The spin only magnetic moment of the $$\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$$ complex can be calculated using the formula:

μ = $$\sqrt {n\left( {n + 2} \right)} $$ Bohr magnetons

Where n is the number of unpaired electrons in the complex. To find the number of unpaired electrons in $$\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$$, we can use the electron configuration of the Mn ion:

For Mn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

From the electron configuration, we can see that Mn²⁺ has 5 unpaired electrons. Thus, the spin only magnetic moment of $$\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$$ is:

μ = $$\sqrt {5\left( {5 + 2} \right)} $$ = $$\sqrt {35} $$ = 5.9 Bohr Magnetons

The nearest integer to 5.9 is 6.

Hence, the spin only magnetic moment of $$\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$$ is 6 Bohr magnetons.