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(i) It cannot explain the formation of molecules like BeF₂, BF₃ in which the central atom has less than 8 electrons in its valence shell. 

(ii) It cannot explain the formation of molecules such as PF₂,SF₆ etc. in which central atom has more than eight electrons in its valence shell.

N₂< SO₂< ClF₃< K₂O < LiF.

It can be explain with 3 types. 

(a) Dipole and Dipole interaction : Dipole – Dipole interaction exists between polar molecules like NH₃, SO₂, HCl… Greater the dipole interaction greater is Vander Waal’s force. Hence such molecules liquefied easily. 

(b) Dipole & induced dipole interaction: A permanent dipole molecule can induce polarity in a non-polar molecular when comes near to each other. As a result between polar of non-polar molecule there exists force of attraction. 

(c) Induced dipole & Induced dipole interaction: Non polar molecules like, O₂ and N₂ having induced polarity on each other which is called Landon forces. When these two molecules comes near to each other the force increases.

O₂ molecule contains two unpaired electrons in the π"2p_x and π"2p_y orbitals.

O₂ would exhibit paramagnetism because it contains one unpaired electron in its Mo configuration.

C₂ is KK (σ_2s)²(σ*_2s)²(π_2px)² = (π_2py)² ;

Bond order = \(\frac{1}{2}\)[8-4] = 2

C₂⁺ is KK (σ_2s)²(σ*_2s)²(π_2px)² = (π_2py)²;

Bond order = \(\frac{1}{2}\)[7 - 4] = 1.5

O₂^- because it contains one unpaired electron in its MO configuration.

Li has a configuration: 1s², 2s¹

There are two s orbitals (1s and 2s) on each atom.

These combine to give four MO's. These are

(1s) + (1s) → σ1s + σ*1s

(2s) + (2s) → σ2s + σ*2s

Thus all the six electrons are accommodated in these four orbitals. The electronic configuration for

Li₂ is (σ1s) 2(σ*1s)² (σ2s)²

N_b = 4, N_a = 2

Bond order in Li₂ = {4 - 2}/{2} = 1

Therefore, Li₂ should be a stable species. Its bond dissociation energy is 105 kJ mol^-1, as compared to 431 kJ mol^-1 for H₂. Thus, the bond in Li₂ is much weaker than that on H₂. This is because 2s orbital of Li (involved in the bonding) is much larger than 1s orbital of H.

Bond order for C₂ → C₂⁺ e^- decreases from 2 to 1.5 but the bond order for O₂ → O₂⁺ + e^- increases from 2 to 2.5.

Along with covalent bond & hydrogen bond, ice form of water exists in the tetrahedral shape from bent structure by leaving vacant space. Due to this vacant space is ice if floats in water.

Covalent bonds and Intermolecular hydrogen bonds.

The [BF₄] has Four covalence.

Two atoms of hydrogen combine to form a molecule. Each hydrogen atom has one electron in its 1s orbital. When the two H atoms come closer to each other their 1s orbitals overlap each other, and both the available electrons are used up in forming a bond between two H-atoms. Since there is no other electron left, the hydrogen molecule (H₂) has no capacity for bonding with more hydrogen atoms. Therefore, only H₂ exists and molecules like H₃, H₄ etc are not formed.

In H₂^- ion, one electron is present in antibonding (σ* 1s) orbital due to which destabilizing effect is more and hence the stability is less than that of H₂⁺ ions.

In H₂^- ion, one electron is present in anti bonding orbital due to which destabilizing effect is more and thus the stability is less than that of H₂⁺ ion.

Covalent bond is stronger than hydrogen bond.

It is because e- present in anti-bonding molecular orbital in H₂^- destabilize the molecule .

(a) AICI₃ − It has sp² hybridisation. 

[AICI₄]⁻ − It has sp³ hybridisation. 

(b) BF₃ − It has sp² hybridisation. 

[BF₄]⁻ − It has sp³ hybridisation.

(c) NH₃ − It has sp³ hybridisation. 

[NH₄]⁺ − It has sp³ hybridisation. 

(d) \(\overset{+}NO_2\) − It has sp hybridisation. 

\(\overset{-}NO_2\) − It has sp² hybridisation.

So, choice (c) NH₃ and [NH4]⁺ have same hybridisation.

When water freezes, water molecules form a crystalline structure maintained by inter molecular hydrogen bonding. The orientation of hydrogen bonds causes molecules to push further apart, which increases the volume of ice. So, density of ice is less than water.

Ne₂ (20) = σ1s² σ*1s², σ2s² σ*2s², σp_x²

π 2p_y² 2π* 2p_y²π*2p_y² 2p_z² σ*

2px²

B.O = 1/2 (10 - 10) = 0

Ne₂ cannot exist because its bond order is zero.

Although H²⁺ and H²⁻ have same bond order i.e,. \(\frac{1}{2}\) Yet, H²⁺ is more stable than H²⁻ because H²⁻ contains one electron in the antibonding orbital which results in repulsion and decreases the stability. However, H²⁺ does not contain any electron in two anti-bonding orbital.

In (a) the lp is present at axial position so there are three 1p – bp repulsions at 90° . Whereas in (b) the lp is in an equatorial position are there are two 1p – bp repulsions. 

Hence, arrangement (b) is more stable than (a).

The bond length may be defined as the average distance between the centres of nuclei of the two bonded atoms. It is expressed in angstrom (Å) units (1 Å = 10^-8 cm) or pm (picometre) 1 pm = 10^-10 cm.

In an ionic compound, bond length between two bonded atoms is obtained by adding up their ionic radii. Similarly, in a covalent- compound, bond length is obtained by adding up the covalent (atomic) radii of the two bonded atoms.

i. The ammonia molecule has sp³ hybridization.The expected bond angle is 109°28′. But the actual bond angle is 107°28′. 

It is due to the following reasons.

• One lone pair and three bond pairs are present in ammonia molecule.
• The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
• Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.

ii. The water molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. 

It is due to the following reasons.

• Two lone pairs and two bond pairs are present in water molecule.
• The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.
• Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.

C) Pyramidal, 107° 48’