The enthalpy change on freezing of 1 mol of water at 5°C to ice at $$-5$$°C is: (Given $$\Delta_{fus}H = 6$$ kJ mol$$^{-1}$$ at 0°C, $$C_p(H_2O, l) = 75.3$$ J mol$$^{-1}$$ K$$^{-1}$$, $$C_p(H_2O, s) = 36.8$$ J mol$$^{-1}$$ K$$^{-1}$$)

We want the enthalpy change when 1 mol of liquid water at $$5^{\circ}\text{C}$$ first cools to $$0^{\circ}\text{C}$$, then freezes at that temperature, and finally the ice cools further to $$-5^{\circ}\text{C}$$. Because each step occurs in sequence, the total enthalpy change will be the algebraic sum of the enthalpy changes of the three individual steps.

Step 1 - Cooling the liquid from $$5^{\circ}\text{C}$$ to $$0^{\circ}\text{C}$$.
The relation between enthalpy change, heat capacity at constant pressure and temperature change is first stated:

$$\Delta H = n\,C_p\,\Delta T$$

Here $$n = 1\ \text{mol}$$, $$C_p(H_2O,l) = 75.3\ \text{J mol}^{-1}\text{K}^{-1}$$ and the temperature change is

$$\Delta T = T_{\text{final}} - T_{\text{initial}} = 0 - 5 = -5\ \text{K}$$

Substituting these numbers, we get

$$$ \Delta H_1 = 1 \times 75.3\ \text{J mol}^{-1}\text{K}^{-1} \times (-5\ \text{K}) = -376.5\ \text{J} $$$

Since $$1\ \text{kJ} = 1000\ \text{J}$$, this is

$$$ \Delta H_1 = -\,0.3765\ \text{kJ} $$$

The negative sign shows that heat is released during cooling.

Step 2 - Freezing the water at $$0^{\circ}\text{C}$$.
The given enthalpy of fusion is $$\Delta_{\text{fus}}H = 6\ \text{kJ mol}^{-1}$$ for melting (ice $$\rightarrow$$ water).
For the reverse process, freezing (water $$\rightarrow$$ ice), we simply change the sign:

$$$ \Delta H_2 = -\,6\ \text{kJ} $$$

Step 3 - Cooling the ice from $$0^{\circ}\text{C}$$ to $$-5^{\circ}\text{C}$$.
Again using $$\Delta H = n\,C_p\,\Delta T$$ but now with the solid heat capacity $$C_p(H_2O,s) = 36.8\ \text{J mol}^{-1}\text{K}^{-1}$$ and the same $$\Delta T = -5\ \text{K}$$, we obtain

$$$ \Delta H_3 = 1 \times 36.8\ \text{J mol}^{-1}\text{K}^{-1} \times (-5\ \text{K}) = -184\ \text{J} = -\,0.184\ \text{kJ} $$$

Now we add the three contributions to get the overall enthalpy change:

$$$ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + \Delta H_3 = (-\,0.3765) + (-\,6) + (-\,0.184)\ \text{kJ} = -\,6.5605\ \text{kJ} $$$

Rounding to three significant figures gives

$$$ \Delta H_{\text{total}} \approx -\,6.56\ \text{kJ mol}^{-1} $$$

The magnitude matches option D, and the negative sign merely indicates that heat is evolved during the process. Examination of the choices shows that the numerical value 6.56 kJ mol$$^{-1}$$ is unique to Option D.

Hence, the correct answer is Option D.