The standard enthalpy of formation of NH$$_3$$ is $$-46.0$$ kJ/mol. If bond enthalpy of H$$_2$$ is $$-436$$ kJ/mol and that of N$$_2$$ is $$-712$$ kJ/mol, the average bond enthalpy of N$$-$$H bond in NH$$_3$$ is:

The standard enthalpy of formation of NH₃ is given as -46.0 kJ/mol. This corresponds to the reaction:

$$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightarrow NH_{3(g)} \quad \Delta H_f = -46.0 \text{ kJ/mol}$$

The bond enthalpies provided are for H₂ and N₂, given as -436 kJ/mol and -712 kJ/mol, respectively. These values represent the enthalpy changes when the bonds are formed from their respective atoms. Therefore, the bond formation enthalpy for H-H is -436 kJ/mol, and for N≡N is -712 kJ/mol.

To find the enthalpy change for the formation of NH₃, we need to consider the atomization of the reactants (breaking bonds) and the formation of bonds in the product. The atomization enthalpy is the reverse of bond formation, so:

Atomization enthalpy for N₂ (breaking N≡N bond) = - (bond formation enthalpy of N₂) = - (-712 $$\text{ kJ/mol}$$) = +712 $$\text{ kJ/mol per mole of N₂}$$

Atomization enthalpy for H₂ (breaking H-H bond) = - (bond formation enthalpy of H₂) = - (-436 $$\text{ kJ/mol}$$) = +436 $$\text{ kJ/mol per mole of H₂}$$

For the reaction, we have:

Atomization of $$\frac{1}{2}$$ mole of N₂: $$\Delta H_1 = \frac{1}{2} \times 712 = 356 \text{ kJ}$$

Atomization of $$\frac{3}{2}$$ moles of H₂: $$\Delta H_2 = \frac{3}{2} \times 436 = 654 \text{ kJ}$$

The total atomization energy is:

$$\Delta H_{\text{atomization}} = \Delta H_1 + \Delta H_2 = 356 + 654 = 1010 \text{ kJ}$$

Now, the formation of NH₃ from atoms (N and 3H) releases energy. Let the enthalpy change for forming one mole of NH₃ from its atoms be $$\Delta H_3$$. This involves forming three N-H bonds.

The overall enthalpy of formation is the sum of the atomization and formation steps:

$$\Delta H_f = \Delta H_{\text{atomization}} + \Delta H_3$$

Substituting the known value:

$$-46.0 = 1010 + \Delta H_3$$

Solving for $$\Delta H_3$$:

$$\Delta H_3 = -46.0 - 1010 = -1056 \text{ kJ/mol}$$

This enthalpy change, $$\Delta H_3 = -1056 \text{ kJ/mol}$$, corresponds to the formation of one mole of NH₃ from one nitrogen atom and three hydrogen atoms, which involves the formation of three N-H bonds.

Therefore, the enthalpy change for forming three N-H bonds is -1056 kJ/mol. The average bond formation enthalpy for one N-H bond is:

$$\frac{-1056}{3} = -352 \text{ kJ/mol}$$

The bond formation enthalpy is negative, indicating energy release. The average bond enthalpy, which is the bond dissociation energy (energy required to break the bond), is the positive value of this. Thus, the average bond enthalpy of the N-H bond is 352 kJ/mol.

Hence, the correct answer is Option D.