Question
The species having bond order different from that in $$CO$$ IS
A.
$$N{O^ - }$$
B.
$$N{O^{ + \,\,}}$$
C.
$$C{N^ - }$$
D.
$${N_2}$$
Answer :
$$N{O^ - }$$
Solution :
Molecular electronic configuration of
$$CO:\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\sigma 2p_x^2} \right.$$
Therefore, bond order $$ = \frac{{{N_b} - {N_a}}}{2} = \frac{{10 - 4}}{2} = 3$$
$$N{O^ + }:\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$ $$\left\{ {\pi 2p_y^2 = \pi p_z^2} \right.$$
Bond order $$ = \frac{{10 - 4}}{2} = 3$$
$$C{N^ - } = \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},$$ $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2} \right.,\sigma 2p_x^2$$
Bond order $$ = \frac{{10 - 4}}{2} = 3$$
$${N_2}:\,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},$$ $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2} \right.,\sigma 2p_x^2$$
Bond order $$\, = \frac{{10 - 4}}{2} = 3$$
$$N{O^ - }:\,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$ $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2} \right.,$$ $$\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^1} \right.$$
Bond order $$\, = \frac{{10 - 6}}{2} = 2$$
∴ $$N{O^ - }$$ has different bond order from that in $$CO.$$