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(A) The total number of electrons in a neutral $N_2$ molecule is $14$ $(7 + 7)$.For the $N_2^+$ ion,the total number of electrons is $14 - 1 = 13$.Acc

Vedclass · Accepted Answer

$\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p)^4 \sigma(2p)^1$

Vedclass · Answer

(A) The total number of electrons in a neutral $N_2$ molecule is $14$ $(7 + 7)$.For the $N_2^+$ ion,the total number of electrons is $14 - 1 = 13$.According to the molecular orbital theory for $N_2$ (where the $\pi(2p_x) = \pi(2p_y)$ orbitals have lower energy than the $\sigma(2p_z)$ orbital),the filling order is: $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2$.For $N_2^+$ ($13$ electrons),we remove one electron from the highest energy orbital,which is $\sigma(2p_z)$.Thus,the configuration becomes: $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p)^4 \sigma(2p)^1$.

From elementary molecular orbital theory,the electronic configuration of the singly positive nitrogen molecular ion $N_2^+$ is:

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