Question

The correct order of boiling point is:

• 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 $H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$

Generally large molecules have more electrons and nuclei that create van der Waal's attractive force. So, 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:
$H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$