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

Solution

Chemistry ma’am · 3 steps

Count the lone pairs on the central atom: zero lone pairs means the H atoms spread out the most.

1. Same starting shape. C, N and O in these hydrides all have 4 electron pairs around them, so each starts from a tetrahedral arrangement of 109.5°.
2. Lone pairs squeeze. A lone pair repels more strongly than a bond pair, so every lone pair pushes the H atoms closer together. In H₂S the sulfur is larger and less electronegative, and its bonds use almost pure p orbitals, so the angle drops close to 90°.

   Molecule	Lone pairs on M	H–M–H angle
   CH₄	0	109.5°
   NH₃	1	107°
   H₂O	2	104.5°
   H₂S	2	about 92°

3. So. CH₄ has no lone pair to squeeze its bonds, so it keeps the full tetrahedral angle of 109.5°, the largest of the four. The answer is (a).

The idea to keep. For hydrides with the same number of electron pairs, each extra lone pair shrinks the bond angle. Going down a group, the angle shrinks again as the central atom gets bigger.

Answer from the NTA final answer key. Solution by PrepWiser.