A) \[{{H}_{2}}\]
B) \[{{O}_{2}}\]
C) \[{{N}_{2}}\]
D) \[CO\]
Correct Answer: B
Solution :
\[{{O}_{2}}\to \]\[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\pi 2{{p}_{x}}^{2}\ \left\{ \begin{matrix} \pi 2{{p}_{y}}^{2} \\ {} \\ \pi 2{{p}_{z}}^{2} \\ \end{matrix} \right.\left\{ \begin{matrix} {{\pi }^{*}}2{{p}_{y}}^{1} \\ {} \\ {{\pi }^{*}}2{{p}_{z}}^{2} \\ \end{matrix} \right.\ \] So two unpaired electron found in \[{{O}_{2}}\] at ground stage by which it shows paramagnetism.
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