Question

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

• A. $H_{2}Te>H_{2}O>H_{2}Se>H_{2}S$
• 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

Answer: (D)

The correct order of boiling point of $H_{2}O,H_{2}S,H_{2}Se$ and $H_{2}Te$ is
$\underset{373K}{H_{2}O}>\underset{269K}{H_{2}Te}>\underset{232K}{H_{2}Se}>\underset{213K}{H_{2}S}$
The high boiling point of water is due to the association of $H_{2}O$ molecules through hydrogen bonding. However, $H$-bonding is not present in other hydrides. The intermolecular forces between the hydrides (except $H_{2}O$ ) are van der Waals' forces. These forces increase with increase in molecular size and therefore, boiling points increase on moving from $H_{2}S$ to $H_{2}Te$.