Question

Show $O_2$ is a paramagnetic.

Answer 1

$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$

$\because$ there are unpaired $e^{-}$ s present in the antibonding $\Pi$ orbital

Thus, $O_2$ is paramagnetic.