Give the electron configuration,magnetic property,bond order,and energy diagram for the oxygen $(O_2)$ molecule.

(N/A) $O_2$ $(Z=8)$ has the configuration $1s^2 2s^2 2p^4$. Total electrons in $O_2 = 16$.
Molecular orbital $(MO)$ configuration for $O_2$:
$KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$
Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 6) = 2$ (Double bond in $O_2$).
Since there are two unpaired electrons in the $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals,the molecule is paramagnetic.
The energy diagram for the $O_2$ molecule is provided below: