Step 1: Total electrons
Fluorine has $Z=9$, so the molecule has $2\times 9 = 18$ electrons.
Step 2: Use bond order
For $F_2$ the bond order is $1$. Let bonding electrons be $N_b$ and antibonding $N_a$. Then $N_b + N_a = 18$ and $(N_b - N_a)/2 = 1$.
Step 3: Solve
$N_b - N_a = 2$, so $N_b = 10$ and $N_a = 8$.
This matches the MO filling: all of the $\pi^*$ orbitals are full and $\sigma^* 2p$ is empty.
Step 4: Answer
Option (C).
Final Answer:
Bonding 10 and antibonding 8.
\[ \boxed{\text{(C)}\ \text{Bonding }10,\ \text{Antibonding }8} \]