1. In an octahedral field, the $d$-orbitals split into: $t_{2g} ext{ (lower)} ext{ and } e_g ext{ (higher)}$

2. We compare the number of unpaired electrons in high-spin and low-spin cases for each given metal ion.

Option A: $Mn^{2+}$

• Atomic number of Mn = 25
• $Mn^{2+} : [Ar]3d^5$

High spin octahedral:

$t_{2g}^3 e_g^2$ Number of unpaired electrons = 5

Low spin octahedral:

$t_{2g}^5 e_g^0$ Number of unpaired electrons = 1

Difference: $5-1=4$ So, not correct.

Option B: $Ni^{2+}$

• Atomic number of Ni = 28
• $Ni^{2+} : [Ar]3d^8$

High spin octahedral:

$t_{2g}^6 e_g^2$ Number of unpaired electrons = 2

Low spin octahedral:

For $d^8$ octahedral, arrangement remains effectively: $t_{2g}^6 e_g^2$ Number of unpaired electrons = 2

Difference: $2-2=0$ So, not correct.

Option C: $Co^{2+}$

• Atomic number of Co = 27
• $Co^{2+} : [Ar]3d^7$

High spin octahedral:

$t_{2g}^5 e_g^2$ Number of unpaired electrons = 3

Low spin octahedral:

$t_{2g}^6 e_g^1$ Number of unpaired electrons = 1

Difference: $3-1=2$ So, this is correct.

Option D: $Fe^{2+}$

• Atomic number of Fe = 26
• $Fe^{2+} : [Ar]3d^6$

High spin octahedral:

$t_{2g}^4 e_g^2$ Number of unpaired electrons = 4

Low spin octahedral:

$t_{2g}^6 e_g^0$ Number of unpaired electrons = 0

Difference: $4-0=4$ So, not correct.

Final conclusion

The metal ion for which the difference in the number of unpaired electrons between high-spin and low-spin octahedral complexes is 2 is: $\boxed{Co^{2+}}$