Question

Give reasons for the following: ${\mathrm{H}}_2\mathrm{Te}$ is more acidic than ${\mathrm{H}}_2\mathrm{S}$.

Answer 1

Bond dissociation enthalpy:

• The bond dissociation energy is the amount of energy necessary to break a bond and generate two atomic or molecular fragments, each having one electron from the original sharing pair
• It is usually an endothermic process.
• Tellurium ($\mathrm{Te}$) is in the fifth period, whereas Sulphur ($\mathrm{S}$) is in the third.
• The bond dissociation enthalpy decreases as we move along the group. As a result, the acidic nature rises.
• The bond dissociation enthalpy of ${\mathrm{H}}_2\mathrm{Te}$ is lower than that of ${\mathrm{H}}_2\mathrm{S}$.
• As a result, less energy is required to break the ${\mathrm{H}}_2\mathrm{Te}$ bond, releasing${\left[\mathrm{H}\right]}^{+}$ is easier, and the acidity of ${\mathrm{H}}_2\mathrm{Te}$ is greater.
• ${\mathrm{H}}_2\mathrm{S}$, on the other hand, has a high bond dissociation energy and so has lower acidity.

Hence, ${\mathrm{H}}_2\mathrm{Te}$ is more acidic than ${\mathrm{H}}_2\mathrm{S}$.