The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na$$^+$$ will be :

First, let us write the definition of the first-ionisation potential of sodium. By definition, the first ionisation process is

$$Na(g) \;\rightarrow\; Na^+(g) + e^-$$

The data in the question tell us that an energy of $$5.1\;\text{eV}$$ must be supplied for this change. In thermodynamic language, the enthalpy (or energy) change for the above forward process is

$$\Delta H_{ionisation}^{\,(\,1\,)} = +5.1\;\text{eV}.$$

Now we come to the quantity asked for in the problem, the electron-gain enthalpy of $$Na^+$$. By its very definition, electron-gain enthalpy (often called electron affinity) refers to the energy change when one electron is added to a gaseous species. The relevant process is therefore the exact reverse of the ionisation step:

$$Na^+(g) + e^- \;\rightarrow\; Na(g).$$

Because this second reaction is simply the reverse of the first, the enthalpy change for it must be the negative of the ionisation enthalpy. Stating the general principle explicitly:

If a reaction $$A \;\rightarrow\; B$$ has an enthalpy change $$\Delta H,$$ then the reverse reaction $$B \;\rightarrow\; A$$ has an enthalpy change $$-\Delta H.$$

Applying this principle here, we have

$$\Delta H_{electron\;gain}(Na^+) = -\Delta H_{ionisation}^{\,(\,1\,)}(Na)$$

Substituting the given numerical value,

$$\Delta H_{electron\;gain}(Na^+) = -(+5.1\;\text{eV}) = -5.1\;\text{eV}.$$

Thus the electron-gain enthalpy of $$Na^+$$ is a negative quantity whose magnitude equals the first ionisation potential of neutral sodium. A negative sign indicates that energy is released when the electron is accepted by $$Na^+$$.

Hence, the correct answer is Option D.