Question

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

• A. $H_2Po < H_2Te< H_2Se < H_2S < H_2O$
• B. $H_2Se < H_2Te< H_2Po < H_2O < H_2S$
• C. $H_2S < H_2O < H_2Se < H_2Te< H_2 Po$
• D. $H_2O < H_2S < H_2Se < H_2Te < H_2Po$

Answer 1

The Correct Option is A

To determine the correct thermal stability order for $H_2E$ (where E represents elements like O, S, Se, Te, Po), we need to consider the bond strength between hydrogen and the chalcogen (oxygen family element).

The thermal stability of hydrides of Group 16 elements decreases from top to bottom in the periodic table. This is because:

• Bond strength: As we move down the group, the size of the chalcogen atom (E) increases, leading to a weaker H–E bond. A weaker bond is more susceptible to breaking down thermally.
• Electronegativity: Oxygen is the most electronegative among these elements, forming the strongest bond with hydrogen and leading to the highest thermal stability for $H_2O$.

Based on this reasoning, the order of thermal stability from least stable to most stable is:

• $H_2Po$ (least stable) < $H_2Te$ < $H_2Se$ < $H_2S$ < $H_2O$ (most stable)

This order aligns with the given correct option: $H_2Po < H_2Te< H_2Se < H_2S < H_2O$.

The reasoning behind this order is the increasing atomic size and decreasing bond strength as we move down the group from oxygen to polonium.