Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: $$$\frac{1}{2}Cl_2(g) \xrightarrow{\frac{1}{2}\Delta_{diss}H^{\ominus}} Cl(g) \xrightarrow{\Delta_{eg}H^{\ominus}} Cl^{-}(g) \xrightarrow{\Delta_{hyd}H^{\ominus}} Cl^{-}(aq).$$$ The energy involved in the conversion of $$\frac{1}{2}Cl_2(g)$$ to $$Cl^{-}(g)$$ (using the data, $$\Delta_{diss}H^{\ominus}_{Cl_2} = 240\ kJ\ mol^{-1},\ \Delta_{eg}H^{\ominus}_{Cl} = -349\ kJ\ mol^{-1},\ \Delta_{hyd}H^{\ominus}_{Cl^{-}} = -381\ kJ\ mol^{-1}$$) will be

For finding the net enthalpy change, add the standard enthalpy of each elementary step that takes $$\tfrac12Cl_2(g)$$ to $$Cl^{-}(aq)$$.

Step 1 - Dissociation of chlorine
$$\tfrac12Cl_2(g)\;\longrightarrow\;Cl(g)$$
Enthalpy change  $$=\;\tfrac12\Delta_{\text{diss}}H^{\ominus}$$
$$=\;\tfrac12(240\;\text{kJ mol}^{-1})=120\;\text{kJ mol}^{-1}$$

Step 2 - Electron-gain by the chlorine atom
$$Cl(g)+e^- \;\longrightarrow\; Cl^{-}(g)$$
Enthalpy change  $$=\;\Delta_{\text{eg}}H^{\ominus}= -349\;\text{kJ mol}^{-1}$$

Step 3 - Hydration of chloride ion
$$Cl^{-}(g)\;\longrightarrow\;Cl^{-}(aq)$$
Enthalpy change  $$=\;\Delta_{\text{hyd}}H^{\ominus}= -381\;\text{kJ mol}^{-1}$$

Total enthalpy change
$$\Delta H^{\ominus}=120+(-349)+(-381)\; \text{kJ mol}^{-1}$$
$$\Delta H^{\ominus}= -610\;\text{kJ mol}^{-1}$$

Hence, the energy involved is $$-610\;\text{kJ mol}^{-1}$$.

Option B which is: $$-610\; \mathbf{kJ\;mol^{-1}}$$