A) $ N_2^{+} $
B) $ O_2^{2+} $
C) $ N_2 $
D) $ N{O^{+}} $
Correct Answer: A
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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