A) \[\text{C}{{\text{N}}^{\text{+}}}\]
B) \[\text{C}{{\text{N}}^{\text{--}}}\]
C) NO
D) CN
Correct Answer: B
Solution :
\[{{\text{(}\sigma \text{1s)}}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(\sigma 2{{p}_{z}})}^{2}},{{(\pi 2{{P}_{x}})}^{2}}\] \[={{(\pi 2{{p}_{y}})}^{2}},{{(\pi *2{{p}_{x}})}^{1}}={{(\pi *2{{p}_{y}})}^{0}}\] \[\text{BO=}\frac{\text{10-5}}{\text{2}}\text{=2}\text{.5}\] \[\text{C}{{\text{N}}^{-}};{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(\pi 2{{p}_{x}})}^{2}}\]\[={{(\pi 2{{p}_{y}})}^{2}},{{(\sigma 2{{p}_{z}})}^{2}}\] \[\text{BO=}\frac{10-4}{2}=3\] \[\text{CN:(}\sigma \text{1s}{{\text{)}}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(\pi 2{{p}_{x}})}^{2}}\]\[={{(\pi 2{{p}_{y}})}^{2}},{{(\sigma 2{{p}_{z}})}^{1}}\] \[\text{BO=}\frac{9-4}{2}=2.5\] \[\text{C}{{\text{N}}^{\text{+}}}:{{(\sigma 1s)}^{2}},{{(\sigma *1s)}^{2}},{{(\sigma 2s)}^{2}},{{(\sigma *2s)}^{2}},{{(\pi 2{{p}_{x}})}^{2}}\]\[={{(\pi 2{{p}_{y}})}^{2}}\] \[BO=\frac{8-4}{2}=2\] Hence, option [b] should be the right answer.
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