Given Reaction Energy Change (in kJ):
Li(s) $$\rightarrow$$ Li(g) : 161
Li(g) $$\rightarrow$$ Li$$^+$$(g) : 520
$$\frac{1}{2}F_2(g) \rightarrow$$ F(g) : 77
F(g) + e$$^-$$ $$\rightarrow$$ F$$^-$$(g) : (Electron gain enthalpy)
Li$$^+$$(g) + F$$^-$$(g) $$\rightarrow$$ LiF(s) : -1047
Li(s) + $$\frac{1}{2}F_2$$(g) $$\rightarrow$$ LiF(s) : -617
Based on data provided, the value of electron gain enthalpy of fluorine would be :

To find the electron gain enthalpy of fluorine, which is the energy change for the reaction F(g) + e⁻ → F⁻(g), we use the given reactions and their energy changes. The overall reaction provided is Li(s) + ½ F₂(g) → LiF(s) with ΔH = -617 kJ. We can break this down into steps using the Born-Haber cycle.

The steps involved are:

1. Sublimation of lithium: Li(s) → Li(g) with ΔH = 161 kJ

2. Ionization of lithium: Li(g) → Li⁺(g) + e⁻ with ΔH = 520 kJ

3. Dissociation of fluorine: ½ F₂(g) → F(g) with ΔH = 77 kJ

4. Electron gain by fluorine: F(g) + e⁻ → F⁻(g) with ΔH = ? (this is the electron gain enthalpy we need to find)

5. Lattice formation: Li⁺(g) + F⁻(g) → LiF(s) with ΔH = -1047 kJ

The sum of these steps equals the overall reaction energy:

ΔH_overall = ΔH_sublimation + ΔH_ionization + ΔH_dissociation + ΔH_electron_gain + ΔH_lattice

Substituting the known values:

-617 = 161 + 520 + 77 + ΔH_electron_gain + (-1047)

Now, simplify the right-hand side:

161 + 520 = 681

681 + 77 = 758

758 - 1047 = -289

So the equation becomes:

-617 = -289 + ΔH_electron_gain

Solve for ΔH_electron_gain:

ΔH_electron_gain = -617 - (-289)

ΔH_electron_gain = -617 + 289

ΔH_electron_gain = -328 kJ

Therefore, the electron gain enthalpy of fluorine is -328 kJ mol⁻¹.

Hence, the correct answer is Option C.