A) \[~{{N}_{2}}\]
B) \[{{O}_{2}}\]
C) \[{{F}_{2}}\]
D) \[C{{O}_{2}}\]
Correct Answer: B
Solution :
Paramagnet shown by\[{{O}_{2}}.\] \[\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{1+1}\] \[\approx \pi 2p_{y}^{1+1}\] \[\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\approx \overset{*}{\mathop{\pi }}\,p_{y}^{1}\overset{*}{\mathop{\sigma }}\,2{{p}_{z}}\] \[\therefore \]It contains two unpaired electrons so, is paramagnetic.
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