Answer:

Correct Answer: B

Solution:

$ {{\text{(}\sigma \text{1s)}}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(\sigma 2p _{z})}^{2}},{{(\pi 2P _{x})}^{2}} $

$ ={{(\pi 2p _{y})}^{2}},{{(\pi *2p _{x})}^{1}}={{(\pi *2p _{y})}^{0}} $

$ \text{BO=}\frac{\text{10-5}}{2}\text{=2}\text{.5} $

$ C{N^{-}};{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(\pi 2p _{x})}^{2}} $

$ ={{(\pi 2p _{y})}^{2}},{{(\sigma 2p _{z})}^{2}} $

$ \text{BO=}\frac{10-4}{2}=3 $

$ \text{CN:(}\sigma 1s{{)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(\pi 2p _{x})}^{2}} $

$ ={{(\pi 2p _{y})}^{2}},{{(\sigma 2p _{z})}^{1}} $

$ \text{BO=}\frac{9-4}{2}=2.5 $

$ C{N^{+}}:{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(\pi 2p _{x})}^{2}} $

$ ={{(\pi 2p _{y})}^{2}} $

$ BO=\frac{8-4}{2}=2 $ Hence, option [b] should be the right answer.