The enthalpy change for X2(g) + 2e- -> 2X-(aq) is the sum of the bond dissociation enthalpy, twice the electron gain enthalpy and twice the hydration enthalpy of X-. Data (kJ mol^-1; bond dissociation, electron gain, hydration): F2: 159, -328, -515; Cl2: 243, -349, -381. The value of Delta H(F2) - Delta H(Cl2) in kJ mol^-1 is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph184

The enthalpy change for $X X_{2} (g) + 2 e X^{-} 2 X X^{-} (aq)$ is the sum of the bond dissociation enthalpy, twice the electron gain enthalpy and twice the hydration enthalpy of $X X^{-}$ . Data ( $kJ mo l^{- 1}$ ; bond dissociation, electron gain, hydration): $F X_{2}$ : 159, $- 328$ , $- 515$ ; $Cl X_{2}$ : 243, $- 349$ , $- 381$ . The value of $Δ H (F X_{2}) - Δ H (Cl X_{2})$ in $kJ mo l^{- 1}$ is

Δ H (F X_{2}) = 159 + 2 (- 328) + 2 (- 515) = - 1527

and

Δ H (Cl X_{2}) = 243 + 2 (- 349) + 2 (- 381) = - 1217

. The difference is

- 1527 - (- 1217) = - 310 kJ mo l^{- 1}

. The much larger hydration enthalpy of

F X^{-}

and the weak F-F bond outweigh the less negative electron gain enthalpy, which is why

F X_{2}

is the stronger oxidant.