The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:Ni2+Fe2+Co2+Mn2+

The metal ion in question is Fe2+.

Here's the step-by-step explanation:

1. First, we need to understand that the number of unpaired electrons in a metal ion depends on its electronic configuration and the nature of the ligand in the complex.

2. In an octahedral complex, a high-spin configuration occurs when the ligand causes weak field splitting, allowing electrons to occupy higher energy orbitals before pairing up in the lower ones. On the other hand, a low-spin configuration occurs when the ligand causes strong field splitting, forcing electrons to pair up in the lower energy orbitals before moving to the higher ones.

3. Now, let's look at the electronic configurations of the given metal ions in their +2 oxidation states:

   • Ni2+: [Ar] 3d8
   • Fe2+: [Ar] 3d6
   • Co2+: [Ar] 3d7
   • Mn2+: [Ar] 3d5

4. In a high-spin octahedral complex, Fe2+ would have four unpaired electrons (↑↑↑↑↓↓), while in a low-spin complex, it would have two unpaired electrons (↑↓↑↓↑↑). So, the difference in the number of unpaired electrons between high-spin and low-spin configurations for Fe2+ is indeed two.

5. For the other ions, the difference in the number of unpaired electrons between high-spin and low-spin configurations is not two, so they can be ruled out.

Therefore, the metal ion that fits the given condition is Fe2+.