A) \[O_{2}^{2-},{{B}_{2}}\]
B) \[O_{2}^{+},N{{O}^{+}}\]
C) \[NO,CO\]
D) \[{{N}_{2}},{{O}_{2}}\]
Correct Answer: A
Solution :
\[O_{2}^{2-}(8+8+2=18)\] \[=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\,\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\overset{{}}{\mathop{\sigma }}\,2p_{z}^{2},\pi 2p_{x}^{2}\] \[\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{2}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{2}\] Bond order (BO)\[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] \[=\frac{10-8}{2}=1\] \[{{B}_{2}}(5+5=10)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\overset{{}}{\mathop{\sigma }}\,2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\] \[\pi 2p_{x}^{1}\approx \pi 2p_{y}^{1}\] \[BO=\frac{6-4}{2}=1\] Thus, \[O_{2}^{2-}\] and \[B_{2}^{{}}\] have the same bond order. Note \[BO\]of \[O_{2}^{+}=2.5,N{{O}^{+}}=3,NO=2.5,\]\[CO=3,{{N}_{2}}=3\]and\[{{O}_{2}}=2\]
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