Chemistry - CHEMICAL BONDING Question with Solution | TestHub

In O₂ the $2s$-$2p$ energy gap is large, so $s$-$p$ mixing is naturally negligible. Following the standard molecular orbital sequence (${\sigma }_{2p_z}<{\pi }_{2p_{x,y}}<{\pi }_{2p_{x,y}}^{\ast }$), the 16 electrons result in a bond order of 2 ($\frac{10-6}{2}$). Because the last two electrons occupy the degenerate ${\pi }^{\ast }$ orbitals singly according to Hund's rule, the molecule is paramagnetic. Thus, both (A) and (C) are correct.