The electron gain enthalpy with negative sign for oxygen `(− 141 KJ mol^(−1))` is numerically less than that for sulphur `(− 200 KJ mol^(−1))`. Give reason.

Explanation

The size of O atom is smaller than that of S. Thus, when an electron is added to neutral gaseous atom, the electon-electron repulsions experienced in the samller 2p-subshell of O are comparatively larger than those encounterned in the bigger 3p-subshell of S. As a result, more energy is released in case of S than
`{:(O(g)+e^(-) rarr O^(-)(g),," "Delta_(eg) = -141 kJ mol^(-1)),(S(g)+e^(-) rarr S^(-)(g),," "Delta_(eg) = -200 kJ mol^(-1)):}`
in case of O. In other words, negative electron gain enthalpy of S is higher than that of O.