In which of the following pairs, the ionisation energy of the first species is less than that of the second

Atomic Number of Sulphur ($\mathrm{S}$) is 16 and its electronic configuration is$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^4$.

Atomic number of phosphorus ($\mathrm{P}$) is 15 and the electronic configuration of phosphorus is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^3.$

It is difficult to remove an electron from the 3p orbital of the phosphorus atom. Because it has a half-filled electronic configuration.

Therefore, the first ionization energy of Sulphur is less than that of phosphorus.

Hence, option(d) is correct.