Question

Consider the following statements regarding hydrides of VIA group elements.
i) The order of volatility
$H_{2}O<H_{2}Te<H_{2}Se<H_{2}S$
ii) The order of Boiling point
$H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$
iii) The order of bond angles
$H_{2}O>H_{2}S>H_{2}Se>H_{2}Te$
Choose the correct option.

• A. All are correct
• B. Only i is correct
• C. ii & iii are correct
• D. i & iii are correct

Answer 1

Answer: (A)

All are correct

$H_{2}O$ have hydrogen bonding with its molecules so it tend to be least volatile and as we go down in group elements tendency of bonding within molecules decreases which increase the their volatility.

Boiling point is inversely proportional to volatility of compounds so its correct order is $H_{2}O>H_{2}Te>H_{2}Se>H_{2}S$.

Due to lone pair - lone pair electron repulsion in water molecule, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions thereby forcing the $H-X-H$ angle to contract slightly. So instead of the $H-O-H$ angle being the perfect tetrahedral angle (${109.5}^o$) it is slightly reduced to ${104.5}^o$. On the other hand, both $H_{2}S$, $H_{2}Se$ and $H_{2}Te$ are not hybridized. That is, $S-H$, $Se-H$ and $Te-H$ bonds use pure p-orbitals from sulphur and selenium respectively. And as $S-H$, $Se-H$ and $Te-H$ bonds used pure p-orbitals so we would expect an $H-X-H$ inter orbital angle of ${90}^o$. So correct order is $H_{2}O>H_{2}S>H_{2}Se>H_{2}Te$