class 12 chemistry board material

“Value Education with Training”
CHAPTER -1 SOLID STATE
TYPOLOGY : KNOWLEDGE
Q.NO. QUESTIONS WITH ANSWERS VALUE
POINTS
ONE MARK QUESTIONS
1. What do you mean by coordination number ?
Ans.
The no of atoms to which central atom is touching 1
2. What is meant by doping in a semiconductor?
Ans.
Doping is defined as addition of impurities from outside to a crystal. 1
3. What are paramagnetic substances.?
Ans.
Substances which are attracted by external magnetic field due to presence of
unpaired electrons . 1
4. Name two type of voids present in a crystal lattice.
Ans.
a) Tetrahedral voids
b) Octahedral void
½
½
5. What is piezoelectricity?
Ans.
Ability of a substance to produce electric current on application of mechanical
stress.
1
6 How many lattice points are there in one unit cell of bcc unit cell ?
Ans.
8 (at corner ) +1 (at body centre) 1
TWO MARK QUESTIONS
7 Define Packing efficiency? What is packing efficiency of hcp.
Ans.
It is the percentage of total space occupied by constituent particles (atoms,
molecules or ions) in a unit cell.
Packing efficiency of hcp is 74%
1
1
8 What are F- centers ? Why are the solids containing the F- centers are
paramagnetic?
Ans .
The free electrons trapped in the anion vacancies are termed as the F- centers.
The solids containing the F-centres are paramagnetic because the electrons
occupying the vacant sites are unpaired.
1
1
9 A unit cell consists of a cube in which there are anions at each corner and one at

the center of the unit cell. The cations are the center of the each face. How many
A) cations and B) anions make up the unit cell C)What is the simplest formula of
the compound ?
Ans.
A) The cation at the center of each face is shared by two unit cells.
Hence no. of cations= 6 X ½ = 3
B) The anion at each corner is shared by 8 unit cells . the anion at the center is not
shared by any other unit cell.
Hence no. of anions= 8 X 1/8 + 1 = 2
C) Since there are 3 cations and 2 anions the simplest formula of the compound is
A3B2.
½
½
1
10. Distinguish between
(i)Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Ans.
(i)
Hexagonal unit cell
For a hexagonal unit cell,
a = b ≠ c
α = β = 90°
γ = 120°
Monoclinic unit cell
For a monoclinic cell,
a ≠ b ≠ c
α = γ = 90°
β ≠ 90°
(ii)
Face-centred unit cell
In a face-centred unit cell, the constituent particles are present at the corners
and one at the centre of each face.
End-centred unit cell
An end-centred unit cell contains particles at the corners and one at the centre
of any two opposite faces.
½
½
½
½
THREE MARK QUESTIONS
11. State any three differences between Schottky and Frankel defect?
Ans.
Schottky defect
I. It occurs due to missing of equal no of cations and anions from lattice
point.
II. It decreases the density of the crystal.
III. It occurs in compounds with high Co-ordination number.
OR
½
½
½

It occurs in compounds in which cations and anions are of similar size.
Examples: NaCl, KCl, KBr, Ag Br CsCl.
Frankel defect
I. It occurs due to missing of cations from their lattice point and occupies
interstitial sites
II. It does not decrease the density of the crystal.
III. It occurs in compounds with low Co-ordination number.
OR
It occurs in compounds in which cations and anions differ in their size to a
large extent. Examples: ZnS, AgCl, AgBr, Agl.
½
½
½
12. Explain the following with suitable example
I. Paramagnetism
II. Ferrimagnetism
III. Antiferromagnetism
Ans.
I. Paramagnetism
The substances that are attracted by a magnetic field are called paramagnetic
substances. Due to one or more unpaired electrons substances get magnetized in a
magnetic field in the same direction, but lose magnetism when the magnetic field is
removed.
Some examples of paramagnetic substances are O2, Cu2+
, Fe3+
, and Cr3+
II. Ferrimagnetism
The substances in which the magnetic moments of the domains are aligned in
parallel and anti-parallel directions, in unequal numbers, are said to have
ferromagnetism.
Examples include Fe3O4 (magnetite), ferrites such as MgFe2O4 and ZnFe2O4.
Schematic alignment of magnetic moments in ferrimagnetic substances
(iii) Antiferromagnetism
Antiferromagnetic substances have domain structures similar to ferromagnetic
substances, but are oppositely-oriented. The oppositely-oriented domains cancel
out each other’s magnetic moments.
Example - MnO
½
½
½
½
½
½

TYPOLOGY : UNDERSTANDING
POINTS
ONE MARK QUESTIONS
13. Why Glass panes fixed to windows of old buildings are invariably found to be
thicker at the bottom than the top ?
Ans.
Glass is a Pseudo solid so behaves like fluid. 1
14. Graphite is soft and generally used as a lubricant. Give reason.
Ans.
Because it has layered structure having weak van der’s wall forces one layer can
slide on another. It is slippery in nature. 1
15. If α=β=ϒ=90 and a≠b≠c, identify the type of crystal system and give one example?
Ans.
Orthorhombic
Example: Rhombic Sulphur
½
½
16. Solid A is soft, a conductor of electricity and has a layered structure. Identify solid A
and its type ?
Ans.
Graphite and covalent ½ ,½
17. What are the number of tetrahedral voids generated if the number of close packed
sphere be N in a crystal?
Ans. 2N 1
TWO MARK QUESTIONS
18. Excess of potassium in the KCl makes the crystal appears violet. Explain why?
Ans.
When KCl is heated in an atmosphere of K metal vapour , the metal K deposits on
the surface of the KCl crystal . The chloride ions diffuse into the surface and
combine with K atoms.
The electrons produced by the ionization of the K Atoms then diffuse into the
crystals and are then trapped in the anion vacancies called F- centers . the excess
of the K+
ions in KCl makes the crystal appear violet
1
1
19. Atom of element B forms hcp lattice and those of the element A occupy 2/3 rd of
tetrahedral void .what would be the formula of the compound ?
Ans.
Suppose number of atom B in hcp = N
No. of Tetrahedral void = 2N
Atom A occupying tetrahedral void = 2/3* 2N
Ratio A:B A4B3
1
1
20. Ferromagnetic and Ferrimagnetic substances become paramagnetic upon heating .
Why?

Ans.
The temperature at which they are changed into paramagnetic is called curie
temperature.
This is because the realignment of electrons spin or their magnetic moments which
are now oriented in one particular direction.
1
1
21. In the mineral spinal; having the formula MgAl2O4. The oxide ions are arranged in
CCP, Mg2+ ions occupy the tetrahedral voids. While Al3+ ions occupy the octahedral
voids.
(i) What percentage of tetrahedral voids is occupied by Mg2+ ions ?
(ii) What percentage of octahedral voids is occupied by Al3+ ions ?
Ans.
According to the formula, MgAl2O4. If there are 4 oxide ions, there will be 1 Mg2+
ions and 2 Al3+. But if the 4 O2– ions are ccp in arrangement, there will be 4
octahedral and 8 tetrahedral voids.
(i) Percentage of tetrahedral voids occupied by Mg2+ = (1 / 8) × 100 = 12.5%
(ii) Percentage of octahedral voids occupied by Al3+ = (2 / 4) × 100 = 50%
1
1
1
22. Classify the following solids in different categories based on the nature of
intermolecular forces operating in them:
Potassium sulphate , benzene, urea, ammonia, water, silicon carbide.
Ans :
Potassium sulphate → Ionic solid
Benzene → Molecular (non-polar) solid
Urea → Polar molecular solid
Ammonia → Hydrogen bonded molecular solid
Water → Hydrogen bonded molecular solid
Silicon carbide → Covalent or network solid
½ each
23. Classify each of the following as being either a p-type or an n-type semiconductor:
(i) Ge doped with In (ii) B doped with Si (iii) Si doped with In
Ans:
(i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole
will be created and the semiconductor generated will be a p-type semiconductor.
(ii)B (a group 13 element) is doped with Si (a group 14 element). So, there will be an
extra electron and the semiconductor
generated will be an n-type semiconductor
(iii ) Si (a group 14 element) is doped with In (a group 13 element). Therefore, a hole
will be created and the semiconductor generated will be a p-type
semiconductor.
1
1
1

TYPOLOGY : APPLICATION
POINTS
ONE MARK QUESTIONS
24. How many spheres are in contact with each other in a single plane of a close
packed structure?
Ans.
Six(6) 1
25. What other elements may be added to silicon to make electrons available for the
conduction of an electric current?
Ans.
Phosphorous or Gallium. 1
26. How many Tetrahedral sites per sphere are there in a cubic closest – packed
( face centered cubic) structure.?
Ans.
Two 1
TWO MARK QUESTIONS
27. Analysis shows that nickel oxide has the formula Ni 0.98 O. What fraction of the
nickel exist as Ni 2+ and Ni 3+.
Ans.
let Ni 2+
= x and Ni 3+
98-x
2*x + 3(98-x) +2= 0
Fraction of Ni 2+ = 94/98*100 =95.9%
Fraction of Ni3+ = 4/98*100 =4.08%
1
1
28. A compound forms hexagonal close packed structure . What is the total number of
void in 0.5 mol of it .How many of these are tetrahedral void ?
Ans.
Number of atom in the close packing = 0.5×6.023×1023
=3.011×1023
Number of octahedral void = 1×3.011×1023
Number of tetrahedral void 2 × 3.011×1023
= 6.033×1023
1
1
29. An element with molar mass 27 g /mol forms a cubic unit cell with edge length
4.05*10-8 cm if its density is 2.7g/cm3 what is the nature of the cubic cell .
Ans.
ȡ = Z×M/a3×NA
2.7 = Z × 27/(4.05*10-8
)3
×6.023 *1023
Z = 1080.29*10-1
/ 27
Z = 4 , FCC
½
½
1

30. Niobium crystallizes in body-centered cubic structure. If density is 8.55 g cm−3
,
calculate atomic radius of niobium using its atomic mass 93 u.
Ans.
It is given that the density of niobium, d = 8.55 g cm−3
Atomic mass, M = 93 gmol−1
As the lattice is bcc type, the number of atoms per unit cell, z = 2
We also know that, NA = 6.022 × 1023
mol−1
Applying the relation:
= 3.612 × 10−23
cm3
So, a = 3.306 × 10−8
cm
For body-centred cubic unit cell:
= 1.432 × 10−8
cm
= 14.32 × 10−9
cm
= 14.32 nm
½
½
1
1
31. Ferric oxide crystallizes in a hexagonal close-packed array of oxide ions with two out
of every three octahedral holes occupied by ferric ions. Derive the formula of the
ferric oxide.
Ans .
Let the number of oxide (O2−
) ions be x.
So, number of octahedral voids = x
It is given that two out of every three octahedral holes are occupied by ferric ions.
So, number of ferric (Fe3+
) ions =2/3x
Therefore, ratio of the number of Fe3+
ions to the number of O2−
ions,
Fe3+
: O2−
= 2 : 3
Hence, the formula of the ferric oxide is Fe2O3.
½
½
1
1
32. Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125
pm.
(i) What is the length of the side of the unit cell?
(ii) How many unit cells are there in 1.00 cm3
of aluminum?
Ans.
(i) For cubic close-packed structure

= 353.55 pm
= 354 pm (approximately)
(ii) Volume of one unit cell = (354 pm)3
= 4.4 × 107
pm3
= 4.4 × 107
× 10−30
cm3
= 4.4 × 10−23
cm3
Therefore, number of unit cells in 1.00 cm3
=
= 2.27 × 1022
½
½
½
½
1

CHAPTER -2 SOLUTIONS
POINTS
ONE MARK QUESTIONS
1. What is molarity?
Ans.
No. of moles of solute dissolved per litre of a solution 1
2. State Henry’s law.
Ans.
The partial pressure of a gas is directly proportional to its mole fraction in solution.
P = KH × x 1
3. What do you mean by vapour pressure?
Ans.
The pressure exerted by vapours above the liquid surface at equilibrium is called vapour
pressure. 1
4. How would you define, an Ideal solution?
Ans.
The solution which obey Raoult’s law at all temperature and pressure is called an ideal
solution which also have ∆mixH and ∆mixV= 0
1
5. Give an example of Ideal solution.
Ans.
n-Hexane and n-Heptane or any other suitable example. 1
6. Define Colligative Properties and give one example.
Ans.
The properties which depends on number of solute particles but independent of its nature.
e.g. Elevation in boiling point / Depression in freezing point
½
½
7. Why are the equimolar solutions of NaCl and glucose not isotonic?
Ans.
Isotonic solutions are those having same concentrations and osmotic pressure but NaCl and
Glucose have not the same osmotic pressure due to the different Van’t Hoff factor .
1
8. What is Van’t Hoff factor?
Ans.
Van’t Hoff factor is the ratio of normal molar mass/abnormal molar mass or any other
suitable formula.
1
9. What is the Van’t Hoff factor in K4[Fe(CN)6] and BaCl2 ?
Ans.
5 and 3 respectively assuming that it is fully dissociated in solution. ½ + ½
10. What is the value of van’t Hoff factor for a compound which undergoes tetramerisation in an
organic solvent?
Ans.
¼ because the four molecules gets associated to one. 1
TWO MARK QUESTIONS
11. What is molal elevation constant or ebullioscopic constant? Write its units.
Ans.
Molal elevation constant or ebullioscopic constant is the elevation in boiling point for one
molal solution.
Its unit is KKgmol-1
1
1

12. Define Raoult’s law for non volatile solute solution. Give its mathematical form.
Ans.
Vapour pressure of a solution is directly proportional to the mole fraction of volatile
component.
Psol= P0
solvent× Xsolvent
1
1
13. Calculate the amount of KCl which must be added to 1Kg of water so that the freezing point is
depressed by 2 kelvin. (Kf for water = 1.86 K Kg mol-1
)
Ans.
ΔTf = i X Kf X m
2 = 2 X 1.86 X ( n/1)
N = 1/ 1.86
= 0.537 mol
½
½
½
½
14. Define Azeotropes? What are maximum and minimum boiling azeotropes ? Explain with
example.
Ans.
Azeotropes are constant boiling mixture which has same composition in liquid phase as well
as in vapour phase.
The non ideal solutions which exhibit negative deviation from ideal solution at a particular
composition are called as maximum boiling azeotropes.
e.g 68% aqueous solution of HNO3 or any other suitable example.
The non ideal solutions which exhibit positive deviation from ideal solution at a particular
composition are called as minimum boiling azeotropes.
e.g 95% aqueous ethanol by volume or any other suitable example.
1
½
½
½
½
TYPOLOGY UNDERSTANDING
ONE MARK QUESTIONS
15. Which is better method for expressing concentration of solution – Molarity or Molality
Ans.
Molality 1
16. Write one example each of solid in gas and liquid in gas solution?
Ans.
Solid in Gas-dust in gas & Liquid in Gas – Moisture in air ½ + ½
17. Which of the following mode of concentration is affected by temperature?
Molarity, Molality, mole fraction and Normality.
Ans.
Molarity and Normality ½ + ½
TWO MARK QUESTIONS
18. Vapour pressure of pure water at 350
C is 31.82 mm Hg when 27.0g of solute is dissolved in
100 g of water (at the same temperature) vapour pressure of the solution thus formed is
30.95mm Hg. Calculate the molecular mass of the solute.
Ans.
PA
0
- PA = XB
PA
0
31.82 - 30.95 = 27/MB ½ + ½

= 0.87 = 27 x 18
31.89 100/18 31.82 MB 100
MB = 177.75 g mol-1
+½ + ½
19. What is the molar concentration of solute particles in human blood if the osmotic pressure is
7.2 atm at normal body temperature, i.e. 370
C?
Ans.
Л = CRT
or C = Л/RT
Л = 7.2 atm
R = 0.0821 L atm K-1
mol-1
T = 370
C = 37 + 273 = 310 K.
Molar concentration (C) = (7.2 atm)
(0.082L Latm K-1
mol-1
) x (310K)
= 0.283 mol-1
= 0.283 M.
½
½
½
½
20. A 1.00 m solution of acetic acid (CH3COOH) in benzene has a freezing point depression of 2.6
K. Calculate the value for i and suggest an explanation for its value.
Ans.
ΔTf = i X Kf X m
2.6 = i X 5.12 X 1
i = 2.6 / 5.12 = 0.502
I < 1 the result suggest that acetic acid undergoes association in organic solvent.
1
1
1
TYPOLOGY APPLICATION
ONE MARK QUESTIONS
21. What happens when blood cells are placed in pure water?
Ans.
Blood cells swell up due to osmosis (plasmolysis). 1
22. Name the substances which are used by deep sea divers to neutralize the toxic effects of
nitrogen dissolved in the blood.
Ans.
Helium (11%) 1
23. Why is Anoxia disease very common at higher altitudes?
Ans.
As concentration of oxygen is less in air at high altitude. 1
24. Name the substance used as cell membrane in reverse osmosis.
Ans.
Cellulose acetate 1
25. During the preparation of solution for intravenous injection which essential factor should be
kept in mind?
Ans.
The solution should be isotonic with RBC. 1
TWO MARK QUESTIONS
26. Why is the freezing point depression of 0.1 M KCl solution nearly twice than that of 0.1 M

sucrose solution?
Ans.
Both have same concentration but in aqueous solution KCl undergoes dissociation whereas
sucrose does not.
KCl on dissociation provides two particles.
1
1
27. The freezing point depression of a 0.10 m solution of HF (aq) solution is -0.201 °C. Calculate
the percent dissociation of HF (aq).
Ans.
ΔTf = i X Kf X m
0.201 = iX 1.86 X 0.1
i = 0.201 / 0.186 = 1.08
α = (i-1) / ( n-1)
=1.08 – 1 = 0.08
% dissociation of HF is 8%
½
½
½
½
28. A 0.2 m aqueous solution of KCl freezes at -0.680
C. Calculate ‘i’ and the osmotic pressure at
00
C. Assume the volume of solution to be that of pure H2O and Kf for H2O is 1.86 KKg/mol.
Ans.
(ΔTf)normal = Kf x m = 1.86 x 0.2= 0.372
i= Observed colligative property/Normal colligative property
= 0.68/0.372
Observed osmotic pressure= i x Normal osmotic pressure
= i x c RT
= 1.83 x 0.2 x 0.082 x 273
= 8.2 atm
1
1
½ + ½
29. Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na2SO4. If
this solution actually freezes at -0.320o
C, what would be the value of Van’t Hoff factor?
Ans.
ΔTf = Kf X m
= 1.86x .0711 = .132 0
C
i = .320/.132 = 2.42
1
1
1
30. At 25o
C, 3g of a solute A in 100 ml of an aqueous solution gave an osmotic pressure of 2.5
atm. What is the nature of the solute (associated or dissociated) if its normal molar mass is
246?
Ans.
Л = i X C X R X T
2.5 = i X ( 3/246) X( 0.0821 / 0.1) X 298
i = 2.01
As i > 1 the solute A undergoes dissociation.
1
1
1

CHAPTER -3 ELECTROCHEMISTRY
TYPOLOGY – KNOWLEDGE
POINTS
ONE MARK QUESTIONS
1. Can absolute value of electrode potential be measured ?
Ans:
No, Absolute value of electrode potentials cannot be measured as electrode
reaction cannot take place by its own.
1
2. What is electrode potential?
Ans:
The potential difference between electrolyte Solution and electrode is called
electrode potential.
1
3. Consider a cell given below :
Cu | Cu2+
|| Cl-
| Cl2 , Pt(s)
Write the reactions that occur at anode and cathode.
Ans :
Cu → Cu2+
+ 2e At anode Cl2 +2e →2 Cl-
At cathode ½ + ½
4. What is the function of salt bridge?
Ans:
(i) To maintain the electrical neutrality of the electrolytes
(ii) To complete the internal circuit of the cell.
1
5. Give an example of secondary cell.
Ans:
Lead storage battery. 1
6. What would happen if the protective tin coating over an iron bucket is
broken in some places?
Ans:
Iron will corrode faster as the oxidation potential of Fe is higher than that of
tin. 1
TWO MARK QUESTIONS
7. Calculate the emf of the cell in which the following reactions takes place
Ni(s) + 2 Ag+
(0.002M) Ni2+
(0.16 M) + Ag(s)
Given that Eo
cell = 1.05 V.
Ans:
Ecell= E0
-0.0591/n x log 0.16/(2 x 10-3
)2
= 1.05 -0.0591/2 x log 16 x 104
/4
=0.91 volt
1
1
8. H2-O2 fuel cell was the most commonly used cell to provide electrical power
in Apollo space programme amongst all the fuel cells. The cell runs
continuously as long as the reactants are supplied.
Answer the following questions.(K)
(i) Write the reactions which occur at cathode.

(ii) Write the reaction which occur at anode.
(iii) Write the name of catalyst that is used in the fuel cell.
(iv) Give the values associated with using fuel cell.
Ans:
(i) At cathode
O2(g)+2H2O(l)+4e-
→4OH-
(aq)
(ii) At anode
2H2(g)+4OH-
(aq)→4H2O+4e-
Overall reaction
2H2(g)+O2(g)→2H2O(l)
(iii) Finely divided platinum and palladium acts as a catalyst.
(iv)
(a)Fuel cell has efficiency of more than 70%.
(b)Fuel cell does not create any environmental pollution.
(c)During the reaction, water vapours are produced, which upon
condensation can be used for drinking purpose.
½
½
½
½
½
½
9. Represent the cell in which the following reaction takes place
Mg(s) + 2Ag+
(0.0001M) Mg2+
(0.130 M) + 2Ag(s)
Calculate its Ecell if Eo
cell = 3.17 V .
Ans:
Ecell = Eo
cell – (RT /2F) x ln{[Mg2+
]/[Ag+
]2
=3.17V - 0.059/2 x log 0.130/(0.0001)2
= 2.96 V
1
1
1
10. Depict the galvanic cell in which the reaction
Zn(s) +2Ag+
(aq) Zn2+
(aq) +2Ag(s) takes place.
Further show:
i) which of the electrode is negatively charged?
ii) the carriers of the current in the cell.
iii) Individual reaction of each electrode.
Ans:
Zn(s)|Zn2+
(aq) || Ag+ (aq) | Ag (s)
I) Zn- electrode
ii) electrons flow from anode to cathode, current flows from cathode to
anode
iii) Reaction at Anode:
Zn (s)  Zn 2+
(aq) + 2e-
Reaction at Cathode:
Ag+
(aq) + e Ag(S)
1
1
½
½
12. Calculate the reduction potential of cell consisting of a platinum electrode
immersed in 2.0 M Fe2+
and 0.02 M Fe3+
solution (Eo
Fe3+
| fe2+
= 0.771 V).
Ans:
E0
Fe
3+
/Fe
2+
- Eo
Fe
3+
/ fe
2+
- 0.0591/1log 2/0.02
= 0.6528 Volt
1
1
1

TYPOLOGY – UNDERSTANDING
TWO MARK QUESTIONS
11. Which solution will show greater conductance of electricity, 1M NaCl at 293K
or 1MNaCl at 323K and Why?
Ans:
1M NaCl at 323K
as the ionic mobilities increase with increase in temperature.
1
1
12. Can a nickel spatula be used to stir a solution of copper sulphate? Justify
your answer.
(E˚Ni
2
/Ni= -0.25V E˚ Cu
2
+/Cu=0.34V)
Ans:
No,
Because Reduction potential of Ni is less than Cu. Ni will replace the Cu from
CuSO4.
1
1
13. Which out of 0.1M HCl and 0.1M NaCl, do you expect have greater ˄∞
mand
why?
Ans:
0.1M HCl will have greater ˄∞
m because
H+
(aq) being smaller in size than Na+
(aq) and have greater mobility.
1
1
14. Which of the following pairs will have greater conduction and why?(E)
(a) Copper wire at 25˚C and Copper wire at 50˚C.
(b) 0.1M acetic acid solution or 1M acetic acid solution?
Ans:
(a) Copper wire at 25˚C because with increase in temperature, resistance
increase, metallic conduction decrease with increase in temp .due to
vibration of kernels.
(b) 0.1M acetic acid solution because with dilution degree of dissociation
increases and hence number of ions.
1
1
15. Consider the electrochemical cell:(APP)
Zn (s)/Zn2+
(aq)// Cu2+
(aq)/Cu. It has an electrical potential of 1.1V when
concentration of Zn2+
and Cu2+
ions is unity.
State the direction of flow of electrons and also specify if Zinc and Copper
are deposited or dissolved at their respective electrodes.
When:
(a) An external opposite potential of 0.8 V is applied.
(b) An external opposite potential of 1.1 V is applied.
(c) An external opposite potential of 1.4 V is applied.
Ans:
(a) Electrons flow from Zn rod to Cu rod.
(b)No flow of electrons and current. No change observed at Zinc and Copper
electrodes (system is equilibrium).
(c) Electrons flow from Cu rod to Zn rod.
Zinc is deposited and Copper gets dissolved.
1
1
1

TYPOLOGY – APPLICATION
TWO MARK QUESTIONS
16. Rusting of iron is quicker in saline water than in ordinary water. Why is it so?
Ans:
In saline water, NaCl helps water to dissociate into H+ and OH-. Greater the
number of H+, quicker will be rusting of iron.
1
1
17. For the reaction:
2Ag+
+ 2Hg → 2Ag + Hg2
2+
E˚Ag+/Ag = 0.80 V E˚Hg2
2+
/Hg = 0.79 V
Predict the direction in which the reaction will proceed if:
[Ag+
] = 10-1
mol/h [Hg2+
] = 10-3
mol/h
Ans:
Cell reaction is:
2Ag+
+ 2Hg → 2Ag + Hg2
2+
Ecell = E˚ cell – 0.0591/2 log [Hg2
2+
]/ [Ag+
]2
= (0.80V-0.79V) -0.0591/2 log 10-3
/ (10-1
)2
= 0.01V- 0.0591/2 (-1) = 0.01+0.0295
= 0.0395V
Since E cell is positive,
the reaction will be spontaneous in the forward direction.
1
½
½
½
½
18. Given that: (R)
CO3+
+ e-
→ CO2+
E˚ = 1.82V
2H2O → O2 + 4H+
+ 4 e-
E˚ = -1.23V
Explain why CO3+
is not stable in aqueous solution?
Ans:
The E˚ cell can be calculated as:
4[CO3+
+ e-
→ CO2+
] E˚= 1.82V
2H20→ O2 + 4H+
+ 4e-
E˚= -1.23V
Cell reaction: 4CO3+
+ 2H2O → CO2+ O
2 + 4H+
E˚ cell = 1.82V-(-1.23V) = 3.05V
Since E˚cell is positive,
the cell reaction is spontaneous. CO3+
iron will take part in the reaction and
hence
½
½
½
½
½
½

CHAPTER - 4 CHEMICAL KINETICS
POINTS
ONE MARK QUESTIONS
1. Define rate of reaction?
Ans.
The change in concentration w.r.t. time is called rate of reaction. 1
2. How would you define activation energy?
Ans.
The minimum excess energy which must be supplied to reacting species to cross the
energy barrier.
1
3. Write relationship between rate constant and frequency factor?
Ans.
K= Ae-Ea/RT
K= rate constant , A = frequency factor
1
4. What is the effect of temperature on rate of a reaction?
Ans.
The rate of a reaction generally increases with increase in temperature. For a chemical
reaction with every rise in temperature by 10 0
,the rate of the most reactions is almost
doubled.
K= Ae-Ea/RT
1
5. What are the units of rate of reaction for gaseous phase?
Ans.
atms-1
or bars-1
1
6. On increasing the concentration of all or any one of the reactants the rate of reaction
does not change. What can you say about the order of reaction?
Ans.
It is a zero order reaction 1
7. What is rate determining step?
Ans.
In complex reactions, the slowest step determines the overall rate of the reaction. This
step is known as rate determining step.
1
TWO MARK QUESTIONS
8. What is the effect of catalyst on rate of reaction with help of graph?
Ans.
The catalyst increases the rate of a reaction by providing an alternate route for the
reaction which is having less activation energy and hence speeds up the reaction rate.
1

1
9. Define effective collision and collision frequency?
Ans.
The collision that leads to the formation of product is called effective collision
and the no. of collisions per second per unit volume is called collision frequency. 1+1
POINTS
ONE MARK QUESTIONS
10. How would you compare the rate constant and rate of reaction?
Ans.
Rate constant has different units for different order of reaction but rate of reaction has
fixed units.
1
11. What do you understand by most probable kinetic energy?
Ans.
K.E of maximum fraction of molecules is called most probable K.E. 1
12. Write the relationship between activation energy and rate constants at two different
temperatures (say T1 and T2)?
Ans.
Log
𝐾2
𝐾1
=
𝐸𝑎
2.303𝑅
[
1
𝑇1
−
1
𝑇2
]
1
13. What are Arrhenius parameters?
Ans.
K= Ae-Ea/RT
A=Frequency factor,Ea = Activation Energy 1
TWO MARK QUESTIONS
14. What do you understand by the half-life a reaction. What is the relationship of between
rate constant and half-life of a zero order reaction?
Ans.
The time required by a reaction to become half of its reactant concentration is known
as half-life of a reaction.
t1/2=
𝐴 0
2𝐾
1
1

15.
Expt. No. [A]
(mol L-1
)
[B2]
(mol L-1
)
Rate
(mol L-1
s-1
)
1 0.50 0.50 1.6×10-4
2 0.50 1.00 3.2×10-4
3 1.00 1.00 3.2×10-4
Write the most probable equation for the rate of reaction giving reason for your
answer.
Ans.
From an examination of above data, it is clear that when the concentration of B2is
doubled, the rate is doubled. Hence the order of reaction with respect to B2is one.
Further when concentration of A is doubled, the rate remains unaltered. So, order of
reaction with respect to A is zero.
OR
The probable rate law for the reaction will be–
𝑑𝑥
𝑥𝑡
= k[B2]1
[A]0
= k[B2] 1
Alternatively Rate = k[B2]x
1.6 × 10-4
= k[0.5]x
3.2 × 10-4
= k[1]x
On dividing we get x = 1
.•. Rate = k[A]0
[B2]1
= k[B2]
1
1
1
1
15. Differentiate between the activation energy and threshold energy ?
Ans.
The energy requied by the reacting species to form the activated complex is called
Activation energy.
Minimum energy associated with reacting species to cross over the potential barrier is
called threshold energy.
Eth= Average K.E + Ea
1
1
POINTS
ONE MARK QUESTIONS
16. The conversion of molecules X to Y follows second order kinetics. If concentration of X is
increased to three times how will it affect the rate of formation of Y?
Ans.
The reaction follows second order kinetics.
Therefore, the rate equation for this reaction will be:
If the concentration of X is increased to three times, then the rate of formation will
increase by 9 times.
1
TWO MARK QUESTIONS
17. In a reaction, Products, the concentration of A decreases from 0.5 mol L-1
to 0.4 mol L-1
to in 10 minutes. Calculate the rate during this interval?

Ans.
Average rate
= 0.005 mol L – 1
min – 1
= 5 ×10 – 3
M min – 1
½
½
½+½
18. Time required to decompose SO2Cl2to half of its initial amount is 60 minutes. If the
decomposition is a first order reaction, calculate the rate constant of the reaction.
Ans.
We know that for a 1st order reaction,
It is given that
= 1.11 x 10-2
min-1
½
½
½+½
19. A first order reaction has a rate constant 1.15×10-3
sec-1
. How long will 5 g of this
reactant take to reduce to 3 g?
Ans.
From the question, we can write down the following information:
Initial amount = 5 g
Final concentration = 3 g
Rate constant =
We know that for a 1st order reaction,
Rate constant = 1.15×10-3
sec-1
We know that for a 1st
order reaction,
½
½

= 444.38 s
= 444 s (approx.
= 444.38 s
= 444 s (approx.)
½
½
20. The time required for 10% completion of first order reaction at 298 K is equal to that
required for its 25% completion at 308K. If the pre-exponential factor for the reaction is
3.56 ×109
s–1
, calculate the energy of activation.
Ans.
Ea = 76.65 kJ/mol
½
½
½
½
½
½
21. The rate law for the reaction, 2Cl2O → 2Cl2 + O2 at 200 o
C is found to be
rate = k[Cl2O]2
(a) How would the rate change if [Cl2O] is reduced to one-third of its original value?
(b) How should the [Cl2O] be changed in order to double the rate?
(c) How would the rate change if [Cl2O] is raised to threefold of its original value?
Ans.
(a) Rate equation for the reaction,
r = k[Cl2O]2
Let the new rate be r'; so
r' = k[(Cl2O)⅓]2
=
1
9
times
(b) In order to have the rate = 2r, let the concentration of Cl2O be x.
So 2r = kx2
.... (i)
We know that r = k[Cl2O]2
.... (ii)
½
½
½

Dividing Eq. (i) by (ii),
2𝑟
𝑟
=
𝑘𝑥2
𝑘 𝐶𝑙2𝑂 2
or 2 =
𝑥2
𝐶𝑙2𝑂 2
or x2
= 2[Cl2O]2
or x = [Cl2 O] × √2
(d) New rate = k[3Cl2O]2
= 9k[Cl2O]2
= 9r
½
½+½
22. The decomposition of NH3 on platinum surface is a zero order reaction. What are the
rate of production of N2 and H2. If k= 2.5 x 10-4
Ans.
2NH3 →N2 + 3H2
-
1𝑑 𝑁𝐻3
2𝑑𝑡
=
𝑑 𝑁2
𝑑𝑡
=
1𝑑 𝐻2
3𝑑𝑡
rate = k x [NH3]0
= 2.5 X 10-4
molL-1
sec-1
𝑑 𝑁2
𝑑𝑡
=
1𝑑 𝐻2
3𝑑𝑡
=
1
2
×2.5×10-4
molL-1
sec-1
d[H2]/dt = - 3/2( d[NH3]/dt) = 3/2 x2.5x10-4
= 3.75x10-4
molL-1
sec-1
Rate = - d[NH3]/dt = k x[NH3]0
= 2.5 x 10-4
molL-1
sec-1
Rate of production of N2 = 2.5X10-4
molL-1
sec-1
½
½
½
½
½+½
23. The following data were obtained during the first order thermal
decomposition of N2O5(g) at constant volume
2N2O5(g) → 2N2O4(g) + O2(g)
S.No. Time/s Total Pressure/(atm)
1. 0 0.5
2. 100 0.512
Calculate the rate constant
Ans.
Let the pressure of N2O5(g) decreases by 2x atm.As two moles of N2O5 decomposes to give
two es of N2O4(g) and one mole of O2(g) ,the pressure of N2O4(g) increases mol by 2x
atm.and that of O2(g) increases by x atm.
2N2O5(g) → 2N2O4(g) + O2(g)
Start t=0 0.5 atm. 0 atm 0 atm.
At time t (0.5 – 2x) atm. 2x atm. x atm.
½
½

pt= p N2O5 +p N2O4 +p O2
= (0.5 – 2x) + 2x + x = 0.5 + x
x = pt– 0.5
p N2O5 = 0.5 - 2x
= 0.5 – 2(pt– 0.5) = 1.5 – 2pt
At t =100 s
pt=0.512 atm
½
½
½+½

CHAPTER -5 SURFACE CHEMISTRY
POINTS
ONE MARK QUESTIONS
1. What is a solid sol ?
Ans.
Colloids which have both dispersed and dispersion medium in solid phase. 1
2. Define adsorption?
Ans.
The accumulation of molecules of a species at the surface rather in the bulk
of a solid or liquid is termed adsorption.
1
3. What is the sign of free energy change during heat of adsorption?
Ans.
Negative 1
4. What are dispersed phase and dispersion medium in cheese?
Ans.
Dispersed phase- liquid ; Dispersion medium solid. ½ + ½
5. What do you understand by reversible and irreversible colloids and why are
these called so?
Ans.
In reversible colloids constituents can be separated easily and colloid is
formed on mixing again while it can not be done in irreversible colloids.
½ + ½
6. A little quantity of egg albumin was shaken with water , Name the type of
colloid formed .
Ans.
Macromolecular lyophilic colloid there is attraction between disperse phase
and dispersion medium.Egg albumin is protein polymer having size of colloidal
dimension.
½ + ½
TWO MARK QUESTIONS
7. What is a multimolecular colloid? Give two examples?
Ans.
Colloidal particle are formed by aggregation of large number of atoms or
molecules.
Gold sol and Sulphur sol.
1
½ + ½
8. What is hydrophilic colloid? How these colloids are prepared?
Ans.
Colloids have dispersion medium is water and the particles of disperse phase
have attraction towards water ,
These are prepared by mixing dispersed phase with water.
1
1

9. 1. Define following.
i) Micelle ii)FOAM iii)Gel
Ans.
i) The substance at low concentration behave as strong electrolytes,
molecules above C.M.C. aggregate and aggregated particle having size of
colloidal range.
ii) The colloid in which dispersed phase is gas and dispersion medium is liquid.
iii) The colloid in which dispersed phase is liquid and dispersion medium
solid.
1
1
1
POINTS
ONE MARK QUESTIONS
10. Why Gold sol can not be prepared by shaking gold with water ?
Ans.
Lyophobic colloid do not form colloid by mixing two phases. 1
11. What is the role of enzyme?
Ans.
These catalyse the biochemical reactions. 1
12. How would you explain the selectivity of catalyst?
Ans.
On the basis of pore size of the catalyst and size of reactant and products 1
13. Name the catalyst in the manufacture of sulphuric acid and give its function?
Ans.
V2O5 accelerates the rate of reaction by the formation of an activated
complex with reactants. ½ +½
14. How would you differentiate adsorption and absorption?
Ans.
Adsorption is a surface phenomena while absorption is bulk phenomena. 1
15. Why is it necessary to remove CO when ammonia is obtained
by Haber’s process?
Ans.
CO acts as poison catalyst for Haber’s process and lowers the
activity of solution therefore it is necessary to remove when NH3
obtained by Haber’s process. 1

TWO MARK QUESTIONS
16. What do you understand about adsorption isotherm? Show graph.
Ans.
The variation in the amount of gas
adsorbed by the adsorbent with pressure
at constant temperature can be expressed
by means of a curve termed as adsorption
isotherm.
1+1
17. How is adsorption of a gas related to its critical temperature? Give reason.
Ans.
Higher the critical temperature of the gas. Greater is the
ease of liquefaction. i.e. greater Vander walls forces of attraction
and hence large adsorption will occur. 1+1
18. What is meant by Shape Selective Catalysis?
Ans.
In the Shape Selective Catalysis, the rate depends upon pore size
of the catalyst and the shape & size of the reactant and products
molecules. 1+1
19. What is hardy Schulz rule? Which of the following electrolyte will coagulate
most easily to positively charged colloid and why?
a )NaCl b)Na2 So4 c)Na3Po4
Ans.
Coagulating power of a coagulating ion is directly proportional to the
charge on the ion.
Na3Po4will coagulate a positively charged colloid most easily because it has
most negative valence.
1
1+1
POINTS
ONE MARK QUESTIONS
20. Name the types of emulsion .Which of these is less stable and how are these
stabilized ?
Ans.
W/O and O/W type. O/W type are less stable .By adding an emulsifying agent
like soap detergent etc.
1
21. Why does sky look blue?
Ans.
Dust particles along with water particles scatter blue light maximum.
1
TWO MARK QUESTIONS
22. How alum removes suspended impurities? ‘
Ans.
Ions Alum neutralize the charge of colloidal soil particles and precipitates.
1

23. Name two food articles which are colloids .
Ans.
Ice cream , Butter ,milk etc. ½ + ½
24. Explain the following:-
i) Delta formation ii) Some medicines are given in colloidal form
Ans.
i) Soil particles in river water is a colloid and is precipitated by electrolytes of
sea water and river stream splits in two streams and so on.
ii) In medicines colloidal form have larger surface area and are effectively
absorbed.
1
1
25. Comment on the statement that “colloid is not a substance but state of a
substance”?
Ans.
Given statement is true. This is because the substance may exist as colloid under
certain conditions and as a crystalloid under certain other conditions.
E.g: NaCl in water behaves as a crystalloid while in benzene, behaves as a
colloid (called associated colloid). It is the size of the particles which matters
i.e. the state in which the substance exist. If the size of the lies in the range
1nm to 1000nm it is in the colloid state.
1
1
26. What is demulsification ? Give two methods of demulsification ?
Ans.
Separating an emulsion in constituent liquids.
Heating and freezing.
1
½ + ½
27. Write short notes on followings:-
(a)Tyndall effect (b)Brownian Movement (c)Hardy Schulze Rule
Ans.
(a)Tyndall effect-scattering of light by colloidal particles by which path of
beam becomes clearly visible. this effect is known as tyndall effect.
(b) Brownian movement-zig-zag motion of colloidal particles.
(c) Hardy Sehulze Law - Coagulating value of a coagulating ion is
directly proportional to the charge on the ion.
e.g: Na +<Ca++< Al 3+ for negatively changed sol.
Cl-
< CO3
2-
< PO4
3-
< [Fe (CN) 6 ]4 -
for positive sol.
1
1
½ + ½

CHAPTER -6 GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS
POINTS
ONE MARK QUESTIONS
1. Name an element which can be refined by zone refining.
Ans.
Si Or Ge. 1
2. Name a method which is used for refining of metal having low melting point.
Ans.
Liquation process. 1
3. Name a method which is used for refining of metal having low boiling point.
Ans.
Distillation method 1
4. Which type of elements are extracted by reduction method?
Ans.
Metals 1
5. Which types of elements are extracted by oxidation method?
Ans.
Nonmetals 1
TWO MARK QUESTIONS
6. What is the principle of vapour phase method ?
Ans.
Metal should form volatile compound with given reagent at low temperature.
Volatile compound of metal should be unstable at high temperature.
1
1
7. Write the principle of following.
i) Zone refining ii) Electrolytic refining
Ans.
ZONE REFINING –Impurities are more soluble in the molten state than the pure
metals i.e. on cooling pure metal crystallize out on cooling while impurities will
remain behind.
ELECTROLYTIC REFINING – Pure metal deposit at cathode from electrolyte by
passing electricity while crude metal from anode decompose to electrolyte.
1
1
8. Name two alloys of copper and write their composition.
Ans.
Brass- Cu 60%,Zn 40%
Bronze – Cu 80 %, Zn 10%, Sn 10 %
1
1

9. (a) What are the two phases of chromatographic method ?
(b) Name a compound which is filled in column during chromatographic method.
(c) What type of metals can be refined by chromatographic method?
Ans.
a) Stationary phase and mobile phase.
b) Alumina
c) This method is used for the extraction of such metals which are present in
minute amount and the impurities are not very different in chemical properties.
1
1
1
10. Write the principle method and equation of electrolytic refining.
Ans.
Principle- On passing electric current pure metal collected at cathode.
Method- Thick plate of impure metal is taken as anode and thin plate of pure
metal is taken as cathode. Electrolyte is an aqueous salt solution of metal. On
passing electric current pure metal is collected at cathode and impurities settle
down as anode mud.
Reaction at anode M Mn+
+ ne-
Reaction at cathode Mn+
+ ne-
M
1
1
½
½
11. What are the different methods used for reduction of metal oxide used for
reduction of metal oxide into metal?
Ans.
I. By heating
II. By chemical – Smelting, Aluminothermite process.
III. Auto reduction
OR
IV. Electrolytic reduction
1
1
1
ONE MARK QUESTIONS
12. Name a method of refining of metal which are available in minute quantities and
impurities are not very different in chemical properties.
Ans.
Chromatographic method.
1
13. Name a metal which is extracted by oxidation method?
Ans.
Au/Ag 1
14. Name an alloy of aluminium used in making aero plane?
Ans.
Duralumin 1
15. What type of metal oxide are reduced by electrolysis?
Ans.
Highly reactive metal or highly electropositive. e.g. Na2O 1

16. Why we add cryolite or fluorspar in alumina during Hall’s Heroults process?
Ans.
To reduce the melting point of alumina and make it more conductive. 1
17. Why graphite rods gradually consume during Hall’s Heroults process?
Ans.
Because oxygen liberates at anode and in presence of oxygen , graphite rod get
consume to produce CO2. 1
TWO MARK QUESTIONS
18. What is the role of iodine in the refining of Titanium? Write equation also?
Ans.
Iodine reacts with titanium to form volatile TiI4 which is thermally decomposable
to give pure Titanium.
Low temp High temp.
Ti + 2I2 ------ [TiI4] ------ Ti + 2I2
1
1
19. Out of molten NaCl and aqueous NaCl, which is used for the extraction of sodium
and why?
Ans.
Molten NaCl is preferred because
Molten NaCl on electrolysis gives sodium but aqueous NaCl produces NaOH
instead of Sodium.
1
1
20. How is chlorine extracted from brine? Write equation?
Ans.
Chlorine is extracted from brine by oxidation method
NaCl(aq) Na+
+ Cl-
At anode
2Cl -
+ 2H2O Cl2 + H2 + 2OH-
At cathode
2H+
+2e-
------ H2
1
1
21. Name 3 types of iron? Write the differences among them.
Ans.
Pig iron, cast iron and wrought iron.
I. Pig iron contains 4% carbon,
II. Cast iron contains 3% iron
III. Wrought iron contains 0.2-0.5% Carbon.
1
1
1
22. Copper can be extracted by hydrometallurgy but not zinc. Explain?
Ans.
E0
(standard reduction potential) of Zn and Fe both are lower than that of copper.
So both can displace copper from salt solution of copper. But being a cheaper iron
scrap is used in hydrometallurgy of Copper.
Fe(s) + Cu2+
(aq) Fe2+
(aq) + Cu(s)
1
1
1

23. Name a metal which can be extracted by oxidation method. Explain this method
with equation.
Ans.
Silver.
Silver is extracted by cyanide process in which it is leached with dilute NaCN
solution in presence of O2, from which the silver is obtained by later displacement
by zinc.
4Ag + 8CN-
+H2O + O2  4[Ag(CN)2]-
+ 4OH-
2[Ag(CN)2]-
+ Zn  2Ag + [Zn(CN)4]2-
1
1
1
ONE MARK QUESTIONS
24. Name a metal used for making foil for chocolate.
Ans.
Aluminium 1
25. Name a metal used in galvanization of iron.
Ans.
Zinc 1
26. Name a metal which is used in making electrical wires and coins.
Ans.
Copper 1
27. How much amount of graphite is consumed to get 2Kg Aluminium?
Ans.
1 Kg 1
TWO MARK QUESTIONS
28. What is German silver? Write its use.
Ans.
German silver is an alloy of Cu. Its composition is Cu=25-30%,
Zn= 25-30%, Ni= 40-50%.
It is used for making artificial jewelry.
1
1
29. Name two factors which are considered in electrolytic reduction during extraction
of metal.
Ans.
a) Reactivity of metal
b) Suitable electrodes
1
1
30. Why copper is used in making
a) Electrical wires
b) Utensils
Ans.
a) Copper is used in making electrical wires because it is ductile and good
conductor of electricity.
b) Copper is used in making utensils because it is good conductor of heat.
1
1

31. Write one alloy for each of the following metal along with its use.
Copper, Aluminium, Iron
Ans.
Copper - BRONZE - COINS
Aluminium - MAGNALUM - PRESSURE COOKER
Iron - STAINLESS STEEL - UTENSILS
½ + ½
½ + ½
½ + ½
32. a) Name the impurities deposit as anode mud during electrolytic refining of
copper.
b) What type of metal can be refined by liquation method.
c) Name the method for refining of metals which are used as semiconductors.
Ans.
a) Sb, Se, Te, Ag, Au, Pt
b) Those metals which have low melting points. e.g. Tin
c) Zone refining
½ + ½
½ + ½
½ + ½

CHAPTER -7 p- BLOCK ELEMENTS
POINTS
ONE MARK QUESTIONS
1. What is laughing gas ?
Ans.
N2O( nitrous oxide) is known as laughing gas 1
2. Name the gas use to preserve biological specimen
Ans.
Liquid N2 1
3. What are the different allotropes of sulphur?
Ans.
Rhombic and monoclinic 1
4. Which form of sulphur is stable at room temperature?
Ans.
Rhombic sulphur 1
5. Write the name of shape of sulphur .
Ans .
Crown shape 1
6. Why is sulphuric acid used as a dehydrating agent?(Level2) 1 mark
Ans .
It has a strong affinity with water. 1
7. Halogens have maximum negative electron gain enthalpy. Why?
Ans.
Halogens have the smallest size in their respective periods 1
8. Why are halogens colored?
Ans.
Due to presence of unpaired electron. 1
9. Fluorine exhibits only –1 oxidation state whereas other halogens exhibit +1, +3,
+5 and +7 oxidation states also. Explain.
Ans.
Fluorine is the most electronegative element and absence of d –orbital. ½+½
10. Name two poisonous gases prepared from chlorine gas.
Ans.
Phosgene gas(COCl2), Mustard gas Cl-C2H4-S-C2H4Cl ½+½

11. Give the reason for bleaching action of Cl2.
Ans.
Cl2+H2O → HCl+[O]
Coloured substance + [O]→ Colourless 1
12. Noble gases have comparatively largest atomic sizes.
Ans.
In noble gases we can measure only Vander Waals radii which are larger than
covalent radii. 1
13. What is the colour of HNO3 in pure form?
Ans.
Yellow 1
14. Noble gases have comparatively largest atomic sizes.
Ans.
In noble gases we can measure only Van der Waals radii which are larger than ionic
radii. 1
15. Noble gases exhibit very high ionization enthalpy.
Ans.
Due to stable electronic configuration these gases exhibit very high ionisation
enthalpy. 1
TWO MARK QUESTIONS
16. How is ozone prepared? State the condition of reaction.
Ans.
3O2 silent electric discharge 2O3
The formation of ozone is an endothermic reaction; it must be carried out at high
temperature
1
1
17. What happens when SO2 reacts with water? (Level2) 2markIt
Ans.
It breaks up in water to give sulphurous acid
SO2 + H2O → H2SO3
Sulphur dioxide Water Sulphurous acid
1
1
18. What is transition temperature for allotropic forms of sulphur?
Ans.
It is a temperature at which both the allotropes of sulphur are stable ; 369K is called
transition temperature.
1+1
19. Two examples to show the anomalous behavior of F2
Ans.
1.It shows (-1) oxidation state only
2. It has less negative electron gain enthalpy than chlorine.
1
1

20. The elements of group 18 are known as noble gases. Give two reasons?
Ans.
The elements present in Group 18 have their valence shell completely filled and,
therefore, react with a few elements only under certain conditions. Therefore, they
are now known as noble gases.
1
1
21. Draw the structure of nitric acid in gaseous state.
Ans.
1 + 1
22. Which chemical compound is formed in the ring test of NO3
-
ions?
Ans.
[Fe(H2O)5NO]SO4 is formed which is pentaaquanitrosyl iron (II) sulphate. 1 + 1
23. Give reasons-
a)Xenon readily forms compounds but Krypton does not form compounds easily.
b) He and Ne does not form compounds with fluorine.
Ans.
a)Xenon has lower ionization energy than Krypton, therefore, Xe forms compounds
b)Due to non-availability of vacant d-orbital.
1
1
24. Why Xe does not forms compounds such as Xe F3 and XeF5.
Ans.
By the promotion of one, two or three electrons from filled p-orbital to the vacant d-
orbital in the valence shell, 2,4 or 6 half filled orbitals are formed. Thus Xe can
combine only with even number of fluorine and not odd.
1
1
25. Explain the formation of sulphuric acid with contact process?
Ans.
The contact process involves three steps.
Step -I:
Production of sulphur dioxide : S + O2 SO2
Step -II:
Formation of sulphur trioxide: SO2 + O2 SO3
Step -III
Conversion of sulphur trioxide into sulphuric acid: The sulphur trioxide formed in the
second step is dissolved in 98% sulphuric acid to give pyrosulphuric acid or oleum.
Oleum is then diluted with water to give sulphuric acid of the desired concentration.
1
1
1

SO3 (Sulphur triioxide) + H2SO4 (Sulphuric acid 98%) →
H2S2O7 (Pyrosulphuric acid -Oleum)
H2S2O7 (Pyrosulphuric acid -Oleum) + H2O (Dilution ) → 2H2SO4(Sulphuric acid)
26. Write the balanced chemical equation for the action of Cl2 with hot and
concentrated NaOH.Is this reaction a disproportionation reaction? Justify.
Ans.
3 Cl2 + 6 NaOHConc →5 NaCl + NaClO3 + 3 H2O
Yes
Chlorine from zero oxidation state is changed to –1 and +5 oxidation states.
1
1
1
27. When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric
chloride. Why?
Ans.
Its reaction with iron produces H2.
2Fe+2HCl→FeCl2+H2
Liberation of H2 prevents the formation of ferric chloride.
1
1
1
28. How are XeF2 XeF4, XeF6, XeO3, XeOF4 prepared?
Ans: (i) Xe (g) + F2 (g) XeF2(s) Temp= 673K, Pressure=1bar
(xenon in excess)
(ii) Xe (g) + 2F2 (g) XeF4(s) Temp= 873 K, Pressure=7 bar
(1:5 ratio)
(iii) Xe (g) + 3F2 (g) XeF6(s) Temp= 573 K, Pressure=60 -70bar
1
1
1

POINTS
ONE MARK QUESTIONS
29. What is the covalence of Nitrogen in N2O5 ?
Ans.
4 1
30. Why ; SO2 acts as a reducing agent?
Ans.
It easily expands its oxidation state form +4 to +6 1
31. Why is ozone thermodynamically unstable?
Ans.
It decomposes to oxygen and nascent oxygen
O3 --------- O2 + [O]
1
32. What happens when SO2 reacts with NaOH?
Ans.
It reacts promptly with sodium hydroxide solution to give sodium sulphite.
SO2 + 2NaOH → Na2SO3 + H2O
1
33. Why is boiling point of sulphuric acid high?
Ans.
It is due to intermolecular hydrogen bonding. 1
34. Why is sulphuric acid known as an oxoacid?(level1) 1 mark
Ans.
It is due the presence of OH group which releases H+
ion. 1
35. What happens when H2SO4 is poured on sugar? (level2) 1 mark
Ans.
White sugar turns into black . 1
36. Deduce the molecular shape of BrF3 on the basis of VSEPR theory.
Ans.
There are three(3) bond pairs and two(2) lone pairs. The two lone pairs will occupy
the position at equatorial which are cause for the distraction of molecule hence it has
a distorted ‘T’ shape structure.
1
37. Why is ICl more reactive than I2?
Ans.
I-Cl bond is weaker than I-I bond 1
38. Why are halogens strong oxidizing agents?
Ans.
Due to low bond dissociation energy, high electronegativity and large negative
electron gain enthalpy 1

39. Although electron gain enthalpy of fluorine is less negative as compared to chlorine,
fluorine is a stronger oxidizing agent than chlorine. Why?
Ans.
It is due to
I. Low enthalpy of dissociation of F-F bond
II. High hydration enthalpy of F
–
½
½
40. What happens when NaCl is heated with sulphuric acid in the presence of MnO2.
Ans.
4NaCl+MnO2+4H2SO4→MnCl2 + 4NaHSO4 +Cl2 +H2O 1
41. Why Noble gases exhibit very high ionization enthalpy.
Ans.
Due to stable electronic configuration (ns2
np6
) these gases exhibit very high
ionization enthalpy.
1
42. Why Noble gases form compounds with fluorine and oxygen only.
Ans.
Because fluorine and oxygen are strong oxidizing agents (most electronegative
elements)
1
43. N2 is considered as a inert gas at room temperature; why ?
Ans.
Due present of triple bond it has very high bond dissociation energy. 1
44. Why metals such as Cr,Al do not dissolve in con. HNO3?
Ans.
Due to the formation of a passive film of oxide on the surface.
1
45. PH3 has lower boiling point than NH3;why
Ans.
There is Inter molecular hydrogen bonding in NH3 1
46. Noble gases form compounds with fluorine and oxygen only.Why?
Ans.
Because fluorine and oxygen are strong oxidizing agents (most electronegative
elements)
1
47. Helium is used for inflating aeroplane tyres & filling balloons for metrological
observations.Why?
Ans.
Helium is a non-inflammable and light gas
1
48. Nitrogen does not form pent halide. Give reason.
Ans.
Nitrogen does not expand its covalence beyond four due to absence of d- orbital
1

TWO MARK QUESTIONS
49. In vapour phase sulphur is paramagnetic; why?
Ans.
It exists in S2 form which has unpaired electron’s in antibinding ‘p’ orbital. 1+1
50. How SO2 reacts with ferric ion and molecular halogens?
Ans.
i) It reduces ferric salts to ferrous salts, and
2Fe3+
+ SO2 + 2H2O → 2Fe2+
+ SO4
2-
+ 4H+
Ferric Sulphurdioxide ferrous salt
ii) it reduces halogens to halogen acids.
X2 + SO2 + 2H2O → SO4
2-
+ 2X-
+ 4H+
Halogen Sulphur water halogen acid
1
1
51. How ozone reacts with i) KI ii) PbS
Ans.
i) 2KI + H2O + O3 → 2KOH + I2 + O2
ii) 4O3 + PbS → 4O2 + PbSO4
1
1
52. Explain why fluorine forms only one oxoacid, HOF.
Ans.
Due to high electronegativity and absence of d-orbitals 1+1
53. Write two uses of ClO2.
Ans.
It is powerful oxidizing agent and chlorinating agent 1+1
54. Noble gases are mostly chemically inert.Why?
Ans.
Their inertness to chemical reactivity is attributed to the following reasons:
(i) The noble gases except helium (1s2
) have completely filled ns2
p6
electronic
configuration in their valence shell.
(ii) They have high ionization enthalpy and more positive electron gain enthalpy
1
1
55. Does the hydrolysis of XeF6 lead to a redox reaction?
Ans.
No, the products of hydrolysis are XeOF4 and XeO2F2 where the oxidation states of all
the elements remain the same as it was in the reacting state.
1+1
56. Noble gases are mostly chemically inert.
Ans.
Their inertness to chemical reactivity is attributed to the following reasons:
(i) The noble gases except helium (1s2
) have completely filled ns2
p6
6 electronic
configuration in their valence shell.
(ii) They have high ionisation enthalpy and more positive electron gain enthalpy.
1
1

57. What happens when SO2 reacts with : i) Cl2 ii) O2
Ans.
i) It reacts with chlorine within the presence of charcoal as an impetus to give
sulphuryl chloride (SO2Cl2).
SO2+ Cl2 → SO2Cl2
Sulphur dioxide Chlorine (Catalyst) Sulphuryl Chloride
ii) Within the sight of vanadium penta oxide ( catalyst) , it gives sulphur trioxide.
V2O5
SO2 + O2 ----- → SO3
Sulphur dioxide Oxygen Sulphur trioxide
½
1
½
1
58. Are all five bonds in PCl5 molecule equivalent? Draw structure.
Ans.
No,
There are three equatorial bonds and two axial bonds. Axial bonds are longer than
equatorial bonds
1
1
1
59. Write the hydrolysis products of XeF2 XeF4, XeF6.
Ans.
i) 2XeF2 (s) + 2H2O(l) → 2Xe (g) + 4 HF(aq) + O2(g
ii) 6XeF4 + 12 H2O → 4Xe + 2Xe03 + 24 HF + 3 O2
iii) XeF6 + 3 H2O → XeO3 + 6 HF
Partial hydrolysis of XeF6 gives oxyfluorides,
XeF6 + H2O → XeOF4 + 2 HF
XeF6 + 2 H2O → XeO2F2 + 4HF
1
1
1
POINTS
ONE MARK QUESTIONS
60. What is the purpose of formation of oleum during contact process?
Ans.
Oleum is diluted with water to give sulphuric acid of the desired concentration. 1
61. Bond dissociation energy of F2 is less than that of Cl2.Why?
Ans.
In F2, large repulsion occur between non bonding electrons on small sized fluorine
atom in the fluorine molecule.
1

62. Why the acid strength of acids increase in the order HF < HCl < HBr < HI.
Ans.
Due to increasing bond length of H-X from HF to HI, It is easy to release H+
ion by HI.
1
63. Why fluorine shows abnormal behavior ?
Ans.
Due to very small size, high electro negativity and high polarizing power.(any two
reasons)
1
64. Why are Inter Halogen compounds more reactive than Halogens?
Ans.
Inter halogen compounds are more reactive than halogens because the X-X bond in
inter halogen compounds is weaker than the X-X bonds in halogen compounds. It
breaks easily and thus compounds react more.
1
65. HF is liquid at room temperature. Give reason
Ans.
Due to intermolecular hydrogen bonding in HF.
1
66. Of the noble gases only Xenon is known to form real chemical compounds.
Ans.
The ionization energy of xenon is relatively low and therefore, it is possible to excite
the paired electrons from np orbital’s to nd sub-shell
1
67. What happens whenPCl5.is heated ?
Ans.
PCl5 PCl3 + Cl2
1
68. It has been difficult to study the chemistry of radon.
Ans.
Radon is radioactive element. 1
69. Which compound led to discovery of compounds of noble gases?
Ans: O2
+
PtF6
-
1
70. How many spheres are in contact with each other in a single plane of a close
packed structure?
Ans.
Six(6) 1
71. What other elements may be added to silicon to make electrons available for the
conduction of an electric current?
Ans.
Phosphorous or Gallium. 1
72. How many Tetrahedral sites per sphere are there in a cubic closest – packed (
face centered cubic) structure.?
Ans.
Two 1

TWO MARK QUESTIONS
73. Why; does NO2 dimerise ? Explain.
Ans.
NO2 Containodd number of valence electrons. It behaves as a typical molecule. In
the liquid and solid state, it dimerise to form stable N2O4 molecule, with even
number of electrons. Therefore, NO2 is paramagnetic, while N2O4 is diamagnetic in
which two unpaired electrons get paired.
1+1
74. Give reasons
(i) Ammonia is a good complexing agent.
(ii) In group 15 elements ,the bond angle H-M-H decreases in the following order
NH3( 107.80
),
PH3 ( 93.60
), AsH3 ( 91.80
).
Ans.
(i) Ammonia has a lone pair of electron therefore it is a good complexing agent.
(ii) With the increase in the size of group 15 element as bond length increases the
bond pair-bond pair repulsion decreases. Therefore the angle H-M-H decreases from
NH3 to AsH3 .
1
1
75. Explain the following:-
(i) Nitrogen exists as diatomic molecule whereas phosphorous exists as tetra atomic
molecule.
(ii) NF3 is an exothermic compound but NCl3 is an endothermic compound.
Ans.
(i) N-N single bond is very weak due to its small size
(ii) F is highly electronegative and N-F bond energy is higher than N-Cl bond energy.
1
1
76. How is monoclinic sulphur prepared?
Ans.
When we take a dish and melt rhombic sulphur in that dish we obtain monoclinic
sulphur after cooling it.
In this process we make two holes in the crust and pour out the remaining liquid.
After this we get colourless needle-shaped crystals of β-sulphur when the crust is
removed
1
1
77. Give any four uses of sulphur.
Ans.
i) Sulphur is used for vulcanization of rubber.
ii) Many of its compounds are used as insecticides in crops.
iii)Many bleaching agents can be manufactured using sulphur.
iv)It is also used in manufacturing of carbon disulphide which in turn is used in skin
ointments and other such products.
½
½
½
½

78. Write balanced reactions of conc. H2SO4 reacts with i) Cu ii)C
Ans.
i) Cu + 2H2SO4 → CuSO4+ SO2 + H2O
ii) C + 2H2SO4 → CO2 + 2SO2 + 2H2O
1
1
79. How does Cl2 react with
(i) cold and dilute NaOH (ii) hot and concentrated NaOH
Ans.
When Chlorine reacts with cold sodium hydroxide, it forms sodium chloride.
(i) 2NaOH + Cl2 = NaCl + NaOCl + H2O
When Chlorine reacts withhot sodium hydroxide, it forms sodium
chloride and other products.
(ii) (ii) 6NaOH + 3Cl2 = 5NaCl + NaClO3 + 3H2O
1
1
80. PCl3 gives fumes in moisture,why?Give equation.
Ans.
It is hydrolysed in moisture and form fumes of HCl.
PCl3 + 3H2O----- 3HCl (Fumes) + H3PO4
1
1
81. Complete the reactions:-
1. S8 + 48 HNO3 -------
2. P4 + 20HNO3 -------- ________________+ 20NO2 + 4H2O
Ans.
1.8H2SO4 + 48NO2 + 16H2O
2. 4H3PO4
1
1
82. Give Reasons-
(i) Noble gases exhibit very high ionization enthalpy.
(ii) Of the noble gases only Xenon is known to form real chemical compounds.
Ans.
(i) Due to stable electronic configuration these gases exhibit very high
ionisation enthalpy.
(ii) The ionization energy of xenon is relatively low and therefore, it is possible
to excite the paired electrons from np orbitals to nd sub-shell
1
1

CHAPTER - 8 d- & f- BLOCK ELEMENTS
POINTS
One Mark Questions
1. Why do transition elements show variable oxidation states?
Ans.
Small energy gap between ns and (n-1) d sub shells, both ns and (n-1) d electrons
take part in bond formation 1
2. Name a transition element which does not exhibit variable oxidation state
Ans.
Scandium(Z=21) 1
3. Name a member of lanthanoids series which is well known to exhibit +4 oxidation
state
Ans.
Cerium(Z=58)
1
Two Mark Questions
4. Define transition elements? Which group of d- block elements is not considered as
transition element and why?
Ans.
Transition elements have partly filled d sub shell in its ground state or any of its
oxidation.
Group 12 elements are not considered as transition elements due to completely
filled d- orbital in ground state as well as oxidation states.
1
1
5. Write electronic configuration of following species.
i) Cr (24) ii) Mn2+
(25)
Ans.
i) Ar(18) 3d5
4s1
ii) Ar(18)3d5
1+ 1
6. Describe the method of preparation of KMnO4 from MnO2.
Ans. It takes place in two steps:
(i)Conversion of MnO2 into K2MnO4
(ii) Conversion of K2MnO4into KMnO4
Following reactions take place
2MnO2+4KOH +O2 2K2MnO4 + 2H2O
3MnO4
2--
+4H+ 2MnO4
-+MnO2 +2H2O
1
1

Three Marks Questions
7. (i)Write the steps involved in the preparation of K2Cr2O7 from chromite ore.
(ii)What is the effect of pH on dichromate ion solution?
Ans.
(i) It takes place in three steps:
Conversion of chromite ore into sodium chromate
4FeCr2O4+ 4Na2CO3 + 7O2 Na2CrO4+ Fe2O3 + 8CO2
Conversion of sodium chromate into sodium dichromate
2Na2CrO4 + 2H+
Na2Cr2O7 + 2Na+
+ H2O
Conversion of sodium dichromate into potassium dichromate
Na2Cr2O7 + 2KCl K2 Cr2O7 + 2NaCl
(ii)Dichromate ion (orange) changes to chromate ion (yellow) in basic solution.
Cr2O7
2-
+ 2 OH-
2CrO4
-2
+ H2O
1
1
1
POINTS
One Mark Questions
8. Explain why Cu+
is not stable in aqueous medium?
Ans.
Cu+ disproportionate into Cu2+, and Cu due to higher hydration enthalpy of Cu++
. 1
9. Actinoid contraction is greater from element to element than
LanthanoidContraction. Why?
Ans.
Due to poor shielding by 5f electrons in the Actinoids than that of the 4f electron in
the Lanthanoids.
1
10. Write electronic configuration Cu+
(Z=29)
Ans.
Ar(18)3d10
1
11. Explain why Ce+4
is a stronger oxidizing agent ?
Ans.
This is because Ce+4
tends to change Ce+3
by losing an electron +3 oxidation state is
more stable.
1
Two Mark Questions
12. Explain giving reason:
(i) Transition metals and their many compounds act as good catalyst.
(ii)Transition metals have a strong tendency to form complexes.
Ans.
(i) Transition metals and their many compounds act as good catalyst It is due to
(a) partially filled(n‐1)d orbital (b) Variable oxidation state (c) Ability to change
oxidation state frequently.
(ii) Transition metals have a strong tendency to form complexes .Most of transition
elements form complex compounds due to -(a) small size (b) high charge (c)
presence of vacant d‐orbital of suitable energy.
1
1

13. What is lanthanide contraction? How does it affect the chemistry of elements,
which follow lanthanoids?
Ans.
Decrease in atomic / ionic radii across lanthanoid series with increase in atomic
number.
Due to lanthanoid contration the atomic/ionic radii 5d Series elements decrease
Consequently the properties of 4d and 5d series elements become similar.
1
1
Three Mark Questions
14. Assign giving suitable reason which of the following pairs exhibits the property
indicated against each.
(i)Sc3+
or Cr3+
exhibits Para magnetism
(ii)V or Mn exhibits more number of oxidation states exhibits
(iii)V4+
or V5+
exhibits colour
Ans.
(i) Cr3+
has three unpaired electron in 3d – sub shell hence it is paramagnetic.
(ii)Mn exhibits more no. of oxidation states as it has five unpaired electrons in (n-1)d
and two electrons in ns orbital take part in bonding.
(iii)V4+
is coloured as it has 3d1
configuration while V5+
has 3d0
configuration, hence
electrons on V4+
undergoes d-d transition.
1
1
1
POINTS
One Mark Questions
15. Why Cobalt (II) is stable in aqueous solution but in the presence of strong ligands it
can be easily oxidized to Co(III) ?
Ans.
Strong ligands force cobalt (II) to lose one more electron from 3d sub shell hence
oxidised into Co (III) to form stable complex by undergoing d2
sp3
hybridisation.
1
16. A tripositive metal ion has electronic configuration 3d5
4s0
in ground state, to which
group of periodic table does this element belong to? Name the element also.
Ans.
It is iron and has atomic no 26.
It belong s to 8th
group.
½
½
17. La(OH)3 is stronger base than Lu(OH)3Why?
Ans.
Due to larger size of La3+ as compared to Lu3+, it has a greater ionic character
hence gives more no. of –OH ions.
½
½

Two Mark Questions
18. Comment on the following:
i) Zr and Hf have similar sizes
ii) Actinones show greater no. of oxidation state than Lanthanones.
Ans.
i) As a consequence of Lanthanoid contraction Zr and Hf have similar sizes.
ii) As energies of 5f, 6d and 7s are comparable therefore electrons can easily be
promoted from lower to higher orbital, hence more no. of oxidation states are
shown. Another reason is that the 5f electrons are available for bonding unlike 4f in
lanthanides.
1
1
19. Give reasons for the following:
(i) Fe has higher melting point than Cu.
(ii) [Ti (H2O)6]3+
is coloured while [Sc(H2O)6]3+
is colourless.
Ans.
(i) This is because Fe (3d6
4s2
) has four unpaired electrons in 3d-subshell. While Cu
(3d10
, 4s1
) only one unpaired electron in 4s shell. Hence metallic bonding is stronger
in Fe than those in Cu.
(ii) The oxidation state of Ti in [Ti (H2O)6]3+
is +3 and its configuration is [Ar] 3d1
i.e
one unpaired electron and hence it is coloured. Whereas the oxidation state of Sc in
[Sc (H2O)6]3+
is +3 and its configuration is [Ar] 3d0
i.e no unpaired electron and hence
it is colourless.
1
½
+
½
20. Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic
number is 25.
Ans.
With atomic number 25, the divalent ion in aqueous solution will have d5
configuration (five unpaired electrons). The magnetic moment, is 5.92 BM
1+1
Three Mark Questions
21. For the first row of transition metals
the E
0
values are:-
E0 values V Cr Mn Fe Co Ni Cu
M2+/M -1.18 -0.91 -1.18 -0.44 -0.28 -0.25 +0.34
Explain the irregularity in the above values.
Ans.
i) From V to Cr E0
value becomes less negative because I.E and enthalpy of
atomisation both increase.
ii)And then from Cr to Mn it becomes more negative due to dip in M.P. of Mn.
Now From Mn to Fe its value again decreases due to increase in Enthalpy of
atomisation and I.E of Fe.
iii) From Fe to Ni the value becomes less and less negative and in the end Cu its
positive, as enthalpy of atomisation and I.E both increase.
1
1
1

22. Give examples and suggest reasons for the following features of the transition metal
chemistry :
(i)The lowest oxide of transition metal is basic, the highest is amphoteric / acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii)Of the d4
species, Cr(II)is strongly reducing while manganese (III) is strongly
oxidizing.
Ans.
(i) In lower oxidation state transition metal has less positive charge as a result it has
a tendency to give electron. While in higher oxidation state due to high positive
charge it shows a tendency to gain electron (Lewis acids).
(ii)A transition metal exhibits higher oxidation states in oxides and fluorides because
oxygen and fluorine are the most electronegative elements and thus easily can
unpair electrons of metal atom.
(iii) Because oxidizing and reducing property depends on E0
value. Since E0value of
Cr3+/Cr2+is negative while that of Mn3+/Mn2+is positive ,as a result Cr(II)act as
reducing agent and Mn(III)is strong oxidizing.
1
1
1
23. Explain giving reasons:
(i)Transition metals and many of their compounds show paramagnetic behaviour.
(ii)The enthalpies of atomization of the transition metals are high.
(iii) The transition metals generally form colored compounds.
Ans.
i) Transition metals and many of their compounds show paramagnetic behavior due
to presence of unpaired electrons in (n‐1) d orbital.
(ii) The enthalpies of atomization of the transition metals are high because of large
number of unpaired electrons in their atoms, they have stronger inter atomic
interaction and hence strong metallic bonding is present between atoms.
(iii) The transition metals generally form coloured compounds due to presence of
unpaired electrons in(n‐1) d orbital and thus they can undergo d‐d transition.
1
1
1

CHAPTER - 9 CO-ORDINATION COMPOUNDS
POINTS
ONE MARK QUESTIONS
1. Give one analytical use of transition metal complexes.
Ans:
Transition metal ions may be estimated by adding reagents to a metal ion solution
to form insoluble precipitate of metal-ligand complex.
E.g-Ni2+is estimated using dimethyl glyoxime. ½ + ½
2. Give two biological applications of complexes.
Ans.
Chlorophyll contains Mg used in photosynthesis .
Vitamin B12 contains cobalt is used to prevent anemia. ½ + ½
3. How complexes are used in metallurgical processes?
Ans.
Au and Ag are extracted by forming cyanide complex and then adding Zn as
reducing agent.
1
4. What is the coordination entity formed when excess of aqueous KCN is added to an
aqueous solution of copper sulphate?
Ans.
[ C u ( C N ) 4 ] 2 -
1
5. What is the hybridisation of Cu in [Cu(NH3)4]2+
?
Ans.
dsp2
1
6. NH3 is strong ligand but NH4
+
isnotwhy?
Ans.
Because NH3 has lone pair . 1
7. Name one complex which is used in medicine.
Ans.
Cisplatin is used in the treatment of cancer.(chemotherapy) 1
TWO MARK QUESTIONS
8. State any two factors which govern the stability of complexes.
Ans.
1 Charge on metal ion
2. Nature of ligands
3.Nature of metal (any two)
1
1

9. Discuss the nature of bonding in metal carbonyls.
Ans
The metal-carbon bonds in metal carbonyls have both σ and π characters. A σ bond
is formed when the carbonyl carbon donates a lone pair of electrons to the vacant
orbital of the metal. A π bond is formed by the donation of a pair of electrons from
the filled metal d orbital into the vacant anti-bonding π*orbital (also known as back
bonding of the carbonyl group).The σ bond strengthens the π bond and vice-versa.
Thus, a synergic effect is created due to this metal-ligand bonding. This synergic
effect strengthens the bond between CO and the metal.
1
1
10. Predict the number of unpaired electrons in the square planar [Pt(CN)4]2−
ion.
Ans.
[Pt(CN)4]2−
, In this complex, Pt is in the +2 state. It forms a square
planar structure.
This means that it undergoes dsp2
hybridization. Now, the
electronic configuration of Pd(+2) is 5d8
.
CN−
being a strong field ligand causes the pairing of unpaired electrons. Hence,
there are no unpaired electrons in[Pt(CN)4]2−
1
1
1
11. Specify the oxidation numbers of the metals in the following coordination
entities:
(i) [Co(H2O)(CN)(en)2]2+
(ii) [PtCl4]2−
Ans. (i)[Co(H2O)(CN)(en)2]2+
Let the oxidation number of Co be x.
The charge on the complex is +2.
1
½

(ii)[PtCl4]2−
Let the oxidation number of Pt be x.
The charge on the complex is −2.
x + 4(−1) = −2
x = + 2
1
½
POINTS
ONE MARK QUESTIONS
12. Define a ligand. Give an example also.
Ans.
Ligand is an atom/ion/molecule which is capable of donating pair of electrons to
the metal atom or ion .
E.g Cl-
1
13. Which type of ligands forms chelates?
Ans.
Polydentate ligands forms chelates 1
TWO MARK QUESTIONS
14. What is meant by the chelate effect? Give an example.
Ans.
When a ligand attaches to the metal ion in a manner that forms a ring, then the
metal- ligand association is found to be more stable. In other words, we can say
that complexes containing chelate rings are more stable than complexes without
rings. This is known as the chelate effect.
For example:
1
1
15. Explain term complex ion. Give Example.

Ans.
Complex ion is eclectically charged species formed by co-ordination of a simple
cation with a number of neutral or charged ligands.
E.g hexacyanoferrate(III)
1
1
16. CuSO4.5H2O is blue in color while CuSO4 is colorless. Why?
Ans.
In CuSO4.5H2O,water acts as ligand as a result it causes crystal field splitting.
Hence d-d transition is possible in field splitting is CuSO4.5H2O and shows color.
In the anhydrous CuSO4 due to the absence of water (ligand), crystal not possible
and hence no color.
1
1
POINTS
ONE MARK QUESTIONS
17. Amongst the following which is the most stable complex & Why?
A) [Fe(H2O)6]
3+
B) [Fe(C2O4)3]
3-
Ans.
B, due to chelation.
½ + ½
18. Why complexes are preferred in the electrolytic bath for electroplating?
Ans.
They dissociate slowly and hence give a smooth and even deposit.
1
19. Why are low spin tetrahedral complexes are not formed?
Ans.
Because for tetrahedral complexes the crystal field stabilization energy is
lower than pairing energy.
1
TWO MARK QUESTIONS
20. Why only transition metals are known to form pi-complexes? Give one
example.
Ans.
Transition metals have empty d-orbitals into which the electron –pairs can
be donated by ligands containing pi electrons.
Example Zeise’s salt K[PtCl3].H2O
1
1
21. Calculate the overall complex dissociation equilibrium constant for the
Cu(NH3)4
2+
ion, given that β4 for this complex is 2.1×10
13
.

Ans.
β4= 2.1×1013
The overall complex dissociation equilibrium constant is the reciprocal of the overall
stability constant,β4.
1/β4=1/ 2.1×10
13
=4.7×10
-14
1
1
22. If to an aqueous solution of CuSO4 in two tubes, we add ammonia solution in
one tube and HCl(aq) to the other tube, how the colour of the solutions will
change ? Explain with the help of reaction.
Ans.
In first case, colour will change from blue to deep blue.
[Cu (H2O)4]2+
+ 4 NH3 [Cu (NH3)4]2+
+ 4 H2O
deep blue
While in second case, its colour will change to yellow.
[Cu (H2O)4]2+
+ 4 Cl–
 [CuCl4]2-
+ 4 H2O
Yellow
1
1
23. On the basis of the following observations made with aqueous solutions,
assign secondary valences to metals in the following compounds:
Formula Moles of AgCl precipitated per mole of
the compounds with excess AgNO3
(i) PdCl2.4NH3 2
(ii) NiCl2.6H2O 2
(iii) PtCl4.2HCl 0
(iv) CoCl3.4NH3 1
Ans.
(i) Secondary valence 4
(ii) Secondary valence 6
(iii) Secondary valence 6
(iv) Secondary valence 6
½
½
½
½
24. Explain the reason behind a colour of some gem stone with the help of
example.
Ans.
The colours of many gem stones are due to the presence of transition metal
ions & colour are produced due to d-d transition. For example the mineral
corundum Al2O3 is colourless when purebut when various M3+
transition
metal ions are present in trace amounts various gem stones are formed.
Ruby is Al2O3 containing about 0.5 – 1% Cr.
1
1
25. Why is the silver plating of copper, K [Ag (CN)2] is used instead of AgNO3 ?
Ans.
This is because if AgNO3 is used Cu will displace Ag+
from AgNO3. The deposit
so obtained is black, soft, non-adhering.
To get a good shining deposit, [Ag (CN)2]–
are used as it is a stable complex,
1

the conc. of Ag+ is very small in the solution. As such no displacement of Ag+
ions with Cu is possible.
1
26. How many EDTA (ethylene diamine tetra acetate) ion are required to make
an octahedral complex with a Ca2+
ion. Why?
Ans.
Only one EDTA ion is required to form octahedral complex.
Because EDTA is a hexadentate ligand
1
1
27. [Fe(CN)6]4−
and [Fe(H2O)6]2+
are of different colours in dilute solutions. Why?
Ans.
The colour of a particular coordination compound depends on the
magnitude of the crystal-field splitting energy, Δ.
This CFSE in turn depends on the nature of the ligand.
In case of [Fe(CN)6]4−
and [Fe(H2O)6]2+
. The colour differs because there is a
difference in the CFSE. Now, CN−
is a strong field ligand having a higher CFSE value as
compared to the CFSE value of water. This means that the absorption of energy for
the intra d-d transition also differs.
Hence, the transmitted colour also differs.
1
1
28. Discuss briefly giving an example in each case the role of coordination compounds
in:
1. Biologicalsystem
2. Medicinal chemistry
3. Analytical chemistry
Ans.
Role of coordination compounds in biological systems:
We know that photosynthesis is made possible by the presence of the chlorophyll
pigment. This pigment is a coordination compound of magnesium. In the human
biological system, several coordination compounds play important roles.
For example, the oxygen-carrier of blood, i.e. haemoglobin is a coordination
compound of iron.
Role of coordination compounds in medicinal chemistry:
Certain coordination compounds of platinum (for example- cis-platin) are used for
inhibiting the growth of tumours.
Role of coordination compounds in analytical chemistry:
During salt analysis, a number of basic radicals are detected with the help of the
colour changes they exhibit with different reagents. These colour changes are a
result of the coordination compounds or complexes that the basic radicals form with
different ligands.
1
1
1
29. Explain on the basis of valence bond theory that [Ni(CN)4]2−
ion with square planar
structure is diamagnetic and the [NiCl4]2−
ion with tetrahedral geometry is
paramagnetic.

Ans.
Ni is in the +2 oxidation state i.e., in d8
configuration.
There are 4 CN−
ions. Thus, it can either have a tetrahedral geometry or square
planar geometry. Since CN−
ion is a strong field ligand, it causes the pairing of
unpaired 3d electrons.
It now undergoes dsp2
hybridization. Since all electrons are paired, it is diamagnetic.
In case of [NiCl4]2−
, Cl−
ion is a weak field ligand. Therefore, it does not lead
to the pairing of unpaired 3d electrons. Therefore, it undergoes sp3
hybridization.
Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.

CHAPTER -10 HALOALKANES AND HALOARENS
POINTS
ONE MARK QUESTIONS
1. The C-X bond in haloalkanes is a polar covalent bond, why ?
Ans.
Due to difference in electronegativity between the carbon and halogen,the shared
pair of electron lies closer to the halogen atom. 1
2. How does the bond length and bond enthalpy changes with size of halogen atom in
halomethanes ?
Ans.
The C-X bond length increases with the increase of size of halogen atom and bond
enthalpy decreases.
1
3. How does the dipole moment in halomethanes changes ?
Ans.
The dipole moment decreases with increase of halogen atom except fluoromethane.
( chloromethane > fluoromethane> bromomethane > iodomethane ) ½ + ½
4. Name the only primary alcohol which gives iodoform test.
Ans.
Ethanol. 1
5. Give the IUPAC name of Carbon Tetra Chloride.
Ans.
1, 1, 1, 1---Tetra Chloro methane. 1
6. What is the difference between Ethylene Chloride and Ethylidene chloride?
Ans.
Cl CH2CH2Cl (Ethylene Chloride) Vicinal Dihalides
CH3CHCl2 (Ethylidene chloride) Gem Dihalides
½ + ½
7. Chlorobenzene is less reactive towards nucleophilic substitution reaction. Give two
reasons.
Ans.
Chlorobenzene is less reactive towards nucleophillic substitution due to –
1. resonance , C- Cl bond acquires a partial double bond character and becomes
stronger than a single bond.
2. sp2
hybridisation in C of C-X bond, the carbon becomes more electronegative and
holds the electron pair of C-X bond more tightly decreasing the bond length
or
Explanation with the help of resonating structure.
½
½

TWO MARK QUESTIONS
8. Arrange the compounds in increasing order of their boiling pts.
(a) CH3CH2CH2CH2Br, CH3CH2CHBrCH3,(CH3)3C Br
(b) CH3Br, CH2Br2, CHBr3
Ans.
(a) (CH3)3C-Br< CH3CH2CHBrCH3< CH3CH2CH2CH2Br
Boiling point decreasing on increasing the branching
(b) CH3Br< CH2Br2< CHBr3
Boiling point increases due increasing molecular mass.
1
1
9. Why 1% ethanol should be added to chloroform sample while storing? Give
equation.
Ans .
Ethanol reacts with phosgene to give harmless diethyl carbonate
COCl2 + 2C2H5OH (C2H5)2CO3 + 2HCl
1
1
10. Write the formula , I.U.P.A.C. name and structure of D.D.T.
Ans .
I.U.P.A.C. Name : 1,1,1-Trichloro-2,2 bis(p-chlorophenyl)ethane
Common name: D.D.T. (p,p'-Dichloro-diphenyltrichloroethane)
½
½
1
11. Differentiate SN1 and SN2 reactions with example.
Ans.
SN1 REACTION
I. SN1/ Nucleophilic Substitution is also called as Unimolecular Nucleophilic
Substitution.
II. In this reaction, only one species is involved in the formation of activated
species
½
½

SN2 REACTION
I. SN2 / Nucleophilic Substitution is also called as Bimolecular Nucleophilic
Substitution.
II. In this reaction, two species are involved in the formation of the activated
species.
1
½
½
1
POINTS
ONE MARK QUESTIONS
12. Cyanide ion acts as an ambident nucleophile. From which end it acts as a stronger
nucleophile in aqueous medium? Give reason for your answer.
Ans.
It acts as a stronger nucleophile from the carbon end because it will lead to the
formation of C – C bond which is more stable (bond between two similar atoms)
than C – N bond.
1
13. Why thionyl chloride is preferred for converting alcohol to haloalkane.
Ans.
Thionyl chloride is preferred for converting alcohol to haloalkane because the by-
products formed are all gases which escape into the atmosphere.
2 ROH +SOCl 2RCl +SO2(g)+HCl(g)
1

14. Why HNO3 is added during iodination of benzene.
Ans.
When benzene is reacted with iodine, the reaction is reversible in nature. It leads to
the formation of reactants back. Therefore and oxidizing agent like HNO3 oxidizes
the HI formed in the reaction to I2 keeps the reaction in forward direction.
1
15. Out of o- and p-dibromobenzene, which one has higher melting point and why?
p-Dibromobenzene has higher melting point than its o-isomer.
Ans.
Due to symmetrical structure has close packing in the crystal.
½
½
TWO MARK QUESTIONS
16. Give reason for the following
(a)Allyl chloride is hydrolysed more readily than n-propyl chloride.
(b)Vinyl chloride is hydrolysed more slowly than ethyl chloride
Ans.
(a) Allyl chloride readily undergoes ionization to produce resonance stabilized allyl
carbocation. Since carbocation are reactive species, therefore allyl carbocation
readily combines with OH ions to form allyl alcohol. In contrast n-propyl chloride
does not undergo ionization to produce stable carbocation.
(b)Vinyl chloride gets ressonance stabilization Carbon-chlorine bond acquires some
double bond character. In contrast in ethyl chloride, the carbon-chlorine bond is a
pure single bond. Thus Vinyl chloride under goes hydrolysis more slowly than ethyl
chloride.
OR
Explanation of both with the help of structures.
1
1
17. What is the correct increasing order of boiling points of the following compounds?
Explain.
1-Bromoethane, 1-Bromopropane, 1-Bromobutane, Bromobenzene
Ans.
1-Bromoethane < 1-Bromopropane < 1-Bromobutane < Bromobenzene
Boiling point increases with increase in molecular mass of the alkyl halide.
1
1
18. Which one of the following has the highest dipole moment, and why?
(a) CH2Cl2 (b) CHCl3 ( c) CCl4

Ans.
CH2Cl2 has the highest dipole moment since both the Cl- atoms are present on one
side (on the head) of c – atom and therefore cause a maximum dipole moment.
In CHCl3 and CCl4, two Cl – atoms and four Cl – atoms cancel out their dipole
moments.
OR
Explain with the help of structure
1
1
19. How will you bring about the following conversions?
(i) But-1-ene to but-2-ene
(ii) Benzene to biphenyl
(iii) 1-Chlorobutane to n-octane
Ans.
HBr
(i)CH3CH2CH꞊CH2 CH3CH2CH(Br)CH3
Markovnikov Addition 2-Bromobutane
KOH(alc),∆ -HBr
CH3CH=CHCH3
But-2-ene
(ii)
(iii)
1
1
1
20. Account for the following
a) Haloarenes are less reactive towards nucleophilic substitution reactions .
b) Grignard reagents should be prepared under anhydrous conditions .
c) Melting point of p- dichlorobenzene is more than O & m isomers.

Ans.
a)due to resonance a partial double bond character is developed between C & X or
any other suitable reason
b)because they are highly reactive towards any source of H+
c) due to more symmetry of p-dichlorobenzene.
1
1
1
POINTS
ONE MARK QUESTIONS
21. Why the dipole moment of fluoromethane is lower than chloromethane?
Ans.
The reason being that although the magnitude of –ve charge on the F is much higher
than that on the Cl atom but due to small size of F as compared to Cl, the C-F bond
distance is so small that the product of charge and distance i.e dipole moment turns
out to be smaller.
1
22. Explain why C – X bond in the haloarenes is extremely less reactive towards
nucleophilic substitution than in haloalkanes.
Ans.
Reactions due to:
(i) Resonance effect: C – X bond acquires a partial double bond character and it
becomes difficult to break C – X bond.
(ii) In C – X bond, C atom attached to halogen is sp2
hybridised. The sp2 hybridised
carbon with a greater s character is more electronegative and can hold the electron
pair of C – X bond more tightly than sp3 hybridised carbon in haloalkane with less 5-
character.
½
½
23. Phenol cannot be converted to chlorobenzene by reacting with HCl.Why?
Ans.
In phenol, due to resonance, the carbon –oxygen bond has a partial double bond
character and is difficult to break being stronger than a single bond. Therefore it can
not be converted to chlorobenzene by reacting with HCl.
1
TWO MARK QUESTIONS
24. Give reasons: (i) C-Cl bond length in chlorobenzene is shorter than C-Cl bond length
in CH3-Cl. (ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl
chloride.

Ans.
(i) In chlorobenzene each carbon atom is sp2 hybridised. Due to resonance
there is a partial double bond character , so bond length is short.
(ii) In chlorobenzene carbon to which chlorine is attached is sp2 hybridised and
and is more electronegative than the corresponding carbon in cyclohexyl
chloride which is sp3 hybridised.
1
1
25. Arrange the following halo alkanes in the increasing order of density. Justify your
answer.
CCl4, CH2Cl2, CHCl3
Ans.
For the same number of carbon atoms (i.e., one) and same halogen (Cl), more the
number of halogen atoms, more is the density.
Thus, the increasing order of density is : CH2Cl2 < CHCl3 < CCl4
OR
Explanation with the help of the structures
1
1
26. How will you test pure chloroform?
Ans.
Impure chloroform give white powder of ethyl carbonate by the addition of ethanol
Chloroform is oxidized by air in presence of light to form Phosgene gas and HCl.
1
1
27. Chlorobenzene is less reactive towards nucleophilic substitution reaction. Give any
three justifications.
Ans.
Chlorobenzene is less reactive towards nucleophillic substitution due to –
1. Resonance , C- Cl bond acquires a double bond character
and becomes stronger than a single bond.
2. SP2
hybridisation in C of C-X bond, the carbon becomes more electronegative and
1

holds the electron pair of C-X bond more tightly decreasing the bond length
3. Instability of phenyl cation.
4. Repulsion for incoming nucleophile from electron rich ring.
1
1
28. How can you distinguish between 1, 1-dichloroethane and 1, 2-dichloroethane ?Give
reactions also.
Ans.
1,1-Dichloroethane on reaction with aq.KOH gives ethanal a pungent smelling
aldehyde, which gives insoluble yellow 2,4-DNP derivative with 2,4-dinitro
phenylhydrazine.
Cl O
CH3 CH Cl CH3 C H Yellow ppt.
1,1-Dichloroethane Ethanal (pungent)
1,2- Dichloroethane on reaction with aq. KOH gives ethane-1,2-diol which is
odourless and gives no ppt. with 2,4 - DNP reagent.
Cl Cl OH OH
2,4 DNP
CH2 CH2 CH2 CH2 No ppt.
1,2 - Dichloroethane Ethane - 1,2-diol
(odorless)
½
1
½
1
+ aq. KOH (2 mol)
-2KCl, -H2O
2,4 - DNP
+ aq. KOH (2 mol)
-2KCl, -H2O

CHAPTER -11 ALCOHOLS PHENOLS AND ETHERS
POINTS
ONE MARK QUESTIONS
1. Name the primary alcohol which gives iodoform test.
Ans
Ethanol.
1
2. Name one reagent which is used for the distinction of primary, secondary and
tertiary alcohols.
Ans.
Lucas reagent (anhyd. ZnCl2 + conc. HCl ) or any other suitable test 1
3. Arrange the following compounds in the decreasing order of acid strength.
Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol
Ans.
2, 4, 6-trinitrophenol> 3,5-dinitrophenol>3-nitrophenol>propan-1-ol 1
4. Diethylether does not react with sodium. Why?
Ans.
Because Diethylether does not contain any active hydrogen. 1
5. Why the dipole moment of diethyl ether (1.18 D) is lower than that of water (1.84
D)?
Ans.
Due to less electro-negativity difference between carbon and oxygen compared to
that of oxygen and hydrogen. C-O bond is less polar than O-H bond.
1
6. How will you know whether a given -OH group is alcoholic or phenolic in nature?
Ans.
Phenolic -OH group gives blue or violet colorations with neutral FeCl3 while
alcoholic -OH group does not.
OR
Phenol gives white ppt with Br2 water where as alcohol does not.
1
7. Write IUPAC name of following CH3-O-CH-CH2-CH2-CH3
CH3
Ans.
2-Methoxypentane
1
8. Write the structure of the molecule of a compound whose IUPAC name is 1-
phenylpropan-2-ol.
Ans.
C6H5CH2 CH(OH)CH3
1

TWO MARK QUESTIONS
9. Arrange the following in order of increasing boiling points. State reason.
(i) CH3CH2CH2CH2OH, (ii) CH3CH2CH2CH3, (iii )CH3CH2OCH2CH3,
(iv)CH3CH2CH2CHO
Ans.
(ii) < (iii) < (iv) < (i)
Reason :This is due to the reason that there is increase in the magnitude of
intermolecular forces in this order, as polarity of C-O bond is increasing ,it’s
tendency to form H-bond also increases and hence boiling point which
depend on the intermolecular forces also increases.
1
1
10. Predict the products of the following reactions
i. Butanol + HCl →
ii. Tertiary butyl bromide + KOH(alc) →
Ans.
i. Butanol + HCl → CH3 CH2 CH2 CH2Cl + H2O
ii. Tertiary butylbromide + KOH (alc) → 2-methylpropene + KBr +H2O
1
1
11. Write the mechanism of acid-catalysed dehydration of ethanol to yield ethene.
Ans.
The mechanism of acid dehydration of ethanol to yield ethene involves the following
three steps:
Step 1:
Protonation of ethanol to form ethyl oxonium ion:
Step 2:
Formation of carbocation (rate determining step):
1
1

Step 3:
Elimination of a proton to form ethene:
The acid consumed in step 1 is released in Step 3. After the formation of ethene, it is
removed to shift the equilibrium in a forward direction.
1
12. Account for the following:
(a) Propanol has higher boiling point than butane
(b) O-nitrophenol is more acidic than o- methoxyphenol
(c) Phenol does not give protonation reaction readily
Ans.
(a) both are of comparable masses but because of intermolecular H-bonding in
propanol , it has higher boiling point.
(b) -NO2 group is an electron withdrawing group and tend to decrease the electron
density on –OH thereby increasing its tendency to lose H+ ions ,consequently
increasing the acidic nature .but in o-methoxyphenol, -OCH3 group has +I effect and
hence less is acidic .
(c ) C-OH in phenol is stabilized due to resonance and electron pair at oxygen atom
in phenol is not readily available to proton ,thus protonation not occurs readily
1
1
1
13. Give names of the reagents to bring about the following transformations:
(i) Ethanoic acid to ethanol
(ii) Propane-1-ol to propanal
(iii)Pent-3-en-2-ol to pent-3-en-2-one
Ans.
(i) LiAlH4 / H3O+
(ii) PCC
(iii) PCC
1
1
1

POINTS
ONE MARK QUESTIONS
14. Why 1% ethanol should be added to chloroform sample while storing?
Ans .
Chloroform forms phosgene gas which is highly poisonous.
Ethanol reacts with phosgene to give harmless diethyl carbonate
COCl2 + 2C2H5OH (C2H5)2CO3 + 2HCl
½
½
15. Phenol has smaller dipole moment than methanol. Why?
OR
Why are dipole moments of phenols smaller than dipole moments of alcohols?
Ans.
Due to electron withdrawing effect (-I effect)of the benzene ring, the C-O bond in
phenol is less polar but in case of methanol due to electron-donating (+I effect)
effect of CH3 group, C-O bond is more polar.
1
16. What is the IUPAC name of the alcohol: HC ≡ C-CH2OH ?
Ans .
Prop-2-yn-1-ol
1
17. Why o & p nitrophenol are more acidic than phenol ?
Ans.
Due to - Ieffect ( electron withdrawing ) of NO2 group.
i)The resulting phenoxide ion is more stable
ii)increase the polarity of O-H bond., hence increases acidic character.
½
½
18. Why the C-O-C bond angle in ethers is higher than the H-O-H angle in water though
oxygen is sp3
-hybridized in both these cases ?
Ans .
It is due to the greater steric repulsions between bulky R(alkyl) groups. 1
TWO MARK QUESTIONS
19. Write the IUPAC name of the following
(i) CH3 C C CH2OH
H3C Br

(ii) C6H5 CH2 CH CH3
OH
Ans.
(i) 2-Bromo-3-methylbut-2-en-1-ol (ii) 1-Phenylpropan-2-ol 1+1
20. a) Why is preparation of ethers by acid catalyzed dehydration of 2° and 3°
Alcohols not a suitable method?
b) Phenol has much less pka than alcohol, explain.
Ans.
a) Because it is accompanied by the formation of alkenes and is not a suitable
method for the preparation of ether from acid catalyzed dehydration of 2° and
3°alcohol, as this reaction involve SN2
mechanism, but dehydration of 2° and 3°
alcohol, it follows SN1
due to steric hindrance and tend to form alkenes
preferably.
b) Because of electron withdrawing nature of benzene, phenoxide ion formed after
removal of H+
is more stable than the alkoxide ion formed. Hence phenol is more
acidic than alcohol having high ka and less pka
1
1
21.
(a) How can we produce nitro phenol from phenol ?
(b)Why are reactions of alcohol/phenol and with acid chloride in the presence of
pyridine?
(c) How is tert-butyl alcohol obtained from acetone?
Ans .
(a) by nitration of phenol:
C6H5OH + dil.HNO3 → o-nitrobenzene + p- C6H4 (NO2)OH
C6H5OH + 3HNO3 (conc. ) → 2, 4, 6- C6H3 (NO2)3
(b) Alcohol/ phenol reacts with acid chloride in the presence of pyridine to form
ester.
R/Ar-OH + R-COCl pyridine R/Ar-OCOR + HCl
(c) tert-butyl alcohol obtained by treating acetone with Grignard’s reagent
CH3COCH3+ CH3MgBr  [(CH3)3C-OMgBr] H+
/H2O (CH3)3C-OH + Mg (OH) Br
½
½
1
1

POINTS
ONE MARK QUESTIONS
22. How will you prepare 1- Butanol from 1-bromobutane
Ans .
CH3CH2CH2CH2Br + (aq KOH) CH3CH2CH2CH2OH 1
23. How will you prepare Phenol from chlorobenzene
Ans .
b) C6H5Cl + NaOH i) (623K, 300atm) ii)H2O/H+
C6H5OH 1
TWO MARK QUESTIONS
24. How is tert-butyl alcohol obtained from acetone?
Ans.
This is obtained by treating acetone with Grignard’s reagent
CH3COCH3+ CH3MgBr  [(CH3)3C-OMgBr] H+
/H2O (CH3)3C-OH + Mg (OH) Br 1+1
25. Give a test to distinguish between
i. phenol and Benzyl alcohol.
ii. ethanol and phenol.
Ans.
(i) Phenols give violet colour with ferric chloride while benzyl alcohol does not give
this colour. Or any other test
(ii) phenol turns blue litmus red, but Ethanol donot have effect on litmus paper
Or any other test
1
1

26. (a)Account for the following
(i) Propanol has higher boiling point than butane
(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol
(iii) Preparation of ethers by acid dehydration of secondary or tertiary
alcoholis not a suitable method
Ans.
(i) It is because propanol can form intermolecular hydrogen bonds.
(ii) It is because –NO2 group is electron withdrawing and –OCH3 group is electron
releasing. Therefore o-nitrophenoxide is more stable than o-methoxyphenoxide
ion.
(iii) It is because secondary and tertiary alcohols on dehydration lead to the
formation of alkene and not ethers due to stability of secondary and tertiary
carbocation.
1
1
1
27. (i) Answer the following
a. What is the order of reactivity of 10
, 20
and 30
alcohols with sodium metal?
b. How will you account for the solubility of lower alcohols in water?
(ii)Give one chemical test to distinguish between the following pair of compounds :
1-propanol and 2-propanol
Ans.
a. 10
> 20
> 30
b. Here—OH group is predominant and the alcohol molecules can form
hydrogen bonds with water molecules.
(ii)Add I2 and NaOH to each (iodoform test)
2- propanol will give yellow ppt. of iodoform (CHI3) whereas 1-propanol will not give
yellow ppt.
1
1
1

class 12 chemistry board material

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