A) \[CO\]
B) \[{{N}_{2}}\]
C) \[{{O}_{2}}\]
D) \[{{H}_{2}}{{O}_{2}}\]
Correct Answer: C
Solution :
Oxygen \[({{O}_{2}})\] contains 2 unpaired electron, hence it is paramagnetic. \[{{O}_{2}}=\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\sigma 2p_{x}^{2}\] \[\pi 2p_{y}^{2}\pi 2p_{z}^{2}{{\pi }^{*}}2p_{y}^{1}{{\pi }^{*}}2p_{z}^{1}\]You need to login to perform this action.
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