1. Identify the central atom geometry in each compound

We compare the bond angle $\angle H-X-H$ in:

• $CH_4$
• $NH_3$
• $H_2O$
• $PH_3$

These are all based on approximately tetrahedral electron-pair arrangement around the central atom.

2. Recall standard bond angles

(A) $CH_4$

Carbon has 4 bond pairs and 0 lone pairs. So the shape is tetrahedral, with ideal bond angle: $109.5^\circ$

(B) $NH_3$

Nitrogen has 3 bond pairs and 1 lone pair. Shape is trigonal pyramidal. Lone pair-bond pair repulsion compresses the angle from $109.5^\circ$ to about: $107^\circ$

(C) $H_2O$

Oxygen has 2 bond pairs and 2 lone pairs. Shape is bent. Two lone pairs compress the angle further to about: $104.5^\circ$

(D) $PH_3$

In $PH_3$, bond angle is even smaller because phosphorus shows very little hybridization; the angle is close to: $93.5^\circ$

3. Compare all bond angles

$CH_4 = 109.5^\circ$ $NH_3 \approx 107^\circ$ $H_2O \approx 104.5^\circ$ $PH_3 \approx 93.5^\circ$

Clearly, $109.5^\circ > 107^\circ > 104.5^\circ > 93.5^\circ$

So the greatest $H-X-H$ bond angle is in: $CH_4$

4. Evaluate options

• A: $CH_4$ ✅ Correct
• B: $NH_3$ ❌ Smaller than $CH_4$
• C: $H_2O$ ❌ Smaller than $NH_3$
• D: $PH_3$ ❌ Smallest among these

5. Final Answer

The compound with the greatest bond angle $H-X-H$ is: $\boxed{CH_4}$