Question

Which one of the following is the correct decreasing order of boiling point?

• A. $H_{2}Te>HzO>H_{2}Se>H_{2}S$
• B. $H_{2}O>H_{2}S>H_{2}Se>H_{2}Te$
• C. $H_{2}Te>H_{2}Se>H_{2}S>H_{2}O$
• D. $H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$

Answer 1

The correct option is D $H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$

Among the hydrides of group 16 elements, the boiling point of $H_{2}O$ is (difference in boiling points of $H_{2}O$ and $H_{2}S$ is around ${200}^{o}C$) because of strong intermolecular hydrogen bonding. After the decrease in boiling point from $H_{2}O$ to $H_{2}S$, from $H_{2}S$ to $H_{2}Te$, it increases due to increase in size of the atoms from S toTe which increases the magnitude of van der Waal's forces among the molecules. So, the correct order of boiling points is
$H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$