Question

The thermal stability of hydrides of oxygen family is in order:
A.\[{H_2}Po < {H_2}Te < {H_2}Se < {H_2}S < {H_2}O\]
B.\[{H_2}Po < {H_2}O < {H_2}Te < {H_2}Se < {H_2}S\]
C.\[{H_2}S < {H_2}O < {H_2}Te < {H_2}Se < {H_2}Po\]
D.\[{H_2}O < {H_2}S < {H_2}Te < {H_2}Se < {H_2}Po\]

Answer 1

Hint: Thermal stability can be understood as the ability of any compound to resist being broken down under heat stress. In simpler terms, it can be understood as the general stability of a given molecule at a constant ambient condition.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
The compounds that are to be compared are all compounds belonging to the oxygen family, i.e. group 16 in the periodic table. The placements of the base elements of these hydrides in the periodic table can be show as follows:
 

Hence, we can see that the atomic number of \[O < S < Se < Te < Po\] . We know that as we move down the group, the atomic radius goes on increasing. Hence, the valence shell moves farther and farther away from the nucleus as it moves down the group. This means that the effective nuclear charge on the valence electrons also decreases as we move down the group due to factors like increased atomic radius and shielding effect.
The thermal stability of a compound depends on the electronegativity difference between the constituent atoms. Now, the electronegativity of an element decreases as we move down a group. Hence in group 16, the oxygen atom has a highest electronegativity while Po has the lowest electronegativity.
Since the compounds contain hydrogen as the constituent element, the only factor for consideration while understanding the thermal stability is the electronegativity of the other constituent atoms. The electronegativity comparison of the other elements can be given as: \[O < S < Se < Te < Po\] .
Hence the hydride formed with oxygen has the highest stability while the hydride formed with polonium has the lowest stability.
Hence, the thermal stability of hydrides of oxygen family is in order \[{H_2}Po < {H_2}Te < {H_2}Se < {H_2}S < {H_2}O\]
Hence, Option A is the correct option

Note: Thermal stability can also define another parameter known as the bond dissociation enthalpy. This enthalpy tells us about the amount of energy required to break the bonds in compounds. Hence, compounds with lower thermal would require lower energy to break bonds and vice versa.