Q26 / 160 · TS EAMCET 2023 (14 May Shift 2) · Chemistry
Among the hydrides of group 16 elements, the hydriae $\mathrm{X}$ has lowest boiling point and the hydride $\mathrm{Y}$ has highest boiling point. $\mathrm{X}$ and $\mathrm{Y}$ respectively are
- A$\mathrm{H}_2 \mathrm{Te}, \mathrm{H}_2 \mathrm{Se}$
- B$\mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{Te}$
- C$\mathrm{H}_2 \mathrm{~S}, \mathrm{H}_2 \mathrm{Te}$
- Correct answerD$\mathrm{H}_2 \mathrm{~S}, \mathrm{H}_2 \mathrm{O}$
Solution
On moving down the group from $\mathrm{S}$ to $\mathrm{Te}$, the boiling point of hydrides increases due to increase in van der wal force of attraction. But $\mathrm{H}_2 \mathrm{O}$ has highest boiling point due to presence of hydrogen bonding.
Hence, the increasing order of boiling point of group 16 hydrides is given as : $\mathrm{H}_2 \mathrm{~S} < \mathrm{H}_2 \mathrm{Se} < \mathrm{H}_2 \mathrm{Te} < \mathrm{H}_2 \mathrm{O}$ Thus, $\mathrm{H}_2 \mathrm{~S}$ has lowest boiling point while $\mathrm{H}_2 \mathrm{O}$ has highest boiling point.
Hence, the correct answer is option: 'D'.