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Chapter –
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It is very difficult to study individually the chemistry of more than one hundred elements known
and their innumerable compounds. This practical problem was felt by scientists and after numerous
attempts the scientists were ultimately successful in arranging the elements. This arrangement of elements
provided a proper classification of elements, which led to the formation of a periodic table.
Periodic table may be defined as the table, which classifies all the known elements in accordance
with their properties in such a way that the elements with similar properties are grouped together in the
same vertical column and dissimilar elements are separated from one another.
The periodic table provides a systematic framework for organizing the large available information
on the chemical behavior of the elements into simple logical patterns. Using the position of an element in
the periodic table, it is possible to predict its chemical behavior.
Several scientists for long have tried to classify the elements and to find patterns in their properties.
Some of them are as:
2.1 DOBEREINER’S TRIADS
He arranged similar elements in groups of three and showed that atomic weights are either
nearly the same or the atomic weight of the middle element is approximately the arithmetic mean of the
other two. e.g.
Element Atomic weight Element Atomic weight Element Atomic weight
Li 7 Ca 40 Cl 35.5
Na 23 Sr 88 Br 80
K 39 Ba 137 I 127
Drawback: This applies only to limited number of elements.
2.2 NEWLAND LAW OF OCTAVE
When lighter elements are arranged in order of their increasing atomic weight, the properties of
every eighth element is similar to the first, like eighth node of a musical scale.
Element Li Be B C N O F
NEED FOR CLASSSIFICATION
1
HISTORICAL DEVELOPMENT OF CLASSIFICATION F ELEMENTS
2
CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Atomic weight 7 9 11 12 14 16 19
Element
Atomic weight
Na Mg Al Si P S
23 24 27 29 31 32
Element
Atomic weight
K Ca
39 40
Drawbacks: (a) It could not be applied for the element of atomic mass greater than 40.
(b) With the discovery of noble gases, the above sequence was disturbed and the
arrangement
was no longer valid.
2.3 LOTHER MEYER ARRANGEMENT
By plotting graph between atomic volume and atomic mass of element it was observed that
elements having similar properties occupy similar positions on the curve e.g. alkali metals occupy maxima
of the curve. Most electronegative elements found ascending portions of the curve and it was found that
physical properties are the periodic function of their atomic weights.
By 1868, Lother Meyer had developed a table of the elements that closely resembles with the
Modern Periodic Table. However, his work was not published until after the work of Dimitri Mendeleev,
the scientist who is generally credited with the development of the Modern Periodic Table.
2.4 MENDELEEV’S PERIODIC TABLE
While Dobereiner initiated the study of periodic relationship, it was Mendeleev who was
responsible for publishing the Periodic law for the first time.
Thus, in the development of the periodic table the work of Dmitri Mendeleev was a
breakthrough. To explain his scheme of classification, he gave a periodic law which is known as
Mendeleev’s periodic law. This law states as follows, the physical and chemical properties of elements
are periodic function of their atomic masses i.e. when the elements are arranged in order of their
increasing atomic masses, elements with similar properties are repeated after certain regular intervals.
On the basis of this periodic law, he constructed a periodic table in such a way that elements
were arranged horizontally in the order of their increasing atomic masses.
Mendeleev’s system of classifying elements was more elaborated than that of Lother Meyer’s. He
fully recognized the significance of periodicity and used broader range of physical and chemical
properties to classify the elements.
2.4.1 Characteristics of Mendeleev’s Periodic Table
1. It has 9 vertical columns called groups I, II, III, IV, V, VI, VII, VIII and zero group. Except VIII and
zero groups, each group is subdivided into two groups designated as
A and B.

2. It has seven horizontal rows called periods. The I, IIand, III periods are short periods, IV and V
periods are long periods and VI period is the longest period.
2.4.2 Importance of Periodic table
1. Classification based on atomic mass facilitates systematic study of the elements.
2. It helped in prediction of new element e.g. Ga, Ge. Keeping his primary aim of arranging the
elements of similar properties in the same group, he proposed that some of the elements
were still undiscovered and, therefore, left several gaps in the table. For example, both
gallium and germanium were unknown at the time when Mendeleev published his Periodic
Table. He left the gap under aluminium and a gap under silicon, and called these elements
Ekaaluminium and Ekasilicon. Mendeleev predicted not only the existence of gallium and
germanium, but also described some of their general physical properties.
3. Correction of doubtful atomic mass. The atomic mass of Be was taken as 13.5 but periodic
classification helped in arriving at its correct atomic mass i.e. 9.
2.4.3 Drawbacks of Mendeleev’s Periodic Table
1. Position of H: H is placed along with alkali metals in group I(A). It can also be placed along
with halogens in group VIIA.
2. Anomalous positions of some elements: Although most of the elements have been arranged
in the increasing order of their atomic masses, but in some cases, the element having higher
atomic mass precedes the element with lower atomic mass. For example, Ar (Atomic mass =
39.9) precedes K (Atomic mass = 39.1) and similarly, Co (Atomic mass = 58.9) precedes Ni
(Atomic mass = 58.7).
3. Position of isotopes: Different isotopes of the same element are not given separate position
in periodic table.
4. No corelation of elements in subgroups: Some dissimilar elements are grouped together
while some similar elements are placed in different groups. Alkali metals
(Li, Na) and coinage metals (Cu, Ag, Au) are placed in the same group although they have
different properties. Copper and mercury have been placed in different groups although they
have similar chemical properties.
5. Position of lanthanides and actinides: These elements could not be accommodated in the
main periodic table. Instead they have been placed in two separate rows at the bottom of the
periodic table.
6. Cause of periodicity: No proper explanation has been offered as to why the elements placed
in a group show resemblance in their properties.
When Mendeleev developed his Periodic Table, chemists knew nothing about the internal structure
of atom. However, the beginning of the 20th
century witnessed profound developments in theories about
sub-atomic particles. In 1913, the English physicist, Henry Moseley observed regularities in the
characteristic X–ray spectra of the elements. A plot of  (where  is frequency of X–rays emitted)
against atomic number (Z) gave a straight line and not by the plot of  vs. atomic mass. Thus, Moseley
MODERN PERIODIC LAW AND LONG FORM OF PERIODIC TABLE
3

observed that atomic number is a more fundamental property of an element than its atomic mass, and
physical and chemical properties of the elements are determined by their atomic numbers. Based on this,
modern periodic law was given which states that physical and chemical properties of the elements are
periodic functions of their atomic numbers. i.e. if the elements are arranged in increasing order of their
atomic numbers, the elements with similar properties are repeated after certain regular intervals.
3.1 CAUSE OF PERIODICITY
When the elements are arranged in an order of increasing atomic number, the periodic repetition of
elements with similar properties is observed after certain regular intervals. This is called periodicity.
The cause of periodicity in properties is due to repetition of similar outer electronic configuration
after certain regular intervals as shown below:
Li (3)  1s2
, 2s1
Na (11)  1s2
, 2s2
, 2p6
, 3s1
K (19)  1s2
, 2s2
, 2p6
, 3s2
, 3p6
, 4s1
3.2 FEATURES OF LONG FORM OF PERIODIC TABLE
3.2.1 Periods
A horizontal row of a periodic table is known as a period. Entire table is divided into
7 periods. The first period contains 2 elements. The subsequent periods consists of 8, 8, 18, 18 and 32
elements, respectively. The seventh period is incomplete and like the sixth period would have a theoretical
maximum (on the basis of quantum numbers) of 32 elements. In this form of the Periodic Table, 14
elements of both sixth and seventh periods (lanthanoids and actinoids, respectively) are placed in separate
panels at the bottom. Thus, 1st
, 2nd
and 3rd
period are short periods, 4th
and 5th
periods are long and 6th
period is the longest and 7th
period is incomplete.
3.2.2 Groups
A vertical column of the periodic table is called a group. It is divided into 18 groups. According to
IUPAC, these groups are numbered from 1 to 18. replacing the older notation of groups I A … VII A, VIII, I
B …. VII B and O.
3.3 ADVANTAGES OF LONG FORM OF PERIODIC TABLE
Long form of periodic table has the following advantages:
1. Same group elements have similar electronic configuration, hence they have similar chemical
properties.
2. Elements are properly divided into reactive metals, heavy metals and non metals.
3. Since same group elements have similar electronic configuration, the study of elements
become very simple.
4. Since, classification is based on the atomic number and not on the atomic mass, the idea of
placing isotopes at one place is fully justified.
3.4. DRAWBACKS OF THE LONG FORM OF PERIODIC TABLE
The main defects of this table are as follows:

1. Position of H: H is still placed along with alkali metals as well as halogens.
2. Separate position of lanthanides and actinides: Like Mendeleev’s periodic table lanthanides
and actinides are placed separately at the bottom of the periodic table.
The entire periodic table can be divided into two broad categories as indicated below:
Periodic Table Classification
On the basis of distribution of
electrons into orbitals of an atom.
On the basis of properties of
the elements.
s-Block elements p-Blockelements d-Blockelements f-Block elements
Metals Non-Metals Metalloids
4.1 CLASSIFICATION ON THE BASIS OF ELECTRONIC DISTRIBUTION OF
ELECTRONS INTO ORBITALS OF AN ATOM
Entire periodic table is divided into four blocks known as s, p, d and f based on the orbital in which
last electron enters.
4.1.1 sBLOCK ELEMENTS: The elements of group 1 (alkali metals) and Group 2 (alkaline earth metals)
which have ns1
and ns2
outermost electronic configuration belong to the s-Block Elements. The
general features of these elements are:
1. The general outer shell electronic configuration of s-block elements is ns12
.
2. They are soft metals with low melting and boiling points.
3. They have low ionization potential and are very reactive.
4. Alkali metals show +1 oxidation state while alkaline earth metals show +2 oxidation state.
5. Most of the elements of this block impart characteristic colour to the flame.
6. They are strong reducing agents.
7. All of them are good conductors of heat and electricity.
4.1.2 p-BLOCK ELEMENTS: The pBlock elements comprise those belonging to group 13 to 18 and
these together with the sBlock elements are called the Representative Elements or Main Group
Elements. The general features of these elements are:
DVISION OF PERIODIC TABLE
4

1. The general outer shell electronic configuration of p-block elements is ns2
, np16
.
2. This block has metals, non-metals and metalloids.
3. Some of the elements show more than one oxidation states.
4. They form mostly covalent compounds.
5. No member of p-block or the salt of it imparts a characteristic color to the flame because the
energy released by excited elements does not appear in the visible region of the spectrum.
Note: Group–18 elements are known as noble gases; Group–17 elements are known as halogens
and the group-16 elements are known as chalcogens.
4.1.3 d-BLOCK ELEMENTS
These are the elements of group 3 to 12 in the centre of the periodic table. These are
characterized by the filling of inner d orbitals by electrons and are therefore referred to as d–
block elements. The general features of these elements are:
1. The general outer shell electronic configuration is (n  1)d110
ns02
.
2. They are also known as transition elements because their properties show a gradual
transition from the properties of sblock to the pblock elements.
3. They are hard, malleable and ductile metals.
4. They form both ionic and covalent compounds.
5. Their compounds are generally coloured and paramagnetic.
6. Most of the elements or their compounds find their use as catalyst due to their tendency to
show variable oxidation states.
7. Most of the transition elements form alloys.
4.1.4 fBLOCK ELEMENTS: The two rows (series) of the elements at the bottom of the periodic table,
called the Lanthanides and Actinides are classified into the f-block elements. The general features
of these elements are:
1. The general outer shell electronic configuration is .
ns
d
)
1
n
(
f
)
2
n
( 2
1
0
14
1 



2. They are known as inner transition elements since they form transition series within the
transition elements of d-block.
3. 58Ce to 71Lu constitutes the 4f inner transition series and are called Lanthanides because they
follow lanthanum in the 5d transition series. They closely resemble lanthanum in their
properties. They are also called rare earth elements since they occur rarely in earth crust.

4. 90Th – 103Lr constitutes the 5f inner transition series and are known as actinides since they
follow actinium in the 6d transition series.
5. All actinides are radioactive elements. Three elements namely Thorium, Protoactinium and
Uranium occur in nature but the remaining 11 elements i.e. from neptunium to Lawrencium
(93Np – 103Lr) have been prepared artificially. These 11 elements are called Transuranium
elements since they follow uranium in periodic table and also have been derived from it
through nuclear reactions.
6. They show variable oxidation states.
7. Their compounds are generally coloured.
8. They have strong tendency to form complexes.
4.2 CLASSIFICATION ON THE BASIS OF PROPERTIES OF THE ELEMENTS.
I. Metals:
Metals comprise more than 78% of all known elements and appear on the left side of the Periodic
Table. The general properties of metals are listed below:
(i) Metals are usually solids at room temperature. Mercury is an exception.
(ii) Metals usually have high melting and boiling points.
(iii) They are good conductors of heat and electricity.
(iv) They are malleable (can be flattened into thin sheets by hammering) and ductile (can be
drawn into wires)
II. Non-metals:
In contrast to metals, non-metals are located at the top right hand side of the periodic table.
Infact, in a horizontal row, the property of elements changes from metallic on the left to non-
metallic on the right and the elements become more metallic as we go down a group. The
characteristics of the non-metals are as follows:
(i) Non-metals are usually solids or gases at room temperature with low melting and boiling
points (boron and carbon are exceptions).
(ii) They are poor conductors of heat and electricity.
(iii) Most non-metallic solids are brittle and are neither malleable nor ductile.
III. Metalloids
Certain elements (Si, Ge, As, Sb, Te, Po and At) show properties that are characteristic of both
metals and non-metals. These elements are called semi-metals or metalloids.
Illustration 1
Question: Which of the following elements given below belong to the same period?
Al, Ca, O, Cs, Si

Solution: Al and Si both belong to the same period.
The effective nuclear charge is the actual charge felt by the valence electrons. Between the outer
most valence electrons and the nucleus of an atom, there exists finite number of shells containing electrons.
Due to the presence of these intervening electrons, the valence electrons are unable to experience the
attractive pull of the actual number of protons in the nucleus. These intervening electrons act as shield
between the valence electrons and protons in the nucleus. Thus, the presence of intervening (shielding)
electrons reduces the electrostatic attraction between the protons in the nucleus and the valence electrons
because intervening electrons repel the valence electrons. The concept of effective nuclear charge allows
us to account for the effects of shielding on periodic properties.
The electronic configuration of the elements shows a periodic variation with increasing atomic
number. Consequently, there are also periodic variations in physical and chemical behaviour of the
elements. Let us now study the variation of some of the atomic properties in the periodic table. These
properties are classified as follows:
(i) Atomic radius
(ii) Ionisation energy
(iii)Electron gain enthalpy
(iv) Electronegativity
7.1 ATOMIC RADIUS
It is defined as the distance from the centre of the nucleus to the outer most shell containing
electrons.
Difficulties in determining radius
(a)Exact size of electron cloud can not be determined as the probability of finding
an electron even at large distance from the nucleus never becomes zero.
(b) It is not possible to isolate a single atom.
(c) The probability of finding an electronin an atom is appreciably affected by the presence of
the other atoms within its molecule or its neighbourhood.
(d) The atomic radius also changes when the atom is present in different bonding state.
Type of Radius
7.1.1 Covalent radius
THE PERIODICITY OF ATOMIC PROPERTIES
6
TRENDS IN PHYSICAL PROPERTIES
7
EFFECTIVE NUCLEAR CHARGE
5

It is defined as one-half of the
distance between the nucleus of two covalently bonded
atoms of the same element in a molecule.
rcovalent =
2
d
where ‘d’ is the
internuclear distance between two bonded atoms.
Bond length
Overlapping of atomic orbitals
to form a covalent bond.
d
A A
Illustration 2
Question: The internuclear distance of a given molecule A2 (dA–A) is 1.4Å. Calculate the atomic (covalent)
radius of an atom A.
Solution: 0.7


 
2
/
4
.
1
2
d
r A
A
A Å.
7.1.2 vander Waal’s radius (collision radius)
It is defined as one half of the distance between two non
bonded isolated atoms or two adjacent atoms of the same
element belonging to the two neighbouring molecules of
the same substance in the solid state. vander Waal’s radius
increases with increasing shell. vander Waal’s radius is
greater than covalent radius. Radius of inert gas is large due
to exactly fully filled configuration and also due to maximum
atomic repulsion between the two atoms. Moreover, in case
of inert gas the atomic size is defined in terms of vander
Waal’s radius instead of covalent radius.
dAB
vander Waal bond
length
v.w. radius =
2
AB
d
Illustration 3
Question: Arrange the given atoms in order of decreasing vander Waal’s radii: Cl, H, O, N.
Solution:
The vander Waal’s radii increase as the number of energy shells increases and decreases as the
nuclear charge increases. Hence the order is Cl > N > O > H.
Illustration 4
Question: Vander Waal’s radius is greater than the covalent radii of the atoms. Explain.
Solution: vander Waal’s radius is measured between two non bonded atoms whereas covalent radii is measured
between two bonded atoms. Hence, the vander Waal’s radius is greater than the covalent radius of the
atoms.
7.1.3 Metallic radius
It is defined as one-half of the internuclear distance between two adjacent metal atoms in the
metallic lattice.

7.1.4 Ionic radius
Ionic radii may be defined as the effective distance from
the nucleus of the ion to the point upto which it exerts
its influence on its electron cloud. Radius of cation is
always smaller than that of the parent atom because it
has fewer electrons while its nuclear charge remains the
same. e.g. the radius of Na+
is lesser than Na. Similarly
the radius of anion is larger than that of the parent atom
because the addition of one or more electrons would
result in increased repulsion among the electrons and a
decrease in effective nuclear charge. e.g. the radius of I 
is greater than I. For isoelectronic species greater the
nuclear charge lesser will be the ionic radius. e.g.
N3
> O2
> F
> Na+
> Mg2+
Cation Anion
A B
Internuclear
distance
Radius of
cation
Radius of
anion
anion
cation
AB r
r
d 

7.1.5 Variation of atomic radii in the periodic table
In general, the atomic radii decreases with increase in the atomic number in a period. It is
because of increase in the effective nuclear charge along the period.
In general, the atomic radii increases from top to bottom within a group. As we move down the
group there is progressive increase in the principal energy level, which increase the distance between
nucleus and outermost electron and the size increases.
7.2 IONIZATION ENERGY
Ionization energy is defined as the minimum energy (in kJ/mole) required to remove the most
loosely bound electron from an isolated gaseous atom in its ground state. Alternatively, ionization
energy is the amount of energy required in kilojoules needed to knock out one mole of electrons from
one mole of isolated gaseous atoms. In this definition, gaseous atoms are specifically used because an
atom in the gas phase is virtually uninfluenced by its neighbours and so, there are no intermolecular forces
to take into account while measuring ionization energy.
The magnitude of ionization energy is a measure of how “tightly” the electron is held in the atom.
Higher is the ionization energy, more difficult it is to remove the electron.
7.2.1 Successive Ionization Energies
For a multielectron atom, the amount of energy required to remove the first electron from the
atom in its ground state is called the first ionization energy (IE1).
X(g) + energy  X+
(g) + e
The second ionization energy (IE2) and the third ionization energy (IE3) are shown in the following
equations:
X+
(g) + energy  X2+
(g) + e
(IE2)
X2+
(g) + energy  X3+
(g) + e
(IE3)
The pattern continues for the removal of subsequent electrons. Theoretically, an atom can have as
many ionization energy values as the number of electrons present in that atom.
The ionization enthalpies required to remove first, second, third etc. electrons from an isolated
gaseous atom are called successive ionization energies. It may be seen that, IE2 is always greater than IE1
and IE3 is always greater than IE2 and so on IE3 > IE2> IE1. After removing first electron, the resultant
force of attraction on the remaining outermost electrons increases and hence, more energy is required to
remove the next electron.

7.2.2 Factors affecting the ionization energy.
Ionization energy depends upon the following factors:
(a) Penetration effect of the electrons. i.e. ionization energy of nf
nd
np
ns 

 , because
sorbital is closest to the nucleus.
(b) Greater the nuclear charge greater will be the ionization energy.
(c) Greater the atomic size lesser will be the ionization energy as the distance of the outermost
electron from the nucleus increases.
(d) Greater the screening effect of inner electrons, lesser will be the ionization enthalpy.
(e) Exactly half filled or completely filled orbitals have higher ionization enthalpy because half
filled and fully filled configuration are more stable than partially filled configuration.
7.2.3 Variation of ionisation energy in the periodic table
In general the value of ionisation energy increases with the increase in atomic number across
the period. This is due to gradual increase in effective nuclear charge and simultaneous decrease in atomic
size. As a result, the attractive force between the nucleus and the electron cloud increases. Consequently,
the electron is more and more tightly bound to the nucleus.
As we move down the group, there is a gradual increase in the atomic size due to progressive
addition of new energy shells. As a result the nuclear hold on the valence electrons decreases gradually and
the ionisation energy also decreases.
Li
Be
B
C
N O
F
Ne
H
Atomic number (Z)
First ionization enthalpies of the second period as a function of atomic number (Z).
Li
H
Atomic number (Z)
Na
K
Rb
Cs
First ionization enthalpies of the first group as a function of atomic number (Z)
From the above graphs, it is clear that the first ionization enthalpy of boron (Z = 5) is slightly less
than that of beryllium (Z = 4) even though the former has a greater nuclear charge. When we consider the
same principal quantum level, an s-electron is attracted to the nucleus more than a p–electron. In beryllium,
the electron removed during the ionisation is a s–electron whereas the electron removed during ionization
of boron is a p–electron. The penetration of a 2s–electron to the nucleus is more than that of a 2p–electron;
hence the 2p–electron of boron is more shielded from the nucleus by the inner core of electrons than the
2s–electrons of beryllium. Therefore, it is easier to remove the 2p–electron from boron compared to the
removal of a 2s–electron from beryllium. Thus, boron has a smaller first ionization enthalpy than
beryllium.

The first ionization enthalpy of oxygen is smaller as compared to nitrogen. This can be attributed
due to stability of half filled orbitals of 2p–subshell of nitrogen.
Illustration 5
Question: The first ionization energy of Al is lower than that of Mg. Explain.
Solution: The electronic configuration of the given elements are:
2
6
2
2
12 s
3
p
2
s
2
s
1
Mg 
1
2
6
2
2
13 p
3
s
3
p
2
s
2
s
1
Al 
In the case of Al, the electron is removed from ‘3p’ orbital, whereas in case of Mg, the electron is
removed from the ‘3s’ orbital.
The ‘3s’ orbital has two electrons which are paired whereas ‘3p’ orbital has only one unpaired
electron. Even though Al atom is smaller than Mg atom, in order to remove an electron from ‘3s’
orbital of Mg, energy has to be spent to unpair the two electrons and to remove one of them from the
nuclear force of attractions. Therefore, 1st
ionization energy of Mg is higher than that of Al.
Illustration 6
Question: Ionisation energy of one H atom is 2.18 × 10–18
J. Calculate the ionization energy of H atom in
kJ mole–1
.
Solution:
I.E.= .
mole
kJ
1310
mole
J
10
31
.
1
mole
atom
10
02
.
6
atom
1
J
10
18
.
2 1
–
1
–
6
23
18





 
Illustration 7
Question: The I, II and III ionization energies of Al are 578, 1817 and 2745 kJ mol1
respectively.
Calculate the energy required to convert all the atoms of Al to Al+3
present in 270 mg of Al
vapours.
Solution:
Total amount of energy required to convert
1
3
mole
5140
2745
1817
578
)
(
to
)
( 




 kJ
g
Al
g
Al
Number of moles of Al in mole
10
27
1
1000
270
mg
270 2




1 mole of Al need for ionization = 5140 kJ of energy
10–2
mole of Al need for ionization = 5140 × 10–2
= 51.40 kJ.
7.3 ELECTRON GAIN ENTHALPY
Electron gain enthalpy of an element may be defined as the energy released when a neutral
isolated gaseous atom accepts an extra electron to form the gaseous negative ion, i.e., anion. It is
denoted by Heg. Higher is the amount of energy released, more is the electron affinity value. This
process may be represented as:
eg
anion
atom
gaseous
neutral
H
H
;
)
g
(
X
e
)
g
(
X 




 

After the addition of one electron, the atom becomes negatively charged and the second electron is
to be added to a negatively charged ion. But the addition of second electron is opposed by electrostatic
repulsion and hence the energy has to be supplied for the addition of a second electron. Thus, the second
electron gain enthalpy of an element is positive.
released)
is
(Energy
1
eg mol
kJ
141
H
;
)
g
(
O
e
)
g
(
O 







 (First electron gain enthalpy)

absorbed)
is
(Energy
1
eg
2
mol
kJ
780
H
);
g
(
O
e
)
g
(
O 








 (Second electron gain enthalpy)
Similarly, the second electron gain enthalpy of S is also positive as can be seen from the data given
below:
released)
is
(Energy
1
eg mol
kJ
200
H
;
)
g
(
S
e
)
g
(
S 







 (First electron gain enthalpy)
absorbed)
is
Energy
(
1
eg
2
mol
kJ
590
H
;
)
g
(
S
e
)
g
(
S 








 (Second electron gain enthalpy)
The elements which have stable configuration have positive first electron gain enthalpy.
Factors on which the electron gain enthalpy depends.
Some important factors on which electron gain enthalpy depends are discussed below:
1. Atomic size
As the size of an atom increases, distance between the nucleus and the last shell which receives
the incoming electron increases. As a result, the force of attraction between the nucleus and the
incoming electron decreases and hence the electron gain enthalpy becomes less negative.
2. Nuclear charge
As the nuclear charge increases, the force of attraction between the nucleus and the incoming
electron increases and hence, the electron gain enthalpy becomes more negative.
3. Electronic configuration
Elements having exactly half-filled or completely filled orbitals are very stable. As a result,
energy has to be supplied to add an extra electron. Hence their electron gain enthalpies have
large positive values since they do not accept additional electron so easily.
7.3.1 Variation of electron gain enthalpy in the periodic table
On moving across the period, the atomic size decreases and effective nuclear charge increases.
Both these factors result into greater attraction for the incoming electron. Therefore, electron gain
enthalpies tend to become more negative as we go from left to right across a period. On moving down a
group, the atomic size as well as nuclear charge increases. But the effect of increase in atomic size is much
more pronounced than that of the nuclear charge and thus, the additional electron feels less attraction.
Consequently, electron gain enthalpy becomes less negative on going down the group.
The electron gain enthalpy of second period element is lower than that of third period elements.
This is probably due to small size of the atom of second period element. The addition of an extra electron
produces high electron charge density in a relatively compact 2p–subshell resulting in strong electron–
electron repulsion. The repulsive forces between electrons imply low electron gain enthalpy. Electron gain
enthalpy of Cl is maximum in periodic table.
Illustration 8
Question: Explain, why the formation of F
(g) from F(g) is exothermic whereas that of O2
(g) from O(g) is
endothermic?
Solution: The addition of an electron to a neutral atom is an exothermic process.
energy
F
e
F 

 



 

O
e
O energy …(i)
The addition of second electron to a monovalent anion O
, to make it O2
is difficult because both
electron and anion have the same charge and experience repulsive forces. Thus, the addition of an
electron to O
requires energy to overcome the force of repulsion.
energy
e
O 
 

 O2–
…(ii)
The energy absorbed in (ii) step is more than the energy released in the (i) step. Hence, the formation
of O2
(g) from O(g) is endothermic in nature.

Illustration 9
Question: The electron affinity values of halogens decrease in the order: F > Cl > Br > I. Comment on the
statement.
Solution:
This statement is wrong because the actual order of electron affinity of halogens is
Cl > F > Br > I
In general, electron affinity decreases down the group but F abnormally shows lower electron affinity
than Cl because of its small size.
7.4 ELECTRONEGATIVITY
The ability of an atom to attract the shared pair of electrons towards itself is known as
electronegativity. It is a unitless quantity because it is relative with respect to F for which the
electronegativity value is fixed as 4. (i) Electronegativity of inert gases is 0. (ii) It is not a property of an
isolated atom but rather property of an atom in a molecule. (iii) Because of electronegativity covalent
molecule acquire some ionic character. (iv) Fluorine is the most electronegative element of the periodic
table.
7.4.1 Factors affecting electronegativity
1. Size of atom: As the size increases, electronegativity decreases because the distance of
electrons from nucleus increases.
2. Oxidation state of element: Electronegativity increases with higher positive oxidation state
because it has more effective nuclear charge. As for example electronegativity of Fe3+
ion is
greater than that of Fe2+
ion.
3. Effective nuclear charge: Electronegativity increases with increase in nuclear charge because
the attraction on shared pair of electrons increases.
7.4.2 Application of electronegativity
1. To predict the nature of bond.
2. To calculate the percentage ionic character.
3. To explain the variation in bond angle.
4. To explain the bond length variation.
7.4.3 Variation of electronegativity in the periodic table
In a period, electronegativity increases on moving from left to right. This is because the effective
nuclear charge increases whereas atomic radius decreases as we move from left to right in a period. In a
group, electronegativity decreases as we move down the group. This is due to the effect of increased
atomic radius.
Illustration 10
Question: Give the correct order of electronegativity of central atom in following compounds:
CH3CH3, CH2=CH2, CHCH
Solution:
Greater the scharacter, greater will be the electronegativity therefore the order is
CHCH > CH2=CH2 > CH3CH3.
Illustration 11
Question: What is the factor on which the polarity of a bond depends?
Solution: The polarity of bond depends upon electronegativity difference of two atoms involved in covalency.
PERIODIC TRENDS IN CHEMICAL PROPERTIES
8

8.1 Periodicity of valency or oxidation states
Valency of representative elements is usually equal to the number of electrons in the outermost orbit
and or equal to eight minus the number of outermost electrons as shown below:
Group 1 2 13 14 15 16 17 18
Number of valence electrons 1 2 3 4 5 6 7 8
Valency 1 2 3 4 3,5 2,6 1,7 0,8
Now a days the term oxidation state is frequently used for valency. The oxidation state of
an element in a particular compound can be defined as the charge acquired by its atom on the basis
of electronegative consideration from other atoms in the molecule.
8.2 Anomalous Properties Of Second Period Elements-Diagonal Relationship
A diagonal relationship is a similarity in properties between diagonal neighbours in the main
groups of the periodic table. Because these properties lie in a diagonal pattern, it is not surprising to find
that the elements within a diagonal band show similar chemical properties. Diagonal relationship helps in
making predictions about the properties of elements and their compounds.
Diagonal relationship is exhibited by lithium and magnesium; beryllium and aluminium.
For example, lithium and magnesium react directly with nitrogen to form nitrides. Beryllium and
aluminum, both are amphoteric i.e. reacts with acids as well as bases.
Reasons for different chemical behavior of the first member of a group of elements in the
s– and p–blocks compared to that of the subsequent members in the same group:
(i) The anomalous behaviour is attributed to their small size, large charge / radius ratio and high
electronegativity of the elements.
(ii) The first member of the group don’t have d orbitals.
(iii)The first member of p-block elements display greater ability to form p-p multiple bonds to
itself (e.g. C=C, CC, N=N, NN) and to other second period elements (e.g. C=O, C=N, CN,
N=O) compared to subsequent members of the same group.
1. Born Haber cycle is used to determine:
a) Lattice energy b)Electron affinity c) Ionization energy d)Either of them
2. The electronic configurations of four elements 𝐿, 𝑃, 𝑄 and 𝑅 are given below,
𝐿 = 1𝑠 , 2𝑠 2𝑝 𝑄 = 1𝑠 , 2𝑠 2𝑝 , 3𝑠 3𝑝
𝑃 = 1𝑠 , 2𝑠 2𝑝 , 3𝑠 𝑅 = 1𝑠 , 2𝑠 2𝑝 , 3𝑠
The formula of the ionic compounds that can be formed between these elements are:
a) 𝐿 𝑃, 𝑅𝐿, 𝑃𝑄, 𝑅 𝑄 b)𝐿𝑃, 𝑅𝐿, 𝑃𝑄, 𝑅𝑄 c) 𝑃 𝐿, 𝑅𝐿, 𝑃𝑄, 𝑅𝑄 d)𝐿𝑃, 𝑅 𝐿, 𝑃 𝑄, 𝑅𝑄
3. The element with strong electropositive nature is:
a) Cu b)Cs c) Cr d)Ba
4. Octet rule is not valid for the molecule:
a) CO b)H O c) O d)CO
5. The correct order of reactivity of halogens is
a) F > Br > Cl > I b)F > Cl > Br > I c) I > Br > Cl > F d)Cl > I > Br > F
6. NH has higher boiling point than expected, because:
a) With water it forms NH OH
b)It has strong intermolecular hydrogen bonds
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 1

c) It has strong intermolecular covalent bonds
d)Its density decreases in freezing
7. The screening effect of 𝑑-electrons is:
a) Equal to the 𝑝-electrons
b)Much more than 𝑝-elecrons
c) Same as 𝑓-electrons
d)Less than 𝑝-electrons
8. Which has the largest first ionisation energy?
a) Li b)Na c) K d)Rb
9. In which of the following molecules are all the bonds not equal?
a) AlF b)NF c) ClF d)BF
10. The bond between two identical non-metal atoms has a pair of electrons:
a) Unequally shared between the two
b)Equally shared between the two
c) Transferred fully from one atom to another
d)None of the above
11. The number of unpaired electrons in a paramagnetic diatomic molecule of an element with
atomic number 16 is:
a) 4 b)1 c) 2 d)3
12. In NO ion, number of bond pair and lone pair electrons are respectively:
a) 2, 2 b)3, 1 c) 1, 3 d)4, 8
13. Which element of second period forms most acidic oxide?
a) Carbon b)Nitrogen c) Boron d)Fluorine
14. The electronic configuration of four elements are given below. Which element does not belong
to the same family?
a) [Xe]4𝑓 5𝑑 6𝑠 b)[Kr] 4𝑑 5𝑠 c) [Ne]3𝑠 3𝑝 d)[Ar] 3𝑑 4𝑠
15. For the four successive transition elements (Cr,Mn, Fe and Co), the stability of +2 oxidation
state will be there in which of the following order?
(At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27)
a) Cr > 𝑀𝑛 > 𝐶𝑜 > 𝐹𝑒 b)Mn > 𝐹𝑒 > 𝐶𝑟 > 𝐶𝑜 c) Fe > 𝑀𝑛 > 𝐶𝑜 > 𝐶𝑟 d)Co > 𝑀𝑛 > 𝐹𝑒 > 𝐶𝑟
16. Which is correct in the following?
a) Radius of Cl atom is 0.99 Å, while that of Cl ion is 1.54 Å
b)Radius of Cl atom is 0.99 Å, while that of Na atom is 1.54 Å
c) The radius of Cl atom is 0.95 Å, while that of Cl ion is 0.81 Å
d)Radius of Na atom is 0.95 Å, while that of Na ion is 1.54 Å
17. The linear structure is possessed by:
a) SnCl b)NCO c) NO d)CS
18. Which of the following has largest ionic radius?
a) Na b)K c) Li d)Cs
19. In the cyanide ion, the formal negative charge is on:
a) C
b)N

c) Both C and N
d)Resonate between C and N
20. The size of ionic species is correctly given in the order:
a) Cl > 𝑆i > 𝑀g > 𝑁a
b)Na > 𝑀g > 𝑆i > 𝐶l
c) Na > 𝑀g > 𝐶l > 𝑆i
d)Cl > 𝑁a > 𝑀g > 𝑆i
21. Which statement is wrong?
a) 2nd ionisation energy shows jump in alkali metals
b)2nd electron affinity for halogens is zero
c) Maximum electron affinity exists for F
d)Maximum ionization energy exists for He
22. Which of the following atoms has minimum covalent radius?
a) Si b)N c) C d)B
23. The second electron affinity is zero for
a) Alkali metals b)Halogens c) Noble gases d)Transition metal
24. For alkali metals, which one of the following trends is incorrect?
a) Hydration energy : Li > Na > K > Rb b)Ionisation energy : Li > Na > K > Rb
c) Density : Li < Na < K < Rb d)Atomic size : Li < Na < K < Rb
25. Na O, MgO, Al O and SiO have heat of formation equal to −416, −602, −1676 and
−911 kJ mol respectively. The most stable oxide is
a) Na O b)MgO c) Al O d)SiO
26. If Aufbau rule is not followed, K-19 will be placed in
a) 𝑠-block b)𝑝-block c) 𝑑-block d)𝑓-block
27. The electronegativity order of O, F, Cl and Br is:
a) F > 𝑂 > 𝐶𝑙 > 𝐵𝑟 b)F > 𝐶𝑙 < 𝐵𝑟 > 𝑂 c) Br > 𝐶𝑙 > 𝐹 > 𝑂 d)F < 𝐶𝑙 < 𝐵𝑟 < 𝑂
28. Which has the minimum bond energy?
a) H − Br b)H − I c) I − I d)H − H
29. The bond angle in H S (for H − S − H) is:
a) Same as that of Cl − Be − Cl in BeCl
b)Greater than H − N − H bond angle in NH
c) Greater than H − Se − H and less than H − O − H
d)Same as Cl − Sn − Cl in SnCl
30. In which of the following arrangements, the sequence is not strictly according to the property
written against it?
a) CO < 𝑆𝑖O < 𝑆𝑛O < 𝑃𝑏O : increasing oxidising power
b)HF < 𝐻𝐶𝑙 < 𝐻𝐵𝑟 < 𝐻𝐼 : increasing acid strength
c) NH > 𝑃H < 𝐴𝑠H < 𝑆𝑏H : increasing basic strength
d)B < 𝐶 < 𝑂 < 𝑁 : increasing first ionisation enthalpy
31. The tenth elements in the Periodic Table resembles with the
a) First period b)Second period c) Fourth period d)Ninth period
32. Which is not the correct order for the stated property?
a) Ba > 𝑆𝑟 > 𝑀𝑔 ; atomic radius b)F > 𝑂 > 𝑁 ; first ionisation enthalpy
c) Cl > 𝐹 > 𝐼; electron affinity d)O > 𝑆𝑒 > 𝑇𝑒; electronegativity
33. The unequal sharing of bonded pair of electrons between two atoms in a molecule gives rise to:

a) Ionic bond
b)Polar covalent bond
c) Non-polar covalent bond
d)None of the above
34. Which of the following oxides is most acidic in nature?
a) BeO b)MgO c) CaO d)BaO
35. In the formation of NaCl by combination of Na and Cl:
a) Sodium and chlorine both lose electrons
b)Sodium and chlorine both gain electrons
c) Sodium loses but chlorine gains electrons
d)Sodium gains but chlorine loses electrons
36. The molecule having three folds of axis of symmetry is:
a) NH b)PCl c) SO d)CO
37. The covalent compound HCl has the polar character because:
a) The electronegativity of hydrogen is greater than that of chlorine
b)The electronegativity of hydrogen is equal to than that of chlorine
c) The electronegativity of chlorine is greater than that of hydrogen
d)Hydrogen and chlorine are gases
38. If the bond has zero percent ionic character, the bond is:
a) Pure covalent b)Partial covalent c) Partial ionic d)Coordinate covalent
39.
a) 𝑠𝑝 b)𝑠𝑝 c) 𝑠𝑝 d)𝑑𝑠𝑝
40. Mendeleef’s Periodic Table is upset by the fact that
a) Many elements has several isotopes b)Noble gases do not form compounds
c)
Some groups stand divided into two sub
groups 𝐴 and 𝐵
d)Atomic weights of elements are not always
whole numbers
41. The incorrect statement among the following is:
a) The first ionization potential of Al is less than the first ionization potential of Mg
b)The second ionization potential of Mg is greater than the second ionization potential of Na
c) The first ionization potential of Na is less than the first ionization potential of Mg
d)The third ionization potential of Mg is greater than the third ionization potential of Al
42. Which one of the following is an amphoteric oxide?
a) ZnO b)Na O c) SO d)B O
43. The shape of ClO ion is:
a) Square planar b)Square pyramidal c) Tetrahedral d) Trigonal bipyramidal
44. Which one is correct?
a) Dinitrogen is paramagnetic
b)Dihydrogen is paramagnetic
c) Dioxygen is paramagnetic
d)Dioxygen is diamagnetic
45. In which one of the following pairs the radius of the second species is greater than that of the
first?
a) Na, Mg b)O , N c) Li , Be d)Ba , Sr

46. Atomic radii of fluorine and neon in angstrom unit are respectively given by:
a) 0.72, 1.60 b)1.60, 1.60 c) 0.72, 0.72 d)1.60, 0.72
47. According to IUPAC nomenclature, a newly discovered element has been named as Uun. The
atomic number of the element is
a) 111 b)112 c) 109 d)110
48. The correct order of increasing electron affinity of halogens is
a) F < 𝐶𝑙 < 𝐵𝑟 < 𝐼 b)I < 𝐵𝑟 < 𝐹 < 𝐶𝑙 c) I > 𝐵𝑟 > 𝐶𝑙 > 𝐹 d)Br > 𝐼 > 𝐹 > 𝐶𝑙
49. Al element 𝑋 has 3 electrons in 𝑝-orbitals and also belongs to III period. Its molecular formula
should be:
a) 𝑋 b)𝑋 c) 𝑋 d)𝑋
50. Which of the following sequence regarding ionisation potential of coinage metal is correct:
a) Cu > 𝐴𝑔 > 𝐴𝑢 b)Cu < 𝐴𝑔 < 𝐴𝑢 c) Cu > 𝐴𝑔 < 𝐴𝑢 d)Ag > 𝐶𝑢 < 𝐴𝑢
51. The bond length is maximum in:
a) H S b) HF c) H O d) Ice
52. Which of the following is the most electropositive element?
a) P b) S c) Mg d) Al
53. Which group of atoms have nearly same atomic radius?
a) Na, K, Rb, Cs b) Li, Be, B, C c) Fe, Co, Ni, Cu d) F, Cl, Br, I
54. Which of the following statements is wrong?
a) Metals are more than non-metals.
b) There are only few metalloids.
c) Hydrogen can be placed with alkali metals as well as with halogen in Periodic Table.
d) Non-metals are more than metals.
55. Which one of the following has the lowest ionisation energy?
a) 1𝑠 2𝑠 2𝑝 b) 1𝑠 2𝑠 2𝑝 3𝑠 c) 1𝑠 2𝑠 2𝑝 d) 1𝑠 2𝑠 2𝑝
56. The set representing the correct order of first ionisation potential is:
a) K > 𝑁𝑎 > 𝐿𝑖 b) Be > 𝑀𝑔 > 𝐶𝑎 c) B > 𝐶 > 𝑁 d) Ge > 𝑆𝑖 > 𝐶
57. Which one of the following belongs to representative group of elements in the Periodic Table?
a) Aluminium b) Chromium c) Argon d) Lanthanum
58. The shape of NO is planar. It is formed by the overlapping of oxygen orbitals with … orbitals of
nitrogen .
a) 𝑠𝑝 -hybridized b) 𝑠𝑝 -hybridized c) Three 𝑝-orbitals d) None of these
59. If a molecule 𝑀𝑋 has zero dipole moment the sigma bonding orbitals used by 𝑀(at. no. < 21) is:
a) Pure 𝑝 b) 𝑠𝑝-hybrid c) 𝑠𝑝 -hybrid d) 𝑠𝑝 -hybrid
60. 1, 3-butadiene has:
a) 6σ and 2𝜋-bonds b) 2σ and 2𝜋-bonds c) 9σ and 2𝜋-bonds d) 6σ and 2𝜋-bonds
61. Which of the following transitions involves maximum amount of energy?
a) 𝑀 (g) → 𝑀(g) b) 𝑀(g) → 𝑀 (g) c) 𝑀 (g) → 𝑀 (g) d) 𝑀 (g) → 𝑀 (g)
62. Which of the following molecular species has unpaired electron(s)?
a) N b) F c) O d) O
63. The element having lowest ionisation energy among the following is:
a) 1𝑠 , 2𝑠 2𝑝 b) 1𝑠 , 2𝑠 2𝑝 , 3𝑠 c) 1𝑠 , 2𝑠 2𝑝 d) 1𝑠 , 2𝑠 2𝑝
64. Which of the following has largest ionic radius?
a) Li b) K c) Na d) Cs

65. Which will not conduct electricity?
a) Aqueous KOH solution
b) Fused NaCl
c) Graphite
d) KCl in solid state
66. The bond order is maximum in:
a) H b) H c) He d) He
67. The isoelectronic species among the following are:
I − CH ; II − NH ; III − NH ; IV − NH
a) I, II, III b) II, III, IV c) I, II, IV d) II, I
68. The screening effect of 𝑑-electros is
a) Equal to that of 𝑝-electrons b) More than that of 𝑝-electrons
c) Same as 𝑓-electrons d) Less than 𝑝-electrons
69. OF is:
a) Linear molecule and 𝑠𝑝-hybridized
b) Tetrahedral molecule and 𝑠𝑝 -hybridized
c) Bent molecule and 𝑠𝑝 -hybridized
d) None of the above
70. Be and Al exhibit diagonal relationship. Which of the following statement about them is/are not true?
I. Both react with HCl to liberate H
II. They are made passive by HNO
III. Their carbides given acetylene on treatment with water
IV. Their oxides are amphoteric
a) (iii) and (iv) b) (i) and (iii) c) (i) only d) (iii) only
71. Which is not linear?
a) CO b) HCN c) C H d) H O
72. In which of the following bond angle is maximum?
a) NH b) NH c) PCl d) SCl
73. The molecule which has pyramidal shape is:
a) PCl b) SO c) CO d) NO
74. The complex ion which has no ′𝑑′ electrons in the central metal atom is:
a) [MnO ] b) [Co(NH ) ] c) [Fe(CN) ] d) [Cr(H O) ]
75. For the formation of covalent bond, the difference in the value of electronegativities should be:
a) Equal to or less than 1.7
b) More than 1.7
c) 1.7 or more
76. Strongest bond is in:
a) NaCl b) CsCl c) Both (a) and (b) d) None of these
77. The formation of the oxide ion O (g) requires first an exothermic and then an endothermic step
as shown below,
O(g) + e → O (g); ∆𝐻 = −142 kJ/mol
𝑂 (g) + 𝑒 → O (g); ∆𝐻 = 844 kJ/mol
These is because:
a) O ion has comparatively larger size than oxygen atom

b) Oxygen has high electron affinity
c) O ion will lead to resist the addition of another electron
d) Oxygen is more electronegative
78. Which among the following has the largest dipole moment?
a) NH b) H O c) HI d) SO
79. The correct order of radii is
a) N < 𝐵𝑒 < B b) F < O < N c) Fe < 𝐹e < 𝐹e d) Na < 𝐿𝑖 < 𝐾
80. Diagonal relationship is for
a) Li-Na b) Be-Mg c) Si-C d) B-Si
81. Bond order of 1.5 is shown by:
a) O b) O c) O d) O
82. Which one of the following is an amphoteric oxide?
a) ZnO b) Na O c) SO d) B O
83. Among, Al O , SiO , P O and SO the correct order of acid strength is
a) SO < P O < SiO < 𝐴l O b) SiO < SO < 𝐴l O < P O
c) Al O < SiO < SO < P O d) Al O < SiO < P O < SO
84. Point out the wrong statement. On moving horizontally from left to right across a period in
the Periodic Table
a) Metallic character decreases
b) Electronegativity increases
c) Gram atomic volume first decreases and then increases
d) Size of the atoms increases for normal elements
85. The correct increasing bond angles order is:
a) BF < 𝑁F < 𝑃F < 𝐶𝑙F
b) ClF < 𝑃F < 𝑁F < 𝐵F
c) BF ≈ NF < 𝑃F < 𝐶𝑙F
d) BF < 𝑁F < 𝑃F > 𝐶𝑙F
86. The incorrect statement among the following is
a) The first ionisation potential of Al is less than the first ionisation potential of Mg
b) The second ionisation potential of Mg is greater than the second ionisation potential of Na
c) The first ionisation potential of Na is less than the first ionisation potential of Mg
d) The third ionisation potential of Mg is greater than that of Al
87. Concept of bond order in the molecular orbital theory depends on the number of electrons in the
bonding and antibonding orbitals. The bond order:
a) Can have a −ve value
b) Has always an integral value
c) Is a non-zero quantity
d) Can assume any +ve value, including zero
88. Which hybridization results non-polar orbitals?
a) 𝑠𝑝 b) 𝑠𝑝 c) 𝑠𝑝 d) 𝑑𝑠𝑝
89. The total number of valency electrons for PO ion is:
a) 32 b) 16 c) 28 d) 30
90. Intramolecular hydrogen bonding is found in:
a) Salicyldehyde b) Water c) Acetaldehyde d) Phenol

91. Amphoteric oxide combinations are in
a) ZnO, K O, SO b) ZnO, P O , Cl O c) SnO , Al O , ZnO d) PbO , SnO , SO
92. Chlorine atom tends to acquire the structure of:
a) He b) Ne c) Ar d) Kr
93. Which of the following ion is the smallest ion?
a) O b) O c) O d) O
94. Variable valency is characteristic of:
a) Noble gas
b) Alkali metals
c) Transition metals
d) Non-metallic elements
95. Which force is strongest?
a) Dipole-dipole forces
b) Ion-ion forces
c) Ion-dipole forces
d) Ion-induced dipole forces
96. Identify the transition element.
a) 1𝑠 , 2𝑠 2𝑝 , 3𝑠 3𝑝 , 4𝑠 b) 1𝑠 , 2𝑠 2𝑝 , 3𝑠 3𝑝 3𝑑 , 4𝑠
c) 1𝑠 , 2𝑠 2𝑝 , 3𝑠 3𝑝 3𝑑 , 4𝑠 4𝑝 d) 1𝑠 , 2𝑠 2𝑝 , 3𝑠 3𝑝 3𝑑 , 4𝑠 4𝑝
97. For a covalent solid, the units which occupy lattice points are:
a) Atoms b) Ions c) Molecules d) Electrons
98. Which is not true in case of ionic bond?
a) It is linear bond
b) It is 100% ionic
c) It is formed between two atoms with large electronegativity difference
99. In the following molecule, the two carbon atoms marked by asterisk (∗) possess the following
type of hybridized orbitals:
a) 𝑠𝑝 -orbital b) 𝑠𝑝 -orbital c) 𝑠𝑝-orbital d) 𝑠-orbital
100. The element which exists in both hard and soft form is:
a) Fe b) Si c) C d) Al
1. Which of the following has the electronic configuration [Ar]3𝑑 ?
a) Cr b) Fe c) Mn d) V
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 2

2. Electron affinity is positive when
a)
O is formed
from O
b)
O is formed
from O
c)
O is formed
from O
d)
Electron affinity
is always a
negative value
3. (𝐴), (𝐵), (𝐶) are elements in the third short period. Oxide of (𝐴) is ionic, that of (𝐵) is
amphoteric and of (𝐶) a giant molecule. (𝐴), (𝐵) and (𝐶) will have atomic number in the
order
a) (𝐴) < (𝐵) < (𝐶) b) (𝐶) < (𝐵) < (𝐴) c) (𝐴) < (𝐶) < (𝐵) d) (𝐵) < (𝐴) < (𝐶)
4. First long period contains …….. elements
a) 8 b) 18 c) 32 d) 2
5. Li resemble Mg. This is called diagonal relationship which is attributed to
a) Same value of
electronegativity
b) Same value of
electron affinity
c) Penetration of
sub-shells
d) Identical
effective nuclear
charge
6. Which of the following does not represent the correct order of the properties indicated?
a)
Ni > Cr >
Fe > Mn
(size)
b)
Sc > Ti > Cr >
𝑀𝑛(size)
c)
Ni < Co <
Fe < Mn
(unpaired
electron)
d)
Fe > Co >
Ni >
Cu (unpaired
electron)
7. If Aufbau and Hund’s rule are not used, then incorrect statement is
a)
K would be
coloured ion
b)
Na will be in
same 𝑠-block (if
these rules are
true)
c)
Cu would be 𝑠-
block element
d)
Magnetic
moment of
Cr(24) would be
zero
8. Representative elements belong to
a) 𝑠- and 𝑝-block b) 𝑑-block c) 𝑑- and 𝑓-block d) 𝑓-block
9. Which of the following oxides is highly basic?
a) Al O b) Cr O c) Na O d) BaO
10. If Aufbau rule is not followed, K-19 will be placed in ……. block
a) 𝑠 b) 𝑝 c) 𝑑 d) 𝑓
11. Which of the following metals forms a amphoteric oxide?
a) Ca b) Ni c) Zn d) Fe
12. Catenation properties of C, Si, Ge, Sn, Pb are in order
a)
C ≫ Si > 𝐺𝑒
≈ 𝑆𝑛 ≫ 𝑃𝑏
b)
C < 𝑆𝑖 < 𝐺𝑒
< 𝑆𝑛 < 𝑃𝑏
c)
C > 𝑆𝑖 > 𝑆𝑛
> 𝐺𝑒 > 𝑃𝑏
d)
None of the
above is correct
13. Select the correct statement
a) More active
metals are on the
left side of the
Periodic Table
b) Less active
metals are on the
left side of the
Periodic Table
c) Reducing power
decreases
moving down
the group
d) All the above are
correct
statements
14. The electronegativities of elements 𝐴 and 𝐵 are 1.2 and 3.4 units respectively. The type of
bond connecting 𝐴 and 𝐵 in compound 𝐴𝐵 is
a) Covalent b) Ionic c) Coordinate
covalent
d) Polar covalent
15. Which of the following will have maximum electron affinity?
a) 1𝑠 2𝑠 2𝑝 b) 1𝑠 2𝑠 2𝑝 c) 1𝑠 2𝑠 2𝑝 3𝑠 3𝑝 d) 1𝑠 2𝑠 2𝑝 3𝑠 3𝑝

16. Select the correct alternate
a)
Due to lanthanide
contraction Zr
and Hf have
almost equal size
b)
Due to
completion of 3𝑑-
subshell, the
electronic charge
density in this
subshell becomes
very high which
increases the
inter-electronic
repulsion hence,
size increases
c)
Both (a) & (b)
are correct
statements
d)
Both of the above
are incorrect
statements
17. Point out the oxide which is amphoteric in nature
a) CO b) Bi O c) PbO d) CO
18. Covalent radius of nitrogen is 70 pm. Hence, covalent radius of boron is about
a) 60 pm b) 110 pm c) 50 pm d) 40 pm
19. Melting points of NaCl, NaBr, NaI and NaF will be in order
a)
NaI < 𝑁𝑎𝐵𝑟
< 𝑁𝑎𝐶𝑙 < 𝑁𝑎𝐹
b)
NaF < 𝑁𝑎𝐶𝑙
< 𝑁𝑎𝐵𝑟 < 𝑁𝑎𝐼
c)
NaBr < 𝑁𝑎𝐹
< 𝑁𝑎𝐶𝑙 < 𝑁𝑎𝐼
d)
NaCl < 𝑁𝑎𝐼
< 𝑁𝑎𝐹 < 𝑁𝑎𝐵𝑟
20. Which pair is different from the others?
a) Li − Mg b) Na − K c) Ca − Mg d) B − Al
21. For the process
𝑋(g) + 𝑒 → 𝑋 (g), ∆𝐻 = 𝑥
and 𝑋 (g) → 𝑋(g) + 𝑒 , ∆𝐻 = 𝑦
Select the correct alternate
a)
Ionization energy
of 𝑋 (g) is 𝑦
b)
Electron affinity
of 𝑋(g) is 𝑥
c)
Electron affinity
of 𝑋(g) is – 𝑦
d)
All the above are
correct
statements
22. The atoms of the elements belonging to the same group of the Periodic Table will have
a) The same
number of
protons
b) The same
number of
electrons in the
valence-shell
c) The same
number of
neutrons
d) The same
number of
electrons
23. EC of Gd (64) is written as
a) [Xe] 4𝑓 5𝑑 6𝑠 b) [Xe] 4𝑓 6𝑠 c) [Xe] 4𝑓 6𝑠 d) [Xe] 4𝑓
24. Numbering of groups as 1, 2,…..18 was adopted by IUPAC in
a) 1986 b) 1906 c) 1908 d) 1988
25. Which of the following forms a stable +4 oxidation state?
a) Lanthanum b) Cerium c) Europium d) Gadolinium
26. A molecule H − 𝑋 will be 50% ionic if electronegativity difference of H and 𝑋 is
a) 1.2 eV b) 1.4 eV c) 1.5 eV d) 1.7 eV
27. Most stable cation of element 113 will be
a) 𝑀 b) 𝑀 c) 𝑀 d) 𝑀
28. Which is incorrect statement?
a)
In solid state O
is stabilized by
neighbouring
cations
b)
Formation of O
from O is
unfavourable in
the gas phase
c)
Electron affinity
of O > S
d)
All of the above
are incorrect
29. With respect to oxygen maximum valency is shown by
a) Halogen family b) Oxygen family c) Nitrogen family d) Boron family

30. The dominant factor in determining the IE of the elements on moving down the groups is
its
a) Atomic radius b) Effective nuclear
charge
c) Both (a) and (b) d) None of the
above
31. Select the correct statement(s)
a) Across a
transition series,
there is only a
small decrease in
atomic radius
from one
element to
another due to
very small
increase in
effective nuclear
charge
b) The rate of
decrease in the
size across the
lanthanide series
is less than the
across the first
transition series
c) Both (a) & (b)
are correct
statements
d) None of the
above statement
is correct
32. Which of the following statement is false?
a) Elements of IB
and IIB groups
are transition
elements
b) Elements of VB
group do not
contain
metalloids
c) Elements of IA
and IIA groups
are normal
elements
d) Elements of IVB
group are
neither strongly
electronegative
nor strongly
electropositive
33. Which has maximum ionization potential?
a) N b) O c) O d) Na
34. Which has maximum IE?
a) Mg b) Mg c) Mg d) Equal
35. 𝑀(g) → 𝑀 (g) + 𝑒 , ∆𝐻 = 100 eV
𝑀(g) → 𝑀 (g) + 2𝑒 , ∆𝐻 = 250 eV
Which is the incorrect statement?
a)
𝐼 of 𝑀(g) is 100
eV
b)
𝐼 of 𝑀 (g) is 150
eV
c)
eV
d)
eV
36. Which is best oxidizing agent?
a) Ge b) Pb c) Sn d) Sn
37. Element with valence shell-electronic configuration as 𝑑 𝑠 is placed in
a) IA, 𝑠-block b) VIA, 𝑠-block c) VIB, 𝑠-block d) VIB, 𝑑-block
38. Which group of elements is analogous to the lanthanides?
a) Halides b) Actinides c) Chalcogenides d) Borides
39. Which will have graded property similar to EC 1𝑠 2𝑠 2𝑝 3𝑠 3𝑝 4𝑠 ?
a) [Ar]3𝑑 4𝑠 b) [Kr]4𝑑 5𝑠 c) [Kr]5𝑠 d) All of these
40. The separation of lanthanides in ion-exchange method is based on
a) Basicity of the
hydroxides
b) Size of the
hydrated ions
c) Size of the
unhydrated ion
d) The solubility of
their nitrates
41. Which of the following will have the most negative electron affinity and which the least

negative?
a) F, Cl b) Cl, F c) Cl, S d) Cl, P
42. Select the correct statement about radius of an atom
a) Values of van der
Waals’ radii are
larger than those
of covalent radii
because the van
der Waals’ forces
are much weaker
than the forces
operating
between atoms
in a covalently
bonded molecule
b) The metallic
radii are smaller
than the van der
Waals’ radii,
since the
bonding forces in
the metallic
crystal lattice are
much stronger
than the van der
Waals’ forces
c) Both (a) & (b)
are correct
d) None of the
above is correct
43. Sodium forms Na ion but it does not form Na because
a)
Very low value of
(IE) and (IE)
b)
Very high value
of (IE) and (IE)
c)
Low value of
(IE) and low
value of (IE)
d)
Low value of
(IE) and high
value of (IE)
44. The electronic configuration
1𝑠 , 2𝑠 , 2𝑝 , 3𝑠 , 3𝑝 , 3𝑑 , 4𝑠 , 4𝑝 4𝑑 , 5𝑠 is for
a) 𝑓-block element b) 𝑑-block element c) 𝑝-block element d) 𝑠-block element
45. Which one of the following has the smallest atomic radius?
a) F b) Cl c) Cs d) Mg
46. Mixture containing aqueous Li , Na , K ions are electrolysed. Cations are discharged at
cathode in the order: (easiest at the end)
𝑀 (𝑎𝑞) + 𝑒 → 𝑀
a) Li , Na , K b) K , Na , Li c) Li , K , Na d) Na , K , Li
47. Screening effect is not observed in
a) He b) Li c) Be d) All of these cases
48. Which has maximum stability?
a) AsCl b) SbCl c) BiCl d) Equal
49. In periodic Table, metallic elements appear
a) In the left-hand
columns
b) In the top rows c) In the right-hand
columns
d) In the bottom
rows
50. For the element (𝑋), student (𝐴) measured its radius as 102 nm, student (𝐵) as 103 nm and
(𝐶) as 100 nm using same apparatus. Their teacher explained that measurements were
correct by saying that recorded values by (𝐴), (𝐵) and (𝐶) are
a) Crystal, van der
Waals and
covalent radii
b) Covalent, crystal
and van der
Waals radii
c) Van der Waals,
ionic and
covalent
d) None is correct
51. The hydration energy of lithium ion is −544 kJ mol which is higher than that of other
alkali metal ions. This is explained in terms of
a) Small size of
lithium
b) Higher IP c) Element of
lowest atomic
d) More reactive
than other alkali

weight metals
52. The ionic radii of O , F , Na and Mg are 1.35, 1.34, 0.95 and 0.66 Å respectively. The
radius of the Ne atom is
a) 1.39 Å b) 1.12 Å c) 0.85 Å d) 0.50 Å
53. Metallic nature increases moving down the group because
a) Nuclear charge
increases
b) Shielding
increases
c) Both (a) and (b) d) None of the
above
54. Which reaction has most negative ∆𝐺° value?
a)
Ga + 2𝑒
→ Ga
b) In + 2𝑒 → In c) Tl + 2𝑒 → Tl d)
Cannot be
predicted
55. Fluorine has the highest electronegativity among the 𝑛𝑠 𝑛𝑝 group on the Pauling scale,
but the electron affinity of fluorine is less than that of chlorine because
a)
The atomic
number of
fluorine is less
than that of
chlorine
b)
Fluorine being
the first member
of the family
behaves in an
unusual manner
c)
Chlorine can
accommodate an
electron better
than fluorine by
utilizing its
vacant 3𝑑-orbital
d)
Small size, high
electron density
and an increased
electron
repulsion makes
addition of an
electron to
fluorine less
favourable than
that in the case of
chlorine
56. Fluorine does not form any polyhalide as other halogens because
a)
It has maximum
ionic character
b)
It has low F − F
bond energy
(38.5 kcal mol )
c)
Of the absence of
𝑑-orbitals in the
valence shell of
fluorine
d)
It brings about
maximum
coordination
number in other
elements
57. Among the following, the number of elements showing only one non-zero oxidation state is
O, Cl, F, N, P, Sn, Tl, Na, Ti
a) 1 b) 2 c) 3 d) 4
a)
Ionic hydrides
are better
reducing agents
b)
Covalent nature
of hydrides
increases across
a period and
decreases down
the group
c)
LiAlH can
reduce carbonyl
compounds to
alcohols
d)
All the above are
correct
statements
59. The following acids have been arranged in the order of decreasing acid strength. Identify
the correct order
ClOH (I) BrOH (II) IOH (III)
a) I > 𝐼𝐼 > 𝐼𝐼𝐼 b) II > 𝐼 > 𝐼𝐼𝐼 c) III > 𝐼𝐼 > 𝐼 d) I > 𝐼𝐼𝐼 > 𝐼𝐼
60. Gd (64) has ……. unpaired electrons with sum of spin……
a) 7, 3.5 b) 8, 3 c) 6, 3 d) 8, 4

61. For Cu , effective nuclear charge felt by a 3𝑑-electron is
a) 14.25 b) 13.95 c) 14.65 d) 29.0
62. The relative thermal stabilities of alkali metal halides are such that
a)
CsCl > 𝑅𝑏𝐶𝑙
> 𝐾𝐶𝑙 > 𝑁𝑎𝐶𝑙
> 𝐿𝑖𝐶𝑙
b)
LiCl > 𝑁𝑎𝐶𝑙
> 𝐾𝐶𝑙 > 𝑅𝑏𝐶𝑙
> 𝐶𝑠𝐶𝑙
c)
CsCl > 𝑅𝑏𝐶𝑙
< 𝐾𝐶𝑙 > 𝑁𝑎𝐶𝑙
< 𝐿𝑖𝐶𝑙
d)
CsCl < 𝑅𝑏𝐶𝑙
> 𝐾𝐶𝑙 < 𝑁𝑎𝐶𝑙
> 𝐿𝑖𝐶𝑙
63. Which of the following order is wrong?
a)
NH < PH <
AsH − acidic
b)
Li < 𝐵𝑒 < 𝐵
< 𝐶 − (IE)
c)
Al O < 𝑀𝑔𝑂 <
Na O < K O −
basic
d)
Li < Na <
K < Cs − ionic
radius
64. The statement is not true for the long form of the Periodic Table
a)
It reflects the
sequence of
filling the
electrons in the
order of the sub-
energy shells
𝑠, 𝑝, 𝑑 and 𝑓
b)
It helps to predict
the stable
valency states of
the elements
c)
It reflects trends
in physical and
chemical
properties of the
elements
d)
It helps to predict
the relative
ionicity of the
bond between
any two elements
65. Element with atomic number 115 has configuration as ……. and with most stable cation as
……
a) [Rn]7𝑠 5𝑑 4𝑓 7
b) [Rn]7𝑠 5𝑑 4𝑓 7
c) [Rn]7𝑠 5𝑑 4𝑓 7
d) [Rn]5𝑑 4𝑓 7𝑝 𝑀
66. In which case effective nuclear charge is minimum?
a) Be b) Be c) Be d) Equal
67. Following graph shows variation of ionization potential (IP) with atomic number in second
period (Li − Ne). Value of ionization potential (IP) of Na (11) will be
a) Above Ne b) Below Ne but
above O
c) Below Li d) Between N and O
68. Recently (in Aug 2003) two new elements have been discovered with atomic numbers
a) 113, 114 b) 114, 115 c) 115, 116 d) 113, 115
69. 𝐴𝐵 is predominantly ionic as 𝐴 𝐵 if (IP stands for ionization potential, EA for electron
affinity and EN for electronegativity)
a) (IP) < (IP) b) (EA) < (EA) c) (EN) < (EN) d) (IP) < (IP)
70. 𝑀 has electronic configuration as [Ar]3𝑑 4𝑠 , hence it lies in
a) 𝑠-block b) 𝑝-block c) 𝑑-block d) 𝑓-block
71. Transition elements have vacant
a) 𝑠-orbital b) 𝑝-orbital c) 𝑑-orbital d) 𝑓-orbital
72. When the following five anions are arranged in order of decreasing ionic radius, the correct

sequence is
a) Se , I , Br , O ,b) I , Se , O , Br ,c) Se , I , Br , F , O
d) I , Se , Br , O ,
73. Match Column I (atomic number of elements) with Column II (position of elements in
Periodic Table) and select the correct answer using the codes given below the Columns
Column I Column II
A
B
C
D
19
22
32
64
1.
2.
3.
4.
𝑝-block
𝑓-block
𝑑-block
𝑠-block
Codes
A B C D
a) 1 2 3 4 b) 4 3 1 2 c) 4 1 3 2 d) 2 1 3 4
74. Ionic radii of:
a) Ti < 𝑀n b)
Cl
< Cl
c) K > 𝐶l d) P > P
75. Covalency is favoured in the following cases
a) A smaller cation b) A larger anion c) Large charges on
cation or anion
d) In all the above
cases
76. Main group elements constitute
a) 𝑠-and 𝑝-block b) 𝑝-and 𝑑-block c) 𝑠-and 𝑑-block d) 𝑑-and 𝑓-block
77. Oxidation energy of Li(𝑠) to Li (𝑎𝑞) is least in group IA elements. This is because of
a)
Maximum heat of
sublimation of
Li(𝑠)
b)
Maximum heat of
hydration of Li
c)
Less negative
heat of hydration
of Li
d)
Maximum
ionization energy
of Li
a) Electron affinity
of nitrogen is
much lower than
that of its
neighbouring
elements carbon
and oxygen
b) Electron affinity
of F is higher
than that of
chlorine
c) Both (a) & (b)
are correct
d) None of the
above is correct
79. Chalcogens are elements of
a) Group 16 b) 𝑝-block c)
𝑛𝑠 𝑛𝑝
configuration
d) All are correct
80. The electron affinities of N, O, S and Cl are such that
a) N < 𝑂 < 𝑆 < 𝐶𝑙 b) O < 𝑁 < 𝐶𝑙 < 𝑆 c) O ≈ Cl < 𝑁 ≈ 𝑆 d) O < 𝑆 < 𝐶𝑙 < 𝑁
81. Ionic radii are
a) Inversely
proportional to
effective nuclear
charge
b) Inversely
proportional to
square of
effective nuclear
charge
c) Directly
proportional to
effective nuclear
charge
d) Directly
proportional to
square of
effective nuclear
charge
82. Out of BeH , CuH , CrH and NaH, covalent hydrides are
a) BeH , NaH b) CuH , CrH c) BeH , CuH , CrH d) All of these
83. Which of the following is a transition element?

a) Al b) As c) Ni d) Rb
84. Elements 𝑋 and 𝑌 have valence shell electron configuration as
𝑋: 𝑛𝑠 𝑛𝑝 ; 𝑌: 𝑛𝑠 𝑛𝑝
Which compound is likely formed from 𝑋 and 𝑌?
a) 𝑋 𝑌 b) 𝑌 𝑋 c) 𝑋𝑌 d) 𝑌𝑋
85. Higher values of ionization energies of the 5𝑑-transition elements are consistent with the
a) Relatively
smaller effective
nuclear charge
b) Relatively
smaller size of
their atoms
c) Relatively
smaller
penetration
d) All of the above
are correct
86. Tendency of I , Br , Cl and F to be oxidized is in order
a)
I > Br > Cl
> F
b)
I < Br < Cl
< F
c)
I < Cl
< F < 𝐵𝑟
d)
I = Br < Cl
= F
87. For which of the following crystals would you expect the assumption of anion-anion contact
to be valid?
a) CsBr b) NlaF c) KCl d) NaI
88. Which of the following represents the correct order of increasing first ionization enthalpy
for Ca, Ba, S, Se and Ar?
a)
Ca < 𝑆 < 𝐵𝑎
< 𝑆𝑒 < 𝐴𝑟
b)
S < 𝑆𝑒 < 𝐶𝑎
< 𝐵𝑎 < 𝐴𝑟
c)
Ba < 𝐶𝑎 < 𝑆𝑒
< 𝑆 < 𝐴𝑟
d)
Ca < 𝐵𝑎 < 𝑆
< 𝑆𝑒 < 𝐴𝑟
89. Valence electrons in the element 𝐴 are 3 and that in element 𝐵 are 6. Most probable
compound formed from 𝐴 and 𝐵 is
a) 𝐴 𝐵 b) 𝐴𝐵 c) 𝐴 𝐵 d) 𝐴 𝐵
90. Which of the following ions has the smallest radius?
a) Ti b) Pt c) Ni d) Zr
91. As we proceed from top to bottom in the Periodic Table
a) Hydroxides are
more basic
b) Oxyacids are less
acidic
c) Neither of the
above
d) Both of the
above
92. Element 113 is produced 𝑣𝑖𝑎
a)
α-decay of
element 115
b)
β-decay of
element 114
c)
α-decay of
element 111
d)
β-decay of
element 112
93. Two new elements (discovered in Aug 2003) with atomic number 113 and 115 are to be
placed in
a) 𝑠-block b) 𝑝-block c) 𝑑-block d) 𝑓-block
94. Nitrogen is found to have higher value of ionization potential because
a)
It has half-filled
𝑝-orbitals
b)
It is chemically
inert
c)
Its shielding
effect overcomes
the nuclear
charge
d)
All of the above
are correct
95. Extent of hydration of Na , Mg , Al is in order
a)
Na < Al
< Mg
b)
Na < Mg
< Al
c)
Al < Mg
< Na
d) Equal
96. Which pair represents incorrect first IE?
a) Be > B b) N > O c) Li > Na d) He > He
97. Of the following pairs, the one containing examples of metalloid elements in the Periodic

Table is
a) Na and K b) F and Cl c) Cu and Ag d) B and Si
98. Which of the following reactions should be most favoured thermodynamically?
a)
Na O + Cl O
→ 2NaClO
b)
Na O + SO
→ Na SO
c)
Na O + P O
→ 2Na PO
d)
Na O + SiO
→ Na SiO
99.
which is the incorrect statements?
a)
𝐼 of Be > 𝐼 of B
but 𝐼 of Be < 𝐼
of B
b)
𝐼 of Be < 𝐼 of B
but 𝐼 of Be < 𝐼
of B
c)
𝐼 of Be < 𝐼 of
B
d)
𝐼 of Be is
abnormally high
100. State, which one of the following has the largest atomic radius?
a) Cs b) Mg c) Ba d) Cr

1) d 2) c 3) b 4) b
5) b 6) b 7) d 8) a
9) c 10) b 11) c 12) d
13) d 14) c 15) b 16) b
17) d 18) d 19) d 20) a
21) c 22) b 23) b 24) c
25) b 26) c 27) a 28) c
29) c 30) c 31) b 32) b
33) b 34) a 35) c 36) a
37) c 38) a 39) c 40) a
41) b 42) a 43) c 44) c
45) b 46) a 47) d 48) b
49) c 50) c 51) a 52) c
53) c 54) d 55) b 56) b
57) a 58) b 59) c 60) c
61) d 62) c 63) b 64) d
65) d 66) a 67) d 68) d
69) c 70) d 71) d 72) b
73) a 74) a 75) a 76) b
77) c 78) b 79) b 80) d
81) d 82) a 83) d 84) d
85) b 86) b 87) d 88) c
89) a 90) a 91) c 92) c
93) b 94) c 95) b 96) b
97) a 98) a 99) c 100) c
1 (d)
Born-Haber cycle inter-relates the various energy terms involved in ionic bonding.
2 (c)
Follow bonding rules.
3 (b)
Alkali metals are most electropositive elements.
4 (b)
In H O, H-atom contains only two electrons.
5 (b)
Fluorine is more reactive than chlorine, bromine and iodine
6 (b)
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 1 (ANSWERS)

Due to H-bonding in NH .
7 (d)
The order of screening effect for a given shell electrons is 𝑠 > 𝑝 > 𝑑 > 𝑓.
8 (a)
The ionisation energy of elements decreases down the group.
9 (c)
Cl in ClF has 𝑠𝑝 𝑑-hybridization
and possesses two axial Cl— Fbonds and one equatorial bond Two lone pairs are at
equatorial position give rise to bent ‘T’ shape to ClF .
10 (b)
In like atoms, electronegativity difference is zero.
11 (c)
S molecule is paramagnetic like O having 2 unpaired electrons.
13 (d)
Along the period acidic strength of oxide increases
14 (c)
In order to belong with the same family, the outer configuration must be the same
15 (b)
Mn is most stable as it has half filled 𝑑-orbitals.
16 (b)
The atomic radius decreases along the period. Also cations are always smaller than their
parent atom and anions are always larger than their parent atom .
17 (d)
S = C = S .
18 (d)
Cation radius increases down the group.
19 (d)
Cyanide ion is,
.
20 (a)
All are isoelectronic species; more is nuclear charge smaller is ionic size.
21 (c)
Electron affinity order for halogens is Cl > 𝐹 > 𝐵𝑟 > 𝐼.
22 (b)
N atom has smallest radius.
23 (b)
Halogens (𝑛𝑠 𝑛𝑝 ) after getting one electron occupy 𝑛𝑠 𝑛𝑝 configuration, thus have 𝐸𝐴
zero
24 (c)
In general, density increases on moving downward in a group but density of potassium (K)

is lesser than that of the sodium (Na). This is because of the abnormal increase in atomic
size on moving from Na (86 pm) to K (227 pm).
Thus, the correct order of density is
Li < 𝐾 < 𝑁𝑎 < 𝑅b
25 (b)
The oxide having maximum heat of formation per oxygen atom (thus energy needed to
break one 𝑀 − O bond will be highest) will be most stable.
MgO is most stable oxide among Na O, SiO , Al O and MgO.
26 (c)
If Aufbau rule is not followed then 19th electron in K enters in 3𝑑 sub-shell, not in 4𝑠
27 (a)
The most electronegative element is F and next to F is O.
28 (c)
Larger is the size of atom, lesser is the tendency for overlapping, lesser is bond energy.
29 (c)
Bond angles in BeCl , NH , H O and SnCl are 180°, 107°, 104.5° and 119° respectively. Also
H S, H O, H Se has 𝑠𝑝 -hybridization and bond angles of hydrides decreases down the
group.
30 (c)
The correct increasing basic strength:
SbH < 𝐴𝑠H < 𝑃H < 𝑁H
NH is the most basic because of its small size, the electron density of electron pair is
concentrated over small region. As the size increases, the electron density gets diffused over
a large surface area and hence the ability to donate the electron pair (basicity) decreases.
31 (b)
Each period consists of a series of elements whose atoms have the same principal quantum
number (𝑛) of the outermost shell, 𝑖𝑒, in second period, 𝑛 = 2, this shell has four orbitals
(one 2𝑠 and three 2𝑝) which can have eight electrons, hence second period contain 8
elements from atomic number 3 to 10
32 (b)
Om moving along a period, ionisation enthalpy increases. Thus, the order of ionisation
enthalpy should be as follow :
F > 𝑂 > 𝑁
But N has half-filled structure, therefore, it is more stable than O, That’s why its ionisation
erthalpy is highper than O. Thus, the correct order of IE is
F > 𝑂 > 𝑁
33 (b)
This give rise to polarity in bonds.
34 (a)
BeO is most acidic in nature amongst the given choices because acidity of oxides increases
with decreases in electropositive character of central atom.
35 (c)
NaCl exist as Na Cl .
36 (a)

NH has pyramidal shape and thus, possesses three folds axis of symmetry.
37 (c)
Larger is the difference in electronegativities of two atom, more is polar character in bond.
38 (a)
Non-polar or pure covalent bond has zero percent ionic character due to the absence of
partial charges on either end.
39 (c)
N in it has three 𝜎-bonds and one lone pair of electron.
40 (a)
Mendeleef failed to assign positions to isotopes on the basis of atomic mass according to his
periodic law
41 (b)
The removal of second electron from Mg takes place from 3𝑠-orbital whereas, the removal
of second electron from Na takes place from 2𝑝-orbital. More closer are shells to the
nucleus, difficult is removal of electron.
42 (a)
ZnO can react with acid and base both
ZnO + 2HClZnCl + H O
ZnO + 2NaOHNa ZnO + H O
43 (c)
ClO has 𝑠𝑝 -hybridization on Cl atom .
44 (c)
O has two unpaired electrons .
45 (b)
O and N both are isoelectronic but differ in the charge possessed by them. As the
negative charge increase, the electrons are held less and less tightly by the nucleus,
therefore ionic radii increases. Hence, ionic radii of N is greater than O .
In a period from left to right atomic radii decreases but in a group on moving downwards it
increases.
46 (a)
Ne has van der Waals radius larger than covalent radius of fluorine.
48 (b)
The value of electron affinity decreases with increase in size of atom, because the nuclear
attraction decreases as the atomic number increases. Fluorine due to its very small size has
lower electron affinity than chlorine. Hence, the increasing order of electron affinity of
halogen is
I < 𝐵𝑟 < 𝐹 < 𝐶𝑙.
49 (c)
The element is P which exists as P .
50 (c)
Atomic size of Ag and Au are closer to each other but nuclear charge is more on Au
51 (a)
S atom is larger in size than O and F.
52 (c)
Electropositive character decreases across the period as metallic character decreases

53 (c)
Due to shielding effect of (𝑛 − 1)𝑑-subshell.
54 (d)
Non-metals are more than metals is the wrong statement.
55 (b)
1𝑠 , 2𝑠 , 2𝑝 , 3𝑠 . It is an alkali metal; hence has least ionisation potential.
56 (b)
The ionisation potential decreases down the group.
58 (b)
N is 𝑠𝑝 -hybridized on NO .
59 (c)
𝑒. g. , BF , a non-polar molecule having 𝑠𝑝 -hybridization.
60 (c)
Butadiene is CH = CH − CH = CH .
61 (d)
𝑀 → 𝑀 , after the removal of 2𝑒 , the nuclear charge per electron increases due to
which high energy is required to remove 3𝑒
62 (c)
O has one unpaired electron in its antibonding molecular orbital.
63 (b)
Removal of electron is easier in the order of shell 4 > 3 > 2 > 1
64 (d)
Ionic radii increases in a group
65 (d)
Ionic compounds conduct current only in fused state.
66 (a)
The bond orders for H , H , He and He are 1.0, 0.5, 0.0 and 0.5 respcetively.
67 (d)
CH and NH both have 8 electrons .
69 (c)
O atom possesses 𝑠𝑝 -hybridization with two lone pair of electron.
70 (d)
Be C + 2H O → CH + 2BeO
Al C + 6H O → 3CH + 2Al O
71 (d)
H O is V shaped.
72 (b)
NH has angle of 109°28′.
73 (a)
Due to 𝑠𝑝 -hybridization on P with one lone pair.
74 (a)
In MnO , the oxidation no. of Mn is +7, 𝑖. 𝑒., all the 4𝑠 and 3𝑑 electrons are lost.
75 (a)
If difference in electronegativity in between two atoms is 1.7, the molecule possesses 50%
covalent +50% ionic nature.
76 (b)
CsCl is most ionic because of most electropositive nature of Cs.
77 (c)
Anion (O ) repels the test electron because of same charge.

78 (b)
It is a fact.
79 (b)
Ionic radii decreases significantly from left to right in a period among representative
elements
80 (d)
B and Si shows the diagonal relationship.
81 (d)
O ∶ 𝜎1𝑠 , 𝜎∗
1𝑠 , 𝜎2𝑠 , 𝜎∗
2𝑠 , 𝜎2𝑝
𝜋2𝑝
𝜋2𝑝
𝜋∗
2𝑝
𝜋∗
2𝑝
∴ B. O. =
10 − 7
2
= 1.5
82 (a)
ZnO can react with acid and base both
ZnO+2HCl→ ZnCl + H O
ZnO+2NaOH→ Na ZnO + H O
83 (d)
While moving along a group from top to bottom, acidic nature of oxides decreases and along
a period left to right acidic nature increases.
amphoteric acidic max. acidic
Thus, Al O < SiO < P O < SO
85 (b)
Bond angles of CIF , PF , NF and BF are (180°, 90°), (101°), (106°) and (120°)
respectively.
86 (b)
IE (II) of Na is higher than that of Mg because in case of Na, the second 𝑒 has to be
removed from the noble gas core while in case of Mg removal of second 𝑒 gives a noble gas
core
Mg has high first ionisation potential than Na because of its stable 𝑛𝑠 configuration
87 (d)
Follow concept of bond order in M.O. theory.
88 (c)
𝑠𝑝 -hybridization leads to tetrahedral geometry.
89 (a)
5 of P + 24 of O + 3 of − ve charge = 32.
91 (c)
SnO , Al O and ZnO are amphoteric oxide.
92 (c)
The inert gas just after chlorine is argon.
93 (b)
Cation has small size than parent atom and anion has larger size than parent atom
94 (c)
Due to the presence of 𝑑-subshell electrons.
95 (b)
Coulombic forces are strongest among all .
96 (b)

Transition elements are those elements which have partially filled 𝑑-subshells in their
elementary form. Therefore, the general electronic configuration of 𝑑-block element is
(𝑛 − 1)𝑑 𝑛𝑠 .
97 (a)
In ionic solids, ions exist at lattice points. In covalent solids atoms lie at lattice points.
98 (a)
Ionic bond are non-directional.
99 (c)
Both carbon atoms have 2 𝜎- and 2 𝜋-bonds
100 (c)
Diamond is hard, graphite is soft.
1) b 2) b 3) a 4) b
5) a 6) a 7) a 8) a
9) d 10) c 11) c 12) a
13) a 14) b 15) c 16) c
17) c 18) b 19) a 20) a
21) d 22) b 23) a 24) d
25) b 26) d 27) c 28) c
29) a 30) a 31) c 32) d
33) c 34) c 35) c 36) b
37) d 38) b 39) c 40) b
41) d 42) c 43) d 44) b
45) a 46) a 47) d 48) c
49) a 50) a 51) a 52) a
53) c 54) c 55) c 56) c
57) b 58) d 59) a 60) d
61) a 62) b 63) b 64) d
65) a 66) a 67) c 68) d
69) c 70) b 71) c 72) d
73) b 74) d 75) d 76) a
77) c 78) a 79) d 80) a
81) a 82) c 83) c 84) c
85) b 86) a 87) d 88) c
89) d 90) c 91) d 92) a
93) b 94) a 95) b 96) d
97) d 98) a 99) b 100) a
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 2 (ANSWERS)

1 (b)
Fe(26): [Ar]3𝑑 4𝑠
Fe : [Ar]3𝑑
3 (a)
Thus, atomic number increases in the order 𝐴 < 𝐵 < 𝐶
4 (b)
First long period starts with 3rd period [K(19)− Kr(36)]
Thus, total =18 elements
6 (a)
Ni is smallest in size
7 (a)
K(19)1𝑠 2𝑠 2𝑝 3𝑠 3𝑝 3𝑑
K (19)1𝑠 2𝑠 2𝑝 3𝑠 3𝑝 Colourless ion due to lack of electron in 𝑑-orbitals
Thus, (a) is incorrect
10 (c)
E.C. of K=19 in the absence of Aufbau rule is
1𝑠 2𝑠 2𝑝 3𝑠 3𝑝 3𝑑
↑
Last-filling electron goes into 𝑑-orbital
Thus, 𝑑-block element
11 (c)
ZnO + 2HCl → ZnCl + H O
Basic oxide
ZnO + 2NaOH → Na ZnO + H O
Acidic oxide
14 (b)
𝐴 (1.2) is electropositive element
𝐵 (3.4) is electronegative element
Thus, ionic bond is formed
15 (c)
(b) and (d) with 𝑠 𝑝 configuration has zero values of EA
Cl(c) with empty 𝑑-orbital has greater EA than F(a)
17 (c)
PbO is soluble in NaOH − an acidic oxide
PbO is soluble in HCl − a basic acid
18 (b)
Radius decreases along a period left to right

Thus, covalent radius of B would be higher than that of N
19 (a)
Smaller the size of anion, smaller the polarizing power and thus larger the ionic character of
NaX. Thus, ionic character
NaI < NaBr < 𝑁𝑎𝐶𝑙 < 𝑁𝑎𝐹
and mp : NaI < 𝑁𝑎𝐵𝑟 < 𝑁𝑎𝐶𝑙 < 𝑁𝑎𝐹
20 (a)
(a) Li − Mg diagonal relationship
(b)
(c)
(d)
Na − K
Ca − Mg
B − Al
same group
21 (d)
𝑋(𝑔) + 𝑒 → 𝑋 (𝑔), ∆𝐻 = 𝑥
(EA of 𝑋(𝑔))
𝑋 (𝑔) → 𝑋(𝑔) + 𝑒 , ∆𝐻 = 𝑦 = IE of 𝑋 (𝑔) = −EA of 𝑋(𝑔)
Thus, (a), (b), (c) true
22 (b)
Elements in a group have same (EC) in valence shell
25 (b)
Ce (58)[Xe] 4𝑓 5𝑑 6𝑠
Thus, Ce has [Xe] configuration
27 (c)
Atomic number 113 belong to 𝑝-block (Group IIIA)
Probable oxidation states are +1, +3, but due to inert-pair effect
stability of +1 > +3
Thus, 𝑀
29 (a)
Cl O Cl = +7
30 (a)
Going down the group effective nuclear charge remains almost constant, hence (IE) is
dependent on radius of two element
32 (d)
(IV) B contain C, Si, Ge, Sn, Pb
Ge, Sn, Pb are strongly electropositive
33 (c)
Isoelectronic N(7) and O (8) have same EC
N(7)
O (8)
1𝑠 2𝑠 2𝑝
↿ ↿ ↿
2𝑝 2𝑝 2𝑝
Most stable due to all unpaired electrons
O ] 1𝑠 2𝑠 2𝑝
⥮ ↿ ↿

Less stable than O and N
Since 𝑍of O > 𝑍 of N
Hence IP of O > 𝑁
Na with only one unpaired electron in 3𝑠 has lowest IP
34 (c)
Mg → Mg + 𝑒 (IE)
(IE) < (IE) < (IE)
35 (c)
𝑀(g) → 𝑀 (g) + 𝑒 , 𝐼 (𝑀) = ∆𝐻 = 100 eV
(a) is correct
𝑀 (g) → 𝑀 (g) + 𝑒 , 𝐼 (𝑀) = 250 − 100 = 150 eV
or 𝐼 (𝑀 ) = 150 eV (b) is correct
𝐼 of 𝑀(g) is 150 eV thus (c) is incorrect
36 (b)
Pb + 2𝑒 → Pb
Stability of Pb > Pb
Hence, Pb is reduced most easily and is thus best oxidizing agent
37 (d)
Valence shell electronic configuration is 𝑑 𝑠
Differentiating electron goes into 𝑑-orbitals. Thus, 𝑑-block element. Group VIB
38 (b)
5𝑓-block elements
39 (c)
EC: 1𝑠 2𝑠 2𝑝 3𝑠 3𝑝 4𝑠
(Total 19 electrons i.e. 𝑍 = 19)
It is K (alkali metals of Group 1)
Graded property will be that of same group
1. [Ar]3𝑑 4𝑠 −Cu group 11
2. [Kr]4𝑑 5𝑠 − Ag group 11
3. [Kr]5𝑠 − Rb group 1
43 (d)
Na
( )
⎯
⎯ Na
[Ne]3s [Ne]
Na
( )
⎯
⎯ Na
Ne
Na has stable inert gas configuration
Thus, (IE) is very high
46 (a)

Size of Li < Na < Rb < Cs
Hydration Li > Na > Rb > Cs
Size of hydrated ion Li > Na > Rb > Cs
Smaller the size, Li < Na < Rb < Cs
Larger the hydration, hence larger the size of the hydrated ion (in aqueous solution)
Heavier the ion, smaller the ionic mobility
Size Li < Na < K
Hydration Li > Na > K
Ionic mobility Li < Na < K
Discharged at cathode Li < Na < K (Most easiest)
47 (d)
He , Li and Be are isoelectronic of H atom (with one electron). Hence, single electron is
not screened
48 (c)
As, Sb, Bi are group 15 having oxidation +3 and +5. Stability is in order
As > Sb > Bi
As < Sb < Bi due to inert-pair effect
Thus, BiCl is most stable
49 (a)
Smaller the (IE), greater the metallic nature
52 (a)
In case of Ne van der Waals’ radius is taken. Hence, it should have maximum size out of the
given option
54 (c)
Due to inert-pair effect, stability of
Ga > In > Tl
and Ga < In < Tl
Thus, Tl + 2𝑒 → Tl
Is most spontaneous. Thus ∆𝐺° < 0 and is most negative
57 (b)
F is the most electronegative element which cannot loose electron to other so it exhibits
only−1 state. Na is alkali metal which can loose only one electron so exhibits only +1 state.
58 (d)
4. NaH + H O → NaOH + H

Reducing nature
59 (a)
Greater the electronegativity of 𝑋(Cl > 𝐵𝑟 > 𝐼) greater the acid strength
Thus, I>II>III
60 (d)
Gd ∶ 4𝑓 5𝑑
Unpaired electrons =8
Sum of spin = 8 × = 4
61 (a)
𝐶𝑢 1𝑠
( )
2𝑠 2𝑝
( )
3𝑠 3𝑝 3𝑑
↓↓↓
𝑆 = 2 × 1 + 8 × 0.85 + 17 × 0.35 = 14.75
(One 𝑑-electron is screened by 17 electrons in 𝑛th, 8 electrons in (𝑛 − 1) and 2 electrons in
(𝑛 − 2)
𝑍 = 29 − 14.75 = 14.25
62 (b)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Size
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Covalent nature MCl
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Lattice energy of MCl
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Thermal stability
63 (b)
(b) Li < B < 𝐵𝑒 < 𝐶 (IE)
Be has paired electron hence its (IE) is larger than that of B
Thus, (b) is incorrect
(c) Correct (d) correct

65 (a)
Atomic number (115) has E.C. has [Rn] 7𝑠 5𝑑 4𝑓 7𝑝
Probable oxidation states are +3, +5
But due to inert pair effect 𝑀 is the most stable cation
66 (a)
Effective nuclear charge 𝑍 = 𝑍 − 𝑆
Where, 𝑍 = atomic number
and 𝑆 = screening constant
= 0.35 per electron for electron in 𝑛th orbit
= 0.85 per electron for electron in (𝑛 − 1)th orbit
= 1.00 per electron for electrons in (𝑛 − 2)th, (𝑛 − 3)th, (𝑛 − 4)th orbit
= 0.30 per electron in 1s-orbital (when alone)
Be 1𝑠 2𝑠 one valence-electron in 2s is screened by one electron in 2𝑠-orbital (𝑛th orbit)
and two electrons in 1𝑠-orbital ((𝑛 − 1)th orbit)
∴ 𝑆 = 0.35 + 2 × 0.85 = 2.05
∴ 𝑍 = 4 − 2.05 = 1.95 (given)
Be 1𝑠 2𝑠 2 𝑠 electron is screened by two electrons in 1𝑠-orbital ((𝑛 − 1)th orbital)
𝑆 = 2 × 0.85 = 1.70
∴ 𝑍 = 4 − 1.70 = 2.30
Be 1𝑠 one electron in 1s-orbital (alone exists) is screened by another electron in same
orbit. Thus,
𝑆 = 0.30
∴ 𝑍 = 4 − 0.30 = 3.70
Be 1𝑠 no-screening (single electron)
Thus, 𝑍 = 4
Thus, 𝑍 : Be < Be < Be < Be
67 (c)
(IP) of Na (11)<(IP) of Li(3)
69 (c)
𝐴 → 𝐴 + 𝑒 (IP) < (IP)
𝐵 + 𝑒 → 𝐵 (EA) > (EA)
(EN) > (EN)
70 (b)
𝑀 : [Ar]3𝑑 4𝑠
𝑀: [Ar]3𝑑 4𝑠 4𝑝
Three electrons have been removed from 4𝑝-suborbit
Thus, 𝑝-block element
72 (d)
O
S
Se
F
Cl
Br
I
Thus,
F < O
F < O
Se > 𝑂
Se > O
I > 𝐵𝑟 > 𝐹
I > Br > F
I > Se
∴ I > Se
Thus, I > Se > Br > O > F

73 (b)
A-19 𝑠-block
B-22 𝑑-block
C-32 𝑝-block
D-64 𝑓-block
79 (d)
Chalcogens are the elements of oxygen family:
Valence shell configuration : 𝑛𝑠 𝑛𝑝
Group: 16
Block: 𝑝
81 (a)
𝑟 =
𝑛 𝑎
𝑍
Thus, 𝑟 ∝
82 (c)
NaH is an ionic hydride, others are covalent hydrides
83 (c)
Ni: [Ar]3𝑑 4𝑠
Last electron enters into 3𝑑-suborbit
Thus, 𝑑-block element
84 (c)
𝑋: 𝑛𝑠 𝑛𝑝
Valency = +3 (as in Al)
It can loss three electrons to attain stable configuration
𝑌: 𝑛𝑠 𝑛𝑝
Valency = −3 (as in N)
It can gain five electrons to attain stable configuration
Thus, 𝑋 𝑌 or 𝑋𝑌
86 (a)
I is oxidized by Br , Cl , F
Br is oxidized by Cl , F
Cl is oxidized by F
89 (d)
ElementValence electrons Possible ion
𝐴 3 𝐴
𝐵 6 𝐵
𝐵 + 2𝑒 → 𝐵
8 electrons in valence shell (stable)
𝐴 → 𝐴 + 3𝑒
Stable
Thus, 𝐴 𝐵 𝐴 𝐵
90 (c)
𝑟 ∝
1
𝑍

Ti (22) > Ni (28)
smaller
Pt (5𝑑-series)
Zr (4𝑑-series)
Have higher number of orbits
Hence, larger size
92 (a)
𝐴 → 𝐵 + α( He)
93 (b)
113 − 32 − 32 − 18 = 31
Thus, 113 and 115 are placed as shown
IIIA VA
Ga 31 33 As
In 49 51 Sb
Tl 81 81 Bi
113 115
Thus, these belong to 𝑝-block
95 (b)
Na , Mg andAl are isoelectronic
Size Na > Mg > Al
Charge +1 +2 +3
Smaller the size of cation,
Larger the charge,
Greater the hydration
Thus, Na < Mg < Al
96 (d)
Thus, IE of N> 0

(IE) due to smaller size of Li
He → He + 𝑒 (IE)
He → He + 𝑒 (IE)
(IE) > (IE)
Thus, (IE) of He > 𝐻𝑒
Thus, (d) is incorrect
97 (d)
5. Na, K – Metals
6. F, Cl – Non-metals
7. Cu, Ag – Metals
8. B, Si – Metalloids
98 (a)
Na O is basic in nature and other oxides are acidic/basic/amphoteric
Thus, greater the acidic nature, greater the tendency of the reaction to occur
Thus, Cl O
100 (a)
𝑟 =
𝑛 𝑎
𝑍
Cs (alkali metal) has largest atomic radius

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