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Intext · Q26

Q.Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2.

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Step 1 — Bond dissociation enthalpy.

ΔdissH(F2)≈159 kJ/mol  (weak — small F atoms, strong lone-pair repulsion)\Delta_{diss}H(F_2) \approx 159\ \text{kJ/mol} \; (\text{weak — small F atoms, strong lone-pair repulsion})

ΔdissH(Cl2)≈242 kJ/mol\Delta_{diss}H(Cl_2) \approx 242\ \text{kJ/mol}

The much lower bond enthalpy of F2F_2 means less energy is required to generate F atoms, favouring F2F_2's oxidising ability.

Step 2 — Electron gain enthalpy.

ΔegH(F)=−333 kJ/mol,ΔegH(Cl)=−349 kJ/mol\Delta_{eg}H(F) = -333\ \text{kJ/mol}, \quad \Delta_{eg}H(Cl) = -349\ \text{kJ/mol}

Taken alone, Cl releases slightly more energy on gaining an electron — this factor alone favours Cl2, but it is only one term in the full cycle.

Step 3 — Hydration enthalpy.

ΔhydH(F−)  is much more negative than  ΔhydH(Cl−)\Delta_{hyd}H(F^-) \; \text{is much more negative than} \; \Delta_{hyd}H(Cl^-)

because F−F^- is a much smaller ion with a higher charge density, so it is hydrated far more strongly, releasing significantly more energy.

Step 4 — Combine all three (thermochemical cycle for 12X2(g)+e−→aqX−(aq)\tfrac12X_2(g) + e^- \xrightarrow{aq} X^-(aq)). …

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