The existence of a molecule can be predicted by its bond order. A molecule with a bond order of zero does not exist.

Bond Order Calculation:

Bond order can be calculated using Molecular Orbital Theory (MOT) as:

$\text{Bond Order}=\frac{1}{2}(\text{Number of electrons in bonding MOs}-\text{Number of electrons in antibonding MOs})$

Option A: He${}_2$

Total electrons in He${}_2$ = 2 (from first He) + 2 (from second He) = 4 electrons.

Electronic configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2$

Number of bonding electrons = 2

Number of antibonding electrons = 2

Bond Order = $\frac{1}{2}(2-2)=0$

Since the bond order is 0, He${}_2$ molecule does not exist.

Option B: F${}_2$

Total electrons in F${}_2$ = 9 (from first F) + 9 (from second F) = 18 electrons.

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

Number of bonding electrons = 2 + 2 + 2 + 2 + 2 = 10

Number of antibonding electrons = 2 + 2 + 2 + 2 = 8

Bond Order = $\frac{1}{2}(10-8)=1$

Since the bond order is 1, F${}_2$ molecule exists.

Option C: C${}_2$

Total electrons in C${}_2$ = 6 (from first C) + 6 (from second C) = 12 electrons.

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

Number of bonding electrons = 2 + 2 + 2 + 2 = 8

Number of antibonding electrons = 2 + 2 = 4

Bond Order = $\frac{1}{2}(8-4)=2$

Since the bond order is 2, C${}_2$ molecule exists.

Option D: N${}_2$

Total electrons in N${}_2$ = 7 (from first N) + 7 (from second N) = 14 electrons.

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

Number of bonding electrons = 2 + 2 + 2 + 2 + 2 = 10

Number of antibonding electrons = 2 + 2 = 4

Bond Order = $\frac{1}{2}(10-4)=3$

Since the bond order is 3, N${}_2$ molecule exists.

Therefore, He${}_2$ does not exist because its bond order is zero.

Correct Answer: (A)