See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Applying Hess’s Law o f sub diss lattice 1 H H H I.E. E.A H 2        – 617 = 161 + 520 + 77 + E.(A) + (-1047) E.(A) = - 167 + 289 = - 328 kJ mol-1 ∴ electron affinity of fluorine = – 328 kJ mol-1