The standard reduction potentials for $$Zn^{2+}/Zn$$, $$Ni^{2+}/Ni$$, and $$Fe^{2+}/Fe$$ are $$-0.76$$, $$-0.23$$ and $$-0.44$$ V respectively. The reaction $$X + Y^{2+} \rightarrow X^{2+} + Y$$ will be spontaneous when:

The reaction given is $$X + Y^{2+} \rightarrow X^{2+} + Y$$.

Here, metal $$X$$ is getting oxidised (loses electrons) and serves as the anode, while the ion $$Y^{2+}$$ is getting reduced (gains electrons) and serves as the cathode.

For any electrochemical cell operated under standard conditions, the standard cell potential is

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$

The process is spontaneous if and only if $$E^\circ_{\text{cell}} \gt 0$$.

The standard reduction potentials provided are
$$E^\circ(Zn^{2+}/Zn) = -0.76\text{ V}$$,
$$E^\circ(Fe^{2+}/Fe) = -0.44\text{ V}$$,
$$E^\circ(Ni^{2+}/Ni) = -0.23\text{ V}$$.

Because the cathode must have the higher (more positive) reduction potential, and the anode the lower (more negative) one, we arrange the three couples in increasing order:

$$Zn^{2+}/Zn\;(-0.76\text{ V}) \lt Fe^{2+}/Fe\;(-0.44\text{ V}) \lt Ni^{2+}/Ni\;(-0.23\text{ V})$$.

Thus, to make $$E^\circ_{\text{cell}}$$ positive we should choose
anode (X) = Zn and cathode (Y^{2+}) = Ni^{2+}.

Calculating to confirm:

Anode (oxidation): $$Zn \rightarrow Zn^{2+} + 2e^-$$, $$E^\circ_{\text{anode}} = -0.76\text{ V}$$.
Cathode (reduction): $$Ni^{2+} + 2e^- \rightarrow Ni$$, $$E^\circ_{\text{cathode}} = -0.23\text{ V}$$.

Therefore
$$E^\circ_{\text{cell}} = (-0.23) - (-0.76) = +0.53\text{ V} \gt 0,$$

which confirms spontaneity.

Only the pair $$X = Zn,\, Y = Ni$$ satisfies this condition. The remaining options give negative cell potentials and are non-spontaneous.

Hence, the correct choice is:
Option D which is: $$X = Zn, Y = Ni$$