1. We compare the $\angle H-M-H$ bond angles in the given hydrides:

• $CH_4$ : $\angle H-C-H$
• $NH_3$ : $\angle H-N-H$
• $H_2O$ : $\angle H-O-H$
• $H_2S$ : $\angle H-S-H$

2. Use VSEPR theory:

• More lone pairs on the central atom compress bond angle.
• Tetrahedral electron geometry gives ideal angle $109.5^\circ$.

3. Now evaluate each molecule:

(A) $CH_4$

• Central atom $C$ has 4 bond pairs and 0 lone pairs.
• Shape: tetrahedral.
• Bond angle: $\angle H-C-H = 109.5^\circ$

(C) $NH_3$

• Central atom $N$ has 3 bond pairs and 1 lone pair.
• Shape: trigonal pyramidal.
• Lone pair-bond pair repulsion decreases bond angle.
• Bond angle: $\angle H-N-H \approx 107^\circ$

(D) $H_2O$

• Central atom $O$ has 2 bond pairs and 2 lone pairs.
• Shape: bent.
• Two lone pairs compress the angle further.
• Bond angle: $\angle H-O-H \approx 104.5^\circ$

(B) $H_2S$

• Central atom $S$ has 2 bond pairs and 2 lone pairs.
• Shape: bent.
• Because of larger size of sulfur and less effective hybridization, the angle is even smaller.
• Bond angle: $\angle H-S-H \approx 92^\circ$

4. Compare all values:

$109.5^\circ > 107^\circ > 104.5^\circ > 92^\circ$

So the largest $H-M-H$ bond angle is in $CH_4$.

5. Therefore, the correct option is:

$\boxed{A\; (CH_4)}$

6. Comparison with stored correct answer:

• Stored correct answer: $A$
• Derived answer: $A$

They match.