NaH is an example of:

The compound under consideration is sodium hydride, written chemically as $$\text{NaH}$$.

We first recall that hydrides are classified into three main types on the basis of the nature of bonding between hydrogen and the other element:

$$\text{1. Saline (ionic) hydrides}$$ - formed when hydrogen combines with highly electropositive $$s$$-block metals (group 1: alkali metals and group 2: alkaline-earth metals except Be and Mg). In such hydrides, hydrogen exists as the hydride ion $$\text{H}^-$$ and the metal exists as a cation, giving the general formula $$\text{M}^+\text{H}^-$$ or $$\text{M}^{2+}(\text{H}^-)_{2}$$. These compounds are crystalline, high-melting, and conduct electricity in the molten state.

$$\text{2. Metallic (interstitial) hydrides}$$ - formed by many transition metals. Here hydrogen atoms occupy interstitial sites in the metallic lattice, leading to non-stoichiometric compositions.

3. Molecular (covalent) or electron-rich hydrides - formed by $$p$$-block elements where hydrogen is covalently bonded, often giving rise to discrete molecules. When these molecules have more electrons than needed for simple covalent bonding, they are specifically called electron-rich hydrides; typical examples include $$\text{NH}_3$$ and $$\text{H}_2\text{O}$$.

Now we analyse $$\text{NaH}$$. Sodium belongs to group 1 of the periodic table and is a highly electropositive alkali metal. On reacting with hydrogen, sodium donates one electron to hydrogen:

$$\text{Na} \rightarrow \text{Na}^+ + e^-$$

$$\tfrac12\,\text{H}_2 + e^- \rightarrow \text{H}^-$$

Combining these half-reactions, we obtain

$$\text{Na} + \tfrac12\,\text{H}_2 \rightarrow \text{Na}^+\text{H}^-$$

The product contains discrete ionic species $$\text{Na}^+$$ and $$\text{H}^-$$. Thus the bonding is predominantly ionic and fits exactly into the description of a saline (ionic) hydride.

Since $$\text{NaH}$$ does not contain a metallic lattice with hydrogen in interstices, option B is eliminated. It is also not a covalent molecular hydride, nor does it possess excess electrons as found in electron-rich hydrides, so options A and D are also ruled out.

Therefore, $$\text{NaH}$$ is correctly categorised as a saline hydride.

Hence, the correct answer is Option C.