A) 2, 2
B) 4, 2
C) 4, 0
D) 2, 4
Correct Answer: A
Solution :
Number of electron in \[H{{e}_{2}}=4\] MO configuration of \[H{{e}_{2}}=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}}\] Here, \[{{N}_{b}}\](bonding electrons) \[=2\] \[{{N}_{a}}\](antibonding electrons) =2You need to login to perform this action.
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