Chemistry - PERIODIC TABLE Question with Solution | TestHub

(i) Electron affinity of second period$\mathrm{p}$-block element is less than third period$\mathrm{p}$-block element due to small size of second period$\mathrm{p}$-block element, due to small size electron density is more and upcoming electron face extra repulsion, so energy release is less. Hence, EA order:$\mathrm{F}<\mathrm{Cl}$.

(ii) Down the group electron affinity decreases due to size increases. EA order:$\mathrm{S}>\mathrm{Se} \mathrm{Li}>\mathrm{Na}$.