Elements of group- 15, 16, and 17 form compounds with hydrogen with general formula \(MH_3\), \(H_2M\), and \(HX\) (called hydrogen halides). The stability of these hydrides, along with \(M-H\) bond strength and basic nature decreases down the group. Boiling point of these hydrides mainly depends on the strength of their intermolecular forces.
The bond angle \(HMH\) among hydrides of group 16 decreases down the group because:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Electronegativity of the central atom decreases down the group
The H–M–H angle falls from \( \ce{H2O} \) (104°) to \( \ce{H2S} \) (92°), \( \ce{H2Se} \) (91°) and \( \ce{H2Te} \) (90°). The cause is the falling electronegativity of the central atom (O 3.5, S 2.5, Se 2.4, Te 2.1):
Oxygen is very electronegative, so the two O–H bond pairs are pulled close to O. Close together, they repel strongly and open the angle to 104°.
S, Se and Te are much less electronegative, so the bond pairs lie farther from the central atom; the repulsion between them is smaller and the angle shrinks towards 90°.
Why the other options are incorrect
Option (2) — Incorrect. The size of the central atom increases down the group (O < S < Se < Te), it does not decrease.
Option (3) — Incorrect. It describes the result (weaker bond pair–bond pair repulsion) of the same effect, tied to the H–M electronegativity difference; the cause asked for is the falling electronegativity of the central atom, given in Option (1).
Option (4) — Incorrect. The two lone pairs are on the central atom throughout the group; a lone pair–bond pair repulsion that increased would close the angle further in \( \ce{H2O} \), the reverse of what is seen (\( \ce{H2O} \) has the largest angle).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.