Given standard reduction potentials

$E^\circ(\text{Zn}^{2+}/\text{Zn})=-0.76\text{ V}$ $E^\circ(\text{Ni}^{2+}/\text{Ni})=-0.23\text{ V}$ $E^\circ(\text{Fe}^{2+}/\text{Fe})=-0.44\text{ V}$

Reaction given

$X + Y^{2+} \rightarrow X^{2+} + Y$

Here:

• $X$ is getting oxidized: $X \rightarrow X^{2+} + 2e^-$
• $Y^{2+}$ is getting reduced: $Y^{2+} + 2e^- \rightarrow Y$

Condition for spontaneity

For a spontaneous reaction under standard conditions,

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} > 0$

Since $Y^{2+}$ is reduced, $Y$ is the cathode metal. Since $X$ is oxidized, $X$ is the anode metal.

Therefore,

$E^\circ_{\text{cell}} = E^\circ(Y^{2+}/Y) - E^\circ(X^{2+}/X)$

So we need:

$E^\circ(Y^{2+}/Y) > E^\circ(X^{2+}/X)$

Check each option

A: $X=\text{Ni},\; Y=\text{Fe}$

$E^\circ_{\text{cell}} = E^\circ(\text{Fe}^{2+}/\text{Fe}) - E^\circ(\text{Ni}^{2+}/\text{Ni})$ $= -0.44 - (-0.23) = -0.21\text{ V}$ Not spontaneous.

B: $X=\text{Ni},\; Y=\text{Zn}$

$E^\circ_{\text{cell}} = -0.76 - (-0.23) = -0.53\text{ V}$ Not spontaneous.

C: $X=\text{Fe},\; Y=\text{Zn}$

$E^\circ_{\text{cell}} = -0.76 - (-0.44) = -0.32\text{ V}$ Not spontaneous.

D: $X=\text{Zn},\; Y=\text{Ni}$

$E^\circ_{\text{cell}} = -0.23 - (-0.76) = +0.53\text{ V}$ Spontaneous.

Conclusion

The reaction is spontaneous only for:

$\boxed{X=\text{Zn},\; Y=\text{Ni}}$

So, the correct option is D.