been gained by two Cr atoms and 6 electrons has been
lost by 6 chlorine atoms.
(+3e-) x 2 = +6 e- (Reduction)
(-1e-) x 6 = -6e- (Oxidation)
         
+1 2 +6 2 -2 7 -1 3 +1 2
+1 -1 +1 -1 +3 -2
0
2 2 7 3 2 2
K Cr O +HCl KCl+CrCl +Cl + H O


 
-1 -1 3
-1
+6 +3 ο
7
2 2 3 2 2
+ H Cl O
Cl
O
HCl + K Cr K +CrCl +Cl +H



225. Electrochemistry
Chapter 10
4. In order to balance the number of electrons lost and gained, multiply HCl with six. In
this way, the 6 electrons lost by 6 Cl

will be gained by 2Cr+6 to give 2Cr+3. But do not
multiply other HCl molecules with anything at this moment.
HCl + K2Cr2O7 + 6HCl KCl + CrCl3 + Cl2 + H2O
5. Let us balance Cr atoms by multiplying CrCl3 by 2. Balance Cl2 on right hand side, whose
oxidation number has changed by multiplying it with 3. In this way, the atoms which
have been oxidized and reduced get balanced.
HCl + K2Cr2O7 + 6HCl KCl + 2CrCl3 + 3Cl2 + H2O
6. To balance K atoms, multiply KCl by 2.
HCl + K2Cr2O7 + 6HCl 2KCl + 2CrCl3 + 3Cl2 + H2O
7. Now balance those atoms of chlorine which have not been oxidized or reduced. There
are 8 such chlorine atoms on the right hand side with 2KCl and 2CrCl3. So, multiply HCl
with eight. This HCl has produced KCl and CrCl3.
8HCl + K2Cr2O7 + 6HCl 2KCl + 2CrCl3 + 3Cl2 + H2O
8. Balance the rest of the equation by inspection method. To balance O-atoms, multiply
H2O with 7.
8HCl + K2Cr2O7 + 6HCl 2KCl + 2CrCl3 + 3Cl2 + 7H2O
or
K2Cr2O7 + 14HCl 2KCl + 2CrCl3 + 3Cl2 + 7H2O
This is the final balanced equation.
Balance the following equation by ion electron method.
Cl
+ -
4
MnO 
 Cl2 + Mn2+
10.1.4 Balancing of Redox Equations by Ion-Electron Method:
The balancing of redox equations by the loss and gain of electrons, usually involves
quite a few ions, which do not undergo change in oxidation number and which are not
really necessary for the process of balancing.
The ion-electron method eliminates all unnecessary ions and retains only those ions
which are necessary.

226. Electrochemistry
Chapter 10
Example: (Acidic medium)
Balance the equation for the reaction of HCI with KMnO4 where CI- is oxidized to
CI2 and MnO4
- is reduced to Mn2+. The skeleton equation which does not contain
either H+ or H2O, is
CI- + MnO4
- CI2 + Mn2+
Solution: It is clear that CI- is oxidized to CI2 and -1
4
MnO is reduced to Mn2+.
i) Splitting the equation into half-reactions,
Oxidation half reaction
Steps to balance the redox equation:
The steps to balance the redox equation by ion-electron method are as follows:
(i) Write skeleton equation in such a way that only represents those substances which
are actually involved in the reaction.
(ii) Split the equation into two half reactions, one showing oxidation half reaction
and other showing the reduction half reaction.
(iii) The element should not be written as free atoms or ion. But it should be written in
the form of free atom or ion if it really exists as such. Otherwise it should be
written as a real molecular or ionic species.
(iv) Balance each half equation with respect to the number of atoms of each element.
 In neutral or acidic solution, H2O is added for balancing oxygen and H+
is
added for balancing hydrogen.
 In alkaline solution, OH-
should be added for balancing oxygen and H2O
should be added for balancing hydrogen.
 Oxygen should be balanced first.
(v) Balance each half reaction as to the number of charges by adding electrons to
either the left or the right side of the equation.
(vi) Multiply each half reaction by a suitable number so that the total number of
electrons lost by the reducing agent equals the number of electrons gained by the
oxidizing agent.
(vii) Add the two half reactions, count the number of atoms of each element on each
side of the equation and also check the net charge on each side, which should
be equal on both sides.

227. Electrochemistry
Chapter 10
CI- CI2
Reduction half reaction
MnO4
- Mn2+
ii) Balancing atoms on both sides of oxidation half reaction.
2CI- CI2 ------------- (1)
Now, balance the reduction half reaction. To balance O-atoms, add 4H2O on
R.H.S. and to balance H-atoms add 8H+ on L.H.S. The reason is that medium is
acidic.
8H+ + MnO4
- Mn2+ + 4H2O --------- (2)
iii) Balancing the charges by adding electrons in equation (1) and (2), we get (3)
and (4).
2CI- CI2 + 2e- ----------------- (3)
8H+ + MnO4
- +5e- Mn2+ + 4H2O --- (4)
For making the number of electrons lost in first equation equal to the number of electron gained
in the second equation, multiply the first equation by 5 and second by 2. After adding both
equations and canceling the common species on both sides, balanced equation is obtained.
[2Cl- Cl2 + 2e-] x 5
[5e- + 8H+ + MnO4
-1 Mn+2 + 4H2O] x 2
- + - 2+
4 2 2
10CI +6H + 2 MnO 5CI + 2Mn +8H O

Example: (Basic medium)
Balancethefollowingequationinbasicaqueoussolutionbyion-electronmethod.
MnO4
-
(aq) + C2O4
2-
(aq) + H2O MnO2(s) + CO2(g) + OH-
(aq)
Solution:
The following steps are involved in the balancing of equation in basic aqueous
solution by ion-electron method.
(i) Identify those elements, which undergo change in oxidation number by writing
oxidation number above each element.
(-2)4 (+3)2(-2)4 (+1)2 (-2)2 (-2)2
+7 -2 +4 +4 2 1
-1 -2 1
4 2 4 2 2 2
(MnO ) +( C O ) + H O MnO CO (OH)
 


  
The elements undergoing a change in oxidation number are Mn and C.
(ii) Split the reaction into two half reactions, the oxidation and reduction half
reactions.
C2O4
-2 CO2 (oxidation half reaction)
MnO4
-1 MnO2 (reduction half reaction)

228. Electrochemistry
Chapter 10
 Balancing of Oxidation Half Reaction:
C2O4
-2 CO2
Balance the Carbon atoms on both sides of the half reaction.
C2O4
-2 2CO2
Balance the charges on both sides of the half reaction by adding the appropriate
number of electrons to the more positive side.
C2O4
-2 2CO2 + 2e- ------------------- (1)
The oxidation half reaction is balanced.

229. Electrochemistry
Chapter 10
MnO4
- MnO2
Balance the Oxygen atoms by adding OH- ions on the side where oxygen is
required. Add two OH- ions for each oxygen atom needed. So, we have to add
4OH- on R.H.S:
MnO4
- MnO2 + 4OH-
Balance the hydrogen, by adding H2O on the other side of the half reaction. Add
one H2O for each two OH- ions. In this way, oxygen and hydrogen atoms are
balanced.
2H2O + MnO4
- MnO2 + 4OH-
Balance the charges by adding three electrons to L.H.S. of equation
3e- + 2H2O + MnO4
- MnO2 + 4OH---(2)
The reduction half reaction is balanced.
Multiply each half reaction by an appropriate number, so that the number of electrons
on both the half reactions becomes equal. For this purpose, multiply the oxidation half
reaction by 3 and the reduction half reaction by 2.
3C2O4
-2 6CO2 + 6e- --------------------(3)
2 x [ 3e- + 2H2O + MnO4
- MnO2 + 4OH-]
6e- + 4H2O + 2MnO4
- 2MnO2 + 8OH------(4)
Add the two half reactions to get the net ionic equation and cancel out anything
appearing on both sides of the equation. For this purpose, add equation (3) and equation
(4).
3C2O4
-2 6CO2 + 6e-
6e- + 4H2O + 2MnO4
- 2MnO2 + 8OH-
3C2O4
-2 + 4H2O + 2MnO4
- 6CO2 +2MnO2 + 8OH-
Hence, the equation is balanced
Balancing of Reduction Half Reaction:

230. Electrochemistry
Chapter 10
Electrolytic Conduction
Conduction is of two types.
(i) Electrolytic conduction
(ii) Electronic or metallic conduction
Electrolytic Conduction:
The conduction in which current is carried by positively and negatively charged ions is
solution or in fused state is called electrolytic conduction.
Electronic Conduction or Metallic Conduction:
The conduction of electricity due to free movement of electrons in the metallic lattice
is called electronic conduction or metallic conduction.
Ionization:
The process of splitting of ionic compounds into charged particles when fused or
dissolved in water is called ionization.
Electrolysis:
The process in which non – spontaneous reaction takes place at the expense of electricity
is called electrolysis.
Electrochemical Cells:
A cell or a system which consists of electrodes dipped into an electrolyte in which a redox
chemical reaction uses or generates electric current is called electrochemical cell.
Types of Electrochemical Cells:

231. Electrochemistry
Chapter 10
There are two types of electrochemical cells.
Electrolytic Cell:
The cell in which non–spontaneous redox reaction takes place at the expense of electrical
energy is called electrolytic cell. In this cell, electrical energy is converted into chemical
energy.
Examples:
Down’s cell and Nelson’s cell are the examples of electrolytic cell.
Voltaic or
galvanic cell
Electrolytic
cell
Electrochemical cells

232. Electrochemistry
Chapter 10
Voltaic or Galvanic Cell:
The electrochemical cell in which spontaneous redox reaction produces electrical energy
is called voltaic or galvanic cell. In this cell, chemical energy is converted into electrical
energy.
Example: Nickel Cadmium cell.
Anode:
The electrode at which oxidation takes place is
called anode. In electrolytic cell, it carries
positive charge and negativeionsmovetowards
it where they release their electrons and get
oxidized. So, oxidation takes place at anode.
Cathode
The electrode at which reduction takes place is called cathode. In electrolytic cell, it
carries negative charge and positive ions move towards it where they gain electrons and
get reduced. So, reduction takes place at cathode.
Products of Electrolysis:
 When molten salts or fused
salts are electrolyzed then
products are predictable.
 When aqueous solutions are electrolyzed then in certain cases hydrogen gas is
obtained at cathode and oxygen gas is obtained at anode. The products formed
from few electrolytes are shown in table.
Explanation of Electrolysis
The process of electrolysis is carried out by using aqueous solution of salts or molten salts.
(a) Electrolysis of Fused Salts
i) Reaction at cathode:
Electrolyte Cathode Anode
PbBr2(molten) Pb(s) Br2(g)
NaCl(molten) Na(s) Cl2(g)
NaCl(aq) H2(g) Cl2(g)
CuCl2(aq) Cu(s) Cl2(g)
CuSO4(aq) Cu(s) O2(g)
KNO3(aq) H2(g) O2(g)
NaOH(aq) H2(g) O2(g)
H2SO4(aq) H2(g) O2(g)
Electrolyte
Copper
cathode
Copper anode
CuSO4(aq) Cu
deposits
Cu(s) dissolves
to form Cu2+ ions
Electrolyte Silver
cathode
Silver anode
AgNO3(aq)
and HNO3(aq)
Ag
deposits
Ag (s) dissolves
to form Ag+ ions

233. Electrochemistry
Chapter 10
When fused salts are electrolyzed, the metal ions called cations carrying positive charge
move towards cathode. Cathode provides electrons to them and they get reduced.
Example:
In the case of fused PbCl2, Pb+2 ions being positively charged move towards cathode and
gain electrons.
   
2
(s)
Pb 2e Pb reduction
 
 

l
i) Reaction at Anode:
The anions move towards anode and they lose their electrons at anode and get oxidized.
Example:
In case of PbCl2, Cl- ions being negatively charged move towards anode and lose
electrons.
2Cl-
(l) + 2e- 
 Cl2(g) (oxidation)
Similarly, for fused NaCl and fused PbBr2, electrolytes are decomposed during electrolysis.
‘Pb’ and Na are produced at cathode while Br2 and Cl2 gases are formed at anode.
ii) Flow of Electrons and Ions
In the fused salt there is electrolytic conduction in the cell while there is electronic
conduction in the external circuit. In, the cell ions move towards anode and cathode to
get oxidized or reduced respectively. The flow of electrons takes place through external
circuit from anode to cathode.
(b) Electrolysis of Aqueous Solutions of Salts
The electrolysis of aqueous solutions of salt is not so simple and is somewhat
complicated. The reason is that water molecules are oxidized as well as reduced that is
why products of electrolysis of aqueous solutions of salts are not precisely predictable.
Example
Let us take the example of aqueous solutions of NaNO3. It dissociates in water as follows:
+1 -1
3 3
NaNO Na +NO
ƒ
Water molecules also get ionized
+ 1
2 2 3 (aq) (aq)
H O+H O H O +OH


i) Reaction at Cathode:
NOTE: Current is conducted by electrons outside the cell in the
external circuit from anode to cathode. Current is conducted by ions
inside the cell from cathode to anode.

234. Electrochemistry
Chapter 10
H3O+ ions being positively charged move towards cathode and get reduced in
competition of Na+1 ions.
 
+1 -
3 2 (l) (g)
H O +1e H O +H reduction


Similarly, one more hydrogen atom is formed. Then
both hydrogen atoms react to form hydrogen gas.
     
2
g g g
H +H H


H2 gas is evolved at cathode. Na+1 ion will remain as such at cathode and donot get
reduced because H+1 ion have greater tendency to gain electrons.
Reaction at anode:
At anode, both nitrate and hydroxide ions are
present.Hydroxide ionsare easier to discharge than
nitrate ions.Nitrate ions willremains in the solution while the electrode reaction isasfollow.
(OH- 
 OH + e-)  4 (oxidation)
The OH groups decompose to give oxygen and water molecules.
4OH 
 O2 + 2HO2
So, oxygen gas is obtained at anode. Na+1 ion and -1
3
NO ions will remain in the solution.
Expectedorderofdischargeofionsmayalsodependupontheirconcentration.Ifanionwith
greater tendency to discharge has low concentration in comparison to ion which has less
tendencytodischarge itself then ion withgreater concentration mayget discharged.
Electrolysis Processes of Industrial Importance
Following are the important industrial applications of electrolysis processes.
(i) Extraction of Sodium:
Extraction of sodium by the electrolysis of fused sodium chloride is carried out in Down’s
cell. In this case, molten NaCl is electrolyzed between graphite anode and iron cathode.
During this process Cl2 gas is obtained as by product with sodium metal. Following
reactions take place at electrodes.
Reactions:
2NaCl(s) ˆ ˆ †
‡ ˆ ˆ 2Na+1 + 2Cl-1 (Ionization)
i) Reaction at Cathode:
2Na+1 2e- 
 2Na (Reduction)
ii) Reaction at Anode:
   
     
g
-1 -
g
2
g g g
2Cl 2Cl +2e
Cl +Cl Cl




 
oxidation
Overall reaction at anode is
2Cl- + 2e- 
 Cl2
Overall reaction taking place at both electrodes is given by adding the reactions taking
place at anode and cathode

235. Electrochemistry
Chapter 10
       
+1 -1 o
2
l l s
2Na + 2C 2Na C

 
l l g
(ii) Industrial preparation of caustic soda:
In industry, caustic soda (NaOH) is prepared by the electrolysis of aqueous concentrated
solution of NaCl(brine) using titanium anode and mercury or steel cathode. This
electrolysis is carried out in Nelson cell and Castner-kellner cell or Hg–cell.
     
2
aq
H O +1 -1
s aq
2NaCl 2Na +2Cl
ˆ ˆ ˆ †
‡ ˆ ˆ ˆ
i) Reaction at Anode:
   
-1 -
2
aq g
2Cl Cl + 2e

  
Oxidaiton
ii) Reaction at Cathode:
 
g
- -
2 ( ) 2 (aq)
2H O + 2e H 2OH

 
l  
Reduction
Overall reaction taking place is
2NaCl(s) + 2H2O( )
l 
    
g g
2 2 (aq)
C +H +2NaOH
l
Chlorine and hydrogen gases are obtained as by products.
(3) Extraction of Magnesium and calcium:
Magnesium and calcium metals are extracted by the electrolysis of their fused
chlorides. Mg and Ca are collected at cathodes while Cl2 at anodes.
(4) Extraction of Aluminum:
Aluminum is extracted by electrolyzing fused bauxite, Al2O3. 2H2O in the presence of
fused cryolite Na3AlF6.This process is called Hall-Beroult process.
(5) Preparation of Anodized Aluminum:
Anodized Aluminum is prepared by making it an anode in electrolytic cell containing
H2CrO4 or H2SO4. As a result, Aluminum anode gets coated with Al2O3 (oxide) layer that
resists the corrosion.
p
Note: Freshly prepared anodized Aluminum is hydrated and can
absorb dyes.

236. Electrochemistry
Chapter 10
(6) Purification of Copper:
Electrolytic cell is used for the purification of copper. Impure copper is made the anode
and thin sheet of pure copper is made the cathode and copper sulphate solution is used
as an electrolyte. The atoms of Cu from impure Cu anode are oxidized and converted into
Cu+2 ions and migrate to cathode which is made of pure Cu and deposit on the pure
Cu(cathode) . So, thickness of pure Cu electrode (cathode) increases. Impurities are left
at anode. Following are the reactions which take place during the purification of copper.
i) Reaction at Anode:
   
+2 -
s aq
Cu Cu +2e

 (Copper dissolves from impure copper anode)
ii) Reaction at Cathode
   
+2 -
aq s
Cu +2e Cu

 (Copper deposits on the surface of copper
cathode)
(7) Electroplating:
Coating of one metal electrolytically on the surface of another metal is called electroplating.
copper, silver, nickel and chromium plating is done by various types of electrolytic cell. By the
processofelectroplating, metalscanbe protected,repaired anddecorated.
Voltaic or Galvanic Cell
The cell in which spontaneous redox reaction proceeds to produce electricity is called
Galvanic cell. In this cell, chemical energy is converted in electrical energy.
Example:
Daniel cell (Zn-Cu cell)
1. Construction of cell:
It consists of two half cells which are electrolytically connected through salt bridge as
shown in the following figure:

237. Electrochemistry
Chapter 10
i) Left Half Cell:
The left half-cell consists of a strip of zinc metal dipped in one molar zinc sulphate
solution. Zn metal and Zn+2 ionsexists in the following equilibrium.
Zn(s) ˆ ˆ †
‡ ˆ ˆ Zn+2
(aq) + 2e-
ii) Right Half Cell:
The right half-cell consists of strip of Cu metal dipped in one molar copper sulphate
solution. Cu metal and Cu+2 ion exists in the following equilibrium.
   
+2 -
s aq
Cu Cu +2e
ˆ ˆ †
‡ ˆ ˆ
Two half cells are connected electrically through salt bridge. Salt bridge is used to avoid
direct mixing of two solutions and direct chemical reactions between two solutions.
Otherwise, two half cells would destroy.
2. Working of cell (Discharging of galvanic cell):
(a) Zinc metal acts as anode (relatively negative):
Zinc metal has greater tendency to lose electrons than Copper and gets negatively
charged relative to Copper. The half-cell reaction is as follows:
Zn(s) ˆ ˆ †
‡ ˆ ˆ
+2
(aq)
Zn ) + 2e- (Oxidation)----------(i)
Accumulation of positive charge in the left beaker (Zinc half-cell):
Zinc atoms after losing electrons dissolve into the solution as Zn+2 ions. Hence, left
beaker becomes positively charged due to the accumulation of Zn+2 ions.
(b) Copper metal acts as cathode (Copper half-cell):
A Galvanic cell consisting of Zn and
Cu electrodes at 25o
C and unit
concentration of electrolytic
solutions.

238. Electrochemistry
Chapter 10
Copper has lesser tendency to lose electrons than Zinc. Instead, copper has greater
tendency to gain electrons and copper electrode gets positively charged relative
to Zinc.
The half-cell reaction is as follows:
+2
(aq)
Cu + 2e- ˆ ˆ †
‡ ˆ ˆ Cu(s) (Reduction)---------(ii)
Accumulation of negative charge in the right beaker (Copper half-cell):
Electrons from Zinc electrode move towards copper electrode through the external
circuit. Cu+2 from the solution gain electrons and deposit on copper strip. As a result,
solution in the right beaker becomes negatively charged due to accumulation of
2
4
SO
ions.
The sum of the two reactions (i) and (ii) taking place at anode and cathode gives
overall reaction.
Zn(s) ˆ ˆ †
‡ ˆ ˆ
+2
(aq)
Zn + 2e-
+2
(aq)
Cu + 2e- ˆ ˆ †
‡ ˆ ˆ Cu(s)
Zn(s) +
+2
(aq)
Cu ˆ ˆ †
‡ ˆ ˆ
+2
(aq)
Zn + Cu(s)
Cell Representation:
Galvanic cell is represented as follows:
       
+2 +2
s aq aq s
Zn / Zn 1M | | Cu 1M/Cu
Salt Bridge:
It is a U-shaped glass tube having a saturated solution of KCl or KNO3 mixed with jelly. It
connects the two half cells.
Note: By convention, oxidation half-cell is
shown on the right while reduction half is
shown on the left in cell representation.
This representation shows
that:
i) Indicates the salt
bridge.
ii) Oxidation occurs at
Zinc (anode).
iii) Reduction occurs at
copper (cathode).

239. Electrochemistry
Chapter 10
Function of salt bridge:
In the left beaker, there is accumulation of positive charges (Zn+2) due to oxidation of
Zn atoms. In the right beaker, there is accumulation of negative charge ( 2
4
SO
) due to
reduction of Cu+2 ions. Salt bridge prevents net charge accumulation in both the half
cells. It allows the negative ions to migrate from right beaker to left beaker by diffusion
through it.
In the absence of salt bridge, there will be accumulation of net charges in the beakers
that will stop the oxidation reduction reactions. As a result, flow of current through
external circuit would stop and cell will stop producing electricity.
Voltaic Cell as Reversible Cells (recharging of galvanic cell):
Voltaic cell can be converted into reversible cell. For this purpose, external circuit of
votaic cell is replaced by the source of electricity,that opposes the reaction of voltaic cell.
Now, the external source will push the electrons in the opposite direction and supply
energy to the cell by which non-spontaneous reverse reaction takes place i.e electrons
moves from Cu to Zn electrode. For the zinc-copper cell, half-cell reactions are reversed
and are as follows:
     
     
+2 -
aq s
+2 -
s aq
Zn +2e Zn reduction
Cu Cu +Ze oxidation




Overall reaction which takes place in reversed cell is
       
+2 +2
aq s s aq
Zn + Cu Zn + Cu



240. Electrochemistry
Chapter 10
During recharging, oxidation occurs at copper electrode and reduction occurs at zinc
electrode. In this cell, electrical energy from external source drives a non-spontaneous
reaction.
Electrode potential
The potential setup when an electrode is in contact with one molar solution of its own ions at 298K is
known as standard electrode potential.It isrepresented asEo.
Explanation:
Whenever a metal rod or strip is placed in a solution of its own ions then there are two
possibilities.
(a) Metal atomsmay dissolve aspositive ions. In this process,the electronsare deposited
on the metal electrode .As a result, electrode carry negative charge and solution
carry positive charge.
(b) Metalionswhicharepresentinthesolutionmaygetelectronsfromthemetalstriporrodand
get discharged as atoms. In this way, positive charge appears on the metal electrode while
negativechargeaccumulateinthesolution.
The electrode potential is basically the measure of tendency of an atom of an
element to gain or lose electrons. As a result of anyone the tendency, a potential
difference is established between an electrode of 1M solution of its ions at 298K.
Example:
When Zn rod is dipped in the solution of its own ions, then Zn rod will bear an
accumulation of negative charge. This is due to greater tendency of Zn atom to lose
electrons. The negative charge on the Zn rod will attract an atmosphere of positively
charged zinc ions around the rod to form an electrical double layer as shown in the
following figure.

241. Electrochemistry
Chapter 10
Standard hydrogen electrode (SHE)
A glass tube provided with finely divided Platinum in which hydrogen at 1atm is passed
and bubbled through 1molar HCl solution is called standard hydrogen electrode.
Construction:
It consists of a piece of platinum foil, which is
coated electrolytically with finely divided
platinum black. Finely divided platinum black
is a used to increase the surface area. It is
suspended in one molar solution of HCl. Pure
hydrogen gas at one atmospheric pressure is
continuously bubbled into 1M HCl solution.
Platinum acts as a conductor and it facilitates
the attainment of equilibrium between the
gas and its ions in solution.
The potential of this electrode is arbitrarily taken as zero because two opposing
reactions are taking place at equal rate.
2H+
(aq) + 2e- H2(g) Eo
rcd = 0.0V
H2(g) 2H+
(aq) + 2e- Eo
oxi = 0.0V
Standard hydrogen electrode
Equilibrium between zinc and its ions in solution Equilibrium
can be represented as Zn(s) ˆ ˆ †
‡ ˆ ˆ Zn+2
(aq) +2e-
Note:
It is used for the measurement of electrode potentials of other
elements.

242. Electrochemistry
Chapter 10
Measurement of Electrode Potential
Electrode potential of single electrode cannot be measured. However, for the
measurement of electrode potential of single electrode, it should be electrically
connected with some standard electrode which is standard hydrogen electrode. For the
measurement of electrode potential of an electrode, it is coupled electrically with the
standard hydrogen electrode.
The solutions of two electrodes are connected through salt bridge to:
i) Provide highly conductive pathway for the diffusion of ions
ii) Keep the two solutions separate
iii) Maintain the electrical neutrality.
The potential difference is measured by a voltmeter gives the potential of electrode as
potential of hydrogen electrode is zero. An oxidation or reduction may take place at
standard hydrogen electrode, depends upon the nature of the electrode which is coupled
with it.
Measurement Of Electrode Potential Of Zinc:
For the measurement of the electrode potential of zinc, a galvanic cell is established
between zinc electrode dipped in IM solution of its own ions and standard hydrogen
electrode as shown in the figure. Under standard conditions,the voltmeter reads0.76volts
and deflection is in such a direction that indicate that zinc has a greater tendency to give off
electronsthan hydrogenby0.76V.
In other words, the half reaction for oxidation of zinc i.e.
Electrode potential of zinc
Zn(s) 
Zn+2 + 2e

has greater tendency to occur than H2(g) 
 2H+ + 2e- by 0.76
volts. The standard electrode potential of zinc is 0.76 volts. It is called oxidation

243. Electrochemistry
Chapter 10
potential of zinc and is given the positive sign. The reduction potential of zinc is -0.76
volts. The reactions which occur at electrodes are:
At anode:
Zn(s) 
 Zn+2
(aq) + 2e- (oxidation)
At cathode:
2H+
(aq) + 2e-

H2(g) (reduction)
Zinc electrode acts as anode because oxidation takes place at zinc electrode, while
hydrogen electrode acts as cathode because reduction takes place at hydrogen
electrode.
Measurement of Electrode Potential of Copper:
The electrode potential of copper can be measured by the establishment of galvanic cell
in which copper electrode is dipped in one molar solution of its own ions and connected
with standard hydrogen electrode (SHE) as shown in the following figure.
Electrode potential of copper
Under standard conditions the voltmeter reads 0.34 volts and the deflection is in such a
direction as to indicate that hydrogen has a greater tendency to give off electrons than
copper by 0.34 volts. In other words, the half reaction H2(g) 
2H+
(aq) + 2e- has greater
tendency to occur than Cu(s) 
 Cu+2
(aq) + 2e- by 0.34 volts. So, this indicates that
copper has greater ability to reduce itself as compared to hydrogen. The standard
reduction potential of copper is +0.34 volts, while the oxidation potential of copper is -
0.34 volts. The reactions which occur at electrodes are as follows.
At anode:
+
2(g) (aq)
H 2H +2 

 e (oxidation)
At cathode:
Cu+2
(aq) + 2 e- 
 Cu(s) (reduction)

244. Electrochemistry
Chapter 10
In this galvanic cell, hydrogen electrode acts as anode because oxidation takes place at
hydrogen electrode while copper electrode acts as cathode because reduction takes
place at copper electrode.
The Electrochemical Series
The list in which elements are arranged in order of their standard electrode potential
(standard reduction potential) on hydrogen scale is called electrochemical series.
Explanation:
In electrochemical series, electrode potential have been given in the reduction mode as
recommended by the IUPAC. Sometimes half reactions are written in the oxidation
mode and the corresponding potential are oxidation potentials, the magnitude of the
potential is not affected by the change in mode but signs are reversed. The important
point to remember in using reduction potential is that they relate only to standard
conditions i.e IM solution of ions at 25oC under one atmospheric pressure.
Change in:
(i) Temperature
(ii) Concentration
(iii) Pressure
will affect the values of reduction potential.
Standard reduction potentials (Eo) of substances at 298K

245. Electrochemistry
Chapter 10
Applications of Electrochemical Series
Following are the important applications of electrochemical series.
(i) Prediction of the feasibility of chemical reaction:
By using electrochemical series, we will come to know whether a particular reaction will
take place or not.
Example:
+2
(aq)
Cu can oxidize Zn(s) but +2
(aq)
Zn cannot oxidize Cu(s).
Zn(s) + +2
(aq)
Cu 
 +2
(aq)
Zn + Cu(s)
By keeping in view the electrochemical series, this reaction can be explained. The
standard reduction potentials of elements are as follows:
+2
(aq)
Cu + 2e- 
 Cu(s) Eo = + 0.34 Volts

246. Electrochemistry
Chapter 10
+2
(aq)
Zn + 2e-

 Zn(s) Eo = -0.76 Volts
Since, zinc is being oxidized in the above reaction, reverse of reduction potential will be
considered.
Zn(s) 
 Zn+2
(aq) + 2e- Eo = 0.76 Volts
+2
(aq)
Cu + Zn(s) 
 Cu(s) + +2
(aq)
Zn Eo
(cell) = 1.10 Volts
The overall positive value for the reaction potential suggests that the process is
energetically feasible. If the sum of Eo values of the two half cell reactions is negative,
then reaction will not be feasible.
(ii) Calculations of the Voltage Or Electromotive Force (emf) of Cells:
The force with which electrons move from one electrode to other electrode in external
circuit is called electromotive force (emf).
Explanation:
The electrode occupying higher position in electrochemical series will act as anode and
oxidation takes place on it and the element which occupying lower position will act as a
cathode and reduction will take place on it.
Let us find out the cell potential or emf of the cell in which ‘Zn’ and Cu act as anode and
cathode respectively.
Zn(s) 
 +2
(aq)
Zn + 2e- (oxidation)
+2
(aq)
Cu + 2e-

 Cu(s) (reduction)
The oxidation potential of Zn is positive and reduction potential of copper is also positive.
So, emf cell is given by:
Eocell = Eo
Oxi + Eo
red
Eocell = 0.76 + 0.34
Eocell = 1.10 volts
(iii) Comparison of Relative Tendency of Metals and Non-metals to Get
Oxidized or Reduced
i) The standard reduction potential of metaltells us about its tendency to lose electrons
and to act as reducing agent. Lesser the value of standard reduction potential,
greater will be tendency of metal to lose electrons and to act as strong reducing
agent.
Example:

247. Electrochemistry
Chapter 10
Li, K, Na, Ca, etc. lie high in the electrochemical series and have high negative value of
standard reduction potential (low value of reduction potential). They have greatest
tendency to lose electrons and act as strongest reducing agent.
ii) The standard reduction potential of non-metal tells us about its tendency to gain
electrons and to act as oxidizing agent. Greater the value of standard reduction
potential, greater will be the tendency of non-metals to gain electrons and to act as
stronger oxidizing agent.
Example:
F2, Cl2, Br2, and I2 lie low in the electrochemical series and have high positive value of
standard reduction potential. They have greater tendency to gain electrons and act as
strong oxidizing agent.
Note Ions like Au+3, Pt+2, Hg+2,
Ag+2, Cu+1 and non-metals
like F2, Cl2, Br2, and I2 lie
below hydrogen in the
electrochemical series.
That is why, they have
strong tendency to gain
electrons to undergo
reduction and act as
strong oxidizing agent.

248. Electrochemistry
Chapter 10
(iv) Relative Chemical Reactivity of Metals:
1. Greater the value of standard reduction potential of a metal, Smaller its tendency to lose
electrons to change into a positive ion and hence lower will be its reactivity.
Example:
i) Cu, Ag and Au are the least reactive because they have +ve reduction potentials.
ii) Metals like Pb, Sn, Ni, Co and Cd which are very close to SHE reacts with steam very
slowly to liberate hydrogen gas.
2. Smaller the value of standard reduction potential of metal, greater is its tendency to lose
electron. So, greater will be its reactivity.
Example:
Metals like, Li, Na, K and Rb are highly reactive because they can lose electrons easily.
(v) Reaction of Metals with Dilute Acids:
1. Greater the value of standard reduction potential of a metal, lesser is its tendency to lose
electrons to form metal ions and so weaker is its tendency to displace hydrogen from acids.
Example:
Au, Pt, Ag and Cu cannot displace hydrogen easily.
2. Metals with very low reduction potentials have greater tendency to lose electrons to form
metal ions. So, greater is their tendency to displace hydrogen from acids.
Example:
Zn, Mg, Ca have greater tendency to lose electrons.
(vi) Displacement of One Metal by Another from its Solution:
Metal will displace another metal from the aqueous solution of its salt if it has lower reduction
potential than the other.
Example:
Fe can displace ‘Cu’ from CuSO4, because Fe has lower reduction potential than Cu but Zn
cannot displace Mg from the solution of MgSO4 because Zn has greater reduction potential
than Mg.

249. Electrochemistry
Chapter 10
1. Caustic soda is obtained on industrial scale by Hg cell which involves the electrolysis of
concentrated aqueous solution of sodium chloride using ____ anode and ______
cathode.
a. Ti and Hg
b. Hg and Ti
c. Hg
d. Cu, Ti
2. A standard hydrogen electrode has zero electrode potential because:
a. It is easy to oxidized hydrogen
b. The electrode potential is arbitrarily taken as zero which serves as standard.
c. Hydrogen atom has only one electron
d. Hydrogen is the lightest element
3. The cell in which copper electrode which is dipped in 1M CuSO4 is connected with SHE
through external circuit has the cell potential:
a. OV
b. +1.10 V
c. +0.34V
d. -0.34V
4. Which of the following element has highest positive standard reduction potential (E0) at
298K?
a. Li
b. Cu
c. Au
d. Al
5. The element which can displace other halogens from their solutions:
a. Cl
b. F
c. Br
d. I
6. The strongest reducing agent of the alkali metal in solution is:
a. Li
b. Na
c. K
d. Cs
Assessment 1
1. Caustic soda is obtained on industrial scale by Hg cell which involves the electrolysis of
concentrated aqueous solution of sodium chloride using ____ anode and ______
cathode.
a. Ti and Hg
b. Hg and Ti
c. Hg
d. Cu, Ti
2. A standard hydrogen electrode has zero electrode potential because:
a. It is easy to oxidized hydrogen
b. The electrode potential is arbitrarily taken as zero which serves as standard.
c. Hydrogen atom has only one electron
d. Hydrogen is the lightest element
3. The cell in which copper electrode which is dipped in 1M CuSO4 is connected with SHE
through external circuit has the cell potential:
a. OV
b. +1.10 V
c. +0.34V
d. -0.34V
4. Which of the following element has highest positive standard reduction potential (E0
) at
298K?
a. Li
b. Cu
c. Au
d. Al
5. The element which can displace other halogens from their solutions:
a. Cl
b. F
c. Br
d. I
6. The strongest reducing agent of the alkali metal in solution is:
a. Li
b. Na
c. K
d. Cs

250. Electrochemistry
Chapter 10
7. The cell voltage or emf measures the force with which
a. Non-spontaneous reaction takes place in voltaic cell
b. It measures the electrode potential of electrode at any temperature called
standard reduction potential.
c. SHE is given standard reduction potential more than zero.
d. Electrons move in the external circuit and therefore measure the tendency of
the cell reaction to takes place.
8. For the feasibility of a redox reaction in a cell, the e.m.f. should be
a. Positive
b. Fixed
c. Zero
d. Negative
9. The standard electrode potential is measured by
a. Electrometer
b. Voltmeter
c. Battery
d. Galvanometer
10. The standard reduction potentials of four elements A, B, C and D are –2.05V, –1.06 V, –
0.70V and +0.80v. The highest oxidizing ability is exhibited by
a. A
b. B
c. C
d. D
7. The cell voltage or emf measures the force with which
a. Non-spontaneous reaction takes place in voltaic cell
b. It measures the electrode potential of electrode at any temperature called
standard reduction potential.
c. SHE is given standard reduction potential more than zero.
d. Electrons move in the external circuit and therefore measure the tendency of the
cell reaction to takes place.
8. For the feasibility of a redox reaction in a cell, the e.m.f. should be
a. Positive
b. Fixed
c. Zero
d. Negative
9. The standard electrode potential is measured by
a. Electrometer
b. Voltmeter
c. Battery
d. Galvanometer
10. The standard reduction potentials of four elements A, B, C and D are –2.05V, –1.06
V, –0.70V and +0.80v. The highest oxidizing ability is exhibited by
a. A
b. B
c. C
d. D

251. Electrochemistry
Chapter 10
1. Which of the following is the application of electrolysis?
a. Deposition of layer of silver on utensils
b. Manufacture of sodium from its chloride
c. Generation of cell electric current by using continuous supply of reactants.
d. Both a and b
2. Which of the following statements is true for an electrochemical cell in which zinc is
coupled with SHE?
a. Oxidation occurs at SHE
b. SHE is anode and Zn is cathode
c. Zn is anode and SHE is cathode
d. Reduction occurs at Zn
3. In a salt bridge, aqueous solution of KCl in gel is used because
a. It has high surface tension.
b. It does not allow the transfer of ions.
c. It allows the diffusional exchange of ions between two solution in contact by it.
d. It is covalent compound having high melting point
4. For the measurement of standard electrode potential, Zn rod is dipped in
a. 1 M ZnO solution
b. 1 M ZnSO4 solution
c. 1M CuSO4 solution
d. 0.1 M ZnSO4 solution
5. In electrolysis of CuCl2(aq) using inert Pt electrode, the specie which is deposited at
cathode is:
a. Cl2
b. Cu
c. O2
d. H2
6. In the given redox reaction, C + 4HNO3(conc.) → CO2 + 4NO2 + 2H2O; the oxidation
number of carbons is changed from:
a. 0 to +4
b. 0 to -4
c. 0 to +2
d. +4 to 0
Assessment 2
1. Which of the following is the application of electrolysis?
a. Deposition of layer of silver on utensils
b. Manufacture of sodium from its chloride
c. Generation of cell electric current by using continuous supply of reactants.
d. Both a and b
2. Which of the following statements is true for an electrochemical cell in which zinc is
coupled with SHE?
a. Oxidation occurs at SHE
b. SHE is anode and Zn is cathode
c. Zn is anode and SHE is cathode
d. Reduction occurs at Zn
3. In a salt bridge, aqueous solution of KCl in gel is used because
a. It has high surface tension.
b. It does not allow the transfer of ions.
c. It allows the diffusional exchange of ions between two solution in contact by it.
d. It is covalent compound having high melting point
4. For the measurement of standard electrode potential, Zn rod is dipped in
a. 1 M ZnO solution
b. 1 M ZnSO4 solution
c. 1M CuSO4 solution
d. 0.1 M ZnSO4 solution
5. In electrolysis of CuCl2(aq) using inert Pt electrode, the specie which is deposited at
cathode is:
a. Cl2
b. Cu
c. O2
d. H2
6. In the given redox reaction, C + 4HNO3(conc.) → CO2 + 4NO2 + 2H2O; the oxidation
number of carbons is changed from:
a. 0 to +4
b. 0 to -4
c. 0 to +2
d. +4 to 0

252. Electrochemistry
Chapter 10
7. Which of the following may act as an oxidizing agent as well as a reducing agent?
a. H2O
b. HNO2
c. H2 gas
d. H2SO4
8. In Thermite process, 2Al + Fe2O3 → 2Fe + Al2O3 ; Aluminum acts as
a. Oxidizing agent
b. Reducing agent
c. Catalyst
d. Either a or b
9. In a given reaction, 3C2O4
-2 → 6CO2 + 6e- ; the oxidation number of carbon is changed
from
a. +3 to +2
b. +3 to +4
c. +2 to +3
d. 0 to +3
10. Which of the following reaction involves neither oxidation nor reduction?
a. CrO4
2- → Cr2O7
2-
b. Cr → CrCl3
c. Na → Na+
d. 2S2O3
2- → S4O6
2-
7. Which of the following may act as an oxidizing agent as well as a reducing agent?
a. H2O
b. HNO3
c. HNO2
d. H2SO4
8. In Thermite process, 2Al + Fe2O3 → 2Fe + Al2O3 ; Aluminum acts as
a. Oxidizing agent
b. Reducing agent
c. Catalyst
d. Either a or b
9. In a given reaction, 3C2O4
-2
→ 6CO2 + 6e- ; the oxidation number of carbon is changed
from
a. +3 to +2
b. +3 to +4
c. +2 to +3
d. 0 to +3
10.Which of the following reaction involves neither oxidation nor reduction?
a. CrO4
2-
→ Cr2O7
2-
b. Cr → CrCl3
c. Na → Na+
d. 2S2O3
2-
→ S4O6
2-

253. Electrochemistry
Chapter 10
1. During the non-spontaneous reaction, the substances are deposited at respective
electrodes and electrolyte is decomposed in:
a. Down’s cell
b. Electrolytic cell
c. Voltaic cell
d. both a and b
2. The oxidation number of Mn is maximum in which of the following:
a. Mn2O3
b. KMnO4
c. MnO2
d. K2MnO4
3. The following reaction is carried out in basic medium:
MnO4
- + 2H2 + x → MnO2 + 4OH-
(X) is the no of moles of electrons added to reduce. The (X) is
a. 1 e-
b. 2 e-
c. 3 e-
d. 4 e-
4. Which of the following reaction involves the gain of five electrons:
a. MnO4
- → Mn2+
b. CrO4
- → Cr3+
c. MnO4
- → MnO2
d. All of these
5. The oxidation state of Cl in CaOCl2 is:
a. 0
b. -1
c. +1, -1
d. +1
Assessment 3
1. During the non-spontaneous reaction, the substances are deposited at respective
electrodes and electrolyte is decomposed in:
a. Down’s cell
b. Electrolytic cell
c. Voltaic cell
d. both a and b
2. The oxidation number of Mn is maximum in which of the following:
a. Mn2O3
b. KMnO4
c. MnO2
d. K2MnO4
3. The following reaction is carried out in basic medium:
MnO4
-
+ 2H2 + x → MnO2 + 4OH-
(X) is the no of moles of electrons added to reduce. The (X) is
a. 1 e-
b. 2 e-
c. 3 e-
d. 4 e-
4. Which of the following reaction involves the gain of five electrons:
a. MnO4
-
→ Mn2+
b. CrO4
-
→ Cr3+
c. MnO4
-
→ MnO2
d. All of these
5. The oxidation state of Cl in CaOCl2 is:
a. 0
b. -1
c. +1, -1
d. +1

254. Electrochemistry
Chapter 10
6. The flow of electrons in electrolytic solution of an electrolyte is
a. From anode to cathode
b. From cathode to anode
c. Does not flow
d. In both ways.
7. Which of the following can reduce others from their solutions?
a) -0.76v b) +0.34 c) +0.80 d) -0.34
a. A
b. B
c. C
d. D
8. The following is the conductor of electricity in solid form:
a. Sulphur
b. Diamond
c. Graphite
d. Sodium chloride
9. The standard cell potential for the following cell is:
Zn I Zn2+(1M) II Cu2+ (1M) I Cu
a. +0.34V
b. +0.76V
c. +1.10V
d. -0.76V
10. A Metal M+1 loses 2 electrons to have an oxidation number
a. 0
b. +3
c. +2
d. -1
1. The flow of electrons in electrolytic solution of an electrolyte is
6. The flow of electrons in electrolytic solution of an electrolyte is:
2. The flow of electrons in electrolytic solution of an electrolyte is
a. From anode to cathode
b. From cathode to anode
c. Does not flow
d. In both ways.
7. Which of the following can reduce others from their solutions?
a) -0.76v b) +0.34 c) +0.80 d) -0.34
a. A
b. B
c. C
d. D
8. The following is the conductor of electricity in solid form:
a. Sulphur
b. Diamond
c. Graphite
d. Sodium chloride
9. The standard cell potential for the following cell is:
Zn I Zn2+
(1M) II Cu2+
(1M) I Cu
a. +0.34V
b. +0.76V
c. +1.10V
d. -0.76V
10. A Metal M+1
loses 2 electrons to have an oxidation number
a. 0
b. +3
c. +2
d. -1

255. Electrochemistry
Chapter 10
1. Which of the following has both positive and negative oxidation states?
a. F
b. Cl
c. Na
d. Ne
2. The following reaction is an example of:
Na + H2O → NaOH + H2
a. Neutralization reaction
b. Oxidation reaction
c. Reduction reaction
d. Redox reaction
3. The oxidation number of carbons in the following compound is:
CH2 Cl2
a. +4
b. -4
c. +3
d. 0
4. The correct order of reducing power of three metals having standard reduction
potential values as (A) +0.05V,(B) -0.76V and (C) +0.34V
a. B > 𝐴 > 𝐶
b. A > 𝐵 > 𝐶
c. C > 𝐴 > B
d. C > 𝐵 > A
5. Which of the following is the most appropriate characteristic of an electrolyte?
a. Conduction of electric current through mobile electrons
b. Dissociation into ions in any solvent
c. Dissociation into ion in suitable solvent
d. Release of gas
6. The main products obtained at anode and cathode during electrolysis of brine solution
(aq. solution of NaCl) are:
(1) (2) Anode (3) Cathode
(4) A (5) H2 (6) Cl2
Assessment 4
1. Which of the following has both positive and negative oxidation states?
a. F
b. Cl
c. Na
d. Ne
2. The following reaction is an example of:
Na + H2O → NaOH + H2
a. Neutralization reaction
b. Oxidation reaction
c. Reduction reaction
d. Redox reaction
3. The oxidation number of carbons in the following compound is:
CH2 Cl2
a. +4
b. -4
c. +3
d. 0
4. The correct order of reducing power of three metals having standard reduction potential
values as (A) +0.05V,(B) -0.76V and (C) +0.34V
a. B > 𝐴 > 𝐶
b. A > 𝐵 > 𝐶
c. C > 𝐴 > B
d. C > 𝐵 > A
5. Which of the following is the most appropriate characteristic of an electrolyte?
a. Conduction of electric current through mobile electrons
b. Dissociation into ions in any solvent
c. Dissociation into ion in suitable solvent
d. Release of gas
6. The main products obtained at anode and cathode during electrolysis of brine solution
(aq. solution of NaCl) are:
(2) Anode (3) Cathode
(4) A (5) H2 (6) Cl2

256. Electrochemistry
Chapter 10
(7) B (8) Cl2 (9) H2
(10) C (11) Na (12) H2
(13) D (14) O2 (15) H2
7. The standard electrode potential involving the following is:
2H+ + 2e- → H2
a. +1
b. -1
c. 0
d. +2
8. Electrolysis of molten sodium chloride results in the release of:
(16) (17) (18) Anode (19) Cathode
(20) A (21) (22) Na (23) Cl2
(24) B (25) (26) Cl2 (27) Na
(28) C (29) (30) Na (31) H2
(32) D (33) (34) O2 (35) H2
9. On the basis of position in the electrochemical series, the metal which does not displace
hydrogen from water and acid is:
a. Hg
b. Pb
c. Al
d. Ba
10. Which of the following is the characteristic of galvanic cell?
a. Chemical energy is changed to electrical energy
b. Electrical energy is changed to chemical energy
c. Non spontaneous reaction is carried out
d. Cell potential remains steady even in absence of salt bridge
(7) B (8) Cl2 (9) H2
(10) C (11) Na (12) H2
(13) D (14) O2 (15) H2
7. The standard electrode potential involving the following is:
2H+ + 2e- → H2
a. +1
b. -1
c. 0
d. +2
8. Electrolysis of molten sodium chloride results in the release of:
(17) (18) Anode (19) Cathode
(20) A (21) (22) Na (23) Cl2
(24) B (25) (26) Cl2 (27) Na
(28) C (29) (30) Na (31) H2
(32) D (33) (34) O2 (35) H2
9. On the basis of position in the electrochemical series, the metal which does not displace
hydrogen from water and acid is:
a. Hg
b. Pb
c. Al
d. Ba
10. Which of the following is the characteristic of galvanic cell?
a. Chemical energy is changed to electrical energy
b. Electrical energy is changed to chemical energy
c. Non spontaneous reaction is carried out
d. Cell potential remains steady even in absence of salt bridge
3.

257. Electrochemistry
Chapter 10
Quiz# 1
1. A.
2. B.
3. C.
4. C.
5. B.
6. A.
7. D.
8. A.
9. B.
10. D.
Quiz #03
1. D.
2. B.
3. C.
4. A.
5. C.
6. B.
7. A.
8. C.
9. C.
10. B.
Quiz #02
1. D.
2. C.
3. C.
4. B.
5. B.
6. A.
8. B.
9. B.
10. A.
Quiz #04
1. B.
2. D.
3. D.
4. A
5. C
6. B
7. C.
8. B
9. A.
10. A
Key
Assessment 1
1. A.
2. B.
3. C.
4. C.
5. B.
6. A.
7. D.
8. A.
9. B.
10. D.
Assessment 2
1. D.
2. C.
3. C.
4. B.
5. B.
6. A
7. C
8. B.
9. B.
10. A.
Assessment 3
1. D.
2. B.
3. C.
4. A.
5. C.
6. B.
7. A.
8. C.
9. C.
10. B
Assessment 4
1. B.
2. D.
3. D.
4. A
5. C
6. B
7. C.
8. B
9. A.
10. A

258. Oldest, Largest and Most Credible Platform
MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org

259. Learning Objectives:
Rate of reaction
Determination of the rate of a chemical reaction
Factors affecting rate of reaction
Specific rate constant or velocity constant
Units of rate constant
Order of reaction and its determination
Reaction Kinetics

260. Reactions Kinetics:
The branch of chemistry which deals with rates of chemical reactions and the factors that affect
the rates of chemical reactions is called reaction kinetics.
Explanation:
Reaction kinetics gives us information about the rates of chemical reactions and the effects of
different parameters like temperature, concentration, pressure and catalyst etc. on the rate of
reaction.
Types of reactions on the basis of rate of reaction:
i) Some reactions proceed at very fast rate.
Example:
A white precipitate of silver chloride is formed immediately on addition of aqueous
silver nitrate solution to aqueous sodium chloride solution.
AgNO3(aq) + NaCl(aq) ⎯⎯
→ AgCl(s) + NaNO3(aq)
White ppt.
ii) Some reactions proceed at a moderate rate.
Example:
Hydrolysis of an ester.
ii) Some reactions are very slow. They proceed at very slow rates.
Examples:
• The rusting of iron.
• The chemical weathering of rocks.
• Stone work of building by acidic gases in the atmosphere.
• Fermentation of sugar.
Rate of Reaction:
The change in concentration of a reactant or a product per unit time is called the rate of
reaction.
Rate of reaction =
changein concentration of substance
timetaken for thechange
Suppose, the change in concentration is represented by ‘∆C’ and time taken for the
change is taken as t. So, rate of reaction =
ΔC
Δt
.
Explanation:
Consider a general reaction
A ⎯⎯
→B
The rate of reaction is given by:

261. Rate of reaction =
dx
dt
For change in concentration of reactants
Rate of reaction = -
d[A]
dt
Negative sign indicates that the concentration of reactants decreases with passage of
time. Where, d[A] is the change in concentration of reactant and dt is the change in
time.
i) For the change in concentration of products.
Rate of reaction = +
d[B]
dt
Positive sign indicates that the concentration of products increases with passage of
time. Where, d[B] is the change in concentration of product and dt is the change in time.
Graphical explanation of rate:
During a chemical reaction, reactants are converted into products. The concentration of
the products increases with the corresponding decrease in the concentration of
reactants. This is explained graphically in the following figure.
Change in the concentration of reactants and
products with time for the reaction A ⎯⎯
→B
From graph it is clear that concentration of reactant ‘A’ goes on decreasing and that of
product ‘B’ goes on increasing.
Rate of reaction is ever changing parameter:
Rate of reaction is changing every moment. It decreases continuously till the reaction
ceases. In the beginning, as it is clear from graph that the concentration of ‘A’ changes
rapidly. So, rate of reaction is fast. With the passage of time, change in concentration is
not so fast. So, rate of reaction decreases.

262. Units of rate of reaction:
The units of rate of reaction are moles.dm-3.sec-1. In case of gas phase reaction, units of
pressure are used in place of molar concentrations i.e atm sec-1.
Instantaneous and Average Rate:
Instantaneous Rate:
The rate of chemical reaction at any instant of time during the course of reaction is
called instantaneous rate.
Instantaneous rate is very fast in the beginning while it is slow at the end.
Average rate:
The rate of reaction between two specific time intervals is called the average rate of reaction.
Comparison of Instantaneous Rate and Average Rate:
i) The average rate and instantaneous rate are equal for only one instant in any time
interval. or
The average rate and instantaneous rate will be equal when the time interval approaches to
zero.
ii) In the beginning of reaction, the instantaneous rate is faster than the average rate
while at the end of the interval the instantaneous rate is slower than the average
rate.
Specific rate constant or velocity constant:
Law of Mass Action:
The rate of chemical reaction is directly proportional to the product of the active masses of
reactants. This is called law of mass action.
Explanation:
Consider the following reaction.
aA + bB ⎯⎯
→cC + dD
Rate of reaction = k [A]a [B]b ------------------(1)
The expression (1) is called rate equation while ‘k’ is proportionality constant called rate
constant for forward reaction. The [ ] represents the concentrations in moles/dm3.
Specific Rate Constant or Velocity Constant:
i) The rate of reaction when the concentrations of reactants are unity is called Specific
rate constant or velocity constant.
Using equation (1):
Rate of reaction = k [A]a [B]b
Suppose [A] = [B] = 1moles/dm3
Rate of reaction = k [1]a [1]b

263. rate of reaction = k (since, [1]a [1]b = 1)
or
ii) Generally, for different concentration of reactants, rate constant can also be defined
as:
The rate of reaction is the ratio of rate of reaction to the product of concentration of
reactants raised to power their number of moles.
Using equation (1):
Rate constant ‘k’ a b
rateof reaction
=
[A] [B]
‘k’ has specific value for a reaction proceeding under some conditions. It means k
remains constant under given conditions but it changes with temperature.
Order of Reactions:
The sum of exponents of concentration terms in the rate equation is called order of reaction
OR
The number of reacting molecules whose concentrations alter as a result of chemical reaction is
called order of reaction.
Explanation:
Consider a general reaction between reactants A and B where ‘a’ moles of ‘A’ and ‘b’ moles of
‘B’ react to form ‘c’ moles of ‘C’ and ‘d’ moles of ‘D’.
aA + bB ⎯⎯
→cC + dD
Rate equation for above reaction is.
R = k [A]a [B]b
The exponent ‘a’ and ‘b’ give the order of reaction with the respect to the individual reactants A
and B respectively. It means that reaction is of order ‘a’ with respect to A and of the order ‘b’
with respect to B. The overall order of reaction is (a + b).
It is important to note that the order of reaction is an experimentally determined quantity and
can not be inferred simply by looking at the reaction equation. The sum of the exponents in
the rate equation may or may not be the same as in balanced chemical equation.
Types of reaction (on the basis of order of reaction):
With respect to order of the reaction, chemical reactions have been classified into
following types.
1. First order reaction
2. Pseudo-first order reaction
3. Second order reaction

264. 4. Third order reaction
5. Fractional order reaction
6. Zero order reaction
The order of reaction provides valuable information about the mechanism of a reaction.
(1) First Order Reaction (usually decomposition reactions):
(i) Decomposition of nitrogen penta-oxide involves the following equations:
Example:
2N2O5 ⎯⎯
→ 2N2O4 + O2
According to balanced chemical equation reaction is considered to be 2nd order but it is
first order. This indicates that reaction mechanism is different from the above
equation.
N2O5
heat
⎯⎯→ N2O4 + ½ O2
Rate of reaction = k [N2O5]
(ii) Thermal decomposition of H2O2 is also first order reaction.
Example:
H2O2
heat
⎯⎯→ H2O + ½ O2
Rate of reaction = k [H2O2]
(2) Pseudo first Order Reaction (usually hydrolytic reactions):
The reactions in which two types of reactants participate but the rate of reaction depends upon
the concentration of only one reactant and is independent of the concentration of second
reactant due to its large excess are called pseudo first order reactions.
Examples:
(i) Hydrolysis of tertiary butyl bromide.
CH3 CH3
| |
CH3 —C — Br + H2O ⎯⎯
→ CH3 — C — OH + HBr
| |
CH3 CH3
The rate equation determined experimentally for this reaction is:
Rate = k [(CH3)3CBr]
The rate of reaction remains effectively independent of the concentration of water because
being a solvent it is present in large excess. Such types of reactions are called pseudo first order
reaction.
(ii) Hydrolysis of sucrose to give fructose and glucose is also pseudo first order reaction.
C12H22O11 + H2O
+
H
⎯⎯
→ C6H12O6 + C6H12O6
Glucose Fructose
Rate of reaction = k [C12H22O11]1

265. (3) Second Order Reaction:
Oxidation of nitric oxide with ozone has been shown to be first order with respect to NO and
first order with respect to O3. The sum of the individual order gives the overall order of reaction
as two.
NO(g) + O3(g) ⎯⎯
→ NO2(g) + O2(g)
Rate = k [NO] [O3]
(4) Third Order Reactions:
(i) The reaction of FeCl3 with KI is a third order reaction. There are eight molecules of
reactants in the balanced chemical equation.
2FeCl3 + 6KI ⎯⎯
→ 2FeI2 + 6KCl + I2
But the rate expression is experimental fact. So, the exponent of FeCl3 is one while that of KI is
two in experimentally determined rate equation.
Rate of reaction = k [FeCl3] [KI]2
Actually, the reaction is taking place in more than one steps as shown in the following
mechanism.
FeCl3(aq) + 2KI(aq)
Slow
⎯⎯⎯
→ FeI2(aq) + 2KCl(aq) + -1
(aq)
Cl
2 KI(aq) + 2 -1
(aq)
Cl Fast
⎯⎯→ 2KCl(aq) + I2
(ii) Formation of nitrosyl chloride is also third order reaction.
2NO + Cl2 ⎯⎯
→ 2NOCl
rate of reaction = k[NO]2 [Cl2]
(5) Fractional Order Reaction:
The reaction of chloroform with chlorine is of order 1.5.
CHCl3(l) + Cl2(g) ⎯⎯
→ CCl(l) + HCl(g)
Rate equation = k[CHCl3]1[Cl2]0.5
Order of reaction = 1+0.5=1.5
(6) Zero Order Reaction:
Such chemical reaction whose rate is independent of concentration of reactants is called
zero order reaction.
Examples:
(1) Photosynthesis Reaction.
6CO2 + 6H2O sunlight
⎯⎯⎯
→C6H12O6 + 6O2
The rate of this reaction depends upon intensity of light instead of concentration.
(2) Photochemical combination of H2 and Cl2 to give HCl when carried out over water surface
is zero order reaction.
H2 + Cl2 ⎯⎯
→ 2HCl

266. Order of Reaction and Unit of Rate Constant:
By using the expression,
k = (Concentration)1-n (time)-1, the unit of rate constant for nth order reaction can be
determined.
Examples:
(i) For zero order reaction, the units of rate constant k are given by:
Units of k = (concentration)1-n (time)-1
For zero Order n = 0
Units of k = (mole/dm3)1-0 (s)-1
Unit of k = (mole/dm3)1 s-1
k = moles.dm-3.s-1
(ii) For first order reaction, the units of rate constant are given by:
Units of k = (concentration)1-n (time)-1
For first order n = 1
Units of k = (moles/dm3)1-1 s-1
Units of k = (moles/dm3)0 s-1
Units of k = s-1
(iii) For second order reaction, the units of rate constants k are given by:
Units of k = (concentration)1-n (time)-1
For second order n = 2
Units of k = (mole/dm3)1-2 (s)-1
Units of k= (mole/dm3)-1 s-1
Units of k = moles-1 dm3 s-1
Units of k = dm3 moles-1 s-1
(iv) For third order reaction, the units of rate constant are given by
Units of k = (concentration)1-n (time)-1
For third ordern = 3
Units of k = (moles / dm3)1-3 (s)-1
Units of k = (mole/dm3)-2 s-1
Units of k = mole-2.dm-6 s-1
Units of k = dm6 moles-2 s-1
Rate Determining Step:
That step of chemical reaction which controls the rate of chemical reaction is the slowest step
and is called rate determining step.
or

267. The slowest step that controls the rate of multi-step chemical reaction is called rate
determining step.
Explanation:
Few chemical reactions occur in single step. While there are many reactions which take place in
more than one steps. If reaction occurs in several steps one of the steps is the slowest. The rate
of this step determines the rate of overall reaction. The slowest step is called rate determining
step or rate limiting step.
Note: The total number of molecules of reacting species taking part in the rate determining
step appear in the rate equation.
Example:
Consider the following reaction.
NO2(g) + CO(g) ⎯⎯
→NO(g) + CO2(g)
The rate equation of this reaction is found to be:
Rate = k[NO2]2
This equation shows that rate of reaction is independent of the concentration of CO. In other
words, the equation gives us information that
i) reaction takes place in more than one step i.e the mechanism of this reaction is
different than as shown in balanced chemical equation.
ii) two molecules of NO2 are involved in the rate determining step.
The proposed mechanism for this reaction is as follows.
NO2(g) + NO2(g)
slow
⎯⎯⎯
→ NO3(g) + NO(g)
NO3(g) + CO(g)
fast
⎯⎯⎯
→ NO2(g) + CO2(g)
The first step is the slowest step and the rate determing step. So, order of reaction is two with
respect to NO2 but it is independent of CO concentration. NO3 which does not appear in
balanced chemical equation is reaction intermediate.
Reaction Intermediate:
A species which has temporary existence and it is unstable relative to the reactants and
products and does not appear in the balanced chemical equation is called reaction
intermediate.
This is a species with normal bonds and may be stable enough to be isolated under
special conditions.
Example:
In above reaction, NO3 is reaction intermediate.
Limitations of balanced chemical equation with respect to reaction kinetics:
Balance chemical equation does not give us exact information about:
i) Rate of reaction.
ii) Order of reaction.

268. iii) Rate equation for reaction.
iv) Mechanism of reaction.
v) No. of steps of reaction.
vi) Reaction intermediate.
vii) Rate determining step.
Method for the Measurement of Rate of Chemical Reactions:
There are two methods for the measurement of rate of chemical reaction. They are as follow:
(a) Physical Methods
(b) Chemical Methods
Physical Methods:
(i) Spectrometry:
This method for the measurement of rate of reaction is only applicable if a reactant or a
product absorbs ultraviolet, visible or infrared radiations. The rate of reaction is
measured by measuring the amount of radiation absorbed.
Note: In this method, rate of change in amount of radiations absorbed is equal
to the rate of reaction.
(ii) Electrical Conductivity method:
The rate of reactions which involve ions is measured by the electrical conductivity
method. The conductivity of such a solution depends upon the rate of change of
concentration of the reacting ions or ions which may form during the chemical reaction.
The conductivity will be directly proportional to the rate of change in the
concentration of such ions.
Note: In this method, rate of change of conductivity of reaction mixture is equal
to the rate of reaction.
(iii) Dilatometric Method:
This method is applicable for the measurement of rate of those reactions which involve
small volume change in the solutions. The volume change is directly proportional to the
extent of reaction.
Note: In this method, rate of change of volume of reaction mixture is equal to
the rate of reaction.
(iv) Refractrometric Method:
The method is applicable to reactions in solutions where there are changes in refractive
indices of the substances taking part in the chemical reactions.
Note: In this method, rate of change of refractive indices of reacting substances
is equal to the rate of reaction.

269. (v) Optical Rotation Method:
In this method, the angle through which plane polarized light is rotated by the reaction
mixture is measured by a polarimeter. The extent of rotation determines the
concentration of optically active substance. If any of the species in the reaction mixture
is optically active, then this method can be followed to find out the rate of reaction.
Note: In this method, rate of change of angle through which plane polarized
light is rotated is equal to the rate of reaction.
Chemical Method:
In chemical method the concentration of reactants and products are analysed
chemically.
Energy of Activation:
The minimum amount of energy which is required for an effective collision leading to
products is called activation energy.
Explanation:
Collision between reactant molecules is a necessary condition for a reaction to take place.
Collisions between reactant molecules are of two types:
i) Ineffective collision:
A collision that does not result in product formation and colliding molecules fly apart as
such is called ineffective collision.
ii) Effective collision:
A collision that results in product formation and colliding molecule fly apart as different
species is called effective collision. Following conditions are required by the colliding
molecules for effective collision.
i) Proper orientation
ii) Proper energy (activation energy = Ea)
The orientation of colliding molecules in which those atoms collide with each other
which are required to form new bonds is called proper orientation. The minimum
amount of energy required by the colliding molecules for effective collision is called
activation energy.
If all the collisions among reacting species at a given temperature are effective forming
the products the reaction is completed in very short time. Most of the reactions are
slow showing that all the collisions are not equally effective.

270. Transition state (Activated Complex):
Let us study a reaction between molecules A2 and B2 to form a new molecule AB. If these
molecules will have energy equal to or more than the activation energy then upon collisions
their bonds will break and new bonds will be formed. The phenomenon is shown in following
figure:
A - A
+
B - B
A ... A
B... B
...
..
.
A A
+
B B
Collisions of molecules, formation of activated complex and formation of products
Activated complex is a unstable combination of all the atoms involved in the reaction for which
the energy is maximum. It is short lived specie and decomposes into the products immediately.
Since, it has a transient existence, that is why it is also called a transition state.
Energy changes at the time of collision:
When the colliding molecules come close to each other at the time of collision, they slow down
collide and fly apart again. If the collision is effective, then the molecules flying apart are
chemically different otherwise the same molecules just bounce back. When the molecules slow
down just before the collision, their kinetic energy decreases and this results in the
corresponding increase in their potential energy. The process can be understood with the help
of a graph between the path of reaction and potential energy of the reacting molecules.
A graph between path of reaction
and the potential energy of the reaction
The reactants reach the peak of curve to form activated complex. Ea is the energy of
activation and it as potential energy hill between reactants and products.
i) Colliding molecules which have proper energy will be able to climb up the hill and will
give product.

271. ii) Colliding molecules whose initial kinetic energy is less than Ea will fail to climb up the hill
fall back chemically unchanged.
Exothermic Reaction:
In exothermic reaction, the potential energy of the products is lower than the potential energy
of the reactants. The difference is denoted byΔH . The value of ΔH depends upon the initial
state (the potential energy of reactants) and final state (the potential energy of products). It
does not depend upon
(i) route (ii) rate and (iii) activation energy of reaction. The reactants require energy
1
a
E which is equal to the activation energy to get the top of barriers. Anyhow, products
require energy 2
a
E which their activation energy to get to top of higher hill to form
activated state. This energy is more than
1
a
E .
1
a
E = energy of activation for forward step
2
a
E = energy of activation for backward step.
The magnitude of ΔH is also the difference of these two energies of activation.
1
a
E – 2
a
E = ΔE or ΔH
From this it is clear that for exothermic reactions the activation energy for forward
reaction is less than the activation energy for backward reaction.
Note:
i)  H is the difference of potential energies of products and reactants.
ii)  H is also the difference of activation energy of reactants and products.
Endothermic Reaction:
In endothermic reactions the potential energy of the products is higher than the
potential energy of the reactants and for such reactions a continuous source of energy is
needed to complete the reaction. In such reactions, the enthalpy change ΔH is positive
since energy is absorbed in the reaction. For endothermic reversible reaction, the
energy of activation for forward reaction is greater than the energy of activation for
backward reaction.
Conclusions
Energy of activation for forward and backward reactions are different for all reactions.
(1) For exothermic reaction, energy of activation for forward reaction is less than the
energy of activation for backward reaction.
(2) For endothermic reaction, energy of activation for forward reaction is greater than
the energy of activation for backward reaction.

272. Finding the order of reaction:
The sum of exponents of the concentration terms in the rate expression of a reaction is
called order of reaction. It can be determined by the following methods.
(i) Method of hit and trial
(ii) Graphical Method
(iii) Differential Method
(iv) Half life method
(v) Method of Large Excess
Half Life Method
1
2
2
1
t
log
t
n=1+
a
log
a
 
 
 
 
 
 
So, if we know the two initial concentrations and two half life values we can calculate the order
of reaction (n).
Method Of Large Excess:
In this method one of the reactants is taken in a very small amount as compared to the rest of
the reactants. The active masses of the substances in large excess remain constant throughout.
The substance taken in small amount controls the rate and the order is noted with respect to
that substance. The reason is that a small change in concentration of a substance taken in very
small amount affects the value of rate more appreciably. In this way the reaction is repeated by
taking rest of the substances in small amounts one by one and overall order is calculated.
Factors Affecting the rate of reaction
All those factors which change the number of effective collisions per second, affect the rate of
chemical reactions. Some of the important factors are as follows
1. Nature of reactants
2. Concentration of reactants
3. Surface area
4. Light
5. Effect of temperature on rate of reaction
Nature of Reactants
The rate of reaction depends upon the nature of reacting substances. The reactivity of
substances depends on the electronic arrangements in their outermost orbitals. The substances

273. which are reactive their rate of reaction is fast while elements which are less reactive their rate
of reaction is show.
Examples:
(i) Elements of IA group have one electron in their outermost s-orbital. They are very
reactive, their rate of reaction with water is very fast.
(ii) Elements of IIA group are less reactive than the IA group elements because they
have two electrons in their outermost shell. Their rate of reaction with water is
slower than the IA group elements.
(iii) The neutralization and double decomposition reactions are very fast as compared
to the reactions in which bonds are rearranged. Oxidation reduction reactions
involve the transfer of electrons and are slower than the ionic reaction.
Concentration of Reactants:
The rate of chemical reactions is directly proportional to concentration of reactants. These
reactions are due to collisions of reactant molecules. The frequency with which the molecules
collide depends upon their concentration. The more crowded the molecules are, the more
likely they are to collide and react with one another. Thus an increase in concentration of
reactants increases the rate of reaction and decrease in concentration of reactants decreases
the rate of reaction.
Surface Area:
With increase in surface area, rate of reactions increases. The reason behind this is that with
increase in surface area, molecules of reacting substances find more chances to collide with
each other.
Examples:
(i) Aluminium foil reacts with NaOH moderately when warmed, but powdered ‘Al’
reacts rapidly with cold NaOH because powdered ‘Al’ has greater surface area.
( )
2 2
4
2Al + 2NaOH + 6H O 2NaAl OH + 3H
⎯⎯
→
(ii) CaCO3 in the powder form reacts with dilute H2SO4 more efficiently than big pieces
of CaCO3
3 2 4 4 2 2
CaCO + H SO CaSO + H O + CO
⎯⎯
→
Light:
Light consists of photons. Each photon of light has specific amount of energy associated with it
depending upon its frequency. When the reactants are irradiated, this energy becomes
available to them and rate of reaction is increased.
Example:
(i) The reaction between CH4 and Cl2 requires light.
(ii) Reaction between H2 and Cl2 at ordinary pressure is negligible in darkness and Very
slow in daylight but very explosive in sunlight.

274. Kinetic energy distributions for a
reaction mixture at two different
temperatures. The size of the shaded areas
under the curves are proportional to the
total fraction of the molecules that possess
to minimum activation energy.
(iii) Light plays a vital role in photosynthesis reaction and rate is influenced by light.
Effect of Temperature on Rate of Reaction:
The rate of any chemical reaction is directly proportional to the temperature. Each and every
factor which increases the number of effective collisions between molecules increases the
rate of reaction. We know that with increase in temperature, energy of reactants increases and
rate of reaction increases. For a collision to be effective, molecules should possess the
activation energy and they must also be properly oriented. For nearly all chemical reactions,
the activation energy is quite large and at ordinary temperature very few molecules are moving
fast enough to have this minimum energy.
All the molecules of reactant do not possess the same energy at a particular temperature. Most
of the molecules possess average energy. A fraction of total molecules will have energy more
than the average energy. This fraction of molecules is indicated by shaded area in the following
figure.
As the temperature increases, the number of molecules
in this fraction which have energy greater than average
energy also increases. The curve at higher temperature
T2 has flattened. It shows that molecules having higher
energies have increased. So, the number of effective
collisions increases and hence, the rate of reaction also
increases.
When the temperature of the reacting gases is raised
By 10oC, the fraction of molecules with energy more
Than activation energy Ea, roughly doubles and so the
Reaction rate also doubles.
Arrhenius Equation:
Arrhenius equation gives us quantitative relationship, between rate constant ‘k’ temperature
‘T’ and activation energy Ea. Arrhenius equation is as follows

275. Ea
-
RT
k=Ae ---------------------(i)
In this equation,
k is rate constant
A is Arrhenius constant
Ea is activation energy
R is general gas constant
T is the absolute temperature
e is the base of natural log
This equation
(i) shows that
i) Rate constant (k) increases with increase in temperature.
ii) Rate constant (k) has smaller value for the reaction whose energy of activation is
greater.
Determination Of Energy Of Activation From Arrhenius Equation:
a
E 1
logk = - + logA---------(iv)
2.303R T
 
 
 
This equation (iv) is equation of straight line y = mx + c
Where, y = log k
Slope , a
E
m = -
2.303R
1
x =
T
and
c = logA
Where ‘m’ is slope of straight line and c is intercept of straight line. Temperature is
independent variable while rate constant is dependent variable and other quantities Ea,
and A are constant.
So, graph is plotted between 1/T on x–axis and log k on y-axis and a straight line is
obtained with a negative slope. The graph is shown in figure below.

276. The slope of this straight line is measured by taking the tan of the angle 'θ' which the
straight line makes with x – axis. To measure the slope, draw a line parallel to x-axis and
measure the angle ‘θ ’. Take tanθ which is slope and this slope is equal to
Slope = a
E
-
2.303R
------------------(v)
Therefore,
Ea = - slope x 2.303 R-------------(vi)
The straight line of different reactions will have different slope and different Ea values.
The unit of slope is Kelvin:
From equation (v), unit of slope can be determined as:
-1
-1 -1
Jmol
slope =
2.303JK mol
Slope = K

277. Assessment 1
1. The specific term (−
𝑑𝑐
𝑑𝑡
) in a rate equation (Rate of reaction = −
𝑑𝑐
𝑑𝑡
) refers to the
a. Concentration of the reactant
b. Decrease in concentration of the reactant with time
c. Increase in concentration of the reactant with time
d. Velocity constant of the reaction
2. Which of the following does not influence the rate of reaction
a. Nature of the reactants
b. Concentration of the reactants
c. Temperature of the reaction
d. Molecularity of the reaction
3. The units for the rate constant of first order reaction is
a. s–1
b. mol dm-3 s–1
c. mol s–1
d. dm3 mol–1 s–1
4. In Arrhenius plot to calculate the energy of activation, intercept is equal to
a. −
𝐸𝑎
𝑅𝑇
b. ln A
c. ln k
d. log10a
5. It is experimentally observed that the rate of chemical reaction is almost double for every 10K
rise in temperature because of
a. Increase in the activation energy
b. Decrease in the activation energy
c. Increase in the number of molecular collisions
d. Increase in the number of activated molecules and effective collisions.
6. The radioactive decay proceeds in the way; A → B + e-, the rate law expression is: rate= k [A].
Which of the following statements is incorrect?
a. The reaction follows first order kinetics
b. The t1/2 of reaction depends on initial concentration of reactants
c. k is constant for the reaction at a constant temperature
d. All of these

278. 7. The reaction in which hydrogen peroxide is decomposed to give water and oxygen has the rate
equation; Rate = k [H2O2]. It is assumed to be
a. Zero order reaction
b. First order reaction
c. Second order reaction
d. Third order reaction
8. Arrhenius has studied the quantitative relationship between:
a. Temperature, energy of activation
b. Mass and energy of activation, rate constant
c. Rate constant, energy of activation and temperature
d. Molar mass, temperature and rate constant
9. The unit of rate constant for photochemical reaction between hydrogen and chlorine is:
a. mol dm–3 s–1
b. dm3 mol–1 s–1
c. dm-6 mol–2 s–1
d. s–1
10. The half life for the reaction; 2N2O5 ⟶ 2N2O4 + O2 is
a. Independent of the initial concentration of the reactant
b. Directly proportional to the initial concentration of the reactants
c. Inversely proportional to the initial concentration of the reactant
d. Directly proportional to the square of the initial concentration of the reactant

279. Assessment 2
1. A first order reaction involves the 50% decomposition of substance in 15 minutes. The time required for
75% n decomposition of same substance will be
a. 45 mint
b. 15 mint
c. 30 mint
d. 60 mint
2. The half life (t1/2) is inversely proportional to the initial concentration of reactants for a reaction having
the order
a. Zero
b. One
c. Two
d. Three
3. The rate of reaction of alkali metals with water is more than those of alkaline earth metals because of
a. Valence electronic arrangement
b. Concentration
c. Atomic volume
d. Temperature involved
4. When a graph is plotted between 1/T on x-axis and log k on y-axis, a straight line is obtained with a
negative slope (
−𝐸𝑎
2.303𝑅
). This line has two ends in
a. I and II quadrant
b. II and III quadrant
c. III and IV quadrant
d. II and IV quadrant
5. The method in which the rate of reaction can be measured by measuring the amount of radiation
absorbed
a. Dilatometric method
b. Optical relation method
c. Refractometric method
d. Spectrophotometer method
6. For the photochemical reaction between hydrogen and chlorine, the order of reaction is
a. 1
b. 2
c. 3
d. 0

280. 7. The given reaction 2NO + O2 → 2NO2 is carried out in two steps; is an example of:
a. First order reaction
b. Second order reaction
c. Third order reaction
d. None of these
8. When a graph is plotted between 1/T on x-axis and log k on y-axis, a straight line is obtained with a
negative slope. The slope of the Arrhenius equation can be represented as:
a. Ea/2.303 RT
b. −Ea/2.303 RT
c. −Ea/2.303 R
d. −Ea/RT
9. The half life for the reaction is assumed to be one hour. After four hours, the %age of initial product left
untreated is:
a. 3.125%
b. 12.5%
c. 6.25%
d. 25%
10. The factor which affects the number of collisions:
a. Pressure applied
b. Concentration of reacting molecules
c. Temperature
d. All the above

281. Assessment 3
1. The physical method in which rate of reaction is determined by measuring the small change in volume
is:
a. Spectrometry
b. Electrical conductivity
c. Dilatometric
d. Refractometeric
2. For the reaction, A + B ⟶ 𝑃𝑟𝑜𝑑𝑢𝑐𝑡.The order of a reaction is determined to be two with respect to a
reactant A when B is taken in large excess. It shows that
a. The rate of the reaction is proportional to [A]
b. The rate of the reaction is proportional to [A]2
c. Two molecules of B are present in the stoichiometric equation
d. The rate of reaction is proportional to [B]2
3. The reaction for which the order of reaction is in fraction?
a. Chlorination of chloroform to form carbon tetrachloride
b. Photochemical reaction of hydrogen and chlorine
c. Oxidation of nitric oxide with ozone
d. Hydrolysis of tertiary butyl bromide
4. The average rate and instantaneous rate of a reaction are equal:
a. When two rates have a time interval equal to zero
b. At start of the reaction
c. At end of the reaction
d. In middle of the reaction
5. The correct form assigned to find the half life for 3rd
order reaction is given as:
a. [ t1/2] =
0.693
𝑘
b. [ t1/2] =
1
𝑘
c. [ t1/2] =
1
𝑘𝑎
d. [ t1/2] =
1.5
𝑘𝑎2
6. Which of the following is not true for the energy of activation?
a. The energy of activation for forward and backward reactions are different for all the
reactions.
b. Energy of activation of forward reaction is more than that of backward reaction for
exothermic reaction.
c. Energy of activation of forward reaction is more than that of backward reaction for
endothermic reaction.
d. Energy of activation of a reaction provides valuable information about the way a
reaction takes place.

282. 7. 2FeCl (aq) + 6KI(aq)→2FeI (aq) + 6KCI(aq) + I2 is an example of:
a. Zero order
b. 1st order
c. 2nd order
d. 3rd order
8. Half life of a first order reaction is 10 mint. What percent of reaction will be completed in 50 mint?
a. 96.87%
b. 75%
c. 50%
d. 25%
9. The rate law of the reaction A + 2B → C is given by
𝑑[𝑑𝐵]
𝑑𝑇
= 𝐾[𝐴][𝐵]2
. If A is taken in excess, the order
of the reaction will be
a. 1
b. 2
c. 3
d. 0
10. The rate of the following reaction is not dependent on
A(aq) → B(aq)
a. Pressure
b. Temperature
c. Concentration
d. None of these

283. Assessment 4
1. The rate equation for the reaction is given as:
Rate= K[A]2
[B]
Which of the following statement corresponds to it?
a. It is second order with respect to A
b. It is first order with respect to B
c. It is third order reaction
d. All of these
2. Catalyst is the specie which accelerate the chemical reaction by
a. Decreasing equilibrium constant
b. Increasing effective collisions
c. Decreasing activation energy
d. Both b and c
3. Which of the following reaction would be the fastest one?
a. NaOH(aq) + HCl(aq) →
b. Al + H2O →
c. CH4 + Cl2 →
d. Both a and c
4. What is the half life of third order reaction if K=1 and a = 1.5
a. 6.66 sec
b. 0.66 sec
c. 1 mint 6 sec
d. 16 mint
5. The activation energy of the reaction as determined by plot given by Arrhenius equation in which slope
= -100K
a. 190 KJ
b. 19KJ
c. 1.9KJ
d. 8.3KJ
6. Which of the following techniques for measuring the rate of reaction measures the changes in
refractive index?
a. Spectrometry
b. Optical rotation method
c. Dilatometric method
d. Refractrometric method

284. 7. The rate at which substance reacts depends upon:
a. Atomic mass
b. Molar mass
c. Active mass
d. Average atomic mass
8. The number of molecules must overcome an energy barrier is given by:
a. A
b. K
c. e−
Ea
RT
d. Ea
9. If the concentration of reactants is increased by a factor (A), the rate constant K becomes:
a. K
b. KX
c. 𝑒
𝑘
𝑥
d. 𝑘/𝑥
10. The activation energy of the reaction is zero. The rate constant of the reaction
a. Increases with increase of temperature.
b. Decrease with decrease of temperature
c. Decrease with increase of temperature
d. Is nearly independent of temperature.

285. KEY
Assessment 1
Assessment 2
1. b
2. d
3. a
4. b
5. d
6. b
7. b
8. c
9. a
10. a
1. c
2. c
3. a
4. d
5. d
6. d
7. c
8. c
9. c
10. d

286. KEY
Assessment 3
Assessment 4
1. c
2. b
3. a
4. a
5. d
6. b
7. d
8. a
9. b
10. a
1. d
2. c
3. a
4. b
5. c
6. d
7. c
8. c
9. a
10. d

287. MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org
Oldest, Largest and Most Credible Platform

288. s-block and p-block Elements (Inorganic chemistry)
Learning Objectives:
Understand the trends in Physical Properties of elements belonging to
period no 2 and period no 3.
Explain the periodicity in physical properties of elements
Atomic Radi
Ionic Radii
Ionization Energy
Electron Affinity
Melting Point
Boiling Point
Electrical Conductivity
Electronegativity
Reaction of Group IA elements with H2O, O2 and Cl2
Reactions of Group IIA elements with H2O, O2 and N2
Reactions of Group IIIIA elements with H2O, O2 and Cl2
S and p - Block Elements

289. Introduction
The Periodic Table provides a basic framework to study the periodic behaviour of physical and
chemical properties of elements as well as their compounds.
Key Features of the modern periodic table which help you to understand the trends in
physical and chemical properties of elements.
 Elements have been arranged on the basis of increasing atomic number. (The idea to
arrange elements on the basis of proton was first floated by Moseley).
 There are 7 horizontal rows called periods (no of shells≈ periods) and 18 vertical
columns called groups.(valence electrons in representative elements ≈ groups).
 Atomic properties gradually change as we move from left to right within same period
but shell no remains same.
 The atomic properties; i.e., ionization energy, electron affinity, density,
electronegativity increases along the period from left to right with few exceptions.
 Every period except first starts with metal and ends with non-metal.
 The prominent periods; 3d transition series, 4d transition series, 5d transition series,
lanthanides and actinides show tremendous variations of properties.
 Within same group, no of valence electrons remain same but no of shells changes as we
move from top to bottom. It accounts that similar elements are placed in same group.
 The position of elements in groups greatly helps in understanding the pattern of
chemical reactivity.
 The famous groups include alkali metals (IA), alkaline earth metals (IIA)noble
gases(VIIIA), Chalcogen (VIA)etc.
 Depending upon the valence electronic configuration, modern periodic table has 4-
block division; s-block, p-block, d-block and f-block.
 Another division of groupscomprises the division of groups into representative
elements (A) and transition elements (B).
Dobereiner
Law of triad
Newlands
Law of Octave
Mendeleev
Periodic table
Moseley
Atomic number
Al Razi
Four group div,

290. Division of Modern Periodic Table into Block
s-block elements:
I-A and II-A group elements and Helium are called s-block because their valence electrons are
present in s-orbital.
p-block elements:
III-A to VIII-A group elements (except He) are called p-block elements because their valence
electrons are present in p-orbital.
d-block elements:
I-B to VIII-B group elements (transition elements) are called d-block elements because their
valence electrons are present in d-orbital.
f-block elements:
Lanthanides and actinides are called f-block elements because their valence electrons are
present in f-orbital.
Periodic trends in physical properties
Atomic radius:
“Half of the distance between the nuclei of two adjacent atoms is considered to be the radius
of the atom.”
 Atoms are assumed to be spherical. Depending upon the type of the compound used
for its measurement, the sizes of atoms are expressed in terms of atomic radii, ionic
radii and covalent radii, etc.
 The atomic radius increases from top to bottom within a group due to increase in
atomic number.
 This is due to the addition of an extra shell of electrons.
Figure 2: Atomic radius trend in alkali metals
Point to Ponder
Factors Affecting Atomic
Radius:
Nuclear Charge:
As it increases, atomic radius
decreases.
No of Shells:
As it increases, atomic radius
increases.

291. Figure 3: Atomic radius trend in halogens
 Within a period from left to right , there is a gradual decrease in atomic radius.
 This gradual decrease in the radius is due to increase in the positive charge in the
nucleus.
Figure 4: Atomic radius in period
 The gradual reduction in the size of Lanthanides is significant and called Lanthanide
Contraction.
 The steady decrease in the size of the atoms and ions of the rare earth elements with
increasing atomic number from lanthanum (atomic number 57) through lutetium
(atomic number 71).
 Atomic radii can be determined, by measuring the distances between the centres of
adjacent atoms with the help of X-rays or by spectroscopic measurements.
Along the Group
Atomic radii increase from top to bottom in a group.
Reasons
1. Increase in the number of shells
2. Increase in the shielding (screening) effect.

292. Along the Period
Atomic radii decrease from left to the right in a period.
Reasons
1. Increase in the nuclear charge.
2. shielding effect remains the same from left to right in a period.
Ionic radius:
Ionic radius is the radius of a monatomic ion in an ionic crystal structure.
 When neutral atom loses an electron one or more electrons, it becomes a positive ion.
 A positive ion is always smaller than the neutral atom from which it is derived.
 Within same group, ionic radius increases down the group.
Figure 5: Ionic radii of IA group elements
 When a neutral atom gains an electron, negative ion is formed which is greater in size
than its corresponding neutral atom.
 Ionic radius of negative ion increases down the group.
Figure 6: Ionic radii of halogens

293.  Within a period, isoelectronic positive ions show a decrease in ionic radius from left to
right, because of the increasing nuclear charge. The same trend is observed for the
isoelectronic negative ions of a period; ionic size decreases from left to right.
Interionic Distance:
The interionic distance ‘R’ in a crystal lattice is the sum of the cationic radius r+ and the anionic
radius r-
R = r+ + r-
For example the interionic distance of potassium chloride (KCl) is:
R = 133pm + 181 pm = 314 pm
Important Points:
 Cations of a period are isoelectronic to one another.
 Anions of a period are isoelectronic to one another as well as noble gas of a same
period.
 Cations of a period are isoelectronic to the anions as well as noble gas of the previous
period.
Test yourself:
Which of the following ion has greater ionic radius than its corresponding
neutral atom?
A. Cl- B. Mg2+ C. Na+ D. H+
A) Cl- B) Mg2+ C) Na+
D) H+

294. Ionization energy:
“The minimum amount of energy which is required to remove an electron from the valence
shell of an isolated gaseous atom to form positive ion.”
Figure 7: Ionization energy
Mg(g) →
Mg+
(g) + 1e-1∆H1 = 738 kjmol-1
Mg+
(g) →
Mg2+
(g) + 1e-1 ∆H2 = 1450 kjmol-1
Mg2+
(g) →
Mg3+
(g) + 1e-1 ∆H3 = 7730 kjmol-1
So 1st I.E. < 2nd I.E. < 3rd I.E. …………… and so on.
 It is measured in KJ/mol or eV.
 3rd IE > 2𝑛𝑑 𝐼𝐸 > 1𝑠𝑡 𝐼𝐸.
 Ionization is the index to metallic character
 Ionization energy decreases down the group.
Figure 8: Ionization energy in a group
 Ionization energy generally increases from left to right in a period.
Mg(g) → Mg+
(g) + 1e-1 ∆H1 = 738 kjmol-1
Mg+
(g) → Mg2+
(g) + 1e-1 ∆H2 = 1450
kjmol-1
Mg2+
(g) → Mg3+
(g) + 1e-1 ∆H3 = 7730
kjmol-1
So 1st I.E. < 2nd I.E. < 3rd I.E. ……………
and so on.

295. Figure 9: Ionization energy in a period
Gripping fact:
Point of ponder
 Ionization energy is an index to metallic character.
 Metals have low ionization energy forming cations.
 Non metals have high ionization energy forming
anions.
 Factors affecting:
 𝐈𝐨𝐧𝐢𝐳𝐚𝐭𝐢𝐨𝐧 𝐞𝐧𝐞𝐫𝐠𝐲 𝐢𝐬 𝐫𝐞𝐥𝐚𝐭𝐞𝐝 𝐭𝐨
𝟏
𝐚𝐭𝐨𝐦𝐢𝐜 𝐬𝐢𝐳𝐞
 Ionization energy increases as nuclear charge increases
 Ionization energy increases with stable electronic
configuration
 Ionization energy increases with more s character
 𝑰𝒐𝒏𝒊𝒛𝒂𝒕𝒊𝒐𝒏 𝒆𝒏𝒆𝒓𝒈𝒚 𝒊𝒔 𝒓𝒆𝒍𝒂𝒕𝒆𝒅 𝒕𝒐
𝟏
𝒔𝒉𝒊𝒆𝒍𝒅𝒊𝒏𝒈 𝒆𝒇𝒇𝒆𝒄𝒕
In second period, Nitrogen does not follow the
general rule of increasing ionization energy, instead it
has greater ionization energy than oxygen which
comes after it because of half filled 3p orbital (stable
electronic configuration).

296. Order of ionization energy is s > p > d > f
Abnormal Trends:
The elements of:
 Group III-A and
 Group VI-A and
have abnormally low values of ionization energy in every period
Ionization energies of 1st 20 elements
Group IIB (Zn, Cd, Hg) and IIIB
(Sc, Y, La) are non-typical
transition elements while all
other are typical transition
elements.
Nitrogen has exceptionally more ionization energy than oxygen due to
A. Atomic size B. Nuclear charge C. Shielding effect D. Stable electronic
configuration
TEST YOURSELF

297. Electron affinity:
“The amount of energy which is released/absorbed on the addition of electron in the valence
shell of an isolated gaseous ion to form anion.”
Energy is released on addition of first electron to neutral atom (exothermic)
Energy is absorbed on addition of 2nd electron to uninegative ion (endothermic) to
overcome repulsion.
 Knowledge of electron affinities can be combined with the knowledge of ionization
energies to predict which atoms can easily lose electrons and which can accept
electrons more readily.
 Electron affinity depends upon atomic size, nuclear charge and vacancy in outermost
shell.
 Electron affinity generally decreases from top to bottom in a group and increases from
left to right in a period.
Figure 10: Electron affinity in a periodic table
 Chlorine has more electron affinity than Fluorine because of greater reduction in size of
fluorine and inter electronic repulsion.
 Noble gases have positive electron affinity as they have no vacancy in valence shell.
 The atom with stable electronic configuration (IIA) or half filled sub shell (VA) may have
low electron affinity values.
 In case of non metals, oxidizing power is proportional to electron affinity.

298.  In case of stable electronic configuration, electron affinity is low.
Factors affecting electron affinity:
 Atomic size, nuclear charge,
 Vacancies in outermost shell.
Melting point and boiling point:
“The temperature at which solid and liquid phase of same substance co-exist is
called melting point while temperature at which external pressure becomes equal to vapour
pressure of liquid.”
 Across the shorter periods, melting point and boiling point increases upto VIA and then
decreases up to noble gases.
Gripping fact:
 MPand BP of IA to IVA ⇒ increase in melting point 𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 binding energy.
 MP and BP of VA to VIIIA ⇒ formation of small covalent molecules and weak interactive
forces.
Figure 11: Trends of MP and BP across the2nd period
 Unique trend of melting point and boiling point in a group is shown as:
 In IA and IIA group, decreases from top to bottom due to increase in atomic size
 In group VIIA, it increases from top to bottom due to increasing polarizability.

299. Figure 12: Trends of MP and BP of IIA group
Magnesium has low melting point because of distorted hexagonal close packing structure in
its crystal lattice.
Figure 13: Melting point of halogens

300. Figure 14: Table showing MP, BP and appearance
Test yourself:
Which of the following element has less melting point?
A) Be B) Mg C) Ca D) Sr

301. Electrical conductivity:
“The ability of the substance to conduct electric current is known as electrical
conductivity.”
 The electrical conductivity of elements is due to:
 Loosely bounded valence electrons.
 Ease of their movement in crystal lattice.
 The electrical conductivity of metals in group IA and IIA generally increases from top to
bottom.
 Coinage metals (IB) have extraordinary electrical conductance.
 The lower elements of group IVA, tin and lead, are fairly good conductors.
 Non-metals, especially of groups VIA and VIIA, show such low electrical conductance
that they can be considered as nonconductors.
Figure 15: Electrical conductivity in 2nd period
Gripping fact:
Carbon in the form of diamond is non-conductor because all of its valence electrons are
tetraherdrally bound and unable to move freely.
Carbon in the form of graphite is conductor because one of its four electrons is free to
move.

302.  Sodium, magnesium and aluminum are all good conductors of electricity.
 Silicon is a semiconductor.
 Electrical conductivity of metals decreases on heating while increment in electrical
conductivity of semiconductor on heating.
 Semi conductors are known as half conductors.
Conductivity of period 3 elements:
Test yourself:
Which of the following is semiconductor?
A) C B) Si C) Mg D) Na

303. Electronegativity:
“The force with which an atom attracts the shared pair of electrons towards itself.”
 Fluorine is the most electronegative element (4.0) while Cesium and Francium are the
least electronegative elements (0.7).
 Electronegativity values increases from left to right in a period. It decreases from top to
bottom in modern periodic table.
Gripping fact:
Electronegativity is about the tendency of an atom to attract a
bondingpair of electrons. Since noble gases do not form covalent bonds, you obviously can't
assign this electronegativity.
Figure 16: Electronegativity values in 2nd period
Figure 17: Electronegativity values in group IA

304. Insight:
Postassium (K) and Rubidium (Rb) have same 0.82 electronegativity due to shielding
effect.
Nature of Bond:
The difference in the electronegativity values of the bonded atoms is an index to the polar
nature of the covalent bond.
 If ∆ E.N > 1.7, the bond is ionic.
 If ∆ E.N < 1.7 but > 0.5, the bond is polar covalent.
 If ∆ E.N ≤ 0.5, the bond is non polar covalent.
 If ∆ E.N = 1.7, shows roughly equal contributions of ionic and covalent bonds.
Reactivity of Alkali metals (Group IA elements)
 Alkali metals have only one electron in ‘s’ orbital of their valence shell.
 They form ionic compounds and show +1 oxidation state.
 There is an increase of atomic and ionic radii (of M+ ions) as we move from lithium to
cesium due to successive addition of shells.
 First ionization energy falls down the group so reactivity increases down the group.
Reactions of group IA with H2O, O2 and Cl2:
 Reaction of Group IA elements with water:
 The elements of IA group become more reactive as we move from top to
bottom.
 Lithium is the metal which releases the most heat during the reaction.
 Potassium, rubidium and caesium are so highly reactive that they react with
ice even at -100°C.
2Li + 2H2O →2LiOH + H2
2Na + 2H2O →2NaOH + H2

305. Lithium's density is only about half that of water so it floats on the surface, gently fizzing and
giving off hydrogen.
Sodium also floats on the surface and it melts almost at once to form a small silvery ball that
dashes around the surface. The hydrogen may catch fire to burn with an orange flame rather
than blue flame due to contamination with sodium compounds
Potassium reacts faster than sodium and enough heat is given off. This time the normal
hydrogen flame is contaminated by potassium compounds and so is coloured lilac (a faintly
bluish pink).
Rubidium is denser than water and so sinks. It reacts violently and immediately, with
everything spitting out of the container again.
Cesium explodes on contact with water, quite possibly shattering the container.
 Reaction of Group IA elements with O2:
 Alkali metals react with O2 or air rapidly and thus get tarnished due to the formation of
their oxide on the surface of the metals.
 Alkali metals are stored in paraffin oil or kerosene.
4LI + O2 →2Li2O (Normal oxide)
4Na + O2(limited) →
2 Na2O (Normal oxide)
2Na + O2(limited) →
Na2O2 (Peroxide)
 Potassium, rubidium and caesium react with oxygen to form superoxides.
K + O2 → KO2 (superoxide)
The normal oxides of alkali metals other than Li2O are not formed by the direct reaction
between the metals and O2 they are formed by indirect methods, eg, by reducing
Peroxides
Nitrites
Nitrates with metal itself
Lithium's density is only about half that of water so it floats on the surface, gently fizzing and giving
off hydrogen.
Sodium also floats on the surface and it melts almost at once to form a small silvery ball that
dashes around the surface. The hydrogen may catch fire to burn with an orange flame rather than
blue flame due to contamination with sodium compounds
Potassium reacts faster than sodium and enough heat is given off. This time the normal hydrogen
flame is contaminated by potassium compounds and so is coloured lilac (a faintly bluish pink).
Rubidium is denser than water and so sinks. It reacts violently and immediately, with everything
spitting out of the container again.
Cesium explodes on contact with water, quite possibly shattering the container.

306. Properties:
Normal oxide (O2-) reacts with water to form hydroxides by proton exchange.
The peroxides and superoxides are strong oxidizing agents and react with water to give H2O2
and O2.
The superoxide ion has a three electron bond as shown below
The presence of one unpaired electron in it makes this paramagnetic and coloured.
 Reaction of Group IA elements with Cl2:
 Lithium and sodium react slowly with chlorine at room temperasture.
 Potassium, rubidium and caesium react vigorously with all the halogens,
forming metal halides.
2M + Cl2 →2MCl
 The chlorides are all white solids at room temperature, and dissolve in water
to make a neutral solution.
 The reaction with chlorine is similar in appearance to the reaction of the
Group 1 metals with oxygen.
 For example, sodium burns with brilliant yellow flame in chlorine in exactly
the same way that it does in pure oxygen.
2Na(s) + Cl2(g) →
2NaCl(s)
 The reactions get more vigorous going down the group.

307. Reactions of group IIA with H2O, O2 and N2:
 Reaction of Group IIA elements with Water:
 The alkaline earth metals form the normal oxides of MO type which are
obtained by heating the metal in O2.
 BeO is amphoteric while other oxides are basic in character.
 BeO and MgO are quite insoluble in H2O while H2O CaO, SrO and BaO react
with H2O to give soluble hydroxides, M(OH)2which are strong bases.
 BeO is covalent which other oxide are ionic.
Calcium sinks in water and reacts with cold water to give its hydroxide and H2 gas.
Strontium and Barium react vigorously with water with effervescence forming their soluble
hydroxides and hydrogen gas.
M + 2H2O →
M (OH)2 + H2
 Reaction of Group IIA elements with Oxygen:
 Alkaline earth metals also react with oxygen, though not as rapidly as Group 1
metals.
These reactions also require heating.
Reactivity increases down the group.
 Be is stable in normal air but oxidized rapidly at 800 oC.
Be + O2
800𝑜𝐶
→ BeO
Mg + O2
𝛥
→ MgO
Mg + N2
𝛥
→ Mg3N2
Potassium superoxide (KO2) has a very interesting use in breathing equipment for mountaineers
and in space craft. It has the ability to absorb carbon dioxide while giving out oxygen at the same time.
4KO2 + 2CO2 → 2K2CO3 + 3O2

308.  Reaction of Group IIA elements with N2:
 Group II-A elements react with nitrogen on heating giving nitrides.
 The nitrides hydrolyse vigorously when treated with water, giving ammonia and
the respective hydroxides.
 Be3N2is volatile while the nitrides of other alkaline earth metals are non-volatile.
6M + 4N2 → 2M3N2
Test yourself:
The oxide which is used by mountaineers to absorb CO2 and released oxygen:
A) BeO B) BaO C) Na2O2 D) KO2
Reactions of group IIIA with H2O, O2 and Cl2:
 Reaction of IIIA elements with H2O:
2B + 3H2O (steam) → B2O3 + 3H2
 Group IIIA elements do not readily react with atmospheric oxygen but when heated,
they form oxides (M2O3) which are readily reduced back to the free metal.
 They react with halogens to form halides (MX3)
 Aluminium foil on heating in dry chlorine form AlCl3.
 If the aluminium powder is heated to 800°C and above, the metal will react with air to
form aluminium oxide, Al2O3, and aluminium nitride, AIN. Do you know?
Because of the ability of
aluminum to combine with both
oxygen and nitrogen, the metal
is often used to remove air
bubbles from molten metals.

309. .
Boron does not react with water but reacts with steam at a high temperature.
2B(s) + 3H2O(g) →
B2O3(s) + 3H2O(g)
Aluminium does not react with water due to protective oxide layer on its surface.
Thallium forms thallium hydroxide in moist air in presence of oxygen.
Tl + 2H2O + O2 →
4TlOH
 Reactions with oxygen:
All the elements of group IIIA react with oxygen at a high temperature to produce trioxides
with the formula M₂O₃, except Thallium, which forms monoxides with formula Tl₂O.
4M(s)+O2(g)→
2M2O
Reactivity of Group IIIA elements increases down the group.
Boron is unreactive towards oxygen in crystalline form but reacts in finely divided amorphous
state to form B₂O₃.

310.  Reaction of Group IIIA with Chlorine:

311. Assessment 1
1. Which of the following is true about the melting point and boiling point?
a. MP and BP tell about the strength of forces present.
b. IVA group elements in 3rd period has higher melting point in the period
c. Zero group elements have lowest mp and bp in the period.
d. All of these
2. The high melting point of group IVA is attributed to:
a. More no of binding electrons.
b. More no of free electrons
c. Giant covalent structure
d. Both a and c
3. Which of the following has greater standard oxidation potential?
a. K
b. Rb
c. Cs
d. All have same
4. The alkali metal having less hydration energy (eV/ion):
a. Na
b. K
c. Rb
d. Cs
5. According to Pauling scale, the electronegativity of lithium is:
a. 4.0
b. 0.7
c. 1.0
d. 3.5
6. There is gradual increase in the density of alkali metals with increase in atomic size. But
there is an unusual character shown by the following element:
a. Cs
b. Na
c. K
d. Rb
7. The origin of flame test given by sodium is due to the following process:
a. Energy absorb to remove the electron
b. Energy released on addition of electron
c. Excitation of electron
d. Complete transfer of electron
Assessment 1
1. Which of the following is true about the melting point and boiling point?
a. MP and BP tell about the strength of forces present.
b. IVA group elements in 3rd period has higher melting point in the period
c. Zero group elements have lowest mp and bp in the period.
d. All of these
2. The high melting point of group IVA is attributed to:
a. More no of binding electrons.
b. More no of free electrons
c. Giant covalent structure
d. Both a and c
3. Which of the following has greater standard oxidation potential?
a. K
b. Rb
c. Cs
d. All have same
4. The alkali metal having less hydration energy (eV/ion):
a. Na
b. K
c. Rb
d. Cs
5. According to Pauling scale, the electronegativity of lithium is:
a. 4.0
b. 0.7
c. 1.0
d. 3.5
6. There is gradual increase in the density of alkali metals with increase in atomic size.
But there is an unusual character shown by the following element:
a. Cs
b. Na
c. K
d. Rb
7. The origin of flame test given by sodium is due to the following process:
a. Energy absorb to remove the electron
b. Energy released on addition of electron
c. Excitation of electron
d. Complete transfer of electron

312. 8. The following alkali metal reacts more vigorously with water is:
a. Sodium
b. Potassium
c. Rubidium
d. Cesium
9. The alkali metals are extremely soft and readily fusible except:
a. Li
b. Na
c. K
d. Rb
10. Potassium is the alkali which is usually kept in:
a. Ethanol
b. Water
c. Kerosene
Key:
1. D 2. D 3. D 4. D 5. C
6. C 7. C 8. D 9. A 10. C
8. The following alkali metal reacts more vigorously with water is:
a. Sodium
b. Potassium
c. Rubidium
d. Cesium
9. The alkali metals are extremely soft and readily fusible except:
a. Li
b. Na
c. K
d. Rb
10. Potassium is the alkali which is usually kept in:
a. Ethanol
b. Water
c. Kerosene

313. Assessment 2
1. Which of the following statement account true for alkali metals?
a. Cation is less stable than the corresponding neutral atom
b. Cation is smaller than the corresponding neutral atom
c. Size of cation and corresponding atom is the same
d. Cation is greater in size than the corresponding neutral atom
2. Sodium reacts more quickly with water than lithium due to the following reason:
a. Its molecular weight is less
b. It is stronger electronegative
c. It is stronger electropositive
d. It is a metal
3. Which of the following is not true about the alkali metals?
a. Alkali metals have only one electron in s orbital of their valence shells.
b. Alkali metals form mono positive ion.
c. Alkali metals have high ionization energy.
d. Alkali metals form ionic compounds and show +1 oxidation state
4. Which of the following shows anomalous properties in group IA of the modern periodic
table?
a. Li
b. Na
c. K
5. The alkali metal normally forming normal oxide on burning in air is:
a. Li
b. Na
c. K
d. All of these
6. The following oxide is formed by the direct reaction of metal with oxygen:
a. Li2O
b. Na2O
c. K2O
d. None of these
7. Which of the following accounts good reason for the reducing property of alkali metal?
a. They are highly electropositive
b. They have high electron affinity
c. Both a and b
d. None of these
Assessment 2
1. Which of the following statement account true for alkali metals?
a. Cation is less stable than the corresponding neutral atom
b. Cation is smaller than the corresponding neutral atom
c. Size of cation and corresponding atom is the same
d. Cation is greater in size than the corresponding neutral atom
2. Sodium reacts more quickly with water than lithium due to the following reason:
a. Its molecular weight is less
b. It is stronger electronegative
c. It is stronger electropositive
d. It is a metal
3. Which of the following is not true about the alkali metals?
a. Alkali metals have only one electron in s orbital of their valence shells.
b. Alkali metals form mono positive ion.
c. Alkali metals have high ionization energy.
d. Alkali metals form ionic compounds and show +1 oxidation state
4. Which of the following shows anomalous properties in group IA of the modern
periodic table?
a. Li
b. Na
c. K
5. The alkali metal normally forming normal oxide on burning in air is:
a. Li
b. Na
c. K
d. All of these
6. The following oxide is formed by the direct reaction of metal with oxygen:
a. Li2O
b. Na2O
c. K2O
d. None of these
7. Which of the following accounts good reason for the reducing property of alkali
metal?
a. They are highly electropositive
b. They have high electron affinity
c. Both a and b
d. None of these

314. 8. The alkali metal when exposed to oxygen in air in presence of moisture, (A) is formed
which when combines with carbon dioxide in the atmosphere to form compound (B).
What are A and B?
a. Oxide and bicarbonate
b. Hydroxide and carbonate
c. Hydroxide and bicarbonate
d. Oxide and carbonate
9. Which of the following pair of alkali metal form oxide in which oxidation number of
oxygen is (−
1
2
)?
a. Li and Na
b. Na and K
c. Li and Rb
d. K and Rb
10. Which of the following statement is not correct about beryllium?
a. Beryllium is the least reactive metal.
b. It oxidizes completely in air at ordinary temperature to form BeO
c. It soon loses the silver appearance on oxidation in air.
d. None of these
1. B 2. C 3. C 4. A 5. A
6. A 7. A 8. D 9. D 10. B
8. The alkali metal when exposed to oxygen in air in presence of moisture, (A) is
formed which when combines with carbon dioxide in the atmosphere to form
compound (B). What are A and B?
a. Oxide and bicarbonate
b. Hydroxide and carbonate
c. Hydroxide and bicarbonate
d. Oxide and carbonate
9. Which of the following pair of alkali metal form oxide in which oxidation number of
oxygen is (−
1
2
)?
a. Li and Na
b. Na and K
c. Li and Rb
d. K and Rb
10. Which of the following statement is not correct about beryllium?
a. Beryllium is the least reactive metal.
b. It oxidizes completely in air at ordinary temperature to form BeO
c. It soon loses the silver appearance on oxidation in air.
d. None of these

315. Assessment 3
1. The compounds which are formed when magnesium is burnt in presence of oxygen:
a. Magnesium oxide
b. Magnesium nitride
c. Magnesium carbonate
d. both a and b
2. Which of the following oxides of alkali metals react with water to give H2O2 and O2 gas?
a. Normal oxide of alkali metals
b. Peroxide of alkali metals
c. Superoxide of alkali metals
d. Both b and c
3. Which of the following statement corresponds to sodium when compared with
potassium:
a. Lower melting point
b. Lower electronegativity
c. Larger atomic radius
d. Higher ionization energy
4. The flame colour imparted by Ba is:
a. Golden yellow
b. Crimson red
c. Apple green
d. Brick red
5. Which of the following elements does not belong to alkaline earth?
a. Be
b. Ba
c. Ra
d. Rn
6. The reaction of an alkali metal oxide with water is the type of reaction:
a. Acid-base reaction
b. Redox reaction
c. Addition reaction
d. Dehydration reaction
7. Which of the following oxide is used in breathing equipments for mountaineer?
a. Na2O
b. H2O2
c. K2O2
d. KO2
Assessment 3
1. The compounds which are formed when magnesium is burnt in presence of oxygen:
a. Magnesium oxide
b. Magnesium nitride
c. Magnesium carbonate
d. both a and b
2. Which of the following oxides of alkali metals react with water to give H2O2 and O2
gas?
a. Normal oxide of alkali metals
b. Peroxide of alkali metals
c. Superoxide of alkali metals
d. Both b and c
3. Which of the following statement corresponds to sodium when compared with
potassium:
a. Lower melting point
b. Lower electronegativity
c. Larger atomic radius
d. Higher ionization energy
4. The flame colour imparted by Ba is:
a. Golden yellow
b. Crimson red
c. Apple green
d. Brick red
5. Which of the following elements does not belong to alkaline earth?
a. Be
b. Ba
c. Ra
d. Rn
6. The reaction of an alkali metal oxide with water is the type of reaction:
a. Acid-base reaction
b. Redox reaction
c. Addition reaction
d. Dehydration reaction
7. Which of the following oxide is used in breathing equipments for mountaineer?
a. Na2O
b. H2O2
c. K2O2
d. KO2

316. 8. Potassium superoxide used in breathing equipments by mountaineers because it:
a. It absorbs carbon dioxide gas
b. It absorbs the whole air
c. It absorbs the oxygen gas
d. It absorbs helium gas
9. Which of the following is the characteristic of the reaction in which superoxide of alkali
metal reacts with water?
a. It is strongly exothermic reaction.
b. Oxygen may also be given in the reaction.
c. Reaction is violent
d. All of these
10. The following cation polarize chloride more effectively:
a. Li
b. Na
c. K
d. Rb
1. D 2. D 3. D 4. C 5. D
6. A 7. D 8. A 9. D 10. A
8. Potassium superoxide used in breathing equipments by mountaineers because it:
a. It absorbs carbon dioxide gas
b. It absorbs the whole air
c. It absorbs the oxygen gas
d. It absorbs helium gas
9. Which of the following is the characteristic of the reaction in which superoxide of
alkali metal reacts with water?
a. It is strongly exothermic reaction.
b. Oxygen may also be given in the reaction.
c. Reaction is violent
d. All of these
10. The following cation polarize chloride more effectively:
a. Li
b. Na
c. K
d. Rb

317. Assessment 4
1. Which of the following has less value of 1st ionization energy?
a. Ra
b. Ba
c. Sr
d. Ca
2. The atomic property which is measured in Pauling scale is:
a. Ionization energy
b. Electronegativity
c. Electron affinity
d. Shielding effect
3. In the following graph, which of the alkaline earth metals appear to have same
electronegativity values?
a. Be and Ca
b. Mg an Ca
c. Ca and Sr
d. Sr and Ba
4. The exceptional low boiling point is observed for the following alkaline earth metal in
the following bar graph:
a. Be
b. Mg
c. Ca
d. Sr
Assessment 4
1. Which of the following has less value of 1st ionization energy?
a. Ra
b. Ba
c. Sr
d. Ca
2. The atomic property which is measured in Pauling scale is:
a. Ionization energy
b. Electronegativity
c. Electron affinity
d. Shielding effect
3. In the following graph, which of the alkaline earth metals appear to have same
electronegativity values?
a. Be and Ca
b. Mg an Ca
c. Ca and Sr
d. Sr and Ba
4. The exceptional low boiling point is observed for the following alkaline earth metal
in the following bar graph:
a. Be
b. Mg
c. Ca
d. Sr

318. 5. The interionic distance ‘R’ in a crystal lattice of potassium chloride is :
(r+ = 133pm and r- = 181)
a. 133 pm
b. 181 pm
c. 314 pm
d. 52 pm
6. Which of the following has highest ionization energy?
a. Carbon
b. Boron
c. Oxygen
d. Nitrogen
7. The elements of which of the following show abnormal trend in ionization energy of the
modern periodic table:
a. IIIA
b. VIA
c. IIA
d. Both b and c
8. Which of the following is not the absolute term of the element?
a. Atomic size
b. Electronegativity
c. Ionization energy
d. Electron affinity
9. At 800 oC, aluminium reacts with normal air to produce:
a. Al2O3
b. AlN
c. Al(OH)3
d. Both a and b
10. The element which is non metallic in nature in group IIIA is?
a. B
b. Al
c. Ga
d. Tl
1. B 2. B 3. C 4. B 5. C
6. D 7. D 8. B 9. D 10. A
5. The interionic distance ‘R’ in a crystal lattice of potassium chloride is :
(r+ = 133pm and r- = 181)
a. 133 pm
b. 181 pm
c. 314 pm
d. 52 pm
6. Which of the following has highest ionization energy?
a. Carbon
b. Boron
c. Oxygen
d. Nitrogen
7. The elements of which of the following show abnormal trend in ionization energy of
the modern periodic table:
a. IIIA
b. VIA
c. IIA
d. Both b and c
8. Which of the following is not the absolute term of the element?
a. Atomic size
b. Electronegativity
c. Ionization energy
d. Electron affinity
9. At 800 oC, aluminium reacts with normal air to produce:
a. Al2O3
b. AlN
c. Al(OH)3
d. Both a and b
10. The element which is non metallic in nature in group IIIA is?
a. B
b. Al
c. Ga
d. Tl

319. Key
Assessment 1
Assessment 2
1. D
2. D
3. D
4. D
5. C
6. C
7. C
8. D
9. A
10 C
1. B
2. C
3. C
4. A
5. A
6. A
7. A
8. D
9. D
10. B

320. Key
Assessment 3
1. D
2. D
3. D
4. C
5. D
6. A
7. D
8. A
9. D
10. A
Assessment 4
1. B
2. B
3. C
4. B
5. C
6. D
7. D
8. B
9. D
10. A

321. 1
Oldest, Largest and Most Credible Platform
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Chemistry
Quick Practice Book
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322. 2
Transition metals:
 Electronic configuration of Transition metals
 General characteristics of transition metals
 Binding energy
 Melting point and boiling point
 Covalent and ionic radii
 Paramagnetism
 Oxidation state
 Colour
 Interstitial compounds
 Alloy formation

323. 3
Electronic Configuration of Transition Metals
Three transition series:
1st Series
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ca Ce As Se Br Kr
2nd Series
Rb Sr Y Zr Nb Mo Te Ru Rh Pd Ag Cd In Sn Sb Te I Xe
3rd Series
Cs Ba La Hf Ta W Mn Re Os Ir Au Hg Ti Pb Ba Po At Ru

324. 4
Test yourself:
The correct and stable electronic configuration of Cr (atomic no=24) is:
A) [Ar]3d44s2
B) [Ar]3d54s1
C) [Ar]3d6
D) [Ar]3d5
General Characteristics of Transition Metals:
Binding Energy:
 The amount of energy that is required to be given to the electron to pull it away from
the attractive (Columbic) force between nucleus and valence electron is called the
binding energy.
 The toughness of these metals indicates strong metallic binding which is due to
participation of valence shell s-electrons and underlying half filled d-orbital.
 Binding energy increases up to VB and VIB due to unpaired electrons and then
decreases until it becomes zero at IIB due to pairing of electrons.

325. 5
Melting and Boiling Point:
 High melting point due to strong binding forces between atoms.
 Melting points increase up to the middle of the series and then decrease to a minimum
level at the end of the series.
Gripping fact:
Covalent and ionic radii:
 Covalent radii decrease rapidly at the start of the series, then become almost constant
and finally begin to increase at the end of the series.
 Changes in the ionic radii along the series are much less regular.
Group IIB (Zn, Cd, Hg) and IIIB (Sc,
Y, La) are non-typical transition
elements while all other are typical
transition elements.

326. 6
Paramagnetism:
 Substances which are weakly attracted by a strong magnetic field are called
paramagnetic substances.
 Paramagnetism of transition metals is due to the presence of unpaired electrons in d
orbitals.
 Paired electrons ~ diamagnetic
 The number of unpaired electrons decreases gradually to zero on both sides of the
transition series.
 Paramagnetic behavior is the strongest for Fe3+ and Mn2+.

327. 7
Insight:
Oxidation state:
 Their variable valency due to the involvement of unpaired s and d orbital electrons in
bond formation.
 All 3d transition elements show oxidation number +2 in addition of higher oxidation
number.
 In 3d series, O.N. increases upto Mn and then decreases due to pairing of electrons.
Did you know?
Color:
 Transition metal complexes are coloured due to d-d transition,
Magnetic moment (
𝝁) is measured in BM
(Bohr magneton).
Among the 3d series, Mn has maximum oxidation states, and goes up to +7.

328. 8
ΔE of d-orbitals varies from ion to ion thus absorbing different wavelengths.
Interstitial compounds:
 Non-stoichiometric compounds
 Also termed as interstitial alloys.
 Small non-metal atoms like H, B, C, N enter the interstices of transition metals and
impart useful features to them.
Alloy formation:
 Alloy is a mixture of two or more than two metals.
 Transition elements have almost similar sizes and atoms of the one metal can easily take
up positions in crystal lattice of the other forming substitution alloys.
[Ti(H2O)6]3+
Violet colour
Yellow light is
absorbed
Blue and red light
is emitted
Appears violet

329. 9
Alloys of
metals
Composition Properties and Uses
Brass Cu= 60-80%
Zn= 20-40%
It is a strong alloy of copper which is soft and
flexible. It does not corrode. Due to low melting
point it is easy to use. It is used to make keys, water
taps, pipes, artificial jewelry, door handles and parts
of machines.
Bronze Cu= 90-95%
Sn= 5-10%
It is strong, brilliant and long lasting. It does not
corrode. It is used to prepare medals, coins, badges
and bullets etc. Besides these, decorative articles are
also made from these.
Nichrome NI= 60%
Cr= 15%
Fe= 25%
It is used in electric heaters and filaments of
furnaces.

330. 10
Assessment 1
1. The elements known as transition elements (d block elements) are placed
a. In between s and p block
b. Below in two rows
c. Extreme right of the periodic table
d. Left side of the periodic table
2. The transition elements which are attracted by magnetic field are:
a. Ferromagnetic
b. Paramagnetic
c. Diamagnetic
d. All of these
3. The characteristic property of transition element is to show:
a. Infinite oxidation number
b. Variable oxidation number
c. Single oxidation number
d. Two oxidation number
4. The property of transition ions to give colour is due to the transition of
electron within subshell:
a. s
b. p to d
c. d
d. f
5. The non Stoichiometric compounds of Titanium are called:
a. Hydrates
b. Hydrides
c. Complex compounds
d. Interstitial compounds
6. Which of the following is a typical transition metal?
a. Co
b. Sc
c. Y
d. Cd
7. The property of formation of interstitial compounds is responsible for:
a. More melting point of metal
b. More malleability
c. More conductivity
d. None of these

331. 11
8.
8. The transition metal and its oxide may used in industry as:
a. Solvent
b. Catalyst
c. Dehydrating agent
d. All of these
9. The accurate reason that accounts an increase in atomic radius of transition
elements at the end of the period:
a. Increase in atomic mass
b. Increase in nuclear charge
c. Increase in electron-electron repulsion
d. Decrease in electron-electron repulsion
10. The following property does not relate to transition elements
a. Variable oxidation number
b. Catalytic ability
c. Low boiling point
d. Malleable
:

332. 12
Assessment 2
1. The type of covalent bond that is exhibited by transition elements:
a. Ionic bond
b. Coordinated covalent bond
c. Covalent bond
d. Metallic bond
2. The transition metal which is comparatively less hard metal given below:
a. Zinc
b. Chromium
c. Tungsten
d. Molybdenum
3. The oxidation number of Mn in Mn2O7 is:
a. +2
b. +7
c. -7
d. 0
4. The valence electronic configuration of scandium is [Ar] 3d1 4s2. Sc3+ is:
a. Paramagnetic
b. Diamagnetic
c. Either paramagnetic or diamagnetic
d. Ferromagnetic and diamagnetic
5. The accurate term that corresponds to the easy formation of alloys by transition
element is:
a. Orbital configuration
b. Very light
c. Atomic size
d. Its binding energy
6. Zn and mercury do not show variable oxidation number due to:
a. Presence of 4s2 sub shell
b. Complete d subshell
c. Inert pair effect
d. All of these
7. The first transition element in modern periodic table is:
a. Copper
b. Nickel
c. Scandium
d. Vanadium

333. 13
8. How Sc ( Z=21) differs from Zinc (z=30)?
a. Both are typical transition elements
b. In case of Sc, 3d subshell is incomplete but in Zn this is completely filled
c. Last electron as assumed to be added to 4s level in case of zinc
d. Both Sc and Zn do not exhibit variable oxidation states
9. Which of the following d block element has half filled d sub shell as well as
half filled valence sub shell?
a. Cr
b. Pd
c. Pt
d. None of these
10. In which of the following there is no d-d transition:
a. Fe3+
b. Ti4+
c. Co2+
d. Ni2+

334. 14
Assessment 3
1. The alloy of copper that contain Zinc is
a. White silver
b. Bronze
c. Nichrome
d. Brass
2. The general electronic configuration of transition element is:
a. (n-1) d1-10
ns0-2
b. nd1-10
ns0-2
c. ns0-2
(n+1) d1-10
d. Any of these
3. The following ion is diamagnetic due to unavailability of unpaired
electrons:
a. Mn2+
b. V2+
c. Sc3+
d. Cr2+
4. The property of transition elements to show variable oxidation numbers is
due to the release of electron from:
a. ns orbitals
b. np orbitals
c. (n-1) d orbitals
d. (n-1) d orbitals and ns orbitals
5. Which of the following has the smallest ionic radii?
a. Ni2+
b. Cr2+
c. Ti2+
d. Mn2+
6. The following ion has the highest paramagnetism?
a. Sc3+
b. Ti3+
c. Mn2+
d. Zn2+
7. Which of the following is diamagnetic due to the presence of zero unpaired
electrons in its d orbitals?
a. Sc3+
b. V3+
c. Zn2+
d. Both a and c
Both a and

335. 15
8. Bronze is the alloy in which major metals are:
a. Cu and Zn
b. Cu and Sn
c. Cu and Bi
d. Cu and Ni
9. Which of the following metal is ferromagnetic?
a. Ni
b. Fe
c. Co
d. All of these
10. The following equation related the magnetic moment to the no of
unpaired electrons:
a. μ = √n(n + 2)
b. μ = √n(n + 1)
c. μ = √(n + 2)
d. μ = √2n(n + 2)

336. 16
Assessment 4
1. In first transition series, the element having lowest melting point belongs to
which group:
a. 3
b. 4
c. 10
d. 12
2. Ag+
ion (Ag = 47) is isoelectronic with
a. Cu+
(Cu = 29)
b. Cd2+
(Cd = 48)
c. Zn2+
(Zn = 30)
d. Pd2+
(Pd =46)
3. Which of the following has maximum no of unpaired electrons in d orbital:
a. Fe+2
b. Fe+3
c. Cr3+
d. Co2+
4. What is the oxidation number of Mn in MnO4
-
?
a. +1
b. +2
c. +7
d. +5
5. The following set of transition ions represent the correct order of
paramagnetism:
a. Cu2+
< V3+
< Mn3+
< Mn2+
b. Cu2+
< V3+
< Mn2+
< Mn3+
c. V2+
< Cu2+
< Mn2+
< Mn3+
d. Cu2+
< Mn3+
< V2+
< Mn2+
6. The transition metal that is widely used as a catalyst in Haber’s process to
prepare ammonia is:
a. Fe
b. Zn
c. Pt
d. Au

337. 17
7. The major metal by %age composition which is present in nichrome is:
a. Ni
b. Cr
c. Fe
d. None of these
8. Which of the following elements have almost same Covalent radii?
a. Mn, Fe, Co, Zn
b. Cr, Mn, Fe, Cu
c. Cr, Mn, Fe, Sc
d. Mn, Fe, Cu, Ni
9. The catalyst of Contact process for converting SO2 to SO3 is:
a. Fe
b. Ni
c. V2O5
d. NO
10. Which alloy is used in preparation of coins?
a. Brass
b. Bronze
c. Nichrome
d. All of these

338. 18
10. C.
Assessment 2
Key
Assessment 1
1. A.
2. B.
3. B.
4. C.
5. D.
6. A.
7. D.
8. B.
9. C.
10. C.
Assessment 2
1. D.
2. A.
3. B.
4. B.
5. C.
6. A
7. C.
8. B.
9. A.
10. B.

339. 19
Assessment 4
Assessment 3
1. D.
2. A.
3. C.
4. D.
5. A.
6. C.
7. D.
8. B.
9. D.
10. A.
1. D.
2. B.
3. B.
4. C.
5. A.
6. A
7. A.
8. B.
9. C.
10. B.

340. 1
MDCAT
Chemistry
Quick Practice Book
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341. 2
Fundamental principles of
organic chemistry:
 Organic chemistry
 Organic compounds
 Synthesis of organic compounds
 Classification of organic compounds
 Functional group
 Isomerism and types

342. 3
FFundamental Principle of Organic Chemistry
Organic chemistry and organic compounds:
 Organic chemistry is the chemistry of hydrocarbons and their derivatives.
 It is the chemistry of living matter or substance which were at one time live.
 Carbon is the essential element of organic compound.
Vital force theory:
 Organic compounds cannot be prepared in laboratory.
 A mysterious force (vital force is required for their production.
 This force is only present in living organisms.
Jacob Berzelius
 Discarded vital force theory
 Synthesized urea (naturally found in urine of mammals) from
ammonium cyanate (inorganic)
 NH4CNO
Heat
→ (NH2)2CO
Fredrick Wohler
CO3
2-
CN-1
HCO3
1-CS2
CO2
CO
NOT ORGANIC

343. 4
Classification of organic compounds on the basis of structure:

344. 5
Functional group:
 Atom, group of atom, double bond or triple bond that gives specific properties to
organic compounds is functional group.
 Alkane does not have any functional group.
 It is the functional part of the organic compound.
 Each functional group defines an organic family.

345. 6
Isomerism:
 Compounds having same molecular formula but different structural formula are called
isomers and the phenomenon is isomerism.
 CH4, C2H6 and C3H8 do not exhibit structural isomerism
 Isomerism is contributed by directional nature of covalent bond.
Types:
Structural Isomerism:
Chain Isomerism:
“Isomerism arises due to different chain lengths”
For example, isomers of butane; n-butane and isobutene

346. 7
Functional group Isomerism:
“Isomerism arises due to different functional groups.”
Position Isomerism:
“Isomerism arises due to the chain in position of functional group attached.”
Metamerism:
“Isomerism arises due to unequal arrangement of carbon chain on either side of
functional group.”

347. 8
Tautomerism:
“Tautomerism arises due to shift of proton within the same molecule.”
Geometrical Isomerism:

348. 9
 Geometrical isomerism or cis-trans isomerism arises due to position of different groups
with respect to carbon carbon double bond.
 The necessary and sufficient condition for a compound to exhibit geometrical isomerism
is that the two groups attached to the same carbon must be different.
 The rotation of two carbon atoms joined by a double bond could happen only if the π
bond breaks. This ordinarily costs too much energy, making geometric isomers possible.
 Trans isomer is more stable than corresponding cis isomer due to the presence of bulky
groups are away from each other avoiding stearic repulsion.
Geometrical isomerism in cyclic compounds:
 Geometrical isomerism is also possible in cyclic compounds because of no rotation
about carbon-carbon single bonds forming a ring.
Geometrical isomerism in 1,2 –dimethylcyclopropane

349. 10
 A requirement for geometrical isomerism in cyclic compounds is that there must be at
least two other groups beside hydrogen on the ring and these must be on different ring
carbon atoms.
 No geometrical isomers are possible for 1,1-dimethylcyclopropane
Chiral Center:
 Chirality is a property present in an object that cannot be superimposed on its mirror
image.
 Chiral objects do not have a plane of symmetry (they are asymmetric).
 Chirality is typically seen in molecules that have a chiral or asymmetric carbon atom.
Plane of symmetry:
 A plane which divides an object into two symmetrical halves, is said to be plane of
symmetry.
 A person’s hand or gloves lack plane of symmetry.
 An object lacking a plane of symmetry is called dissymmetric or chiral.
 A symmetric object is referred to as achiral.
Cis-1,2-dimethylcyclopropane
both the methyl groups are on
the same side of ring
trans-1,2-dimethylcyclopropane
the two methyl groups are on the
opposite side of ring

350. 11
Carbon-based Chiral Centers:
 A carbon atom which is bonded to four different groups is called an asymmetric
carbon atom.
 The term asymmetric carbon atom means the carbon which is bonded to four
different groups and a molecule lacks plane of symmetry.
Optical Activity
 Ordinary light = waves vibration in different planes

351. 12
 Plane polarized light = waves vibrating only in one plane
 When the ordinary lamp light is passed through Nicol prism (Calcite, CaCO3) or
Polaroid lens, light is found to vibrate in only one plane and is called plane polarized
light.
 Optical activity is measured by means of polarimeter in terms of degree of rotation.

352. 13
Optical isomers:
 There are two optical isomers, dextrorotatory and levorotatory
 The isomer which rotates the plane of polarized light to the right (clockwise
direction) is known as Dextrorotatory isomer or (+) isomer.
 The isomer which rotates the plane of polarized light to the left (anticlockwise
direction) is known as levorotatory isomer or (-) isomer.
Optical isomerism of lactic acid:

353. 14
 It has one asymmetric carbon
 Two non- super imposable images to each other.
 Non-super imposable mirror images are known as enantiomers.
 Three forms of lactic acids are known
(+) lactic acid-rotates the plane polarized light in clockwise direction
(-) lactic acid-rotates the plane polarized light in anticlockwise direction
(±) Does not rotate the plane polarized light and is optically inactive. It is an equimolar
mixture of (+) and (-) forms. (Racemic mixture)
Optical isomerism of Tartaric acid:
 Two asymmetric carbons
 Four forms of tartaric acids are known. Two of them are optically active and two are
optically inactive

354. 15
Assessment 1
1. Which of the following compound is not an organic compound?
a. imine ((R)2C=N-R)
b. Calcium carbide (CaC2)
c. Hydrazine (N2H2)
d. Sucrose (C12H22O11)
2. Which of the following structural feature is not true for iso-pentane
a. It has three CH3 groups
b. It has four carbons in sequence and fifth carbon is attached as branch.
c. It has one Carbon to which only one hydrogen is attached.
d. It has a carbon to which no hydrogen is bonded
3. Cyclobutane has the chemical formula similar to the following
a. Butene
b. Butane
c. Butylene
d. Pentane
4. Which of the following statements does not relate to Naphthalene and anthracene?
a. These are alicyclic compounds.
b. These are aromatic compounds.
c. They may contain more than one benzene ring systems.
d. Naphthalene contains two benzene rings while anthracene contains three
benzene rings.
5. The compound in which carbon of benzene ring is substituted by nitrogen is called:
a. Amine
b. Urea
c. Pyridine
d. Benzopheone
6. Which of the following is example of heterocyclic compound?
a. Pyridine
b. Furan
c. Pyrrole
d. All of these
Assessment 1
1. Which of the following compound is not an organic compound?
a. imine ((R)2C=N-R)
b. Calcium carbide (CaC2)
c. Hydrazine (N2H2)
d. Sucrose (C12H22O11)
2. Which of the following structural feature is not true for iso-pentane
a. It has three CH3 groups
b. It has four carbons in sequence and fifth carbon is attached as branch.
c. It has one Carbon to which only one hydrogen is attached.
d. It has a carbon to which no hydrogen is bonded
3. Cyclobutane has the chemical formula similar to the following
a. Butene
b. Butane
c. Butylene
d. Pentane
4. Which of the following statements does not relate to Naphthalene and anthracene?
a. These are alicyclic compounds.
b. These are aromatic compounds.
c. They may contain more than one benzene ring systems.
d. Naphthalene contains two benzene rings while anthracene contains three
benzene rings.
5. The compound in which carbon of benzene ring is substituted by nitrogen is called:
a. Amine
b. Urea
c. Pyridine
d. Benzopheone
6. Which of the following is example of heterocyclic compound?
a. Pyridine
b. Furan
c. Pyrrole
d. All of these

355. 16
7. The carboxyl functional group (– COOH) is present in
a. picric acid
b. Acetic acid
c. carbolic acid
d. All of these
8. In which organic compound, there is –OH present as functional group
a. Ethylene glycol
b. Ethoxy propane
c. Propanone
d. Ethanal
9. The isomer of diethyl ether is
a. (CH3)3C-OH
b. CH3CH2CHO
c. CH3CH2COCH3
d. CH3COCH3
10. The type of isomerism that arises in dimethyl ether and ethyl alcohol is:
a. Metamerism
b. Cis- trans
c. Functional group isomerism
d. Position isomerism
7. The carboxyl functional group (– COOH) is present in
a. picric acid
b. Acetic acid
c. carbolic acid
d. All of these
8. In which organic compound, there is –OH present as functional group
a. Ethylene glycol
b. Ethoxy propane
c. Propanone
d. Ethanal
9. The isomer of diethyl ether is
a. (CH3)3C-OH
b. CH3CH2CHO
c. CH3CH2COCH3
d. CH3COCH3
10. The type of isomerism that arises in dimethyl ether and ethyl alcohol is:
a. Metamerism
b. Cis- trans
c. Functional group isomerism
d. Position isomerism

356. 17
Assessment 2
1. Geometrical isomerism only arise in:
a. 2- methyl propene
b. 2-butene
c. 2- methyl -2- butene
d. 1-propene
2. The isomerism that arises due to the shift of proton is shown by
a. Aminoacid
b. Butane
c. Diethyl ether
d. Phenol
3. Positional isomerism is not exhibited by:
a. Alcohols
b. Aldehydes
c. Ketones
d. Halo alkanes
4. The molecule which does not have any functional group is:
a. Phenol
b. Ether
c. Acetone
d. Propane
5. Ali is asked to draw the structure of the molecule which does not exhibit structural
isomerism. Which of the following is the correct option he should draw?
a. C5H12
b. C6H14
c. C3H8
d. C4H10
6. The organic compound having >C=O is:
a. Aldehyde
b. Ketone
c. Ester
d. All of these
Assessment 2
1. Geometrical isomerism only arise in:
a. 2- methyl propene
b. 2-butene
c. 2- methyl -2- butene
d. 1-propene
2. The isomerism that arises due to the shift of proton is shown by
a. Aminoacid
b. Butane
c. Diethyl ether
d. Phenol
3. Positional isomerism is not exhibited by:
a. Alcohols
b. Aldehydes
c. Ketones
d. Halo alkanes
4. The molecule which does not have any functional group is:
a. Phenol
b. Ether
c. Acetone
d. Propane
5. Ali is asked to draw the structure of the molecule which does not exhibit structural
isomerism. Which of the following is the correct option he should draw?
a. C5H12
b. C6H14
c. C3H8
d. C4H10
6. The organic compound having >C=O is:
a. Aldehyde
b. Ketone
c. Ester
d. All of these

357. 18
7. Which of the following is acetamide?
a.
b.
c.
d.
8. The optical isomer which rotate the plane of polarized light in clockwise direction is:
a. Mesomer
b. Racemic mixture
c. Dextro rotatory
d. Levo rotatry
9. Which of the following cyco alkane exhibit geometrical isomerism?
a. 1,1-dimethyl cyclopropane
b. 1,1-dimethyl cyclobutane
c. 1,2-dimethyl cyclopropane
7. Which of the following is acetamide?
a.
b.
c.
d.
8. The optical isomer which rotate the plane of polarized light in clockwise direction
is:
a. Mesomer
b. Racemic mixture
c. Dextro rotatory
d. Levo rotatry
9. Which of the following cyco alkane exhibit geometrical isomerism?
a. 1,1-dimethyl cyclopropane
b. 1,1-dimethyl cyclobutane
c. 1,2-dimethyl cyclopropane

358. 19
d. None of these
10. A carbon atom which is bonded to four different groups is called:
a. Chiral carbon
b. Asymmetric carbon
c. Symmetircal carbon
d. Both a and b
d. None of these
10. A carbon atom which is bonded to four different groups is called:
a. Chiral carbon
b. Asymmetric carbon
c. Symmetircal carbon
d. Both a and b

359. 20
Assessment 3
1. Which of the following is true about the optical isomerism of tartaric acid:
a. It contains two asymmetric carbon atoms.
b. It exhibit meso form having plane of symmetry.
c. Meso form of tartaric acid is optically inactive.
d. All of these
2. n-propyl alcohol and isopropyl alcohol are compounds having same molecular formula
but different:
a. Functional group
b. Chain
c. Position of functional group
d. Both b and c
3. Which of the following are known as stereoisomers?
a. Metamers
b. Tautomers
c. Geometrical isomers
d. Functional group isomers
4. The concept of functional group is very importance in organic chemistry because:
a. It serves as the basis for nomenclature.
b. It serves as the basic for classification
c. It contributes the specific class or chemical reactions
d. All of these
5. The chemical formula of methane thiol is:
a. CH3-SH
b. CH3-OH
c. CH3-S-CH3
d. C2H5-SH
6. The number of ether metamers represented by general formula
C4H10Ois:
a. 1
b. 2
c. 3
d. 4
Assessment 3
1. Which of the following is true about the optical isomerism of tartaric acid:
a. It contains two asymmetric carbon atoms.
b. It exhibit meso form having plane of symmetry.
c. Meso form of tartaric acid is optically inactive.
d. All of these
2. n-propyl alcohol and isopropyl alcohol are compounds having same molecular formula
but different:
a. Functional group
b. Chain
c. Position of functional group
d. Both b and c
3. Which of the following are known as stereoisomers?
a. Metamers
b. Tautomers
c. Geometrical isomers
d. Functional group isomers
4. The concept of functional group is very importance in organic chemistry because:
a. It serves as the basis for nomenclature.
b. It serves as the basic for classification
c. It contributes the specific class or chemical reactions
d. All of these
5. The chemical formula of methane thiol is:
a. CH3-SH
b. CH3-OH
c. CH3-S-CH3
d. C2H5-SH
6. The number of ether metamers represented by general formula
C4H10Ois:
a. 1
b. 2
c. 3
d. 4

360. 21
7. The restricted rotation about carbon carbon double bond in 2-butene is due to:
a. Sidewise Overlap of two p orbitals
b. Parallel overlap of sp2 and p orbital
c. Overlap of sp2-sp2 orbitals
d. Head to head overlap of sp2-sp2 orbitals
8. The following alkene does not show geometrical isomerism:
a. 1,2-dichloro-1-pentene
b. 1,3-dichloro-2-pentene
c. 1,1-dichloro-1-pentene
d. 1,4-dichloro-2-pentene
9. Two Isomers of a compound (X) must have same:
a. Structural formula
b. chemical properties
c. Molecular formula
d. Both a and b
10. A similarity between optical isomerism and geometrical isomerism is:
a. Double bond
b. Chiral carbon
c. Spatial arrangement
d. All of these
7. The restricted rotation about carbon carbon double bond in 2-butene is due to:
a. Sidewise Overlap of two p orbitals
b. Parallel overlap of sp2
and p orbital
c. Overlap of sp2
-sp2
orbitals
d. Head to head overlap of sp2
-sp2
orbitals
8. The no of structural isomers given by pentane is=
a. 1
b. 2
c. 3
d. 4
9. Two Isomers of a compound (X) must have same:
a. Structural formula
b. chemical properties
c. Molecular formula
d. Both a and b
10. A similarity between optical isomerism and geometrical isomerism is:
a. Double bond
b. Chiral carbon
c. Spatial arrangement
d. All of these

361. 22
Assessment 4
1. The molecular formula C5H12 contains how many isomeric alkanes:
a. 1
b. 2
c. 3
d. 4
2. A fat on hydrolysis would yield the compound A and soap. The compound A has the
following functional group:
a. –COOH
b. –CHO
c. –OH
d. >C=C<
3. Which of the following is independent of the functional group?
a. Chain isomerism
b. Position isomerism
c. Functional group isomerism
d. Both A and B
4. The hydrocarbon which cannot be considered for alicyclic class of hydrocarbon:
a. compound having 3-C atoms
b. compound having 1-C atoms
c. compound having 2-C atoms
d. Both b and c
5. Which of the following compound has functional group with no oxygen atom?
a. Alcohol
b. Ether
c. Esters
d. Amines
6. Friedrick Wohler rejected the vital force theory by heating:
a. Urea
b. Ammonium cyanate
c. Carbamate
d. Ammonia
Assessment 4
1. The molecular formula C5H12 contains how many isomeric alkanes:
a. 1
b. 2
c. 3
d. 4
2. A fat on hydrolysis would yield the compound A and soap. The compound A has the
following functional group:
a. –COOH
b. –CHO
c. –OH
d. >C=C<
3. Which of the following is independent of the functional group?
a. Chain isomerism
b. Position isomerism
c. Functional group isomerism
d. Both A and B
4. The hydrocarbon which cannot be considered for alicyclic class of hydrocarbon:
a. compound having 3-C atoms
b. compound having 1-C atoms
c. compound having 2-C atoms
d. Both b and c
5. Which of the following compound has functional group with no oxygen atom?
a. Alcohol
b. Ether
c. Esters
d. Amines
6. Friedrick Wohler rejected the vital force theory by heating:
a. Urea
b. Ammonium cyanate
c. Carbamate
d. Ammonia

362. 23
7. Which of the following are tautomers to each other?
a. Vinyl alcohol and acetaldehyde
b. Diethyl ether and methyl-n-propyl ether
c. Ethanol and acetaldehyde
d. All of these
8. The following compounds belongs to which of the class of organic compounds:
a. Alcohol
b. Ether
c. Ester
d. Sulphones
9. Fructose is the fruit sugar and belongs to following class of organic compound:
a. Aldehyde
b. Ketones
c. Ester
d. Carboxylic acid
10. Which of the following class of compounds are isomers of alkene?
a. Alicyclic hydrocarbons
b. Alicyclic unsaturated hydrocarbons
c. Alicyclic saturated hydrocarbons
d. All of these
7. Which of the following are tautomers to each other?
a. Vinyl alcohol and acetaldehyde
b. Diethyl ether and methyl-n-propyl ether
c. Ethanol and acetaldehyde
d. All of these
8. The following compounds belongs to which of the class of organic compounds:
a. Alcohol
b. Ether
c. Ester
d. Sulphones
9. Fructose is the fruit sugar and belongs to following class of organic compound:
a. Aldehyde
b. Ketones
c. Ester
d. Carboxylic acid
10. Which of the following class of compounds are isomers of alkene?
a. Alicyclic hydrocarbons
b. Alicyclic unsaturated hydrocarbons
c. Alicyclic saturated hydrocarbons
d. All of these

363. 24
Key
Assessment 1
1. c
2. d
3. a
4. a
5. c
6. d
7. b
8. a
9. a
10. c
Assessment 2
1. b
2. a
3. a
4. d
5. c
6. d
7. b
Key
Assessment 1
1. C
2. D
3. A
4. A
5. C
6. D
7. B
8. A
9. A
10. C
Assessment 2
1. B
2. A
3. A
4. D
5. C
6. D
7. B
8. C
9. C
10. D

364. 25
Key
Assessment 3
1. d
2. c
3. c
4. d
5. a
6. c
7. a
8. c
9. c
10. c
Assessment 4
1. c
2. c
3. a
4. d
5. d
6. b
7. a
8. c
9. b
10. c
Key
Assessment 3
1. D
2. C
3. C
4. D
5. A
6. C
7. A
8. C
9. C
10. c
Assessment 4
1. C
2. C
3. A
4. D
5. D
6. B
7. A
8. C
9. B
10. C

365. Oldest, Largest and Most Credible Platform
MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org

366. Learning Objectives:
➢ Open chain and closed chain hydrocarbons
➢ Nomenclature of alkanes, alkenes and alkynes
➢ Benzene: Properties, structure, modern
representation, reactions, resonance method,
electrophilic substitution
➢ The molecular orbital treatment of benzene.
Chemistry of Hydrocarbons

367. Chemistry of Hydrocarbons:

368. Alkane: (Nomenclature)
Common or trivial name:
IUPAC Nomenclature of AlkaneSel
ct
the longest continuous chain
Select the longest continuous chain.
Assign the number to carbon from end closer to attached
substituent.
Assign position of substituent, name of substituent and
then parent name.
Each substituent is assigned the separate number no matter
if attached to same carbon atom of long chain.
Use of prefix di, tri etc in case of more than one substituent
of similar kind.
In case of more than one long chains, long chain with
greater no of substituent is selected.

369. IUPAC Nomenclature of Alkene
Select the longest parent chain containing C=C bond.
Assign number to carbon atoms of long chain from end
closer to C=C bond.
Designate the location of the double bond by using the
number of the first atom of the double bond as a
prefix.
Assign the number to the position of the substituent.
Use prefix alkadiene, alkatriene etc for more than
double, triple double bonds

370. IUPAC Nomenclature of Alkynes
The largest continuous carbon chain containing triple
bond is selected..
The position of triple bond is shown by numbering.
It is named as alkadiyne and triyne, etc. depending on
the number of triple bonds.
Lowest possible number is assigned to a double or a
triple bond irrespective of whether ene or yne gets the
lower number.
In case a double and a triple bond are present at
identical positions, the double bond is given the lower
number.

371. Practice the Nomenclature:
1.
2.
3.
4.

372. Decarboxylation
From Grignard’s
reagent
Reduction of carbonyl compounds
5.
Preparation of Alkane:
R-CH=CH-R + H2
𝑵𝒊/𝟐𝟎𝟎𝟎𝑪
→ R-CH2-CH2-R
➢ Pt or Pd catalyst (expensive & room temp
➢ Hydrogenation of vegetable oil produces
vegetable ghee
Hydrogenation
R-X + Zn + HX → R-H + ZnX2
Also by Hydrogenolysis (hydrogenation accompanied
by bond cleavage) by heating
R-X + H2
𝑷𝒅/𝑪
→ R-H + H-X
From Alkyl halide
NaOH + CaO = Soda lime
Kolbe’s method
➢ Free radical mechanism
➢ Use of electric current
➢ Methane cannot be prepared

373. Free radical Mechanism:
Clemensen reduction and Wolf Kishner’s reduction:
Alkanes react with chlorine and bromine in the
presence of sunlight or UV light or at high
temperature resulting in the successive
replacement of hydrogen atoms with halogens
called halogenation.
Extent of halogenation depends upon
the amount of halogen used.
The order of reactivity of halogens is
F2>Cl2>Br2>I2.

374. Mechanism of Halogenation:
Initiation Step
Propagation Step
Termination Step

375. Ethene:
➢ A pi-bond is a weaker bond as compared to a sigma-bond.
➢ During a reaction it breaks comparatively easily rendering alkenes as reactive group
of compounds.
➢ The loosely held p-electrons are more exposed to attack by the electrophilic
reagents.
➢ Alkenes, therefore, undergo electrophilic reactions very easily.

376. Preparation of Alkene:
Preparation
of
Alkene
Dehydration of Alcohols Dehalogenation of RX
The ease of Dehydration of alcohols
10
< 20
< 30
Vicinal dihalides have two halogens on adjacent
carbon atoms.
It requires anhydrous solvent such as methanol or
acetic acid.
Zinc removes two halogens from the adjacent
carbon atoms.
Hydrogenation:
➢ Addition of hydrogen to unsaturated hydrocarbons in presence of
catalyst
➢ Exothermic and Heat of hydrogen for each C=C bond in most of
alkanes is 120 KJ/mole
Raney nickel:
Ni-Al + NaOH + H2O → Ni + NaAlO2 +
𝟑
𝟐
H2
CH2=CH2 + H2
𝑵𝒊
→ CH3 – CH3
➢ Most alkenes are hydrogenated over Raney nickel at about 100°C and
upto 3-atmosphere pressure.
Raney Nickel
(amorphous)

377. In the addition of an unsymmetrical reagent to an unsymmetrical alkene, the negative part
of the adding reagent goes to that carbon, constituting the double bond, which has least
number of hydrogen atoms.
Hydrohalogenation:
Hydration:
➢ Addition of water to ethene by treating cold concentration sulphuric
acid.
When alkenes are treated with cold concentrated sulphuric acid, they are dissolved
because they react by addition to form alkyl hydrogen sulphate.
Alkylhydrogen sulphates on boiling with water decompose to give corresponding alcohols.
Ethane hydrogen sulphate

378. Halohydration involves halogenations of alkene which is carried out in an aqueous solution.
Haloalcohol is formed called Halohydrin
In it, solvent becomes reactants too.
Halogenation:
➢ Polarization of bromine when approached to ethane.
➢ Formation of a bromonium ion
➢ A trans product is formed.
Halohydration:
The alkenes on
treatment with halogen
in an inert solvent like
carbon tetrachloride at
room temperature give
vicinal dihalides or 1,2
dihalogenated products.
This is the test which is
used to detect
unsaturation.

379. Small organic molecules (monomers) combine together to form larger molecules known as
Polymers.
Preparation of alkynes using elimination reactions:
It involves:
➢ Vicinal dihalide as starting reactant
➢ Strong base
eg Alcoholic KOH
Drastic conditions are required for removal of second HX molecule.
Polymerization:
Ethene at 400°C and 100 atm pressure, polymerize to polythene or polyethylene.
A good quality polythene is obtained, when ethene is polymerized in the presence
of aluminium triethyl Al(C2H5)3 and titanium tetrachloride catalysts (TiCl4).
Dehydrohalogenation of Vicinal dihalide:

380. Active metals are required for dehalogenation of tetrahaide
Chemistry of Alkynes:
Dehalogenation of Tetrahalide:
Hydrogenation:
Hydrohalogenation:
𝐂𝐇 ≡ 𝐂𝐇 + H-Br → CH2 = CH-Br (vinyl bromide)
CH2 = CH-Br + HBr
𝑴𝒂𝒓𝒌𝒐𝒘𝒏𝒊𝒌𝒐′𝒔𝒓𝒖𝒍𝒆
→ CH3 – CH (Br)2 (1,1-dibomoethane)
Addition of four hydrogen atoms to ethyne to form ethane.
It is carried out I the presence of catalyst like Ni, Pt or Pd

381. 75 oC is required.
Vinyl alcohol is an unstable enol
Enol has hydroxy group attached to doubly bonded carbon atom and isomerizes to
acetaldehyde
Acidic nature of alkynes:
Hydration:
Acidic character of Alkyne:
Silver and copper acetylides react with acids to regenerate alkynes.
Addition of H-Br to
vinyl bromide is in
accordance with
“Markownikov’s rule”

382. The sp hybridized carbon atom of a terminal alkyne pulls the electrons more strongly
making the attached hydrogen atom slightly acidic.
An sp hybrid orbital has 50% s-character in it and renders the carbon atom more
electronegative than sp2 and sp3 hybridized carbons.
Reactivity of Alkane, Alkene and Alkyne:
The general order of decreasing reactivity is given as:
𝑨𝒍𝒌𝒆𝒏𝒆 > 𝑨𝒍𝒌𝒚𝒏𝒆 >Alkane
Because
➢ Alkanes are least reactive towards Electrophilic attack because
these contain sigma bonds which are inert
➢ 𝜋-bond in alkene is weak.
➢ Electrons are more exposed to an attack by Electrophilic reagent.
➢ In case of alkynes, because of shortening of bond length and
electrons are less exposed for the Electrophilic attack.

383. Structure of Benzene
➢ Empirical formula (CH) by elemental analysis
➢ Molecular mass (78.108) by vapour density
method
➢ Molecular mass determined to be C6H6
Open chain unsaturated structures are rule out
Because
➢ Benzene is stable to KMnO4 solution
➢ It adds 3 hydrogen molecules.
➢ It gives substitution reaction with conc.
HNO3 and Conc. Sulphuric acid
➢ It gives only one monosubstituted product
➢ Its general formula (CnHn) does not
corresponds to alkane, alkene and alkyne
Kekule structure:
➢ Structure of benzene remained mystery
for 40 years.
➢ Problem was solved by A.Kekule in 1865
➢ He purposed hexagonal cyclic structure of
benzene
Kekule supporting evidences
➢ Gives only one
monosubsituted product
➢ Gives three disubstituted
product
➢ Adds 3H2 molecules
➢ Adds 3Cl2 molecules
➢ Three double bonds
alternating with three
single bonds
X-ray analysis of benzene:
Regular hexagonal planar structure as

384. Atomic Orbital Treatment
Each carbon in benzene is sp2
Hybridized.
Ground state Excited state Hybridization
There is availability of 3 sp2 hybrid
orbitals and one unhybridized
orbital
The three sp2hybrid orbitals on each carbon
are utilized to form three 𝜎-bonds, two with
adjacent carbon atoms and one with
hydrogen.
The unhybridized 2pz orbitals remain at right
angle to these sp2 orbitals.
All the carbon and hydrogen atoms are
coplanar.
Overlap of unhybridized orbitals of six carbons
form continuous sheath below and above the
benzene ring.
Since each 2pzorbital is overlapped by the 2pz
orbitals of adjacent carbon atoms, therefore,
this overlapping gives, 'diffused' or 'delocalized'
electron cloud.

385. Stability and Resonance energy of Benzene
Stability of benzene is due to extensive
delocaization of electron density.
The extent of stability can be understood by
measuring the heat of hydrogenation of 1,3,5-
cyclohexatriene.
The difference between amount of heat actually
released and that calculated on the basis of the
Kekule’s structure is now called the ‘Resonance
energy’ of the compound.
“The possibility of different pairing schemes of valence electrons of atoms is called resonance”
and the different structures thus arranged are called “Resonance structures”.
Dewar structures are energetically unfavourable.
Therefore, benzene molecule can be represented by either of the two Kekule’s structure.

386. Electrophilic substitution reaction of benzene
In alkanes, C-C = 1.54 A0
In alkenes, C=C = 1.34 A0
In Alkynes, C ≡ C = 1.20 A0
In benzene C-C = 1.397 A0
Halogenation:
Strong electrophile = Cl+
𝐹𝑙𝑢𝑜𝑟𝑖𝑛𝑎𝑡𝑖𝑜𝑛 > 𝑐ℎ𝑙𝑜𝑟𝑖𝑛𝑎𝑡𝑖𝑜𝑛 ≅
𝑏𝑟𝑜𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 >iodination
Alky group attached to benzene ring is
substitued by free radical mechanism
Benzyl chloride, benzal chloride and
benzotrichloride are formed
Nitration:
50-55 oC, Electrophile is Nitronium ion (NO2
+)
Sulphonation:

387. Friedel Craft Alkylation:
AlCl3 is catalyst form carbocation. Anhydrous
conditions
Friedel craft Acylation:
(R-C+=O) = Electrophile , C6H5COCH3 (Acetophenone)

388. Addition Reactions Of Benzene And Methyl Benzene
Addition reactions:

389. Oxidation of side chain:
Alkyl groups attached to benzene are oxidized by acidified KMnO4
Whatever the length of an alkyl group may be it gives only one carboxyl
group.
This is the test for alkyl benzenes.

390. Orientation in Electrophilic Substitution reactions
The group present in the mono-substituted benzene ring has the directive effect and thus
determines the position or orientation for the new incoming groups.
Ortho para directing group
➢ Electron donating
➢ Increases the reactivity of benzene
toward Electrophilic attack at ortho and
para position.
Examples of ortho para directing group:
N(CH3)2, -NH2, -OH, -OCH3, -Cl, -Br, -I
Meta directing group
➢ Electron withdrawing groups
➢ Decreases the chemical reactivity of
benzene
➢ The ortho and para positions are
rendered more electron deficient than
the meta position.
➢ The incoming electrophile will prefer
to attack on meta position rather than
ortho and para positions.
Other examples of meta directing groups:
N+R3 ,-C≡N, -COOH, -CHO,- COR

391. Comparison Of Reactivities Of Alkanes, Alkenes And Benzene
➢ Alkanes are unreactive due to inert nature of sigma bond
Undergoes substitution reactions
➢ Alkenes are reactive due to weakness of pi bond and availability of pi electrons for the electrophilic
reagent.
➢ Undergo electrophilic addition reactions easily.
➢ Alkenes undergo polymerization reactions and they are also readily oxidized.
➢ Benzene is highly unsaturated and also stable
➢ It undergoes electrophilic substitution reactions.
➢ Resistant to oxidation.
➢ Does not undergo polymerization.

392. Assessment 1
1. Alkanes give substitution reactions. Halogenation of alkanes in diffused light proceeds by:
a. Generation of Carbanions
b. Generation of Carbocations
c. Generation of alkenes
d. Generation of Free rdicals
2. The initiation step of free radical mechanism of photochemical chlorination of alkane involves:
a. Oxidation
b. Substitution
c. Homolysis
d. Peroxidation
3. The principle industrial application of hydrogenation of alkene involves:
a. Formation of alcohols
b. Vegetable ghee from vegetable oil
c. Hydrogenolysis to form HCl
d. Forming sodium salts
4. C2H4 when treated with Br2 in the presence of CCl4 forms which of the following product
a. 1, 2-dibromoethane
b. Acetylene
c. 1,1-dibromoethane
d. Both a and b
5. The general reactions which are usually preferred by alkene under suitable conditions:
a. Electrophilic
b. Substitution
c. Elimination
d. All of these
6. Which of the following hydrocarbon is not considered for the product of the reaction obtained
by Sabatier and Senderen's reaction?
a. C2H6
b. CH4
c. C3H8
d. All of these
7. Which of the following method involves the formation of ethene from ethanol
a. By dehydrohalogenation
b. By dehydrogenation
c. By dehydration with conc. H2SO4 at 170 0C
d. By reduction with hydrogen iodide

393. 8. Which of the following process is an ideal and easy way to convert olefin into paraffin:
a. Halogenation
b. Dehydration
c. Hydrogenation
d. Hydrolysis
9. Markownikov’s rule governs the addition of HBr over:
a. CH2=CH2
b. CH3-CH3
c. CH3-CH=CH2
d. All of these
10. C2H4 is the alkene with planar structure. The C=C bond length is:
a. 1.54 Å
b. 1.34 Å
c. 1.10 Å
d. 2.40 Å

394. Assessment 2
1. Polyethylene is obtained from C2H4 by a reaction called:
a. Oxidation
b. Isomerization
c. Polymerization
d. Substitution
2. Markownikoff's rule is applicable when there is requirement of addition of:
a. Symmetrical reagent to symmetrical alkene
b. Unsymmetrical reagent to unsymmetrical alkene
c. Unsymmetrical reagent to symmetrical alkene
d. Symmetrical reagent to unsymmetrical alkene
3. Which is the catalyst used when HCl is added to vinyl acetylene, chloroprene is obtained which
readily polymerize to neoprene, used as synthetic rubber:
a. HgSO4 + H2SO4
b. Cu2Cl2
c. Cu2Cl2 + NH4Cl
d. Cu2Cl2 + NH4OH
4. If HCl is added to vinyl acetylene, chloroprene is obtained. According to IUPAC, the name of
Chloroprene is
a. 2-chloro-1, 3-butadiene
b. 3-chloro-2, 3-butadiene
c. 2, 3-dichlorobutadiene
d. None of these
5. Benzene is the polymer produced by the polymerization of:
a. Methane
b. Ethane
c. Ethylene
d. Acetylene
6. Which of the following compound cannot be classified as an aromatic compound?
a. Benzene
b. Naphthalene
c. Pyridine
d. 1,3 cyclohexadiene
7. Which of the following group substitute the hydrogen of benzene ring to form aniline?
a. Nitro group
b. Amino group
c. Hydroxyl group
d. Nitrogen

395. 8. Which of the following has the adjacent rings have common carbon to carbon bonds?
a. Biphenyl
b. Diphenyl methane
c. Toluene
d. Anthracene
9. Which of the following comes first in priority order of groups while giving the name to
disubstituted benzene?
a. –R
b. –OH
c. –CHO
d. –COOH
10. The x-rays studies of benzene reveals that carbon atoms in benzene molecule is inclined at an
angle of:
a. 1200
b. 180o
c. 109o
d. 107.5o

396. Assessment 3
1. According to hybridization approach, each carbon in benzene undergoes:
a. sp hybridization
b. sp2 hybridization
c. sp3 hybridization
d. None of these
2. The extraordinary stability of benzene molecule is largely contributed by its
a. Hexagonal planar ring
b. Dewar structure
c. Resonance structures
d. Chemical bonding
3. The reaction which involves treatment of benzene with halogen in presence of catalyst gives
which type of product to retain its characteristic stability:
a. Elimination product
b. Addition product
c. Substitution product
d. Oxidation product
4. Which of the following electrophile is generated when concentrated nitric acid is made to
react with concentrated sulphuric acid?
a. NO2
+
b. NO
c. NO2
-
d. NO3
-
5. The catalytic oxidation of benzene at 450 oC in presence of V2O5 give:
a. Oxalic acid
b. Succinic acid
c. Maleic acid
d. Formic acid
6. What is the end product which is obtained on the nitration of toluene:
a. o-nitrotoluene
b. p-nitrotoluene
c. 2, 4-dinitrotoluene
d. 2, 4, 6-trinitrotoluene
7. Benzene is highly stable molecule which in presence of sunlight adds:
a. One molecule of chlorine
b. Two molecules of chlorine
c. Three molecules of chlorine
d. Four atoms of Chlorine

397. 8. Which of the following is/are not the characteristics reaction of benzene?
a. Polymerization
b. Substitution
c. Oxidation
d. Addition
9. Consider the electrophilic substitution reaction of benzene given as:
Which of the following is the final product?
a.
b.
c.
d. All of these
10. meta-directing substituents in aromatic substitution are given below. Which one is most
deactivating?
(a) –C ≡ N
(b) -SO3H
(c) -COOH
(d) -NO2

398. Assessment 4
1. The following is the correct reactivity order of R-H bond of alkane is:
a. 30
> 20
> 10
b. 20
> 30
> 10
c. 10
> 20
> 30
d. 30
= 20
= 10
2. Which of the following alcohol is easily dehydrated?
a. Tertiary butyl alcohol
b. 2-propanol
c. 1-butanol
d. Ethanol
3. The following compounds on reaction with each other undergoes dehydrohalogenation to
form alkene:
a.
b.
c.
d.
4. Markonwikov’s rule is applicable and decisive to the addition of HX to the following alkene:
a.
b.
c.
d. All of these

399. 5. The addition of Br(aq) to alkene results in the formation of:
a. Vicinal dihalide
b. Gem dihalide
c. Halohydrin
d. Dihalide
6. Which of the carbon atom is asymmetric carbon atom in the following compound?
a. Carbon 1
b. Carbon 2
c. Carbon 3
d. Carbon 4 or Carbon 5
7. An equimolar mixture of (+) and (-) forms of lactic acid is called:
a. Enantiomer
b. Optical isomer
c. Racemic
d. Dextro rotator
8. The following compound has more acidic hydrogen:
a. 𝐻 − 𝐶 ≡ 𝐶 − 𝐻
b. 𝐶𝐻3 − 𝐶 = 𝐶 − 𝐶𝐻3
c. 𝐶𝐻3 − 𝐶 ≡ 𝐶 − 𝐻
d. 𝐶𝐻2 = 𝐶𝐻2
9. Which of the following statement is true about the reduction of alkynes in presence of
Na/NH3?
a. The reaction proceeds via single transfer
b. Trans alkene is the product
c. It is limited to alkene formation only.
d. All of these
10. The final product formed by the hydration of 𝐶𝐻 ≡ 𝐶𝐻 by mercuric sulphate dissolved in
sulphuric acid at 75 oC is:
a. Vinyl alcohol
b. Acetaldehyde
c. Acetone
d. Propanone

400. Key
Assessment 1
Assessment 2
1. d
2. c
3. b
4. a
5. a
6. b
7. c
8. c
9. c
10. a
1. c
2. b
3. c
4. a
5. d
6. d
7. b
8. d
9. d
10. a

401. Assessment 3
Assessment 4
1. b
2. c
3. c
4. a
5. c
6. d
7. c
8. a
9. b
10. d
1. a
2. a
3. a
4. a
5. c
6. c
7. c
8. a
9. d
10. c

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403. Halogalk:anes
Alkyl halides
Classification
Nomenclature
Reactions
Mechanism of nucleophilic substitution reaction SN1,SN2,E1
andE2reaction Learning
Alkyl Halides

404. Halogalkenes:
Those organic compounds which contain –X(F,Cl,Br,I) as functional group are called alkyl halide or
halogalkanes.
Classification
Primary alkyl halide is alkyl halide
in which carbon to which halogen
is attached is linked to only one
carbon.
Secondary alkyl halide in which
carbon to which halogen is
attached is further linked to two
carbon atoms.
Tertiary alkyl halide is one in which
carbon to which halogen is
attached is further linked to three
carbon atoms

405. Nomenclature of alkyl halide:
Common names:
 These are termed as haloalkanes
 Name as:
Alkyl + halide
Ethyl bromide Methyl chloride
n-propyl bromide
iso-propyl bromide
IUPAC Nomenclature:
 Selection of longest continuous chain
 Assign lowest possible number to
carbon with which halogen is attached
 Use of prefix di, tri etc for presence of
more than one halogen
2-bromo-3-methyl butane

406. Structure of alkyl halide:
 Alkyl halide molecule consists of two parts; an alkyl group and halide group
 Carbon of alkyl group bears the partial positive charge while negative charge appears on
halogen atom.
Reactivity of alkyl halide:
 Two factors govern the reactivity of alkyl halide; bond energy and bond length
Bond energy:
The order of reactivity of alkyl halide is:
R— I > R— Br > R— Cl > R— F
Bond Polarity
 If an electrophile is the attacking reagent then this difference suggests that alkyl
fluorides would be the most reactive one.

407. So the overall order of reactivity of alkyl halides for a particular alkyl group is:
Iodide > Bromide > Chloride > Fluoride
Overall order of reactivity:
 Experiments have shown that the strength of carbon halogen bond is the main factor
which decides the reactivity of alkyl halides.
 The overall order of reactivity of alkyl halides for a particular alkyl group is:
Iodide > Bromide > Chloride > Fluoride
Reactions of Alkyl halides:
Nucleophilic Substitution Reactions:
Important terms to remember:
 Nucleophile:
 Electron rich species
 Having unshared pair of electrons available for bonding
 Mostly basic in character
 Negatively charged or neutral

408.  Electrophile:
 Electron deficient specie
 Carbon attached to halogen bears partial positive charge and is Electrophilic in
character
 Electrophile may be neutral or positively charged.
 Leaving group:
 Leaving group is the group which departs with an unshared pair of electrons.
 It departs when comparatively stronger nucleophile approaches.
 Good leaving groups include: Cl-, Br-, I- and HSO4
-
 Poor leaving groups include: OH-, OR- and NH2
-
 Iodide ion is a good nucleophile as well as leaving group.
 Substrate:
 The alkyl halide molecule on which a nucleophile attacks is called a substrate
molecule.
Nucleophilic Substitution Bimolecular (SN2)
 Single step
 The extent of bond formation is equal to the
extent of bond breakage.
 Transition state exists
 Coming nucleophile always attack from side
opposite to leaving group.
 The substrate carbon atom changes its state
of hybridization from tetrahedral sp3to
planar sp2 in transition state.
 The attack of the nuclephile, the change in
the state of hybridization and the departure
of the leaving group, everything occurs at
the same time.
 Inversion of configuration (Stereochemistry)
 Rate = k [Alkyl halide]1 [Nucleophile]1
 2nd order reaction (Chemical kinetics).
Nucleophilic Substitution Unimolecular (SN1)
 Two step process; first step involves reversible
ionization and second step involves the attack of
nucleophile
 Intermediate state exists
 Coming nucleophile attacked from both sides
 Ionization is carried out in aqueous acetone or
aqueous ethanol.
 First step is the rate determining step.
 Racemic mixture (50% retention of configuration
and 50% inversion of configuration) is formed
(stereochemistry)
Nucleophilic Substitution Bimolecular (SN2)
 Single step
 The extent of bond formation is equal to
the extent of bond breakage.
 Transition state exists
 Coming nucleophile always attack from
side opposite to leaving group.
 The substrate carbon atom changes its
state of hybridization from tetrahedral
sp3to planar sp2 in transition state.
 The attack of the nuclephile, the change in
the state of hybridization and the
departure of the leaving group, everything
occurs at the same time.
 Inversion of configuration
(Stereochemistry)
 Rate = k [Alkyl halide]1 [Nucleophile]1
 2nd order reaction (Chemical kinetics).
Nucleophilic Substitution Unimolecular (SN1)
 Two step process; first step involves
reversible ionization and second step
involves the attack of nucleophile
 Intermediate state exists
 Coming nucleophile attacked from both sides
 Ionization is carried out in aqueous acetone
or aqueous ethanol.
 First step is the rate determining step.
 Racemic mixture (50% retention of
configuration and 50% inversion of
configuration) is formed (stereochemistry)
 Rate = k [Alkyl halide]
 1st order reaction (chemical kinetics)

410. Wurtz synthesis:
Alkyl halides react with sodium in ether solvent to give alkanes. The reaction is particularly
useful for the preparation of symmetrical alkanes.
Reduction of Alkyl halide:
Alkyl halides can be reduced with zinc in the presence of an aqueous acid such as HCI or
CH3COOH.
Reaction with sodium lead alloy:
Methyl chloride and ethyl chloride react with sodium lead alloy giving tetramethyl lead and
tetraethyl lead,respectively. These compounds are important anti-knock agents and are used in
gasoline.

411. 1. Which of the following alkyl halide is secondary alkyl halide?
a. Isopropyl chloride
b. Isobutyl chloride
c. n-propyl chloride
d. n-butyl chloride
2. The correct name given to alkyl halide containing 5 carbon and to 1st carbon chlorine and at 2nd
carbon, methyl group is attached according to IUPAC is:
a. 2-methyl-1-chlorobutane
b. 1-chloro-2-methylhexane
c. 1-chloro-2-methylpentane
d. 2-methyl-1-chlorohexane
3. Which of the following carbon atom is tertiary carbon atom?
a. Carbon no 1
b. Carbon no 2
c. Carbon no 3
d. Carbon no 4
4. Depending upon the bond energy, which of the following is the least reactive alkyl halides?
a. Alkyl iodides
b. Alkyl chlorides
c. Alkyl bromides
d. Alkyl fluorides
5. Which of the following is/are correctly related to each other?
a. C-H bond, 413 KJ/mole
b. C-I bond, 413 KJ/mole
c. C-F bond, 228 KJ/mole
d. C-Cl bond, 290 KJ/mole
Assessment 1

412. 6. Alkyl fluorides do not react under ordinary conditions because of:
a. Stronger C-H bond
b. Weaker C-F bond
c. Stronger C-F bond
d. Stronger C-H bond
7. When the attacking specie is electrophile, which of the following alkyl halide is more reactive
depending upon the electronegativity difference between carbon and halogen is:
a. CH3-CH2-Br
b. CH3-CH2-Cl
c. CH3-CH2-I
d. CH3-CH2-F
8. Which of the following is not true about the nucleophile?
a. It has an unshared electron pair available for bonding.
b. In most cases it is basic in character.
c. It may be positively charged.
d. Nucleophile means nucleus loving.
9. Product formation in SN1 reaction given by tertiary alkyl halide proceeds by the intermediate:
a. Free radical
b. Carboanion
c. Carbocation
d. Molecular anion
10. The rate of SN1 reaction involving alkyl halide and nucleophile OH- depends only upon the
concentration of:
a. Nucleophile
b. OH-
c. Alkyl halide
d. Both a and b

413. Assessment 2
1. Which of the following may undergo both SN1 and SN2 reaction?
a. CH3-CH2-Cl
b. CH3-CH2-CH2-Cl
c. CH3-CH(Cl)-CH3
d. CH3-C(CH3)(Cl) –CH3
2. When the nucleophile attacks on hydrogen instead of carbon to which halogen is attached,
the expected product is:
a. Alcohol
b. Aldehyde
c. Alkene
d. Esters
3. Which of the following is the prominent feature of Wurtz reaction carried out by alkyl halide?
a. This method is particularly useful for the preparation of unsymmetrical alkene
b. This method is particularly useful for the preparation of symmetrical alcohols
c. This method is particularly useful for the preparation of symmetrical alkene
d. This method is particularly useful for the preparation of symmetrical alkyne.
4. The reduction of alkyl halide is carried out by zinc in the presence of:
a. Water
b. Molten zinc
c. All of these
d. HCl(aq)
5. The formation of tetraethyl lead, an antiknocking agent involves the reaction of
a. Ethyl chloride and lead
b. Ethane, NaCl and lead
c. Ethyl chloride and sodium lead alloy
d. Methyl chloride and sodium lead alloy
6. The Reaction of ethyl chloride with sodium in presence of ether proceeds to form:
a. Ethane
b. Propane
c. n-butane
d. n-pentane
7. Which of the following is boiled with ethyl chloride to form ethyl alcohol:
a. Alcoholic KOH
b. Aqueous KOH
c. NaOH
d. H2O

414. 8. . Which of the following alkyl halides cannot undergo both E1 mechanism:
a. Primary alkyl halide
b. Secondary alkyl halide
c. Tertiary alkyl halide
d. None of these
9. C2H5MgBr reacts with CNCl in presence of ether to produce intermediate product which on
hydrolysis in the presence of acid to give the final product:
a. 2-propanol
b. Propanoic acid
c. Ethyl nitrile
d. None
10. The removal of hydrogen halide experienced by adjacent carbon atoms of alkyl halide is
generally called:
a. Dehydrogenation
b. Hydrogenation
c. Dehydrohalogenation
d. Hydrohalogenation

415. Assessment 3
1. For which mechanism shown by alkyl halide, the first step is carried out in same fashion:
a. E1 and E2
b. E2 and SN2
c. E1 and SN1
d. SN1 and SN2
2. Tertiary butyl chloride is practically non reactive to proceed under SN2 mechanism due to:
a. Its Insolubility
b. Its Instability
c. Its Inductive effect
d. Presence of Bulky alkyl group/steric hinderance
3. Which of the following is an example of good nucleophile as well as good leaving group?
a. F-
b. Br-
c. I-
d. Cl-
4. A carbon atom bearing positive charge and attached to three other carbon atoms or alkyl
groups is called:
a. Carbanion
b. Carbonium ion
c. Oxonium ion
d. hydroxo ion
5. 50% inversion of configuration and 50% retention of configuration is possible in SN1 reaction
given by alkyl halide due to:
a. Tetrahedral geometry of carbocation
b. Hexagonal geometry of carbocation
c. Trigonal planar geometry of carbocation
d. Linear geometry of carbocation
6. Which one of the following is not a nucleophile?
a. H2O
b. Br-
c. BF3
d. NH3
7. Which of the following haloalkane is a vicinal dihalide?
a. ClCH2CH2Cl
b. CH3CHCl2
c. BrCH2CH2CH2Br
d. BrCH2CH2CH2CH2Br

416. 8. The increasing order of nucleophilicity would be?
a. Cl– < Br– < I–
b. I– < Cl– < Br–
c. Br– < Cl– < F–
d. I– < Br– < Cl–
9. The treatment of ethanol with phosphorous pentachloride produces:
a. C2H5Cl and POCl
b. C2H5Cl and HCl
c. C2H5Cl
d. C2H5Cl, POCl3 and HCl
10. Which of the following is formed by the reaction of Ethyl bromide with ethyl amine?
a. Ethylamine
b. Diethylamine
c. Triethylamine
d. nitroethane

417. Assessment 4
1. The first step of SN1 reaction involves the ionization of alkyl halide. Which is more stable
carbocation:
a. Primary
b. Secondary
c. Tertiary
d. All have same stability
2. A mixture of1-chloropropane and 2-chloropropane on treatment with alcoholic KOH gives
a. 1-propene
b. 2-propene
c. Isopropylene
d. Hexane
3. The final product of reaction of ammonia with excess of ethyl chloride is:
a. Methyl amine
b. Dimethyl amine
c. Trimethyl amine
d. Tetraethyl ammonium chloride
4. The reagent used for dehalogenation of 1,2-dichloropropane is:
a. Zn-Cu
b. Zn-Hg
c. Zn dust
d. Na
5. Propyl bromide and isopropyl bromide are:
a. Metamers
b. Functiona group isomers
c. Position isomers
d. Chain isomers
6. Which of the following is not the characteristic of SN2 reaction?
a. 50% inversion of configuration.
b. It is favoured in non-polar solvents
c. It is bimolecular reaction.
d. It is single step reaction
7. The conversion of an alkyl halide into an alcohol involves:
a. Addition
b. Elimination
c. Substitution
d. Dehydrogenation

418. 8. Which of the following factor favour the elimination reaction over substitution?
a. Crowding within substrate
b. Weak base
c. Increasing solvent polarity
d. Decrease in temperature
9. Reactivity of order of alkyl halides for dehydrohalogenation is:
a. 𝑅 − 𝐹 > 𝑅 − 𝐶𝑙 > 𝑅 − 𝐵𝑟 > 𝑅 − 𝐼
b. 𝑅 − 𝐼 > 𝑅 − 𝐵𝑟 > 𝑅 − 𝐶𝑙 > 𝑅 − 𝐹
c. 𝑅 − 𝐼 > 𝑅 − 𝐶𝑙 > 𝑅 − 𝐵𝑟 > 𝑅 − 𝐹
d. 𝑅 − 𝐹 > 𝑅 − 𝐼 > 𝑅 − 𝐵𝑟 > 𝑅 − 𝐶𝑙
10. In the following, which atom is acting as electrophile?
a. Br
b. C
c. H
d. None of these

419. 1. A
2. C
3. C
4. D
5. A
6. C
7. D
8. C
9. C
10. C
KEY
Assessment 1
Assessment 2
1. C
2. C
3. C
4. D
5. C
6. C
7. B
8. A
9. B
10. C

420. Assessment 3
1. C
2. D
3. C
4. B
5. C
6. C
7. C
8. A
9. D
10. B
Assessment 4
1. C
2. A
3. D
4. C
5. C
6. A
7. C
8. A
9. B
10. B

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422. Learning Outcomes:
• Nomenclature and structure of alcohols
• Reactivity of Alcohols
• Chemistry of Alcohols by preparation of
ethers and esters
• Nomenclature and Structure of Phenols
• Reactivity of Phenols and their chemistry
by electrophilic aromatic substitution
• Difference between Alcohols and Phenols
Alcohols and Phenols
Learning Objectives:
• Nomenclature and Structure of Alcohols
• Reactivity of Alcohols
• Chemistry of Alcohols by preparation of Ethers
• And Esters
• Nomenclature and Structure of Phenols
• Reactivity of Phenol and their chemistry by
• Electrophilic Substitution
• Differentiate between Alcohols and Phenols
Alcohols and Phenols

423. Alcohol
• Structure of alcohol and phenols resembles with water as:
o
• Both; alcohol and phenol contain hydroxyl (-OH) group so they may also be termed as
hydroxy derivatives of alkanes and benzene respectively.
• Alcohols are represented by a general formula ROH where R is an alkyl group which may
be CH3— ,CH3CH2 — , (CH3)2C H — and C6H5 — CH2— ,etc.
Classification:
Depending on the number of –OH groups:
1) Alcohols with one –OH group are called monohydric alcohol.
2) Alcohol with more than one –OH groups are called polyhydric alcohols:
Depending upon Nature of carbon:

424. Nomenclature of Alcohols:
• Common or trivial system:
• Lower and simpler alcohols are usually known by their common or trivial names,
obtained by adding the name of alcohol after the name of the alkyl group to
which the OH group is attached, e.g.,
➢ IUPAC Nomenclature:
• Select the longest chain containing –OH group.
• The carbon chain is numbered, starting from the end where carbon atom
attached with OH group gets the lowest possible number.
• If more than one OH groups are attached, they are indicated by an appropriate
suffix diol, triol, etc.
• The unsaturated alcohols are numbered in such a way that hydroxyl group rather
than the point of unsaturation gets the lower number.
• When hydroxyl group is not a preferred functional group as in hydroxy acids,
aldehydes and ketones, the substituent name hydroxy is used as a prefix to
indicate the position of OH group.
Sample examples:

425. Reactivity of Alcohols:
▪ Alcohols undergo the following two types of reactions:
▪ Reactions in which C — O bond breaks
▪ Reactions in which O — H bond breaks
▪ If a nucleophile attacks, it is the C — O bond which breaks. On the other hand, if an
electrophile attacks on alcohol, it is the O — H bond which breaks.

426. ▪ The order of reactivity of alcohols when C — O bond breaks:
▪ The order of reactivity of alcohols when O — H bond breaks:
Chemistry of alcohol by preparation of ether:
• It proceeds via acid catalyzed condensation of alcohols
• Reagent is H2SO4 and and heat
• Typically limited to symmetrical ethers of primary alcohol
• The substitution involves the (O) of nucleophile of one alcohol attacking the
Electrophilic carbon in other displacing a water molecule.
Mechanism:
There are three key steps.
▪ First of all, one equivalent of alcohol is protonated to its conjugate acid – which has
the good leaving group, OH2 (water, a weak base).
▪ Next, another equivalent of the alcohol can now perform nucleophilic attack at carbon
(SN2), leading to displacement of OH2 (water) and formation of a new C-O bond. This is
an SN2 reaction.
▪ The final step is deprotonation of the product by another equivalent of solvent (or other
weak base), resulting in our ether product.

427. Chemistry of alcohol by preparation Esters:
➢ Alcohols react with organic and inorganic acids to form their respective esters.
➢ Glycerine on reaction with a mixture of HNO3 and H2SO4 to give an ester called
nitroglycerine or gycerltrinitrate. (Explosive liquid, when mixed with fine sand and
moulded into sticks called dynamite).
➢ Esters are also formed by treating acid chlorides with sodium alkoxides.

428. Phenol
Physical properties:
• Phenol is a colourless, crystalline, deliquescent solid with characteristic phenolic odour.
• It has melting point 41°C and boiling point 182°C.
• It is sparingly soluble in water forming pink solution at room temperature but
completely soluble above 68.5°C.
• It is poisonous and used as a disinfectant in hospitals and washrooms.
• It causes blisters on skin.
Structure:
• The alcohol functional group consists of an O atom bonded to an sp2 hybridized
aromatic C-atom and a H-atom via 𝛿 𝑏𝑜𝑛𝑑𝑠.
• Both C-O and O-H bonds are polar due to high electronegativity of the O atom.
• Conjugation exists between an unshared electron pair on oxygen and the aromatic ring.
• This result is, compared to simple alcohols:
o A shorter carbon oxygen bond distance
o A more basic hydroxyl oxygen
o A more acidic hydroxyl proton (-OH)
Nomenclature of phenol:
➢ Phenols are named just like other derivative of benzene. Most of the members of this
family are given special names.
➢ In IUPAC system, -OH group is represented by name hydroxyl. It is used as prefix, while
the benzene part of the molecule is used as suffix.

429. Acidity of phenol:
• Phenol is much more acidic than alcohols but less acidic than carboxylic acids.
• It dissolves readily in alkalies but it is too weak to affect the litmus paper or to evolve
CO2 from carbonates.
• Its dissociation constant (Ka) is 1.3xl0-10.
• Phenol is partially soluble in water and its solution has a pH of around 5 or 6. This makes
phenol different from aliphatic alcohols.
• The reason why phenol is acidic lies in the nature of the phenoxide ion.
• The negative charge on oxygen atom can become involved with the p-electron cloud on
the benzene ring.
• The negative charge is thus delocalized in the ring and the phenoxide ion becomes
relatively stable.
➢ Relative acidic strength of alcohol, phenol, water and carboxylic acid is as follows:

430. Reactions of phenol:
• Phenols are potentially very reactive towards Electrophilic substitution reaction.
• This is because the hydroxyl group, -OH is a strongly activating, ortho para directing
substituent.
➢ Nitration of phenol:
➢ Sulphonation of phenol:
➢ Halogenation of phenol

431. ➢ Alkylation of phenol:
➢ Acylation of phenol:
➢ Nitrosation of phenol:

432. Differentiate between Alcohols and Phenols:
The compounds in which –OH is attached to
an alkyl group.
Alcohols are hydroxyl derivatives of alkanes.
The general formula of alcohol is R-OH.
Lower alcohols are generally colorless
liquids.
Alcohols have a characteristics sweet smell
and burning taste.
They are readily soluble in water but
solubility decreases in higher alcohols.
Alcohols react with other reagents in two
ways, either in which C-O bond breaks or in
which O-H bond breaks.
The compounds in which hydroxyl group is
attached to an aryl group.
Phenols are derivatives of benzene.
The general formula of phenol is C6H5OH. It
is also known as carbolic acid.
They are colorless, crystalline and
deliquescent solids.
Phenolate ions have resonance structures
but alcohols do not have such type
structures.
Phenols
Alcohols

433. Assessment 1
1. Which of the following is not considered as a class of organic compounds which are much
closer to water in structure and considered as derivatives of water?
a. Alcohol
b. Phenol
c. Ether
d. Ester
2. The chemical formula of benzyl alcohol is:
a. C6H5-OH
b. C6H5-CH=CH-OH
c. C6H5-CH2-OH
d. C2H5-OH
3. Glycerol is one of the examples of which type of alcohol?
a. Monohydric alcohol
b. Primary alcohol
c. Dihydric alcohol
d. Trihydric alcohol
4. Cyclohexanol has a chemical formula C6H11-OH and is an example of:
a. Primary alcohol
b. Secondary alcohol
c. Tertiary alcohol
d. Phenol
5. Absolute alcohol (C2H5OH) is the term generally given to:
a. 100% pure ethanol
b. 95% alcohol + 5% H2O
c. Ethanol + water + phenol
d. 95% ethanol and 5% methanol
6. Which of the following does not correspond to the solubility of alcohol?
a. They are readily soluble in water.
b. Solubility increases in higher alcohols
c. The solubility of alcohols is due to hydrogen bonding which is prominent in lower
alcohols
d. None of these

434. 7. Ethyl alcohol is liquid but ethane is a gas. The more boiling point of ethyl alcohol than ethane
is due to:
a. Presence of C-C bonds
b. Presence of hydrogen bonding
c. More molar mass of ethanol
d. All of these
8. The oxidation of compound (A) bearing molecular formula C3H8O which on oxidation produces
C3H6O (ketone). The compound (A) is:
a. Secondary alcohol
b. Alkene
c. Aldehyde
d. Tertiary alcohol
9. Which of the following on reaction with ethanol produces hydrogen gas?
a. Ether
b. ThO2
c. SOCl2
d. Na
10. Esterification which produces ethyl acetate and water in presence of mineral acid. It involves
the reaction of:
a. Ethanol with Grignard’s reagent
b. Ethanol with SO2
c. Ethanol with acetic acid
d. Ethanol with water

435. Assessment 2
1. Which of the following alcohols are resistant to oxidation?
a. Primary alcohol
b. Secondary alcohol
c. Tertiary alcohol
d. All of these
2. The alcohol that produces turbidity immediately with ZnCl2 + conc. HCl at room temperature:
a. 1-hydroxybutane
b. 2-hydroxybutane
c. 2-hydroxy-2-methylpropane
d. 1-hydroxy-2-methylpropane
3. An unknown compound ‘X’ on oxidation first oxidizes to aldehyde and then acetic acid by acid
dichromate. The unknown compound ‘X’ may be labeled as:
a. CH3CHO
b. CH3CH2OH
c. CH3CH2CH3
d. CH3-O-CH3
4. Ethanol on reaction with con. H2SO4 at different temperatures give the:
a. Ethene and Diethyl ether
a. Ethene and ethane
b. Diethyl ether
c. Ethene only
5. The dehydration reaction of ethanol with conc. sulphuric acid does not produces:
a. Ethene
b. Diethyl ether
c. Acetylene
d. Ethyl hydrogen sulphate
6. Which of the following gives oily layer on heating by reacting with mineral acid in presence of
ZnCl2?
a. Secondary
b. Tertiary
c. Primary
d. All equal

436. 7. The quantity of hydrogen gas liberated on reaction of 23 g of sodium with 1 mole of ethanol.
a. 1 mole
b. 2.016g
c. 1.008g
8. Which of the reagent can be used to convert acetic acid into primary alcohol?
a. LiAlH4 + Ether
b. Na + Alcohol
c. H2 + Pt
d. Sn + HCl
9. Which of the following test involves the identification of ethanol from methanol?
a. Lucas test
b. Bromine water test
c. Iodoform test
d. Beyer’s test
10. The chemical formula of Picric acid is:
a. Trinitroaniline
b. Trinitrotoluene
c. A volatile liquid
d. 2, 4, 6 trinitrophenol

437. Assessment 3
1. Which of the following aromatic hydrocarbon is called Carbolic acid:
a. Phenol
b. Phenyl benzoate
c. Phenyl acetate
d. Benzene
2. It is sparingly soluble in water forming pink solution at room temperature but completely
soluble above
a. 34 0C
b. 56 oC
c. 68.5 0c
d. 100 oC
3. Reaction of phenol with dil. HNO3 gives:
a. p and m-nitrophenols
b. o-and p-nitrophenols
c. Picric acid
d. o-and m-nitrophenols
4. Phenol on treatment with bromine water and shaken form the white precipitate of:
a. m-bromophenol
b. 2, 4-dibromophenol
c. 2, 4, 6-tribromophenol
d. A mixture of o-and p-bromophenols
5. The dissociation constant (Ka) for phenol is:
a. 1.3x1010
b. 1.3x10-10
c. 1.3
d. 1.3x10-5
6. Phenol (C6H5OH) at 25 oC exist in the form:
a. White crystalline solid
b. Transparent liquid
c. Pink Solution
d. Brown gas

438. 7. Which of the following not gives effervescence with NaHCO3
a. Phenol
b. Benzoic acid
c. Hydrocloric acid
d. None of these
8. Phenol is partially soluble in water and its solution has a pH of around:
a. 3-4
b. 2-3
c. 4-5
d. 5-6
9. The reason why phenol is acidic lies in the fact that:
a. Negative charge on oxygen atom
b. Stability of phenol
c. Stability of phenoxide due to delocalization of positive charge
d. Stability of phenoxide due to delocalizaiaton of negative charge
10. Relative acidic strength of alcohol, phenol, water and carboxylic acid is as follows:
a. Carboxylic acid > 𝑤𝑎𝑡𝑒𝑟 > 𝑝ℎ𝑒𝑛𝑜𝑙 > 𝑎𝑙𝑐𝑜ℎ𝑜𝑙
b. Carboxylic acid > 𝑝ℎ𝑒𝑛𝑜𝑙 > 𝑤𝑎𝑡𝑒𝑟 > 𝑎𝑙𝑐𝑜ℎ𝑜𝑙
c. Alcohol > 𝑤𝑎𝑡𝑒𝑟 > 𝑝ℎ𝑒𝑛𝑜𝑙 > 𝐶𝑎𝑟𝑏𝑜𝑥𝑦𝑙𝑖𝑐 𝑎𝑐𝑖𝑑
d. Water > 𝐶𝑎𝑟𝑏𝑜𝑥𝑦𝑙𝑖𝑐 𝑎𝑐𝑖𝑑 > 𝑝ℎ𝑒𝑛𝑜𝑙 > 𝑎𝑙𝑐𝑜ℎ𝑜𝑙

439. Assessment 4
1. In order to prepare phenyl acetate (ester) using sodium hydroxide, phenol is made to react
with:
a. Acetyl chloride
b. Acetic acid
c. Ethanol
d. Ammonium chloride
2. Phenol on distilling with zinc dust produces:
a. Cyclohexanol
b. Cyclohexane
c. Benzene
d. Benzyl alcohol
3. Which of the following polymer is formed by the condensation of phenol and formaldehyde?
a. Polyvinyl chloride
b. Bakelite
c. Styrene
d. Polyethylene
4. Which of the following method is used to separate 1 : 1 mixture of ortho and para
nitrophenols :
a. Distillation
b. Sublimation
c. Crystallization
d. Chromatography
5. Which of the following is the characteristic of alcohols?
a. These are soluble in water.
b. These are sweet in smell.
c. They have burning sensation in taste.
d. All of these
6. The major product that is formed by the reaction of toluene with chlorine in the presence of
FeCl3:
a. O- and p-chlorotoluene
b. meta chloro toluene
c. benzyl chloride
d. All of these

440. 7. Phenol on reduction with H2 in the presence of Ni catalyst gives
a. benzene
b. toluene
c. cyclohexane
d. Cyclohexanol
8. Which of the following is more reactive when O-H bond breaks?
a. Tertiary alcohol
b. Secondary alcohol
c. Primary alcohol
d. Methanol
9. What is the IUPAC name of the following compound?
a. 2-buten-4-ol
b. 2-buten-1-ol
c. 1-butenol
d. 2-butenol
10. Which of the following is more reactive alcohol when C-O bond breaks:
a. Methanol
b. Primary alcohol
c. Secondary alcohol
d. Tertiary alcohol

441. KEY
Assessment 1
Assessment 2
1. D
2. C
3. D
4. B
5. A
6. B
7. B
8. A
9. D
10. C
1. C
2. C
3. B
4. A
5. C
6. C
7. C
8. A
9. C
10 D

442. KEY
Assessment 3
Assessment 4
1. A
2. C
3. B
4. C
5. B
6. A
7. A
8. D
9. D
10. B
1. A
2. C
3. B
4. A
5. D
6. A
7. D
8. D
9. B
10. D

443. Oldest, Largest and Most Credible Platform
MDCAT
Chemistry
Quick Practice Book
www.nearpeer.org

444. Aldehydes and Ketones
Learning Objectives:
• Nomenclature and structure of aldehydes and
ketones.
• Preparation of aldehydes and ketones.
• Reactivity of aldehydes and ketones and their
comparison.
• Chemistry of aldehydes and ketones by their
reduction to alcohols
• Oxidation reactions of aldehydes and ketones.

445. Aldehydes and Ketones
Aldehydes:
Common names:
• The common names of aldehydes are obtained from the common names of
carboxylic acids containing the same number of carbon atoms.
• The ending -ic acid in the common name of the acid is replaced by the word
aldehyde.
• The positions of other groups on the chain are indicated by Greek letters
(𝛼, 𝛽, 𝛾, 𝛿). Lettering starts on the carbon adjacent to the carbonyl group.
IUPAC Names:
• The letter - e in the name of the alkane is replaced with al.
• The positions of other groups on the chain are indicated by using numbers.
• Numbering starts from the carbonyl carbon.
• Aromatic aldehydes are not given IUPAC names.

446. Ketone:
Common names:
• The common names of ketones are obtained by separately writing the names of the
alkyl groups attached to the carbonyl carbon.
• The word ketone is then added as a separate word.
• The names of the alkyl groups are written alphabetically.
• When the two alkyl groups are the same, the prefix di - is added before the name of the
alkyl group.
• The positions of other groups are indicated by Greek letters, the 𝛼 - carbon atom being
the one adjacent to the carbonyl group.
• If the two alkyl groups in a ketone are the same, the ketone is said to be symmetrical, if
unlike, unsymmetrical.
IUPAC Names:
• The IUPAC names of ketones are derived from the names of alkanes having the
same number of carbon atoms.
• The letter e in the name of alkane is replaced with the suffix -one.
• The positions of the carbonyl group and of other groups on the chain are
indicated by numbers.
• Numbering is started from that end which is nearest to the carbonyl group.
• Aromatic ketones are not given IUPAC names.

447. Preparation Of Aldehyde And Ketone:
Preparation of formalin:
Laboratory method:
• Passing a mixture of methyl alcohol vapours and air over platinised asbestos or
copper or silver catalyst at 300 0C.
• Methyl alcohol is oxidized to gaseous formaldehyde which is absorbed in water.
The resulting mixture is called formalin.
• Formalin is a mixture of 40 % formaldehyde, 8% methyl alcohol and 52 % water.
Industrial method:
• Formaldehyde is manufactured by passing a mixture of methanol vapours and air
over iron oxide-molybdenum oxide or silver catalyst at 500 °C.

448. Preparation of acetaldehyde:
Laboratory preparation:
• Acetaldehyde is prepared in the laboratory by the oxidation of ethyl alcohol with
acidified sodium dichromate solution.
• A mixture of ethyl alcohol and sodium dichromate solution is run into boiling
dilute sulphuric acid. Immediately a vigorous reaction takes place and the
acetaldehyde formed in liquid state is immediately distilled off.
• This prevents the oxidation of acetaldyde to acetic acid.
• Ethyl alcohol remains in solution until it is oxidized.
• Pure acetaldehyde is obtained by redistillation.
Acetaldehyde can also be prepared by the dry distillation of a mixture of calcium salts of
formic acid and acetic acid.

449. Industrial method:
• Acetaldehyde is prepared industrially by air oxidation of ethylene.
• Using palladium chloride catalyst with a cupric chloride promoter.
Preparation of acetone:
Acetone is prepared by dry distillation of calcium acetate.
Reactivity of carbonyl compounds:
• The carbonyl group has a 𝜎-bond and a 𝜋-bond. Thus it can undergo addition reactions.
• Most reagents react with the carbonyl group by adding to it.
• As oxygen is more electronegative, it tends to attract the 𝜋 electrons to it. This
attraction makes the carbonyl group a polar group.
• The oxygen atom has a partial negative charge on it and is nucleophilic, whereas the
carbon atom has a partial positive charge and is Electrophilic.
Nucleophilic addition reactions:
• As a result of the unsymmetrical electronic distribution about the carbonyl group,
the nucleophilic reagent can start the initial attack on the carbon.

450. • It appears that whether the initial attack is to be by a nucleophilic reagent or by an
electrophilic reagent depends upon a particular reaction and upon the conditions under
which that reaction is carried out.
• In these reactions of aldehydes and ketones, the negative part of the reagent combines
with the electrophilic carbon of the carbonyl group, whereas the positive part, which is
usually hydrogen goes to the oxygen.
• The nucleophilic addition reactions of carbonyl group are catalysed by bases or acids.
• Remember that whether the addition is base-catalysed or acid-catalysed, the adduct is
the same.
• A base catalyst increases the nucleophilic character of the reagent, while an acid-
catalyst promotes the nucleophilic attack by increasing the positive character
(electrophilic character) of the carbonyl carbon atom.
Nucleophilic addition reaction:
Base catalyzed addition reaction:
• A base-catalysed nucleophilic addition reaction will take place with a strong
nucleophilic reagent.
• The base reacts with the reagent and generates the nucleophile.
• The addition is initiated by the attack of a nucleophile on the electrophilic carbon of
the carbonyl group.
Addition of HCN:
• Hydrogen cyanide adds to aldehydes and ketones to form cyanohydrins.
• The reaction is carried out by adding slowly a mineral acid to an aqueous solution of
sodium cyanide.
• The acid generates HCN from sodium cyanide in situ.

451. The cyano group, — C ≡ N is hydrolysed by an aqueous acid into a carboxylic acid through
an acid amide.
• The reaction is used in the synthesis of 𝛼-hydroxy acids that contain one carbon
atom more than the number of carbon atoms in the starting aldehydes or ketones.
• Hydrogen cyanide itself is not very nucleophilic and does not ionize to from cyanide
to a significant extent.
• Thu,a source of cyanide ion such as NaCN or KCN is used.
• The hydroxide ion liberated in the formation of cyanohydrin reacts with
undissociated hydrogen cyanide and produces more cyanide ions, which in turn
react with more carbonyl compound.
Addition of Grignard’s reagent:
• Grignard reagents add to aldehydes and ketones to form adducts which on
hydrolysis with a dilute mineral acid (HCI, H2SO4) give alcohols.

452. Addition of sodium bisulphite:
Aldehydes and small methyl ketones react with a saturated aqueous solution of sodium
bisulphite to form a crystalline white precipitate of sodium bisulphite adduct.
Bisulphite on heating with a dilute mineral acid (HCl or H2SO4), regenerates the parent
aldehyde or ketone.

453. The reaction is used for the separation and purification of carbonyl compounds from non-
carbonyl compounds such as alcohols.
Mechanism:
Sodium bisulphite ionises to form sulphite ions.
The sulphite ion acts as a nucleophile, since the sulphur atom is more nucleophilic than
oxygen, a C—S bond is formed.
Proton is attached to the negatively charged oxygen atom to form bisulphite addition
product.
Ketones in which both alkyl groups are larger than methyl do not react with sodium
bisulphite.
Condensation reactions:
The reactions, in which two molecules of the same or different compounds combine to form
a new compound with or without the elimination of a small molecule like H2O or NH3, are
called condensation reactions.
Aldol condensation:
• Aldehydes and ketones possessing 𝛼-hydrogen atoms react with a cold dilute
solution of an alkali to form addition products known as aldols.
• The name ‘aldol’ is given to the product because it contains both aldehyde and
alcohol functional groups.
• Two molecules of the same carbonyl compound condense to form an aldol.

454. • The aldol compound readily loses water on heating in the presence of dilute acid to
form an unsaturated carbonyl compound.
• A carbon-carbon double bond is formed between the 𝛼- and 𝛽- carbon atoms.
Mechanism of aldol condensation:

455. Cannizaro’s reaction:
• Aldehydes that have no 𝛼-hydrogen atoms undergo Cannizzaro’s reaction.
• It is a disproportionation (self oxidation-reduction) reaction.
• Two molecules of the aldehyde are involved, one molecule being converted into the
corresponding alcohol (the reduced product) and the other into the acid in the salt
form (the oxidation product).
• The reaction is carried out with 50 percent aqueous solution of sodium hydroxide at
room temperature.
Mechanism:
The hydroxide ion acts as a nucleophile. It attacks on the electrophilic carbonyl carbon to
form a complex anion.
The anion transfers a hydride ion to second molecule of formaldehyde.

456. The methoxide ion acts as a base and abstracts a proton from formic acid to form methanol
and formate ion.
The formate ion in the presence of alkali gives a salt of the acid.
Cannizaro’s reaction by benzaldehyde:
Haloform reaction:
It is given by:
• Acetaldehyde
• Methyl ketones
• Ethanol (only primary alcohol)
• Secondary alcohols containing the hydroxyl group on the second carbon
atom.
➢ The term haloform is used for the reaction because a haloform (chloroform,
bromoform or iodoform) is one of the products.
➢ From a synthetic point of view the haloform reaction affords a convenient method
for converting a methyl ketone to a carboxylic acid containing one carbon atom less
than the parent compound.

457. Iodoform test:
➢ The haloform reaction using iodine and aqueous sodium hydroxide is called the
iodoform test.
➢ It results in the formation of water insoluble iodoform which is a yellow solid.
➢ Iodoform test is used for distinguishing methyl ketones from other ketones. It is also
used to distinguish ethanol from methanol and other primary alcohols.
➢ It can be used to distinguish acetaldehyde from other aldehydes.
Acid catalyzed reactions:
➢ The acid catalysed nucleophilic addition reaction will take place with a weak nucleophilic
reagent.
➢ The addition is initiated by the proton (H+) liberated by the acid.
➢ The proton combines with the carbonyl oxygen atom and increases the electrophilic
character of the carbonyl carbon.
➢ As a result, the attack of the weaker nucleophile on the electrophilic carbon becomes
easier.
General mechanism:

458. Acid catalyzed reactions:
Polymerization of Aldehyde:
Both formaldehyde and acetaldehyde polymerize in the presence of dil. H2SO4 to give
metaformaldehyde and paraldehyde respectively.
Reactions of Ammonia Derivatives:
➢ The reaction is known as condensation reaction or addition - elimination reaction
because water is lost after addition occurs.
➢ Compounds containing the group, and water are formed.
➢ General reaction is given as:

459. ➢ Some commonly used ammonia derivatives are hydroxylamine, NH2OH, hydrazine,
NH2NH2, phenylhydrazine, C6H5NHNH2, semicarbazide, NH2NHCONH2, and 2,4-
dinitrophenylhydrazine, NH2NHC6H3(NO2)2.
Reaction with hydroxylamine (H2NOH):
Aldehyde or ketone + hydroxylamine
𝐻+
→ Oxime
Reaction with phenyl hydrazine (H2N-NH2):
Aldehyde/ketone + Hydrazine
𝐻+
→ Hydrazones

460. Reaction with Phenyl hydrazines (C6H5-NH-NH2)
Aldehyde/ketone + Hydrazine
𝐻+
→ Phenyl hydrazones
Reaction with 2,4-Dinitrophenylhydrzine (2,4-DNPH):
Aldehyde/ketone + 2,4-DNPH
𝐻+
→ 2,4-dinitrophenylhydrazones
➢ It is identification test for Aldehyde and ketone because 2,4-dinitrophenylhydrazones are
usually yellow or orange crystalline solids.
Addition of alcohols:
The hydrogen chloride gas acts as a catalyst.
Both the alcohol and the hydrogen chloride gas must be dry.

461. ➢ The reaction may be used to protect the aldehyde group against alkaline oxidising
agents.
➢ To regenerate aldehyde, the acetal is hydrolysed in the presence of an acid.
➢ Ketones do not react under these conditions.
Reduction of Aldehyde and ketone to alcohols:
• Aldehydes are reduced to primary alcohols whereas ketones to secondary alcohols.
• The carbonyl group is converted into an alcohol.
• The reducing agents LiAlH4 and NaBH4 act as source of H-(hydride ion).
• Overall 2 H atoms are added across the to give H-C-O-H.
• Hydride reacts with carbonyl group in Aldehyde/ketone to give alcohols.
• Sodium borohydride reduces the carbon-oxygen double bond but not the carbon-
carbon multiple bond.

462. Mechanism:
➢ The tetrahydridoborate (III) ion, BH4- is source of hydride ion, H-.The hydride ion acts as
a nucleophile.
➢ It attacks on the electrophilic carbon of the carbonyl group to give an alkoxide ion.
➢ The alkoxide ion is protonated with water to give an alcohol.
Catalytic reduction:
➢ Aldehydes and ketones on reduction with hydrogen in the presence of a metal catalyst
like Pd, Pt or Ni form primary and secondary alcohols respectively.
Oxidation of Aldehyde:
• Aldehydes are easily oxidised by mild oxidising agents like Tollen's reagent, Fehling's
solution and Benedict’s solution.
• They are oxidised to carboxylic acids by strong oxidising agents such as K2Cr2O7 / H2SO4,
KMnO4 / H2SO4, and dilute nitric acid.
• The hydrogen atom attached to the carbonyl group in aldehydes is oxidised to OH
group.

463. ➢ The carboxylic acid has the same number of carbon atoms as are present in the parent
aldehyde.
Oxidation of Ketones:
• Ketones do not undergo oxidation easily because they require breaking of strong
carbon - carbon bond.
• They give no reaction with mild oxidising agents. They are only oxidised by
strong oxidising agents such as K2Cr2O7/ H2SO4 , KMnO4 / H2SO4, and conc.
HNO3.
• In oxidation of ketones, only the carbon atoms adjacent to the carbonyl group
are attacked.
• The carbon atom joined to the smaller number of hydrogen atoms is
preferentially oxidized.
• In case of symmetrical ketones only one carbon atom adjacent to the carbonyl
group is oxidised and a mixture of two carboxylic acids is always obtained.
• However, in case of unsymmetrical ketones, the carbon atom joined to the smaller
number of hydrogen atoms is preferentially oxidized and the carbonyl group remains
with the smaller alkyl group.
Identification of carbonyl compounds
Detection tests for aldehydes and Ketones.
2,4 DNPH Test:
Aldehydes and ketones form a yellow or red precipitate with 2,4 dinitrophenylhdrazine
solution.

464. Sodium Bisulphite Test:
Aldehydes and small methyl ketones form a crystalline white precipitate with saturated
sodium bisulphite solution.
Tollen's Test [Silver Mirror Test]:
• Aldehydes form silver mirror with Tollen’s reagent (ammoniacal silver nitrate solution).
• Add Tollen’s reagent to an aldehyde solution in a test tube and warm.
• A silver mirror is formed on the inside of the test tube.
• High quality mirrors are manufactured by using this principle.
• Ketones do not give this test.
Fehling’s Solution Test [an alkaline solution containing a cupric tartrate complex ion]:
• Aliphatic aldehydes form a brick-red precipitate with Fehling’s solution.
• To an aldehyde solution, add Fehling’s solution and boil.
• A brick red precipitate of cuprous oxide is formed.
• Ketones do not give this test.
Benedict's Solution Test |an alkaline solution containing a cupric
citrate complex ion]:
• Aliphatic aldehydes form a brick-red precipitate with Benedicts's solution.
• To an aldehyde solution, add Benedict's solution and boil.
• A brick-red precipitate of cuprous oxide is formed.
• Ketones do not give this test.
Sodium Nitroprusside Test:
• Ketones produce a wine red or orange red colour on adding alkaline sodium
nitroprusside solution dropwise.
• Aldehydes do not give this test.

465. • Assessment 1
1. In aldehydes and ketones, carbon of carbonyl group undergoes the hybridization of type:
a. sp3
b. sp2
c. sp
d. sp3 and sp
2. Acetone may have the same molecular formula but different structural formula as that of:
a. Propanal
b. Diethyl ether
c. propanol
d. Propionic acid
3. What is the name of the compound if two valencies of carbon of carbonyl group are satisfied
by two alkyl groups:
a. Aldehyde
b. Ketone
c. Acid
d. Acid chloride
4. Which of the following is an example of unsymmetrical ketone:
a. Pentanone
b. Acetophenone
c. Benzophenone
d. All of these
5. Which of the following statement is true about acetone and acetaldehyde?
a. These are Positional isomers to each other.
b. These are Functional group isomers to each other.
c. These are carbonyl compounds but are not isomers to each other.
d. These are carbonyl compounds but are Chain isomers to each other.
6. Ketones can be prepared by hydration of alkynes in the presence of:
a. Water
b. Acid
c. Ammonia
d. Acetone

466. •
7. Laboratory preparation of formaldehyde involves the passage of mixture of ______ over Pt-
asbestos at 300 oC.
a. Ethyl alcohol (vapors) and air
b. Methyl alcohol (liquid) and air
c. Methyl alcohol (vapors) and air
d. Rectified spirit only.
8. The catalytic oxidation of methyl alcohol is carried out when a mixture of methyl alcohol
vapours and air at 300 oC is passed over:
a. Nickel
b. Cr2O3+SiO2
c. Pt-asbestos
d. Al2O3
9. Methyl alcohol is oxidized to gaseous formaldehyde which on absorption in water produces
the mixture called:
a. Formaldehyde
b. Methanol
c. Methylene
d. Formalin
10. The accurate percentage composition of formalin is given as:
a. 52 % water,40% formaldehyde and 8% methyl alcohol
b. 52 % formaldehyde,40% water and 8% methyl alcohol
c. 52 % methyl alcohol ,40% formaldehyde and 8% water
d. 52 % water,40% methyl alcohol and 8% formaldehyde

467. •
Assessment 2
1. When a mixture of C2H5OH and Na2Cr2O7 solution is run into boiling H2SO4(dil), a vigorous
reaction is carried out in which the product (X) is immediately distilled off. The product (X) is:
a. Acetic acid
b. Acetaldehyde
c. Ethanol
d. Water
2. Acetaldehyde can also be prepared by the dry distillation of a mixture of:
a. Formic acid and acetic acid
b. Calcium salt of formic acid only
c. Calcium salt of formic acid and acetic acid
d. Ethyl alcohol
3. Dry distillation of calcium acetate results in the preparation of:
a. Acetaldehyde
b. Acetone
c. Ethanol
d. Formalin
4. The carbonyl group has a sigma bond and a pi bond. Thus it can undergo:
a. Electrophilic substitution reaction
b. Nucleophilic substitution reaction
c. Nucleophilic addition reaction
d. All of these
5. In > 𝐶 = 𝑂 , which of the following is Nucleophilic in nature?
a. Carbon
b. Alkyl
c. Hydrogen
d. Oxygen
6. Hydrogen cyanide when adds to aldehydes and ketones form the final product:
a. Sodium cyanide
b. Hydrogen cyanide
c. Cyanohydrin
d. Hydrogen

468. •
7. Acetaldehyde undergoes Nucleophilic addition reaction to produce acetaldehyde cyanohydrin
which on acid hydrolysis produces which of the following, the final product:
a. Acetone
b. Acid chloride
c. Ether
d. Lactic acid
8. The compound (X) forms the adduct (Y) on addition with Grignard’s reagent. The adduct (Y) on
hydrolysis with a dilute mineral acid give primary alcohol. The compound (X) is:
a. CH3COCH3
b. CH3CHO
c. NH3
d. HCHO
9. Which of the following carbonyl compounds do not give crystalline white ppt. with a saturated
aqueous solution of sodium bisulphite?
a. CH3CHO
b. (CH3-CH2-CH2)2CO
c. CH3COCH3
d. None of these
10. Which of the following aldehyde or ketone undergo aldol condensation reaction?
a. HCHO
b. C6H5CHO
c. CH3CHO
d. All of these

469. • Assessment 3
1. In aldol condensation, the hydroxide ion acts as a base which removes
a. proton from any carbon of carbonyl compound
b. proton from α carbon of carbonyl compound
c. proton from β carbon of carbonyl compound
d. H- from α carbon of carbonyl compound
2. Which of the following is dispropotionation reaction of aldehyde?
a. Nucleophilic addition of Sodium bisulphite
b. Cannizaro’s reaction
c. Aldol condensation
d. Iodoform reaction
3. In Cannizaro’s reaction, OH- attacks on carbonyl carbon of reactant molecule to form complex
anion which transfers _____ to second reactant molecule.
a. H+
b. H
c. H-
d. O2-
4. Iodoform is the test which is applied to distinguish:
a. Methyl ketones from other ketones
b. Methanol from other Ethanol
c. Acetaldehyde from other aldehyde
d. All of these
5. In acid catalyzed Nucleophilic addition reaction, proton combines with the carbonyl oxygen
atom and increases the Electrophilic character of:
a. Oxygen atom
b. Hydrogen atom
c. Carbonyl carbon
d. Carbon of alkyl group
6. The polymerization of formaldehyde in the presence of dilute H2SO4:
a. Metaformaldehyde
b. Paraldehyde
c. Oxime
d. None of these

470. •
Assessment 1
1. Positive rays are also termed as canal rays which are produce by
a. By combustion of gas.
b. By cooling of the gas.
c. By the ionization of gas by cathode rays.
d. Anode electrode as cathode rays is produced.
2. Positive rays are also termed as canal rays. These rays give flash on screen coated with
a. AgCl
b. ZnO
c. AgNO3
d. ZnS
3. Positive rays are also termed as canal rays or anode rays. Which of the following is not true:
a. Their e/m ratio is constant
b. They are deflected by electrical and magnetic field
c. They are produced by ionization of molecules of the residual gas
d. Their e/m ratio depends on nature of residual gas
4. e/m for positive rays changes with the change in gas to be filled in glass discharge tube. This
value is maximum for which gas
a. Helium
b. Helium
c. Oxygen
d. Hydrogen
5. The canal ray with one proton bears the following charge:
a. -1.602x10-19C/kg
b. -1.602x10-19C
c. +1.602x10-19C/mol
d. +1.602x10-19C
6. Max planks proposed the quantum theory in 1900 to explain the emission and absorption of
radiation. Which is true among these
a. Energy travels in continuous form
b. Energy is emitted or absorbed continuously
c. Energy is not emitted or absorbed continuously
d. In case of light, energy packet is called quanta
7. The acid catalysed reaction of CH3CHO with hydroxylamine produces:
a. Ethanoate
b. Ethanaloxime
c. Acetaldehyde hydrazone
d. Ethanoic acid
8. The final product in the given reaction is:
CH3CHO + NH2NHC6H5
𝐻+
→
a. Ethanaloxime
b. Phenylhydrazine
c. Phenylhydrazone
d. Paraldehyde
9. 1,1-diethoxyethane (acetal) is formed by the reaction of acetaldehyde in presence of an acid
with:
a. Acetone
b. Ethanoic acid
c. Grignard’sreagent
d. Ethanol
10. The tetrahydrideoborate (III) ion, BH4 is a source of:
a. Proton(H+)
b. Hydride (H-)
c. Hydroxyl (OH-)
d. Oxide (O2-)

471. •
Assessment 4
1. Which of the following reagent oxidize the Aldehyde to carboxylic acid?
a. Tollen’s reagent
b. Fehling’s solution
c. K2Cr2O7/H2SO4
d. All of these
2. Ketones do not give reaction with the following oxidizing agent?
a. K2Cr2O7/H2SO4
b. KMnO4/H2SO4
c. Conc. HNO3
d. Fehling’s solution
3. Which of the following statement is considered incorrect about the oxidation of ketones?
a. In oxidation of ketones, only the carbon atoms adjacent to the carbonyl group are
attacked.
b. The carbon atom joined to the larger number of hydrogen atoms is preferentially
oxidised.
c. In case of symmetrical ketones only one carbon atom adjacent to the carbonyl group is
oxidised and a mixture of two carboxylic acids is always obtained.
d. None of these
4. Aliphatic Aldehydes form a brick red precipitate with Benedict’s solution. Which of the
following form the brick red precipitate?
a. Ag2O
b. Cu2O
c. Ag
d. Cu(OH)2
5. The compound which gives brick red precipitate with Fehling’s solution:
a. Benzaldehyde
b. Acetone
c. Butanone
d. Acetaldehyde
6. Tollen’s reagent used in silver mirror test to detect the aldehydes is:
a. 2,4-dinitrophenyl hydrazine
b. Alkaline solution containing cupric citrate complex ion
c. Ammonical silver nitrate solution
d. Alkaline solution containing cupric tartrate complex ion

472. •
7. Which of the following test cannot be used to identify Aldehyde?
a. Fehling’s solution test
b. Benedict’s solution test
c. Sodium nitroprusside test
d. Tollen reagent test
8. The oxidation of butanone by mild oxidizing agent(Fehling’s solution) produces:
a. Acetic acid only
b. Formic acid and acetic acid
c. Formic acid only
d. Does not oxidize
9. The possible product in the oxidation of acetone by strong oxidizing agent (K2Cr2O7/H2SO4) is:
a. Acetic acid and formic acid
b. Acetic acid only
c. Formic acid only
d. Ethanol and formic acid
10. The following reactions are given by carbonyl compounds (aldehydes and ketones):
a. Oxidation
b. Polymerization
c. Nucleophilic addition
d. All of these

473. •
Key
Assessment 1
Assessment 2
1. b
2. a
3. b
4. b
5. c
6. b
7. c
8. c
9. d
10. a
1. b
2. c
3. b
4. c
5. d
6. c
7. d
8. d
9. b
10. c

474. •
Key
Assessment 3
Assessment 4
1. b
2. b
3. c
4. d
5. c
6. a
7. b
8. c
9. d
10. b
1. d
2. d
3. b
4. b
5. d
6. c
7. c
8. d
9. a
10. d

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476. Learning Objectives:
Nomenclature
Classification
Physical Properties
Preparation
Reactivity
Carboxylic Acid

477. Nomenclature of Carboxylic acid:
The aliphatic monocarboxylic acids are commonly called fatty acids because higher
members of this series such as palmitic acid (C15H31COOH) stearic acid (C17H35COOH),
etc. are obtained by the hydrolysis of fats and oils.
Common names:
▪ The common names of carboxylic acids were derived from the source from which they
are isolated.
▪ The irritation caused by an ant bite is due to formic acid (Latin word formica, ant).
▪ It was first isolated by the distillation of red ants.
▪ Acetic acid was first isolated from vinegar and butyric acid was named after butyrum
means butter.
▪ The positions of other groups attached with the chain containing the carboxyl group are
indicated by the Greek letters 𝛼, 𝛽, 𝛾, 𝛿 etc.
▪ The carbon adjacent to carboxylic group is 𝛼 carbon.
The IUPAC Nomenclature:
▪ The IUPAC names of saturated monocarboxylic acids are alkanoic acids.
▪ These are derived from the names of the alkanes containing the same number of
carbon atoms as the acid.
▪ The ending “e” of the alkane name is dropped and suffix-oic acid is added.
▪ Thus acetic acid gets the name ethanoic acid.

478. Trimethyl acetic acid n-valeric acid (Pentanoic acid)
Oxalic acid Malonic acid
(Ethanedioic acid) (propanedioic acid)
Succinic acid (Butandioic acid) Glutaric acid (pentandioic acid)
Adipic acid (Hexandioic acid)

479. Benzoic acid Phthalic acid (1,2-benzendioic acid)
Physical properties:
▪ The polar nature of both the O-H and C=O bonds results in the formation of strong
hydrogen bonds with other carboxylic acid molecule or other hydrogen bonding
systems.
▪ The first three aliphatic acids i.e. formic acid, acetic acid and propionic acid are
colourless liquids and have pungent smell. The next three acids C4 to C6 are colourless
liquids with somewhat unpleasant smell.
▪ Among the aliphatic acids, the first four members are very soluble in water due to
hydrogen bonding.
▪ The boiling points of carboxylic acids are relatively high due to intermolecular hydrogen
bonding.
▪ The molecular mass determination in non-polar solvent like benzene shows that
Carboxylic acids exist as cyclic dimers.
▪ It has been observed that the melting points of carboxylic acids containing even number
of carbon atoms are higher than the next lower and higher members containing odd
number of carbon atoms

480. Preparation of carboxylic acid:
From Primary Alcohols and Aldehydes:
▪ Primary alcohol/Aldehyde
𝑂𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛
→ carboxylic acid
▪ K2Cr2O7/H2SO4 is an oxidizing agent.
▪ Aldehydes are easily oxidised to corresponding carboxylic acids even by mild oxidizing
agents such as Tollen’s Reagent (Ammonical silver nitrate).
From alkyl nitrile:
▪ Compounds having a cyanide ( - C ≡ N) group are called nitriles.
▪ Hydrolysis of an alkanenitrile on boiling with mineral acids or alkalis yields
corresponding carboxylic acid.
▪ Alkanenitriles can be prepared by treating alkyl halide with alcoholic potassium cyanide.

481. ▪ It may be noted that acid produced has one carbon atom more than the original alkyl
halide.
From Grignard’s reagent:
▪ Carboxylic acids can be prepared by the action of Grignard reagent with carbon dioxide.
▪ This reaction is either carried out by passing carbon dioxide through the ethereal
solution of corresponding Grignard reagent or by adding Grignard reagent to crushed
dry ice suspended in ether.
▪ The addition product on reaction with a mineral acid produces carboxylic acid.
By the hydrolysis of ester:
▪ The appropriate ester on boiling with concentrated sodium hydroxide yields sodium salt
of the acid.
▪ This resulting salt when treated with dilute HCl gives the free carboxylic acid.

482. By the oxidative cleavage of alkenes:
▪ Alkenes when heated with alkaline KMnO4 are cleaved at the double bond to form
carboxylic acids.
Reactivity of carbonyl group:
Carboxylic acids undergo the following type of reactions:
▪ The reaction in which hydrogen atom of the carboxyl group is involved.
▪ The reaction in which OH group is replaced by another group.
▪ The reactions involving carboxyl group as a whole.
Reactions involving H atom of carboxyl group:
➢ Carboxylic acids are weaker acids than mineral acids. They furnish H+ when dissolved in
water.
➢ In the presence of water (H2O), the proton breaks away as H3O+ ion.
Reactions with Bases:
➢ Carboxylic acid + Bases ⟶ Salts and water
Reactions with carbonates and bicarbonates:
➢ Carboxylic acid + Carbonates/bicarbonates ⟶ CO2 evolved with effervescence

483. Reaction with metals:
➢ Carboxylic acid + Active metals(Na, K, Ca and Mg)⟶ H2 gas
Reactions Involving the OH Group of Carboxylic Acids:
➢ Carbonyl group of aldehydes and ketones, is susceptible to attack by a nucleophile.
➢ The addition of a nucleophile to the carboxyl group is always followed by the
displacement of the OH group by some other group, producing a carboxylic acid
derivative.
➢ The OH group can thus be replaced by X, OR and NH2 to form halides, esters and
amides,respectively.
Preparation of acyl chloride:
➢ Acyl chlorides are prepared by treating the carboxylic acid with thionyl chloride (SOCl2)
in the presence of base.
Mechanism:

484. Formation of Ester:
➢ When carboxylic acids are heated with alcohols in the presence of concentratedH2SO4,
esters are formed.
Mechanism:
• Protonation of Carboxylic Acid:
• Attack of CH3 -CH2OH:
• Hydrogen Ion Transfer:
• Elimination of Water and H+
This reaction is also known as Fischer esterification.
Esters are obtained by refluxing the parent carboxylic acid with the appropriate alcohol with
an acid catalyst.
The equilibrium can be driven to completion by using an excess of either the alcohol or the
carboxylic acid, or by removing the water as it forms.
Alcohol reactivity order: CH3OH>10 >20>30

485. Esters can also be made from other carboxylic acid derivatives, especially acyl halides and
anhydrides, by reacting them with appropriate alcohol in the presence of a weak base.
Esters have fruity smell and are used as artificial flavours.
Formation of amide:
➢ Carboxylic acids react with ammonia to form ammonium salts which on heating produce
acid amides.
Mechanism:

486. Formation of Acid Anhydride:
➢ Carboxylic acids are dehydrated on heating strongly in the presence of phosphorus
pentoxide.
Reactions involving carboxyl group:
Partial Reduction to Alcohols:
➢ Carboxylic acids on reaction with lithium aluminium hydride (LiAIH4) are reduced to
alcohols.
Complete reduction to alkanes:
➢ Carboxylic acids on reduction with HI and red phosphorus give alkanes.

487. Assessment 1
1. Which of the following is dicarboxylic acid having chemical formulaH2 C2O4?
a. Acetic acid
b. Lactic acid
c. Malonic acid
d. Oxalic acid
2. Which of the following is obtained by the hydrolysis of fats/oils?
a. Stearic acid
b. Oxalic acid
c. Palmitic acid
d. Both a and c
3. The carboxylic acid in which two –COOH groups are attached to benzene ring is called:
a. Phthalic acid
b. Benzoic acid
c. Stearic acid
d. Palmitic acid
4. Propanedioic acid is the carboxylic acid which is commonly known as:
a. Oxalic acid
b. Methanoic acid
c. Malonic acid
d. Phthalic acid
5. The main acid which was first isolated from vinegar and butter are respectively:
a. Butyric acid and formic acid
b. Oxalic acid and acetic acid
c. butyric acid and acetic acid
d. Acetic acid and butyric acid
6. The oxidation of primary alcohol and Aldehyde to corresponding carboxylic acid is usually
carried out in:
a. Dry conditions
b. Acidic medium
c. Basic medium
d. Neutral

488. 7. The compound (A) is made to react with NaOH on boiling forming sodium salt of acid which on
treatment with HCl yields acetic acid. The compound A is:
a. Ethyl magnesium bromide
b. Ethanol
c. Ethyl acetate
d. Acetaldehyde
8. Acetic acid (CH3COOH) which is dissolved in benzene shows a molecular mass of:
a. 30
b. 60
c. 120
d. 240
9. The carboxylic acids which are colourless liquids with somewhat unpleasant smell:
a. C1-C2
b. C2-C3
c. C1-C3
d. C4-C6
10. The significant solubility of aliphatic acids in water due to hydrogen bonding is exhibited by:
a. Formic acid only
b. Acetic acid only
c. Formic acid, acetic only
d. Formic acid, acetic acid, propanoic acid and butanoic acid

489. Assessment 2
1. Which of the following carboxylic acid has higher melting point?
a. 1C containing carboxylic acid
b. 3C containing carboxylic acid
c. 4C containing carboxylic acid
d. 5C containing carboxylic acid
2. The gas which is released when acetic acid reacts with sodium carbonate:
a. Hydrogen gas
b. Carbon dioxide gas
c. Nitrogen gas
d. None of these
3. Which of the following is not the typical chemical property of aliphatic acid as an acid:
a. Reaction with carbon dioxide to release carbon dioxide gas
b. Reaction with Sodium metal to release hydrogen gas
c. Reaction with bases to form salts and water
d. All of these
4. Which of the following is not the product of reaction of acetic acid with phosphorous
pentachloride?
a. CH3COCl
b. POCl3
c. PCl3
d. HCl
5. The addition of a nucleophile to the carboxyl group is always followed by the:
a. Displacement of –OH group by other group producing alkene.
b. Displacement of –OH group by other group producing carboxylic acid derivative.
c. Displacement of –OH group by other group producing ester.
d. Displacement of –OH group by other group producing alkyl nitrile.
6. The reaction of acetic acid with thionyl chloride results in the release of gas:
a. Chlorine
b. Sulphur dioxide
c. Carbon dioxide
d. Hydrogen

490. 7. Which of the following is not correctly related to each other?
a. Amylacetate-Banana
b. Octylacetate-Orange
c. Benzylacetate-Raspberry
d. Amylbutyrate-Apricot
8. Carboxylic acids react with ammonia to form ammonium salts which on heating produce:
a. CH3CONH2
b. CH3COONH2
c. (NH2)2 CO
d. All of these
9. Which of the following forms acetic acid when undergoes hydrolysis?
a. Ethanol
b. Acetamide
c. Methyl nitrile
d. Methanol
10. Which of the following is the product when two molecules of acetic acid are heated with P2O5:
a. 2 mole of ethyl alcohol
b. 1 mole Formic anhydride
c. 1 mole of Acetic anhydride
d. 2 moles of methyl cyanide

491. Assessment 3
1. Acetylene is treated with 20% H2SO4 and 1.0% HgSO4 at 80°C to give compound (X) which is
then oxidised using V2O5 to give product (Y). What is the name of compound (Y)?
a. Ethanal
b. Ethane
c. Acetic acid
d. Acetaldehyde
2. Acetic acid due to formation of hydrogen bonding is miscible with:
a. Water
b. Alcohol
c. Ether
d. All of these
3. The group which is present in waxes:
a. −𝑂 −
b. – 𝐶𝑂𝑂𝐻
c. −𝐶𝑂𝑂 −
d. – 𝐶𝑂 −
4. When acetamide is hydrolyzed by boiling with acid to form the product. The product obtained
is:
a. Ethylamine
b. Ethanol
c. Acetic acid
d. Acetone
5. The following is the general representation of acid anhydride:
a. 𝑅 − 𝐶𝑂𝑂 − 𝑅
b. 𝑅 − 𝐶𝑂 − 𝑅
c. 𝑅 − 𝑂 − 𝑅
d. (𝑅𝐶𝑂)2𝑂
6. The following is the most reactive one in the following list:
𝐴 = (𝑅𝐶𝑂)2𝑂, 𝐵 = 𝑅𝐶𝑂𝑋, 𝐶 = 𝑅𝐶𝑂𝑁𝐻2, 𝐷 = 𝑅𝐶𝑂𝑂𝑅
a. A
b. B
c. C
d. D

492. 7. The reaction of which of the following compound with sodium metal results in evolution of
hydrogen gas:
a. Ethanol
b. Acetic acid
c. Ethane
d. Both a and b
8. Two aromatic compounds; phenol and benzaldehyde are present in a solution. Which of the
following is mostly used to separate benzaldehyde from phenol in a given solution?
a. NaHCO3
b. Lime
c. NaOH
d. All of these
9. Acetic acid is manufactured by the fermentation of:
a. Methanol
b. Acetone
c. Ethanol
d. Ethanal
10. A fruity smell is produced by the reaction of ethanol in presence of mineral acid with:
a. Phosphorous pentachloride
b. Acetic acid
c. Acetaldehyde
d. Acetic anhydride

493. Assessment 4
1. Which of the following carboxylc acid can not be yielded by the hydrolysis of an alkanenitrile
on boiling with mineral acids or alkalis :
a. Formic acid
b. Acetic acid
c. Propionic acid
d. None of these
2. In order to prepare carboxylic acid, Ethereal solution of Grignard reagent is made to react with
the following with the addition of mineral acid:
a. Water
b. Ammonia
c. Carbon dioxide
d. Formaldehyde
3. 2-Methylbut-2-ene on oxidation with hot alkaline KMnO4 solution form two products; A and B.
The product A and B are:
a. Methanoic acid and ethanoic acid
b. Ethanoic acid and Ethanoic acid
c. Ethanoic acid propanoic acid
d. Ethanoic acid and propanone
4. The statement which is true about the phthalic acid and oxalic acid is:
a. Both are aromatic
b. Both are dicarboxylic acids
c. Both are strong acids
d. All of these
5. Which of the following carboxylic acid is strong acid in water?
a. Methanoic acid
b. Ethanoic acid
c. Propanoic acid
d. All have same acidic strength
6. The aromatic compound which does not contain –COOH group is:
a. Benzoic acid
b. Phthalic acid
c. Carbolic acid
d. 2-chloro benzoic acid

494. KEY
Assessment 1
Assessment 2
1. D
2. D
3. A
4. C
5. D
6. B
7. C
8. C
9. D
10. D
1. C
2. B
3. D
4. C
5. B
6. B
7. C
8. A
9. C
10. C

495. KEY
Assessment 3
Assessment 4
1. C
2. D
3. C
4. C
5. D
6. B
7. D
8. A
9. C
10. B
1. A
2. C
3. D
4. B
5. A
6. C
7. C
8. B
9. B
10. D

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497. Learning Objectives:
Structure of proteins
Classification of Proteins
Role of various proteins in the body
Role of enzymes in biocatalysts
Macromolecules

498. Proteins
➢ The molecules which yield amino acids on complete hydrolysis are called proteins.
➢ Proteins are the basis for the major structural components of animal and human tissue.
➢ Proteins are natural polymers consisting of amino acid units.
➢ Elements of protein = Carbon, hydrogen, oxygen and nitrogen
General Role of protein:
➢ Biological catalysts (enzymes)
➢ Structural components of animal tissue
➢ Participate in cell signal and recognition factors
➢ Acts as molecule of immunity
➢ Can be a source of fuel
Classification of proteins:
Three classes of protein; simple protein, compound or conjugated protein and derived
protein
Simple proteins:
These proteins on hydrolysis yield only amino acids or their derivatives
Albumin:
Blood (serumbumin)
Milk (lactabumin)
Egg white (ovolbumin)
Lentils (legumelin)
Kidney beans (Phaeolin)
Wheat (leucosin)
Sources
Soluble in water and dilute salt solution
Precipitated by saturation with ammonium sulphate
solution
Coagulated by heating
Usually found in plant and animal tissue
Properties

499. Globulin:
Glutelins
Histones:
Blood (serum globulin)
Muscle (myosin)
Potato (tubrin)
Brzil nut (excelsin)
Lentils (legumin)
Hemp (Edestin)
Sources
Sparingly soluble in water and in neutral solution
Precipitated by dilute ammonium sulfate and
coagulated by heat
Distributed both in plant and animal tissue
Properties
Wheat (glutenin)
Rice (oryzenin)
Sources
Insoluble in water and dilute salt solution
Soluble in dilute acids
Found in grains and cereals
Properties
Thymus gland
Pancreas and nucleoproteins
(nucleohistones)
Sources
Soluble in water, salt solution and dilute
acids
Insoluble in ammonium hydroxide
Yield large amount of lysine and arginine;
combined wih nucleic acids within cells.
Properties

500. Conjugate protein:
Those proteins which are attached or conjugated to some non-protein group which are
called prosthetic group. Few examples are given below:
Nucleoprotein:
Mucoprotein:
Glycoprotein:
Cytoplasm of cells (ribonucleoprotien)
Nucleus of chromosomes
(deoxyribonucleo protein)
Virus and bacteriophages.
Sources
It contains nucleic acid, nitrogen and
phosphorous
It is present in chromosomes and in all
living forms as a combination of proteins
with either DNA or RNA.
Properties
Saliva (mucin)
Egg white (ovomucoid)
Sources
Proteins combined with amino sugar,
sugar acids and sulfates.
Properties
Bone (osseomucoid),
tendons (tendomucoid) and
cartilage (chondromucoid)
Sources
Containing more than 4% hexosamine,
mucoproteins
If less than 4% then glyoprotein
Properties

501. Phosphoprotein:
Derived protein:
Proteans:
Proteoses:
Milk (Casein)
Egg yolk (ovovitellin)
Sources
Phosphoric acid joined in ester linkage to
protein.
Properties
Edestan (from elastin)
Myosin (myosin)
Sources
Insoluble in water
It results from short action of acids or
enzymes.
Properties
Intermediate products of
protein digestion.
Sources
Soluble in water
Not coagulated by heat
Precipitated by saturated ammonium
sulfate
Result from partial digestion of protein by
pepsin or trypsin
Properties

502. Peptones:
Peptides:
Structure of protein:
Primary Structure:
Intermediate product of
protein digestion.
Sources
Same property as that of proteases
except they cannot be salted out
Smaller molecular weight than proteases
Two or more amino acids joined by a
peptide linkage
Hydrolyzed to individual amino acids
Properties
The sequence of amino acids in a peptide chain is called primary structure.
Amino acids are linked with one another through peptide bond.

503. Secondary Structure:
Tertiary Structure:
Quaternary Structure:
The peptide chains may acquire spiral shape or may be present in zig
zag manner.
This coiling or zig zagging of polypeptide is called secondary structure
of protein.
It is due to hydrogen bonding.
Twisting or folding of polypeptide chains represents tertiary structure of
protein.
Quaternary means four
Quaternary protein is the arrangement of multiple folded protein or
coiling protein molecules in multi-subunit complex.
A variety of bonding interactions including hydrogen bonding salt
bridges and disulphide bonds hold the various chains into a particular
geometry.

504. Properties of Proteins:
• Proteins are involved in processes such as food digestion, cell structure, catalysis,
movement, energy manipulation.
• They are complex, huge associations of molecular sub units that appear impossibly
difficult to understand.
• Proteins are polymers, composed of smaller subunits- the amino acids joined together
in long chains.
• There are 20-22 common amino acids found in most proteins. All but one of these small
molecules has the same common structure, but varies in the nature one chemical
group-termed the “R-group”.
• Amino acids are joined together in long chains called “polypeptides.”
Importance of proteins:
• Play important role in the formation of protoplasm.
• Nucleoproteins are carrier of heredity from one generation to the other.
• Enzymes are biological catalysts.
• Hemoglobin as carrier of oxygen
• Hormones which regulate the functions of body
• Tanning of hides by precipitation of proteins by tannic acid
• Gelatin which is obtained by heating bones, skins and tendons in water. It is used in
bakery goods.
• Casein used in the manufacture of buttons and buckles.
• Proteins obtained from the soyabean are used for the manufacture of plastics.

505. Enzymes
• Greek words; En means in and Zyme means yeast
• Biocatalysts accelerating the metabolic activities in living bodies
• Complex proteins
• Specific in action
• Sensitive to temperature and pH
Role of enzymes as biocatalysts:
➢ Metabolism is the set of biochemical reactions that occur in living organs in order to
maintain life.
➢ These process allow an organism to:
• Grow
• Reproduce
• Maintain the structure
• Respond to their environment
➢ Anabolism: Biochemical reaction in which smaller molecules combine to form larger
molecule. Energy is consumed.
➢ Catabolism: Biochemical reaction in which larger molecules are broken to form small
molecules. Energy is released.
➢ Biocatalysts speed up and regulate the metabolic pathways.
➢ Enzymes are proteins which do not change during the chemical reactions.
➢ The substance at which enzymes act are called substrates and enzymes convert them
into different molecules, called products.

506. How does an enzyme work?
Two working models of enzymes are:
Industrial application of enzymes:
• Food industry: Starch
𝐸𝑛𝑧𝑦𝑚𝑒
→ 𝑆𝑢𝑔𝑎𝑟. These are used in the production of white bread,
buns etc.
• Brewing industry: Enzymes break starch and protein to produce product which is used
in the fermentation to produce alcohol.
• Paper industry: Enzyme break starch to lower its viscosity that aids in making paper.
• Biological detergent: Protease enzyme are used for the removal of protein stains from
clothes. Amylase enzyme are used in dish washing to remove resistant starch residues.
Lock and Key Model
Presented by Emil Fisher in1894
Name it to lock and key model
Both enzyme and substance have
specific shapes that fit into each other.
This explains the enzymes specificity.
Induced Fit Model
Given by Daniel Koshland in 1958
Modification of lock and key model
According to this model, active site is not a rigid
structure rather is moulded into the required
shape to perform its function.
More acceptable than lock and key model

507. Additional notes:
Factors affecting enzymes:
• Optimum temperature favours the maximum metabolism
• At high temperature globular structure of enzyme is lost because its atoms vibrate with
greater velocity.(Denaturing of enzymes)
Effect of substrate concentration:

508. • When the active sites of all enzymes are occupied (at high substrate conc.), any more
substrate molecule do not find free active site. This state is called saturation of active
site and reaction does not increase.
Effect of pH
• Slight change causes the retardation of reaction
• Pepsin works in low pH in stomach.
• Trypsin works well in high pH in small intestine
• Change in pH can affect the ionization of amino acids at the active site.
Inhibitors:
• The chemical reactant which block the active site and thus reduces the enzymatic
activity
• Irreversible inhibitor: Inhibitor which form covalent linkages. It may occupy the
active site or destroy the globular structure of enzyme.
• Reversible inhibitor: It forms Weak linkages with enzymes. Its effect can be
decreased by increasing the conc. of substrate.

509. Example:(Competitive inhibitor)

510. Assessment 1
1. Proteins are biomolecules in which amino acids are linked together by:
a. Ionic bond
b. Peptide bond
c. Glycosidic bond
d. Dative bond
2. The helical structure of protein is generally stabilized by which of the bonds:
a. Hydrogen bonds
b. Peptide bond
c. Dipeptide bonds
d. Ether bond
3. Haemoglobin is the protein which acts as:
a. Building protein
b. Transport protein
c. Genetic protein
d. Antibiotic
4. Which of the following protein is not considered as conjugated proteins?
a. Lipoproteins
b. Glycoproteins
c. Albumin
d. Mucoproteins
5. Which of the following statements is true about proteins?
a. Proteins are made up of amino acids.
b. Proteins are essential for the development of skin, teeth and bones.
c. Protein is the only nutrient that can build, repair and maintain body tissues.
d. All of the above
6. Which of the following cannot be the part of protein?
a. Zn
b. S
c. N
d. None of these

511. 7. Which of the following is not protein?
a. Glycine
b. Globulin
c. Collagen
d. Legumin
8. Simple proteins constitute the body protein upto:
a. 25%
b. 25-35%
c. 20%
d. 50-60%
9. Lecithin is the protein of which type:
a. Simple protein
b. Compound protein
c. Derived protein
d. None of these
10. The protein which is classified as derived protein is:
a. Peptones
b. Proteases enzymes
c. Oligopeptides
d. All of these

512. Assessment 2
1. The sequence of the amino acids combined in a peptide chain is referred to as:
a. Primary structure
b. Secondary structure
c. Tertiary structure
d. Quaternary structure
2. The three dimensional twisting and folding of the polypeptide chain results in the:
a. Primary structure
b. Secondary structure
c. Tertiary structure
d. Quart airy structure
3. Which of the following is protein in nature?
a. Nucleoproteins
b. Enzymes
c. Hormones
d. All of these
4. Which of the following protein is used in the manufacture of buttons and buckles?
a. Gelatin
b. Caesein
c. Tannic acid
d. Insuline
5. How many amino acids make up a protein?
a. 10
b. 20
c. 30
d. 50
6. Triglycerides are easily hydrolyzed by enzymes to fatty acids and glycerols. The enzyme is:
a. Urease
b. Ligase
c. Lipase
d. Amylase

513. 7. The biomolecules which on hydrolysis yield amino acids are termed as
a. Carbohydrates
b. Proteins
c. Vitamins
d. Lipids
8. Which of the following is true about a dipeptide?
a. 2 aminoacids and 1 peptide bond
b. 2 amioacids and 2 peptide bonds
c. 2 aminoacids and 4 peptide bonds
d. 3 aminoacids and 2 peptide bonds
9. The “lock and key” model of enzyme action illustrates that a particular enzyme molecule
a. Forms a permanent enzyme-substrate complex
b. May be destroyed and resynthesized several times
c. Interacts with a specific type of substrate molecule
d. Reacts at identical rates under all conditions
10. Which of the following is true about the enzyme?
𝐸𝑛𝑧𝑦𝑚𝑒 − 𝐶𝑜 𝐸𝑛𝑧𝑦𝑚𝑒 = ?
a. Iso-enzyme
b. Co-factor
c. Apoenzyme
d. All of these

514. Assessment 3
1. The non protein part of enzyme is known as:
a. Apo enzyme
b. Co factor
c. Co enzyme
d. Both a and b
2. The protein part of an enzyme is known as:
a. Apo enzyme
b. Co factor
c. Co enzyme
d. All of these
3. The enzymes which catalyze the addition of ammonia, water or carbon dioxide to double
bonds or removal of these to form double bonds:
a. Lyases
b. Hydrolases
c. Isomerases
d. Transferases
4. The conversion of fumaric acid to maleic acid is carried out in the presence of fumarase
enzyme. This enzyme belongs to:
a. Ligases
b. Lyases
c. Isomerases
d. Transferases
5. Hexokinase is the enzyme which catalyses the conversion of which of the following to 6-
phosphate derivative?
a. Glucose
b. Fructose
c. Mannose
d. All of these
6. The protein which maintains blood sugar level in the human body:
a. Haemoglobin
b. Amylase
c. Insulin
d. Thyroxin

515. 7. The enzymes from the same organisms which catalyze the same reaction but are chemically
and physically distinct from each other:
a. Apo enzymes
b. Co enzymes
c. Iso enzymes
d. Co factor
8. The rate of reaction is also directly proportional to the:
a. Concentration of substrates.
b. Square root of concentration of enzymes.
c. Square root of substrates.
d. Both a and b
9. Which of the following enzymes is raised in heart diseases?
a. Lactic dehydrogenase
b. Alkaline phosphatase
c. Lactic hydrogenase
d. L-asparaginas
10. Thrombin is an enzyme which is:
a. Used to treat blood cancer in children
b. Raised in heart attack
c. Used locally to stop bleeding
d. Raised in rickets

516. Assessment 4
1. Which of the following is not the function of proteins?
a. Participate in cell signals and recognition factor
b. Act as molecules of immunity.
c. Acts as source of fuel
d. All of these
2. The following simple protein is present in milk and lentils:
a. Albumin
b. Globulin
c. Glutelins
d. Histones
3. The following does not correspond to the Glutelin:
a. Glutelin is present in wheat.
b. Soluble in Dilute acids
c. Insoluble in Dilute salt solution
d. All of these
4. In Mucoprotein, proteins are combined with
a. Amino sugars
b. Sugar acids
c. Sulfates
d. All of these
5. Casein is phosphoprotein which is present in milk. Which of the following acid is joined in
ester linkage to protein?
a. Hydrochloric acid
b. Phosphoric acid
c. Phosphorous acid
d. Nitric acid
6. Quaternary structure of protein experiences the following type of interactions:
a. Hydrogen bonding
b. Salt bridges
c. Disulphide bonds
d. All of these

517. 7. Amino acid is the building block of protein. There are 20-22 different aminoacids found in
most of proteins. The following statement does not corresponds to the different types of
amino acids:
a. They have same common structure
b. They have same R-group.
c. They have amino group as well as carboxylic group.
d. They are classified as acidic, basic and neutral aminoacids.
8. Consider the following diagram and choose the name of the labeled (A) part:
a. Enzyme
b. Substrate
c. Apo-enzyme
d. Co-enzyme
9. The following statement best relates to Koshland Induced fit model:
a. Enzyme changes its shape
b. Substrate changes it shape
c. Both enzyme and substrate changes its shape
d. Both are rigid
10. The small region of an enzyme to which catalytic activity is restricted is called:
a. Active site
b. Substrate site
c. Apo enzymes
d. All of these
A

518. KEY
Assessment 1
Assessment 2
1. B
2. A
3. B
4. C
5. D
6. D
7. A
8. B
9. B
10. D
1. A
2. C
3. D
4. B
5. B
6. C
7. B
8. A
9. C
10. C

519. KEY
Assessment 3
Assessment 4
1. D
2. A
3. A
4. C
5. D
6. C
7. C
8. D
9. A
10. C
1. D
2. A
3. C
4. D
5. B
6. D
7. B
8. B
9. A
10. A

521. MDCAT-Chemistry Formulas
Chapter 1
No. of atoms (Mass/Atomic mass) * NA
No. of Molecules (Mass/ Molecular Mass) * NA
No. of ions (Mass/ Ionic Mass) * NA
Percentage Yield (Actual Yield/Theoretical yield) * 100
% of hydrogen (2.016/18.00) * (Mass of H2O/Mass of organic compound)
*100
% of carbon (12.00/44.00) * (Mass of CO2/Mass of organic compound)
*100
% of oxygen 100 - (% of Carbon + % of hydrogen)
Molecular formula n * Empirical formula
n Molecular mass/ Empirical formula mass
Average atomic mass [(Relative abundance of 1st isotope * atomic mass of 1st
isotope) + (Relative abundance of 2nd isotope * atomic mass
of 2nd isotope)]/100
% of an element (Mass of an element in 1 mole of compound/mass of the
compound) *100
No. of moles of gas Mass of gas in gram/ molar mass of gas
Particles, Atoms, Ions Mole * NA (6.02*1023)
Mole Particles, Atoms, Ions / NA
Mass (g) Mole * Molar mass
Mole mass / molar mass
Atoms moles * NA
moles Atoms/ NA
No. of atoms of an element No. of gram atoms * NA
No. of molecules of compound No. of gram mole * NA
No. of formula unit in ionic
compound
No. of gram formula * NA
No. of ions No. of gram ions * NA
No. of atoms of an element in
a molecule
No. of molecules * No. of atoms of element in molecule
Volume of gas at STP No. of moles of gas * 22.414 dm3
No. of atom (Mass in g / molar mass) * NA
Percentages
% w/w (mass of solute/mass of solution) * 100
% w/v (mass of solute/volume of solution) * 100

522. % v/w (volume of solute/mass of solution) * 100
% v/v (volume of solute/volume of solution) * 100
% w/v (Molarity * Molar mass)/ 10
1 atm
1 atm = 760 torr , 101325 Nm-2 , 760 mmHg , 101325 pascal ,
76cmHg , 1.013 bar , 14.7 psi
Molarity (M)
M No. of moles of solute/ dm3 of solution
Molarity [(%w/v) * 10] / Molar mass
If moles and dm3 given M = No. of moles/Volume in dm3
1ml = 1cm3 1 liter = 1 dm3
If moles and cm3 given M = No. of moles (1000 /Volume in cm3 )
Mass in g and dm3 M = (Mass in (g) / Molar mass) * 1/ Vol. in dm3
Mass in g and cm3 M = (Mass in (g) / Molar mass) * 1000/ Vol. in cm3
N m/M = N/NA = V/Vm
m/M = N/NA = V/Vm
N/NA = V/Vm
Molality (M)
M No. of moles of solute/ mass of solvent in kg
Mole Fraction
Ratio mole of components/ Total No. of moles = n
XA + XB 1
XB 1- XA
XA 1- XB
Mole % Mole fraction * 100
XA nA / (nA + nB + nC)
XA + XB + XC 1

523. Parts Per Million
Ppm [solute (g) / solution (g) ] * 106
m/e H2r2/2e
General Gas Equation
V ∝ 1/P V ∝ T , V ∝ n
V nRT/P
PV nRT
PV (m/M) * RT , n = m/M
PM (m/v) * RT
PM dRT
D PM/RT
r ∝ 1/√d, r ∝ 1/√M
r1/r2 √ (d2/d1 ) = √ (M2/M1)
P1V1/T1 P2V2/T2
P ∝ 1/V V ∝ 1/P , V= k/P
P1V1 = P2V2 P1 / P2 = V2 / V1
V kT
V1/T1 V2/T2
VT V0 (1 + (t oC/273))
∆oC ∆oK
K.E1/T1 K.E2/T2
K.E1/K.E2 T1/T2
V ∝ n V = kn , k = V/n
V1/n1 V2/n2
Vol. of ideal gas Moles* 22.4
Temperature Conversions
K C +273.16
C 5/9 (F-32)
F 9/5 (C + 32)

524. Angular Momentum
Mvr nh / 2π
To calculate No. of spectral lines
Shell N
Subshells L
No. Of Electrons in a shell 2n2
No of orbitals in a shell N2
No of electron in a subshell 2(2l+1) = 4l +2
No of orbitals in a subshell 2l +1
Magnetic quantum number 2l + 1
Mass number Nucleon number
Energy of a subshell α n+l
value
When two subshells have the same n+ l value, then energy
of subshell is given by:
Energy of subshell α Value of n
For Neutral Atom A
XZ
Z No. of protons
No. of protons No of electrons
No. of neutrons A-Z
No of electrons in a valence shell Group number
Principle number of valence shell Period number
Ionization Energy
Atomic radius of atom is inversely related with the ionization energy
Nuclear charge or proton number is inversely related with the ionization energy
Shielding effect of inner electrons is inversely related with the ionization energy
Nature of orbital (s > p > d > f)
Reactivity of metal is inversely related with the ionization energy
3rd I.E > 2nd I.E > 1st I.E
Abnormal behavior
(IIA > IIIA)
( VA > VIA)

525. Percentage of a bond
%age of covalent bond = (No of covalent bond/Total bond) * 100
%age of Co-ordinate covalent bond = (No of co-ordinate covalent bond/Total bond) * 100
Hybridization
Calculate no of lone pairs on central atom + no of sigma bonds formed by central atom
If answer is 4, it is sp3 hybridization
If answer is 3, it is sp2 hybridization
If answer is 2, it is sp hybridization
Type of molecule
Calculate [
𝐺𝑟𝑜𝑢𝑝 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑎𝑡𝑜𝑚+𝑁𝑜 𝑜𝑓 𝑏𝑜𝑛𝑑𝑒𝑑 𝑎𝑡𝑜𝑚𝑠 𝑤𝑖𝑡ℎ 𝑠𝑖𝑛𝑔𝑙𝑒 𝑏𝑜𝑛𝑑−𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑐ℎ𝑎𝑟𝑔𝑒
2
]
If answer is 4, it is AB4 type
If answer is 3, it is AB3 type
If answer is 2, it is AB2 type
Dipole Movement (μ)
Μ q*r
% ionic Character
% ionic character (μobs / μionic ) * 100
Internal energy
E K.E + P.E

526. 1st
Law of Thermodynamics
ΔE q +w (work on system)
ΔE q -w (work by system)
ΔE q + (PΔV)
If piston is fixed ΔE= qv
ΔH HP – HR (Enthalpy)
Enthalpy
ΔH ΔE + PΔV
ΔH ΔE
PΔV ΔnRT
ΔH ΔE + ΔnRT
At constant pressure
ΔE q – PΔV
ΔH q – PΔV
ΔH Q
Δn No. of moles of product - No. of moles of
reactants
qp – qv ΔnRT
Hess’s Law
(Σ ΔH = 0), ΔH ΔH1 + ΔH2
ΔHf ΔHx + ΔHLattice
Other Formulas
Q (Glass Calorimeter) m * S * ΔT

527. Q (Bomb Caloriemeter) c * ΔT
Oxidation Potential (Cell Potential) - (Reduction potential)
CHEMICAL EQUILIBRIUM
Equilibrium Constant Kc = [product]/[reactant]
Equilibrium Constant Kc = Kf/Kr
Units of Kc = (concentration) ∆n
where ∆n= no of moles of products – No of moles of reactants
Units of Kp = (atm) ∆n
Concentration = No of moles / volume
Kp = Kc(RT) ∆n ; Kc = Kp(RT) -∆n ; Kc = Kp/(RT) ∆n
Kp = Kc(RT) ∆n
Possibilities if np = nr (Kp=Kc)
if np > nr (Kp>Kc)
if np < nr (Kp < Kc)
pH = -log [H+]
pOH = -log [OH-]
For neutral water [H+] = [OH-] = 10-7
pH = 1/2pKw
H+ α T ; pH α 1/T
Kw α T ; pKw α 1/T
pOH α Acidic strength α 1/Basic strength
pH α Basic strength α 1/Acidic strength
For monoprotic acid → H+ = 10-pH
For monohydroxy base → OH- = 10-pOH
Ka * Kb = 14
pKa + pKb = 14
pH = pKa + log
[𝑠𝑎𝑙𝑡]
[𝑎𝑐𝑖𝑑]
pH = pKa - log
[𝑎𝑐𝑖𝑑]
[𝑠𝑎𝑙𝑡]
pH = pKa + log
[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
pH = pKa - log
[𝑎𝑐𝑖𝑑]
[𝑏𝑎𝑠𝑒]
pH = pKa + log
[𝑠𝑎𝑙𝑡]
[𝑎𝑐𝑖𝑑]
𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠 → 𝐼𝑓 [𝑠𝑎𝑙𝑡] = [𝑎𝑐𝑖𝑑] 𝑡ℎ𝑒𝑛 𝑝𝐻 = 𝑝𝐾𝑎
𝐼𝑓 [𝑠𝑎𝑙𝑡] > [𝑎𝑐𝑖𝑑] 𝑡ℎ𝑒𝑛 𝑝𝐻 > 𝑝𝐾𝑎
𝐼𝑓 [𝑠𝑎𝑙𝑡] < [𝑎𝑐𝑖𝑑] 𝑡ℎ𝑒𝑛 𝑝𝐻 < 𝑝𝐾𝑎
For a reaction AxBy ⇌ xA+y + yB-x
Ksp = [A+y]x + [B-x]y

528. Cell Voltage
Eo
cell = Eo
oxidation + Eo
reduction
Eo
cell = Eo
red(cathode) – Eo
red(anode)
Reaction Kinetics
Rate of reaction =
−d[A]
𝑑𝑡
=
+d[B]
𝑑𝑡
Units of K = (concentration)1-n * sec-1
Where n = order of reaction
T1/2 α 1/an-1
Half life time = Total time / No of half life
Arrhenius equation K = Ae-Ea/RT