Question

Which is the correct thermal stability order of ${{\text{H}}_2}{\text{E}}\left( {{\text{E = O,S,Se,Te}}{\text{ and Po }}} \right)$ ?
A) ${{\text{H}}_2}{\text{S < }}{{\text{H}}_2}{\text{O}} < {{\text{H}}_2}{\text{Se < }}{{\text{H}}_2}{\text{Te < }}{{\text{H}}_2}{\text{Po}}$
B) ${{\text{H}}_2}{\text{O}} < {{\text{H}}_2}{\text{S}} < {{\text{H}}_2}{\text{Se < }}{{\text{H}}_2}{\text{Te < }}{{\text{H}}_2}{\text{Po}}$
C) ${{\text{H}}_2}{\text{Po < }}{{\text{H}}_2}{\text{Te}} < {{\text{H}}_2}{\text{Se < }}{{\text{H}}_2}{\text{S < }}{{\text{H}}_2}{\text{O}}$
D) ${{\text{H}}_2}{\text{Se < }}{{\text{H}}_2}{\text{Te}} < {{\text{H}}_2}{\text{Po < }}{{\text{H}}_2}{\text{O < }}{{\text{H}}_2}{\text{S}}$

Answer 1

Hint : All the elements are of Oxygen family .On going down the group, the thermal stability of ${{\text{H}}_2}{\text{E}}$ decreases as the ${\text{M - H}}$ bond energy decreases where M represents metals. So, ${{\text{H}}_2}{\text{O}}$ will be the most thermally stable molecule.

Complete Step By Step Answer:
${\text{O,S,Se,Te,Po}}$ are the members of the oxygen family.

On going down the group, the thermal stability of ${{\text{H}}_2}{\text{E}}$ decreases where E represents members of the oxygen family. This is because the ${\text{M - H}}$ bond energy decreases as the size of the central atom increases. On going down the group,
The size of the central atom increases (due to increase in number of electrons).
The electrons of the most outer shell are far distance from the nucleus so they can be easily given which decreases the thermal stability.
The ${\text{M - H}}$ atoms when combining their orbitals overlap. Thus they form a lower energy state and become stable.
This means that the two atoms come so close they penetrate each other's orbital and a new hybridized orbital is formed which is stable.
So the greater the overlapping, the greater is the bond energy. ${{\text{H}}_2}{\text{O}}$ has greater overlapping than the other ${{\text{H}}_2}{\text{E}}$ .
So the order is ${{\text{H}}_2}{\text{Po < }}{{\text{H}}_2}{\text{Te}} < {{\text{H}}_2}{\text{Se < }}{{\text{H}}_2}{\text{S < }}{{\text{H}}_2}{\text{O}}$ .
Hence the correct answer is ‘C’.

Note :
The ${\text{M - H}}$ bond energy is the energy required to break the bond formed between the M and H atom. This energy is affected by the following-
1. Atomic radius- with the increase in atomic radius, the bond energy between M and H decreases.
2. Nuclear charge-as the nuclear charge between the nucleus and the outer shell decreases, the bond energy also decreases.