Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The molecular orbital (MO) configuration of O₂ is $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p_z}{)}^2({\pi }_{2p_x}{)}^2({\pi }_{2p_y}{)}^2({\pi }_{2p_x}^{\ast }{)}^1({\pi }_{2p_y}^{\ast }{)}^1$.

A. The Highest Occupied Molecular Orbital (HOMO) is ${\pi }_{2p}^{\ast }$. This is correct.

B. There are two unpaired electrons in the ${\pi }_{2p}^{\ast }$ (antibonding) orbitals. These are degenerate and are filled according to Hund's rule. This is correct.

C. The ionization energy of O₂ (1207 kJ/mol) is less than that of an oxygen atom (1314 kJ/mol). This option is incorrect.

D. Bond order = $1/2$ (number of bonding electrons - number of antibonding electrons) = $1/2(10-6)=2$. The bond consists of one $\sigma$ bond and two half $\pi$ bonds, which effectively makes one $\pi$ bond. This is correct.