Provide the electron configuration,magnetic property,bond order,and molecular orbital energy diagram for the fluorine $(F_2)$ molecule.

(N/A) The atomic number of fluorine $(F)$ is $Z=9$,and its electronic configuration is $1s^2 2s^2 2p^5$.
There are $7$ valence electrons in each $F$ atom,so the total number of electrons in the $F_2$ molecule is $14$.
The molecular orbital $(MO)$ configuration for $F_2$ is: $KK(\sigma_{2s})^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^2(\pi^* 2p_y)^2$.
Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 8) = 1$.
This indicates a single bond between the two fluorine atoms $(F-F)$.
Magnetic property: Since all electrons are paired,the $F_2$ molecule is diamagnetic.
The energy level diagram is provided below: