A) \[N_{2}^{+}\]
B) \[O_{2}^{2+}\]
C) \[{{N}_{2}}\]
D) \[N{{O}^{+}}\]
Correct Answer: A
Solution :
B.O. for \[N_{2}^{+}\] =\[\frac{1}{2}[{{N}_{b}}-{{N}_{a}}]\]=\[\frac{1}{2}[9-4]=\frac{5}{2}=2.5\].
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