NEET Solved Paper 2018 Question 42
Question: Consider the following species $ :C{N^{+}}\text{, C}{N^{–}}\text{, NO and CN} $ . Which one of these will have the highest bond order? [NEET - 2018]
Options:
A) $ C{N^{+}} $
B) $ C{N^{–}} $
C) NO
D) CN
Show Answer
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.
