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 INDEX

S. No. Chapter Name Page No.

1. Classification of Elements and Periodicity in Properties

Theory and Exercise 1.1 to 1.5 1 – 29

Exercise 2 30 – 34

Exercise 3 35 – 41

Exercise 4 (Previous Year’s Questions) 42 – 45

Answer Key 46 – 47

2. Chemical Bonding

Theory and Exercise 1.1 to 1.7 49 – 86

Exercise 2 87 – 92

Exercise 3 93 – 96

Exercise 4 (Previous Year’s Questions) 97 – 104

Answer Key 105 – 106

Classification of Elements and Periodicity in Properties CHEMISTRY

Chapter CLASSIFICATION
OF ELEMENTS
1 AND PERIODICITY
IN PROPERTIES

Introduction
Chapter Overview

• Introduction
• Periodicity
 Prediction of Period,
Group and Block of a
Given Element
• Development of Modern
Periodic Table
• Long Form of Periodic
Table
• Screening Effect (σ) and
Effective Nuclear Charge
(Zeff)
• Atomic Radius
• Ionisation Energy
• Electron Gain Enthalpy
• Electron-Negativity (EN)
• Metallic Property
• Periodic Trends and
Chemical Reactivity
• Some Important Tables &
Graph of Periodic
Properties from NCERT
• Summary

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Classification of Elements and Periodicity in Properties CHEMISTRY

Points to Ponder

Periodicity

Along a Period
Along a Group
Max. Z
Constant Z* (effective due to screening)
Max.
Cation smaller than atom, anion larger than atom
rn (atomic, ionic radius)
Fe, Co, Ni have size, also Zr, Hf

Be > B, N > O, inert gas max, I1 < I2 < I3 …...…

(I.P.)
Inert gas zero
(Electronegativity)
Inert gas zero, Cl > F, S > O, P > N
(Electron-affinity)

(Hydration and Hydration Energy)

(for cations)
(Size of Hydrated ion)

(Polarising Power of cation) Na+ < Mg2+ < Al3+

N3– < O2– < F–

(Polarising Power of anion)
K+ < Na+ < Li+

(Acidic nature of oxides)

(Basic nature of oxides)

(Reactivity)

(Reducing Nature)

• Isoelectronic ions have different size.

• Reducing nature of hydride increases in a group and decreases in a period.

Facts to be remembered

1. Seven Periods:
st
1 Period : 2 elements : Shortest period
nd rd
2 & 3 Period : 8 elements : Short period
th th
4 & 5 period : 18 elements : Long period
th
6 period : 32 elements : Longest period
th
7 period : 32 elements : complete period (longest)
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CHEMISTRY Classification of Elements and Periodicity in Properties
2. 18 groups:
st
Alkali Metals : IA / 1 group
nd
Alkaline earth metals : IIA / 2 group
th
Coinage Metals : IB / 11 group
th
Pnicogens : VA / 15 group
th
Chalcogens : VIA / 16 group
th
Halogens : VIIA / 17 group
th
Noble gases or inert gases or zero group : 18 group
3. Configurations:
1-2
s–Block : ns
2 1–6
p–Block : ns np
1–10 0-2
d-Block : (n-1)d ns
1–14 0-1 2
f–Block : (n–2)f (n–1)d ns
4. Nonmetal liquid at room temperature = Bromine
5. Metal liquid at room temperature = Mercury
° º º
6. Metals with very low Melting point (a) Ga = 29.8 C; (b) Cs = 28.5 C; (c) Fr = 27.0 C
7. Tungsten is (a) Most tensile
(b) Highest boiling metal
(c) Highest melting metal
8. Oxygen is the most abundant element & Al is the most abundant metal
9. Carbon is the highest boiling non-metal. (4827ºC)
10. Osmium and Iridium are the High-density elements & Lithium is the lightest metal: d = 0.54gm/cc
11. Silver is the best conductor of electricity. Copper is the second-best conductor of electricity
12. Diamond is the hardest substance known.
13. Osmium & Ruthenium shows maximum Oxidation state: +8
14. Tin (Sn) has maximum number of isotopes: 10 isotopes
15. Fluorine is the most electronegative element & Cesium is the most electropositive element.
16. Diagonal relationship is between:
Li Be B C

Na Mg Al Si
17. Elements of 3rd period are known as Typical elements.

Prediction of Period, Group and Block of a Given Element

The period, group and block of an element can be easily predicted from its electronic configuration as follows:
• The period of an element corresponds to the principal quantum number of the valence shell.
• The block of an element corresponds to the type of orbital which receives the last electron.
• The group of an elements is predicted from the number of electrons in the valence shell or/and
penultimate shell (last but one, i.e. n–1) as follows:
(a) For s-block elements, group number is equal to the number of valence electrons.
(b) For p-block elements, group number is equal to 10 + number of electrons in the valence shell.
(c) For d-block elements, group number is equal to the number of electrons in (n–1) d-subshell +
number of electrons in valence shell (nth shell).

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Classification of Elements and Periodicity in Properties CHEMISTRY
Development of Modern Periodic Table
(a) Dobereiner’s Triads:
He arranged similar elements in the groups of three elements called as triads, in which the atomic mass
of the central element was merely the arithmetic mean of atomic masses of other two elements.
Li Na K
7 + 39
7 23 39 = 23
2
It was restricted to few elements, therefore, discarded

(b) Newland’s Law of Octave:

He was the first to correlate the chemical properties of the elements with their atomic masses.
According to him if the elements are arranged in the order of their increasing atomic masses the eighth
element starting from given one is similar in properties to the first one.
This arrangement of elements is called as Newland’s law of Octave.
Li Be B C N O F
Na Mg Al Si P S Cl
K Ca
This classification worked quite well for the lighter elements but it failed in case of heavier elements
and, therefore, discarded

(c) Lothar Meyer’s Classification:

He determined the atomic volumes by dividing atomic mass with its density in solid states. He plotted
a graph between atomic masses against their respective atomic volumes for a number of elements. He
found the following observations.
80
75 Cs
70
65
60
Rb
Atomic volume cm3

55
50 K
45
Ba
40
35
Ca IV Sr V VI
30
25 Na I
Br
Period Cl
20 Sb La Nd
II III S Se Zr ln Te
15 As Sn
I Li Mg P Fe Zn Nb Ru Cd
10 Si
Al V Mo Rh
5 Be Co Cu
C
0
Atomic mass, amu
Lothar Meyer’s Curve

(i) Elements with similar properties occupied similar positions on the curve.
(ii) Alkali metals having larger atomic volumes occupied the crests.
(iii) Transitions elements occupied the throughs.
(iv) The halogens occupied the ascending portions of the curve before the inert gases.
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CHEMISTRY Classification of Elements and Periodicity in Properties
(v) Alkaline earth metals occupied the positions at about the mid points of the descending portions
of the curve.
On the basis of these observations, he concluded that the atomic volumes (a physical property) of the
elements are a periodic function of their atomic masses. It was discarded as it lacks practical utility.
(d) Mendeleev’s Periodic Table:
Mendeleev’s Periodic’s Law:
According to him the physical and chemical properties of the elements are a periodic function of their
atomic masses.
He arranged known elements in order of their increasing atomic masses considering the facts that
elements with similar properties should fall in the same vertical columns and leaving out blank spaces
where necessary.
Predication of new elements: It gave encouragement to new elements as some gaps were left in it.
Sc (Scandium) Ga (Gallium) Ge (Germanium) Tc (Technetium)
These were the elements for whom position and properties were well defined by Mendeleev even before
their discoveries and he left the blank spaces for them in his table.
The table is divided into vertical columns called groups and horizontal rows called periods.
The table is divided into nine vertical columns called groups and seven horizontal rows called periods.
Periods No. of Elements Called as
(1)st n = 1 2 Very short period
(2)nd n = 2 8 Short period
(3)rd n = 3 8 Short period
(4)th n = 4 18 Long period
(5 )thn = 5 18 Long period
(6)th n = 6 32 Very long period
(7 )thn = 7 19 Incomplete period
The groups were numbered as I, II, III, IV, V, VI, VII, VIII and Zero group

Merits of Mendeleev’s Periodic table:

(i) It has simplified and systematised the study of elements and their compounds
(ii) It has helped in predicting the discovery of new elements on the basis of the blank spaces given in
its periodic table
(iii) Atomic weights of elements were corrected. Atomic weight of Be was calculated to be 3 × 4.5 =
13.5 by considering its valency 3, was correctly calculated considering its valency 2 (2 × 4.5 = 9)
Demerits in Mendeleev’s Periodic Table:
(i) Position of hydrogen is uncertain. It has been placed in 1A and VIIA groups because of its
resemblance with both the groups.
(ii) No separate positions were given to isotopes.
(iii) It is not clear whether the lanthanides and actinides are related to IIIA or IIIB group.
(iv) Order of increasing atomic weights is not strictly followed in the arrangement of elements in the
periodic table. For e.g.–Ar (At.wt.39.94) is placed before K (39.08) and Te (127.6) is placed before I
(126.9)
(v) Similar elements were placed in different groups Au (IB) and Pt (VIII) and the elements with
different properties were placed in same groups alkali metals (IA) and coinage metals (IB)
(vi) It didn’t explain the cause of periodicity.
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Long Form of Periodic Table
Modern Periodic Law (Moseley’s Periodic Law):
Physical and chemical properties of elements are the periodic functions of their atomic number. If the
elements are arranged in order of their increasing atomic number, after a regular interval, element
with similar properties are repeated.
Cause of Periodicity:
Repetition of the properties of the elements is due to the recurrence of similar valence shell electronic
configuration after certain regular intervals. For example, alkali metals have same electronic
1
configuration ns , therefore, have similar properties.
The long form of periodic table is the contribution of Range, Werner, Bohr and Bury
This table is also referred to as Bohr‘s table since it follows Bohr’s scheme of the arrangements of
elements into four types based on electronic configuration of elements
The periodic table is divided into 18 groups and 7 periods.
(i) Each period consists of a series of elements having same valence shell.
(ii) Each period corresponds to a particular principal quantum number of the valence shell present in it.
1
(iii) Each period starts with an alkali metal having outermost electronic configuration ns .
2 6
(iv) Each period ends with a noble gas with outermost electronic configuration ns np except Helium
2
having outermost electronic configuration 1s .
2 6
(v) The number of elements in a period is equal to the number of necessary electrons to acquire ns np
configuration in the outermost shell of first element (alkali metal) of the periodic. First period
contains two elements.
(vi) Each period starts with the filling of new energy level.

p–Block Elements
S–Block Elements

1 18
IA VIII A

H
1
2 Transition Metals (d –Block Elements) 13 14 15 16 17
He
2
II A III A IV A VA VI A VII A
1.007 4.002
3 4 5 6 7 8 9 10
Li Be B C N O F Ne
6.941 9.012 10.811 12.011 14.006 15.999 18.998 20.179
11 12 13 14 15 16 17 18
3 4 5 6 7 8 9 10 11 12
Na Mg Al Si P S Cl Ar
III B IV B VB VI B VII B VIII VIII VIII IB II B
22.98 24.30 26.981 28.085 30.973 32.006 35.452 39.948
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
39.08 40.078 44.959 47.88 50.9415 51.996 54.938 55.84 55.933 58.693 63.546 65.39 69.723 72.61 74.921 78.96 79.904 83.80
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
85.46 87.62 88.905 91.224 92.906 95.94 98 101.07 102.905 106.42 107.868 112.411 114.82 118.710 121.757 127.60 126.904 132.29
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
132.90 137.27 138.905 178.49 180.947 183.85 186.207 190.2 192.22 195.08 196.666 200.59 204.383 207.2 207.980 209 210 222
87 88 89 104 105 106 107 108 109 110
114
Fr Ra Ac** Rf Ha Sg Bh Hs Mt Uun
Uuq
223 226 227 261.11 262.114 263.118 262.12 265 266 269

Inner - Transition Metals (f-Block elements)

58 59 60 61 62 63 64 65 66 67 68 69 70 71
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
*Lanthanides 140.115 140.907 144.24 145 150.36 151.965 157.25 158.925 162.50 164.930 167.26 168.934 173.04 174.967
90 91 92 93 94 95 96 97 98 99 100 101 102 103
**Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
232.038 231 238.028 237 244 243 247 247 251 252 257 258 259 260

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Points to Ponder
NCERT questions (solve yourself)
Q. 1 What would be the IUPAC name and symbol for the element with atomic number 120?
Q. 2 How would you justify the presence of 18 elements in the 5th period of the Periodic Table?
Q. 3 The elements Z = 117 and 120 have not yet been discovered. In which family / group would
you place these elements and also give the electronic configuration in each case.
Q. 4 Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory (ii) Seaborg’s group?
Q. 5 Consider the following species:
+ 2+ 3+
N3–, O2–, F–, Na , Mg and Al
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Q. 6 Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Q. 7 Assign the position of the element having outer electronic configuration
2 4 2 2
(i) ns np for n=3 (ii) (n-1)d ns for n=4, and
7 1 2
(iii) (n-2) f (n-1)d ns for n=6, in the periodic table.
Q. 8 Considering the atomic number and position in the periodic table, arrange the following
elements in the increasing order of metallic character: Si, Be, Mg, Na, P.

Exercise 1.1
1. The fourteen elements collectively placed 4. The places that were left empty by
rd th
in 3 group and 7 period are called:- Mendeleef were, for:-
(1) Typical elements (1) Aluminium & Silicon
(2) Representative element (2) Galium and germinium
(3) Actinides (3) Arsenic and antimony
(4) Lanthanoids (4) Molybdenum and tungstun

2. Which of the following is/are Doeberiners 5. Elements which occupied position in the
triad : lother meyer curve, on the peaks, were :–
(a) P, As, Sb (b) Cu, Ag, Au (1) Alkali metals
(c) Fe, Co, Ni (d) S, Se, Te (2) Highly electro positive elements
Correct answer is :- (3) Elements having large atomic volume
(1) a and b (2) b and c (4) All
(3) a and d (4) All
6. Which of the following statement is wrong :
3. Which of the following sets of elements th
(1) Og gas is present in 7 period
follows Newland's octave rule :- rd
(2) 3 period contains 18 elements
(1) Be, Mg, Ca st
(2) Na, K, Rb (3) 1 period contains two non-metals
(3) F, Cl, Br (4) In p–block, metal, nonmetal and
(4) B, Al, Ga metalloids are present

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Classification of Elements and Periodicity in Properties CHEMISTRY
7. In the Doberiener's triad all three element 13. The element with atomic number Z = 115
have same:- will be placed in :-
th
(1) Electronic configuration (1) 7 period, IA group
th
(2) Properties (2) 8 period, IVA group
th
(3) Number of shells (3) 7 period, VA group
th
(4) (1) & (2) both (4) 6 period, VB group

14. In 6th period of the modern periodic table,

8. Which statement is wrong for the long form
electronic energy levels is in the order :-
of periodic table :- (1) 6s, 4f, 5d, 6p (2) 6s, 6p, 4f, 5d
(1) Number of periods are 7 and groups 18 (3) 4f, 5d, 6s, 6p (4) None
(2) No. of valence shell electrons in a
period are same 15. Out of first 100 elements no. of elements
rd having electrons in 3d orbitals (in their
(3) III B group contains 32 elements
complete electronic configuration) are :-
(4) Lanthanides and actinides are placed in (1) 80 (2) 100
same group (3) 40 (4) 60

9. Which pair of successive elements follows 16. The atom having the valence shell electronic
configuration 4s2 4p2 would be in:-
increasing order of atomic weight in
(1) Group II A and period 3
mendeleev's periodic table
(2) Group II B and period 4
(1) Argon and potassium (3) Group IV A and period 4
(2) Lithium and Berrilium (4) Group IV A and period 3
(3) Cobalt and nickel
(4) Tellurium and iodine 17. The electronic configuration of transition
elements is exhibited by :-
0-2 1-10 2 10
10. Choose the s-block element from the (1) ns (n-1)d (2) ns (n - 1) d
10 2 2 5
following: (3) (n - 1)d s (4) ns np
2 2 6 2 6 5 1
(1) 1s , 2s , 2p , 3s , 3p , 3d , 4s 18. Which of the following electronic
2 2 6 2
(2) 1s , 2s , 2p , 3s , 3p , 3d , 4s
6 10 1
configurations in the outermost shell is
2 2 6 2 6 1 characteristic of alkali metals
(3) 1s , 2s , 2p , 3s , 3p , 4s 2 6 2 1 2 6 10 1
(4) all of the above (1) (n–1) s p ns p (2) (n–1) s p d ns
2 6 1 2 6 10
(3) (n–1) s p ns (4) ns np (n–1)d
11. If there were 10 periods in the periodic 19. An element which is recently discovered is
table then how many elements would this th th
placed in 7 period and 10 group. IUPAC
period can maximum comprise of. name of the element will be :-
(1) 50 (2) 72 (1) Unnilseptium (2) Ununnilium
(3) 32 (4) 98 (3) Ununbium (4) None

20. Which of the following statement is wrong

12. The electronic configuration of an element
2 2 6 2 4
for the transition elements :-
is 1s 2s 2p 3s 3p . The atomic number rd
(1) Transition elements are placed from 3
and the group number of the element ‘X’ th
to 6 period.
which is just below the above element in (2) Last electron enters in ( n –1)d orbital
the periodic table are respectively. (3) Exhibits variable valency
(1) 24 & 6 (2) 24 & 15 (4) General electronic configuration is
(3) 34 & 16 (4) 34 & 8 1–10
(n-1)d ns
1-2

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CHEMISTRY Classification of Elements and Periodicity in Properties
21. In the general electronic configuration - 26. The electronic configuration of an element
1–14 0–1 2 2 2 6 2 6 10 1
(n–2)f (n–1)d ns , if value of n = 7 the is 1s 2s 2p 3s 3p 3d 4s .
configuration will be of - What is the atomic number of next element
(1) Lanthenides of the same group which is recently
(2) Actinides discovered :-
(3) Transition elements (1) 20 (2) 119
(4) None (3) 111 (4) None

22. Element with the electronic configuration 27. From atomic number 58 to 71, elements are
given below, belong to which group in the
placed in :-
periodic table th
2 2 6 2 6 10 2 6 10 2 3 (1) 5 period and III A group
1s , 2s 2p , 3s 3p 3d , 4s 4p 4d , 5s 5p th
rd th (2) 6 period and III B group
(1) 3 (2) 5
th th (3) separate period and group
(3) 15 (4) 17 th
(4) 7 period and IV B group
3 2
23. 4d 5s configuration belongs to which
28. Element X belongs to 4th period. It contains
group :-
18 and 1 electron in the penultimate and
(1) IIA (2) IIB
ultimate orbit. The X should be :
(3) V B (4) III B
(1) normal element
24. Which of the following general electronic (2) transition element
configuration for transition elements is not (3) inert gas
correct (4) inner - transition element
1–2 1–10
(1) (n + 1) s nd
(2) ns
1–2
(n – 1)d
1 – 10
(Where n = 2, 3, 4 .......) 29. Element with atomic number 56 belong to
(3) ns
0,1,2
(n –1)s p d
2 6 1–10 which block?
1–10 0–2 (1) s (2) p (3) d (4) f
(4) (n – 1)d ns

30. Outer electronic configuration of K, Cu, and

25. Which of the following electronic
Cr are respectively
configuration belongs to inert gas elements:- 1 10 5 2 10 4
2 10 2 2 6 (1) 4s , 3d , 3d (2) 4s , 3d , 3d
(1) ns (n – 1)d (2) ns (n – 1)s p
1 9 4 1 9 4
(3) ns np
2 6
(4) None (3) 4s , 3d , 3d (4) 4s , 3d , 3d

Screening Effect (σ) and Effective Nuclear Charge (Zeff)

–
(i) Valence shell e suffer force of attraction due to nucleus and force of repulsion due to inner shell
electrons.
– –
(ii) The decrease in force of attraction on valence e due to inner shell e is called screening effect or
shielding effect. (i.e. total repulsive force is called shielding effect.)
–
(iii) Due to screening effect valence shell e experiences less force of attraction exerted by nucleus.
i.e. total attraction force experienced by valence electrons represented by a number is Zeff.
(iv) There is a reduction in nuclear charge due to screening effect. Reduced nuclear charge is called
effective nuclear charge.
(v) If nuclear charge = Z, effective nuclear charge = Zeff, σ (Sigma) = Screening constant or shielding
constant. So, Z=
eff (Z – σ)
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Classification of Elements and Periodicity in Properties CHEMISTRY
For same shell shielding effect has the order as s > p > d > f (die to penetration effect) Zeff for different
ions of an element
positive charge (i) Zeff for different ions of an element
Zeff ∝
negative charge (ii) Zeff for isoelectronic species.
(i) Zeff for different ions of an element
Example:
+ –
N > N > N = Zeff
(ii) Zeff of isoelectronic species
Example:
– + +2 +3 –
H < Li < Be < B (2e species)
–3 –2 – + +2 –
N < O < F < Na < Mg (10e species)

Atomic Radius
Probability of finding the electron is never zero even at large distance from the nucleus. Based on
probability concept, an atom does not have well defined boundary. Hence exact value of the atomic
radius can’t be evaluated. Atomic radius is taken as the effective size which is the distance of the closet
approach of one atom to another atom in a given bonding state. Atomic radius can be
(a) Covalent radius: X X
It is one-half of the distance between the centres of two nuclei (of like
atoms) bonded by a single covalent bond.
A B

1
AB = rcov alent
2
(of element X)

(b) Vander Waals radius (Collision radius): X X X X

It is one-half of the internuclear distance
between two adjacent atoms in two nearest A B
neighboring molecules of the substance in solid
state.

1
AB = rvander Wasls
2
(of element X)
(c) Metallic radius (Crystal radius):
It is one-half of the distance between the nuclei of two adjacent X X
metal atoms in the metallic crystal lattice.
C D
• rcovalent < rcrystal < rvander Walls

1
CD = rcrystal
2
(of element X)

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CHEMISTRY Classification of Elements and Periodicity in Properties
Ionic Radius:
The effective distance from the centre of nucleus of the ion up to which it has an influence in the
ionic bond is called ionic radius.
• The d and f orbitals do not shield the nuclear charge very effectively. Therefore, there is significant
reduction in the size of the ions. Just after d or f orbitals have been filled completely. This is called
a lanthanide contraction. Atomic radii of Zr and Hf are almost identical due to lanthanide
contraction.
n2a0
♦ Atomic radius / Ionic radius in the nth orbit is rn =
Z*
º
(a0 = Bohr’s radius of H–atom = 0.529 A )

• The species containing the same number of electrons but differ in the magnitude of their nuclear
charges are called as isoelectronic species.
1
Pauling’s empirical formula: Ionic radius ∝
effective nuclear charge (Z*)
Points to Ponder
Some Important Increasing Order
Atomic/Ionic size Ionic Radii in water
2+ + – 2– 3– + + + + +
(1) Mg , Na , F , O , N (1) CS , Rb , K , Na , Li
(Hint : Iso electronic series) +
(2) Li , Be
2+

(2) B, Be, Li, Na + +2

(3) Na , Mg , Al
+3
– 2–
(3) F, O, F , O
Hydration of ions/Hydration Energy Molar Conductivity in Water
+2 +2 +2 +2 +2 + + + + +
(1) Ba , Sr , Ca , Mg , Be Li , Na , K , Rb , Cs

NCERT questions (solve yourself)

Q. 1 2+
Which of the following species will have the largest and the smallest size? Mg, Mg , Al, Al .
3+

Q. 2 Consider the following species:

3– 2– – + 2+ 3+
N , O , F , Na , Mg and Al
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Q. 3 Explain why cation are smaller and anions larger in radii than their parent atoms?

Exercise 1.2
1. Atomic radii of fluorine and neon in 3.
+
Na is smaller than Na atom because:
angstroms units are respectively: (1) nucleus in each case contains different
(1) 1.60 and 1.60 (2) 0.72 and 1.60 nucleons
(3) 0.72 and 0.72 (4) none of these (2) sodium atom has an electron lesser
than sodium ion
(3) sodium atom has 11 electrons and
2. The difference between ions and atoms is of:
sodium ion has 10 electrons
(1) relative size (2) configuration +
(4) the force of attractions is less in Na
(3) presence of charge (4) all of the above
than in Na atom
12 Sarvam Career Institute
Classification of Elements and Periodicity in Properties CHEMISTRY
4. The ionic radii of a cation w. r. t. parent 12. The screening effect of d- electrons is: -
atom is always: (1) Equal to the p - electrons
(1) less than atomic radii (2) Much more than p - electrons
(2) more than atomic radii (3) Same as f - electrons
(3) equal to atomic radii (4) Less than p - electrons
(4) cannot be predicted
13. If the difference in atomic size of :
Na – Li = x, Rb – K = y, Fr – Cs = z
5. Which of the following has largest size?
+
Then correct order will be:-
(1) Na (2) Na (1) x = y = z (2) x > y > z
2+
(3) Mg (4) Mg (3) x < y < z (4) x < y << z

6. An element M has an atomic number 9 and 14. The correct order of size would be:-
atomic mass 19. Its ion will be represented (1) Ni < Pd ≈ Pt (2) Pd < Pt < Ni
by (3) Pt > Ni > Pd (4) Pd > Pt > Ni
2+
(1) M (2) M
– 2–
15. Which of the following order of radii is
(3) M (4) M correct
+ + –
(1) Li < Be < Mg (2) H < Li < H
7. Which one of the following is smallest in + – –2
(3) O < F < Ne (4) Na > F > O
size ?
+ 2–
(1) Na (2) O 16. Which of the following is not different for
(3) N
3–
(4) F
– an atom and its corresponding ion :
(1) Number of electrons
8. Which of the following is the smallest (2) Nuclear charge
cation? (3) Ionization energy
+ 2+ (4) Size
(1) Na (2) Mg
2+ 3+
(3) Ca (4) Al 17. Which group of atoms have nearly same
atomic radius:
9. In the isoelectronic species, the ionic radii (1) Na, K, Rb, Cs (2) Li, Be, B, C
3– 2– –
(A) of N , O and F are respectively given (3) Fe, Co, Ni (4) F, Cl, Br, I
by
(1) 1.36, 1.40, 1.71 (2) 1.36, 1.71, 1.40 18. Which of the following has largest radius :
2 2 6 2
(3) 1.71, 1.40, 1.36 (4) 1.71, 1.36, 1.40 (1) 1s , 2s , 2p , 3s
2 2 6 2 1
(2) 1s , 2s , 2p , 3s , 3p
2 2 6 2 3
10. Chloride ion and potassium ion are (3) 1s , 2s , 2p , 3s , 3p
isoelectronic. Then: 2 2 6 2
(4) 1s , 2s , 2p , 3s , 3p
5

(1) their sizes are same

–
(2) Cl ion is bigger than K ions
+ 19. Which has the lowest anion to cation size
+ ratio-
(3) K ion is relatively bigger
(1) LiF (2) NaF
(4) their sizes depend on other cation and (3) CsI (4) CsF
anion
20. Arrange the elements in increasing order of
11. In sodium atom the screening is due to: - atomic radius Na, Rb, K, Mg :–
2 6 1
(1) 3s , 3p (2) 2s (1) Na, K, Mg, Rb (2) K, Na, Mg, Rb
2
(3) 1s , 2s , 2p
2 6
(4) 1s , 2s
2 2 (3) Mg, Na, K, Rb (4) Rb, K, Mg, Na

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CHEMISTRY Classification of Elements and Periodicity in Properties
21. Which of the following order of atomic/ionic 27. In an anion :-
radius is not correct :– (1) Number of proton decreases
– + +2 + –
(1) I > I > I (2) Mg > Na > F (2) Protons are more than electrons
+5 +3
(3) P < P (4) Li > Be > B (3) Effective nuclear charge is more
(4) radius is larger than neutral atom
22. In which of the following compound,
distance between two nuclei is maximum:–
28. Maximum size of first member of a period is
(1) CsF (2) KI (3) CsI (4) LiI
due to
23. In the lithium atom screening effect of (1) Maximum number of shells
valence shell electron is caused by- (2) Maximum screening effect
(1) Electrons of K and L shell (3) Minimum Zeff
(2) Electrons of K shell
st nd (4) All
(3) Two electrons of 1 and one of 2 shell
(4) None
29. Which of the following ion has largest size:-
– +3
24. The radius of potassium atom is 0.203 nm. (1) F (2) Al
The radius of the potassium ion in (3) Cs
+
(4) O
–2

nanometer will be :–
(1) 0.133 (2) 0.231 30. In which of the following pair radii of
(3) 0.234 (4) 0.251
second species is smaller than that of first
25.
–2
S is not isoelectronic with :- species :-
– – +3 + –
(1) Ar (2) Cl (3) HS (4) Ti (1) Li, Na (2) Na , F
–3 +3 +7 +4
(3) N , Al (4) Mn , Mn
26. The best reason to account for the general
tendency of atomic diameters to decrease
as the atomic numbers increase within a 31. Which of the following orders of ionic radii
period of the periodic table is the fact that are correct: -
(1) Outer electrons repel inner electrons (a) Li < Be < Na (b) Ni < Cu < Zn
(2) Closer packing among the nuclear (c) Ti > V > Cr (d) Ti > Zr > Hf
particles is achieved
Correct answer is: -
(3) The number of neutrons increases
(4) The increasing nuclear charge exerts a (1) All (2) a, b
greater attractive force on the electrons (3) b, c (4) b, d

Ionisation Energy
Ionisation energy (IE), sometimes also called ionisation potential (IP), of an element is defined as the
amount of energy required to remove an electron from an isolated gaseous atom of that element
resulting in the formation of positive ion.
+ –
M(g) →
(IE)
M (g) + e

(IE)1, (IE)2, (IE)3...are respectively first, second, third ...ionisation energies required to remove first,
second , third.....electron from the isolated gaseous atoms.
(IE)1 < (IE)2 < (IE)3 < ..............

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Classification of Elements and Periodicity in Properties CHEMISTRY
Factors Affecting Ionisation Energy:
(IE) variation in a period and group may or not be regular and can be influenced by:

(a) Size of the Atom:

In a small atom, the electrons are tightly held while in a larger atom, the electrons are less strongly
held, the Coulombic force of attraction F being
1
I.E. ∝
size
(b) Effective Nuclear Charge (Z*):
As given above the force of attraction between the nucleus and the outermost electron increases with
increase in nuclear charge. Greater the effective nuclear charge, greater energy required to pull the
electrons from the atom. Here (IE) increases with increase in nuclear charge.
I.E. ∝ Zeffective
(c) Shielding effect (s):
The electrons in the inner shells act as a screen or shield between the nucleus and the electrons in the
outermost shell. This is called shielding effect. The larger the number of electrons in the inner shells,
greater is the screening effect and smaller the force of attraction and thus (IE) decreases.
1
I.E. ∝
σ
(d) Penetration Effect:
An s electron penetrates nearer to the nucleus, and is therefore more tightly held than a ‘d‘electron,
and a ‘d’ electron is more tightly held than f electron. Other factors being equal, ionisation energies are
in the order s > p > d > f.
I.E. ∝ penetration effect
(e) Electronic Configuration:
If an atom has full filled or half-filled orbitals, its (IE) is higher than expected normally from its position
in the periodic table.

Z 3 4 5 6 7 8 9 10
Z* 1.3 1.95 2.6 3.25 3.9 4.55 5.2 5.85
n
rn
F
(IE)
Li Be B C N O F Ne
3 1.3 Na Z*–atomic number
11 2.2 Z*–effective nuclear charge due to screening
K n–orbits , rn – radius
19 2.2
Rb F–force of attraction between electron and nucleus
37 2.2
Cs (IE)–ionization energy
55 2.2 Increases in the given direction
87 2.2 Fr
constant
Variation of (IE) in a group period

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CHEMISTRY Classification of Elements and Periodicity in Properties

Points to Ponder

Some Important Increasing Order

Ionization Energy
st
(1) Li, B, Be, C, O, N, F, Ne, He (I I.P.)

NCERT questions (solve yourself)

Q. 1 The first ionization enthalpy (∆iH) values of the third period elements, Na, Mg and Si are
respectively 496, 737 and 786 kJ mol–1. Predict whether the first ∆iH value for Al will be
–1
closer to 575 or 760 kJ mol ? Justify your answer.
Q. 2 –18
Energy of an electron in the ground state of the hydrogen atom is –2.18×10 J. Calculate
–1
the ionization enthalpy of atomic hydrogen in terms of J mol .
Hint: Apply the idea of mole concept to derive the answer.

Q. 3 How would you explain the fact that the first ionization enthalpy of sodium is lower than
that of magnesium but its second ionization enthalpy is higher than that of magnesium?

Exercise 1.3
1.
st
Correct orders of I I.P. are :- 5. In which of the following pairs, the
(a) Li < B < Be < C ionisation energy of the first species is less
(b) O < N < F than that of the second :-
- 2-
(c) Be < N < Ne (1) O , O (2) S, P
+
(1) a, b (2) b, c (3) N, P (4) Be , Be
(3) a, c (4) a, b, c
6. Least ionisation potential will be of: -
3+
2. The second ionisation potentials in electron (1) Be (2) H
+2 +
volts of oxygen and fluorine atoms are (3) Li (4) He
respectively given by :-
7. Ionisation energy increases in the order: -
(1) 35.1, 38.3 (2) 38.3, 38.3 (1) Be, B, C, N (2) B, Be, C, N
(3) 38.3, 35.1 (4) 35.1, 35.1 (3) C, N, Be, B (4) N, C, Be, B
3. A sudden large jump between the values of 2nd 8. Mg forms Mg(II) because of :-
and 3rd IP of an element would be associated (1) The oxidation state of Mg is + 2
with the electronic configuration :- (2) Difference between I.P1 and I.P2 is
2 2 6
(1) 1s , 2s 2p , 3s
1 greater than 16.0 eV
2 2 6 2 5 (3) There are only two electrons in the
(2) 1s , 2s 2p , 3s 3p
2 2 6 2 2
outermost energy level of Mg
(3) 1s , 2s 2p , 3s 3p (4) Difference between I.P1 and I.P2 is less
2 2 6 2
(4) 1s , 2s 2p 3s than 11 eV

4. Compared to the first ionisation potential, 9. IP1 and IP2 of Mg are 178 and 348 K. cal mol–1.
the value of second ionisation potential of The enthalpy required for the reaction
2+ –
an element is :- Mg → Mg + 2e is :-
(1) Negligible (2) Smaller (1) + 170 K.cal (2) + 526 K.cal
(3) Greater (4) Double (3) - 170 K.cal (4) - 526 K.cal

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Classification of Elements and Periodicity in Properties CHEMISTRY
10. IP is influenced by - 17. The ionisation energy of B and Al as
(1) Size of atom compared to Be and Mg are :–
(2) Effective nuclear charge (1) Lower (2) Higher
(3) Electrons present in inner shell (3) Equal (4) None of these
(4) All
nd
18. II IP of which of the element is maximum–
11. Highest ionisation potential in a period is (1) Lithium (2) Oxygen
shown by (3) Nitrogen (4) Fluorine
(1) Alkali metals
(2) Noble gases 19. Which has the lowest IE :-
(3) Halogens (1) 3d
2
(2) 4s
1

(4) Representative elements 3 6

(3) 3p (4) 2p
12. Which of the following decreases in going
20. The energy needed to remove one electron
down the halogen group: -
(1) Ionic radius from unipositive ion is abbreviated as :-
st rd
(2) Atomic radius (1) I I.P. (2) 3 I.P.
nd st
(3) Ionisation potential (3) 2 I.P. (4) 1 E.A.
(4) Boiling point nd st
21. Which of the following has 2 IP < I IP
13. Minimum first ionisation energy is shown (1) Mg (2) Ne
by which electronic configuration: - (3) C (4) None
2 2 5
(1) 1s , 2s , 2p
2 2 6 2 2 22. Among the following elements (Whose
(2)1s , 2s , 2p , 3s , 3p
2 2 6 1 electronic configuration is given below) the
(3) 1s , 2s , 2p , 3s
2 2 6 one having the highest ionisation energy is
(4) 1s , 2s , 2p 2 3 2 4
(1) (Ne) 3s 3p (2) (Ne) 3s 3p
2 5 10 2 2
14. The IP1, IP2, IP3, IP4 and IP5 of an element are (3) (Ne) 3s 3p (4) (Ar) 3d 4s 4p
7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. + +2 –2 –3
The element is likely to be: - 23. Out of Na , Mg , O and N , the pair of
(1) Na (2) Si species showing minimum and maximum IP
(3) F (4) Ca would be.
+ +2 +2 –3
(1) Na , Mg (2) Mg , N
15. With reference to ionisation potential which –3
(3) N , Mg
+2 –2
(4) O , N
–3

one of the following sets is correct: -

(1) Li > K > B (2) B > Li > K 24. If the graph is plotted between atomic
(3) Cs > Li > K (4) Cs < Li < K numbers and ionisation potential. Which
group of elements occupy the lowest
16. Successive ionisation energies of an
position on the curve:–
element 'X' are given below (in K. Cal)
(1) Alkaline earth metal
IP1 IP2 IP3 IP4
(2) Inert gas
165 195 556 595
(3) Actinides
Electronic configuration of the element 'X'
(4) Alkali metals
is:-
2 2 6 2 2
(1) 1s , 2s 2p , 3s 3p 25. The element having highest I.P. in the two
2 1
(2) 1s , 2s series C, N, O and Si, P, S :–
2 2
(3) 1s , 2s 2p
2
(1) P (2) N
2 2 6
(4) 1s , 2s 2p , 3s
2 (3) S (4) O

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CHEMISTRY Classification of Elements and Periodicity in Properties
26. Out of I, Fe, H and Cl the pair of species 34. In which of the following the energy change
showing minimum and maximum IP would be: corresponds to first ionisation potential :-
(1) H, Fe (2) I, Cl +
(1) X(g) → X (g) + e
– +
(2) X2(g) → X (g) + e
–

(3) Fe, H (4) I, H +

(3) X(s) → X (g) + e
– +
(4) X(aq) → X (aq)+e
–

27. Factor which does not affects the ionisation 35. Which of the following electronic
potential configuration belongs to least and most
(1) Atomic size metallic character respectively:-
(2) Bond order 2
(a) 1s 2s
1 2
(b) 5s 5p
5

(3) Effective nuclear charge 2 6

(c) 3s 3p 4s
1 2 2
(d) 1s 2s 2p
5

(4) Shielding effect (1) a, b (2) d, c

(3) b, a (4) c, d
28. Lowest IP will be shown by the element
having the configuration: – 36. In the given process which oxidation state
(1) [He] 2s
2
(2) 1s
2
is more stable.
+
2
(3) [He] 2s 2p
2 2
(4) [He] 2s 2p
5
M(g) → M(g) IE1 = 7.9 eV
+ +2
M(g) → M(g) IE2 = 15.5 eV
29. The strongest reducing agent among the (1) M
+
(2) M
+2
(3) Both (4) None
following is :–
3– –2 +
(1) Na (2) Mg 37. Triad - I [N , O , Na ]
(3) Al (4) K + + +
Triad - II [ N , C , O ]
Choose the species of lowest IP from triad–
30. Which ionisation potential (IP) in the I and highest IP from triad–II respectively
following equations involves the greatest 3– + + +
(1) N , O (2) Na , C
ammount of energy: - 3– + – +
+ +2 – + +2 – (3) N , N (4) O , C
(1) K → K + e (2) Li → Li + e
+
(3) Fe → Fe + e
– + +2
(4) Ca → Ca + e
–
38. The correct values of ionization energies
–1
(in kJ mol ) of Be, Ne, He and N
31. Values of first four ionisation potential of respectively are
an elements are 68, 370, 400, 485. It (1) 786, 1012, 999, 1256
belongs to which of the following electronic (2) 1012, 786, 999, 1256
configuration:- (3) 786, 1012, 1256, 999
2 1 2 2 1
(1) 1s 2s (2) 1s 2s 2p (4) 786, 999, 1012, 1256
2 2 6 1
(3) 1s 2s 2p 3s (4) (1) and (3) both 39. Following graph shows variation of I.P. with
atomic number in second period (Li – Ne).
32. In which case the maximum energy is
Value of I.P. of Na (11) will be :-
needed in the formation of monopositive
Ne
gaseous ion :-
(1) 1 mole of Li atoms N F
C
(2) 1 mole of Na atoms
(IP)

Be O
(3) 1 mole of Cs atoms
(4) 1 mole of Be atoms B
Li
– +
3 4 5 6 7 8 9 10 11
33. (a) M (g) → M(g) (b) M(g) → M (g) Z
+ +2
(c) M (g) → M (g)
+2 +3
(d) M (g) → M (g) (1) Above Ne
Minimum and maximum I.P. would be of :- (2) Below Ne but above O
(1) a, d (2) b, c (3) Below Li
(3) c, d (4) d, a (4) Between N and O

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Classification of Elements and Periodicity in Properties CHEMISTRY
40.
+ –
M(g) → M (g) + e , ∆H = 100 eV, 44. The electronic configuration of some
2+ –
M(g) → M (g) + 2e , ∆H = 250 eV neutral atoms are given below: –
2 1 2 2 3
which is incorrect statements: - (A) 1s 2s (B) 1s 2s 2p
(1) IE1 of M(g) is 100 eV 2
(C) 1s 2s 2p
2 5 2 2
(D) 1s 2s 2p 3s
6 1

(2) IE2 of M(g) is 150 eV

In which of these electronic configurations
(3) IE2 of M(g) is 250 eV
would you expect to have highest (i) IE1
(4) none
(ii) IE2 respectively:
41. In the plot of the first ionization energy (1) C, A (2) B, A
against atomic number the peaks are (3) C, B (4) B, D
occupied by: -
(1) Inert gases 45. Which one of the following statements is
(2) Alkali metals incorrect in relation to ionization enthalpy?
(3) Halogens (1) Ionization enthalpy increases for each
(4) Transition elements successive electron.
(2) The greatest increase in ionization
42. Which one of the following has highest
ionization potential: – enthalpy is experienced on removal of
(1) Li
+
(2) Mg
+
(3) He (4) Ne electron from core noble gas
configuration.
43. In which of the following pairs, the (3) End of valence electrons is marked by a
ionization energy of the first species is less big jump in ionization enthalpy.
than that of the second (4) Removal of electron from orbitals
+
(1) N, P (2) Be , Be bearing lower n value is easier than
– +
(3) N, N (4) Ne, Ne from orbital having higher n value.
Electron Gain Enthalpy (∆egH)
• When an electron is added to a neutral gaseous atom (X) to convert it into a negative ion, the
enthalpy change accompanying the process is defined as the Electron Gain Enthalpy (∆egH).
• The negative of the enthalpy change for the process s defined as the Electron Affinity (EA) of the
atom under consideration. If energy is released when an electron is added to an atom, the electron
affinity is taken as positive, contrary to thermodynamic convention. If energy has to be supplied to
add an electron to an atom, then the electron affinity of the atom is assigned a negative sign.
Electron affinity, (EA) is the amount of energy released when an additional electron is added to the
outermost shell of an isolated gaseous atom to form anion.
→ X (−g ) + E.A. ( i )
X (g ) + e− 
+Energy EA ( ii)
X (−g ) + e−  → X 2(g−)
• EA (i) is exothermic whereas EA (ii) is endothermic. In the process of adding further electron, the
second electron is added to gaseous anion against the electrostatic repulsion between the
electron being added and the gaseous anion.
• The electron affinity may also be called as the energy required to remove electron from the
gaseous negative ions.
• The greater the energy released in the process of taking up the extra electron, the greater is the
electron affinity resulting in the stability of anion.
• (EA) values of metals are low while those of non–metals are high.
• The electron affinity of an atom measures the tightness with it binds an additional electron to
itself.
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CHEMISTRY Classification of Elements and Periodicity in Properties
Exceptions:
(a) Electron affinity values of second period elements (N, O, F) are smaller than the electron affinity
values of third period elements (P, S, Cl). The electron being added experiences comparatively
more repulsion because of high value of charge densities of second period elements due to much
smaller size.
(b) Electron affinity of alkaline earth metals (group-2) are almost zero. It is because of Completely
filled s–orbitals in their valence shells.
(c) Nitrogen has low electron affinity values. The atom of these elements have stable exactly half-
filled p–orbitals.
(d) Electron affinity of noble gases is zero. It may be attributed to the stable completely filled valence
shell electronic configurations.

Points to Ponder
Some Important Increasing Order
Electron Affinity:
(1) I, Br, F, Cl (2) O, Po, Te, Se, S

NCERT questions (solve yourself)

Q. 1 Which of the following will have the most negative electron gain enthalpy and which the
least negative?
P, S, Cl, F. Explain your answer.
Q. 2 Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F (ii) F or Cl
Q. 3 Would you expect the second electron gain enthalpy of O as positive, more negative or less
negative than the first? Justify your answer.

Exercise 1.4
1. In which case the energy released is 4. Which of the following statements is wrong
minimum:- for fluorine :-
(1) Cl → Cl
–
(2) B → B
–
(1) It's standard reduction potential is
(3) N → N
–
(4) C → C
– highest
(2) It is most electronegative element
2. In the formation of a chloride ion, from an (3) Bond energy of F2 < Cl2
isolated gaseous chlorine atom, 3.8 eV (4) Fluorine has highest electron affinity
energy is released, which would be equal to:-
(1) Electron affinity of Cl
– 5. Electron addition would be easier in :-
+ – +2
(2) Ionisation potential of Cl (1) O (2) O (3) O (4) O
(3) Electronegativity of Cl
– + I II
(4) Ionisation potential of Cl 6. Process Na  → Na(g)  → Na(s)
(1) In (I) energy released, (II) energy
3. The correct order of electron affinity is :- absorbed
(1) Be < B < C < N (2) In both (I) and (II) energy is absorbed
(2) Be < N < B < C (3) In both (I) and (II) energy is released
(3) N < Be < C < B (4) In (I) energy absorbed, (II) energy
(4) N < C < B < Be released
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Classification of Elements and Periodicity in Properties CHEMISTRY

7. In the process Cl(g) + e– 

→ Cl–(g), ∆H is ∆H 14. The electron affinity
(1) Positive (1) Of carbon is greater than oxygen
(2) Negative (2) Of fluorine is less than iodine
(3) Zero (3) Of F is less than Cl
(4) None (4) Of S is less than oxygen

8. Process in which maximum energy is

15. The process requiring the absorption of
released:-
–2 + +2
energy is.
(1) O → O (2) Mg → Mg –
– –
(1) F → F
(3) Cl → Cl (4) F → F –
(2) Cl → Cl
– 2– 2–
9. O(g) + 2e → O (g); ∆Heg = 744.7 KJ/mole. (3) O → O
The positive value of ∆Heg is due to :- (4) H → H
–

(1) Energy is released to add on 1 e– to O–1

(2) Energy is required to add on 1 e– to O–1 16. Which of the following configuration will
–
(3) Energy is needed to add on 1e to O have least electron affinity.
(4) None of the above is correct 2 5 2 2
(1) ns np (2) ns np
2 3 2 4
10. Which of the following is energy releasing (3) ns np (4) ns np
process
— –
(1) X → X (g) + e 17. Energy absorbed in second electron
— – 2–
(2) O (g) + e → O addition in an atom is called.
+ –
(3) O (g) → O (g) + e (1) 1 IP
st nd
(2) 2 EA
– –
(4) O (g) + e → O (g) (3) 1 EA
st nd
(4) 2 IP

11. In which of the following process energy is 18. The amount of energy released for the
liberated:- process
+
(1) Cl → Cl + e
X(g) + e → X– (g) is minimum and maximum
–
+ –
(2) HCl → H + Cl
–
respectively for :–
(3) Cl + e → Cl
– -2
(a) F (b) Cl
(4) O + e → O (c) N (d) B
12. Element of which atomic number has Correct answer is :–
highest electron affinity:- (1) c & a (2) d & b
(1) 35 (2) 17 (3) a & b (4) c & b
(3) 9 (4) 53
19. Which of the following electronic
13. Second electron affinity of an element is :– configuration is expected to have highest
(1) Always exothermic electron affinity:-
(2) Endothermic for few elements 2 0 2 2
(1) 2s 2p (2) 2s 2p
(3) Exothermic for few elements 2 3 2 1
(4) Always endothermic (3) 2s 2p (4) 2s 2p

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CHEMISTRY Classification of Elements and Periodicity in Properties
Electro-Negativity (EN)
Electronegativity is a measure of the tendency of an element to attract electrons to itself in a covalently
bonded molecule.
The magnitude of electronegativity of an element depends upon its ionisation potential & electron
affinity. Higher ionisation potential & electron affinity values indicates higher electronegativity value.
There is no direct method to measure the value of electronegativity, however, there are some scales to
measure its value.
The electronegativity of any given element is not constant. It varies depending on the element to which
it is bound.
(a) Pauling’s Scale:
This scale determines electronegativity of the element X w.r.t hydrogen H
| χ X=
− χH | 0.208 ∆H− x
–1
Where (X) represents electronegativity, ∆H–x is the extra bond energy in kcal mol
and ∆=
H− x EH− X − EH− x × EX − X
EH – x, EH – H and DX–x represents bond energy of H–X, H–H and X–X bonds respectively
(b) Mulliken’s scale:
Electronegativity (EN) can be regarded as the average of the ionisation energy (IE) and the
electron affinity (EA) of an atom.
IE + EA
(EA )
m
=
2
(EN) m
( )
= 2.8 EN
p

If both (EA) and (IE) are determined in eV units then paulings’s electronegativity (EN)P is related to
Mulliken’s electronegativity Mulliken’s values were about 2.8 times larger than the Pauling’s values.
(c) Allred–Rochow’s Electronegativity
Allred and Rochow defined electronegativity as the force exerted by the nucleus of an
atom on its valence electrons:
0.359Zeffective
(EN)AR = + 0.744
r2
where Zeffective is the effective nuclear charge and r the covalent radius (in).

Points to Ponder

SOME IMPORTANT INCREASING ORDER

Electro Negativity
(1) I, Br, Cl, F (2) C, N, O, F

NCERT questions (solve yourself)

Q. 1 What is the basic difference between the terms electron gain enthalpy and electronegativity?

Q. 2 How would you react to the statement that the electronegativity of N on Pauling scale is 3.0
in all the nitrogen compounds?

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Classification of Elements and Periodicity in Properties CHEMISTRY
Exercise 1.5
1. Which of the following is amphoteric in 9. Correct order of electronegativity of N, P, C
nature? and Si is :
(1) SiO2 (2) SO2 (1) N < P < C < Si (2) N > C > Si > P
(3) Al2O3 (4) Fe2O3 (3) N = P > C = Si (4) N > C > P > Si
2. Which of the following can be used to
separate Al2O3 and Fe2O3 ? 10. Polarity of a bond can be explained by :-
(1) HCl (2) NaOH (1) Electron affinity
(3) Both (4) None (2) Ionisation potential
(3) Electronegativity
3. Most basic oxide is : (4) All of the above
(1) Sb2O3 (2) Bi2O3
(3) B2O3 (4) Cr2O3
11. Least electronegative element is :-
4. Incorrect match is : (1) Ι (2) Br (3) C (4) Fr
Compound Type/Nature
(1) Mn2O7 Acidic 12. 1 2 3 4
(2) NO2 Neutral H3C - CH = C = CH2
(3) Mn3O4 Mixed oxide In the given compound which carbon atom
(4) SnO2 Amphoteric will show maximum electronegativity -
5. Soluble in excess of NaOH : (1) Fourth
(1) Fe2O3 (2) MgO (2) First
(3) CuO (4) BeO (3) Third
(4) EN of all the carbon atoms is same
6. Which one is incorrect order of acidic
strength ? 13. The nomenclature of ICl is iodine chloride
(1) H3PO2 < H3PO3 < H3PO4
because
(2) HClO2 < HClO3 < HClO4
(1) Size of I < Size of Cl
(3) SiO2 < P2O5 < SO3 < Cl2O7
(4) Na2O < ZnO < CO2 (2) Atomic number of I > Atomic number of Cl
(3) E.N. of I < E.N. of Cl
7. Nature of hydroxide A-O-H would be : (4) E. A. of I < E. A. of Cl
(a) Generally acidic if A is more EN element
(b) Generally basic if A is less EN element 14. Among the following least and most polar
(1) Only (a) is correct bonds are respectively :-
(2) Only (b) is correct (a) C - I (b) N - O
(3) Both (a) & (b) are correct
(c) C - F (d) P - F
(4) Both (a) & (b) are incorrect
(1) d and c (2) a and d
8. Which of the following is affected by stable (3) b and d (4) b and c
configuration of an atom :
(a) Electronegativity 15. If the ionisation potential is IP, electron
(b) Ionisation potential affinity is EA and electronegativity is x then
(c) Electron affinity which of the following relation is correct :-
Correct answer is :-
(1) 2X - EA - IP = 0
(1) Only electronegativity
(2) 2EA - X - IP = 0
(2) Only ionisation potential
(3) 2IP - X - EA = 0
(3) Electron affinity and ionisation potential
(4) All of the above (4) All of the above

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CHEMISTRY Classification of Elements and Periodicity in Properties
16. The properties which are not common to 17. The X - X bond length is 1.00 Å and C - C
both groups 1 and 17 elements in the bond length is 1.54 Å. If electronegativities
periodic table are :
of 'X' and 'C' are 3.0 and 2.0 respectively,
(1) Electropositive character increases
down the groups the C - X bond length is likely to be :
(2) Reactivity decreases from top to bottom (1) 1.27 Å
in these groups (2) 1.18 Å
(3) Atomic radii increase as the atomic
(3) 1.08 Å
number increases
(4) Electronegativity decreases on moving (4) 1.28 Å
down a group

Metallic Property
Metals have the tendency to form cations by loss of electrons and this property makes the elements as
electropositive elements or metals.
M(g) → M(g)
+
+ e−

Effective Nuclear Shielding Tending to Reducing Metallic

Charge make + ve stength Property

But the shielding

increases

increases

increases

increases
Constant

effect over weighs so⇒

Increase
Effective Nuclear charge
Shielding Increase

But the nuclear charge over weighs ⇒

decreases L to R
Tendency to form +ve ions
decreases L to R
Reducing strength
decreases L to R
Metallic property

Periodic Trends and Chemical Reactivity

We have observed the periodic trends in certain fundamental properties such as atomic and ionic radii,
ionization enthalpy, electron gain enthalpy. We know by now that the periodicity is related to electronic
configuration. That is, all chemical and physical properties are a manifestation of the electronic
configuration of elements. We shall now try to explore relationships between these fundamental
properties of elements with their chemical reactivity.
The atomic and ionic radii, as we know, generally decrease in a period from left to right. As a
consequence, the ionization enthalpies generally increase and electron gain enthalpies become more
negative across a period. In other words, the ionization enthalpy of the extreme left element in a period
is the least and the electron gain enthalpy of the element on the extreme right is the highest negative.
This results into high chemical reactivity at the two extremes and the lowest in the centre. Thus, the
maximum chemical reactivity at the extreme left (among alkali metals) is exhibited by the loss of an
electron leading to the formation of a cation and at the extreme right (among halogens) shown by the
gain of an electron forming an anion. This property can be related with the reducing and oxidizing

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Classification of Elements and Periodicity in Properties CHEMISTRY
behavior of the elements. It can be directly related to the metallic and non-metallic character of
elements. Thus, the metallic character of an element, which is highest at the extremely left decreases
and the non-metallic character increases while moving from left to right across the period. The
chemical reactivity of an element can be best shown by its reactions with oxygen and halogens. We
shall consider the reaction of the elements with oxygen only. Elements on two extremes of a period
easily combine with oxygen to form oxides. The normal oxide formed by the element on extreme left is
the most basic (e.g., Na2O), whereas that formed by the element on extreme right is the most acidic
(e.g., Cl2O7). Oxides of elements in the centre are amphoteric (e.g., Al2O3, As2O3) or neutral (e.g., CO,
NO, N2O). Amphoteric oxides behave as acidic with bases and as basic with acids, whereas neutral
oxides have no acidic or basic properties.

SOME IMPORTANT INCREASING ORDER Basic Nature of Oxides/Hydroxide

Acidic Nature of Oxides/Hydroxide (1) LiOH, NaOH, KOH, RbOH, CsOH
(1) Li2O, BeO, B2O3, CO2, N2O5 (2) Be(OH)2, Mg(OH)2, Ca(OH)2, Sr(OH)2, Ba(OH)2
(2) BaO, SrO, CaO, MgO, BeO (3) Al2O3, Ga2O3, In2O3, Tl2O
(3) V2O3, V2O4, V2O5
(4) CrO, Cr2O3, CrO3
(5) AS2O3, P2O3, N2O3

NCERT questions (solve yourself)

Q. 1 Using the Periodic Table, predict the formulas of compounds which might be formed by the
following pairs of elements; (a) silicon and bromine (b) aluminium and sulphur.
Q. 2 2+
Are the oxidation state and covalency of Al in [AlCl(H2O)5] same?
Q. 3 Show by a chemical reaction with water that Na2O is a basic oxide and Cl2O7 is an acidic oxide.
Q. 4 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs whereas
that among group 17 elements is F > CI > Br > I. Explain.
Q. 5 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is,
Q. 6 Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is,
Q. 7 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in
terms of oxidizing property is,

Some Important Tables & Graphs of Periodic Properties from NCERT

• Notation for IUPAC Nomenclature of Elements
Digit Name Abbreviation
0 nil n
1 un u
2 bi b
3 tri t
4 quad q
5 pent p
6 hex h
7 sept s
8 oct o
9 enn e

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CHEMISTRY Classification of Elements and Periodicity in Properties
• Nomenclature of Elements with Atomic Number Above 100
Atomic Number Name according to Symbol IUPAC Official Name IUPAC Symbol
IUPAC nomenclature
101 Unnilunium Unu Mendelevium Md
102 Unnilbium Unb Nobelium No
103 Unniltrium Unt Lawrencium Lr
104 Unnilquadium Unq Rutherfordium Rf
105 Unnilpentium Unp Dubnium Db
106 Unnilhexium Unh Seaborgium Sg
107 Unnilseptium Uns Bohrium Bh
108 Unniloctium Uno Hassium Hs
109 Unnilennium Une Meitnerium Mt
110 Ununnillium Uun Darmstadtium Ds
111 Unununnium Uuu Rontgenium Rg
112 Ununbium Uub Copernicium Cn
113 Ununtrium Uut Nihonium Nh
114 Ununquadium Uuq Flerovium Fl
115 Ununpentium Uup Livermorium Mc
116 Ununhexium Uuh Tennessine Lv
117 Ununseptium Uus Tennessine Ts
118 Ununoctium Uuo Oganesson Og

Table : Electronegativity Values (on Pauling scale) Across the Periods

Atom (Period II) Li Be B C N O F
Atomic radius 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Atom (Period III) Na Mg Al Si P S Cl
Atomic radius 0.9 1.2 1.5 1.8 2.1 2.5 3.0

Table : Electronegativity Values (on Pauling scale) Down a Family

Atom Group (I) Electronegativity value Atom (Group 17) Electronegativity Value
Li 1.0 F 4.0
Na 0.9 Cl 3.0
K 0.8 Br 2.8
Rb 0.8 I 2.5
Cs 0.7 At 2.2

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Classification of Elements and Periodicity in Properties CHEMISTRY

160
Li

140

Atomic radius/pm
120
Be

100
B
80 C N
O F
4 6 8 10 60 2
Atomic number (Z)
Variation of atomic radius with atomic number across the sec. period

300

250 Cs(262)
Rb(244)
K(231)
200
Atomic radius/pm

Na(186)
150
Li(152)
I(133)
100 Br(114)
Cl(99)
50
F(72)

Atomic number (Z)

Variation of atomic radius with atomic number for alkali metals and halogens

2500 Hc Na

2000

Ar
∆, H/Kj mol–1

1500 Kr
Xe
1000

500 Li Na
Rb Cs K
0
0 10 20 30 40 50 60
Atomic number(Z)
Variation of first ionization enthalpies (∆iH) with atomic number for elements with Z = 1 to 60

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CHEMISTRY Classification of Elements and Periodicity in Properties

2500

Ne
2000 (2080)

∆, H/kJ mol–1
F
1500 N (1681)
(1402)
Bc O
C (1314)
1000 (899)
Li (1086)
B
(520) (801)
500
0 2 4 6 8 10
Atomic number(Z)
First ionization enthalpies (∆iH) of elements of the second period as a function of atomic number (Z)

550
Li(520)
500 Na(496)
∆, H/kJ mol–1

450

K Rb(403)
400
(419)
Cs(374)
350
0 10
20 30 40 50 60
60 Atomic number(Z)
∆iH of alkali metals as a function of Z.

Electron Gain Enthalpy

Ionization Enthalpy
Atomic Radius

Electron Gain Enthalpy

Ionization Enthalpy
Electronegativity

Atomic Radius

Electronegativity

The periodic trends of elements in the periodic table

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Classification of Elements and Periodicity in Properties CHEMISTRY
SUMMARY
1. Introduction to Periodic Classification:
Elements are arranged based on increasing atomic number. Periodic Law: "The physical and
chemical properties of elements are periodic functions of their atomic numbers."

2. History of Periodic Table:

Dobereiner’s Triads (1829): Grouped elements in sets of three based on similar properties.
Newlands’ Law of Octaves (1866): Every eighth element had similar properties.
Mendeleev’s Periodic Table (1869): Arranged elements by atomic mass; left gaps for undiscovered
elements.
Modern Periodic Table (by Moseley, 1913): Based on increasing atomic number instead of atomic
mass.

3. Modern Periodic Table (Long Form):

Structure:
18 vertical columns (Groups)
7 horizontal rows (Periods)
Blocks in the Periodic Table:
s-Block: Group 1 (Alkali metals), Group 2 (Alkaline earth metals)
p-Block: Groups 13-18 (Includes metalloids, non-metals, and noble gases)
d-Block: Transition elements (Groups 3-12)
f-Block: Lanthanides and Actinides (Inner transition elements)

4. Periodic Trends:
Atomic Radius:
Decreases across a period (left to right) due to increased nuclear charge.
Increases down a group due to the addition of new electron shells.
Ionization Enthalpy:
Increases across a period (due to increased nuclear attraction).
Decreases down a group (due to increased atomic size).
Electron Gain Enthalpy (Electronegativity):
Becomes more negative across a period (non-metals attract electrons more).
Becomes less negative down a group (size increases, nuclear attraction decreases).
Metallic and Non-Metallic Character:
Metals (left side of the periodic table) lose electrons easily.
Non-metals (right side) gain electrons easily.
Metallic character decreases across a period and increases down a group.
Electronegativity:
Increases across a period (atoms attract electrons more strongly).
Decreases down a group (atomic size increases, reducing electron attraction).

5. Special Periodic Properties:

Lanthanide Contraction: Steady decrease in atomic size of lanthanides due to poor shielding of f-
electrons.
Diagonal Relationship: Similar properties between elements of different groups but adjacent
periods (e.g., Li & Mg, Be & Al).
Inert Pair Effect: Heavier p-block elements show lower oxidation states due to reluctance of s-
electrons to participate in bonding.
This chapter is crucial for understanding the systematic classification of elements and predicting
their chemical behavior based on periodic trends.
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CHEMISTRY Classification of Elements and Periodicity in Properties

Exercise 2
1. The element with the highest ionisation 7. The incorrect statement among the
potential is following is
(1) oxygen (2) nitrogen (1) the first ionisation potential of Al is less
than the first ionisation potential of Mg
(3) carbon (4) boron
(2) the second ionisation potential of Mg is
greater than the second ionisation
2. The first ionisation potentials in electron potential of Na
volts of nitrogen and oxygen atoms are (3) the first ionisation potential of Na is less
respectively given by : than the first ionisation potential of Mg
(1) 14.6, 13.6 (4) the third ionisation potential of Mg is
greater than third ionisation potential of Al
(2) 13.6, 14.6
(3) 13.6, 13.6 8. Identify the least stable ion amongst the
(4) 14.6, 14.6 following
– – – –
(1) Li (2) Be (3) B (4) C
3. The maximum ionisation potential in a
period is shown by 9. Which of the following elements has the
maximum electron affinity ?
(1) alkali metals
(1) oxygen (2) chlorine
(2) inert gases
(3) fluorine (4) nitrogen
(3) representative elements
(4) halogens 10. Electron affinity of X would be equal to
–
(1) electron affinity of X
–
4. Ionisation energy of nitrogen is more than (2) ionisation potential of X
oxygen because: (3) ionisation potential of X
(1) nucleus has more attraction for electron (4) none of the above
(2) half filled 'p' orbitals are more stable
11. Which of the following species has the
(3) nitrogen atom is small highest electron affinity?
(4) more penetration effect (1) F
–
(2) O (3) O
–
(4) Na
+

5. Amongst the following elements (whose 12. Increasing order of electron affinity is
electronic configuration are given below) (1) N < O < Cl < Al (2) O < N < Al < Cl
the one having highest ionisation energy is (3) N < Al < O < Cl (4) Cl < N < O < Al
2 1 2 3
(1) [Ne] 3s 3p (2) [Ne] 3s 3p 13. The electronegativity of the following
2 2 10 2 3
(3) [Ne] 3s 3p (4) 3d , 4s 4p elements increases in the order of :
(1) C, N, Si, P (2) N, Si, C, P
6. A sudden large jump between the values of (3) Si, P, C, N (4) P, Si, N, C
second and third ionisation energies of an
14. In the series carbon, nitrogen, oxygen and
element would be associated with the fluorine, electronegativity
electronic configuration (1) decreases from carbon to fluorine
2 2 6 1
(1) 1s , 2s 2p , 3s (2) remains constant
(2) 1s2, 2s2 2p6, 3s2 3p1 (3) decreases from carbon to oxygen and
then increases
(3) 1s2, 2s2 2p6, 3s2 3p2
2 2 6 2
(4) generally, increases from carbon to
(4) 1s , 2s 2p , 3s fluorine
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Classification of Elements and Periodicity in Properties CHEMISTRY
15. Which of the following elements, which one 23. Lanthanoids are: -
has the highest electronegativity? (1) 14 elements in the seventh period
(1) Ι (2) Br (3) Cl (4) F (Atomic no.= 58 to 71) that are filling 4f
sub level
16. Which one of the following configurations (2) 14 elements in the sixth period (atomic
represents a metallic character ? no.= 90 to 103) that are filling 4f
(1) 2, 8, 2 (2) 2, 8, 4 sublevel
(3) 2, 8, 7 (4) 2, 8, 8 (3) 14 elements in the seventh period
(atomic no.= 90 to 103) that are filling
17. The most non-metallic element among the
5f sublevel
following is
2 2 6 2 2 5
(4) 14 elements in the sixth period (atomic
(1) 1s , 2s 2p (2) 1s , 2s 2p no=58 to 71) that are filling 4 f sublevel
2 2 4 2 2 3
(3) 1s , 2s 2p (4) 1s , 2s 2p
24. For electron affinity of halogens which of
18. Atoms of which of the following group lose the following is correct:-
electrons most easily ?
(1) Br > F (2) F > Cl
(1) Li, Na, K (2) Cl, Br, Ι
(3) Br > Cl (4) F > I
(3) O, S, Se (4) N, P, As
25. The liquidified metal expanding on solidification is
19. Which of the following statement is
(1) Ga (2) Al
incorrect for an atom having electronic
configuration 2, 8, 7 : (3) Zn (4) Cu
(1) It forms diatomic molecules
(2) It is a non-metal element 26. Four successive members of the first-row
(3) Its valency is 1 transition elements are listed below with
(4) It forms basic oxide their atomic numbers. Which one of them is
expected to have the highest third
20. Electronegativity is the measurement of ionization enthalpy:-
capacity of an atom by which: (1) Vanadium (Z = 23)
(1) Electrons get repelled (2) Manganese (Z = 25)
(2) Electrons get attracted (3) Chromium (Z = 24)
(3) Gain of electron (4) Iron (Z = 26)
(4) Lose of proton
27. The pair of amphoteric hydroxide is
21. A transition element X has the configuration (1) Al(OH)3, LiOH (2) Be(OH)2, Mg(OH)2
[Ar] 3d4 in its +3-oxidation state. Its atomic (3) B(OH)3, Be(OH)2 (4) Be(OH)2, Zn(OH)2
number is
(1) 25 (2) 26 (3) 22 (4) 19
28. Which of the following is the most basic
oxide?
22. Ionic radii are :-
(1) SeO2 (2) Al2O3
(1) Directly proportional to square of
(3) Sb2O3 (4) Bi2O3
effective nuclear charge
(2) Inversely proportional to effective
nuclear charge 29. The correct order regarding the
(3) Inversely proportional to square of electronegativity of hybrid orbitals of
effective nuclear charge carbon is?
2 3 2 3
(4) Directly proportional to effective (1) sp < sp < sp (2) sp > sp < sp
nuclear charge 2
(3) sp > sp > sp
3 2
(4) sp < sp >sp
3

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CHEMISTRY Classification of Elements and Periodicity in Properties
30. Which one of the following orders is not in 37. Which one of the following ions has the
accordance with the property stated highest value of ionic radius :
against it? (1) O
2–
(2) B
3+

(1) F2 > Cl2 > Br2 > I2 : bond dissociation (3) Li

+
(4) F
–

energy
(2) F2 > Cl2 > Br2 > I2 : oxidising power
38. Which among the following factors is the
(3) HI > HBr > HCI > HF : acidic property in
most important in making fluorine the
water
strongest oxidizing halogen?
(4) F2 > Cl2 > Br2 > I2 : electronegativity
(1) Hydration enthalpy
(2) Ionization enthalpy
31. Which of the following electronic
(3) Electron affinity
configuration an atom has the lowest
(4) Bond dissociation energy
ionisation enthalpy?
2 2 3 2 2 6 1
(1) 1s 2s 2p (2) 1s 2s 2p 3s
2 2 6 2 2 5
39. Lanthanide contraction relates to decrease is:
(3) 1s 2s 2p (4) 1s 2s 2p (1) Atomic radii
+3
(2) Atomic as well as M radii
32. The correct order of decreasing second
(3) Valence electron
ionisation enthalpy of Ti(22), V(23), Cr(24)
(4) Oxidation state
and Mn(25) is:
(1) Mn > Cr > Ti > V (2) Ti > V > Cr > Mn
40. Mg & Li are similar in their properties due to:
(3) Cr > Mn > V > Ti (4) V > Mn > Cr > Ti
(1) Same e/m ratio
(2) Same electron affinity
33. Among the following which is the strongest
(3) Same group
oxidising agent?
(1) Cl2 (2) F2 (3) Br2 (4) I2 (4) Same ionic potential

34. Which of the following oxides is not 41. Which of these have no unit :
expected to react with sodium hydroxide? (1) Electronegativity
(1) BeO (2) B2O3 (2) Electron affinity
(3) CaO (4) SiO2 (3) Ionisation energy
(4) Excitation potential
35. Amongst the elements with following
electronic configurations, which one of 42. In the following which has maximum
them may have the highest ionization ionisation potential :-
energy? (1) H (2) He (3) Li (4) Be
2 1 2 3
(1) Ne[3s 3p ] (2) Ne[3s 3p ]
2 2 10 2 3 43. Which of the following conjugate base has
(3) Ne[3s 3p ] (4) Ar[3d 4s 4p ]
maximum charge density :-
–2 –2 –2 –2
36. Atomic radii of alkali metals (M) follow the (1) O (2) S (3) Te (4) Se
order Li < Na < K < Rb but ionic radii in
aqueous solution follow the reverse order 44. Element having maximum EN is :
Li+ > Na+ > K + > Rb+. The reason of the reverse (1) Li (2) Be (3) C (4) O
order is :
(1) Increase in the ionisation energy 45. Which of the following has electron affinity
(2) Decrease in the metallic bond character less than zero ( ∆Heg = +ve) :
(3) Increase in the electropositive character (1) O
–2
(2) S
–2

(4) Decrease in the amount of hydration (3) (1) & (2) both (4) O
+

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Classification of Elements and Periodicity in Properties CHEMISTRY

46.
2 2
The compound has 3d 4s configuration is 54. Following statements regarding the
related to which block: - periodic trends of chemical reactivity of the
(1) s - block (2) P-block alkali metals and the halogens are given.
(3) d-block (4) f-block Which of these statements gives the correct
picture?
47. Total elements in fifth period are: (1) Chemical reactivity increases with
(1) 8 (2) 10 (3) 18 (4) 32 increase in atomic number down the
group in both the alkali metals and
48. Ionization energy of nitrogen is more than halogens
oxygen due to: - (2) In alkali metals the reactivity increases
(1) Increased attraction of electron but, in the halogens, it decreases with
towards nucleus increase in atomic number down the
(2) Extra stability of half filled 'p' orbital group
(3) Small size of N2 (3) The reactivity decreases in the alkali
(4) None of the above metals but increases in the halogens
with increase in atomic number down
49. Element having highest I.P. is: the group
(1) H (2) Li (3) B (4) Na (4) In both the alkali metals and the
halogens, the chemical reactivity
50. Which of the following oxides is amphoteric decreases with increases in atomic
in character? number down the group
(1) CaO (2) CO2
(3) SiO2 (4) SnO2 55. Which one of the following is isoelectronic:
+3 +3 +3 +2
(1) Fe , Co (2) Fe , Mn
51. In which of the following arrangements the +3 +3 +3 +3
order is NOT according to the property (3) Co , Sr (4) Sc , Ti
indicated against it?
(1) Al3+ < Mg2+ < Na+ < F– - increasing ionic size 56. Which of the following aqueous acid is most
(2) B < C < N < O - increasing first ionization acidic
energy (1) HCl (2) HF (3) HI (4) HBr
(3) I < Br < F < Cl - increasing electron gain
ethalpy (with negative sign) 57. Which of the following valence electron
(4) Li < Na < K < Rb - increasing metallic configuration belongs to halogen family
radius 2 2
(1) s p
2 3
(2) s p
2 4
(3) s p
2 5
(4) s p

52. Which one of the following sets of ions 58. The correct order of first ionization
represents a collection of isoelectronic enthalpy of B, C, O & N is:
species? (1) B < C < N < O (2) B < C < O < N
3– 2– – 2–
(1) N , O , F , S (3) O < N < B < C (4) N < O < C < B
+ + +2 +2
(2) Li , Na , Mg , Ca
+ – +2
(3) K , Cl , Ca , Sc
+3
59. Which of the following is isoelectronic to :
+2 +2 +2 +2 –
(4) Ba , Sr , K , Ca CN
+ –
(1) O2 (2) S (3) CO (4) F2
53. The increasing order of the first ionization
enthalpies of the elements B, P, S and F 60. The correct order of atomic radius of
(lowest first) is following elements :-
(1) B < P < S < F (2) B < S < P < F (1) N > P > Si (2) N < P < Si
(3) F < S < P < B (4) P < S < B < F (3) N < P > Si (4) Si > N > P

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CHEMISTRY Classification of Elements and Periodicity in Properties

61. In alkaline metals, correct order of 1

st
67. Which one of the following orders presents
ionisation energy is :- the correct sequence of the increasing
(1) Li > Na > K > Rb (2) K > Rb > Li > Na basic nature of the given oxides?
(3) Li > Na > Rb > K (4) Rb > Li > K > Na (1) Na2O < K2O < MgO < Al2O3
(2) K2O < Na2O < Al2O3 < MgO
62. In which of the following have maximum (3) Al2O3 < MgO < Na2O < K2O
electron affinity: - (4) MgO < K2O < Al2O3 < Na2O
(1) H (2) Cl
(3) I (4) Na
68. The correct order of electron gain enthalpy
with negative sign of F, Cl, Br and I, having
63. Which one has the minimum value of
atomic number 9, 17, 35 and 53
ionisation potential :-
(1) Na (2) K respectively, is :-
(3) Li (4) Cs (1) I > Br > Cl > F
(2) F > Cl > Br > I
64. Which of the following has maximum value (3) Cl > F > Br > I
of EA:- (4) Br > Cl > I > F
2 2 5
(1) 1s , 2s ,2p
2 2 6 2 5
(2) 1s , 2s 2p ,3s 3p 69. Correct order of IP is :
2
(3) 1s (1) Na < Mg > Al < Si
(4) 1s 2s
2 2
(2) Na < Mg < Al < Si
(3) Na > Mg > Al > Si
65. Correct order of EN is :- (4) Mg < Al > Na > Si
(1) F > N < O > C (2) O > F > N > C
(3) F > O > C > N (4) N > F > O > C 70. The electronic configuration of the outer
+ 2+ most shell of the most electronegative
66. Among K, K , Sr , Ar, which atom/ion will
element is :
have the smallest radius? 2 5 2 5
+ 2+ (1) 2s 2p (2) 3s 3p
(1) K (2) Sr
2 5 2 5
(3) Ar (4) K (3) 4s 4p (4) 5s 5p

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Classification of Elements and Periodicity in Properties CHEMISTRY

Exercise 3
1. Consider the isoelectronic species, Na ,
+
7. The period number in the long form of the
2+ – 2–
Mg , F and O . The correct order of periodic table is equal to
increasing length of their radii is (1) Magnetic quantum number of any
___________. element of the period.
– 2– 2+ + (2) atomic number of any element of the
(1) F < O < Mg < Na
2+ + – 2–
period.
(2) Mg < Na < F < O (3) Maximum Principal quantum number of
2– – + 2+
(3) O < F < Na < Mg any element of the period.
2– – 2+ +
(4) O < F < Mg < Na (4) Maximum Azimuthal quantum number
of any element of the period.
2. Which of the following is not an actinoid ?
(1) Curium (Z = 96) 8. The elements in which electrons are
(2) Californium (Z = 98) progressively filled in 4f-orbital are called
(3) uranium (Z = 92) (1) actinoids
(2) transition elements
(4) Terbium (Z = 65)
(3) lanthanoids
(4) halogens
3. The order of screening effect of electrons of
s, p, d and f orbitals of a given shell of an 9. Which of the following is the correct order
atom on its outer shell electrons is : of size of the given species.
(1) s > p > d > f (2) f > d > p > s – + + –
(1) I > I > I (2) I > I > I
(3) f < d < s < p (4) f > p > s > d + – – +
(3) I > I > I (4) I > I > I
4. The electronic configuration of gadolinium 10. The element with atomic number 57 belongs to
(Atomic number 64) is (1) s-block (2) p-block
3 5 2 7 2 1
(1) [Xe] 4f 5d 6s (2) [Xe] 4f 5d 6s (3) d-block (4) f-block
7 1 2 8 6 2
(3) [Xe] 4f 5d 6s (4) [Xe] 4f 5d 6s
th
11. The last element of the p-block in 6 period
5. The statement that is not correct for is represented by the outermost electronic
periodic classification of elements is : configuration.
2 6 14 10 2 0
(1) The properties of elements are periodic (1) 7s 7p (2) 5f 6d 7s 7p
14 10 2 6 14 10 2 4
function of their atomic numbers. (3) 4f 5d 6s 6p (4) 4f 5d 6s 6p
(2) Non-metallic elements are less in
number than metallic elements. 12. Which of the elements whose atomic
(3) For transition elements, the 3d-orbitals number are given below, cannot be
are filled with electrons after 3p- accommodate in the present set up of the
orbitals and before 4s-orbitals. long form of the periodic table ?
(4) The first ionisation enthalpies of (1) 107 (2) 118
elements generally increase with (3) 126 (4) 102
increase in atomic number as we go
along a period. 13. The electronic configuration of the element
which is just above the element with atomic
number 43 in the same group is ________.
6. Among halogens, the correct order of 2 2 6 2 6 5 2
amount of energy released in electron gain (1) 1s 2s 2p 3s 3p 3d 4s
2 2 6 2 6 5 3 6
(electron gain enthalpy) is : (2) 1s 2s 2p 3s 3p 3d 4s 4p
2 2 6 2 6 6 2
(1) F > Cl > Br > I (2) F < Cl < Br < I (3) 1s 2s 2p 3s 3p 3d 4s
2 2 6 2 6 7 2
(3) F < Cl > Br > I (4) F < Cl < Br < I (4) 1s 2s 2p 3s 3p 3d 4s

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CHEMISTRY Classification of Elements and Periodicity in Properties
14. The element with atomic numbers 35, 53, 20. Which of the following statements are
and 85 are all ____________ . correct?
(1) noble gases (2) halogens (A) Helium has the highest first ionisation
(3) heavy metals (4) light metals enthalpy in the periodic table.
(B) Chlorine has less negative electron gain
15. Electronic configurations of four elements
enthalpy than fluorine.
A, B, C and D are given below:
2 2 6 2 2 4 (C) Mercury and bromine are liquids at
(A) 1s 2s 2p (B) 1s 2s 2p
2 2 6 1 2 2 5 room temperature.
(C) 1s 2s 2p 3s (D) 1s 2s 2p
Which of the following is the correct order (D) In any period, atomic radius of alkali
of increasing tendency to gain electron: metal is the highest.
(1) A < C < B < D (2) A < B < C < D (1) A and B (2) B and C
(3) D < B < C <A (4) D < A < B < C (3) A, B and C (4) All

16. Which of the following elements can show

21. Which of the following sets contain only
covalency greater than 4?
(A) Be (B) P isoelectronic ions?
2+ 2+ 3+ 3+
(C) S (D) B (A) Zn , Ca , Ga , Al
(1) A and B (2) B and C + 2+
(B) K , Ca , Sc , Cl
3+ –

(3) C and D (4) B and D

(C) P3− ,S2− ,Cl − ,K +

17. Those elements impart colour to the flame (D) Ti4 + , Ar,Cr 3+ , V 5+
on heating in it, the atoms of which require (1) A and B (2) B and C
low energy for the ionisation (i.e., absorb (3) C and D (4) B and D
energy in the visible region of spectrum).
The elements of which of the following
groups will impart colour to the flame? 22. In which of the following options order of
(A) 2 (B) 13 arrangement agree with the variation of
(C) 1 (D) 17 property indicated against it?
(1) A and B (2) B and C (1) Mg2+ < Al 3+ < Na+ < F − (increasing ionic
(3) C and D (4) A and C
size)
(2) B < C < O < N (increasing first ionisation
18. Which of the following sequences contain
atomic numbers of only representative enthalpy)
elements? (3) I < Br < F < Cl (increasing electron gain
(A) 3, 33, 53, 87 (B) 2, 10, 22, 36 enthalpy)
(C) 7, 17, 25, 37, 48 (D) 9, 35, 51, 88 (4) Li < Na < Rb < K (increasing metallic
(1) A and B (2) B and C radius)
(3) A and D (4) B and D
23. Which of the following have no unit?
19. Which of the following elements will gain
(A) Electronegativity
one electron more readily in comparison to
(B) Electron gain enthalpy
other elements of their group?
(A) S (g) (B) Na (g) (C) Ionisation enthalpy
(C) O (g) (D) Cl (g) (D) Metallic character
(1) A and D (2) B, C and D (1) A and D (2) B and C
(3) C and D (4) All (3) C and D (4) All

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Classification of Elements and Periodicity in Properties CHEMISTRY
24. Match the column: 27. Match the correct ionisation enthalpies and
electron gain enthalpies of the following
Column-I Column-II elements.
(Order) (Properties)
Elements ∆ H1 ∆ H2 ∆eg H
(I) B > Tl > Ga (A) Electron (i) Most A. 419 3051 – 48
> Al > In Affinity (EA) reactive
(II) Cl > F > Br (B) Electronegativity non metal
>I (EN) (ii) Most B. 1681 3374 – 328
(III) F > Cl > Br (C) Ionisation reactive
>I Energy (IP) metal
(IV) B < Ga < Al (D) Atomic Radius (iii) Least C. 738 1451 – 40
< In < Tl reactive
element
(1) I – A, II – B, III – D, IV – C (iv) Metal D. 2372 5251 + 48
(2) I – A, II – B, III – C, IV – D forming
(3) I – C, II – A, III – B, IV – D binary
(4) I – D, II – C, III – B, IV – A
halide

25. An element belongs to 3rd period and (1) i → B ; ii → A ; iii → C ; iv → D

group-13 of the periodic table. Which of the (2) i → B ; ii → A ; iii → D ; iv → C
following properties will be shown by the (3) i → A ; ii → B ; iii → D ; iv → C
element? (4) i → B ; ii → C ; iii → D ; iv → A
(A) Good conductor of electricity
(B) Liquid, metallic 28. Electronic configuration of some elements
(C) Solid, metallic is given in Column I and their electron gain
(D) Solid, non-metallic enthalpies are given in Column II. Match
(1) A and B (2) B and C the electronic configuration with electron
(3) C and D (4) A and C gain enthalpy.

26. Match the correct atomic radius with the Column (I) Column (II)
element. Electronic Electron gain
configuration enthalpy/
Element Atomic radius (pm) kJ mol –1
(a) Be (p) 74 (I) 2 2
1s 2s sp
6
(A) – 53
(b) C (q) 88
(II) 2 2
1s 2s 2p 3s
6 1
(B) – 328
(c) O (r) 111
(III) 2 2
1s 2s 2p
5
(C) – 141
(d) B (s) 77
(e) N (t) 66 (IV) 2 2
1s 2s 2p
4
(D) + 48

(1) a → r ; b → s ; c → t ; d → q ; e → p (1) i → B ; ii → A ; iii → C ; iv → D

(2) a → s ; b → r ; c → t ; d → q ; e → p (2) i → D ; ii → A ; iii → B ; iv → C
(3) a → r ; b → s ; c → p ; d → q ; e → t (3) i → A ; ii → B ; iii → D ; iv → C
(4) a → r ; b → s ; c → t ; d → p ; e → q (4) i → D ; ii → A ; iii → C ; iv → B

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CHEMISTRY Classification of Elements and Periodicity in Properties
29. Assertion (A) : Generally, ionisation 31. Assertion (A) : Electron gain enthalpy
enthalpy increases from left to right in a becomes less negative as we go down a
period. group.
Reason (R) : Size of the atom increases on
Reason (R) : When successive electrons are
going down the group and the added
added to the orbitals in the same principal electron would be farther from the nucleus.
quantum level, the shielding effect of inner (1) Assertion and reason both are correct
core of electrons does not increase very statements but reason is not correct
much to compensate for the increased explanation for assertion.
attraction of the electron to the nucleus. (2) Assertion and reason both are correct
statements and reason is correct
(1) Assertion is correct statement and
explanation for assertion.
reason is wrong statement. (3) Assertion and reason both are wrong
(2) Assertion and reason both are correct statements.
statements and reason is correct (4) Assertion is wrong statement but
explanation of assertion. reason is correct statement.
(3) Assertion and reason both are wrong
32. Assertion: F atom has a less negative
statements. electron affinity than Cl atom.
(4) Assertion is wrong statement and Reason: Additional electrons are repelled
reason is correct statement. more effectively by 3p electrons in Cl atom
than by 2p electrons in F atom.
(1) Both Assertion and Reason are true and
30. Assertion (A) : Boron has a smaller first
Reason is the correct explanation of
ionisation enthalpy than beryllium.
Assertion
Reason (R) : The penetration of a 2s (2) Both Assertion and Reason are true but
electron to the nucleus is more than the 2p Reason is not the correct explanation of
electron hence 2p electron is more shielded Assertion
by the inner core of electrons than the 2s (3) Assertion is true but Reason is false.
(4) Assertion is false but Reason is true.
electrons.
(1) Assertion and reason both are correct
33. Assertion: The first ionization energy of Be
statements but reason is not correct is greater than that of B.
explanation for assertion. Reason: 2p orbital is lower in energy than 2s.
(2) Assertion is correct statement but (1) Both Assertion and Reason are true and
reason is wrong statement. Reason is the correct explanation of
Assertion
(3) Assertion and reason both are correct
(2) Both Assertion and Reason are true but
statements and reason is correct
Reason is not the correct explanation of
explanation for assertion. Assertion
(4) Assertion and reason both are wrong (3) Assertion is true but Reason is false.
statements. (4) Assertion is false but Reason is true.

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34. Assertion (A): Be and Al show diagonal 37. Assertion (A): F is most electro negative
relationship. element of periodic table.
Reason (R): Cl is having highest electron
Reason (R): Be and Al are exhibit similar
affinity
charge/radius ratio.
(1) Both (A) & (R) are true and the (R) is the
(1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
correct explanation of the (A) (2) Both (A) & (R) are true but the (R) is not
(2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(3) (A) is true but (R) is false
the correct explanation of the (A)
(4) Both (A) and (R) are false
(3) (A) is true but (R) is false
(4) Both (A) and (R) are false 38. Assertion (A): Cu, Ag, Au are known as
coinage metal.
35. Assertion (A): The first ionization energy of Reason (R): Coinage metals are d–block
metals.
Al is lower than magnesium.
(1) Both (A) & (R) are true and the (R) is the
Reason (R): Atomic radius of Al is smaller
correct explanation of the (A)
then magnesium. (2) Both (A) & (R) are true but the (R) is not
(1) Both (A) & (R) are true and the (R) is the the correct explanation of the (A)
correct explanation of the (A) (3) (A) is true but (R) is false
(4) Both (A) and (R) are false
(2) Both (A) & (R) are true but the (R) is not
the correct explanation of the (A)
39. The correct order of Electronegativity
(3) (A) is true but (R) is false of underlined element.
(4) Both (A) and (R) are false (1) C2H4 < C2H2 (2) OF2 < Na2O
(3) HCl < H2 (4) H2 < NaH

36. Assertion (A): Electron affinity of oxygen is

40. Assertion (A): According to Mendeleev,
lower than sulphur. periodic properties of elements is a
Reason (R): Number of valence orbitals function of their atomic mass.

containing electrons are different Reason (R): Atomic number is equal to the
number of protons.
(1) Both (A) & (R) are true and the (R) is the
(1) Both (A) & (R) are true and the (R) is the
correct explanation of the (A)
correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not (2) Both (A) & (R) are true but the (R) is not
the correct explanation of the (A) the correct explanation of the (A)
(3) (A) is true but (R) is false (3) (A) is true but (R) is false
(4) Both (A) and (R) are false
(4) Both (A) and (R) are false

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CHEMISTRY Classification of Elements and Periodicity in Properties
41. Assertion (A): Atomic number of the 44. Assertion (A): Atomic size decreases along
element ununbium is 112. a period.

Reason (R): Name for digits 1 and 2 is un– Reason (R): Effective nuclear charge

and bi respectively in latin words. increases as the atomic number increases

resulting in the increased attraction of
(1) Both (A) & (R) are true and the (R) is the
electrons to the nucleus.
correct explanation of the (A)
(1) Both (A) & (R) are true and the (R) is the
(2) Both (A) & (R) are true but the (R) is not
correct explanation of the (A)
the correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not
(3) (A) is true but (R) is false
the correct explanation of the (A)
(4) Both (A) and (R) are false (3) (A) is true but (R) is false
(4) Both (A) and (R) are false
42. Assertion (A): Helium is placed in group 18
along with p-block elements. 45. Assertion (A): Second ionization enthalpy

Reason (R): It shows properties similar to will be higher the first ionization enthalpy.

18 Group Elements. Reason (R): Ionization enthalpy is a

quantitative measure of the tendency of an
(1) Both (A) & (R) are true and the (R) is the
element to lose electron.
correct explanation of the (A)
(1) Both (A) & (R) are true and the (R) is the
(2) Both (A) & (R) are true but the (R) is not
correct explanation of the (A)
the correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not
(3) (A) is true but (R) is false
the correct explanation of the (A)
(4) Both (A) and (R) are false (3) (A) is true but (R) is false
(4) Both (A) and (R) are false
43. Assertion (A): Hydrogen can be placed in
group 1. 46. Assertion (A): Alkali metals have least
value of ionization energy within a period.
Reason (R): Hydrogen can gain an electron
Reason (R): Alkali Metals have the
to achieve a noble gas arrangement.
maximum size in a period.
(1) Both (A) & (R) are true and the (R) is the
(1) Both (A) & (R) are true and the (R) is the
correct explanation of the (A)
correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not
(2) Both (A) & (R) are true but the (R) is not
the correct explanation of the (A) the correct explanation of the (A)
(3) (A) is true but (R) is false (3) (A) is true but (R) is false
(4) Both (A) and (R) are false (4) Both (A) and (R) are false

40 Sarvam Career Institute

Classification of Elements and Periodicity in Properties CHEMISTRY
47. Assertion (A): Electron gain enthalpy can 49. Given below are to statements:
be exothermic or endothermic. Assertion (A): The decrease in the first
Reason (R): Electron gain enthalpy ionization enthalpy from B to Al is much
larger than that from Al to Ga.
provides a measure of the ease with which
Reason (R): The d orbitals in Ga are
an atom adds an electron to form anion.
completely filled.
(1) Both (A) & (R) are true and the (R) is the In the light of the above statements, choose
correct explanation of the (A) the most appropriate answer from the
(2) Both (A) & (R) are true but the (R) is not options given below
the correct explanation of the (A) (1) Both (A) & (R) are true and the (R) is the
(3) (A) is true but (R) is false correct explanation of the (A)
(4) Both (A) and (R) are false (2) Both (A) & (R) are true but the (R) is not
the correct explanation of the (A)
(3) (A) is true but (R) is false
48. Assertion (A): Smaller the size of an atom (4) Both (A) and (R) are false
greater is the electronegativity.
Reason (R): Electronegativity refers to the 50. Match the column I with column II.
tendency of atom to attract electrons from Column I Column II
other atom. (i) Newland’s octave rule (a) 1862
(1) Both (A) & (R) are true and the (R) is the (ii) A.E.B. de chancourtois (b) 1865
(iii) Moseley (c) 1829
correct explanation of the (A)
(iv) Dobereiner’s Triad (d) 1913
(2) Both (A) & (R) are true but the (R) is not
(1) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
the correct explanation of the (A) (2) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d)
(3) (A) is true but (R) is false (3) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)
(4) Both (A) and (R) are false (4) (i)-(d), (ii)-(a), (iii)-(b), (iv)-(c)

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CHEMISTRY Classification of Elements and Periodicity in Properties

Exercise 4 (Previous Year's Questions)

1. Among the elements Ca, Mg, P and Cl, the 6.
st
The 1 Ionisation enthalpy of Na, Mg and Si
order of increasing atomic radii is:- –1
are 496, 737, 776 kJ mol respectively then
(AIPMT 2010) st
what will be the 1 ionisation enthalpy of Al
(1) Cl < P < Mg < Ca (2) P < Cl < Ca < Mg –1
in kJmol : (AIIMS 2013)
(3) Ca < Mg < P < Cl (4) Mg < Ca < Cl < P –1
(1) > 766 kJmol
–1
2. The correct order of the decreasing ionic (2) > 496 and <737 kJmol
–1
radii among the following isoelectronic (3) > 737 and < 766 kJmol
–1
species is:- (AIPMT 2010) (4) 496 kJmol
2+ – 2–
(1) K+ > Ca > Cl > S
2+ + 2– – 7. Which of the following orders of ionic radii
(2) Ca > K > S > Cl
– 2– 2+ + is correctly represented? (AIPMT 2014)
(3) Cl > S > Ca > K – + + – 2–
2– – + 2+ (1) H > H > H (2) Na < F < O
(4) S > Cl > K > Ca – 2– + 3+ 2+ 3–
(3) F > O > Na (4) Al > Mg > N

3. Which of the following represents the 2+

correct order of increasing electron gain 8. Be is isoelectronic with which of the
enthalpy with negative sign for the following ions? (AIPMT 2014)
+ +
elements O, S, F and Cl? (AIPMT 2010) (1) H (2) Li
+ 2+
(1) S < O < Cl < F (2) Cl < F < O < S (3) Na (4) Mg
(3) O < S < F < Cl (4) F < S < O < Cl
9. Reason of lanthanoid contraction is:
4. What is the value of electron gain enthalpy (AIPMT 2014)
+
of Na if IE1 of Na = 5.1 eV:- (1) Negligible screening effect of ‘f’ orbitals
(AIPMT MAIN 2011) (2) Increasing nuclear charge
(1) +0.2 eV (2) –5.1 eV (3) Decreasing nuclear chareg
(4) Decreasing screening effect
(3) –10.2 eV (4) +2.55 Ev
2+
5. Identify the wrong statement in the 10. The number of d-electrons in Fe (Z=26) is
following: (AIPMT PRE 2012) not equal to the number of electrons in
(1) Atomic radius of the elements increases which one of the following? (AIPMT 2015)
as one moves down the first group of (1) p-electrons in Cl (Z = 17)
the periodic table (2) d-electrons in Fe (Z = 26)
(3) p-electrons in Ne (Z = 10)
(2) Atomic radius of the elements
(4) s-electrons in Mg (Z = 12)
decreases as one-moves across from
nd
left to right in the 2 period of the
11. Because of lanthanoid contraction, which
periodic table
of the following pairs of elements have
(3) Amongst isoelectronic species, smaller
nearly same radii? (Numbers in the brackets
the positive charge on the cation,
are atomic numbers). (AIPMT 2015)
smaller is the ionic radius
(1) Zr (40) and Nb (41)
(4) Amongst isoelectronic species, greater (2) Zr (40) and Hf (72)
the negative charge on the anion, larger (3) Zr (40) and Ta (73)
is the ionic radius (4) Ti (22) and Zr (40)

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Classification of Elements and Periodicity in Properties CHEMISTRY

12. The formation of the oxide ion, O (g), from

2–
17. The element Z = 114 has been discovered
oxygen atom requires first an exothermic recently. It will belong to which of the
following family/group and electronic
and then an endothermic step as shown
configuration? (NEET – 2017)
below : 14 10 2 2
– – –1
(1) Carbon family, [Rn] 5f 6d 7s 7p
O(g) + e → O (g) ;∆f HΘ = – 141 kJ mol 14 10 2 4
– – –1
(2) Oxygen family, [Rn] 5f 6d 7s 7p
O (g) + e → ; ∆fHΘ = + 780 kJ mol 14 10
(3) Nitrogen family, [Rn] 5f 6d 7s 7p
2 6

2–
Thus, process of formation of O in gas 14 10
(4) Halogen family, [Rn] 5f 6d 7s 7p
2 5

2–
phase is unfavorable even though O is
isoelectronic with neon. It is due to the fact 18. Which of the following oxides is most acidic
that, (RE-AIPMT 2015) in nature? (NEET 2018)
(1) MgO (2) BeO
(1) Oxygen is more electronegative
(3) BaO (4) CaO
(2) Addition of electron in oxygen results in
larger size of the ion 19. Magnesium reacts with an element (X) to
(3) Electron repulsion outweighs the form an ionic compound. If the ground state
2 2 3
stability gained by achieving noble gas electronic configuration of (X) is 1s 2s 2p
configuration ,the simplest formula for this compound is :
–
(4) O ion has comparatively smaller size (NEET 2018)
(1) Mg2X3 (2) MgX2
than oxygen atom.
(3) Mg2X (4) Mg3X2

13. Which is the correct order of increasing 20. The correct order of atomic radii in group 13
energy of the listed orbitals in the atom of elements is : (NEET 2018)
titanium? (At. no. Z = 22) (RE-AIPMT 2015) (1) B < Al < In < Ga < Tl
(1) 3s 3p 3d 4s (2) 3s 3p 4s 3d (2) B < Al < Ga < In < TI
(3) 3s 4s 3p 3d (4) 4s 3s 3p 3d (3) B < Ga < Al < Tl < In
(4) B < Ga < Al < In < TI

14. Electronic configuration of Al with 21. Which of the following is an amphoteric

excluding bonded electron in aluminate hydroxide? (NEET-UG 2019)
ion. (AIIMS 2015) (1) Sr (OH)2 (2) Ca (OH) 2
(1) [Ne] (2) [Ar] (3) Mg (OH)2 (4) Be (OH)2
2 2
(3) [Ne]3s (4) [Ar]4s
22. For the second period elements the correct
increasing order of first ionisation enthalpy is
15. The biggest gap in electronegativity is : (NEET-UG 2019)
(AIIMS 2016) (1) Li > Be > B > C < N < O < F < Ne
(1) B → Al (2) Al → Ga (2) Li < B < Be < C < O < N < F < Ne
(3) Ga → In (4) In → TI (3) Li > B > Be > C < N < O < F < Ne
(4) Li > Be > B > C < O < N < F < Ne
16. Which one of the following orders is correct
for the bond dissociation enthalpy of 23. 4d, 5p, 5f and 6p orbitals are arranged in
halogen molecules? (NEET-I 2016) the order of decreasing energy. The correct
option is:- (NEET-UG 2019)
(1) I2 > Br2 > Cl2 > F2
(1) 5f > 6p > 5p > 4d
(2) Cl2 > Br2 > F2 > I2 (2) 6p > 5f > 5p > 4d
(3) Br2 > I2 > F2 > Cl2 (3) 6p > 5f > 4d > 5p
(4) F2 > Cl2 > Br2 > I2 (4) 5f > 6p > 4d > 5p

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CHEMISTRY Classification of Elements and Periodicity in Properties
24. Match the oxide given in column A with its 26. Identify the incorrect match.
[NEET-2020]
property given in column B:
Name IUPAC Official
[NEET – 2019 (ODISHA)] Name
(a) Unnilunium (i) Mendelvium
Column-A Column-B
(b) Unniltrium (ii) Lawrencium
(i) Na2O (a) Neutral
(c) Unnilhexium (iii) Seaborgium
(ii) Al2O3 (b) Basic
(d) Unununnium (iv) Darmstadtium
(iii) N2O (c) Acidic

(iv) Cl2O7 (d) Amphoteric (1) (c), (iii) (2) (d), (iv)
(3) (a), (i) (4) (b), (ii)

27. Match the element in column I with that in

Which of the following options has all column II.
correct pairs? [NEET 2020 Exam (Covid Re-exam)]
Column-I Column-II
(1) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c) (a) Copper (i) Non-metal
(2) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d) (b) Fluorine (ii) Transition
metal
(3) (i)-(a), (ii)-(d), (iii)-(b), (iv)-(c) (c) Silicon (iii) Lanthanoid
(4) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c) (d) Cerium (iv) Metalloid

Identify the correct match :

(1) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
25. Match the following : [NEET-2020] (2) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(3) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(4) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
Oxide Nature 28. Zr (Z = 40) and Hf(Z = 72) have similar
atomic and ionic radii because of :
(a) CO (i) Basic [NEET-2021]
(1) having similar chemical properties
(b) BaO (ii) Neutral (2) belogning to same group
(3) diagonal relationship
(c) Al2O3 (iii) Acidic (4) lanthanide contraction

29. Gadolinium has a low value of third

(d) Cl2O7 (iv) Amphoteric ionization enthalpy because of:
[NEET-2022]
(1) high exchange enthalpy
Which of the following is correct option? (2) high electronegativity
(3) high basic character
(1) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
(4) small size
(2) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
30. The IUPAC name of an element with atomic
(3) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) number 119 is : [NEET-2022]
(1) ununennium (2) unnilennium
(4) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) (3) unununnium (4) ununoctium

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Classification of Elements and Periodicity in Properties CHEMISTRY
31. The element expected to form largest ion to 34. The correct decreasing order of atomic radii
achieve the nearest noble gas configuration (pm) of Li, Be, B and C is: [RE-NEET-2024]
is: [NEET-2023] (1) Be > Li > B > C (2) Li > Be > B > C
(1) F (2) N (3) C > B > Be > Li (4) Li > C > Be > B
(3) Na (4) O
35. Match List-I with List-II:
32. Arrange the following elements in [RE-NEET-2024]
increasing order of electronegativity: N, O, List-I List –II
F, C, Si Choose the correct answer form the (Block/group (Element)
options given below : [NEET-2024] in periodic
(1) O < F < N < C < Si table)
(2) F < O < N < C < Si A. Lanthanoid I. Ce
(3) Si < C < N < O < F B. d-block II. As
(4) Si < C < O < N < F element
C. p-block III. Cs
33. Arrange the following elements in element
increasing order of first ionization D. s-block IV. Mn
element
enthalpy: Li, Be, B, C, N Choose the correct
answer from the options given below: Choose the correct answer from the options
[NEET-2024] given below:
(1) Li < Be < C < B < N (1) А-І, В-II, C-IV, D-III
(2) Li < Be < N < B < C (2) A-I, B-IV, C-III, D-II
(3) Li < Be < B < C < N (3) A-I, B-IV, C-II, D-III
(4) Li < B < Be < C < N (4) A-IV, B-I, C-II, D-III

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CHEMISTRY Classification of Elements and Periodicity in Properties

ANSWER KEY

Exercise 1.1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 3 1 2 4 2 2 2 2 3 2 3 3 1 1 3 1 3 2 1
Que. 21 22 23 24 25 26 27 28 29 30
Ans. 2 3 3 2 3 3 2 2 1 1

Exercise 1.2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 2 4 3 1 1 3 1 4 3 2 3 4 2 1 2 2 3 1 4 3
Que. 21 22 23 24 25 26 27 28 29 30 31
Ans. 2 3 2 1 4 4 4 3 3 3 3

Exercise 1.3
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 3 4 3 2 2 2 4 2 4 2 3 3 2 2 4 1 1 2 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 4 3 3 4 2 3 2 1 4 2 3 4 1 1 2 2 1 3 3 3
Que. 41 42 43 44 45
Ans. 1 1 4 1 4

Exercise 1.4

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Ans. 3 4 2 4 4 3 2 3 2 4 3 2 4 3 3 3 2 4 2

Exercise 1.5

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Ans. 3 2 2 2 4 1 3 3 4 3 4 3 3 2 1 2 2

Exercise 2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 2 1 2 2 2 4 2 2 2 2 4 3 3 4 4 1 2 1 4 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 2 4 4 1 2 4 4 3 1 2 3 2 3 2 4 1 1 2 4
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 1 2 1 4 3 3 3 2 1 4 2 3 2 2 2 3 4 2 3 2
Que. 61 62 63 64 65 66 67 68 69 70
Ans. 1 2 4 2 1 2 3 3 1 1

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Classification of Elements and Periodicity in Properties CHEMISTRY

Exercise 3
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 2 4 1 3 3 3 3 3 4 3 3 3 1 2 1 2 4 3 1 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 2 2 1 3 4 1 2 2 2 3 2 3 3 1 2 3 2 2 1 2
Que. 41 42 43 44 45 46 47 48 49 50
Ans. 2 1 2 1 2 1 2 2 3 1

Exercise 4 (Previous Year's Questions)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 4 3 2 3 2 2 2 1 1 2 3 2 1 1 2 1 2 4 4
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
Ans. 4 2 1 4 4 2 2 4 1 1 2 3 4 2 3

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 Notes

Sarvam Career Institute

Chapter
CHEMICAL
2 BONDING

Chapter Overview Introduction

• Introduction The attractive force which holds various constituents such as

• Octet Rule atoms, ions etc., together in different chemical species is called
• Classification of Chemical
a chemical bond.
Bonds
• Electrovalent or Ionic Bond Octet Rule
 Fajan’s Rule (Covalent “Tendency of atoms to have eight electrons in their outermost
Character in Ionic shell is known as Lewis octet rule". To achieve inert gas
Compounds) configuration atoms, lose, gain or share electrons.
• Covalent Bond (i) It has been observed that atoms of noble gases have little
 Lewis Dot Structures or no tendency to combine with each other or with atoms
 Formal Charge of other elements.
 Resonance (ii) It means that these atoms must have a stable electronic
• Valence Bond Theory configuration.
 Orbital Overlap Concept (iii) These elements (noble gases) have 8 electrons (ns np )
2 6

• Hybridisation 2
except helium which has 2 electrons (1s ) in their outer
• Valence Shell Electron Pair most shell.
Repulsion (VSEPR) Theory 2 6
(iv) It is therefore concluded that ns np configuration in the
 Bond Length Comparison
outer energy level constitutes a structure of maximum
 Bond Angle
stability or minimum energy.
• Molecular Orbital Theory
(MOT)
The Octet rule can be understood by considering the formation of
• Polarity of Bonds & Dipole
the chlorine molecule, Cl2. The Cl atom with electronic
Moment 10 2 5
• Hydrogen Bond configuration, [Ne] 3s 3p , is one electron short of the argon
• Summary of Chemical
configuration. The formation of the Cl2 molecule can be
Bonding understood in terms of the sharing of a pair of electrons between
• Summary the two chlorine atoms, each chlorine atom contributing one
electron to the shared pair. In the process both

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 CHEMISTRY Chemical Bonding

Cl + Cl Cl Cl
– –
8e 8e
or Cl — Cl
Covalent bond between two Cl atoms
Figure : chlorine atoms attain the outer shell octet of the nearest noble gas (i.e., argon). The dots
represent electrons. Such structures are referred to as Lewis dot structures.

Limitations of the Octet Rule:

1. The incomplete octet of the central atom
In some compounds, the number of electrons surrounding the central atom Is less than eight.
Examples are LiCl, BeH2 and BCl3. BeF2 , BF3 , AlCl3
2. Odd-electron molecules
In molecules with an odd number of electrons.
e.g. NO, ClO2 , NO2
3. The expanded octet, Super octet or Hypervalent molecules
In a number of compounds of these elements there are more than eight valence electrons around the
central atom. This is termed as the expanded octet. Obviously, the octet rule does not apply in such
cases.
F F O
F F F
F P S H O S O H
F F F
F F O
PF5 SF6 H2SO4
10 electrons around the 12 electrons around the 12 electrons around the
P atom S atom S atom
4. It is clear that octet rule is based upon the chemical inertness of noble gases. However, some noble
gases (for example xenon and krypton) also combine with oxygen and fluorine to form a number of
compounds like XeF2 , KrF2 , XeOF2 etc.,
5. This theory does not account for the shape of molecules.
6. It does not explain the relative stability of the molecules being totally silent about the energy of a
molecule

Classification of Chemical Bonds

Chemical Bond

Strong Bond Weak Force

Electrovalent Covalent Coordinate Metallic Hydrogen Vander

or ionic bond Bond Bond Bond Bond Wall’s Force

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Electrovalent or Ionic Bond
The chemical bond formed between two or more atoms as a result of complete transfer of one or more
electrons from one atom to another is called ionic or electrovalent bond.
Electrovalent bond is not possible between similar atoms. This type of bonding requires two atoms of
different nature. One atom should have the tendency to loose electrons i.e. electropositive in nature
and the other atom should have the tendency to accept electrons i.e. electronegative in nature.
Electropositive atom loses electrons
Electronegative atom gains electron
Ex.: IA and VII A group elements form strong ionic compound.
Na + Cl → Na+ + Cl–
2, 8, 1 2, 8, 7 2, 8 2, 8, 8
(Ne configuration) (Ar configuration)

More the distance between two elements in the periodic table more will be the ionic character of the
bond.
Total number of electrons lost or gained is called Electrovelency.

Ex.: Ca + 2Cl → CaCl2

+2 –
Ca Ca Cl Cl
2, 8, 8 2, 8, 8, 2 2, 8, 7 2, 8, 8
1 e–
Cl Cl
–

1e – 2, 8, 7 2, 8, 8

Factors favouring formation of Ionic bonds:

Formation of ionic bond depends upon three factors :

(1) Ionisation energy (IE):

Amount of energy required to remove an electron from the outermost orbit of an isolated gaseous atom
to form positive ion or cation is called ionization energy [energy is absorbed so it is an endothermic
process]
M + I.E. → M+ + e–
Less Ionisation energy ⇒ Greater tendency to form cation.
Na+ > Mg+2 > Al+3
Ex.: Cation formation tendency
Cs+ > Rb+ > K+ > Na+ > Li+

(2) Electron gain enthalpy:

Amount of energy released when an electron is added to an isolated gaseous atom to form negative ion
or anion is called electron affinity [energy is released so it is an exothermic process]
X + e– → X– + EA

High Electron gain enthalpy ⇒ Greater tendency to form anions

– – –
Cl > F > Br > I–
– –2 –3
Anion formation tendency
F > O > N

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 CHEMISTRY Chemical Bonding
(3) Lattice Enthalpy:
The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole
of a solid ionic compound into gaseous constituent ions.
Factors affecting L.E.:
1 + –
(i) Lattice energy (L.E.) α r = r + r = interionic distance
r
+ –
(ii) L.E. α Z , Z
+
Z ⇒ charge on cation
–
Z ⇒ charge on anion
(iii) Charge density of cation ↑ L.E. ↑
(a) Al2O3 > MgO > Na2O (b) NaCl < MgCl2
(c) NaCl < MgO (d) NaCl > KCl
Characteristics of ionic compounds:
(i) Physical state:
(a) Ionic compounds are generally crystalline, hard & brittle in nature.
(b) These compounds are generally made from ions which are arranged in a regular way as a
lattice structure.
(c) Thus electrovalent compounds exist as three-dimensional solid aggregates.
(d) Normally each-ion is surrounded by a number of oppositely charged ions and this number is
called co-ordination number.
(ii)Electrical conductivity:
It depends on ionic mobility. In solid state there are no free ions so they are bad conductors of
electricity and in fused state or aqueous solution free ions are present so they are good
conductors of electricity
Conductivity order = Solid state < Molten state < Aqueous solution
(iii) Boiling point and melting point:
High boiling point and melting points are due to strong electrostatic force of attraction.
(iv) Solubility: Ionic compounds are soluble in polar solvent like H2O, HF etc.
To explain solubility of ionic compound, consider an example of NaCl in water.
H2O is polar solvent. (Dielectric constant ≈ 81)

Solvation or Hydration:
Whenever any compound generally ionic or polar covalent is dissolved in an polar solvent or in water
then., different ions of the compound will get separated and will get surrounded by polar solvent
molecules. This process is known as solvation or hydration. Energy released in this process is known as
solvation energy or hydration energy. The ionic compound will be soluble only if solvation energy (H.E.)
is more than the lattice energy
Applications of Hydration energy:
(a) Size of the hydrated ions: Greater the hydration of the ion greater will be its hydrated radii
+ +
Li (aq) > Na (aq)
(b) Mobility of the ion: More is the hydration energy smaller will be the mobility of the ions
1
Mobility of the ion ∝
Hydrated radii
+ + + + +
Li (aq) < Na (aq) < K (aq) < Rb (aq) < Cs (aq).
(c) Electrical conductance is related to mobility of ion so follows the same order.

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 Fajan’s Rule (Covalent Character in Ionic Compounds):

When anion and cation approach each other, the valence shell of anion is pulled towards cation nucleus
and thus shape of anion is deformed. This phenomenon of deformation of anion by a cation is known as
polarisation and the ability of cation to polarize a nearby anion is called as Polarizing power of cation.

Cation anion

Fajan’s pointed out that greater is the polarization of anion in a molecule, more is covalent character
in it.
More distortion of anion, more will be the polarisation and covalent character increases.
According to Fajan’s rules which govern the covalent character in the ionic compounds, which are as
follows:
1
(i) Size of cation : Size of cation α
polarisation power
Ex.: BeCl2 MgCl2 CaCl2 SrCl2 BaCl2

Size of cation ↑ Polarisation ↓ Covalent character ↓

(ii) Charge on cation : polarisation α Charge on cation.

Ex.: NaCl MgCl2 AlCl3
+ 2+ 3+
Na Mg Al

(iii) Size of anion : polarisation α Size of anion.

Ex.: LiF LiCl LiBr Lil

(iv) Charge on anion : polarisation α Charge on anion.

Ex.: AlF3 Al2O3 AlN
– 2– 3–
F, O , N

(v) Pseudo inert gas configuration of cation : Cation having pseudo inert gas configuration has more
polarizing power than the cation that has inert gas configuration. Thus, NaCl having inert gas
configuration will be more ionic whereas CuCl having pseudo inert gas configuration will be more
covalent in nature.
+ 10 + 2 6
Cu = [Ar] 3d Na = [He]2s 2p
– –
18e 8e
Pseudo inert gas configuration inert gas configuration
(Poor Shielding of d- electron) (Strong Shielding of s and p-electron)

NCERT questions (solve yourself)

Q. 1 Explain the formation of a chemical bond.
Q. 2 Define octet rule. Write its significance and limitations.
Q. 3 Write the favourable factors for the formation of ionic bond.

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 CHEMISTRY Chemical Bonding

Exercise 1.1
1. The compound which contains both ionic 8. The electronic configuration of metal 'M' is
and covalent bonds is: 2 1 2 2
1s 2s and of the nonmetal 'X' is 1s 2s 2p .
3

(1) Al4C3 (2) HCl When these two elements combine, the
(3) NH4Cl (4) KCl formula of the compound will be :
(1) MX3 (2) M3X (3) M2X3 (4) M3X2
2. The electronic structure of four elements
a,b,c and d are : 9. This is not the characteristics of ionic
2 2 2 2 2 5
a = 1s2, b = 1s , 2s 2p , c = 1s 2s 2p , compound:
2 2
d = 1s 2s 2p
6 (1) Brittle nature
The tendency to form electrovalent bond is (2) Solubility in polar solvent
greatest in : (3) Directional bond
(1) a (2) b (3) c (4) d (4) Conduction of electricity in fused state

3. Least ionic bond is: 10. The electronic configuration of metal M is

2 2 6 1
(1) P—Cl (2) S—Cl 1s 2s 2p 3s . The formula of its oxide will
(3) I—Cl (4) Cl—Cl be :
(1) MO (2) M2O (3) M2O3 (4) MO2
4. Conditions for ionic bond formation is/are :
(a) Small cation, large anion 11. Consider two elements with atomic no. 37
(b) Low IP of cation, high electron affinity and 53, the bond between their atoms
of anion would be :
(c) Large cation, small anion and less (1) Covalent (2) Ionic
charge (3) Co-ordinate (4) Metallic
(d) Less lattice energy
Correct answer is: 12. Which of the following pairs will form the
(1) a, d (2) b, c and d most stable ionic bond ?
(3) b and c (4) a, b (1) Na and Cl (2) Mg and F
(3) Li and F (4) Na and F
5. Formula of a metal oxide is MO; formula of
its phosphate will be :
13. Which one is the correct statement with
(1) M3(PO4)2 (2) MPO4
reference to solubility of MgSO4 in water:
(3) M2(PO4)2 (4) M2(PO4)3
(1) Hydration energy of MgSO4 is higher in
comparison to its lattice energy
6. Electrovalent compounds do not show 2+
stereoisomerism. The reason is : (2) Ionic potential of Mg is very low
(1) Presence of ions (3) SO24− ion mainly contributes towards
(2) Strong electro static force of attraction hydration energy
(3) Brittleness 2+
(4) Size of Mg and SO24− are similar
(4) Non - directional nature of ionic bond
14. The force responsible for dissolution of
7. Nature of bond formed when two elements
ionic compound in water is :
react, depends on :
(1) Dipole – dipole forces
(1) Ionisation potential
(2) Ion – dipole force
(2) Electronegativity
(3) Ion – ion force
(3) Oxidation potential
(4) Hydrogen bond
(4) Electron - affinity
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Chemical Bonding CHEMISTRY

15.
+2
The hydration energy of Mg is greater than 22. Which of the following is not a correct
the hydration energy of : statement about an ionic compound:
(1) Al
+3
(2) Mg
+3 (1) Higher the lattice energy, higher is
+ +2 melting point
(3) Na (4) Be
(2) Higher the dipole moment of solvent,
more the solubility
16. LiCl is soluble in organic solvent while NaCl
(3) Higher the lattice energy, more the solubility
is not because :
(4) More difference in electronegativity,
(1) Lattice energy of NaCl is less than that
more is the ionic nature
of LiCl
(2) Ionisation potential of Li is more than 23. Pick out the wrong statement :-
that of Na (1) LiF has less solubility in water than LiI
(3) Li+ has more hydration energy than Na+ ion (2) Lattice energy of MgO is greater than Na2O
(4) LiCl is more covalent compound than (3) LiH is more stable than KH
that NaCl (4) KO2 is diamagnetic and colourless

17. Which of the compound is least soluble in 24. Among the following which compounds will
water? show the highest lattice energy ?
(1) AgF (2) AgCl (1) KF (2) NaF (3) CsF (4) RbF
(3) AgBr (4) AgI
25. The lattice energy of the lithium is in the
+
18. Ionic conductances of hydrated M ions are following order :
in the order : (1) LiF > LiCl > LiBr > LiI
(1) Li+(aq)>Na+(aq)>K+(aq)>Rb+(aq) > Cs+(aq) (2) LiCl > LiF > LiBr > LiI
(2) Li+(aq)>Na+(aq)<K+(aq)<Rb+(aq)<Cs+(aq) (3) LiBr > LiCl > LiF > LiI
(3) Li+(aq)>Na+(aq)>K+(aq)>Rb+(aq)<Cs+(aq) (4) LiI > LiBr > LiCl > LiF
(4) Li+(aq)<Na+(aq)<K+(aq)<Rb+(aq)<Cs+(aq)
26. Which of the following compound has
19. Which of the following substance will have highest Lattice energy?
highest b.p. ? (1) AlF3 (2) Na2S (3) Al2O3 (4) CaF2
(1) He (2) CsF
(3) NH3 (4) CHCl3 27. The correct expected order of decreasing
lattice energy is :
20. As compared to covalent compounds (1) CaO > MgBr2 > CsI (2) MgBr2 > CaO > CsI
electrovalent compounds generally possess (3) CsI > MgBr2 > CaO (4) CsI > CaO > MgBr2
(1) High m.p. and high b.p.
(2) Low m.p. and low b.p. 28. The pair of elements which on combination
(3) Low m.p. and high b.p. are most likely to form an ionic compound is :
(4) high m.p. and low b.p. (1) Na and Ca (2) K and O
(3) O and Cl (4) Al and I
21. In which of the following pair of
compounds? Their melting point are closest 29. Out of the following the compound with
to each other : maximum ionic nature are :
(1) LiCl, NaCl (2) RbCl, LiCl (1) Metal oxide (2) Metal chloride
(3) CsCl, NaCl (4) LiCl, CsCl (3) Metal phosphide (4) Metal sulphide

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 CHEMISTRY Chemical Bonding
30. The most covalent halide is : 39. On the basis of concept of ionic potential
(1) AlF3 (2) AlCl3 (φ), the tendency to form covalent bond in a
(3) AlBr3 (4) AlI3 group :
(1) increases
31. Ionic potential (φ) of electropositive element (2) decreases
will be highest in which of the following (3) remains unchanged
compound: (4) shows erratic/irregular change
(1) CsCl (2) MgCl2
(3) AlF3 (4) SF6 40. According to Fajan’s rules necessary
condition to form covalent bond is :
32. AlCl3 is covalent, while AlF3 is ionic. This is (1) small cation and large anion
justified by : (2) small cation and small anion
(1) Molecular orbital theory (3) large cation and large anion
(2) Valency bond theory (4) large cation and small anion
(3) Fajan rule
(4) Lattice energy 41. Which is most ionic ?
(1) P2O5 (2) MnO
33. Correct order of covalent character of (3) CrO3 (4) Mn2O7
alkaline earth metal chloride in :
(1) BeCl2 < MgCl2 < CaCl2 < SrCl2 42. Which of the following order is incorrect?
(2) BeCl2 < CaCl2 < SrCl2 < MgCl2 (1) Ionic character = MCl < MCl2 < MCl3
(3) BeCl2 > MgCl2 > CaCl2 > SrCl2 (2) Polarizibility = F–< Cl– < Br– < I
–

(4) SrCl2 > BeCl2 > CaCl2 > MgCl2 (3) Polarising power = Na+< Ca+2 < Mg+2 <Al+3
(4) Covalent character = LiF < LiCl < LiBr < LiI
34. Which pair in the following has maximum
and minimum ionic character respectively?
(1) LiCl, RbCl (2) RbCl, BeCl2 43. According to Fajan’s rules, electrovalent-
bond formation is favoured by :
(3) BeCl2, RbCl (4) AgCl, RbCl
(1) low positive charge, and small size of
35. Maximum covalent character will be shown cations and large size of anions
by the compound is : (2) low positive charge, and large size of
(1) SiCl4 (2) AlCl3 (3) HgCl2 (4) NaCl cations and small size of anions
(3) high negitive charge, and large size of
36. CCl4 is more covalent than LiCl because: cations and large size of anions
(1) There is more polarization of Cl in CCl4 (4) high positive charge, and small size of
(2) There is more polarization of Cl in LiCl cations and small size of anions
(3) CCl4 has more weight
44. The electrovalency of the element is equal
(4) None of above
to the:
37. Among LiCl, BeCl2, BCl3 and CCl4, the (1) Number of electrons lost
covalent bond character follows the order : (2) Numbers of electrons gained
(1) LiCl < BeCl2 > BCl3 > CCl4 (3) Number of electrons transferred
(2) LiCl > BeCl2 < BCl3 < CCl4 (4) Number of electrons lost or gained by
the atom of the element during the
(3) LiCl < BeCl2 < BCl3 < CCl4
formation of ions of ionic compounds
(4) LiCl > BeCl2 > BCl3 > CCl4

38. The correct order of decreasing polarisable 45. The ionic mobility of alkali metal ions in
ions is : aqueous solution is maximum for
+ +
– – –
(1) Cl , Br ,I , F
– – – –
(2) F , I , Br , Cl
– (1) Li (2) Na
+ +
– – –
(3) F , I , Br , Cl
– – – –
(4) I , Br ,Cl , F
– (3) K (4) Rb

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Chemical Bonding CHEMISTRY

Covalent Bond
A chemical bond formed by sharing of electrons between two elements is called as covalent bond.
A − A (Single bond) : When 2 electrons are shared between the two combining elements.
A = A (Double bond) : When 4 electrons are shared between the two combining elements.
A ≡ A (Triple bond) : When 6 electrons are shared between the two combining elements.

H H O O N N

H2 molecule O2 N2
H—H O=O N≡N
 Lewis Dot Structures:
(i) The total number of electrons are obtained by adding the valence electrons of combining atoms.
(ii) For Anions, we need to add one electron for each negative charge.
(iii) For cations, we need to subtract one electron for each positive charge.
(iv) After then the central atom is decided.
To decide, Central atom, following steps are followed:
In general, the least electronegative atom occupies the central position in the molecule/ion.

 Formal Charge:
1
Formal charge =Total VE − lonepair e− − sharedpair e−
2
1

O
2 3
O O
The atoms have been numbered as 1. 2 and 3. The formal charge on:
1
The central O atom marked 1 = 6 –2 – (6) = + 1
2
1
The terminal O atom marked 2 = 6 – 4 – (4) = 0
2
1
The terminal O atom marked 3 = 6 – 6 – (2) = – 1
2
Hence, we represent O3 along with the formal charges as follows:
+
O
–
O O
 Resonance:
It is often observed that a single Lewis structure is inadequate for the representation of a molecule in
conformity with its experimentally determined parameters. For example, the ozone, O3 molecule can
be equally represented by the structures I and II shown below:
O O O

O O O O O O
I II III
Resonance in the O3 molecule
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 CHEMISTRY Chemical Bonding
Structures I and II represent the two canonical forms.
According to the concept of resonance, whenever a single Lewis structure cannot describe a molecule
accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding
pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule
accurately.
– – 2–
O O O O
C C C ≡ C
O
–
O
–
O
–
O O O
–
O O

Characteristics of covalent compounds:

(i) Physical state:
(a) Under normal temperature and pressure, they exist as gases or liquids of low boiling points.
(b) If their molecular masses are high, they exist as soft solids
(c) This is due to the fact that very weak forces of attraction (Vander Waal’s forces) exist between
the molecules due to which molecules are far from each other.
Ex. Sulphur, phosphorus and iodine

(ii) Crystal structures:

Various types of structures that are present in a covalent compound are as follows.
(a) Covalent solid – In this type of structure every atom is bonded to four other atoms by single
covalent bonds resulting in the formation of a giant structure e.g. SiC, AIN and diamond these
crystals are very hard and possess high melting point.
(b) Molecular solids: They are formed when one atom combines with another by a covalent bond
and then the molecule combines with another similar molecule with the help of Vander waal’s
force of attraction or hydrogen bond
Ex.: CH4(Solid), dry ice, ice

(iii) Solubility:
(a) Non polar compounds are soluble in non-polar solvents.
(b) Polar compounds are soluble in polar solvents

(iv) Electrical conductivity:

(a) In general, covalent substances are bad conductors of electricity. Since they do not contain
charged particles or free electrons.
(b) Substances which have polar character like HCl in a solution, can conduct electricity.
(c) Graphite can conduct electricity since electrons can pass from one layer to other.
(d) Some show conductivity due to self-ionisation. example Liq. NH3

NH3 + NH3 → NH4+ + NH2− (Amide ion)

+ –
H2O + H2O → H3O + OH

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Chemical Bonding CHEMISTRY
Valence Bond Theory

Valence bond theory was introduced by Heitler and London (1927) and developed further by Pauling
and others. A discussion of the valence bond theory is based on the knowledge of atomic orbitals,
electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridization of
atomic orbitals and the principles of variation and superposition. Consider two hydrogen atoms A and
B approaching each other having nuclei NA and NB and electrons present in them are represented by eA
and eB. When the two atoms are at large distance from each other, there is no interaction between
them. As these two atoms approach each other, new attractive and repulsive forces begin to operate.
EA
–

– E
A
HA + +H
B A + + B
HA HB
Old forces
– – E
EB New forces B

Attractive Forces Repulsive Forces

Fig- Forces of attraction and repulsion during the formation of H2 molecule.
Energy(kJ/mol)

0
Distance of separation

Bond
435.8 Energy
Bond Length 74pm Internuclear distance

Fig. The potential energy curve for the formation of H2 molecule as a function of internuclear distance
of the H atoms. The minimum in the curve corresponds to the most stable state of H2.

 Orbital Overlap Concept:

In the formation of hydrogen molecule, there is a minimum energy state when two hydrogen atoms are
so near that their atomic orbitals undergo partial interpenetration. This partial merging of atomic
orbitals is called overlapping of atomic orbitals which results in the pairing of electrons. The extent of
overlap decides the strength of a covalent bond. In general, greater the overlap the stronger is the bond
formed between two atoms. Therefore, according to orbital overlap concept, the formation of a
covalent bond between two atoms results by pairing of electrons present In the valence shell having
opposite spins.

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 CHEMISTRY Chemical Bonding
Overlapping of Atomic Orbitals:
Positive overlap Negative overlap
– + + – Z – + – + Z
PZ PZ PZ PZ
(A) (E)

– + + Z + – + Z
PZ PZ
(B) (F)

+ + + –
Z Z

– – – +

Px Px Px Px
(C) (G)

Z Z

Py Py Py Py
(D) (H)

Zero overlap:

+ +

+ Z Z

– S –

Px Px
Py
(I) (II)
Figure: Positive, negative and zero overlaps of s and p atomic orbitals

Types of Overlapping and Nature of Covalent Bonds:

The covalent bond may be classified into two types depending upon the types of overlapping:
(i) Sigma (σ) bond, and (ii) pi (π) bond

(i) Sigma (σ) bond:

This type of covalent bond is formed by the end to end (hand-on) overlap of bonding orbitals along the
internuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of
the following types of combinations of atomic orbitals.

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 Chemical Bonding CHEMISTRY
 s-s overlapping: In this case, there is overlap of two half-filled s-orbitals along the internuclear axis as
shown below:

S–orbital S–orbital S–S overlapping

 s-p overlapping: This type of overlap occurs between half-filled s-orbitals of one atom and half-filled
p-orbitals of another atom.

S–orbital P–orbital S–P orbital

 p-p overlapping: This type of overlap takes place between half-filled p-orbitals of the two approaching
atoms.

P–orbital P–orbital P–P overlapping

(iii) pi(π) bond :
In the formation of π bond the atomic orbitals overlap in such a way that their axes remain parallel to
each other and perpendicular to the internuclear axis. The orbitals formed due to sidewise overlapping
consists of two saucer type charged clouds above and below the plane of the participating atoms.

+ Or

P–orbital P–orbital P–P overlapping

Strength of Sigma and pi Bonds:
Basically, the strength of a bond depends upon the extent of overlapping- In case of sigma bond, the
overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond
where the extent of overlapping occurs to a smaller extent. Further, it is important to note that pi bond.
between two atoms is formed in addition to a sigma bond. It is always present in the molecules
containing multiple bond (double or triple bonds)

NCERT questions (solve yourself)

Q. 1 Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.
Q. 2 Write Lewis symbols for the following atoms and ions:
2– 3+ –
S and S ; Al and Al ; H and H
Q. 3 Draw the Lewis structures for the following molecules and ions :
H2S, SiCl4, BeF2, CO23− , HCOOH
Q. 4 Explain the important aspects of resonance with reference to the CO23− ion.
Q. 5 H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be
taken as the canonical forms of the resonance hybrid representing H3PO3? If not, give
reasons for the same.
H H O P O H
(1) H O P O H (2) O
O H

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 CHEMISTRY Chemical Bonding

Q. 6 Write the resonance structures for SO3, NO2 and NO3− .

Q. 7 Use Lewis symbols to show electron transfer between the following atoms to form cations
and anions : (a) K and S (b) Ca and O (c) Al and N.
Q. 8 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are
shown incorrectly. Write the correct Lewis structure for acetic acid.
H O

H C C O H

H
Q. 9 Considering x-axis as the internuclear axis which out of the following will not form a sigma
bond and why? (a) 1s and 1s ; (b) 1s and 2px ; (c) 2py and 2py ; (d) 1s and 2s.
Q. 10 Write the significance of a plus and a minus sign shown in representing the orbitals.

Exercise 1.2
1. In a triple bond there is sharing of :– 6. Which overlapping is involved in HCl
(1) 3–electrons (2) 4–electrons molecule :
(3) Several electrons (4) 6–electrons (1) s–s overlap (2) p–p overlap
← (3) s–d overlap (4) s–p overlap
2. The triple bond in C = O is made up of :–
(1) Three sigma bonds
7. Which of the following bonds will have
(2) Three π–bonds
directional character
(3) One sigma and two π –bonds
(1) Ionic bond
(4) Two sigma and one π –bond (2) Metallic bond
(3) Covalent bond
3. The strength of bonds by 2s - 2s, 2p - 2p and
(4) Both covalent & metallic
2p -2s overlapping has the order :-
(1) s – s > p – p > s – p (2) s – s > p – s > p – p
(3) p – p > s – p > s – s (4) p – p > s – s > p – s 8. Which is not characteristic of p-bond:-
(1) π- bond is formed when a sigma bond
4. Which of the following configuration shows already formed
second excitation state of Iodine:- (2) π- bond are formed from hybrid orbitals
(1) (3) π-bond may be formed by the
overlapping of p-orbitals
(2)
(4) π-bond results from lateral overlap of
(3) atomic orbitals

(4) 9. Co-axial overlapping of 2 hybrid orbitals

can lead to the formation of :
5. A sigma bond is formed by the overlapping of: (1) Ionic bond
(1) s-s orbital only (2) π –bond
(2) s and p orbitals only
(3) σ–bond
(3) s–s, s–p or p–p orbitals along internuclear axis
(4) (2) and (3) both
(4) p–p orbital along the sides

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Chemical Bonding CHEMISTRY
10. Which of the following statements is not 17. Which of the following element does not
correct? form diatomic molecule.
(1) Double bonds is shorter than a single (1) Iodine (2) Oxygen
bond (3) Phosphorus (4) Nitrogen
(2) σ –bond is weaker than a π bond 18. The F-F bond is weak because :
(3) Double bond is stronger than a single (1) The repulsion between the nonbonding
bond pairs of electrons of two fluorine atom
(4) Covalent bond is stronger than a hydrogen is large
bond (2) The ionization energy of the fluorine
atom is very low
11. In N2H4 molecule type of overlapping (3) The length of the F-F bond much larger
present between N–H bond :– than the bond lengths in other halogen
2 molecules
(1) s – p (2) sp – p
3 2
(4) The F-F bond distance is small and
(3) sp – s (4) sp – s hence the internuclear repulsion
between the two F atoms is very low
12. The type of bond formed between two
electronegative atoms would be 19. Which of the following compound does not
(1) Covalent (2) Ionic exist?
(3) Co-ordinate (4) All the above (1) PF3 (2) PCl5
(3) PBr5 (4) PI5
13. Nitrogen does not form NF5 because
20. The correct order of decreasing bond
(1) Nitrogen is member of V group energy is :
(2) It contains no empty d-orbital (1) O – O > S – S > Se – Se
(3) The bond energy of N — N is very high (2) C – C > Si – Si > Ge – Ge
(4) Inert pair effect exists in the molecule (3) F – F > O – O > N – N
(4) F – F > Cl – Cl > Br – Br
14. The correct order of bond length is
21. Correct order of extent of overlapping is :
(1) C − C < C ≡ C < C = C
(1) 1s – 1s < 2s – 2s < 2s – 2p
(2) C ≡ C < C = C < C − C
(2) 2p – 2p(axial) < 2p – 2p (colateral)
(3) C − C < C = C < C ≡ C
(3) 1s –1s > 2p – 2p (axial) > 2s – 2p > 2s – 2s
(4) C = C < C − C < C ≡ C (4) 1s –1s > 2s – 2p > 2s – 2s > 2p – 2p(axial)

15. Linear combination of two hybridized 22. One of the resonating structures of SO–2
4 is
orbitals belonging to two atoms and each O
having one electron leads to a :
Θ Θ
(1) Sigma bond O S O
(2) Double bond
(3) Co-ordinate covalent bond O
Which set of formal charge on oxygen and
(4) pi bond
bond order is correct
(1) –0.5 and 1.5 (2) 1.5 and 3
16. CO2 is a gas, while SiO2 is a solid but both
(3) 2 and 3 (4) 1.5 and 1.5
are-
(1) Covalent containing π–bond 23. The correct order of the O–O bond length in
O2, H2O2 and O3 is :–
(2) Molecules having p π– dπ bonding
(3) Acidic (1) O3 > H2O2 > O2 (2) O2 > H2O2 > O3
(4) Discrete molecules (3) O2 > O3 > H2O2 (4) H2O2 > O3 > O2
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 CHEMISTRY Chemical Bonding
24. Higher is the bond order, greater is - 30. Which one of the following oxides is expected
(1) Bond dissociation energy to exhibit paramagnetic behaviour:
(2) Covalent character (1) CO2 (2) ClO2
(3) Bond length (3) SO2 (4) SiO2
(4) Paramagnetism
31. Which of the following structures is the
25. In which of the following lewis dot most preferred and hence is of lowest
energy for SO3?
structure is written with incorrect formal
charge? O O
–2 S
0 +1 +1 –1 (1) S (2)
(1) N N N (2) C O
O O O O
0 +1 0 –1 +1 –1
(3) O N O (4) N N N
O
26. The species having no pπ-pπ bond but has S S
bond order equal to that of O2 : (3) O O (4) O O
O
(1) ClO −
3 (2) PO 3−
4

(3) SO24− (4) XeO3 32. According to the valence bond theory, when
a covalent bond is formed between two
27. Cl–O, Bond order and formal charge on reacting atoms, the potential energy of the
system becomes:
Oxygen atom of perchlorate ion are :
(1) negative (2) positive
(1) 1.75 and –0.25 (2) 1.75 and –0.33
(3) minimum (4) maximum
(3) 1.66 and –0.33 (4) 1.50 and –0.50
33. The total number of lone pairs in a chlorine
28. Nitrogen form N2 and phosphorous do not molecule is
form P2, but it converts into P4, at an instant (1) six (2) three
the reason is - (3) four (4) seven
(1) Triple bond present between
34. Consider the following statements :
phosphorous atom
I. A sigma (σ) bond is formed when two
(2) pπ – pπ bonding is weak s-orbitals overlap
(3) pπ – pπ bonding is strong II. A pi (π) bond is formed when two
(4) Multiple bond form easily p-orbitals axially overlap
III. A σ- bond is weaker than π-bond
29. Which of the following statements is not Which of the above statements is/are
correct for sigma and pi bond formed correct ?
between two carbon atoms? (1) I and II (2) II and III
(3) I alone (4) II alone
(1) Free rotation of atoms about a sigma-
bond is allowed but not in case of a pi-
35. π bond is formed :
bond
(1) By overlapping of hybridized orbital
(2) Sigma-bond determines the direction
(2) Overlapping of s-s orbital
between carbon atoms but a pi-bond (3) Head on overlapping of p-p orbital
has no primary effect in this regard (4) By p-p collateral overlapping
(3) Sigma-bond is stronger than a pi-bond
(4) Bond energies of sigma– and pi– bonds 36. Which of the following bonds is strongest ?
are of the order of 264 kJ/mol and 347 (1) 1s-1s (2) 2p-2p
kJ/mol. respectively (3) 2s-2p (4) 1s-2p
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Chemical Bonding CHEMISTRY
Hybridisation
– Hypothetical concept Introduced by Pauling and Slater.
– Atomic orbitals combine to form new set of equivalent orbitals know as hybrid orbitals.
– This phenomenon is known hybridization.
– Process of Intermixing of the atomic orbitals of equal or slightly different energies in the formation
of new set of orbitals of equivalent energies and shape is known as hybridization.
Salient features of hybridisation: The main features of hybridisation are as under:
1. The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.
2. The hybridised orbitals are always equivalent in energy and shape.
3. The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
4. These hybrid orbitals are directed in space in some preferred direction to have minimum repulsion
between electron pairs and thus a stable arrangement is obtained. Therefore, the type of hybridisation
indicates the geometry of the molecules.
Determination of Hybridisation state of an atom:
Method (I) : Number of hybrid orbital = number of σ bond + number of lone pair [surrounding the central atom]
Method (II) : To predict hybridization following formulae may be used:
1
No. of hybrid orbital = [Ve– +M+C - A]
2
– –
Ve = Total number of valence e in the central atom
M = total number of monovalent atoms; C = Positive Charge; A = Negative Charge
If no. of hybrid orbital is
2
Two - sp hybridization Three - sp hybridization
3 3
Four - sp hybridization Five - sp d hybridization
3 2 3 3
Six - sp d hybridization Seven - sp d hybridization
1 3 1 3
Ex.: NH4+ = [5 + 4 – 1] = 4 sp hybridization. SF4 = [6 + 4] = 5 sp d hybridization.
2 2
1 3 1 2
4 =
SO2– [6 + 2] = 4 sp hybridization. NO–3 = [5 + 1] = 3 sp hybridization
2 2
('O' is divalent so add only charge on anion)
S. Formula Total pair of e– Hybridization Geometry/Shape Example
No. bp lp
1. AB2 2 0 sp Linear BeCl2, CO2
2. AB3 3 0 Sp
2
Trigonal Planar BCl3, BF3
3. AB4 4 0 Sp
3
Tetrahedral CH4, CCl4
4. AB5 5 0 3
Sp d Trigonal bipyramidal PCl5
5. AB6 6 0 3 2
Sp d Octahedral SF6
(Square bipyramidal)
6. AB7 7 0 3 3
Sp d Pentagonal IF7
bipyramidal
Position of lone pair & multiple bond
2 3 3 3 2
(i) sp/sp /sp = Any where; (ii) sp d = equatorial; (iii) sp d = axial (defined first)

(iv) Lone pair = 1 then equatorial

sp3d3
Lone pair = 2 then axial
sp d hybridization Axial bond length > equatorial bond length
3

(v) Terminal atom same

sp3d3 hybridization Axial bond length < equatorial bond length
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 CHEMISTRY Chemical Bonding
Valence Shell Electron Pair Repulsion (VSEPR) Theory
(i) The shape of a molecule depends upon the number of valence shell electron pairs [bonded or
nonbonded) around the central atom.
(ii) Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
(iii) These pairs of electrons tend to occupy such positions in space that minimize repulsion and thus
maximise distance between them.
(iv) The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at
maximum distance from one another.
(v) A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a
multiple bond are treated as a single super pair.
(vi) Where two or more resonance structures can represent a molecule, the VSEPR model is applicable
to any such structure.
The repulsive interaction of electron pairs decreases in the order:
[lone pair (p) - lone pair (p)] > [lone pair ( p) - bond pair (bp)] > [bond pair (bp) -bond pair (bp)]
Shape (molecular geometry) of Some Simple Molecules / ions with central atom / ion having no Lone
Pairs of Electrons (E).
Number of Arrangement of Molecular geometry Example
electron pairs electron pairs
180° B—A—B
2 Linear BeCl2, HgCl2
A
Linear
B
120°
120°
3 A BF3, BCl3
A
Trigonal planar B B
Trigonal planar
109.5° B
A
A B +
4 B CH4, NH4

Tetrahedral B
Tetrahedral
B

90°
90°
A B B
5 120° A PCl5
120° B

Trigonal bipyramidal B
Trigonal bipyramidal
B
90°
B
90°
A B
A
6 B SF5
B
Octahedral B
Octahedral

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Chemical Bonding CHEMISTRY
Shape (molecular geometry) of Some Simple Molecules/Ions with central atom / ions having One
or more Lone Pairs of Electrons (E).

Molecule No. of No. of Arrangement Shape Examples

type Bonding lone of electron

pairs pairs pairs

AB2E 2 1 A Bent SO2, O3

B B

AB3E 3 1 A Trigonal pyramidal NH3, PF3

B B
B

AB2E2 2 2 A Bent H2O

B
B

B
B
AB4E 4 1 A See saw SF4
B
B

AB3E2 3 2 B A T-shape Cl F3

AB2E3 2 3 A Linear I3− , ICl 2− , XeF2

B
AB5E 5 1 B B Square pyramidal BrF5
A
B B

B B
AB4E2 4 2 A Square planar XeF4
B B

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 CHEMISTRY Chemical Bonding
Shapes of Molecules containing Bond Pair and Lone Pair
Molecule No. of No. of Arrangement of Shape Reason for the shape acquired
type Bonding lone electron pairs
pairs pairs
S It is found to be bent or v-shaped.
O 119.5° O The reason being the lone pair-bond
pair repulsion is much more as
AB2E 4 1 Bent
compared to the bond pair-bond pair
repulsion. So, the angle is reduced
S
O O to 119.5° from 120°.

N
H 107° H
It is found to be trigonal pyramidal
H due to the repulsion between lp-bp
Trigonal
AB3E 3 1 (which is more than bp-bp
pyramidal
repulsion) the angle between bond
N pairs is reduced to 107° to 109.5°.
H H
H
O
H The shape is distorted tetrahedral or
H 104.5°
angular. The reason is lp-lp
AB2E2 2 2 Bent repulsion is more than lp-lp
repulsion. Thus, the angle is
O
reduced to 104.5° from 109.5°.
H H
(i) F In (i) the lp is present at axial
F S position so there are three lp-bp
F repulsion at 90°. In (ii) the lp is an
F equatorial position, and there are
AB4E 4 1 F See saw two lp-bp repulsions. Hence,
F arrangement (ii) is more stable. The
(ii) S shape shown in (ii) is called as a
F distorted tetrahedron, a folded
F square or a see-saw.
(More stable)
F In (i) the lone pairs are at equatorial
position (120º) so there are less lp–
bp repulsions as compared to others
AB3E2 3 2 (i) Cl F T-shape
in which the lp are at axial positions.
So structure (i) is most stable.
F (T – shaped).
F F
(iii)
(ii) F Cl
Cl F
F
F
I –
AB2E3 2 3 I Linear I3− , ICl 2− , XeF2

I
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 Bond Length Comparison:
Bond Length
(i) Size of atom (see along the group) ∝ bond length
HF < HCl < HBr < HI
F – F < Cl – Cl < Br – Br < Ι – Ι
CH4 < SiH4 < GeH4 < SnH4

(ii) Multiplicity of bond (nearly same period element)

single bond > double bond > triple bond
C—C > C=C > C≡C
F—F > O=O > N≡N
(iii) Electronegativity difference (See along the period)
H—C > H—N > H—O > H—F

NCERT questions (solve yourself)

Q. 1 Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
Q. 2 How do you express the bond strength in terms of bond order?
Q. 3 Define the bond length.
Q. 4 Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while
that of CO2 is linear. Explain this on the basis of dipole moment.
Q. 5 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid
orbitals.
Q. 6 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to
equatorial bonds?

Exercise 1.3
1. In the protonation of H2O, change occurs in 4. Among the following species identify the
(1) Hybridisation state of oxygen isostructural pairs:-
+
(2) Shape of molecule NF3, NO–3 , BF3, H3O , HN3
(3) Hybridisation and shape both +
(1) [NF3, NO–3 ] and [BF3, H3O ]
(4) None
(2) [NF3, HN3] and [ NO–3 , BF3]
+
2. The d-orbitals involved in sp3d hybridisation is:- (3) [NF3, H3O ] and [ NO–3 , BF3]
+
(1) dx2 − y 2 (2) dz2 (4) [NF3, H3O ] and [HN3, BF3]
(3) dxy (4) dxz
5. Which of the set of species have same
3
hybridisation state but different shapes:-
3. A sp hybrid orbital contains:-
(1) NO2+ , NO2, NO–2
3 1
(1) s- character (2) p – character (2) ClO–4 , SF4, XeF4
4 4
+
3 1 (3) NH4+ , H3O , OF2
(3) p - character (4) s - character
4 2 (4) SO–2
4 , PO 4 , ClO 4
–3 –

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6. Which of the following elements can not 13. Hybridisation state of I in ICl2+ is
3
exhibit sp d hybridisation state:- (1) dsp
2
(2) sp
(a) C (b) P (c) Cl (d) B 3 2
(3) sp (4) sp
Correct answer is:-
(1) a, c (2) a, d 14. A s bonded molecule MX3 is T–shaped. The
(3) b, c (4) b, d number of non–bonding pairs of electrons
is
7. Which of the following species are expected (1) 0
to be planar:- (2) 2
(a) NH3 (b) NH23+ (3) 1
(c) CH3+ (d) PCl3 (4) Can be predicted only if atomic number
of M is known.
The correct answer is:-
(1) b and c (2) c and d
15. The regular geometry of XeO2F2 is :–
(3) b and d (4) a and d
(1) Plane triangular
(2) Trigonal bipyramidal
8. In which following set of compound/ion has
(3) Square planar
linear geometry:- (4) Tetrahedral
(1) CH4, NH4+ , BH–4 (2) CO–2
4 , NO 4 , BF3
–

(3) NO2+ , CO2, XeF2 (4) BeCl2, BCl3, CH4 16. Incorrect code regarding shape is :-
(1) Linear : N3− , (CN)2, ICl2−
9. Which of the following set is not correct:– (2) Pyramidal : CH3− , NH3, XeO3
(1) SO3, O3, NH4+ all have coordinate bonds •
(3) Trigonal planar : CH3⊕, CH3 , CH3Θ
(2) H2O, NO2, ClO–2 , all are 'V' shape
(4) Tetrahedral : SiH4, XeO4, PCl 4+
molecules
–
(3) I3 , ICl–2 , NO2+ ; all are linear molecules
17. Amongst CO23− , AsO33− , XeO3, ClO3− , BO33−
(4) SF4, SiF4 XeF4 are tetrahedral in shape
and SO23− the non-planar species are :-
10.
3
Hybridisation in XeOF2, XeO2F2 is sp d. But (1) XeO3, ClO3− , SO23− , AsO33−
–2
shape will be respectively: - (2) AsO33− , XeO3, CO3
(1) T, 'V' shape (3) BO33− , CO23− , SO23−
(2) T shape, See-Saw
(3) Both have T shape (4) AsO33− , BO33− , CO23−
(4) T shape, irregular octahedral
18. The type of hybrid orbitals used by chlorine
–
11. The shape of IF4+ will be :- atom in ClO , ClO2− , ClO3− and ClO4− is/are :-
2 3 3
(1) Square planar (1) sp, sp , sp and sp d
3
(2) Tetrahedral (2) sp and sp
(3) Pentagonal bipyramidal (3) Only sp
3

(4) Distorted tetrahedral (4) Only sp

12. In which of following compound, has four 19. Descending order of electronegativity of
3 2
bond pair and one lone pair:- sp , sp & sp hybridised orbitals
2 3 3 2
(1) NH4+ (2) ICl–4 (1) sp , sp, sp (2) sp , sp , sp
2 3 3 2
(3) SF4 (4) XeF4 (3) sp, sp . sp (4) sp, sp , sp
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20. The central atom of which of the following 28. Molecular shapes of SF4, CF4 and XeF4 are :-
molecule is different from other three (1) The same, with 2, 0 and 1 lone pairs of
(1) H2O (2) SO2 electrons on central atom respectively
(3) Cl2O (4) OF2 (2) The same, with 1, 1 and 1 lone pairs of
electrons on central atom respectively
21. Hybridization involves (3) Different, with 0, 1 and 2 lone pairs of
(1) Addition of an electron pair electrons on central atom respectively
(2) Mixing up of atomic orbitals (4) Different, with 1, 0 and 2 lone pairs of
(3) Removal of an electron pair electrons on central atom respectively.
(4) Separation of orbitals
29. Which one of the following is a correct pair
22. Which is having maximum % of s character with respect to molecular formula of xenon
(1) C—C (2) C = C compound and hybridization state of xenon
(3) C ≡ C (4) None in it:
3
(1) XeF4, sp (2) XeF2, sp
23. The AsF5 molecule is trigonal bipyramidal. 3
(3) XeF2, sp d (4) XeF4, sp
2

The hybrid orbitals used by the atoms for

bonding are: 30. The xenon compound(s) that are iso-
(1) dx2 − y2 ,dz2 ,s,px ,p y (2) dxy ,s,px ,p y ,pz structural with IBr2− and BrO3− respectively
(3) s,px ,p y ,pz ,dz2 (4) dx2 − y2 , s,px ,p y are :
(1) Linear XeF2 and pyramidal XeO3
24. Among the following compounds the one (2) Bent XeF2 and pyramidal XeO3
that is polar and has the central atom with (3) Bent XeF2 and planar XeO3
3
sp hybridisation is : (4) Linear XeF2 and tetrahedral XeO3
(1) H2CO3 (2) SiF4
(3) BF3 (4) HClO2 31. If equatorial plane in PCl5 molecule is the
X–Y plane, the orbital hybridizing to
25. Which of the following molecule has produce axial bonds :
regular geometry – (1) pz , dz2 (2) dx2 − y2 , dxy
(1) H2O (2) PF3
(3) py, px (4) dz2 , dxz
(3) SF6 (4) XeF6
32. VSEPR theory does not state :
26. In sulphate ion the oxidation state of
(1) the order of repulsion between
sulphur is +6 and hybridization state of
different pair of electrons is lp – lp > lp
sulphur is :
2 3 2
– bp > bp – bp (lp = lone pair electrons,
(1) sp (2) sp d bp = bond pair of electrons)
3 3
(3) dsp (4) sp (2) as the number of lone pair of electrons
on central atom increase, the deviation
27. The hybridization states of the central in BA from normal BA (Bond-Angle)
atoms of the ions I3− , ICl 4− and ICl 2− are also increase
respectively : (3) the number of lone pair on O in H2O is 1
2 2
(1) sp , dsp , sp
3
while on N in NH3 is 2.
3 3 2 3
(2) sp d, sp d and sp d (4) the structure of Xenon-fluorides and
3 3
(3) sp d, sp d, dsp
2 Xenon-oxyfluorides could be explained
2 on the basis of VSEPR theory.
(4) sp, sp, dsp
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33. Which of the following molecules is having 38. Which of the following is not isostructural
2pπ–3pπ bond as well as 2pπ–3dπ bond? with SiCl4?
(1) SO3 (2) NO3− (1) NH4+ (2) SCl4
(3) SO 2−
4 (4) PO 3−
4 (3) SO24− (4) PO34−

34. Which is the hybridisation on the central

39. In which of the following pairs of
atom of SiO2? 2
2 molecules/ions, the central atoms have sp
(1) sp (2) sp
3 3
hybridization :
(3) sp (4) sp d
(1) BF3 and NH2− (2) NO2− and NH3
35. Which of the following has pπ – dπ bonding: (3) BF3 and NO2− (4) NH2− and H2O
(1) NO3− (2) SO3−2
40. Considering the state of hybridization of
(3) BO3−3 (4) CO3−2
carbon atoms, find out the molecule among
36. Among the following pair which of the two the following which is linear:
species are not isostructural: (1) CH3–CH=CH–CH3
(1) PF6− and SF6 (2) CH3–C≡C–CH3
(2) SiF4 and SF4 (3) CH2=CH–CH2C≡CH
(3) IO and XeO3
− (4) CH3–CH2–CH2–CH3
3

(4) BH4− and NH4+

41. Which one of the following pairs is
37. Which of the following molecules has isostructural (i.e. having the same shape
trigonal planer geometry: and hybridization):
(1) NH3 (2) BF3 (1) NF3 and BF3 (2) BF4− and NH4+
(3) PCl3 (4) IF3 (3) BCl3 and BrCl3 (4) NH3 and NO3−

 Bond Angle:
1. Hybridisation:
2 3 3 2
sp > sp > sp > sp d
180° 120° 109°28’ 90°
2. Number of lone pair:
If hybridisation of the central atom is same but number of lone pair is different than more is the number
of lone pair less is the bond angle.
Ex. CH4 NH3 H2O
3 3 3
hybridisation sp sp sp
lone pair .P. = 0 .P. = 1 .P. = 2
Bond angle 109°28’ 107° 104.5°
3. Size or electronegativity of central atom:
When hybridisation is same and no. of lone pair is same but central atom is different than see the
electronegativity of central atom. More is the electronegativity more is the bond angle.
Ex. NH3 PH3 AsH3 SbH3
3
hybridisation sp no no no
Bond angle 107° 93° 92° 91°
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4. Size or electronegativity of terminal atom:
Hybridisation same, lone pair same, central atom same but terminal atom is different than greater is
the size of the terminal atom greater will be the bond angle. Only in case of fluorine the
electronegativity factor is considered, due to greater electronegativity of the fluorine atom the bond
angle for it comes out to be smallest (due to smaller bond-bond pair repulsions)

Ex. PF3 PCl3 PBr3 PI3

3 3 3 3
hybridisation sp sp sp sp
B.A. 98° 100° 101° 102°

NCERT questions (solve yourself)

Q. 1 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in
water is less than that of ammonia. Discuss.

Exercise 1.4
1. When the hybridization state of carbon 5. Among the following orbital/bonds, the
3 2
atom changes from sp , sp and sp, the angle is minimum between:
angle between the hybridized orbitals. 3
(1) sp bonds
(1) decrease considerably (2) px and py orbitals
(2) increase progressively
(3) H–O–H bond in water
(3) decrease gradually
(4) sp bonds
(4) all of these

6. In which of the following change, adjacent

2. Which order of decreasing bond angle is
correct: - bond angle increases?
–
(1) CCl4 > BF3 > NO2+ (1) BeF2 + 2F 
→ BeF4−2
(2) NH3 > NCl3 > NBr3 (2) SiF4 + 2F 
–
→ SiF6−2
(3) Br2O > Cl2O > OF2 –
(3) BF3 + F 
→ BF4−
(4) PCl3 > PBr3 > PI3
+
(4) NH3 + H 
→ NH4+
3. By the hybridization of one s & one p
orbitals it will be obtained 7. Which part(s) has same bond angles?
(1) Two orbitals mutually at 90ºangle
(a) BF3, BCl3
(2) two orbitals mutually at 180ºangle
(3) Two orbitals mutually at 120ºangle (b) PO4−3 , SO4−2
(4) Two orbitals mutually at 150º angle (c) BF3, PF3
⊕
(d) NO2 , N2O
4. In compounds X, all the bond angles are
exactly 109º28’, X is : Correct option are :
(1) Chloromethane (1) a, b, d
(2) Carbon tetrachloride (2) b, d
(3) Iodoform (3) b, c, d
(4) Chloroform (4) a, d
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Molecular Orbital Theory (MOT)
Molecular orbital (MO) theory was developed by F. Hund and R.S. Mulliken in 1932. The salient features of
this theory are :
(i) The electrons in a molecule are present in the various molecular orbitals as the electrons of
atoms are present in the various atomic orbitals.
(ii) The atomic orbitals of comparable energies and proper symmetry combine to form molecular
orbitals.
(iii) While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is
influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus,
an atomic orbital is monocentric while a molecular orbital is polycentric.
(iv) The number of molecular orbital formed is equal to the number of combining atomic orbitals.
When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding
molecular orbital while the other is called antibonding molecular orbital.
(v) The bonding molecular orbital has lower energy and hence greater stability than the
corresponding antibonding molecular orbital.
(vi) Just as the electron probability distribution around a nucleus in an atom is given by an atomic
orbital, the electron probability distribution around a group of nuclei in a molecule is given by a
molecular orbital.
(vii) The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle
obeying the Paulis exclusion principle and the Hunds rule. But the filling order of these
molecular orbitals is always experimentally decided, there is no rule like (n + l) rule in case of
atomic orbitals.

Formation of Molecular Orbitals : Linear Combination of Atomic Orbitals(LCAO):

Antibonding orbital
Higher energy than
that of atomic orbitals
Molecular
Increasing energy

Atomic orbital Atomic

orbital orbital
σ*=ΨA+ΨB

ΨA ΨB

σ=ΨA+ΨB
Bonding orbital
lower energy than
that of atomic orbitals

Fig.: Formation of bonding (σ) and antibonding (σ*) molecular orbitals by the linear combination of
atomic orbitals ΨA and ΨB centered on two atoms A and B respectively.

The linear combination of atomic orbitals to form molecular orbitals takes place only if the following
conditions are satisfied:
1. The combining atomic orbitals must have the same or nearly the same energy.
2. The combining atomic orbitals must have the same symmetry about the molecular axis.
3. The combining atomic orbitals must overlap to the maximum extent.

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Energy Level Diagram for Molecular Orbitals:
The energy levels of molecular orbitals have been determined experimentally from spectroscopic data
for homonuclear diatomic molecules of second row elements of the periodic table. The increasing order
of energies of various molecular orbitals for O2 and F2 is given below :
<
σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) (π*2px = π*2pz) < σ*2pz.

The increasing order of energies of various molecular orbitals for Be2, B2, C2, N2 etc., is :
σ 1s < σ* 1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz

The important characteristic feature of this order is that the energy of σ2pz molecular orbital is higher
than that of π2px and π2py molecular orbitals.
Molecular
Antibonding sigma
orbitals
Molecular orbital
Atomic Atomic . . . .
+ – +
Energy

σ*1s ++
orbital orbital 1s 1s σ*1s
Bonding sigma
1s 1s
Molecular orbital
. . . .
σ1s + – + ++
1s 1s σ1s
(a)

Molecular
Antibonding sigma
orbitals
Molecular orbital
Atomic Atomic – –+ – + –– –– + – –+
Energy

orbital σ*2pz orbital 2pz 2pz σ*1s

2pz 2pz Bonding sigma

Molecular orbital
σ2pz
– –+ + + –– – – +– –
(b) 2pz 2pz σ2pz

Molecular Antibonding sigma

orbitals Molecular orbital
Atomic Atomic +. – +. +. .–
Energy

π*2px – – – +
orbital orbital
2px 2px π*2px
Bonding sigma
2px 2px Molecular orbital
+. – +. .+.
π2px – – –
2px 2px π2px
(c)
Fig. Bonding and antibonding molecular orbitals formed through combinations of
(a) 1s atomic orbitals; (b) 2pz atomic orbitals and (c) 2px atomic orbitals.

Electronic configuration and molecular behaviour:

Bond order:
Bond order (B.O.) is defined as one half the difference between the number of electrons present in
the bonding and the antibonding orbitals i.e., Bond order = ½ (Nb – Na)
A positive bond order (i.e., Nb > Na) means a stable molecule while a negative (i.e., Nb < Na) or zero
(i.e., Nb = Na) bond order means an unstable molecule.

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 CHEMISTRY Chemical Bonding
Nature of the bond:
Integral bond order values of 1, 2 or 3 correspond to single, double or triple bonds respectively.
Bond-length:
The bond order between two atoms in a molecule may be taken as an approximate measure of the bond
length. The bond length decreases as bond order increases.
Magnetic nature:
If all the molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic (repelled
by magnetic field) e.g., N2 molecule. However, if one or more molecular orbitals are singly occupied it
is paramagnetic (attracted by magnetic field), e.g., O2 molecule.
2 2 2 2 2 2 2
Nitrogen molecule (N2): (σ1s) (σ*1s) (σ2s) (σ*2s) (π2p x = π2p y ) (σ2pz)

σ*2pz

π*2px = π*2py
2p 2p

σ2pz

π2px = π2py

σ*2s

2s 2s
N(AO) N(AO)

σ2s
N2(MO)
M.O. Energy level diagram for N2 molecule
N2 has a triple bond according to both the Lewis and the molecular orbital models.
The bond order of N2 is 1/2(10 – 4) = 3. It contains one sigma and two π bonds.
2 2 2 2 2 2 2 1 1
Oxygen molecule (O2): O2 : (σ1s) (σ*1s) (σ2s) (σ*2s) (σ2pz) (π2p x = π2p y ) (π*2px = π*2p y)

σ*2pz

π*2px = π*2py

π2px = π2py

σ2pz

σ*2s

A(AO) O(AO)

σ2s
O2(MO)
M.O. Energy level diagram for O2 molecule
– +
Note: Observed B.O. of CO, CN NO = 3
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NCERT questions (solve yourself):

Q. 1 Write the important conditions required for the linear combination of atomic orbitals to form
molecular orbitals.
Q. 2 Use molecular orbital theory to explain why the Be2 molecule does not exist.
Q. 3 Compare the relative stability of the following species and indicate their magnetic
properties;
O2 , O2+ , O2− (superoxide), O22− (peroxide)
Q. 4 What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O–2 .

Exercise 1.5
1. The ion that is isoelectronic with CO and 7. In which of the following set, the value of
having same bond order is :- bond order will be 2.5:-
– +2
(1) CN (2) O2+ (1) O2+ , NO, NO , CN
+2 –
(3) O2− (4) N2+ (2) CN, NO , CN , F2
+2 –
(3) O2+ , NO , O2+2 , CN
2. Bond order in C2+ is:- (4) O2−2 , O2− , O2+ , O2
1 2
(1) (2)
2 3 8. Which of the following ion is diamagnetic: -
3 (1) O2−1 (2) O2−2 (3) O2 (4) O2+1
(3) (4) 1
2
9. When two atomic orbitals combine, it forms -
3. Which of the following molecules have (1) One molecular orbital
unpaired electron: - (2) Two molecular orbitals
(1) H2 (2) H3O
+
(3) Two bonding molecular orbitals
(3) H2O (4) HeH (4) Two anti-bonding molecular orbitals

4. In the following which of the two are 10. The paramagnetic property of oxygen is
paramagnetic well explained by: -
(a) N2 (b) CO (1) Molecular orbital theory
(c) B2 (d) NO2 (2) Resonance theory
(3) Valence bond theory
Correct answer is:-
(4) VSEPR theory
(1) a and c (2) b and c
(3) c and d (4) b and d
11. Which of the following species which has
+ the highest bond order and shortest bond
5. Increasing order of bond length in NO, NO
– length:
and NO is:- + 2+ –
– + + – NO, NO , NO , NO
(1) NO > NO > NO (2) NO < NO < NO
+ – + –
(1) NO only
(3) NO < NO < NO (4) NO < NO = NO (2) Bond order of NO is highest and bond
2+
length of NO is shortest
6. Bond order of Li2 is:- +
(3) NO only
(1) 0 (2) 1 2+
(3) 2 (4) 3 (4) NO only

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12. The bond order of CO molecule on the basis 19. In which of the following pairs of
of molecular orbital theory is molecules/ions, both the species are not
(1) Zero (2) 2 likely to exist?
(3) 3 (4) 1
(1) H22+ , He2

13. The energy of σ2s orbital is greater, than (2) H2−2 ,He22+
σ1s* orbital because (3) H2+ ,He22+
(1) σ2s orbital is bigger than σ*1s orbital (4) H2− ,He22−
(2) σ2s orbital is a bonding orbital whereas
σ*1s is an antibonding orbital 20. Which of the following species is
(3) σ2s orbital has a greater value of n than paramagnetic with one unpaired electron?
σ*1s orbital (1) BaO2 (2) CaC2
(4) None (3) O2 (4) KO2

14. The molecule having one unpaired electron

21. Which of the following LCAO represent
is
formation of bonding molecular orbital?
(1) NO (2) CO
(3) CN
–
(4) O2 (1) + — +
1s 1s

15. The no. of antibonding electron pair in O2− (2) – + + – +

2pz 2pz
is
(1) 4 (2) 3 + –
(3)
(3) 8 (4) 10 – — +
2Px 2Px
16. Maximum bond energy will be shown by the – +
(4) + +
species 2pz
2S
(1) O2+ (2) O2
(3) O–2 (4) O2−2 22. Which of the following is not the correct
statement regarding C2 molecule?
17. N2 and O2 are converted into monoanions, (1) It has total 12 electrons, out of which
– –
N2− and O2− respectively. Which of the 8e occupy bonding orbital and 4e anti
bonding orbitals
following statements is wrong ?
(2) C2 molecules has been found to exist in
(1) In N2− , N-N bond weakens
vapour phase
(2) In O2− , O-O bond order increases
(3) C2 molecule containing double bonds
(3) In O2− , O-O bond order decreases
and both are π-bonds
(4) N2− becomes paramagnetic (4) C2 Molecule is paramagnetic

18. In a homonuclear molecule which of the 23. Stability of the species Li2, Li2− and Li2+
following set of orbitals are degenerate?
increases in the order is :
(1) σ2s and σ1s
(1) Li2− < Li2 < Li2+
(2) π2p and π2py
x (2) Li2 < Li2+ < Li2−
(3) σ2px and σ2pz (3) Li2− < Li2+ < Li2
(4) σ2pz and π *2px
(4) Li2 < Li2− < Li2+

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Polarity of Bonds & Dipole Moment
In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond
between two atoms, there is some ionic character.
When a covalent bond is formed between two similar atoms, for example in H2, O2, Cl2, N2 or F2 the
shared pair of electrons is equally attracted by the atoms. As a result, electron pair is situated exactly
between the two identical nuclei. The bond so formed is called nonpolar covalent bond. Contrary to this
in case of a heteronuclear molecule like HF, the shared electron pair between the two atoms gets
displaced more towards fluorine since the electronegativity of fluorine is far greater than that of
hydrogen. The resultant covalent bond is a polar covalent bond.
As a result of polarisation, the molecule possesses the dipole moment which can be defined as the
product of magnitude of the partial charge (δ+ or δ–) developed on any of the covalently bonded atoms
and the distance between two atoms.
Dipole moment (µ) = Magnitude of charge (q) × distance of separation (d)
Dipole moment is usually expressed in Debye units (D).
The conversion factors are
–30
• 1 D = 3.33564 × 10 C-m, where C is coulomb and m is meter.
–18
• 1 Debye = 1 × 10 e.s.u. cm.
Further dipole moment is a vector quantity and is depicted by a small arrow with tail on the positive
centre and head pointing towards the negative centre. For example, the dipole moment of HF may be
represented as
H F
The shift in electron density is represented by crossed arrow ( ) above the Lewis
structure to indicate the direction of the shift.

(A) Diatomic molecules:

(i) Dipole moment is a vector quantity so, it has magnitude as well as direction. It is often represented
by an arrow with its tail at the positive centre and head pointing towards the negative end
(+ → –)
(ii) As a polar diatomic molecule posse only one polar bond, the dipole moment of that molecule is
equal to the dipole moment of the polar bond e.g. in case of HCl, the molecular dipole moment is
equal to the dipole moment of H–Cl bond i.e. 1.03 D. Thus.
+δ –δ
H Cl µ = 1.03 D
(B) Polyatomic molecules :
(i) As a polyatomic molecule has more than one polar bond, the dipole moment is equal to the
resultant dipole moment of all the individual bonds.
(ii) For example, dipole moment of H2O is 1.84 D which is equal to the resultant dipole moment of two
O–H bonds.
δ–
O
δ+H δ+
H
µ=1.84D
(iii) Think about CO2 molecule though C = O bond is polar due to electronegativity difference but the
resultant dipole moment of molecule is zero as the individual dipole moments are of equal moment
and opposite sign. This shows that CO2 is a linear molecule.

O C O

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(iv) Again, in case of symmetrical molecules such as BF3, CH4 and CCl4 the molecular dipole moment
is found to be zero. This is due to the fact that individual dipole moments cancel out on account of
symmetry of the molecule.
δ+ δ–
H Cl
δ–
F
δ–
F B δ– C
δ+
C
δ– δ+
F δ+ δ– Cl
δ+ H Cl
H δ+
δ–
Cl
H

Some important points about dipole moment :

Type of Example Dipole Geometry
Molecule Moment, µ(D)
Molecule (AB) HF 1.78 Linear
HCl 1.07 Linear
HBr 0.79 Linear
HI 0.39 Linear
H2 0 Linear
Molecule (AB2) H2O 1.85 Bent
H2S 0.95 Bent
CO2 0 Linear
Molecule (AB3) NH3 1.47 Trigonal-pyramidal
NF3 0.23 Trigonal-pyramidal
BF3 0 Trigonal-planar
Molecule (AB4) CH4 0 Tetrahedral
CHCl3 1.04 Tetrahedral
CCl4 0 Tetrahedral

NCERT questions (solve yourself)

Q. 1 Write the significance/applications of dipole moment.

Exercise 1.6
1. Which statement is correct: - 2. Which of the following species are
(1) All the compounds having polar bonds, symmetrical: -
have dipole moment (a) XeF4 (b) XeF6
(2) SO2 is non-polar
(c) SO2 (d) NH1+
4
(3) H2O molecule is non-polar, having polar
Correct answer is:-
bonds
(1) a and b (2) b and c
(4) PH3 is polar molecule having non-polar
(3) c and d (4) a and d
bonds
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3. Which of the following molecule have zero 11. Species having zero dipole moment :-
dipole moment: - (1) XeF4
(1) BF3 (2) CH2Cl2 (2) 1,2,4 trichloro benzene
(3) NF3 (4) SO2 (3) SF4
(4) CH2Cl2
4. The dipole moment of NH3 is:-
(1) Less than dipole moment of NCl3 12. PCl5 is non-polar because :-
(2) Higher than dipole moment of NCl3 (1) P – Cl bond is non-polar
(3) Equal to the dipole moment of NCl3 (2) Its dipole moment is zero
(4) None of these (3) P – Cl bond is polar
(4) P & Cl have equal electronegativity
5. Which of the following species are polar :
13. Which contains both polar and non-polar
(a) C6H6 (b) XeF2
bonds?
(c) SO2 (d) SF4
(1) NH4Cl (2) HCN
(e) SF6
(3) H2O2 (4) CH4
Correct answer is :-
(1) (b) and (d) (2) (a), (b) and (e) 14. The correct order of dipole moment is :
(3) (a) and (e) (4) (c) and (d) (1) CH4 < NF3 < NH3 < H2O
(2) NF3 < CH4 < NH3 < H2O
6. Which of the following order of polar
molecules is correct:- (3) NH3 < NF3 < CH4 < H2O
(1) HF > NH3 > PH3 (4) H2O < NH3 < NF3 < CH4
(2) CH4 > NH3 > H2O
15. Arrange the following compound in order of
(3) CH3Cl < CH2Cl2 < CHCl3
increasing dipole moment :
(4) BF3 > BeF2 > F2
(I) 1, 3, 5-Trichloro benzene
(II) 1, 2, 4-Trichloro benzene
7. Which set of molecules is polar :- (III) 1, 2, 3, 4-Tetrachloro benzene
(1) XeF4, IF7, SO3 (IV) P-dichloro benzene
(2) PCl5, C6H6, SF6 (1) I = IV < II < III (2) IV < I < II < III
(3) SnCl2, SO2, NO2 (3) IV = I < III < II (4) IV < II < I < III
(4) CO2, CS2, C2H6
16. Which of the following are arranged in the
8. An example of a polar covalent molecule is decreasing order of dipole moment.
(1) CH3Cl, CH3Br, CH3F
(1) S8 (2) H— O—H
+ – (2) CH3Cl, CH3F, CH3Br
(3) Na Cl (4) F — F
(3) CH3Br, CH3Cl, CH3F
9. Which of the following molecules has polar (4) CH3Br, CH3F, CH3Cl
character?
(1) CO2 (2) CH4 17. H2O is polar, whereas BeF2 is not because:
(3) PF5 (4) NH3 (1) H2O is angular and BeF2 is linear
(2) The electronegativity of F is greater
10. Which of the following has symmetrical than that of O
structure: (3) H2O involves hydrogen bonding
(1) PCl3 (2) CH2Cl2 whereas BeF2 is a discreate molecule
(3) CHCl3 (4) CCl4 (4) H2O is linear and BeF2 is angular

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Hydrogen Bond
Nitrogen, oxygen and fluorine are the highly electronegative elements. When they are attached to a
hydrogen atom to form covalent bond, the electrons of the covalent bond are shifted towards the more
electronegative atom. This partially positively charged hydrogen atom forms a bond with the other
more electronegative atom. This bond is known as hydrogen bond and is weaker than the covalent bond.
For example, in hf molecule, the hydrogen bond exists between hydrogen atom of one molecule and
fluorine atom of another molecule as depicted below:
– – – H δ+ – F δ– – – – H δ+ – F δ– – – – H δ+ – F δ–
Here, hydrogen bond acts as a bridge between atoms which holds one atom by covalent bond and the
other by hydrogen bond. Hydrogen bond is represented by a dotted line (– – –) while a solid line
represents the covalent bond. Thus, hydrogen bond can be defined as the attractive force which binds
hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule.

Cause of Formation of Hydrogen Bond:

When hydrogen is bonded to strongly Electronegative element ‘x’, the electron pair Shared between
the two atoms moves far away From hydrogen atom. As a result the hydrogen Atom becomes highly
electropositive with Respect to the other atom ‘x’. Since there is Displacement of electrons towards x,
the Hydrogen acquires fractional positive charge (δ+) while ‘x’ attain fractional negative charge (δ¯).
This results in the formation of a polar molecule having electrostatic force of attraction which can be
represented as :
Hδ+–Xδ– ––– Hδ+ – Xδ– –––Hδ+ – Xδ–
The magnitude of H-bonding depends on the physical state of the compound. it is maximum in the solid
state and minimum in the gaseous state. thus, the hydrogen bonds have strong influence on the
structure and properties of the compounds.

Types of H-Bonds:
There are two types of H-bonds
(i) Intermolecular hydrogen bond (ii) Intramolecular hydrogen bond

(1) Intermolecular Hydrogen Bond:

it is formed between two different molecules of the same or different compounds. For example, H-bond
in case of HF molecule, alcohol or water molecules, etc.

(2) Intramolecular Hydrogen Bond: it is formed when hydrogen atom is in between the two highly
electronegative (F, O, N) atoms present within the same molecule. For example, in o-nitrophenol the
hydrogen is in between the two oxygen atoms.
O
N
O
H
O
Fig. : Intramolecular hydrogen bonding in o-nitrophenol molecule

NCERT questions (solve yourself)

Q. 1 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?

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Exercise 1.7
1. The hydrogen bond is strongest in:- 6. Glycerol is more viscous than glycol the
(1) O – H - - - S (2) S – H - - - O reason is :-
(3) F – H - - - F (4) O – H - - - O (1) Higher molecular wt.
(2) More covalent
2. H2O boils at higher temperature than H2S, (3) More extent of hydrogen bonding
because it is capable of forming:- (4) Complex structure
(1) Ionic bonds
(2) Covalent bonds 7. The correct order of volatility is:-
(3) Hydrogen bonds (1) NH3 < H2O
(4) Metallic bonds (2) p– nitro phenol < o– nitro phenol
(3) CH3OH > CH3 – O– CH3
3. Intermolecular hydrogen bonds are not
(4) HF > HCl
present in:-
(1) CH3CH2OH (2) CH3COOH
8. Maximum number of H–bonding is shown
(3) C2H5NH2 (4) CH3OCH3 by
(1) H2O (2) H2Se
4. In which of the following molecule, the
shown hydrogen bond is not possible:- (3) H2S (4) HF

H H 9. Intramolecular H–bond :-
N H N H (1) Decreases Volatility
(1)
(2) Increases melting point
H H
(3) Increases viscosity
OH (4) Increases vapour pressure
O
10. The incorrect order of decreasing boiling
(2) C H points is
(1) NH3 > AsH3 > PH3
(2) H2O > H2Se > H2S
O H
(3) Br2 > Cl2 > F2
(4) CH4 > GeH4 > SiH4
(3)
11. Incorrect order of viscosity :-
O N O (1) H2SO4 > HNO3
Cl H
(2) H2O > CH3OH
O
(4) Cl C C H (3) o-nitro phenol > p-nitro phenol
O (4) Glycol > ether
Cl H
12. The crystal lattice of ice is mostly formed
5. Correct order of volatility is:- by:-
(1) HF > HCl > HBr > HI (1) Ionic forces
(2) HCl > HBr > HI > HF (2) Covalent bonds
(3) HI > HBr > HCl > HF (3) Inter molecular H–bonds
(4) HBr < HCl < HI < HF (4) Covalent as well as H–bonds
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13. Which of the following has strongest intra 20. Strongest hydrogen bond is shown by :
molecular hydrogen bonding :- (1) Water
OH OCH3 (2) Ammonia
COOH COOH (3) Hydrogen fluoride
(1) (2) (4) Hydrogen sulphide

OH OCH3 21. The boiling point of p-nitrophenol is higher

than that of o-nitrophenol because :
(1) NO2 group at p-position behaves in a
(3) (4)
different way from that at o-position
(2) Intramolecular hydrogen bonding
COOH COOH exists in p-nitrophenol
(3) There is intermolecular hydrogen
14. Acetic acid exists as dimer in benzene due to:- bonding in p-nitrophenol
(1) Condensation reaction (4) p-nitrophenol has a higher molecular
(2) Hydrogen bonding weight than o-nitrophenol
(3) Presence of carboxyl group
(4) None of the above 22. NH3 has abnormally high boiling point
because it has :
15. Increasing strength of H–bonding (X---H – X) (1) Alkaline nature (2) Distorted shape
in S, O, F, Cl, N is :– 3
(3) sp - Hybridisation (4) Hydrogen bonding
(1) Cl, S, N, O, F (2) N, Cl, S, O, F
(3) S, Cl, N, O, F (4) S, N, Cl, O, F 23. Which of the following is soluble in water ?
(1) CS2 (2) C2H5OH
16. Which of the following does not form a (3) CCl4 (4) CHCl3
hydrogen bond with water
(1) (CH3)2CO (2) CH3CN 24. KF combines with HF to form KHF2. The
(3) CH3OH (4) C2H6 compound contains the species :
+ — + + —
(1) K , F and H (2) K , F and HF
17. In which of the following compounds + — +
(3) K and [HF2] (4) [KHF] and F2
intramolecular hydrogen bond is present
(1) Ethyl alcohol (2) water
25. Which of the following compounds show
(3) Salicylaldehyde (4) Hydrogen sulphide
intramolecular hydrogen bonding :
18. Hydrogen bonding is formed in compounds (A) o - nitrophenol
containing hydrogen and _________ : (B) p - nitrophenol
(1) Highly electro-negative atoms (C) phenol
(2) Highly electro-positive atoms (D) salicylaldehyde
(3) Metal atoms with d-orbitals occupied (1) A & B (2) A & C
(4) Metalloids (3) A & D (4) B & C

19. The intermolecular force in hydrogen 26. The boiling point of a compound is raised
fluoride is due to : by:
(1) Dipole-induced dipole interactions (1) intermolecular hydrogen bonding
(2) Dipole-dipole interactions (2) High volatility
(3) Hydrogen bond (3) intramolecular hydrogen bonding
(4) Dispersion interaction (4) non-polarity
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SUMMARY
1. Introduction to Chemical Bonding:
Atoms combine to form molecules due to the tendency to attain a stable electronic configuration (octet
rule).
Types of chemical bonds: Ionic bond, Covalent bond, and Coordinate bond.

2. Ionic Bond (Electrovalent Bond):

Formed by the transfer of electrons from a metal to a non-metal.
Ex.: NaCl (Sodium chloride) → Na⁺ + Cl⁻
Factors affecting ionic bond formation:
Ionization Enthalpy: Lower ionization energy favors cation formation.
Electron Gain Enthalpy: More negative value favors anion formation.
Lattice Energy: Higher lattice energy stabilizes the ionic compound.

3. Covalent Bond:
Formed by the sharing of electrons between two non-metallic atoms.
Ex.: H₂, O₂, Cl₂, CH₄.
Types of Covalent Bonds:
Single Bond (Ex., H₂, Cl₂)
Double Bond (Ex., O₂)
Triple Bond (Ex., N₂)
Polarity in Covalent Bonds: Due to the difference in electronegativity (Ex., HCl).

4. Lewis Structure (Electron Dot Structure):

Representation of valence electrons in molecules.
Octet rule: Atoms tend to complete their octet (8 valence electrons) except for H (duplet).
Exceptions to the octet rule: Incomplete octet (BeCl₂, BCl₃), Expanded octet (SF₆, PCl₅).

5. VSEPR Theory (Valence Shell Electron Pair Repulsion Theory):

Predicts the shape of molecules based on electron repulsions.
Shapes of Molecules:
Linear (180°) – BeCl₂, CO₂
Trigonal Planar (120°) – BF₃
Tetrahedral (109.5°) – CH₄
Trigonal Bipyramidal (90°, 120°) – PCl₅
Octahedral (90°) – SF₆

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6. Valence Bond Theory (VBT) and Hybridization:
Covalent bonds form due to atomic orbital overlap.
Types of Hybridization:
sp (Linear, Ex:- CO2, BeCl₂)
sp² (Trigonal planar, Ex:- BF₃)
sp³ (Tetrahedral, Ex:- CH₄)
sp³d (Trigonal bipyramidal, Ex:- PCl₅)
sp³d² (Octahedral, Ex:- SF₆)

7. Molecular Orbital Theory (MOT):

Molecular orbitals form by the combination of atomic orbitals.
Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
Higher bond order means greater bond strength and stability.
Example: O₂ has a bond order of 2 (double bond).

8. Hydrogen Bonding:
A weak electrostatic attraction between a hydrogen atom and an electronegative atom (N, O, F).
Types:
Intermolecular (between molecules, Ex:- H₂O, HF)
Intramolecular (within a molecule, Ex:- o-nitrophenol)
Important in determining properties like boiling points and solubility.
This chapter lays the foundation for understanding molecular structures, chemical interactions, and
bonding theories essential for advanced chemistry topics.

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Exercise 2
1. Pick out the incorrect statement :- 6. Select the correct statement(s) :
(1) sp d hybridisation involves dx2 − y2 orbital
3 (1) Bond angle H–N–H in NH3 is less than
H–O–H bond angle in water
(2) Generally Hybridised orbital form σ- (2) In PCl5 molecules, axial and equatorial
bond when overlaps with other orbitals. bonds are of equal length
(3) SF2 molecule is more polar than CS2 (3) H2S is having high boiling point than H2O
(4) o-nitrophenol is more volatile than p- (4) N2 has higher bond order than N2+ .
nitrophenol.
7. Which of the following compound exist –
(1) B2F6 (2) B2H6
2. The nature of π-bond in perchlorate ( ClO4− )
(3) Al2F6 (4) B2Cl6
ion is:-
(1) O(dπ) — Cl(pπ) (2) O(pπ) — Cl(pπ) 8. In which of the following molecule electron
(3) O(pπ) — Cl(dπ) (4) O(dπ) — Cl(dπ) of 2s in carbon is not promoted to 2p sub
shell:
(1) HCN (2) CO2 (3) CS2 (4) CO
3. The true statements from the following are
(a) PH5, NCl5 and BiCl5 do not exist 9. Which of the following pair has electron
(b) I3+ has bent geometry deficient compounds :
(1) B2H6, AlCl3 (2) C2H6, Al2Cl6
(c) XeF4 is non-polar molecule
(3) SF2, Cl2O (4) NaBH4, ICl
(d) CO and C22− has same bond order
(e) MgCl2 has more ionic nature than MgO 10. An oxide of chlorine which is an odd
(1) a, b, c, d (2) b, c, d electron molecule is :
(3) a, b, c (4) a, b, c, d, e (1) ClO2 (2) Cl2O6
(3) Cl2O7 (4) Cl2O
4. In which of the following compounds ionic
11. Which of the following does not act as lewis
& covalent bonds are present
acid?
(1) KCl (2) SO2 (1) BF3 (2) SnCl4 (3) CCl4 (4) SF4
(3) NaOH (4) CH4
12. Boron compounds behave as Lewis acids
5. Which of the following statement is not because of their :
(1) Acidic nature
correct?
(2) Covalent nature
(1) N2H4 is pyramidal around each N-atom
(3) Electron defficient character
(2) The shape of N(SiH3) is trigonal planar (4) Ionising property
around N-atom
(3) In ClO3− , NH3 and XeO3 the hybridisation 13. Which of the following is not correctly
matched?
and the number of lone pairs on central (1) ClF3 excited state I of Cl
atoms are same (2) XeO2F2 excited state II of Xe
(4) The B-F bond length in BF3 is 1.35 Å
(3) BeF4−2 excited state I of Be
much longar than 1.30 Å in BF . −
4 (4) SF6 excited state II of S
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14. Which is the weakest among the following 21. Which force is least sensitive for distance?
types of bonds ? (1) Ion–dipole attraction
(1) Debye force (2) Dipole–induced dipole
(2) Metallic bond (3) Ion-induced dipole
(3) Dipole-dipole bond (4) Dispersion force
(4) Hydrogen bond
22. Which species do not exists :
15. In which molecule the Vander Waals force (1) Al 36− (2) BF4−
(dispersion force) is likely to be the most (3) BeF4−2 (4) CCl 26−
important in determining the m.pt. and b.p.
(1) Br2 (2) CO 23. Which of the following contains Co-
(3) H2S (4) HCl ordinate and covalent bonds :
+ +
(a) N2H5 (b) H3O
16. Covalent molecules are usually held in a (c) HCl (d) H2O
crystal structure by : Correct answer is :
(1) Dipole-dipole attraction (1) a & d (2) a & b
(2) Electrostatic attraction (3) c & d (4) Only a
(3) Hydrogen bond
(4) Van-der waal’s attraction 24. In Co-ordinate bond, the acceptor atoms
must essentially contain in its valency shell
17. In solid argon the atoms are held together : an orbital :
(1) by ionic bonds (1) With paired electron
(2) by hydrogen bonds (2) With single electron
(3) by vander Waals forces (3) With no electron
(4) by hydrophobic bonds (4) With three electron

18. Which of the following is incorrectly 25. Correct statement regarding this reaction
matched? BF3 + NH3 → [F3B ← NH3]
(1) KCl + H2O = dipole –dipole attraction. (1) Hybridization of N is changed
(2) CH3 C CH3 + CH3 C N (2) Hybridization of B is changed
(3) B–F bond length increases
O
(4) (2) & (3) both
Keesom attraction
(3) Xe + H2O–Debye attraction
26. The pair of compounds which can form a
(4) CF4 + CF4–London force
co–ordinate bond is :
(1) (C2H5)3B and (CH3)3N
19. Which substance has the strongest London (2) HCl and HBr
dispersion forces? (3) BF3 and NH3
(1) SiH4 (2) CH4
(4) (1) & (3) both
(3) SnH4 (4) GeH4
27. In the neutralization process of NH3 and
20. Which among the following attraction is AlCl3 the compound formed will have the
strongest? bonding :
+
(1) HF.... H2O (2) Na .... HCl (1) Ionic (2) Covalent
(3) H2O....Cl2 (4) Cl–Cl...Cl–Cl (3) Co-ordinate (4) Hydrogen

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28. In the co-ordinate valence : 33. The correct order of B-F bond length
(1) electrons are equally shared by the follows the sequence :
atoms (1) BF3 < BF2–OH < BF2–NH2 < BF4−
(2) electrons of one atom are shared by two
(2) BF2–NH2 < BF2–OH < BF3 < BF4−
atoms
(3) hydrogen bond is formed (3) BF3 < BF4− < BF2–OH < BF2–NH2
(4) none of the above (4) BF3 < BF2–NH2 < BF2–OH < BF4−

29. The compound containing co-ordinate bond 34. Which of the following is not possible due
is : to back bonding:
(1) H2SO4 (2) O3 (1) State of hybridisation may change
(3) SO3 (4) All of these (2) Bond order increases
(3) Bond angle always decreases
30. In BF3, the B-F bond length is 1.30 Å, when (4) Lewis acidic strength decreases
BF3 is allowed to be treated with Me3N, it
35. Among following molecule N-Si bond
forms an adduct, Me3N → BF3, the bond
length is shortest :
length of B-F in the adduct is :
(1) N(SiH3)3
(1) Greater than 1.30 Å
(2) Smaller than 1.30 Å (2) NH(SiH3)2
(3) Equal to 1.30 Å (3) NH2(SiH3)
(4) None of these (4) All have equal N-Si bond length

31. Which of the following structure correctly 36. The geometry with respect to the central
represents the boron trifluoride molecule : atom of the following molecules are :
F F F N(SiH3)3, Me3N, (SiH3)3P
(1) (1) Planar, pyramidal, planar
B B B
(2) planar, pyramidal, pyramidal
F F F F F F (3) pyramidal, pyramidal, pyramidal
F F
(4) Pyramidal, planar, pyramidal
(2)
B B
F F F F 37. In which of the following compounds B-F
F F F bond length is shortest ?
(1) BF4− (2) BF3 ← NH3
(3) B +
B +
B +

– – – – (3) BF3 (4) BF3 ← N(CH3)3

F F F F F F
–
F F F 38. Back bonding is absent in :
(4) (1) (SiH3)3N (2) H3Si–N=C=O
B B+ B
F F
– (3) O(SiH3)2 (4) P(SiH3)3
F F F F

39. Oxygen molecule is :

32. Trisilylamine has a :
(1) Planar geometry (1) Diamagnetic with no unpaired electrons
(2) Tetrahedral geometry (2) Diamagnetic with two unpaired electrons
(3) Pyramidal geometry (3) Paramagnetic with two unpaired electrons
(4) None of these (4) Paramagnetic with no unpaired electron

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40. Among the following ions, the pπ - dπ 48. The ONO angle is maximum in :-
– –
overlap could be present in : (1) NO3 (2) NO2
+
(1) NO3− (2) PO34− (3) NO2 (4) NO2
(3) CO23− (4) NO2−
49. Which of the following would have a
permanent dipole moment
41. Nitrogen form N2 and phosphorous form P2, but (1) BF3 (2) SF4
it convert into P4, at a instant the reason is - (3) SiF4 (4) XeF4
(1) Triple bond present between
phosphorous atom 50. The correct sequence of increasing
covalent character is represented by
(2) pπ – pπ bonding is weak (1) BeCl2 < NaCl < LiCl
(3) pπ – pπ bonding is strong (2) NaCl < LiCl < BeCl2
(4) Multiple bond form easily (3) BeCl2 < LiCl < NaCl
(4) LiCl < NaCl < BeCl2
42. Which of the following molecule has
highest bond energy ? 51. Which of the following is the electron
(1) F — F (2) N — N deficient molecule –
(3) C — C (4) O — O (1) C2H6 (2) SiH4
(3) PH3 (4) B2H6
43. Paramagnetic species is –
(1) KO2 (2) SiO2 52. The surface tension of which of the
(3) TiO2 following liquid is maximum :
(4) BaO2
(1) H2O (2) C6H6
(3) CH3OH (4) C2H5OH
44. In BrF3 molecule, the lone pairs occupy
equatorial positions to minimize : 53. Among the following molecules
(1) Lone pair lone pair repulsion only (i) XeO3 (ii) XeOF4 (iii) XeF6
(2) Lone pair-bond pair repulsion only Those having same number of lone pairs on
(3) Bond pair - bond pair repulsion only Xe are:
(4) Lone pair-lone pair repulsion and lone (1) (i) and (ii) only (2) (i) and (iii) only
pair-bond pair repulsion (3) (ii) and (iii) only (4) (i), (ii) and (iii)

45. Among the following pair in which of the 54. The number of unpaired electrons in a
paramagnetic diatomic molecule of an
two species are not isostructural is :-
element with atomic number 16 is :
(1) PF6− and SF6 (2) SiF4 and SF4 (1) 1 (2) 2 (3) 3 (4) 4
(3) IO3− and XeO3 (4) BH4− and NH4+
55. Which of the following is not a correct
statement?
46. In a regular octahedral molecule, MX6 the
(1) Multiple bonds are always shorter than
number of X – M – X bonds at 180° is :- corresponding single bonds
(1) Four (2) Three (2) The electron-deficient molecules can
(3) Two (4) Six act as Lewis acids
(3) The canonical structures have no real
47. Shape of O2F2 is similar to that of : existence
(1) C2F2 (2) H2O2 (4) Every AB5 molecule does in fact have
square pyramid structure.
(3) H2F2 (4) C2H2
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56. Which of the following species has a linear 62. The angular shape of ozone molecule (O3)
shape? consists of :
(1) O3 (2) NO–2 (3) SO2 (4) NO+2 (1) 1 sigma and 1 pi bonds
(2) 2 sigma and 1 pi bonds
57. The electronegativity difference between N (3) 1 sigma and 2 pi bonds
and F is greater than that between N and H (4) 2 sigma and 2 pi bonds
yet the dipole moment of NH3 (1.5 D) is
63. The correct order of increasing bond angles
larger than that of NF3 (0.2 D). This is
in the following triatomic species is:-
because: (1) NO+2 <NO2 <NO–2 (2) NO+2 <NO–2 <NO2
(1) in NH3 the atomic dipole and bond
(3) NO+2 <NO–2 <NO2 (4) NO–2 <NO2 <NO+2
dipole are in the opposite directions
whereas in NF3 these are in the same 64. In which of the following molecules/ions
direction BF3, NO2− , NH2− and H2O, the central atom is
(2) in NH3 as well as in NF3 the atomic 2
sp hybridized?
dipole and bond dipole are in the same
direction (1) BF3 and NO2− (2) NO2− and NH2−
(3) in NH3 the atomic dipole and bond (3) NH2− and H2O (4) NO2− and H2O
dipole are in the same direction
whereas in NF3 these are in opposite 65. According to MO theory which of the
directions following lists ranks the nitrogen species in
(4) in NH3 as well as NF3 the atomic dipole terms of increasing bond order?
and bond dipole are in opposite (1) N2− < N22− < N2 (2) N2− < N2 < N22−
directions (3) N22− < N2− < N2 (4) N2 < N22− < N2−

58. In which of the following molecules are all 66. In the case of alkali metals, the covalent
the bonds not equal? character decreases in the order :
(1) NF3 (2) CIF3 (3) BF3 (4) AIF3 (1) MI > MBr > MCl > MF
(2) MCl > MI > MBr > MF
59. The correct order of C–O bond length (3) MF > MCl > MBr > MI
among CO, CO-2 (4) MF > MCl > MI > MBr
3 , CO2 is :

(1) CO < CO-2

3 < CO2 (2) CO-2
3 < CO2 < CO 67. What is the dominant intermolecular force
(3) CO < CO2 < CO -2
3 (4) CO2 < CO < CO -2
3
or bond that must be overcome in
converting liquid CH3OH to a gas?
60. In which of the following hydration energy (1) London or dispersion force
is higher than lattice energy : (2) Hydrogen bonding
(1) MgSO4 (2) RaSO4 (3) Dipole-dipole interaction
(3) BaSO4 (4) SrSO4 (4) Covalent bonds

61. Four diatomic species are listed below in 68. The straight chain polymer is formed by :
different sequences. Which of these (1) Hydrolysis of (CH3)2SiCl2 followed by
presents the correct order of their condensation polymerisation
increasing bond order ? (2) Hydrolysis of (CH3)3SiCl followed by
(1) C2- + – condensation polymerisation
2 <He2 <NO < O2
(3) Hydrolysis of CH3SiCl3 followed by
(2) He+2 < O–2 <NO < C2–
2
condensation polymerisation
(3) O–2 <NO < C22– <He+2 (4) Hydrolysis of Si (CH3)4 by addition
(4) NO < C2- – +
2 < O2 <He2 Polymerisation
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69. Among the following which is the strongest 79. Alkaline earth metal nitrates on heating
oxidising agent ? decompose to give :
(1) Cl2 (2) F2 (3) Br2 (4) I2 (1) M(NO2) and O2 only
70. N2 and O2 are converted into monocations, (2) MO, N2 and O2
(3) MO, NO2 and O2
N2+ and O2+ respectively. Which of the
(4) MO and NO2 only
following is wrong ?
(1) In N2+ , N-N bond weakens 80. Which of the following halide can be
(2) In O , the O-O bond order increases
+ converted into anhydrous form on heating?
2
(1) BeCl2.H2O (2) KCl.6H2O
(3) In O2+ , paramagnetism decreases
(3) AlCl3.6H2O (4) FeCl3.6H2O
(4) N2+ becomes diamagnetic
81. Which of the following is not correct for IIA
71. In a molecule of water following bonds are
present- metals?
(1) Two hydrogen bond (1) Thermal stability of carbonate, sulphate
(2) Two ionic bond increases on moving down the group
(3) One covalent and one ionic bond (2) Thermal stability of all type of oxides
(4) Two covalent bond decrease on moving down the group
(3) Solubility of Hydroxides increases on
72. In dry ice the bond present between two
moving down the group
molecules is
(4) Bicarbonate do not exist in solid state.
(1) Ionic bond (2) Covalent bond
(3) Hydrogen bond (4) Vander Waal 82. The correct order of increasing thermal stability
73. Which of the following halides has the of K2CO3, MgCO3 CaCO3 and BeCO3 is:
highest melting point ? (1) BeCO3 < MgCO3 < CaCO3 < K2CO3
(1) NaCl (2) KCl (3) NaBr (4) NaF (2) MgCO3 < BeCO3 <CaCO3 < K2CO3
74. Which of the following does not give an (3) K2CO3 < MgCO3 < CaCO3 < BeCO3
oxide on heating : (4) BeCO3 < MgCO3 < K2CO3 < CaCO3
(1) MgCO3 (2) Li2CO3 (3) ZnCO3 (4) K2CO3
83. Which of the following alkaline earth metal
75. Which decomposes on heating ?
sulphates has hydration enthalpy higher
(1) NaOH (2) KOH
than the lattice enthalpy?
(3) LiOH (4) RbOH
(1) SrSO4 (2) CaSO4
76. Which of the following forms metal oxide (3) BeSO4 (4) BaSO4
on heating?
(1) Na2CO3 (2) Li2CO3 84. The products obtained on heating LiNO3
(3) K2SO4 (4) NaHCO3 will be:
(1) LiNO2 + O2 (2) Li2O + NO2 + O2
77. Increasing order of stability of :
I. K2CO3 II. MgCO3 (3) Li3N + O2 (4) Li2O + NO + O2
III. Na2CO3 85. Which of the following metal has stable
(1) I < II < III (2) II < III < I carbonates ?
(3) II < I < III (4) I < III < II (1) Na (2) Mg (3) Al (4) Si
78. Which of the following carbonate will not
decompose on heating ? 86. The most stable carbonate is :
(1) BaCO3 (2) ZnCO3 (1) Li2CO3 (2) BeCO3
(3) Na2CO3 (4) Li2CO3 (3) CaCO3 (4) BaCO3
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Exercise 3
1. Isostructural species are those which have 9. Which of the following statement is not
the same shape. Among the given species correct from the view point of molecular
identify the isostructural pairs. orbital theory?
(1) [NF3 and BF3] (2) [ BF4− and NH4+ ] (1) Be2 is not a stable molecule.

(3) [BCl3 and BrCl3] (4) [NH3 and NO3− ] (2) He2 is not stable but He2+ is expected to
exist.
2. The types of hybrid orbitals of nitrogen in (3) Bond strength of N2 is maximum
NO2+ , NO3− and NH4+ respectively are amongst the homonuclear diatomic
molecules belonging to the second
expected to be :
3 2 2 3 period.
(1) sp, sp and sp (2) sp, sp and sp
2 3 2 3
(4) The order of energies of molecular orbitals
(3) sp , sp and sp (4) sp , sp and sp in N2 molecule is : σ2s < σ*2s < σ2pz <
(π2px = π2py) < (π*2px = π*2py) < σ2pz
3. In NO3− ion, the number of bond pairs and
lone pairs of electrons on nitrogen atom are: 10. Which of the following options represents
(1) 2, 2 (2) 3, 1 the correct bond order :
(3) 1, 3 (4) 4, 0
(1) O2− > O2 > O2+ (2) O2− < O2 < O2+
(3) O2− > O2 < O2+ (4) O2− < O2 > O2+
4. Which of the following species has
tetrahedral geometry ?
11. Which of the following statements is
(1) BH4− (2) NH2− (3) CO23− (4) H3O+
correct?
(1) In the formation of dioxygen from
5. Which molecule/ion out of the following
oxygen atoms 10 molecular orbitals will
does not contain unpaired electrons ?
be formed.
(1) N2+ (2) O2 (3) O22− (4) B2 (2) All the molecular orbitals in the
dioxygen will be completely filled.
6. In which of the following substances will (3) Total number of bonding molecular
hydrogen bond be strongest? orbitals will not be same as total
(1) HCl (2) H2O number of anti-bonding orbitals in
(3) HI (4) H2S dioxygen.
(4) Number of filled bonding orbitals will
7. Which of the following angle corresponds be same as number of filled anti
2 bonding orbitals.
to sp hybridisation?
(1) 90º (2) 120º
12. Which of the following molecular orbitals
(3) 180º (4) 109º
has maximum number of nodal planes?
(1) σ*1s (2) σ*2pz
8. Which of the following order of energies of
molecular orbitals of N2 is correct ? (3) π2px (4) π*2py
(1) (π2py) < (σ2pz) < (π*2px) ≈ ( π*2py)
13. Which of the following pair is expected to
(2) (π2py) > (σ2pz) > (π*2px) ≈ ( π*2py)
have the same bond order?
(3) (π2py) < (σ2pz) > (π*2px) ≈ ( π*2py) (1) O2, N2 (2) O2+ , N2−
(4) (π2py) > (σ2pz) < (π*2px) ≈ ( π*2py) (3) O2− , N2+ (4) O2− , N2−

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14. In which of the following molecules, σ2pz 21. CO is isoelectronic with

+
molecular orbitals is filled after π2px and (A) NO (B) N2
π2py molecular orbitals? (C) SnCl2 (D) NO2−
(1) O2 (2) Ne2 (3) N2 (4) F2 (1) A and B (2) B and C
(3) C and D (4) A, B and C
15. Polarity in a molecule and hence the dipole 22. Which of the following species have the
moment depends primarily on electronegativity same shape?
of the constituent atoms and shape of a
(A) CO2 (B) CCl4
molecule. Which of the following has the
highest dipole moment? (C) O3 (D) NO2−
(1) CO2 (2) HI (3) H2O (4) SO2 (1) A and B (2) B and C
(3) C and D (4) A, B and C
16. Hydrogen bonds are formed in many
23. Which of the following statements are
compounds e.g., H2O, HF, NH3. The boiling
correct about CO23− ?
point of such compounds depends to a large 3
extent on the strength of hydrogen bond (A) The hybridisation of central atom is sp .
and the number of hydrogen bonds. The (B) Its resonance structure has one C–O
correct decreasing order of the boiling single bond and two C=O double bonds.
points of above compounds is : (C) The average formal charge on each
(1) HF > H2O > NH3 (2) H2O > HF > NH3 oxygen atom is 0.67 units.
(D) All C–O bond lengths are equal.
(3) NH3 > HF > H2O (4) NH3 > H2O > HF
(1) A and B (2) B and C
17. In PO34− ion the average formal charge on (3) C and D (4) All
the oxygen atom of P–O bond is : 24. Dimagnetic species are those which contain
(1) +1 (2) –1 no unpaired electrons. Which among the
(3) –0.75 (4) +0.75 following are dimagnetic?
(A) N2 (B) N22−
18. Number of π bonds and σ bonds in the
(C) O2 (D) O22−
following structure is–
H H (1) A and B (2) B and C
H (3) C and D (4) A and D
25. Species having same bond order are :
H H (A) N2 (B) N2−
H H
(C) F2+ (D) O2−
(1) 6, 19 (2) 4, 20
(1) A and B (2) B and C
(3) 5, 19 (4) 5, 20
(3) C and D (4) A, B and C
19. Which of the following have identical bond order?
– +
26. Which of the following statements are not
(A) CN (B) NO correct?
(C) O2− (D) O22− (A) NaCl being an ionic compound is a good
(1) A and B (2) B and C conductor of electricity in the solid state.
(3) C and D (4) A, B and C (B) In canonical structures there is a difference
in the arrangement of atoms.
20. Which of the following attain the linear (C) Hybrid orbitals form stronger bonds than
structure: pure orbitals.
+
(A) BeCl2 (B) NCO (D) VSEPR Theory can explain the square
(C) NO2 (D) CS2 planar geometry of XeF4 .
(1) A and B (2) B and C (1) A and B (2) B and C
(3) C and D (4) A, B and D (3) C and D (4) A, B and C
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27. Assertion (A) : Sodium chloride formed by 30. Match the species in Column I with the type
the action of chlorine gas on sodium metal of hybrid orbitals in Column II.
is a stable compound.
Column I Column II
Reason (R) : This is because sodium and (Molecules) (Hybridisation)
chloride ions acquire octet in sodium
(a) SF4 (i) sp3d2
chloride formation. (b) IF5 (ii) d2sp3
(1) A and R both are correct, and R is the (c) NO2+ (iii) sp3d
correct explanation of A (d) NH4+ (iv) sp3
(2) A and R both are correct, but R is not the (v) sp
correct explanation of A.
(3) A is true but R is false. (1) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
(2) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
(4) A and R both are false.
(3) (a)-(iii), (b)-(i), (c)-(v), (d)-(iv)
(4) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
28. Assertion (A) : Though the central atom of
3 31. Match the species in Column I with the
both NH3 and H2O molecules are sp
geometry/shape in Column II.
hybridised, yet H–N–H bond angle is
greater than that of H–O–H. Column I Column II
Reason (R) : This is because nitrogen atom (Species) (Shape/Geometry)
has one lone pair and oxygen atom has two (a) H3O+ (i) Linear
lone pairs. (b) HC ≡ CH (ii) Angular
(1) A and R both are correct, and R is the (c) ClO2− (iii) Tetrahedral

correct explanation of A (d) NH4+ (iv) Trigonal

bipyramidal
(2) A and R both are correct, but R is not the
(v) Pyramidal
correct explanation of A.
(3) A is true but R is false. (1) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
(4) A and R both are false. (2) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
(3) (a)-(iii), (b)-(i), (c)-(v), (d)-(iv)
(4) (a)-(v), (b)-(i), (c)-(ii), (d)-(iii)
29. Assertion (A): Among the two O–H bonds in
H2O molecule, the energy required to break 32. Match the species in Column I with the
bond order in Column II.
the first O–H bond and the other O–H bond
is the same. Column I Column II
Reason (R) : This is because the electronic (Species) (Bond order)
environment around oxygen is the same (a) NO (i) 1.5
even after breakage of one O–H bond. (b) CO (ii) 2.0
(c) O2− (iii) 2.5
(1) A and R both are correct, and R is the
(d) O2 (iv) 3.0
correct explanation of A
(2) A and R both are correct, but R is not the (1) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
correct explanation of A. (2) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
(3) A is true but R is false. (3) (a)-(iii), (b)-(i), (c)-(v), (d)-(iv)
(4) A and R both are false. (4) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

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33. Match the items given in Column I with 34. Match the shape of molecules in Column I
examples given in Column II. with the type of hybridisation in Column II.
Column I Column II
Column I Column II (Shape of (Hybridisation)
(a) Hydrogen (i) C molecules)
bond (a) Tetrahedral (i) sp2
(b) Resonance (ii) LiF (b) Trigonal (ii) sp
(c) Ionic solid (iii) H2 (c) Linear (iii) sp3
(d) Covalent solid (iv) HF
(d) Trigonal (iv) sp3d
(v) O3
bipyramidal

(1) (a)-(i), (b)-(iv), (c)-(ii), (d)-(v) (1) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
(2) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) (2) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
(3) (a)-(iii), (b)-(i), (c)-(v), (d)-(iv) (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
(4) (a)-(iv), (b)-(v), (c)-(ii), (d)-(i) (4) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

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Exercise 4 (Previous Year's Questions)

1. Some of the properties of the two species, 8. The correct order of increasing bond length
NO–3 and H3O+ are described below. Which of C–H, C–O, C–C and C=C is :-
one of them is correct :- [AIPMT 2010] [AIPMT Pre.-2011]
(1) Isostructural with same hybridization (1) C–H < C=C < C–O < C–C
for the central atom. (2) C–C < C=C < C–O < C–H
(2) Isostructural with different hybridization (3) C–O < C–H < C–C < C=C
for the central atom. (4) C–H < C–O < C–C < C=C
(3) Similar in hybridization for the central
atom with different structures. 9. Which of the following structures is the
(4) Dissimilar in hybridization for the most preferred and hence of lowest energy
central atom with different structures. for SO3? [AIPMT MAINS 2011]
O O
2. In which of the following molecules the (1) S (2) S
central atom does not have sp3
hybridization :- [AIPMT 2010]
(1) SF4 −
(2) BF4 (3) NH4+ (4) CH4 O
S
(3) (4) S
3. Which one of the following species does not O
exist under normal conditions?
[AIPMT 2010] 10. Decreasing order of bond angle is :
(1) Li2 +
(2) Be2 (3) Be2 (4) B2 [AIIMS 2011]
(1) BeCl2 > NO2 > SO2 (2) BeCl2 > SO2 > NO2
4. The correct order of increasing bond angle (3) SO2 > BeCl2 > NO2 (4) SO2 > NO2 > BeCl2
in the following species is :- [AIPMT 2010]
11. Which of the following species contains
(1) ClO2− < Cl2O < ClO2 (2) Cl2O < ClO2 < ClO2−
three bond pairs and one lone pair around
(3) ClO2 < Cl2O < ClO2− (4) Cl2O < ClO2– < ClO2 the central atom? [AIPMT Pre.- 2012]
−
(1) NH2 (2) PCl3
5. In which one of the following species the
(3) H2O (4) BF3
central atom has the type of hybridisation
which is not the same as that present in the
12. The pair of species with the same bond
other three ? [AIPMT 2010]
order is: [AIPMT Pre.- 2012]
(1) PCl5 (2) SF4
(1) NO, CO (2) N2, O2
(3) I3− (4) SbCl52− +
(3) O22− , B2 (4) O2+ , NO
6. Which of the following has the minimum
bond length? [AIPMT Pre.-2011] 13. Bond order of 1.5 is shown by:
[AIPMT Pre.- 2012]
(1) O2
+
(2) O2
–
(3) O2–
2 (4) O2 2−
(1) O2 (2) O2
7. Which of the two ions from the list given (3) O2+ (4) O2−
below that have the geometry that is
explained by the same hybridization of 14. In the replacement reaction
orbitals, NO–2 , NO–3 , NH2– , NH+4 , SCN– ? Cl + MF CF + MI
The reaction will be most favourable if M
[AIPMT Pre.-2011]
happens to be : [AIPMT MAINS-2012]
(1) NO2 and NO3
– –
(2) NH+4 and NO–3
–
(1) Rb (2) Li
(3) SCN and NH+2 (4) NO–2 and NH+2 (3) Na (4) K

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15. During change of O2 to O2− ion, the electron 24. Which of the following molecules has the
adds on which one of the following orbitals? maximum dipole moment? [AIIMS 2013]
[AIPMT MAINS-2012] (1) CO2 (2) CH4
(1) σ* orbital (2) σ orbital (3) NH3 (4) NF3
(3) π* orbital (4) π orbital
25. Which one of the following species has
16. Which of the following is electron- plane triangular shape? [AIIMS 2014]
−
deficient? [NEET-UG 2013] (1) N3 (2) NO–3
(1) PH3 (2) (CH3)2 (3) NO–2 (4) CO2
(3) (SiH3)2 (4) (BH3)2
26. Decreasing order of stability of O2, O–2 , O+2
17. Which one of the following molecules and O2–
2 is : [RE-AIPMT 2015]
contains no π bond ? [NEET-UG 2013] (1) O2 > O > O
+ 2–
>O –
2 2 2
(1) NO2 (2) CO2
(2) O > O
–
2
2–
2 > O > O2
+
2
(3) H2O (4) SO2
(3) O > O2 > O–2 > O2–
+
2 2

18. XeF2 is isostructural with : [NEET-UG 2013] (4) O22– > O–2 > O2 > O+2
(1) BaCl2 (2) TeF2
27. The correct bond order in the following
(3) ICl–2 (4) SbCl3
species is : [AIPMT 2015]
(1) O2 < O2 > O2
2+ – +
(2) O2 < O–2 < O2+
+
2
19. Dipole induced dipole interactions are
(3) O–2 < O+2 < O22+ (4) O2+
2 < O2 < O2
+ –
present in which of the following pairs :
[NEET-UG 2013]
(1) SiF4 and He atoms (2) H2O and alcohol 28. Which of the following pairs of ions are
isoelectronic and isostructural?
(3) Cl2 and CCl4 (4) HCl and He atoms
[AIPMT 2015]
(1) ClO3 , CO3
– 2–
(2) SO2–
3 , NO3
–

20. Which of the following is a polar molecule?

(3) ClO–3 , SO2– (4) CO2–
3 , SO3
2–
[NEET-UG 2013] 3

(1) XeF4 (2) BF3

(3) SF4 (4) SiF4 29. Solubility of the alkaline earth’s metal
sulphates in water decreases in the
sequence : [AIPMT 2015]
21. Which of the following is paramagnetic?
(1) Ca > Sr > Ba > Mg (2) Sr > Ca > Mg > Ba
[NEET-UG 2013]
+ (3) Ba > Mg > Sr > Ca (4) Mg > Ca > Sr > Ba
(1) NO (2) CO
–
(3) O2
–
(4) CN 30. SF4 & XeF2 shape respectively are
[AIPMT 2015]
22. Identify the correct order of solubility in (1) Linear and distorted tetrahedral
aqueous medium : [NEET-UG 2013] (2) See-saw and linear
(1) Na2S > ZnS > CuS (2) CuS > ZnS > Na2S (3) T-shape and trigonal bipyramidal
(3) ZnS > Na2S > CuS (4) Na2S < CuS > ZnS (4) Tetrahedral and linear

23. Total number of Antibonding electrons 31. In CIF3 lone pair are present at approx.
present in O2 will be : [AIIMS 2013] [AIPMT 2015]
(1) 6 (2) 8 (1) 180º (2) 120º
(3) 4 (4) 2 (3) 90º (4) 60º
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32. The correct geometry and hybridization for 37. Predict the correct order among the following
XeF4 are : [NEET-II 2016] : [NEET-I 2016]
(1) Planar triangle, sp d
3 3 (1) Ione pair - lone pair > lone pair - bond
3 2 pair > bond pair - bond pair
(2) square planar, sp d
3 2
(2) lone pair- lone pair > bond pair-bond
(3) octahedral, sp d pair > lone pair - bond pair
3
(4) trigonal bipyramidal, sp d (3) bond pair - bond pair > lone pair - bond
pair > lone pair - lone pair
33. Among the following which one is a wrong (4) lone pair - bond pair > bond pair - bond
statement? [NEET-II 2016] pair > lone pair - lone pair
(1) SeF4 and CH4 have same shape
(2) I+3 has bent geometry 38. Match the compounds given in column I
with the hybridization and shape given in
(3) PH5 and BiCl5 do not exist
column II and mark the correct option.
(4) pp-dp bond is present in SO2 [NEET-I 2016]

Column I Column II
34. Which one of the following compounds
(a) XeF6 (i) Distorted
shows the presence of intramolecular
Octahedral
hydrogen bond? [NEET-II 2016]
(b) XeO3 (ii) Square planar
(1) Cellulose
(2) Concentrated acetic acid (c) XeOF4 (iii) Pyramidal
(3) H2O2 (d) XeF4 (iv) Square Pyramidal
(4) HCN
(a) (b) (c) (d)
(1) (i) (iii) (iv) (ii)
35. Which of the following fluoro-compounds is (2) (i) (ii) (iv) (iii)
most likely to behave as a Lewis base? (3) (iv) (iii) (i) (ii)
[NEET-II 2016] (4) (iv) (i) (ii) (iii)
(1) CF4 (2) SiF4
(3) BF3 (4) PF3 39. Minimum lone pair on central atom will be
present in: [AIIMS 2016]
36. Consider the molecules CH4, NH3 and H2O. (1) XeO3 (2) XeF4
Which of the given statements is false? (3) CIF3 (4) HOCl
[NEET-I 2016]
(1) The H–C–H bond angle in CH4, the 40. Which of the following has paramagnetic
character in gaseous state: [AIIMS 2016]
H–N–H bond angle in NH3, and the H–O–H 2–
(1) S (2) S2
bond angle in H2O are all greater than 90°
(3) S8 (4) S6
(2) The H–O–H bond angle in H2O is larger
than the H–C–H bond angle in CH4.
41. In which of the following molecule dipole
(3) The H–O–H bond angle in H2O is smaller moment is more than zero but less than one?
than the H–N–H bond angle is NH3. [AIIMS 2016]
(4) The H–C–H bond angle in CH4 is larger (1) NH3 (2) NF3
than the H–N–H bond angle in NH3. (3) BeF2 (4) BF3
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42. Select the incorrect statement : 49. Match the interhalogen compounds of
[AIIMS 2016] column-I with the geometry in column II
(1) Excitation of electron is not essential and assign the correct code: [NEET 2017]
for hybridisation.
(2) Only half-filled atomic orbital are Column I Column II
participate in hybridisation. (a) XX’ (i) T–shape
(3) Bond energy of hybrid orbitals are more (b) XX’3 (ii) Pentagonal
than unhybrid atomic orbital bipyramidal
(4) Without hybridisation geometry of (c) XX’5 (iii) Linear
molecule is uncertain.
(d) XX’7 (iv) square –pyramidal
43. Which of the following has tetrahderal (v) Tetrahedral
structure? [AIIMS 2016]
Code :
(1) XeF4 (2) H3PO4
(a) (b) (c) (d)
(3) SF4 (4) CIF3
(1) (iii) (iv) (i) (ii)
(2) (iii) (i) (iv) (ii)
44. Which of the following statement is true
(3) (v) (iv) (iii) (ii)
regarding H-bond? [AIIMS 2016]
(4) (iv) (iii) (ii) (i)
(1) It is a directional bond
(2) bond strength is similar to O-H bond
50. Which one of the following pairs of species
(3) It is attraction between hydrogen and
electronegative atom like as F, O, N have the same bond order? [NEET 2017]
+
(4) It does not require minimum one lone (1) CO, NO (2) O2, NO
pair of electrons at electronegative atom (3) CN , CO
–
(4) N2, O2−

45. Which of the following is correct?

51. Consider the following species :
[AIIMS 2016] + –
(1) N2O is a coloured gas CN , CN , NO and CN
Which one of these will have the highest
(2) CO is an acidic oxide
bond order? [NEET 2018]
(3) CO2 is not absorbed in CsOH –
(1) NO (2) CN
(4) N2O is a neutral oxide +
(3) CN (4) CN
46. Which of the following molecule has more
than one lone pair on central atom. 52. Which molecule pair do not have identical
[AIIMS 2016] structure: [AIIMS 2018]
–
(1) SO2 (2) XeF2 (1) I3 BeF2 (2) HClO, SO2
(3) PCl5 (4) IF7 (3) BF3, ICI3 (4) BrF4– ,XeF4

47. The species, having bond angles of 120° is : 53. Which contain at least one e in σ2p
–

[NEET 2017]
bonding MO: [AIIMS 2018]
(1) PH3 (2) CIF3
(1) O2 (2) B2
(3) NCl3 (4) BCl3
(3) C2 (4) Li2

48. Which of the following pairs of compounds

is isoelectronic and isostructural? 54. In which of the following shape is same but
[NEET 2017] hybridization is different: [AIIMS 2018]
–
(1) BeCl2, XeF2 (2) TeI2, XeF2 (1) ICI2 , XeF2 (2) SO2, NO+2
(3) IBr2− , XeF2 (4) IF3, XeF2 (3) SO2, NH–2 (4) CO2, SO2

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Chemical Bonding CHEMISTRY
55. % s-character of N-H bond is maximum in : 62 Which one of the following elements is
[AIIMS 2018] unable to from MF63− ion? [NEET 2018]
(1) N2H2 (2) N2H4 (1) Ga (2) Al
(3) NH3 (4) NH+4 (3) B (4) In

63. In the structure of ClF3, the number of lone

56. For N–3 which statement is wrong
pairs of electrons on central atom 'Cl' is:
[AIIMS 2018]
[NEET 2018]
(1) Isoelectronic with CO2
(1) One (2) Two
(2) NH2OH and N–3 have same O.N. on (3) Four (4) Three
nitrogen atom
(3) N–N bond length are same 64. Which of the following diatomic molecular
(4) HN3 have linear shape species has only π bonds according to
Molecular Orbital Theory? [NEET 2019]
57. Which pair of diatomic species have same (1) O2 (2) N2
bond order ? [AIIMS 2018] (3) C2 (4) Be2
–
(1) B2 , C2 (2) O2 , F2–
2–

65. Which of the following species is not

(3) N+2 , O–2 (4) B2–
2 , C2
stable? [NEET 2019]
2– 2–
(1) [SiF6] (2) [GeCl6]
58. CIF2– , CIF4– find out number of lone pair and 2– 2–
(3) [Sn(OH)6] (4) [SiCl6]
geometry. [AIIMS
2018]
66. Identify the incorrect statement related to
(1) 3 – Linear, 2 – Square planar
PCl5 from the following: [NEET 2019]
(2) 3 – Square planar, 2 – Linear
(1) Three equatorial P-Cl bonds make an
(3) 0 – Linear, 3 – Square planar
angle of 120° with each other
(4) 2 – Linear, 2 – Square planar
(2) Two axial P-Cl bonds make an angle of
180° with each other
59. Which have corrected order of dipole
(3) Axial P-Cl bonds are longer than
moment: [AIIMS 2018]
equatorial P-Cl bonds
(1) SO2 > H2O (2) NF3 > NH3 (4) PCl5 molecule is non-reactive
(3) BF3 < NH3 (4) SO2 < SO3
67. Match the Xenon compounds in Column-I
60. Among CaH2, BeH2, BaH2, the order of ionic with its structure in Column-II and assign
character is [NEET 2018] the correct code: [NEET 2019]
(1) BeH2 < CaH2 < BaH2
(2) CaH2 < BeH2 < BaH2 Column I Column II
(a) XeF4 (i) Pyramidal
(3) BeH2 < BaH2 < CaH2
(b) XeF6 (ii) Square planar
(4) BaH2 < BeH2 < CaH2
(c) XeOF4 (iii) Distorted octahedral
61. Magnesium reacts with element (X) to form (d) XeO3 (iv) square pyramidal
an ionic compound. If the ground state
2 2 3 Code:
electronic configuration of (X) is 1s 2s 2p ,
(a) (b) (c) (d)
the simplest formula for this compound is
(1) (i) (ii) (iii) (iv)
[NEET 2018] (2) (ii) (iii) (iv) (i)
(1) Mg2X3 (2) MgX2 (3) (ii) (iii) (i) (iv)
(3) Mg2X (4) Mg3X2 (4) (iii) (iv) (i) (ii)
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 CHEMISTRY Chemical Bonding
68. Which is the correct thermal stability order 74. Match the compounds of Xe in column I
for H2E (E=O, S, Se, Te and Po)? with the molecular structure in column II.
[NEET II-2020]
[NEET 2019]
(1) H2S < H2O < H2Se < H2Te < H2Po Column I Column II
(2) H2O < H2S < H2Se < H2Te < H2Po (a) XeF2 (i) Square planar
(b) XeF4 (ii) Linear
(3) H2Po < H2Te < H2Se < H2S < H2O
(c) XeO3 (iii) Square pyramidal
(4) H2Se < H2Te < H2Po < H2O < H2S
(d) XeOF4 (iv) Pyramidal

69. Which of the following is the correct order (1) (a)-(ii) (b)-(i) (c)-(iii) (d)-(iv)
of dipole moment? [NEET 2019(ODISHA)] (2) (a)-(ii) (b)-(iv) (c)-(iii) (d)-(i)
(3) (a)-(ii) (b)-(iii) (c)-(i) (d)-(iv)
(1) NH3 < BF3 < NF3 < H2O
(4) (a)-(ii) (b)-(i) (c)-(iv) (d)-(iii)
(2) BF3 < NF3 < NH3 < H2O
(3) BF3 < NH3 < NF3 < H2O 75. BF3 is planar and electron deficient compound.
(4) H2O < NF3 < NH3 < BF3 Hybridization and number of electrons around
the central atom, respectively are
[NEET-2021]
70. The number of hydrogen bonded water 3 3
(1) sp and 4 (2) sp and 6
molecule(s) associated with CuSO4.5H2O is: 2 2
(3) sp and 6 (4) sp and 8
[NEET 2019(ODISHA)]
(1) 3 (2) 1 76. Match List-I with List–II [NEET-2021]
(3) 2 (4) 5 List-I List-II
(a) PCl5 (i) Square pyramidal
71. Identify a molecule which does not exist (b) SF6 (ii) Trigonal planar
[NEET-2020] (c) BrF5 (iii) Octahedral
(1) O2 (2) He2 (d) BF3 (iv) Trigonal
(3) Li2 (4) C2 bipyramidal

Choose the correct answer from the options

72. Which of the following set of molecules will given below.
have zero dipole moment? [NEET-2020] (1) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(1) Boron trifluoride, beryllium difluoride, (2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
carbon dioxide, 1,4-dichlorobenzene (3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(2) Ammonia, beryllium difluoride, water, (4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
1,4-dichlorobenzene
77. The correct sequence of bond enthalpy of
(3) Boron trifluoride, hydrogen fluoride,
'C-X' bond is [NEET-2021]
carbon dioxide, 1,3-dichlorobenzene (1) CH3–F < CH3–Cl < CH3–Br < CH3–I
(4) Nitrogen trifluoride, beryllium difluoride, (2) CH3–F > CH3–Cl > CH3–Br > CH3–I
water, 1,3-dichlorobenzene (3) CH3–F < CH3–Cl < CH3–Br < CH3–I
(4) CH3–Cl < CH3–F < CH3–Br < CH3–I
73. Among the compounds shown below which
one revealed a linear structure? 78. which of the following molecules in non-
[NEET II-2020] polar in nature? [NEET-2021]
(1) NO2 (2) HOCl (1) POCl3 (2) CH2O
(3) O3 (4) N2O (3) SbCl5 (4) NO2

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Chemical Bonding CHEMISTRY
79. Given below are two statements: 83. Amongst the following, the total number of
Statement I: The boiling points of the species NOT having eight electrons around
following hydrides of group 16 elements central atom in its outer most shell, is NH3,
increases in the order AlCl3, BeCl2, CCl4, PCl5 : [NEET-2023]
H2O < H2S < H2Se < H2Te
(1) 3 (2) 2
Statement II: The boiling points of these
(3) 4 (4) 1
hydrides increase with increase in molar
mass.
In the light of the above statements, choose 84. The correct order of dipole moments for
the most appropriate answer from the molecules NH3, H2S, CH4 and HF, is:
options given below: [NEET-2022] [NEET- Manipur-2023]
(1) Both Statement I and Statement II are (1) CH4 > H2S > NH3 > HF
correct
(2) Both Statement I and Statement II are (2) H2S > NH3 > HF > CH4
incorrect (3) NH3 > HF > CH4 > H2S
(3) Statement I is correct but Statement II (4) HF > NH3 > H2S > CH4
is incorrect
(4) Statement I is incorrect but Statement 85. Which one of the following statements is
II is correct
incorrect related to Molecular Orbital
Theory? [NEET- Manipur-2023]
80. Amongst the following which one will have
maximum 'lone pair-lone pair' electron (1) The π * antibonding molecular orbital
repulsions? [NEET-2022] has a node between the nuclei.
(1) ClF3 (2) IF5 (2) In the formation of bonding molecular
(3) SF4 (4) XeF2 orbital, the two electron waves of the
bonding atoms reinforce each other.
81. Which amongst the following is incorrect (3) Molecular orbitals obtained from 2Px
statement? [NEET-2022] and 2Py orbitals are symmetrical around
(1) The bond orders of O2 , O2 , O–2 and O2–
+
2
the bond axis.
are 2.5, 2, 1.5 and 1, respectively. (4) A π-bonding molecular orbital has larger
(2) C2 molecule has four electrons in its two electron density above and below the
degenerate π molecular orbitals. internuclear axis.
(3) H2+ ion has one electron.
(4) O2+ ion is diamagnetic. 86. Intramolecular hydrogen bonding is
present in [NEET-2024]
82. The correct order of energies of molecular NO2
orbitals of N2 molecule, is: [NEET-2023] (1)
* *
(1) σ 1s < σ 1s < σ 2s < σ 2s < (π2px = π
* * *
2py) < σ 2pz < (π 2px = π 2py) < σ 2pz OH
* *
(2) σ 1s < σ 1s < σ 2s < σ 2s < σ 2pz < (π (2) HF
* *
2px = π 2py) < (π 2px = π 2py) < σ 2pz
* NO2
* * * (3)
(3) σ 1s < σ 1s < σ 2s < σ 2s < σ 2pz < σ
* *
OH
2pz < (π 2px = π 2py) < (π 2px = π 2py) NO2
* *
(4) σ 1s < σ 1s < σ 2s < σ 2s < (π2px = π (4)
* * *
2py) < (π 2px = π 2py) < σ 2pz < σ 2pz HO

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87. Match List I with List II [NEET-2024] 90. Identify the correct answer. [NEET-2024]
(1) Dipole moment of NF3 is greater than
List-I List-II
that of NH3.
Compound Shape/geometry
(2) Three canonical forms can be drawn for
A. NH3 I Trigonal
pyramidal CO23− ion.
B. BrF5 II Square planar (3) Three resonance structures can be
C. XeF4 III Octahedral drawn for ozone
(4) BF3 has non-zero dipole moment.
D. SF6 IV Square
pyramidal
91. Which of the following molecules has "NON
Choose the correct answer form the options
ZERO" dipole moment value ?
given below
[RE-NEET-2024]
(1) A-III, B-IV, C-I, D-II (2) A-II, B-III, C-IV, D-I
(1) CCI4
(3) A-I, B-IV, C-II, D-III (4) A-II, B-IV, C-III, D-I
(2) HI
88. Match List I with List II. [NEET-2024] (3) CO2
List I List II (4) BF3
(Molecule) (Number and types
of bond/s between 92. Arrange the following compounds in
two Corbon atoms) increasing order of their solubilities in
A. ethane I one σ-bond and chloroform: [RE-NEET-2024]
two π-bonds NaCl, CH3OH, cyclohexane, CH3CN
B. ethene II two π-bonds (1) NaCl <CH3CN < CH3OH < Cyclohexane
C. Carbon III one σ-bond (2) CH3OH <CH3CN <NaCl < Cyclohexane
molecule, C2 (3) NaCl <CH3OH < CH3CN <Cyclohexane
D. ethyne IV one σ-bond and (4) Cyclohexane < CH3CN <CH3OH < NaCl
one π-bond
Choose the correct answer from the options 93. Identify the incorrect statement about PCl5
given below: [RE-NEET-2024]
(1) A-III, B-IV, C-II, D-I (2) A-III, B-IV, C-I, D-II (1) PCl5 possesses two different Cl-P-Cl
(3) A-I, B-IV, C-II, D-III (4) A-IV, B-III, C-II, D-I bond angles.
89. Given below are two statements: (2) All five P-Cl bonds are identical in
length.
[NEET-2024]
(3) PCl5 exhibits sp³d hybridisation
Statement I: The boiling point of hydrides
of Group 16 elements follows the order H2O (4) PCl5 consists of five P-Cl (sigma) bonds.
> H2Te > H2Se > H2S.
Statement II: On the basis of molecular 94. Identify the incorrect statement.
mass, H2O is expected to have lower boiling [RE-NEET-2024]
point than the other members of the group (1) PEt3 and AsPh3 as ligands can form dπ-
but due to the presence of extensive H- dπ bond with transition metals.
bonding in H2O, it has higher boiling point. (2) The N-N single bond is as strong as the
In the light of the above statements, choose the P-P single bond.
correct answer from the options given below: (3) Nitrogen has unique ability to form
(1) Statement I is true, but Statement II is false. pπ -pπ multiple bonds with nitrogen,
(2) Statement I is false, but Statement II is true. carbon and oxygen.
(3) Both Statement I and Statement II are true. (4) Nitrogen cannot form dπ-pπ bond as
(4) Both Statement I and Statement II are false. other heavier elements of its groups

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ANSWER KEY

Exercise 1.1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 3 4 3 1 4 2 2 3 2 2 2 1 2 3 4 4 4 2 1
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 4 3 4 2 1 3 1 2 1 4 4 3 3 2 1 1 3 4 2 1
Que. 41 42 43 44 45
Ans. 2 1 2 4 4

Exercise 1.2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 3 3 3 3 4 3 2 3 2 3 1 2 2 1 3 3 1 4 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Ans. 3 1 4 1 2 4 1 2 4 2 1 3 1 3 4 1

Exercise 1.3
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 2 2 3 3 3 2 1 3 4 2 4 3 3 2 2 3 1 3 3 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 2 3 3 4 3 4 2 4 3 1 1 3 1 3 2 2 2 2 3 2
Que. 41
Ans. 2

Exercise 1.4
Que. 1 2 3 4 5 6 7
Ans. 2 3 2 2 2 4 1

Exercise 1.5
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 3 4 3 2 2 1 2 2 1 3 3 3 1 2 1 2 2 1 4
Que. 21 22 23
Ans. 3 4 3

Exercise 1.6
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Ans. 4 4 1 2 4 1 3 2 4 4 1 2 3 1 1 2 1

Exercise 1.7
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 3 4 3 2 3 2 1 4 4 3 4 1 2 3 4 3 1 3 3
Que. 21 22 23 24 25 26
Ans. 3 4 2 3 3 1

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Exercise 2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 3 1 3 4 4 2 4 1 1 3 3 2 1 1 4 3 1 3 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 4 2 3 4 4 3 2 4 1 1 1 1 3 3 2 3 4 3 2
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 2 3 1 4 2 2 2 4 2 2 4 1 4 2 4 4 3 2 3 1
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans. 2 2 4 1 3 1 2 1 2 4 4 4 4 4 3 2 2 3 3 2
Que. 81 82 83 84 85 86
Ans. 2 1 3 2 1 4

Exercise 3
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 2 2 4 1 3 2 2 1 4 2 1 4 2 3 2 2 3 3 1 4
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34
Ans. 1 3 3 4 3 1 1 1 4 3 4 2 4 3

Exercise 4 (Previous Year's Questions)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 1 3 4 4 1 1 1 1 1 2 3 4 1 3 4 3 3 4 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 3 1 1 3 2 3 3 3 4 2 2 2 1 1 4 2 1 1 1 2
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 2 2 2 3 4 2 4 3 2 3 2 3 1 3 1 4 4 1 3 1
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans. 4 3 2 3 4 4 2 3 2 2 2 1 4 4 3 1 2 3 2 4
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94
Ans. 4 1 1 4 3 3 3 1 3 2 2 3 2 2

106 Sarvam Career Institute

 Notes

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 Notes

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