Chemical Bonding and Molecular Structure

Bond lengths of O $-$ O in O₂ is 1.21 Å, in H₂O₂ is 1.48 Å and in O₃ is 1.28Å. Therefore, correct order of the O $-$ O bond length is H₂O₂ > O₃ > O₂ or O₂ < O₃ < H₂O₂

Hydrogen bonding in H₂O > C₂H₅OH > CH₃OH

Hence, H₂O has maximum surface tension.

IF₃ has three bond pairs and two lone pairs, thus, it has sp³d hybridization.



PCl₃ has three bond pairs and one lone pair thus, it has sp³ hybridization.



NH₃ has three bond pairs and one lone pair thus, it has sp³ hybridisation.


BF₃ has three bond pairs and no lone pair thus, it has sp² hybridisation.

Lone pair - Lone pair repulsion > lone pair - bond pair repulsion > bond pair-bond pair repulsion.

The lone pairs of electrons are arranged on equatorial position as they are at maximum distance at these positions and hence they have minimum repulsion at these positions.

In H₂O the central atom oxygen has two lone pairs of electrons and two bond pairs of electrons. Thus, the shape of H₂O is bent or V-shape. Thus, the two dipoles of H-O bonds have some resultant value as they are not exactly in the opposite direction to cancel out each other. While in BeF₂, there are two bond pairs on central atom Be, hence it is linear in shape. Thus, the two dipoles of bonds are exactly in opposite direction to each other and cancel each other to give net zero dipole moment.

The MX₆ molecule with regular octahedral geometry is as follows :
Thus, there are three X—M—X bonds with bond angle 180^o.

SiF₄ has symmetrical tetrahedral shape which is due to sp³ hybridisation of the central silicon atom in its excited state configuration.

SF₄ has distorted tetrahedral or sea-saw geometry which arises due to sp³d hybridisation of the central sulphur atom and due to the presence of one lone pair of electrons in one of the equatorial hybrid orbital.

As sigma bond is stronger than the $\pi $(pi) bond, so it must be having higher bond energy than $\pi $(pi) bond.

${O^ - }\hbox{---}\mathop N\limits^{\mathop \uparrow \limits^O } = O$

In NO₃^- ion, nitrogen has 4 bond pair of electrons and no lone pair of electrons.

Electronic structure of S atom in excited state


In 'S' unhybride d- orbital is present, which will involved in bond formation with oxygen atom.

In oxygen two unpaired p- orbital is present in these one is involved in $\sigma $ bond formation while other is used in $\pi $ bond formation.

So, in SO₃^2$-$, p$\pi $ $-$ d$\pi $ bonding present.

In CO, the number of electrons

$ \Rightarrow $ 6 + 8 = 14 [ Z of C = 6 and O = 8]

Electronic configuration of molecular orbital of CO :

${\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}$

CN^- also get (6 + 7 + 1) 14 electrons and the configuration is similar to that CO.

So, CN and CO is isoelectronic.

For strong $\pi $-bonding, p$\pi $ - p$\pi $ bonding should be strong. In case of P, due to larger size as compared to N-atom, p$\pi $ - p$\pi $ bonding is not so strong.

Compounds with same shape and same hybridisation are known as isostructural.

XeF₂, IF₂^- $ \to $ both are sp³ hybridised linear molecules.

Bond angle is maximum in NH₄⁺ tetrahedral molecule with bond angle 109°.

As X and Y both are electronegative elements thus both attracts the electron density from H thus, electron density on H decreases and on X it increases.

In PO₄^3–, P atom has vacant d-orbitals, thus it can form p$\pi $ - d$\pi $ bond. ‘N’ and ‘C’ have no vacant ‘d’ orbital in their valence shell, so they cannot form such bond.

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(A) N₂⁺ has 13 electrons.

Moleculer orbital configuration of N₂⁺

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

Here 1 unpaired electrons present, so it is paramagnetic.

(B) Molecular orbital configuration of $O_2^ - $ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

Here 1 unpaired electrons present, so it is paramagnetic.

(A) CO has 14 electrons.

Moleculer orbital configuration of CO

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

Here 0 unpaired electrons present, so it is diamagnetic.

(D) Molecular orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

Here 1 unpaired electrons present, so it is paramagnetic.

XeF₄ is sp³d² hybridised and it is square planar :

$N_2$ has 14 electrons.

Moleculer orbital configuration of $N_2$

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

N₂⁺ has 13 electrons.

Moleculer orbital configuration of N₂⁺

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

$\therefore\,\,\,\,$ N_b = 9

N_a = 4

$\therefore$ BO = ${1 \over 2}\left[ {9 - 4} \right]$ = 2.5

As the bond order in N₂ is more than N₂⁺ so the dissociation energy of N₂ is higher than N₂⁺.