Consider the following species :
$CN^+, CN^–, NO$ and CN
Which one of these will have the highest bond order?
$CN^+$
$CN^–$
NO
CN
Explanation:
NO : $(\sigma 1s)^2, (\sigma^*1s)^2, (\sigma 2s)^2, (\sigma^*2s)^2, (\sigma 2p_z)^2, (\pi2p_x)^2$
$=(\pi 2p_y)^2, (\pi^*2p_x)^1=(\pi^*2p_y)^0$
$BO =\dfrac{10-5}{2}=2.5$
$CN^-:(\sigma 1s)^2, (\sigma^*1s)^2, (\sigma 2s)^2, (\sigma^*2s)^2, (\pi2p_x)^2$
$=(\pi2p_y)^2, (\sigma2p_z)^2$
$BO=\dfrac {10-4}{2}=3$
$CN:(\sigma 1s)^2, (\sigma^*1s)^2, (\sigma 2s)^2, (\sigma^*2s)^2, (\pi2p_x)^2$
$=(\pi2p_y)^2, (\sigma2p_z)^1$
$BO=\dfrac {9-4}{2}=2.5$
$CN^+:(\sigma 1s)^2, (\sigma^*1s)^2, (\sigma 2s)^2, (\sigma^*2s)^2, (\pi2p_x)^2$
$(\pi2p_y)^2$
$BO=\dfrac{8-4}{2}=2$
Hence, option(2) should be the right answer.
NO : $(\sigma 1s)^2, (\sigma^*1s)^2, (\sigma 2s)^2, (\sigma^*2s)^2, (\sigma 2p_z)^2, (\pi2p_x)^2$
$=(\pi 2p_y)^2, (\pi^*2p_x)^1=(\pi^*2p_y)^0$
$BO =\dfrac{10-5}{2}=2.5$
$CN^-:(\sigma 1s)^2, (\sigma^*1s)^2, (\sigma 2s)^2, (\sigma^*2s)^2, (\pi2p_x)^2$
$=(\pi2p_y)^2, (\sigma2p_z)^2$
$BO=\dfrac {10-4}{2}=3$
$CN:(\sigma 1s)^2, (\sigma^*1s)^2, (\sigma 2s)^2, (\sigma^*2s)^2, (\pi2p_x)^2$
$=(\pi2p_y)^2, (\sigma2p_z)^1$
$BO=\dfrac {9-4}{2}=2.5$
$CN^+:(\sigma 1s)^2, (\sigma^*1s)^2, (\sigma 2s)^2, (\sigma^*2s)^2, (\pi2p_x)^2$
$(\pi2p_y)^2$
$BO=\dfrac{8-4}{2}=2$
Hence, option(2) should be the right answer.