Question

Arrange in decreasing order of boiling point $H_{2}S,H_{2}O,H_{2}Se,H_{2}Te$

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

Answer 1

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

One of the general trends in boiling point is that:

Boiling point $\propto$ molecular mass

Thus, the order should have been $H_{2}Te>H_{2}Se>H_{2}S>H_{2}O$

Note that O, S, Se, Te are all members of same group starting with O at the top.

But in case of $H_{2}O$, high electronegativity and small size of O facilitates H-bonding.

This intermolecular H-bonding leads to high boiling point of $H_{2}O$.

Thus, the order is $H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$

More molecular weight $\Rightarrow$ more electrons in one molecule

$\Downarrow$

more Van der Waals forces $\Rightarrow$ More boiling point