Solution

DPP - CHEMICAL BONDING

Class 11 - Chemistry
1. The Structure of atoms, ions or molecules in which outer shell electrons are shown as dots surrounding the symbol. These
symbols are known as Lewis symbols or Lewis structures.
H2 − H : H

BeF2- : F¨ : Be : F¨ :
⋅⋅ ⋅⋅

2. i. Water is a polar solvent. Ionic compounds consist of positively and negatively charged ions which get separated in water. But
covalent compounds are generally non-polar so they are insoluble in water (like dissolves like).
ii. Because of the presence of strong electrostatic forces of attraction between oppositely charged ions, the melting points of ionic
compounds are higher than that of covalent compounds in which electrons are shared between the atoms.
3. i. When the atoms combine together to form a molecule, there is always the release of energy. Thus, the potential energy of a
molecule is less than that of uncombined atoms and therefore, the molecule is more stable.
ii. Solid sodium chloride has a crystalline structure in which the ions are not free to move. The ions become mobile when it is in
a molten state and thus, the electricity can be conducted.
iii. Oxygen has a high electronegativity than sulphur. As a result, H2O forms hydrogen bonding. Consequently, molecules of
water come nearer to each other through hydrogen bonding. This result is higher boiling point of water and hence it is a liquid.
4.
(d) Sharing of electrons contributed by one atom only
Explanation: Covalent bond is formed by two atoms sharing a pair of electrons. The atoms are held together because the
electron pair is attracted by both of the nuclei.
In the formation of a simple covalent bond, each atom supplies one electron to the bond. But a co-ordinate bond (also called a
dative covalent bond) is a covalent bond (a shared pair of electrons) in which both electrons come from the same atom.
5.
(d) 4,0
Explanation: The nitrate ion is formed by the loss of the hydrogen ion, and so its structure is:

Around the central nitrogen there are 4 pairs of shared electrons, and no remaining lone pair. The original lone pair has now
become a bonding pair. Two of those pairs make up a double bond. The double bond unit and the two single bonds arrange
themselves as far apart as possible in a trigonal planar arrangement - exactly the same as the carbonate ion.
6. i. Structure of C2H4

Orbital picture of ethene molecule

ii. Structure of C2H2

Orbitals picture of ethyne

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 7. Electro-negativity: Electro-negativity is the tendency of an atom to attract shared pair of electrons. It is the property of bonded
atom.
Whereas electron gain enthalpy is the tendency of an atom to attract outside electron. It is the property of an isolated atom.
8. The lone pairs of electrons are localized on the central atom, whereas bond pairs are shared between two atoms. Consequently, the
lone pair of electrons in a molecule occupy more space as compared to the bond pairs. Thus, there is greater repulsion between
lone pairs of electrons as compared to the lone pair-bond pair and bond pair-bond pair repulsions. The effects of these repulsions
are deviations from idealized shapes and alterations in bond angles in molecules. Due to which lone pair-lone pair repulsions
stronger than the lone pair-bond pair repulsion.
9. NO and CO2 are sp hypbridised. therefore, they have linear shape.
+
2

+
O = N = O, O = C = O
Linear Linear

NO
−

2
and O3 are sp2 hybridused. Both have lone pair of electrons but in former case it is not donated but in later case it is
dominated.
Thus, their shape is angular.

10. The VSEPR theory is used to determine the geometry of the individual molecules. It is developed by Gillespie-Nyholm.The main
postulates of valence shell Electron pair repulsion (VSEPR) theory are:-
(i) The shape of a molecule depends upon the no. of electron pairs around the central atom.
(ii) There is a repulsive force between the electron pairs, which tend to repel one another
(iii) The electron pairs in space tend to occupy such positions that they are at a maximum distance so, that the repulsive force will
be minimum.
(iv) A multiple bond is treated as if it is single bond and the remaining electron pairs which constitute the bond may be regarded
as a single super pair.
11. In NH3 and NF3 the difference in electro-negativity is nearly same but the dipole moment of NH3 (1.46 D) and NF3 are (0.28 D).
In NH3 the dipole moments of the three N-H bonds are in the same direction to the lone pair of electron. But in NF3 , the dipole
moments of the three N-F bonds are is the direction opposite to that of the lone pair. Therefore, the resultant dipole moment in
NH3 is more than its NF3 .

12. Molecules Lewis symbol No. of bond pairs No. of lone pairs Type Shape

BeCl2 2 0 AB2 Linear

BCl3 3 0 AB3 Triangular planar

SiCl4 4 0 AB4 Tetrahedral

ASF5 5 0 AB5 Trigonal Bipyramidal

H2S 2 2 AB2L2 Bent/V-shaped

PH3 3 1 AB2L Trigonal pyramidal

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13. (a) Isoelectronic
Explanation: Isoelectronic species are elements or ions that have the same, or equal number of electrons. Although
isoelectronic species have the same number of electrons, they are different in their physical and chemical properties. All of
given species have 14 electrons.
14. (a) C-O > C=O > C ≡ O
Explanation: The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond
order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between
two atoms is approximately the sum of the covalent radii of the two atoms. CO has a triple bond so has minimum bond length.
15. i. CO2,
ii. CH4,
iii. H2O,
iv. C2H2
16. i. C C I 4 (109.5
∘ ∘
), N H3 (107 ) , H2 O(104.5
∘
), SF6 (90 )
∘

T etrahedral P yramidal Angular Octahedral

ii. BF 3 (120
∘ ∘ ∘
), C H4 (109.5 ), N H3 (107 ), H2 O (104.5 )
∘

P lanar T etrahedral P yramidal Angular

iii. BeH 2 (180

∘
), AI C I3 (120
∘
), H2 O(104.5 ),
∘
H2 S(100 ),
∘

Linear P lanar Angular Angular but S is less

electronegative T han O

17. Bond order is defined as half of the difference between the number of electrons present in bonding and antibonding molecular
orbitals.
Bond order (B.O) = [N − N ] 1

2
b a

If the bond order is positive (Nb > Na) the molecule or ion will be stable. It is zero or negative than the molecule or ion does not
exist.If it is 0.5 than the molecule or ion will be unstable and to retain its stability it reacts with other species.
18. (a) All of these
Explanation: All of these
19.
(d)

Explanation: Out of phase overlap occurs in

Thus, it shows negative overlap

20. Plus and minus sign is used to identify the nature of electrons wave. Plus (+ve) sign denotes crest while (-ve) sign denotes trough.
21. There are 9 σ - bonds (3 between C – C and 6 between C – H) and 2 π - bonds.The complete structural formula is

22.
(d) sp2 , sp , sp3
Explanation: Hybridization of orbitals of N atom in NO , NO and NH −

3
+

2 4
+
are sp2, sp, sp3 which can be explained by their
Lewis structures. The empty p- orbitals of N take part in hybridization.
23.
(b) BF 3 → B F4
−

Explanation: BF3 has sp2 hybridization and its shape is a triangular planar. While BF4- has sp3 hybridization and shape is
tetrahedral.

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24. The actual structure is in between of these two contributing structures (i.e. the third structure) and it is known as the resonance
hybrid and the different individual structures are termed as resonating structures or canonical forms. This phenomenon is called
resonance. Double headed arrows are used in between canonical structures to separate them as shown below.

Resonance is responsible for equal bond lengths in the ozone molecule.

25. In BF3, B having 3 valence electrons, shares 1 valence electron with each F. So there are 3 bond pairs and 0 lone pair in BF3, so

boron is sp2 hybridised and in NH3, N having 5 valence electrons shares 1 valence electron with each H, so there are 3 bond pairs

and 1 lone pair, so nitrogen is sp3 hybridised. After the reaction, hybridisation of boron changes to sp3 as now it forms 4 bonds.
But hybridisation of nitrogen remains the same because N shares its lone pair with electron deficient B.

26.

27. BCl3 - sp2 hybridization - Trigonal planar

CH4 - sp3 hybridization - Tetrahedral.

CO2 - sp hybridization - Linear.

NH3 - sp3 hybridization - Pyramidal.

28. Hybridization: This concept was put forward by Pauling. He suggested that the atomic orbitals mix together to generate a new set
of equivalent orbitals, called as hybrid orbitals or hybridized orbitals.
Types of hybridization in the carbon atom
a. Diagonal or sp-hybridization- Carbons involved in forming C ≡ C (triple bond) like in ethyne (C2H2) exhibit sp-
hybridization.
b. Trigonal or sp2-hybridization- Carbons involved in forming C = C (double bond) like in ethene (C2H4) exhibit sp​2
hybridization.
c. Tetrahedral or sp3 -hybridization- Carbons involved in forming single bonds only like in ethane (C2H6 ) exhibit sp 3
hybridization.
29. The structure of the compound NH4NO3 can be represented as:

It is evident from the structure that NH ion contains covalent bonds and
+

4
NO
−

3
also contains covalent bonds. Also, the structure
clearly indicates that NH is a cation and NO is an anion and hence the bond between them is ionic. It has coordinate bond
+

4
−
3

also as the lone pair of electron on nitrogen is donated to H. All type of bonds are present in it.
The N of NH ion is sp3 hybridised and is tetrahedral and the N of NO ion is sp2 hybridised and is planar.
+

4
−

30. A single Lewis structure of CO ion is inadequate for the representation of all the properties of this ion. It can be represented as
2−

a resonance hybrid of the following structures:

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 If it were represented only by one structure, there should be two types of bonds, i.e., C = O double bond, and C-O single bonds but
actually, all bonds are found to be identical with the same bond length and same bond strength.

31.

Resonance in, CO32- I, II and III represent the three canonical forms.

i. In these structures, the position of nuclei is the same.

ii. All the three forms have the almost equal energy
iii. The same number of paired and unpaired electrons they differ only in their position.
32. a. The hybridisation of marked carbons is same i.e. sp2 as each C atom forms 3 sigma bonds with neighbouring atoms.

b. The hybridisation of the marked carbon atom is sp3 as the C atom forms 4 sigma bonds with neighbouring atoms.

c. The hybridisation of the marked carbon atom is sp2 as the C atom forms 3 sigma bonds with neighbouring atoms.

d. The hybridisation of the marked carbon atom is sp3 as the C atom forms 4 sigma bonds with neighbouring atoms.

e. The hybridisation of the marked carbon atom is sp as the C atom forms 2 sigma bonds with neighbouring atoms.

33. The atomic orbitals mix together to generate a new set of equivalent orbitals, called the hybrid orbitals or hybridised orbitals.
These orbitals are used in bond formation. The phenomenon is called hybridisation.
Types of hybridization in carbon atom.
i. a. Diagonal or sp-hybridisation- All compounds of carbon containing C ≡ C (double bond) like ethene (C2H2).

b. Trigonal or sp2-hybridisation- All compounds of carbon containing C = C (double bond) like ethene (C2H4).

c. Tetrahedral or sp3-hybridisation- All compounds of carbon containing C-C single bonds only like ethane (C2H6).
O

||
∗

ii. a. C H 2 = CH − C
∗
− O − H ;
2 2
sp sp

b. C H 3 − C H2 − OH ;
3
sp

||

c. C H 3 − C H2 − C
∗
− H ;
2
sp

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 ∗

d. C H 3 = C H = C H − C H3
3
sp

e. C H 3 − C ≡ CH
sp

34.
(d) A is false but R is true.
Explanation:

In OF2, O is sp3 hybridised.

35.
(c) A is true but R is false.
Explanation: σ bond is stronger than π bond because in σ bond the overlap is axial while in π bond. It is lateral.
Due to the lateral overlap in π bond atoms cannot rotate freely in π bond.
So, the assertion is true and the reason is false.
36.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation: In the formation of ethyne molecule, both the carbon atoms undergo sp-hybridisation having two unhybridised
orbital i.e., 2py and 2px. One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon
atom to form a C–C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half-filled s
orbital of hydrogen atoms forming sigma bonds.
37. Atomic orbitals with comparable energies and proper orientation overlap to form molecular orbits.
There are two types of overlap:
i. σ (Sigma overlap) - This involves head-on overlap of two s orbitals (s-s), one s and one p orbital (s-p) or two p -orbitals (p-p)
ii. π (Pi overlap) - This involves sideways overlap of two p orbitals
38. In H2, the first electron has to be removed from σ (1s) which has lower energy than 1s orbital of H atom and hence it is more
stable than an individual hydrogen atom. In O2, the first electron has to be removed from π(2p x) or π(2py ) orbital which has
higher energy than 2p-orbital of an individual oxygen atom and hence it is less stable than an individual oxygen atom
39. The electrons placed in bonding molecular orbitals tend to hold the nuclei together and hence greater stability than the
corresponding antibonding molecular orbital where the mutual repulsion of the electron is more than attraction.
40. π(2px) and π*(2px) have one and two nodes respectively.
H H

| |

41. i. is H − C = C − CI (5 σ bonds, one π -bonds)

ii. is

(9 σ bonds, two π -bonds)

H H H

| | |

iii. is C − C − C = −C ≡ C − H
|

(10 σ bonds, 3π−bonds)

42. Calculate bond order of given species in the same way as given in table on pages 200.
The order of stability is Li2, > Li > Li . +
2
−
2

This is because presence of electrons in antibonding orbital decreases the stability.

43.
(c) paramagnetic character decreases and the bond order increases
Explanation: For O2: σ 1s2, σ *1s2, σ 2s2, σ *2s2, σ 2pz2

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 (π 2py2 = π 2px2), (π *2py1 = π *2px1)σ *2pz
For O2: σ 1s2, σ *1s2, σ 2s2, σ *2s2, σ 2pz2

(π 2py2 = π 2px2), (π *2py1 = π *2px0)σ *2pz

Nb − Na
Bond order = 2
10−6
For O2 = 2
=2

O2+ =
10−5
For 2
= 2.5

44.
(d) O 2−

Explanation:

There is no unpaired electrons.

45. O 2−

2
< O
−

2
< O2 < O
+

46. i. Bond energy of N = Bond energy of N because bond order is the same in both species. [However, N is slightly more
+

2
−

2
+

stable than N as antibonding electrons number is higher in N than in N ]

−
2
−
2
+
2

ii. a. Both F2 and O 2−

2
have some bond order some bond length and are diamagnetic
b. These are isoelectric species, possess the same bond order and same bond length.
47. The electronic configurations are
+ 1
H : σls
2

2 ∗ 2 2 ∗ 2 2 2 2 ∗ 1 ∗ 1
O2 : σ1s , σ 1s , σ2s , σ 2s , σ2pz , (π2px = π2py ), (π 2px = π 2py )
2+ 2 ∗ 2 2 ∗ 2 2 2 2
O : σ1s , σ 1s , σ2s , σ 2s , σ2pz , (π2px = π2py )
2

It is observed from the electronic configuration that only O 2+

2
has all the electrons paired and hence it is diamagnetic.
48. Sequence o energy levels
σ 1s < σ *1s <σ 2s < σ *2s < π 2px = π 2py < σ 2pz

For N2 molecule the M.O. configuration is:

σ1s2 σ *1s2 σ 2s2 σ *2s2 π2p = π2p σ2p 2

x
2
y
2
z

B.O. = (10 − 4) = 3, diamagnetic in nature.

1

N : 1s2 σ *1s2 σ 2s2 σ *2s2 π2p = π2p σ2p

+
2
2
x
2
y
1
z

B.O. = (9 − 4) = 2.5 paramagnetic in nature

1

N : 1s2 σ *1s2 σ 2s2 σ *2s2 π2p = π2p σ2p π ∗ 2p

−

2
2
x
2
y
2
z
1
x

B.O. = (10 − 5) = 2.5, paramagnetic in nature.

1

N
2
: 1s2 σ *1s2 σ 2s2 σ *2s2 π2p = π2p
2+ 2
x
2
y

B.O. = (8 − 4) = 2, diamagnetic in nature

1

Stability order: N2 > N −

2
>N +

2
>N 2+

(N −
2
has more bonding electrons as compared to N ) +
2

49.
(d) If both Assertion and Reason are false statements.
Explanation: B2​ molecule has no. of electrons =10
Molecular orbital configuration:
2 2 2 2 1 1
σ(1s) σ ∗ (1s) σ(2s) σ ∗ (2s) π2px ​π2py ​

Due to the unpaired electron, it is paramagnetic.

The highest occupied MO is of π- type.
Hence, both the assertion and reason are false.
50.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation: The bond order of H2 molecule 1. This means that the two hydrogen atoms are bonded together by a single
covalent bond.
51.
(c) 104.50

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 Explanation: Due to the presence of two lone pairs on O in the H2O bond angle reduce to 104.50 from 1090.

52. Viscosity of a fluid is its resistance to flow. It depends upon the strength of intermolecular forces of the liquid.
There is greater intermolecular hydrogen bonding in H2O than that in HF as each H2O molecule forms four H-bonds with other
water molecules whereas HF forms only two H-bonds with other HF molecules. Greater the intermolecular H-bonding, greater is
the viscosity. Hence, HF is less viscous than H2O.
53. When hydrogen is attached to a highly electronegative element (F, O, N) in a covalent bonding the electrons of the covalent bond
are shifted towards the more electronegative atom. Thus partially positively charged hydrogen atom forms a bond with the other
more electronegative atom. This bond is known as a hydrogen bond. The hydrogen bond is stronger than the van der Waal's
forces.
54. (a) Both A and R are true and R is the correct explanation of A.
Explanation: The amides generally have high boiling points and melting points. It is due to strong intermolecular hydrogen
bonding in their molecules.
These characteristics and their solubility in water result from the polar nature of the amide group and hydrogen bonding.
Hence, both assertion and reason are correct and the reason is the correct explanation for the assertion.
55.
(c) A is true but R is false.
Explanation: Due to the presence of the hydroxyl group (-OH), there is extensive hydrogen bonding between the ethanol
molecules (C H OH) . But there is no such Hydrogen bonding in dimethyl ether (due to absence of - OH group). So the
2 5

boiling point of dimethyl ether is much lower than ethanol.

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