Animated Solution for Chemistry - s and p-Block Elements: In view of the signs of ${\Delta }_r{G}^{\circ }$ for the following reactions $\mathrm{Pb}{\mathrm{O}}_2+\mathrm{Pb}⟶2PbO,{\Delta }_r{G}^{\circ }<0$ $\mathrm{Sn}{\mathrm{O}}_2+\mathrm{Sn}⟶2SnO,{\Delta }_r{G}^{\circ }>0$ Which oxidation states are more characteristic for lead and tin?

When tackling problems involving the stability of oxidation states, thermodynamics is our most reliable guide. The standard Gibbs free energy change, denoted as ${\Delta }_r{G}^{\circ }$, acts as the ultimate judge of a reaction's spontaneity.

A fundamental rule of chemistry states that if ${\Delta }_r{G}^{\circ }<0$, the reaction is spontaneous in the forward direction, meaning the products are thermodynamically more stable than the reactants. Conversely, if ${\Delta }_r{G}^{\circ }>0$, the forward reaction is non-spontaneous, and the reactants are actually the more stable species.

A Downhill Journey

Let's analyze the first reaction provided in the problem:

If we assign oxidation states, we see that lead in $\mathrm{Pb}{\mathrm{O}}_2$ is in the $+4$ state, elemental lead is $0$, and lead in the product $\mathrm{PbO}$ is in the $+2$ state. The problem explicitly states that for this reaction, ${\Delta }_r{G}^{\circ }<0$.

Because the Gibbs free energy change is negative, this reaction naturally wants to proceed forward. Lead is actively seeking to transition from the $+4$ state down to the $+2$ state. This thermodynamic preference clearly indicates that for lead, the $\mathrm{P}{\mathrm{b}}^{2+}$ oxidation state is significantly more stable than $\mathrm{P}{\mathrm{b}}^{4+}$.

An Uphill Battle

Now, let's look at the analogous reaction for tin, which sits just above lead in Group 14 of the periodic table:

Just like before, tin is attempting to go from the $+4$ state to the $+2$ state. However, there is a massive difference here: ${\Delta }_r{G}^{\circ }>0$.

A positive Gibbs free energy change means this forward reaction is an uphill battle; it is non-spontaneous. Tin does not want to form $\mathrm{SnO}$. Instead, the reverse reaction is spontaneous. This tells us that tin is much happier staying in the $+4$ oxidation state ($\mathrm{S}{\mathrm{n}}^{4+}$) rather than reducing to $+2$.

The Inert Pair Effect

Why do two elements in the exact same group behave so differently? The answer lies in a classic inorganic chemistry phenomenon known as the Inert Pair Effect.

As we descend Group 14 from carbon down to lead, the intervening $d$ and $f$ orbitals are filled. These orbitals are notoriously poor at shielding the outer valence electrons from the pull of the nucleus. By the time we reach heavy elements like lead, the effective nuclear charge felt by the valence $n{s}^2$ electrons is exceptionally high.

Because these $6s^2$ electrons are pulled so tightly toward the nucleus, they become chemically "inert" and refuse to participate in bonding. Consequently, lead prefers to only lose its two $6p$ electrons, resulting in a highly stable $+2$ oxidation state. Tin, being lighter, does not experience this effect as drastically, so it readily loses all four valence electrons to form stable $+4$ compounds.

By combining our thermodynamic data with our understanding of periodic trends, the conclusion is crystal clear. The $+2$ oxidation state is the hallmark of lead, while the $+4$ oxidation state is characteristic of tin. Therefore, the correct option is (d).