Chemical Bonding & Molecular Structure

Option (A) : According to molecular orbital theory, the arrangement of the electrons in the molecular orbitals is as follows :

For $C_2^{2 - }$, the total number of electrons is 14. The molecular orbital configuration is

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

There is no unpaired electron and thus it is diamagnetic.

Option (B) : For $O_2^{2 + }$, the total number of electrons is 14. The molecular orbital configuration 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^0}^ * \,\, = \pi _{2p_y^0}^ * $

Thus the bond order = (10 $-$ 4)/2 = 3;

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}^ * $

So O$_2$ has 2 unpaired electrons. for O₂, bond order = 2.

As bond order is inversely proportional to bond length, thus the bond length of $O_2^{2 + }$ is less than the bond length of O₂.

Option (C) : For $N_2^ + $, the total number of electrons is 13. The molecular orbital configuration is

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

Bond order of $N_2^ + $ = (9 $-$ 4)/2 = 2.5

   $N_2^{ - }$ has 15 electrons.

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

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^0}^ *$

$\therefore\,\,\,\,$N_a = 5

N_b = 10

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

Thus, the bond orders are the same.

Option (D) : As some energy is released during the formation of $He_2^ + $ from two isolated He atoms, thus it has lesser energy as compared to two isolated He atoms.

Structure of each of the compound is represented as :

(A) BrF₅



Structure : Square pyramidal

(i) One lone pair

(ii) Four bond pair

(B) ClF₃



Structure : T-Shaped

(i) Two lone pair

(ii) Three bond pair

(C) XeF₄



Structure : Square planar

(i) Two lone pair

(ii) Four bond pair

(D) SF₄



Structure : See-saw

(i) One lone pair

(ii) Four bond pair

Hydrogen bond is a type of strong electrostatic dipole- dipole intersection and dependent on the inverse cube of distance between the molecular ion-dipole -

interaction $ \propto {1 \over {{r^3}}}.$

Write the molecular orbital electronic configuration keeping in mind that there is no $2 s-2 p$ mixing, then if highest occupied molecular orbital contain unpaired electron then molecule is paramagnetic otherwise diamagnetic.

Note: Since there is no mixing of $2 s$ and $2 p$ orbitals of each of the atom in a diatomic molecule, the energy of bonding molecular orbital $2 p z$, i.e., $2 p$ is less than that of bonding molecular orbital of $2 p x$ and $2 p y$, i.e., $\left(\pi_{2 p x}^*\right.$ and $\left.\pi_{2 p y}^*\right)$.

$ \therefore $ The order of molecular orbital energy levels, i.e., bonding and anti-bonding energy levels is as follows:

$ \begin{aligned} \left(\sigma_{1 s}\right)\left(\sigma_{1 s}^*\right)\left(\sigma_{2 s}\right)\left(\sigma_{2 s}^*\right)\left(\sigma_{2 p z}\right)\left(\pi_{2 p x}\right. & \left.\equiv \pi_{2 p y}\right) \left(\pi_{2 p x}^*\right. \left.\equiv \pi_{2 p y}^*\right)\left(\sigma_{2 p z}^*\right) \end{aligned} $

(A) Total number (no.) of electrons in $\mathrm{Be}_2=4+4=$ 8. The electronic configuration of $\mathrm{Be}_2$ is:

$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2 $

All the electrons in $\mathrm{Be}_2$ are paired up, $\mathrm{Be}_2$ is diamagnetic.

(B) Total number (no.) of electrons in $\mathrm{B}_2=5+5=$ 10. The electronic configuration of $B_2$ is:

$\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2$

All the electrons in $B_2$ are paired up, $B_2$ is diamagnetic.

(C) Total no. of electrons in $\mathrm{C}_2=6+6=12$ The electronic configuration of $\mathrm{C}_2$ is:

$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2\left(\pi_{2 p x}^{* 1} \equiv \pi_{2 p y}^{* 1}\right) $

There are two unpaired electrons in doubly degenerate.

$\pi_{2 p x}$ and $\pi_{2 p y} \mathrm{C}_2$ is paramagnetic.

(D) Total no. of electrons in $\mathrm{N}_2=7+7=14$

The electronic configuration of $\mathrm{N}_2$ is:

$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2\left(\pi_{2 p x}^{* 2} \equiv \pi_{2 p y}^{* 2}\right) $

All the electrons in $\mathrm{N}_2$ are paired up; $\mathrm{N}_2$ is diamagnetic.

(a) Ice floats in water due to the low density of ice as compare to water which is due to open cage like structure (formed by intermolecular H-bonding).

(b) Basic strength of RNH₂ > R₃N. It is also explained by hydrogen bonding.


(c)
(d) Dimerisation of acetic acid in benzene is due to intermolecular hydrogen bonding.

$H_2^{2+}$ , $He_2$

Both have zero bond order. Thus, they do not exist.

Li₂ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,$

$ \therefore $ Bond order = ${1 \over 2}\left( {4 - 2} \right)$ = 1

$Li_2^+$ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^1}}} \,$

$ \therefore $ Bond order = ${1 \over 2}\left( {3 - 2} \right)$ = 0.5

$Li_2^-$ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,\,\sigma _{1{s^1}}^ * \,$

$ \therefore $ Bond order = ${1 \over 2}\left( {4 - 3} \right)$ = 0.5

The bond order of $Li_2^+$ and $Li_2^-$ is same but $Li_2^+$ is more stable than $Li_2^-$ because $Li_2^+$ is smaller in size and has 2 electrons in antibonding orbitals whereas $Li_2^-$ has 3 electrons in antibonding orbitals. Hence $Li_2^+$ is more stable than $Li_2^-$.

Those species which have unpaired electrons are called paramagnetic species.

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

(A)   $N_2$ has 14 electrons.

Moleculer orbital configuration of $N_2$

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

Here no unpaired electron present, so it is diamagnetic.

(B)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ 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}^ * $

Here 2 unpaired electrons present, so it is paramagnetic.

(C)    $S_2$ has 32 electrons.

It also has 2 unpaired electrons like $O_2$, so it is paramagnetic.

(D)   $C_2$ has 12 electrons.

Moleculer orbital configuration of $C_2$

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

Here no unpaired electron present, so it is diamagnetic.

PF₅ $ \Rightarrow $ sp³d hybridized and shape is trigonal bipyramidal.

BrF₅ $ \Rightarrow $ 5 bond pair and 1 lone pair present so sp³d² hybridized and shape is square pyramidal.

The electronic configuration of Xe is 5s² 5p⁶; and that of Xe in excited state is 5s² 5p⁵ 5d¹. The hybridisation is sp³d and geometry is see-saw. The actual shape is trigonal bipyramidal but due to the preaence of lone pair it gets distorted to see-saw structure.

Charge of cation/Size of cation is called polarising power.

$ \therefore $ (i) Polarising power $ \propto $ charge of cation

(ii) Polarising power $ \propto $

1

size of cation

Here the AlCl₃ is satisfying the above two conditions i.e., Al is in +3 oxidation state and also has small size. So it has more covalent character.

The structure of IF₇ pentagonal bipyramidal having sp³d³ hybridisation.

$\mathrm{OSF}_2$ has $s p^3$ hybridisation. According to VSEPR theory, we can conclude that the shape of $\mathrm{OSF}_2$ is a trigonal pyramid.

Sulphur has a non-bonding pair of electrons and three sulphur-oxygen ( $\mathrm{S}-\mathrm{O})$ bond pairs. $\mathrm{SO}_2$ is $s p^2$ hybridised with three bond pairs (two sigma and one pi) and one lone pair. It is replace trigonal by bent.

$\mathrm{BrF}_3$ is $s p^3 d$ hybridized with three bond pair and two lone pairs. It is T shaped.

$\mathrm{SiO}_3^{2-}$ is $s p^2$ hybridised with four bond pairs (three sigma and one pi) and no lone pair. It has trigonal planar shape.

Molecular orbital configuration of B₂ (10 electrons) is

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

Here in B₂, 2 unpaired electrons present.

Since the Hund's rule is violated, two electrons are placed in $\pi 2 p_x$ molecular orbital, so

$ \mathrm{B}_2(10): {\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} $

Thus, bond order $=\frac{6-4}{2}=1$.

As there are no unpaired electrons, the nature is diamagnetic.

Note :

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

(2) $\,$ 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

(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$_2^ {2+ }$ (14 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^0}^ * \, = \,\pi _{2p_y^o}^ * $

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

N_a = 4

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

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

As Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

So $O_2^{2+}$ has the shortest bond length.

(A) ${B_2} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\pi 2p_x^1 = \pi 2p_y^1$

It is paramagnetic due to two unpaired electrons. The bond order is

${{{N_b} - {N_a}} \over 2} = {{6 - 4} \over 2} = 1$

The gain of electron increases bond order, so reduction is possible.

(B) ${N_2} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\pi 2p_x^2 = \pi 2p_y^2\sigma 2p_z^2$

There are no unpaired electrons. Bond order is

${{{N_b} - {N_a}} \over 2} = {{10 - 4} \over 2} = {6 \over 2} = 3$

Mixing of 2s and 2p orbitals is possible because of similar energies.

(C) $O_2^ - \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\sigma 2p_z^2\pi 2p_x^2 = \pi 2p_y^2\,\pi * 2p_x^2\,\pi * 2p_y^1$

The molecule is paramagnetic due to presence of unpaired electron. Bond order is less than 2.

${{{N_b} - {N_a}} \over 2} = {{10 - 7} \over 2} = {3 \over 2}$

Loss of electron increases the bond order so oxidation is possible.

(D) $O_2^{} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\sigma 2p_z^2\pi 2p_x^2 = \pi 2p_y^2\,\pi * 2p_x^1\,\pi * 2p_y^1$

The molecule is paramagnetic due to presence of unpaired electrons. Bond order is

${{{N_b} - {N_a}} \over 2} = {{10 - 6} \over 2} = {4 \over 2} = 2$

Loss of electron causes increase in bond order, so it undergoes oxidation.

Number of electron in NO⁺ = number of electron in CN^– = 14 electrons.

As both have same number of electrons so their bond order is equal.

Moleculer orbital configuration of NO⁺ (14 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^0}^ * = \,\,\pi _{2p_y^0}^ * $

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

Moleculer orbital configuration of CN^- (14 electrons) 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 $ B.O = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

In BF₃, B is sp² hybridised and by Back Bonding method strong p$\pi $-p$\pi $ bond is created between filled p-orbital of F and vacant p-orbital of B which leads to shortening of B–F bond length which results in higher bond dissociation energy of the B–F bond. However in CF₄, C does not have any vacant p-orbitals to undergo $\pi $-bonding.

Among F, O and N, F is most electronegative so F pulls bond pair of electron in F - H towards itself and develops highly positive charge on H atom.

This highly positive charged H atom creates stongest hydrogen bonding by taking lone pair of electron form electronegative atom F/N/O. Hence among the given options F – H ...... F is the strongest bond.

Note : F – H ...... N has strongest hydogen bonding among F – H ...... F, F – H ...... O and F – H ...... N because N is least electronegative among F, O and N and can easily donate lone pair of electron.

As charge/size ratio of a cation determines its polarizing power so high charge and small size of the cations increases polarisation.

As the size of the given cations decreases as

K⁺ > Ca²⁺ > Mg²⁺ > Be²⁺

Hence, polarising power decreases as

K⁺ < Ca²⁺ < Mg²⁺ < Be²⁺

Those species which have unpaired electrons are called paramagnetic species.

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

(a)    $O_2^{2−}$ has 18 electrons.

Moleculer orbital configuration of $O_2^{2−}$ 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}^ * $

Here is no unpaired electron so it is diamagnetic.

(b)   NO has 15 electrons.

Moleculer orbital configuration of NO 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^0}^ * $

Here is 1 unpaired electron, So it is Paramagnetic.

(c)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ 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}^ * $

Here 2 unpaired electron present, so it is paramagnetic.

(d)   $O_2^{+}$ has 15 electrons.

Moleculer orbital configuration of $O_2^{+}$ 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^0}^ * $

Here 1 unpaired electron present, so it is paramagnetic.

(A) Moleculer orbital configuration of $C_2$ (12 electrons)

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

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

N_b = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {8 - 4} \right] = 2$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $C_2^+$ (11 electrons)

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

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

N_b = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {7 - 4} \right] = 1.5$

Here 1 unpaired electron present, so it is paramagnetic.

(B) Moleculer orbital configuration of $N_2$ (14 electrons)

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

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

N_b = 10

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

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $N_2^+$ (13 electrons)

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

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

N_b = 9

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

Here 1 unpaired electron present, so it is paramagnetic.

(C) Moleculer 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^0}^ * $

$\therefore\,\,\,\,$N_a = 5

N_b = 10

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

Here is 1 unpaired electron, So it is paramagnetic.

Moleculer orbital configuration of NO⁺ (14 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^0}^ * = \,\,\pi _{2p_y^0}^ * $

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

N_b = 10

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

Here is no unpaired electron, So it is diamagnetic.

(D) 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 = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Here 2 unpaired electrons present, so it is paramagnetic.

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

Here 1 unpaired electrons present, so it is also paramagnetic.

Species having the same number of electrons will have the same bond order.

Number of electrons in CO are 14 (8 from oxygen and 6 from carbon).

NO$^-$ has 16 electrons (7 from nitrogen, 8 from oxygen, add 1 for negative charge).

NO$^+$ has 14 electrons (7 from nitrogen, 8 from oxygen, subtract 1 for positive charge).

CN$^-$ has 14 electrons (7 from nitrogen, 6 from carbon, add 1 for negative charge).

N$_2$ has 14 electrons (7 from each nitrogen).

$\begin{array}{ll}\text { (a) } \mathrm{Na}_2 \mathrm{O}_2 & \rightarrow \mathrm{Na}^{+} \quad 8 \mathrm{e}^{-} \text {all shells are filled. } \\ \mathrm{O}_2^{2-} & \left.\rightarrow \quad \begin{array}{lll}1 s^2 & 2 s^2 & 2 p^5 \\ 1 s^2 & 2 s^2 & 2 p^5\end{array}\right] 10 e^{-}\end{array}$

(c) $\mathrm{N}_2 \mathrm{O}$

$ : \mathrm{N}=\mathrm{N} \rightarrow \ddot{\mathrm{O}}: $

all electrons are paired.

(d) $\mathrm{KO}_2 \rightarrow \mathrm{~K}^{+} \mathrm{O}_2^{-}$

$ \left.\mathrm{O}_2^{-}=\begin{array}{ccc} 1 s^2 & 2 s^2 & 2 p^4 \\ 1 s^2 & 2 s^2 & 2 p^5 \end{array}\right] 9 e^{-} $

Statement 1: Boron always forms covalent bonds.

→ Boron has a small size and high ionization energy. It does not easily lose electrons to form $ B^{3+} $ ions, since removing three electrons requires a lot of energy.

→ Therefore, boron compounds (like BF₃, BCl₃, BH₃, etc.) are covalent in nature.

✅ Statement 1 is True.

Statement 2: The small size of $ B^{3+} $ favors formation of covalent bond.

→ According to Fajan’s rules, small cations with high charge density have strong polarizing power, which distorts the electron cloud of the anion, leading to covalency.

→ $ B^{3+} $ is extremely small and highly charged, therefore its compounds are predominantly covalent.

✅ Statement 2 is True.

Furthermore, Statement 2 gives the correct reason why boron compounds are covalent — due to the small size and high charge of $ B^{3+} $.

✅ Correct Option: A

Answer: Option A — Both statements are true, and Statement 2 is the correct explanation for Statement 1.

(a)   XeF₄  is sp³d² hybridised with 4 bond pairs and 1 lone pair and structure is square planar. Here all the bond lengths are equal.



(b) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral. Here all the bond lengths are equal.



(c)   SF₄ is sp³d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles equal to < 120^o and 179^o instead of the expected angles of 120^o and 180^o respectively. Here axial and equitorial both bonds are presents. And we know axial bonds are longer and weaker.



(d) SiF₄ is sp³ hybridisation and regular tetrahedral geometry. Here all the bond lengths are equal.

(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}^ * $

So O $_2^{2 - }$ has no unpaired electrons.

(B) Molecular orbital configuration of B₂ (10 electrons) is

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

Here in B₂, 2 unpaired electrons present.

(C) Moleculer orbital configuration of $N_2^{ + }$ (13 electrons)

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

Here in $N_2^{ + }$, 1 unpaired electron present.

(D) 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}^ * $

So O$_2$ has 2 unpaired electrons.

Nitrogen(N) atom is smaller in size so it's lone pair is relatively unstable. That is why repulsion between lone pairs present on nitrogen atom and bonded pairs of electrons is relatively high.

But size of Sb is more so its lone pair is relatively stable. That is why it does not want to perticipate in bond pair lone pair repulsion and decreases lp-bp repulsion.

The lattice energy of an ionic compound is a measure of the strength of the bonds in that ionic compound. Specifically, it is the energy required to separate one mole of an ionic solid into its constituent ions in the gaseous state. The lattice energy depends on several factors, primarily the charge on the ions and the size (or radius) of the ions.

1. Charge on the ions: The lattice energy is directly proportional to the product of the charges of the cation and the anion. Higher charges result in stronger electrostatic attraction between the ions, thereby increasing the lattice energy. This relationship can be derived from Coulomb's law, which states that the force of attraction between two charged particles is directly proportional to the product of their charges.

Mathematically:

$U \propto \frac{q_1 \cdot q_2}{r}$

where $U$ is the lattice energy, $q_1$ and $q_2$ are the charges on the ions, and $r$ is the distance between the centers of the ions.

2. Size of the ions: The lattice energy is inversely proportional to the distance between the ions, which depends on the sum of their ionic radii. Smaller ions can come closer together, increasing the electrostatic attractions between them, thereby increasing the lattice energy. Larger ions, being further apart, result in a lower lattice energy.

Considering these two factors, the correct choice is:

Option D: Charge on the ion and size of the ion

TIPS/Formulae :

A diamagnetic substance contains no unpaired electron.

${H_2}$ is diamagnetic as it contains all paired electrons

$\mathop {{H_2} = \sigma _b^2}\limits_{\left( {diamagnetic} \right)} \,\,,\,\,\mathop {H_2^ + = \sigma _b^1,}\limits_{\left( {paramagnetic} \right)} \,\,\mathop {H_2^ - = \sigma _b^2,}\limits_{\left( {paramagnetic} \right)} $

$\mathop {\sigma _a^{ * 1};He_2^ + }\limits_{\left( {paramagnetic} \right)} = \mathop {\sigma _b^2,\sigma _a^{ \ne 1}}\limits_{\left( {paramagnetic} \right)} $

To determine which species has the maximum number of lone pairs of electrons on the central atom, we'll examine the electron configuration of each molecule's central atom:

Option A: $[ClO_3]^-$

The central atom is chlorine (Cl).

Chlorine has 7 valence electrons. In the $[ClO_3]^-$ ion, chlorine forms three bonds with oxygen atoms and there is an additional negative charge (one extra electron). Thus, chlorine has:

$7 + 1 - 3 = 5$ valence electrons left, which form 2.5 lone pairs. However, due to the requirement to minimize formal charges, typically chlorine in $[ClO_3]^-$ is involved in resonance structures reducing the average lone pairs per resonance structure to 1.

So, $[ClO_3]^-$ has 1 lone pair on the central atom.

Option B: $XeF_4$

The central atom is xenon (Xe).

Xenon has 8 valence electrons. In $XeF_4$, xenon forms four bonds with four fluorine atoms. Thus, xenon has:

$8 - 4 = 4$ valence electrons left, which form 2 lone pairs.

So, $XeF_4$ has 2 lone pairs on the central atom.

Option C: $SF_4$

The central atom is sulfur (S).

Sulfur has 6 valence electrons. In $SF_4$, sulfur forms four bonds with four fluorine atoms. Thus, sulfur has:

$6 - 4 = 2$ valence electrons left, which form 1 lone pair.

So, $SF_4$ has 1 lone pair on the central atom.

Option D: $[I_3]^-$

The central atom is iodine (I).

Iodine has 7 valence electrons. In the $[I_3]^-$ ion, iodines form a linear structure where the central iodine forms two bonds with two other iodine atoms and there is an additional negative charge (one extra electron). Thus, the central iodine has:

$7 + 1 - 2 = 6$ valence electrons left, which form 3 lone pairs.

So, $[I_3]^-$ has 3 lone pairs on the central atom.

Hence, the species with the maximum number of lone pairs of electrons on the central atom is:

Option D: $[I_3]^-$

Here eight 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1, $\angle $5M1, $\angle $5M2, $\angle $5M3, $\angle $5M4.

Here twelve 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1, $\angle $5M1, $\angle $5M2, $\angle $5M3, $\angle $5M4, $\angle $6M1, $\angle $6M2, $\angle $6M3, $\angle $6M4.

Here four 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1.

Here six 90^o angles between bond pair and bond pair. Those angles are $\angle $1M3, $\angle $1M4, $\angle $1M5, $\angle $2M3, $\angle $2M4, $\angle $2M5.

Structure of  H₃BO₃  is



Here Boron has 3 sigma bond. So it is sp² hybridised.

And oxygen has two sigma bond and two lone pair. So it sp³ hybridised.

1. In H₂S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H₂S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(S) atom and size of S^$-$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

2. In NH₃, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107^o.



3. $\,\,\,\,$ BF₃ has sp² hybridization. So bond angle is 120^o.



4.
Here Si is sp³ hybridised and shape is regular tetrahedral and bond angle 109^o 28'

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}^ * $

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

N_a = 5

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

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

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

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

N_a = 4

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

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

So, bond length in NO > NO⁺

Regular Tetrahedral structure is possible in sp³ hybridization where central atom has 4 bond pair and no lone pair.

(a)   XeF₄  is sp³d² hybridised and structure is square planar.



(b)   [Ni(CN)₄]^$-$2 is coordinate compound and oxidation number of Ni is +2.

Electronic configuration of Ni⁺² is $=$ [Ar]3d⁸



But because of CN^$-$ ion which is a strong field ligand , it can perform pairing of electron.



And the structure of dsp² hybridization is square planar.



(C) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral.



(d)   

SF₄ is sp³d hybridised and structure is see-saw.

Here $\mu $_total = $\mu $₁ + $\mu $₂ $ \ne $ 0

$\mu $_total = $\mu $₁ $-$ $\mu $₂ = 0

$\mu $_total = $\mu $₁ + $\mu $₂ $ \ne $ 0

$\mu $₁ + $\mu $₂ + $\mu $₃ = $\mu $₄

   $\mu $_total = 0

In NO₂ and O₃ have permanent dipole moment as both of them are not symmetric molecule.

In CO₂ and SiF₄ each individual bond c = O and S- $-$ F have dipole moment but because of symmetric structure individual dipole moment get's cancelled.