Which of the following molecule(s) show/s paramagnetic behavior?
A. O₂
B. N₂
C. F₂
D. S₂
E. Cl₂
Choose the correct answer from the options given below:

Solution

Chemistry ma’am · 3 steps

Fill the MO diagram and look only at the π^* orbitals: two half-filled means paramagnetic.

1. What decides magnetism. A molecule is paramagnetic only if it has at least one unpaired electron. In these diatomics, any unpaired electrons sit in the two degenerate π^* orbitals, so count the electrons there.
2. Same group, same pattern. S₂ is built like O₂, and Cl₂ like F₂. Each pair has the same number of valence electrons, just from the next shell, so each fills its MO diagram the same way.

   Molecule	Valence electrons	Electrons in π^*	Unpaired	Behaviour
   O₂	12	2, one in each π^* (Hund's rule)	2	paramagnetic
   N₂	10	0	0	diamagnetic
   F₂	14	4, both π^* full	0	diamagnetic
   S₂	12	2, one in each π^* (Hund's rule)	2	paramagnetic
   Cl₂	14	4, both π^* full	0	diamagnetic

3. So. Only O₂ (A) and S₂ (D) have unpaired electrons, so the answer is (c), A & D only.

The idea to keep. For homonuclear diatomics, 12 valence electrons (like O₂ and S₂) leaves two unpaired π^* electrons. Any element from the same group behaves the same way.

Answer from the NTA final answer key. Solution by PrepWiser.