Question

The decreasing order of bond angles in hydrides of chalcogens is:

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

Answer 1

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

We know that electronegativity of the central atom decreases down the group. In $H_{2}O$, the bond pair of electrons is more attracted towards the oxygen atom and as a result there is more bond pair-bond pair repulsion and thus the bond angle increases. So, $H_{2}O$ has the highest bond angle. As size increases down the group, the bond angle decreases.