A) \[O_{2}^{2-}>O_{2}^{+}>O_{2}^{-}>{{O}_{2}}\]
B) \[O_{2}^{+}>O_{2}^{2-}>O_{2}^{-}>{{O}_{2}}\]
C) \[O_{2}^{+}>{{O}_{2}}>O_{2}^{-}>O_{2}^{2-}\]
D) \[{{O}_{2}}>O_{2}^{-}>O_{2}^{2-}>O_{2}^{+}\]
Correct Answer: C
Solution :
\[{{O}_{2}}(8+8=16)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,{{s}^{2}},\sigma {{s}^{2}},\sigma 2p_{z}^{2}\], \[\pi \,2p_{x}^{2}\approx \pi \,2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] \[BO=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-6}{2}=2\] \[O_{2}^{+}(8+8-1=15)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2}\], \[\pi 2_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\] \[BO=\frac{10-5}{2}=2.5\] \[O_{2}^{-}(8+8+1=17)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma 2{{s}^{2}}}}\,\], \[\sigma 2p_{z}^{2},\pi p_{x}^{2}\approx \pi 2p_{y}^{2},\,\,\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] \[BO=\frac{10-7}{2}=1.5\] \[O_{2}^{2-}(8+8+2=18)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\], \[\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2}\approx 2p_{x}^{2}\approx \pi \,2p_{y}^{2}\] \[BO=\frac{10-8}{2}=1\] Thus, the correct order of bond order is \[O_{2}^{2-}<O_{2}^{-}<{{O}_{2}}<O_{2}^{+}\]
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