Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is _________.
(A) α-sulphur
(B) β-sulphur
(C) S₂-form

Solution

Chemistry ma’am · 4 steps

Look at the molecule inside: S₈ rings are fully paired, but S₂ is like O₂.

1. What paramagnetism needs. A substance is paramagnetic only if its molecules have unpaired electrons. So check which allotrope is built from molecules with unpaired electrons.
2. Rhombic and monoclinic. Both α-sulphur (rhombic) and β-sulphur (monoclinic) are made of puckered S₈ rings. Each S atom forms two single bonds and keeps two lone pairs, so every electron is paired.
3. The S2 form. S₂ exists in sulphur vapour at high temperature. Like O₂, it has two unpaired electrons, one in each of its two π^* orbitals, so it is paramagnetic.

   Allotrope	Molecular unit	Unpaired electrons	Magnetic nature
   α-sulphur	S₈ ring	0	diamagnetic
   β-sulphur	S₈ ring	0	diamagnetic
   S₂ form	S₂ molecule	2 (in π^*)	paramagnetic

4. So. Only the S₂ form is paramagnetic, so the number is 1.

The common slip: 0. Treating S₂ as an ordinary diamagnetic molecule gives 0. Fill its MO diagram like O₂ and the two unpaired π^* electrons appear.

The idea to keep. Allotropes of the same element can differ magnetically because the molecular unit differs. S₈ is diamagnetic, while S₂, like O₂, has two unpaired π^* electrons.

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