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Kelvi

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Which is the correct thermal stability order for $H_2E$ (E = O, S, Se, Te and Po)?

medium

Chemical bonding and Molecular structure

2019

chemistry

$H_2S < H_2O < H_2Se < H_2Te < H_2Po$

$H_2O < H_2S < H_2Se < H_2Te < H_2Po$

$H_2Po < H_2Te < H_2Se < H_2S < H_2O$

$H_2Se < H_2Te < H_2Po < H_2O < H_2S$

Explanation

To determine the thermal stability order of $H_2E$ (where $E = O, S, Se, Te,$ and $Po), \\$

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we need to consider the bond strength between hydrogen and the chalcogen (E).

\\

The thermal stability of these hydrides is primarily influenced by the bond dissociation energy.

\\

As we move down the group in the periodic table, the size of the chalcogen atom increases.

\\

This leads to a decrease in bond strength due to poorer overlap between the hydrogen 1s orbital

\\

and the larger, more diffuse orbitals of the heavier chalcogens.

\\

Therefore, the bond strength decreases as we move from

H_2O

to

H_2Po. \\

This results in a decrease in thermal stability in the same order.

\\

Thus, the correct thermal stability order is:

\\ H_2Po < H_2Te < H_2Se < H_2S < H_2O \\

This corresponds to Option 3.