The compound that has the largest $$H-M-H$$ bond angle ($$M = N, O, S, C$$) is:

We recall VSEPR theory, which states that the shape of a molecule is decided by the total number of electron pairs (bond pairs + lone pairs) surrounding the central atom. The ideal geometries and the corresponding bond angles can be predicted once we know this number. Larger bond angles arise when there are fewer lone-pair-bond-pair repulsions, because lone pairs occupy more space than bond pairs.

Let us examine each molecule one by one.

First, $$\text{H}_2\text{O}$$. The central atom is oxygen. Oxygen has $$6$$ valence electrons. With two $$\text{O-H}$$ bonds, two electrons are used in each bond pair, so $$4$$ electrons remain as two lone pairs. Thus, total electron pairs $$= 2$$ bond pairs $$+ 2$$ lone pairs $$= 4$$. According to VSEPR, $$4$$ electron pairs give a tetrahedral arrangement. However, because there are two lone pairs, the molecular shape becomes bent (angular). Lone-pair-lone-pair and lone-pair-bond-pair repulsions compress the $$\text{H-O-H}$$ angle below the ideal $$109.5^\circ$$. Experimentally,

$$\angle \text{H-O-H}=104.5^\circ.$$

Next, $$\text{NH}_3$$. The nitrogen atom has $$5$$ valence electrons. Three electrons take part in three $$\text{N-H}$$ bonds, leaving $$2$$ electrons as one lone pair. Total electron pairs $$= 3$$ bond pairs $$+ 1$$ lone pair $$= 4$$, again a tetrahedral arrangement of electron pairs. Only one lone pair is present, so the compression is less than in water. The measured bond angle is

$$\angle \text{H-N-H}=107^\circ.$$

Now, $$\text{H}_2\text{S}$$. Sulphur has $$6$$ valence electrons like oxygen, so the count is identical to water: two bond pairs and two lone pairs. Because sulphur is larger and the $$\text{S-H}$$ bonds are longer, the bond-pair-bond-pair repulsion is even weaker, allowing lone pairs to spread out more, which further contracts the bond angle. The observed value is approximately

$$\angle \text{H-S-H}\approx92^\circ.$$

Finally, $$\text{CH}_4$$. Carbon has $$4$$ valence electrons, each forming a $$\text{C-H}$$ bond. There are

$$4\text{ bond pairs} + 0\text{ lone pairs} = 4\text{ electron pairs}.$$

With no lone pairs, the geometry remains the ideal tetrahedron, and no compression occurs. Therefore,

$$\angle \text{H-C-H}=109.5^\circ.$$

Comparing all the angles we obtained:

$$\angle\text{H-C-H}=109.5^\circ > \angle\text{H-N-H}=107^\circ > \angle\text{H-O-H}=104.5^\circ > \angle\text{H-S-H}\approx92^\circ.$$

Clearly, the largest $$\text{H-M-H}$$ bond angle is in methane, $$\text{CH}_4$$, which corresponds to Option D.

Hence, the correct answer is Option D.