A) \[{{O}_{2}}\]
B) \[O_{2}^{+}\]
C) \[O_{2}^{-}\]
D) \[O_{2}^{2-}\]
Correct Answer: B
Solution :
Higher the bond order, maximum is the bond strength. [a] \[{{O}_{2}}(16)\] \[=\sigma 1{{s}^{2}},\sigma *1{{s}^{2}},\sigma 2{{s}^{2}},\sigma *2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\] \[\approx \pi 2p_{y}^{2},\pi *2p_{x}^{1}\approx \pi *2p_{y}^{1}\] Thus, \[BO=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-6}{2}=2\] [b] \[O_{2}^{+}(8+8-1=15),\] Thus, \[BO=\frac{10-5}{2}=2.5\] [c] \[O_{2}^{-}(8+8+1=17),\] Thus, \[BO=\frac{10-7}{2}=1.5\] [d] \[O_{2}^{2-}(8+8+2=18),\] \[BO=\frac{10-8}{2}=1\] Hence, bond strength is maximum in \[O_{2}^{+}\].
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