1. H ₂ S < H ₂ O < H ₂ Se < H ₂ Te < H ₂ Po 2. H ₂ O < H ₂ S < H ₂ Se < H ₂ Te < H ₂ Po 3. H ₂ Po < H ₂ Te < H ₂ Se < H ₂ S < H ₂ O 4. H ₂ Se < H ₂ Te < H ₂ Po < H ₂ O < H ₂ S

Topic: General trends, Group 15 and 16· p-Block Elements

Which is the correct thermal stability order for H_2E (E = O, S, Se, Te and Po) ?

1. H\(_2\)S < H\(_2\)O < H\(_2\)Se < H\(_2\)Te < H\(_2\)Po

2. H\(_2\)O < H\(_2\)S < H\(_2\)Se < H\(_2\)Te < H\(_2\)Po

3. H\(_2\)Po < H\(_2\)Te < H\(_2\)Se < H\(_2\)S < H\(_2\)O

4. H\(_2\)Se < H\(_2\)Te < H\(_2\)Po < H\(_2\)O < H\(_2\)S

Doubtly Asked 30 Oct 2026 NEET UG 2019 Practice test neet Save WhatsApp Report

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Answer: 3. H\(_2\)Po < H\(_2\)Te < H\(_2\)Se < H\(_2\)S < H\(_2\)O

Thermal stability of hydrides of group 16 depends on the E-H bond strength, which falls as the size of E increases down the group.

So stability: \(\mathrm{H_2O > H_2S > H_2Se > H_2Te > H_2Po}\). Written in increasing order of stability, that is the reverse, and the given option (3) is \(\mathrm{H_2Po < H_2Te < H_2Se < H_2S < H_2O}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Which is the correct thermal stability order for H_2E (E = O, S, Se, Te and Po) ?

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