QJEEM23Chdef1: JEE Main 2023 Chemistry - Class 11 JEEM23Chdef1 Solution: According to MO theory the bond orders for , and respectively,…

Species	B.O.
O $_2^{2 - }$	1
CO	3
NO $^ \oplus$	3

Note :

(1) $\,$ 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



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

(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 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

(B)    CO has 14 electrons.

Moleculer orbital configuration of CO 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_b = 10

N_a = 4

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

(C)   NO⁺ has 14 electrons.

Moleculer orbital configuration of NO⁺ is

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

$\therefore$ N_b = 10

N_a = 4

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