1. Identify the oxidation state and d-electron count

In all complexes, $\mathrm{H_2O}$ is a neutral ligand, so the metal ion is in the $+2$ oxidation state.

Thus:

• $\mathrm{Fe^{2+}} : [\mathrm{Ar}]\,3d^6$
• $\mathrm{Mn^{2+}} : [\mathrm{Ar}]\,3d^5$
• $\mathrm{Co^{2+}} : [\mathrm{Ar}]\,3d^7$
• $\mathrm{Cr^{2+}} : [\mathrm{Ar}]\,3d^4$

2. Nature of ligand

$\mathrm{H_2O}$ is a weak-field ligand, so these octahedral complexes are high-spin.

Hence electron distributions in octahedral field are:

• $d^4 : t_{2g}^3 e_g^1$
• $d^5 : t_{2g}^3 e_g^2$
• $d^6 : t_{2g}^4 e_g^2$
• $d^7 : t_{2g}^5 e_g^2$

3. Count unpaired electrons

• A: $[\mathrm{Fe(H_2O)_6}]^{2+}$ is high-spin $d^6$

  Unpaired electrons $=4$

• B: $[\mathrm{Mn(H_2O)_6}]^{2+}$ is high-spin $d^5$

  Unpaired electrons $=5$

• C: $[\mathrm{Co(H_2O)_6}]^{2+}$ is high-spin $d^7$

  Unpaired electrons $=3$

• D: $[\mathrm{Cr(H_2O)_6}]^{2+}$ is high-spin $d^4$

  Unpaired electrons $=4$

4. Compare paramagnetism

Paramagnetic behaviour increases with the number of unpaired electrons.

So the complex with the least paramagnetic behaviour is the one with the minimum number of unpaired electrons.

Among the options:

• A: $4$
• B: $5$
• C: $3$
• D: $4$

Minimum is 3 for $[\mathrm{Co(H_2O)_6}]^{2+}$.

5. Final answer

$\boxed{\left[\mathrm{Co(H_2O)_6}\right]^{2+}}$

So, Option C is correct.