In the molecular orbital diagram for the mole | vedclass

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Question

Total electrons in  $N_2^ +  = (7 	imes 2) - 1 = 13$

$N_2^ +  	o {\sigma _{1{s^2}}},\sigma _{1{s^2}}^*,{\sigma _{2{s^2}}},\si

Vedclass · Accepted Answer

$1$

Vedclass · Answer

Total electrons in  $N_2^ +  = (7 	imes 2) - 1 = 13$

$N_2^ +  	o {\sigma _{1{s^2}}},\sigma _{1{s^2}}^*,{\sigma _{2{s^2}}},\sigma _{2{s^2}}^*,[\pi _{2{p_x}}^2 = \pi _{2{p_y}}^2]\sigma _{2{p_z}}^1$

Number of electron in ${\sigma _2}_{{p_z}}$ is $1$

In the molecular orbital diagram for the molecular ion, $N_2^ + $ , the number of electrons in the ${\sigma _{2p}}$ molecular orbital is

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