QJEEM21Ch255c: JEE Main 2021 Chemistry - Class 11 JEEM21Ch255c Solution: The difference between bond orders of CO and NO is where x =…

Bond order of CO = 3

Bond order of NO⁺ = 3

Difference = 0 = ${x \over 2}$

$\Rightarrow$ x = 0

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)    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

(B)   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