JEE Main 2024 — Chemical Bonding and Molecular Structure Question with Solution

Bond enthalpy, also referred to as bond energy or bond dissociation energy, is the energy required to break a particular covalent bond in one mole of molecule in a gaseous state.

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, Both A and R are true and R is the correct explanation of A.