Atomic orbitals involved

For each atom, the orbitals considered are:

• one $2s$ orbital
• three $2p$ orbitals: $2p_x, 2p_y, 2p_z$

So, for a diatomic molecule, these atomic orbitals combine pairwise to form molecular orbitals.

Formation from $2s$ orbitals

Two $2s$ atomic orbitals combine to give:

• one bonding MO: $\sigma_{2s}$
• one antibonding MO: $\sigma^*_{2s}$

Thus, antibonding orbitals from $2s$ = $1$.

Formation from $2p$ orbitals

The three $2p$ orbitals from one atom combine with the three $2p$ orbitals from the other atom.

• Along internuclear axis: one $\sigma_{2p}$ and one $\sigma^*_{2p}$
• Perpendicular to internuclear axis: two $\pi_{2p}$ and two $\pi^*_{2p}$

Thus, antibonding orbitals from $2p$ = $1\,(\sigma^*) + 2\,(\pi^*) = 3$

Total antibonding molecular orbitals

Therefore, $\text{Total antibonding MOs} = 1 + 3 = 4$

Final answer

$\boxed{4}$

Comparison with stored answer

Stored correct answer = $4$

This matches the derived answer.