A) \[(2\pi _{x}^{*}){{(2{{\pi }_{y}})}^{1}}\]
B) \[{{({{\pi }^{*}}2{{p}_{y}})}^{1}}({{\pi }^{*}}2p_{z}^{1})\]
C) \[{{(2\sigma _{s}^{*})}^{1}}{{(2{{\pi }_{y}})}^{1}}\]
D) \[{{(2{{\sigma }_{s}}^{*})}^{1}}{{(2{{\pi }_{y}})}^{1}}\]
Correct Answer: B
Solution :
Key Idea: Write configuration of oxygen according to molecular orbital theory \[{{O}_{2}}=16=\sigma 1{{s}^{2}},\,\,\sigma {{1}^{*}}{{s}^{2}},\,\,\sigma 2{{s}^{2}},\,\,{{\sigma }^{*}}2{{s}^{2}},\,\,\sigma 2{{p}_{x}}^{2},\] \[\pi 2{{p}_{y}}^{2},\,\,\pi 2{{p}_{z}}^{2},\,\,{{\pi }^{*}}2{{p}_{y}}^{1},\,\,{{\pi }^{*}}{{p}_{z}}^{1}\] \[\therefore \]It has two unpaired electrons. \[\therefore \]\[{{O}_{2}}\] is paramagnetic.You need to login to perform this action.
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