Chemistry - CHEMICAL BONDING Question with Solution | TestHub

To determine diamagnetic behavior, we examine the molecular orbital (MO) configuration for unpaired electrons. Diamagnetic molecules have all electrons paired.

A. C₂ (12 electrons): $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p}{)}^4({\sigma }_{2p}{)}^2$. All electrons are paired.

B. N₂ (14 electrons): $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p}{)}^4({\sigma }_{2p}{)}^2$. All electrons are paired.

C. O₂ (16 electrons): $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p}{)}^2({\pi }_{2p}{)}^4({\pi }_{2p}^{\ast }{)}^2$. Contains two unpaired electrons in ${\pi }_{2p}^{\ast }$ orbitals.

D. S₂ (32 electrons): Similar to O₂, it has two unpaired electrons in ${\pi }_{3p}^{\ast }$ orbitals.

Therefore, N₂ is diamagnetic.

The final answer is $\boxed{B}$