QJEEM21Ch255b: JEE Main 2021 Chemistry - Class 11 JEEM21Ch255b Solution: According to molecular orbital theory, the species among the…

Note :

According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.

We know, 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

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

$\,\,\,$ Configuration of $He_2^ +$ (3 electrons) is = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^1}}^ *$

$\therefore\,\,\,$ Bond order = ${1 \over 2}$ (2 $-$ 1) = 0.5

$\,\,\,$ Configuration of $He_2^ -$ (5 electrons) is = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * {\sigma _{2{s^1}}}\,$

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

$\,\,\,$ Configuration of $Be_2$ (4 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^2}}^ *$

$\therefore$ Bond order = ${1 \over 2}$ (2 $-$ 2) = 0