To determine the number of antibonding electrons in ${\mathrm{N}}_2$, we first need to write its molecular orbital (MO) electronic configuration.

Step 1: Determine the total number of electrons.
Nitrogen (N) has 7 electrons. So, ${\mathrm{N}}_2$ has $2\times 7=14$ electrons.

Step 2: Write the molecular orbital configuration for ${\mathrm{N}}_2$.
For diatomic molecules with 14 or fewer electrons (like ${\mathrm{N}}_2$), the order of filling molecular orbitals is:

${\sigma }_{1s}<{\sigma }_{1s}^{\ast }<{\sigma }_{2s}<{\sigma }_{2s}^{\ast }<{\pi }_{2p_x}={\pi }_{2p_y}<{\sigma }_{2p_z}<{\pi }_{2p_x}^{\ast }={\pi }_{2p_y}^{\ast }<{\sigma }_{2p_z}^{\ast }$

Filling 14 electrons according to this order:

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

Step 3: Count the antibonding electrons.
Antibonding orbitals are denoted by an asterisk (*). From the configuration, the antibonding orbitals are ${\sigma }_{1s}^{\ast }$ and ${\sigma }_{2s}^{\ast }$.

• Electrons in ${\sigma }_{1s}^{\ast }$ = 2
• Electrons in ${\sigma }_{2s}^{\ast }$ = 2

Total antibonding electrons = $2+2=4$.

Option Analysis:

• A) 4: This is the correct number of antibonding electrons as calculated from the MO configuration.
• B) 10: This represents the number of bonding electrons (2 in ${\sigma }_{1s}$, 2 in ${\sigma }_{2s}$, 4 in ${\pi }_{2p}$, 2 in ${\sigma }_{2p_z}$).
• C) 12: This is incorrect.
• D) 14: This is the total number of electrons in ${\mathrm{N}}_2$, not just antibonding electrons.

Correct Answer: (A)