A) zero
B) one
C) two
D) four
Correct Answer: C
Solution :
The molecular orbital elctronic configuration of\[{{O}_{2}}\]is - \[{{O}_{2}}(16{{e}^{-}})=\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\sigma 2p{{x}^{2}}\] \[\pi 2p{{y}^{2}}\pi 2p{{z}^{2}}{{\pi }^{*}}2p{{y}^{1}}{{\pi }^{*}}2p{{z}^{1}}\] Hence, it has two unpaired Iectrons and B paramagnetic.You need to login to perform this action.
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