Question

Bond angle in $H_{2}O$(104.5${}^{\circ }$) is higher than the bond angle of $H_{2}S$ (92.1${}^{\circ }$). The difference is due to:

• A. O is diatomic and S is tetra-atomic
• B. difference in electronegativity of S and O
• C. difference in oxidation states of S and O
• D. difference in shapes of hybrid orbitals of S and O

Answer 1

The correct option is B difference in electronegativity of S and O

The difference in the bond angles is linked with the electronegativities of oxygen and sulphur atoms. As we move down the group from O to Te, the size of the central atom goes on increasing and its electronegativity goes on decreasing.

As a result, the position of the two bond pairs shifts away and away from the central atom as we move from $H_{2}O$ to $H_{2}Te$. Consequently, the repulsion between the bond pairs decreases as we move from $H_{2}O$ to $H_{2}Te$ and therefore, the bond angle decreases in the order.

Hence, the correct answer is option $\text{B}$.