Question

What is the correct order of boiling point?

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

Answer 1

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

Due to van der Waal's attractive force, the heavier compounds generally have higher boiling points than similar compounds made up of smaller molecules. But here in $H_{2}O$, due to the high electronegativity of $O-$ atoms, intermolecular $H-$ bonding is present. Due to the presence of $H-$ bonding $H_{2}O$ has the highest boiling point.
Among $H_{2}S,H_{2}Se$ and $H_{2}Te$ the correct order is $H_{2}Te>H_{2}Se>H_{2}S$, because on moving down the group the size of the atoms increases.
Hence the correct order is:
$H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$