Chemical Bonding & Molecular Structure

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.

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.

$\mathrm{OSF}_2$ 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.

(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.