Question

Consider the molecular orbital electron configurations for${\mathit{N}}_{\mathbf{2}}$ , ${\mathit{N}}_{\mathbf{2}}^{\mathbf{+}}$, and ${\mathit{N}}_{\mathbf{2}}^{\mathbf{-}}$. For each compound or ion, fill in the table below with the correct number of electrons in each molecular orbital.

MO	${\mathit{N}}_{\mathbf{2}}$	role="math" localid="1663764752673" ${\mathit{N}}_{\mathbf{2}}^{\mathbf{+}}$	role="math" localid="1663764766917" ${\mathit{N}}_{\mathbf{2}}^{\mathbf{-}}$
role="math" localid="1663764800461" ${\mathit{\sigma }}_{\mathbf{2}{\mathbf{p}}^{\mathbf{*}}}$
role="math" localid="1663764811953" ${\mathit{\pi }}_{\mathbf{2}{\mathbf{p}}^{\mathbf{*}}}$
role="math" localid="1663764822519" ${\mathit{\sigma }}_{\mathbf{2}\mathbf{p}}$
role="math" localid="1663764834336" ${\mathit{\pi }}_{\mathbf{2}\mathbf{p}}$
role="math" localid="1663764966754" ${\mathit{\sigma }}_{\mathbf{2}{\mathbf{s}}^{\mathbf{*}}}$
role="math" localid="1663764978558" ${\mathit{\sigma }}_{\mathbf{2}\mathbf{s}}$

Short answer

When two atomic orbitals of different elements combine, they form molecular orbital. By molecular orbital configuration, we will get the number of electrons in each orbital that is:

MO	$N_2$	$N_2^{+}$	$N_2^{-}$
${\sigma }_{2p^{*}}$	0	0	0
${\pi }_{2p^{*}}$	0	0	1
${\sigma }_{2p}$	2	2	2
${\pi }_{2p}$	4	4	4
${\sigma }_{2s^{*}}$	2	2	2
${\sigma }_{2s}$	2	2	2

Step-by-step solution

Molecular orbital configuration for N2

In N₂, there are a total of 14 electrons. So, the molecular orbital configuration will be

$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^2$

Molecular orbital configuration for N2+

In $N_2^{+}$, there are a total of 13 electrons as a positive charge is present on the molecule there will be one less electron.

$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^1$

Step 3: Molecular orbital configuration for N2-

In $N_2^{-}$, there is total of 15 electrons as a negative charge is present on the molecule, there will be one extra electron.

$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^2({\pi }^{*}2p_x{)}^1$