A) 2, 2
B) 4, 2
C) 4, 0
D) 2, 4
Correct Answer: A
Solution :
Number of electrons in\[H{{e}_{2}}=4\] MQ configuration of\[H{{e}_{2}}=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}}\] Here,\[{{N}_{b}}\](bonding electrons) = 2 \[{{N}_{a}}\] (antibonding electrons) =2
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