A) \[{{H}^{2+}}\]
B) \[H{{e}^{2+}}\]
C) \[{{O}_{2}}\]
D) \[{{N}_{2}}\]
Correct Answer: D
Solution :
: \[{{N}_{2}}(14):KK,\sigma 2{{s}^{2}},\sigma *2{{s}^{2}},\pi 2p_{x}^{2}=\pi 2p_{y}^{2},\sigma 2p_{z}^{2}\] It does not have any unpaired electrons. Hence, it is diamagnetic. \[H_{2}^{+}\] (1 electron), \[He_{2}^{+}\] (3 electrons) and \[{{O}_{2}}\] (16 electrons) are paramagnetic.
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