Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The incorrect trend regarding group $16$ hydrides $\left( H _{2} M\right)$ is

The p-Block Elements - Part2

Solution:

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. Consequently, the position of bond pairs of electrons shifts more and more away from the central atom in moving from $H _{2} O$ to $H _{2} Te$. For example, the bond pair in $O - H$ bond is closer to oxygen than the bond pair in $S - H$ bond. As a result, the force of repulsion between bonded pairs of electrons in $H _{2} O$ is more than in $H _{2} S$. In general, the force of repulsion between the bonded pairs of electrons decreases as we move from $H _{2} O$ to $H _{2} Te$ and therefore, the bond angle decreases in the same order as :

$\underset{\text{Bond angle:}}{} \,\, \underset{104.5^{\circ}}{H_2O} >\underset{92.1^\circ}{H2S} > \underset{91^\circ}{H_2Se} > \underset{90^\circ}{H_2Te}$