1. Find the oxidation state and electronic configuration of Mn in the complex

The complex is: $\left[\mathrm{Mn}(\mathrm{H_2O})_6\right]^{2+}$

Since $\mathrm{H_2O}$ is a neutral ligand, the oxidation state of Mn is $+2$.

Mn has atomic number $25$.

Ground state electronic configuration of Mn: $\mathrm{Mn}: [\mathrm{Ar}]\,3d^5 4s^2$

For $\mathrm{Mn}^{2+}$, remove two electrons from $4s$ first: $\mathrm{Mn}^{2+}: [\mathrm{Ar}]\,3d^5$

2. Determine the nature of the ligand field

$\mathrm{H_2O}$ is a weak field ligand, so in an octahedral complex it gives a high-spin configuration.

For $d^5$ high-spin octahedral: $t_{2g}^3 e_g^2$

So, number of unpaired electrons: $n = 5$

3. Use the spin-only magnetic moment formula

$\mu = \sqrt{n(n+2)}\ \text{B.M.}$

Substitute $n=5$: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{B.M.}$

4. Nearest integer

$5.92 \approx 6$

Final Answer

The spin-only magnetic moment is: $\boxed{6\ \text{B.M.}}$