A) \[\text{N}_{2}^{+}\]
B) \[C{{N}^{-}}\]
C) \[CO\]
D) \[N{{O}^{+}}\]
Correct Answer: A
Solution :
\[O_{2}^{+}(15{{e}^{-}})=K\,{{K}^{*}}{{({{\sigma }^{*}}2s)}^{2}}\] \[{{(\sigma 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{(\pi 2{{p}_{z}})}^{2}}\] \[{{({{\pi }^{*}}2{{p}_{y}})}^{1}}{{({{\pi }^{*}}2{{p}_{z}})}^{0}}\] Hence, bond order \[=\frac{1}{2}(10-5)=2.5\] \[N_{2}^{+}(13{{e}^{-}})=K\,{{K}^{*}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{({{\sigma }^{*}}2{{p}_{x}})}^{2}}\] \[{{(\pi 2{{p}_{y}})}^{2}}{{(\pi 2{{p}_{z}})}^{1}}\] Hence, bond order \[=\frac{1}{2}(9-4)=2.5\]
You need to login to perform this action.
You will be redirected in
3 sec
