QJEEAdv16Chccd0: JEE Advanced 2016 Chemistry - Class 11 JEEAdv16Chccd0 Solution: According to Molecular Orbital Theory, which of the following…

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.