A) 2
B) 1.5
C) 3
D) 3.5
Correct Answer: A
Solution :
Electronic configuration of \[{{O}_{2}}\] is \[{{O}_{2}}={{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}\] \[(\pi 2p_{x}^{2}\equiv \pi 2p_{y}^{2})\ ({{\pi }^{*}}2p_{x}^{1}\equiv {{\pi }^{*}}2p_{y}^{1})\] Hence bond order \[=\frac{1}{2}\left[ {{N}_{b}}-{{N}_{a}} \right]\] \[=\frac{1}{2}[10-6]=2\].
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