Animated Solution for Chemistry - s and p-Block Elements: 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

Have you ever wondered why some materials are attracted to magnets while others are completely indifferent? The secret lies deep within their molecular structure, specifically in how their electrons are arranged. Today, we are going to dive into the fascinating world of sulphur allotropes and uncover which of them possesses the hidden power of paramagnetism.

Sulphur is a versatile element that exists in several allotropic forms depending on the temperature and conditions. The most common ones you will encounter are $\alpha$-sulphur (rhombic) and $\beta$-sulphur (monoclinic). However, when you heat sulphur to extremely high temperatures, it vaporizes and forms a diatomic molecule known as $S_2$.

Our mission is to determine how many of these three forms—$\alpha$-sulphur, $\beta$-sulphur, and $S_2$—are paramagnetic.

Let's start with the familiar solid forms. Both $\alpha$-sulphur and $\beta$-sulphur consist of $S_8$ molecules. Imagine eight sulphur atoms holding hands in a puckered, crown-like ring.

In these $S_8$ rings, every single valence electron is perfectly paired up in stable covalent bonds. Because paramagnetism requires the presence of unpaired electrons, the complete pairing in $S_8$ means that both $\alpha$ and $\beta$ forms are strictly diamagnetic. They will not respond to a magnetic field.

Now, let's turn up the heat! Above $1000\text{ K}$, the $S_8$ rings break apart, and sulphur exists primarily as $S_2$ gas.

If you look at the periodic table, sulphur sits right below oxygen in Group 16. This means $S_2$ is the heavier cousin of the oxygen molecule, $O_2$. Just like $O_2$, the $S_2$ molecule hides a magnetic secret that can only be revealed by Molecular Orbital (MO) Theory.

The Ultimate Proof

To understand the magnetism of $S_2$, we need to look at its valence $p$-electrons. Each sulphur atom has $4$ electrons in its $3p$ orbitals, giving us a total of $8$ $p$-electrons to distribute in the $S_2$ molecule.

When these atomic orbitals combine, they form molecular orbitals. We fill them in order of increasing energy: first the ${\sigma }_{3p_z}$ orbital takes $2$ electrons, then the ${\pi }_{3p_x}$ and ${\pi }_{3p_y}$ bonding orbitals take $4$ electrons.

We have $2$ electrons left. Where do they go? They enter the degenerate anti-bonding orbitals, ${\pi }_{3p_x}^{\ast }$ and ${\pi }_{3p_y}^{\ast }$. According to Hund's Rule of Maximum Multiplicity, these electrons will occupy separate orbitals and remain unpaired.

It is the presence of these two unpaired electrons in the ${\pi }^{\ast }$ orbitals that gives $S_2$ its paramagnetic character.

So, out of the three allotropes we analyzed, only the $S_2$ form is paramagnetic. The $\alpha$ and $\beta$ forms are diamagnetic. Therefore, the number of paramagnetic allotropic forms is exactly $1$.