The correct molecular orbital diagram for F$$_2$$ molecule in the ground state is

The relative energies of the molecular orbitals (MOs) formed from the 2s- and 2p-orbitals depend on the atomic number of the two atoms.

For $$Z \ge 8$$ (that is, for $$O_2,\,F_2,\,Ne_2$$) the order is

$$\sigma_{2s}\;,\;\sigma^{*}_{2s}\;,\;\sigma_{2p_z}\;,\;\pi_{2p_x}=\pi_{2p_y}\;,\;\pi^{*}_{2p_x}=\pi^{*}_{2p_y}\;,\;\sigma^{*}_{2p_z} \quad -(1)$$

Fluorine has $$Z = 9$$, so $$F_2$$ must follow the sequence given in $$(1)$$.

Each fluorine atom supplies 7 valence electrons. Total valence electrons in $$F_2$$:

$$7 + 7 = 14$$

Placing these 14 electrons into the MOs of $$(1)$$:

$$\sigma_{2s}^2\,\sigma^{*}_{2s}^2\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi^{*}_{2p_x}^2\,\pi^{*}_{2p_y}^2\,\sigma^{*}_{2p_z}^0$$

Number of bonding electrons $$N_b = 8$$ (those in $$\sigma_{2s},\,\sigma_{2p_z},\,\pi_{2p_x},\,\pi_{2p_y}$$).
Number of antibonding electrons $$N_a = 6$$ (those in $$\sigma^{*}_{2s},\,\pi^{*}_{2p_x},\,\pi^{*}_{2p_y}$$).

Bond order

$$\text{B.O.} = \frac{N_b - N_a}{2} = \frac{8 - 6}{2} = 1$$

All electrons are paired, so $$F_2$$ is diamagnetic.

The diagram that shows the energy sequence of $$(1)$$, with $$\pi^{*}_{2p_x}$$ and $$\pi^{*}_{2p_y}$$ completely filled, $$\sigma^{*}_{2p_z}$$ empty, and no unpaired electrons, is Option C.

Therefore, the correct molecular orbital diagram for $$F_2$$ in its ground state is given by Option C.