A) \[\text{O}_{\text{2}}^{\text{+}}\]
B) \[\text{O}_{\text{2}}^{-}\]
C) \[\text{O}_{\text{2}}^{2-}\]
D) \[\text{O}_{\text{2}}^{{}}\]
Correct Answer: B
Solution :
MO configuration of \[O_{4}^{+}(8+8-1=15)\] \[=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2s,\sigma 2p_{z}^{2},\] \[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{0}\] \[BO=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] (where, \[{{N}_{b}}=\] number of electrons in bonding molecular orbital \[{{N}_{a}}=\] number of electrons in anti-bonding molecular orbital \[\therefore \] \[BO=\frac{10-5}{2}=2.5\]Similarly, \[O_{2}^{-}(8+8+1=17)\] so\[BO=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-7}{2}=1.5\] \[O_{2}^{2=}(8+8+2=18)\] \[BO=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-8}{2}=1\] \[{{O}_{2}}(8+8=16)\] \[BO=\frac{10-6}{2}=2\] Thus, \[O_{2}^{-}\] shows the bond order 1.5.
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