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Question

Chemistry Question on p -Block Elements

Based on data provided, the value of electron gain enthalpy of fluorine would be :

A

−300 kJ mol−1- 300 \, kJ \, mol^{-1}

B

−350 kJ mol−1- 350 \, kJ \, mol^{-1}

C

−328 kJ mol−1- 328 \, kJ \, mol^{-1}

D

−228 kJ mol−1- 228 \, kJ \, mol^{-1}

Answer

−328 kJ mol−1- 328 \, kJ \, mol^{-1}

Explanation

Solution

Applying Hess's Law ΔfH∘=Δsub H+12Δdiss H+I.E+E.A+Δlattice H\Delta_{f} H^{\circ}=\Delta_{\text {sub }} H+\frac{1}{2} \Delta_{\text {diss }} H+I . E+E . A+\Delta_{\text {lattice }} H −617=161+520+77+-617=161+520+77+ E.A. +(−1047)+(-1047) E. A.=−617+289=−328 kJ mol−1A .=-617+289=-328 \,kJ \,mol ^{-1} ∴\therefore electron affinity of fluorine =−328 kJ mol−1=-328\, kJ\, mol ^{-1}