Lattice enthalpy and enthalpy of solution of NaCl are $$788\,\text{kJ mol}^{-1}$$ and $$4\,\text{kJ mol}^{-1}$$, respectively. The hydration enthalpy of $$NaCl$$ is:

For dissolving an ionic solid such as sodium chloride, we use the thermodynamic relation

$$\Delta H_{\text{solution}} \;=\; \Delta H_{\text{lattice}} \;+\; \Delta H_{\text{hydration}}.$$

First, let us note the sign convention: $$\Delta H_{\text{lattice}}$$ is positive because energy must be supplied to break the crystal into free gaseous ions, whereas $$\Delta H_{\text{hydration}}$$ is usually negative because energy is released when those ions become surrounded by water molecules.

We have from the data of the question

$$\Delta H_{\text{solution}} = +4\,\text{kJ mol}^{-1},$$

$$\Delta H_{\text{lattice}} = +788\,\text{kJ mol}^{-1}.$$

Substituting these numerical values into the formula gives

$$+4 \;=\; +788 \;+\; \Delta H_{\text{hydration}}.$$

Isolating $$\Delta H_{\text{hydration}}$$, we subtract $$+788$$ from both sides:

$$\Delta H_{\text{hydration}} \;=\; +4 \;-\; 788.$$

Carrying out the subtraction,

$$\Delta H_{\text{hydration}} \;=\; -784\,\text{kJ mol}^{-1}.$$

Thus the hydration enthalpy of sodium chloride is a large negative quantity, indicating that considerable energy is released when Na+ and Cl− ions are solvated by water.

Hence, the correct answer is Option C.