Question

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

F_{2}

and

{Cl}_{2} .

Accepted Answer

The oxidising power of an element(Fluorine and Chlorine) is determined by three factors. They are:
(1) Bond dissociation enthalpy
(2) Electron gain enthalpy
(3) Hydration enthalpy

The electron gain enthalpy of chlorine $(- 394 kJ {mol}^{- 1})$ is more negative than that of fluorine $(- 333 kJ {mol}^{- 1})$ whereas the bond dissociation energy of fluorine $(- 158.8 kJ {mol}^{- 1})$ is much lesser than that of chlorine $(- 242.6 kJ {mol}^{- 1})$ . Also, due to the small size of fluorine, the hydration energy of fluorine is much higher than that of chlorine $(- 381 kJ {mol}^{- 1})$ . Therefore, bond dissociation enthalpy and hydration enthalpy compensate more than the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidising agent than chlorine.

Telangana Board Solutions for Chapter: p-Block Elements, Exercise 13: Intext Questions

Telangana Board Chemistry Solutions for Exercise - Telangana Board Solutions for Chapter: p-Block Elements, Exercise 13: Intext Questions

Questions from Telangana Board Solutions for Chapter: p-Block Elements, Exercise 13: Intext Questions with Hints & Solutions

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