Animated Solution for Chemistry - s and p-Block Elements: Statement I: Sodium hydride can be used as an oxidising agent. Statement II: The lone pair of electrons on nitrogen in pyridine makes it basic. Choose the correct answer from the options given below.

Redox and Basicity

In competitive exams like JEE, multi-statement questions are designed to test your conceptual clarity across completely different domains of chemistry. In this problem, we are tasked with evaluating an inorganic redox property and an organic structural property. Let's dissect them one by one.

Statement I claims that Sodium hydride ($\text{NaH}$) can be used as an oxidizing agent. To verify this, we must look at the oxidation states of the constituent atoms.

Sodium hydride is an ionic compound composed of a sodium cation (${\text{Na}}^{+}$) and a hydride anion (${\text{H}}^{-}$). Because sodium is an alkali metal, it strictly exhibits an oxidation state of $+1$. Consequently, hydrogen is forced into an oxidation state of $-1$.

Now, we must ask ourselves: what are the limits of hydrogen's oxidation states? Hydrogen can exhibit oxidation states of $+1$, $0$, and $-1$. The $-1$ state is its absolute minimum. Because the hydride ion (${\text{H}}^{-}$) is already at its lowest possible oxidation state, it cannot accept any more electrons; it cannot undergo further reduction.

Instead, it can only lose electrons to reach a higher oxidation state, such as $0$ in hydrogen gas (${\text{H}}_2$). Since the hydride ion undergoes oxidation, it inherently acts as a reducing agent. Therefore, Statement I is fundamentally false.

Statement II shifts our focus to organic chemistry, specifically the basicity of pyridine (${\text{C}}_5{\text{H}}_5\text{N}$). Pyridine is a six-membered heterocyclic aromatic compound.

The nitrogen atom in pyridine is $s{p}^2$ hybridized. It forms two $\sigma$-bonds with adjacent carbon atoms and possesses one lone pair of electrons. The critical factor determining its basicity is the spatial orientation of this lone pair.

Unlike pyrrole, where the nitrogen's lone pair is housed in an unhybridized $p$-orbital and delocalized into the aromatic $\pi$-system, the lone pair in pyridine resides in an $s{p}^2$ hybridized orbital. This orbital lies flat in the plane of the ring, making it completely orthogonal (perpendicular) to the $p$-orbitals that form the aromatic $\pi$-cloud.

Because of this orthogonality, the lone pair cannot participate in resonance. It remains strictly localized on the nitrogen atom, making it highly available to accept an incoming proton (${\text{H}}^{+}$). This localized nature makes pyridine a very effective Lewis base. Thus, Statement II is absolutely true.

Combining our rigorous conceptual analysis, we find that Statement I is false, while Statement II is true. This perfectly aligns with option (d).