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 = No of electrons in bending molecular orbital

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

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

JEE Main 2019 (Online) 8th April Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 186 English Explanation 1

JEE Main 2019 (Online) 8th April Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 186 English Explanation 1

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

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

JEE Main 2019 (Online) 8th April Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 186 English Explanation 2

JEE Main 2019 (Online) 8th April Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 186 English Explanation 2

Here N_a = 10

and N_b = 10

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

Then 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 = $${1 \over 2}\left[ {10 - 6} \right] = 2$$

Here 2 unpaired electrons are present so it is paramagnetic.

(D) 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

Here no unpaired electrons are present so it is diamagnetic.

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

Here no unpaired electron present, so it is diamagnetic.

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

Here 2 unpaired electron present, so it is paramagnetic.

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

so among two diamagnetic ions $$C_2^{2 - }$$ and $$O_2^{2 - }$$, bond order of $$C_2^{2 - }$$ is more so it will have shorter bond length.