A) \[{{N}_{2}}>N_{2}^{+}>N_{2}^{-}>N_{2}^{2-}\]
B) \[N_{2}^{+}>N_{2}^{-}>N_{2}^{2-}>{{N}_{2}}\]
C) \[{{N}_{2}}>N_{2}^{+}=N_{2}^{-}>N_{2}^{2-}\]
D) \[N_{2}^{-}>N_{2}^{{}}=N_{2}^{+}>N_{2}^{2-}\]
Correct Answer: C
Solution :
[c] Bond order of various species \[{{N}_{2}}=3;N_{2}^{+}=2.5;N_{2}^{-}=2.5;\] \[{{N}_{2}}^{2-}=2;\] Bond order\[\propto \] bond energy. Hence, order is \[{{N}_{2}}>N_{2}^{+}=N_{2}^{-}>N_{2}^{2-}\]
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