Chemical Bonding & Molecular Structure

(1) Upto 14 electrons, molecular orbital configuration is ->


N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(2) After 14 electrons to 20 electrons molecular orbital configuration is ->



Here N_a = 10

and N_b = 10

Now from question, in F atom 9 electrons present, so in F₂, 9 $ \times $ 2 = 18 electrons present.

Molecular orbital configuration of F₂(18 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^2}^ * $

The bond angle of $\mathrm{SO}_2$ is more than $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ because $\mathrm{SO}_2$ is $s p^2$-hybridised while $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ are $s p^3$-hybridised. Hence, in $\mathrm{SO}_2$ bond angle will be $120^{\circ}$ while in $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ it will be approx. $109.5^{\circ}$.

Out of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3, \mathrm{NH}_3$ has more bond angle as in $\mathrm{NH}_3$ there is one lone pair and in $\mathrm{H}_2 \mathrm{O}$ there are two lone pair of electrons. Hence, repulsion is less in $\mathrm{NH}_3$ and bond angle is more.

Thus, the order of bond angles will be

$ \mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\mathrm{SO}_2 $

Bond order $=\frac{1}{2}$ [No. of bonding electrons - No. of anti-bonding electrons]

$ =\frac{1}{2}\left[N_b-N_a\right] $

Molecular orbital configuration

$ \begin{aligned} & C_2=\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2 \\ & \text { Bond order }=\frac{1}{2}[8-4]=2 \end{aligned} $

$ \begin{aligned} &\begin{array}{rl} \mathrm{N}_2=\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma * & 2 s^2, \pi 2 p_x^2 \\ & \approx \pi 2 p_y^2, \sigma 2 p_z^2 \end{array}\\ &\begin{aligned} & \text { Bond order }=\frac{1}{2}[10-4]=3 \\ & \begin{array}{l} \mathrm{He}_2=\sigma \mathrm{l} s^2, \sigma^* 1 s^2=\frac{1}{2}[2-2]=0 \\ \mathrm{~B}_2=\sigma \mathrm{l} s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^1, \pi 2 p_y^1 \\ \quad=\frac{1}{2}[6-4]=1 \end{array} \end{aligned}\\ &\text { Thus, order of bond order will be }\\ &\mathrm{He}_2<\mathrm{B}_2<\mathrm{C}_2<\mathrm{N}_2 \end{aligned} $

Molecular orbital configuration of $\mathrm{C}_2$ is

$ \sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2, \pi^* 2 p_y^2 $

So, $\mathrm{C}_2$ molecule contains only $\pi$-bonds according to MOT.

$\mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+}$

$\mathrm{NH}_3$ is $s p^3$-hybridised and it has 3 -bond pairs and 1-lone pair.

$\mathrm{NH}_4^{+}$is also $s p^3$-hybridised, it has 4 -bond pair and no lone pair.

$ \begin{aligned} \text { Lone pair } & =6 \\ \text { Bond pair } & =3 \\ & 6: 3 \text { or }(2: 1) \end{aligned} $

The Lewis dot structures of the given compounds are as follows:

Thus, there are 4 compounds in which central atom has one lone pair of electron. i.e. $\mathrm{PbCl}_2, \mathrm{PH}_3, \mathrm{SF}_4$ and $\mathrm{SnCl}_2$.

Given molecules	No. of lone pair of electrons on central atom	Hybridisa-tion of central atom	No. of d-orbital involved in the hybridisation of central atom
1. ${\mathrm{ClF}}_3$${\mathrm{ClF}}_3$ClF_(3)	$3bp-2/p$$3bp-2/p$3bp-2//p	$s{p}^{3}d$$s{p}^{3}d$sp^(3)d	$1d$$1d$1d-orbital
2. ${\mathrm{PCl}}_5$${\mathrm{PCl}}_5$PCl_(5)	5bp-0/p	$s{p}^{3}d$$s{p}^{3}d$sp^(3)d	$1d$$1d$1d-orbital
3. ${\mathrm{BrF}}_5$${\mathrm{BrF}}_5$BrF_(5)	$5bp-1/p$$5bp-1/p$5bp-1//p	$s{p}^3{d}^2$$s{p}^3{d}^2$sp^(3)d^(2)	$2d$$2d$2d-orbitals
4. ${\mathrm{SF}}_4$${\mathrm{SF}}_4$SF_(4)	$4bp-1/p$$4bp-1/p$4bp-1//p	$s{p}^{3}d$$s{p}^{3}d$sp^(3)d	$1d$$1d$1d-orbital

Thus, $\mathrm{SF}_4$ molecules has same number of lone pair of electrons on central atom and $d$-orbitals involved in the hybridisation of central atom i.e. one.

Species containing same number of electrons will have same bond order.

I. $\mathrm{CN}^{-}-14$ electrons ( 6 from carbon, 7 from nitrogen add 1 for negative charge).

II. $\mathrm{N}_2^{+}-13$ electrons ( 7 from each N -atom, subtract 1 for positive charge).

III. $\mathrm{O}_2^{2-}-18$ electrons ( 8 from each O -atom, add 2 for ( -2 ) negative charge).

IV. $\mathrm{NO}^{+}-14$ electrons ( 8 from O -atom, 7 from N -atom, subtract 1 for positive charge).

Hence, I and IV will have same bond order.

The correct match is A-III, B-IV, C-I, D-III

Lone pair-lone pair (lp-lp) repulsion is the repulsion between various lone pair present in molecule. More the number of lone pair of electrons in molecule, greater will be the lone pair-lone pair repulsion.

1. 1.  $\mathrm{ClF}_3$ - Central atom Cl has 3 bond pairs and 2 lone pairs.

2. 2.  $\mathrm{IF}_5-\mathrm{I}$ has 5 bond pairs and 1 lone pair

3. 3.  $\mathrm{SF}_4-\mathrm{S}$ has 4 bond pairs and 1 lone pair

4. 4.  $\mathrm{XeF}_2-\mathrm{Xe}$ has 2 bond pairs and 3 lone pairs.

Since, lone pairs are maximum in XeF 2. Hence, it has maximum lone pair-lone pair repulsion.

The correct order of dipole moment is $\mathrm{BF}_3<\mathrm{NF}_3<\mathrm{NH}_3$

$\mathrm{NH}_3$ and $\mathrm{NF}_3$ have pyramidal geometry where central nitrogen has 3 bond pairs and 1 lone pair. Hence, both are $s p^3$-hybridised.

$\mathrm{NH}_3$, dipole vector of bond and lone pair are in same direction while in $\mathrm{NF}_3$, they are in opposite direction. Hence, dipole moment of $\mathrm{NH}_3$ is larger than $\mathrm{NF}_3$.

$\mathrm{BF}_3$ is asymmetrical molecule where dipole vectors are arranged at $120^{\circ}$. Therefore, the resultant of two dipole vectors cancel out with the third dipole vector. Hence it's dipole moment will be zero.

Bond Order Calculation

The bond order (BO) of a molecule is calculated using the formula:

$ \text{BO} = \frac{1}{2} [\text{Number of bonding electrons} - \text{Number of antibonding electrons}] $

Oxygen Molecule Configurations and Calculations

For $\mathrm{O}_2^{+}$:

Electronic configuration:

$ \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2, \pi^* 2p_x^1 \approx \pi^* 2p_y^0 $

Bond order calculation:

$ \mathrm{BO} = \frac{1}{2} [10 - 5] = \frac{5}{2} $

Given $\mathrm{BO}$ of $\mathrm{O}_2^+$ is $x$. Therefore:

$ \frac{5}{2} = x \quad \text{...(i)} $

For $\mathrm{O}_2^{-}$:

Electronic configuration:

$ \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2, \pi^* 2p_x^2 \approx \pi^* 2p_y^1 $

Bond order calculation:

$ \mathrm{BO} = \frac{1}{2} [10 - 7] = \frac{3}{2} $

Using equation (i), the bond order of $\mathrm{O}_2^-$ is expressed as:

$ \mathrm{BO} = \frac{3x}{5} $

For $\mathrm{O}_2^{2+}$:

Electronic configuration:

$ \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2 $

Bond order calculation:

$ \mathrm{BO} = \frac{1}{2} [10 - 4] = 3 $

From equation (i), substituting for $x$, we can express it as related fractions of bond order in terms of $x$.

By understanding these calculations, you can determine the bond orders of $\mathrm{O}_2^-$ and $\mathrm{O}_2^{2+}$ based on the given $x$ value and their respective electronic configurations.

The correct order of bond angles is

$ \mathrm{P}_4<\mathrm{S}_6<\mathrm{S}_8<\mathrm{O}_3 $

In $\mathrm{P}_4$, the four $s p^3$ hybridised P -atom lie at corner of regular tetrahedron with $\angle \mathrm{PPP}=60^{\circ}$.

In $\mathrm{S}_8$, the eight S -atoms arranged in crown shaped structure with $\angle S S S=107^{\circ}$.

In $\mathrm{S}_6$, the six S -atoms arranged in chair form with

$ \angle S S S=102.2^{\circ} $

In $\mathrm{O}_3$, hybridisation of oxygen atom is $s p^2$. So, angle between 0 -atom will be $120^{\circ}$ but due to presence of one lone pair, the bond angle decreases from $120^{\circ}$ to $116.8^{\circ}$.

The set of molecules in option (d), consisting of SF₄, XeF₄, and CH₄, has central atoms with different types of hybridization.

SF₄:

Calculation: $ 4 + \frac{1}{2} \times (6 - 4) = 4 + 1 $

Hybridization: $ \text{sp}^3d $

XeF₄:

Calculation: $ 4 + \frac{1}{2} \times (8 - 4) = 4 + 2 $

Hybridization: $ \text{sp}^3d^2 $

(Square planar shape)

CH₄:

Calculation: $ 4 + \frac{1}{2} \times (4 - 4) = 4 $

Hybridization: $ \text{sp}^3 $

This demonstrates that the hybridizations are $ \text{sp}^3d $, $ \text{sp}^3d^2 $, and $ \text{sp}^3 $ for SF₄, XeF₄, and CH₄, respectively, indicating distinct hybridization states among the central atoms.

Thus, statements II and IV are incorrect while statement I is correct. Also; from the structure of compound $Z$ i.e., $\mathrm{XeO}_3$ it is clear that statement III is also correct

$3 \pi$-bonds and 1 lone pair of electrons.

In $\mathrm{BrF}_3$ molecule, $\mathrm{Br}$ is $s p^3 d$ hybridised, but geometry is T-shaped due to distortion of geometry from trigonal-bipyramidal to T-shaped by the involvement of lone pair-lone pair repulsion. The lone pair occupies equitorial position in order to minimize lone pair-lone pair and lone pair-bond pair repulsion.

Here,

$\begin{array}{l}l p-l p \text { repulsion }=1 \\ l p-b p \text { repulsion }=2 \\ b p-b p \text { repulsion }=2\end{array}$

$\mathrm{I}_3^{-}$ ion is made up of an $\mathrm{I}_2$ molecule with an $\mathrm{I}^{-}$ bonded to it by means of a coordinate bond in which $\mathrm{I}_2$ is lone pair acceptor (Lewis acid) and $I^{-}$ is lone pair donor (Lewis base).

There are two bond pairs and three lone pairs in the outer shell of central atom. To minimise the repulsive forces the three lone pairs occupy the equatorial position.

The ion is therefore, linear in shape with a bond angle of exactly $180^{\circ}$.

$ \underset{\substack{\text { Lewis } \\ \text { acid }}}{\mathrm{I}_2}+\underset{\substack{\text { Lewis } \\ \text { base }}}{\mathrm{I}^{-}} \longrightarrow[\mathrm{I}-\mathrm{I} \leftarrow \mathrm{I}]^{-}$

Similarly, $\mathrm{N}_3^{-}$ ion is also linear in shape. $\mathrm{I}_3^{+}$ does not have linear structure.

(A) $\mathrm{BCl}_{3}$ - Even electron molecules
       $\mathrm{SF}_{6}$ - Expended octet molecules

(B) $\mathrm{NO}$ - Odd electron molecules
      $\mathrm{H}_{2} \mathrm{SO}_{4}$ - Expanded octet

(C) $\mathrm{SF}_{6}-$ Even electron molecules
      $\mathrm{H}_{2} \mathrm{SO}_{4}$ - Expanded octet

(D) $\mathrm{BCl}_{3}-$ Even electron molecules
      $\mathrm{NO}$ - odd electron molecules

The number of lone pair of electrons in the central atom of $\mathrm{SCl}_{2}, \mathrm{O}_{3}, \mathrm{ClF}_{3}$ and $\mathrm{SF}_{6}$ are 2, 1, 2 and $\mathrm{0}$ respectively

Their structures are as,

$\Psi_{A}-\Psi_{B}=\Psi_{M O}$ is anti-boding molecular orbital

$ \begin{aligned} &\mu=Q \times r \text { is dipole moment } \\\\ &\frac{N_{b}-N_{a}}{2}=\text { bond order } \\\\ &\Psi_{\mathrm{A}}+\Psi_{\mathrm{B}}=\Psi_{\mathrm{MO}} \text { is boding molecular orbital. } \end{aligned} $

$\mathrm{O}_{2}, \mathrm{Cu}^{2+}$ and $\mathrm{Fe}^{3+}$ have 2,1 and 5 unpaired electrons respectively, so these are the paramagnetic species. Hence, they are attracted by magnetic field.

$\mathrm{NaCl}$ and $\mathrm{H}_{2} \mathrm{O}$ are the diamagnetic species so they are repelled by the magnetic field.

From Fajan's rule, for a given metal ion, as the size of anion increases, polarizability of anion increases and hence covalent character of the given ionic compound increases.

Hence, the increasing order of covalent character is $\mathrm{CaF}_{2}<\mathrm{CaCl}_{2}<\mathrm{CaBr}_{2}<\mathrm{Cal}_{2}$

(A) $\mathrm{BrF}_{5}-$ square pyramidal

(B) $\left[\mathrm{CrF}_{6}\right]^{3-}-$ octahedral

(C) $\mathrm{O}_{3}-$ bent

(D) $\mathrm{PCl}_{5}$ - trigonal bipyramidal

$\mathrm{XeO}_{3}-s p^{3}$, Pyramidal

$\mathrm{XeF}_{2}-s p^{3} d$, linear

$\mathrm{XeOF}_{4}-s p^{3} d^{2}$, Square Pyramidal

$\mathrm{XeF}_{6}-s p^{3} d^{3}$, distorted octahedral

All are tetrahedral and each have 10 electrons.

Steric no. = 5 (sp³d), lone pair = 2

Bent T shape.

SF₄ → sp³d hybridisation.


The lone pair of electrons on S is in an equatorial position and there are two lone pair-bond pair repulsions at 90°.

PCl₅


All three equatorial bonds in a plane

sp³d hybridization

Trigonal bipyramidal

Axial bonds are weaker than equatorial bonds.

Due to the high difference in electronegativity of H and N the H-bond strength of NH₃ is highest. There is no H-bond in CH₄.

CH₄ < HCN < NH₃

Note :

(1) $\,\,\,\,$ Bond strength $ \propto $ Bond order

(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = Number of electrons in bonding molecular orbital

N_a $=$ Number of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

Then in $O_2^ + $ no of electrons = 15

in $O_2^ - $ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (16 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^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ + $ (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}^ * $

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

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

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}^ * $

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

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Molecular orbital configuration of O $_2^{2 - }$ (18 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^2}^ * $

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

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

So, correct order of Bond order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $

If an electron is removed from the anti-bonding orbital, then it will tend to increase the bond order. The HOMO in NO and O₂ is antibonding molecular orbital .

Hence, in NO and O₂ bond order will increase on loss of electron.

Electron deficient species have less than 8 electrons (or two electrons for $\mathrm{H}$ ) in their valence (incomplete octet).

$\mathrm{B}_{2} \mathrm{H}_{6}, \mathrm{BCl}_{3}$ have incomplete octet.

Note :

(1) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(2) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(3) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

in $O_2^{2 - }$ no of electrons = 18

(A) Molecular orbital configuration of O $_2^{2 - }$ (18 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^2}^ * $

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

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

(B)   $C_2^{2 - }$ has 14 electrons.

Moleculer orbital configuration of $C_2^{2 - }$ is

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

$\therefore\,\,\,\,$N_a = 4

N_b = 10

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

(C)   $N_2^{2 - }$ has 16 electrons.

Moleculer orbital configuration of $N_2^{2 - }$ is

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

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

The correct order of bond orders of ${C_2}^{2 - }$, ${N_2}^{2 - }$ and ${O_2}^{2 - }$

${O_2}^{2 - }$ < ${N_2}^{2 - }$ < ${C_2}^{2 - }$