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4.Chemical Bonding and Molecular Structure
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In the molecular orbital diagram for the molecular ion, $N_2^ + $ , the number of electrons in the ${\sigma _{2p}}$ molecular orbital is
A
$0$
B
$2$
C
$3$
D
$1$
(JEE MAIN-2018)
Solution
Total electrons in $N_2^ + = (7 \times 2) – 1 = 13$
$N_2^ + \to {\sigma _{1{s^2}}},\sigma _{1{s^2}}^*,{\sigma _{2{s^2}}},\sigma _{2{s^2}}^*,[\pi _{2{p_x}}^2 = \pi _{2{p_y}}^2]\sigma _{2{p_z}}^1$
Number of electron in ${\sigma _2}_{{p_z}}$ is $1$
Standard 11
Chemistry
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