Question

Give a reason why ${\mathrm{H}}_2\mathrm{S}$ is less acidic than ${\mathrm{H}}_2\mathrm{Te}$.

Answer 1

The reason for the less acidic nature of ${\mathrm{H}}_2\mathrm{S}$ than ${\mathrm{H}}_2\mathrm{Te}$.

1. On moving from top to bottom in the group the size of the atom increases.
2. With the increase in the size of the atom, there is a decrease in ionization enthalpy.
3. As ionization enthalpy decreases the dissociation of the bond becomes easy.

In the case of ${\mathrm{H}}_2\mathrm{Te}$(Hydrogen telluride):

1. As the size of Te (Tellurium) is much larger than S (Sulphur), the ionization enthalpy of Te is less than S.
2. Due to the small ionization enthalpy, the bond dissociation becomes easy.
3. By this, the removal of a hydrogen ion from the compound becomes easy.
4. With the ease in the removal of hydrogen ions, the acidic strength of the compound increases.

In the case of ${\mathrm{H}}_2\mathrm{S}$(Hydrogen sulfide):

1. As the size of S (Sulphur) is much smaller than Te (Tellurium), the ionization enthalpy of S is more than Te.
2. Due to the large value of ionization enthalpy, the bond dissociation becomes difficult.
3. By this, the removal of a hydrogen ion from the compound becomes difficult.
4. As the removal of hydrogen from the compound becomes difficult, the acidic strength of the compound decreases.

Hence, due to difficulty in removal of hydrogen ion in ${\mathrm{H}}_2\mathrm{S}$ the acidic strength of ${\mathrm{H}}_2\mathrm{S}$ is less acidic than ${\mathrm{H}}_2\mathrm{Te}$.