A) Paramagnetic and bond order<\[{{O}_{2}}\]
B) Paramagnetic and bond order>\[{{O}_{2}}\]
C) Dimagnetic and bond order<\[{{O}_{2}}\]
D) Dimagnetic and bond order>\[{{O}_{2}}\]
Correct Answer: B
Solution :
\[{{O}_{2}}\]: \[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2{{p}_{x}}^{2}\left\{ \begin{matrix} \pi 2{{p}_{y}}^{2} \\ {} \\ \pi 2{{p}_{z}}^{2} \\ \end{matrix} \right.\ \left\{ \begin{matrix} {{\pi }^{*}}2{{p}_{y}}^{1} \\ {} \\ {{\pi }^{*}}2{{p}_{z}}^{1} \\ \end{matrix} \right.\] Bond order \[=\frac{10-6}{2}=2.0\] (Two unpaired electrons in antibonding molecular orbital)\[O_{2}^{+}:\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2{{p}_{x}}^{2}\left\{ \begin{matrix} \pi 2p{{y}^{2}} \\ \pi 2p{{z}^{2}} \\ \end{matrix} \right.\left\{ \begin{matrix} {{\pi }^{*}}2p{{y}^{1}} \\ {{\pi }^{*}}2p{{z}^{0}} \\ \end{matrix} \right.\]Bond order \[=\frac{10-5}{2}=2.5\] (One unpaired electron in antibonding molecular orbital so it is paramagnetic)
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