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

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

AMUAMU 2000

Solution:

The correct order of boiling point of $H _{2} O , H _{2} S , H _{2} Se$ and $H _{2} Te$ is
$\underset{373\,K}{H_2O} \,\, > \,\, \underset{269\,K}{H_2Te}\,\, > \,\, \underset{232\,K}{H_2Se} \,\, > \,\, \underset{213\,K}{H_2S}$
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 $H _{2} Te$.