To find which of the given options has the maximum number of unpaired electrons, we first need to determine the electronic configurations of the ions and then count the number of unpaired electrons in each. Let's evaluate each option:

Option A: Mg²⁺

Magnesium (Mg) has an atomic number of 12, so its ground state electronic configuration is $1s^2 2s^2 2p^6 3s^2$. The $Mg^{2+}$ ion loses two electrons:

$$\text{Configuration of } Mg^{2+}: 1s^2 2s^2 2p^6$$

All the electrons in $Mg^{2+}$ are paired, so there are 0 unpaired electrons.

Option B: Ti³⁺

Titanium (Ti) has an atomic number of 22, and its ground state electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2$. The $Ti^{3+}$ ion loses two electrons from the 4s orbital and one from the 3d orbital:

$$\text{Configuration of } Ti^{3+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^1$$

As there is one electron in the 3d orbital, $Ti^{3+}$ has 1 unpaired electron.

Option C: V³⁺

Vanadium (V) has an atomic number of 23, and its ground state electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3$. The $V^{3+}$ ion loses two electrons from the 4s orbital and one electron from the 3d orbital:

$$\text{Configuration of } V^{3+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^2$$

There are 2 unpaired electrons in the 3d orbital of $V^{3+}$.

Option D: Fe²⁺

Iron (Fe) has an atomic number of 26, and its ground state electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$. The $Fe^{2+}$ ion loses two electrons from the 4s orbital:

$$\text{Configuration of } Fe^{2+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$$

In the $Fe^{2+}$ ion, there are 4 unpaired electrons in the 3d subshell.

Conclusion: Among the given ions, $Fe^{2+}$ has the maximum number of unpaired electrons (4 unpaired electrons), making Option D the correct choice.