Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is ______.
(A) $$\alpha$$-sulphur
(B) $$\beta$$-sulphur
(C) $$S_2$$-form

We need to determine which allotropic forms of sulphur are paramagnetic. Paramagnetism arises from the presence of unpaired electrons.

$$\alpha$$-sulphur exists as $$S_8$$ molecules (crown-shaped ring). All electrons in $$S_8$$ are paired, making it diamagnetic.

$$\beta$$-sulphur also exists as $$S_8$$ molecules (with a different crystal packing). Since the molecular structure is the same $$S_8$$ ring, all electrons are paired and it is diamagnetic.

The $$S_2$$ form of sulphur is analogous to $$O_2$$. Like molecular oxygen, $$S_2$$ has two unpaired electrons in its antibonding $$\pi^*$$ molecular orbitals based on molecular orbital theory. This makes $$S_2$$ paramagnetic.

Therefore, among the given allotropes, only $$S_2$$ is paramagnetic.

The number of allotropic forms showing paramagnetism is $$\textbf{1}$$.