Animated Solution for Chemistry - s and p-Block Elements: A s-block element (M) reacts with oxygen to form an oxide of the formula ${\text{MO}}_2$. The oxide is pale yellow in colour and paramagnetic. The element (M) is

The problem presents us with a fascinating chemical mystery. We have an unknown s-block element, let's call it $M$, which reacts with oxygen to form an oxide with the formula $M{O}_2$.

We are given two critical clues about this oxide: it is pale yellow in color, and it is paramagnetic. Our mission is to identify the element $M$ from the given options: Magnesium, Sodium, Calcium, or Potassium.

Let's start by analyzing the formula $M{O}_2$. In the s-block of the periodic table, elements form different types of oxides depending on their group and size.

If $M$ is an alkaline earth metal (Group 2) like Magnesium or Calcium, it forms a $+2$ cation ($M^{2+}$). To form a compound with the formula $M{O}_2$, the oxygen part must be a peroxide ion, $O_2^{2-}$.

However, if $M$ is an alkali metal (Group 1) like Sodium or Potassium, it forms a $+1$ cation ($M^{+}$). In this case, to balance the charge and form $M{O}_2$, the oxygen part must be a superoxide ion, $O_2^{-}$.

The most powerful hint in the question is that the oxide is paramagnetic. Paramagnetism arises when a molecule or ion contains unpaired electrons. Let's investigate the magnetic nature of the possible oxygen ions using Molecular Orbital (MO) Theory.

A peroxide ion ($O_2^{2-}$) has a total of 18 valence electrons. When we fill these electrons into the molecular orbitals, all of them pair up perfectly. Therefore, peroxides are diamagnetic. This immediately eliminates the Group 2 metals (Magnesium and Calcium) and Sodium (which primarily forms the peroxide $N{a}_2{O}_2$).

Now, let's look at the superoxide ion ($O_2^{-}$). It has a total of 17 valence electrons. When we distribute these electrons according to MO Theory, the highest occupied molecular orbitals are the ${\pi }^{\ast }$ antibonding orbitals.

Because there are 3 electrons in the degenerate ${\pi }^{\ast }$ orbitals, one of them must remain unpaired. This single unpaired electron makes the superoxide ion paramagnetic! Furthermore, electronic transitions involving this unpaired electron are responsible for the characteristic pale yellow color of the compound.

Since the compound must be a superoxide to fit the paramagnetic and color criteria, the element $M$ must be an alkali metal capable of forming a stable superoxide.

Among the options, Potassium ($K$) is the only element that reacts with excess oxygen to form a superoxide:

Therefore, the mystery element is Potassium.

The Role of Lattice Energy

You might wonder, why doesn't Sodium or Lithium form superoxides? This is a classic JEE concept rooted in Lattice Energy.

A fundamental rule of solid-state chemistry is that a large cation effectively stabilizes a large anion. The superoxide ion ($O_2^{-}$) is quite bulky. Small cations like $L{i}^{+}$ or $N{a}^{+}$ have high positive charge densities and cannot form a stable crystal lattice with the large superoxide ion.

Potassium ($K^{+}$), being larger, provides the perfect spatial fit to stabilize the $O_2^{-}$ ion, leading to the formation of $K{O}_2$. Always keep this size-compatibility principle in mind when predicting the stability of ionic compounds!