Q.Classify each of the following as an ionic, a covalent or an interstitial hydride, giving a reason in each case:

The three classes of hydride are distinguished chiefly by the nature of the element bonded to hydrogen and by the resulting stoichiometry:

(a) $\text{NaH}$ is an ionic (saline) hydride. Sodium is one of the most strongly electropositive elements in the periodic table, so on combining with hydrogen it transfers its single valence electron essentially completely, giving a crystalline ionic solid built from $\text{Na}^+$ cations and hydride anions, $\text{H}^-$. It has a fixed 1:1 stoichiometry, a high melting point, and conducts electricity when molten.

(b) $\text{CH}_4$ is a covalent (molecular) hydride. Carbon is a nonmetal of comparable electronegativity to hydrogen, so the four C-H bonds are formed by genuine electron sharing rather than electron transfer, giving a discrete covalent molecule. Being a simple molecular substance, methane is a gas at room temperature, held together in bulk only by weak intermolecular forces.

(c) $\text{PdH}_{0.6}$ is an interstitial (metallic) hydride. The tell-tale sign is the non-integer subscript $0.6$: this is not a fixed-ratio chemical compound at all, but hydrogen atoms diffused into and occupying a variable fraction of the empty interstitial spaces within the palladium metal's crystal lattice. Because the metal lattice is essentially unaltered, palladium hydride retains palladium's own metallic lustre and conductivity.

Classify each hydride by the electronegativity/character of the non-hydrogen element (electropositive metal -> ionic; nonmetal sharing electrons -> covalent) and by whether the stoichiometry is a fixed whole-number ratio or not (non-integer -> interstitial).

Classifying PdH0.6 as 'ionic' because it involves a metal, missing that transition-metal interstitial hydrides are structurally quite different from Group 1/2 ionic hydrides; overlooking the non-integer subscript as the key clue for interstitial hydrides.