A) \[NO<C_{2}^{2-}<O_{2}^{-}<He_{2}^{+}\]
B) \[C_{2}^{2-}<He_{2}^{+}<NO<O_{2}^{-}\]
C) \[He_{2}^{+}<O_{2}^{-}<NO<C_{2}^{2-}\]
D) \[O_{2}^{-}<NO<C_{2}^{2-}<He_{2}^{+}\]
Correct Answer: C
Solution :
B.O. of \[NO=\frac{1}{2}(8-3)=\frac{5}{2}=2.5\] B.O. of \[C_{2}^{-}=\frac{1}{2}(8-2)=\frac{6}{2}=3\] B.O. of \[O_{2}^{-}=\frac{1}{2}(8-5)=\frac{3}{2}=1.5\] B.O. of \[He_{2}^{+}=\frac{1}{2}(2-1)=\frac{1}{2}=0.5\] \[\Rightarrow \]B.O. is, \[C_{2}^{-}>NO>O_{2}^{-}>He_{2}^{+}\]
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