Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of $F_{2} a n d C l_{2}$ .

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Solution

Oxidising power of $F_{2} & C l_{2}$

The oxidising powers of both the members of halogen family are expressed in terms of their electron accepting tendency and can be compared with their standard reduction potential values.

$F_{2} + 2 e^{-} \to 2 F^{-}; E^{\circ} = 2.87 V$

$C l_{2} + 2 e^{-} \to 2 C l^{-}; E^{\circ} = 1.36 V$

Since, $E_{F_{2} / F^{-}}^{^{\circ}} \to E_{C l_{2} / C l^{-}}^{^{\circ}}$ , $F_{2}$ is a stronger oxidising agent.

Three factors contribute towards the oxidation potentials of both the halogens.

(i) Bond dissociation enthalpy: Bond dissociation enthalpy of $F_{2} (158 k J m o l^{- 1})$ is less compared to that of $C l_{2} (242.6 k J m o l^{- 1})$

(ii) Electron gain enthalpy: The negative electron gain enthalpy of $F (- 332.6 k J m o l^{- 1})$ is slightly less than of $C l (- 348.5 k J m o l^{- 1})$ .

(iii) Hydration enthalpy: The hydration enthalpy of $F^{- i o n} (515 k J m o l^{- 1})$ is much higher than that of $C l^{- i o n} (381 k J m o l^{- 1})$ due to its smaller size.

From the available data, we may conclude that lesser bond dissociation enthalpy and higher hydration enthalpy compensate lower negative electron gain enthalpy of fluorine as compared to chlorine.

Consequently, $F_{2}$ is a more powerful oxidising agent than $C l_{2}$ .

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

Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of $F_{2} a n d C l_{2}$ .