The spin only magnetic moment of $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ complexes is _________ B.M. (Nearest integer)

(Given : Atomic no. of Mn is 25)

<p>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:</p> <p>μ = $\sqrt {n\left( {n + 2} \right)}$ Bohr magnetons</p> <p>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:</p> <p>For Mn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ </p> <p>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:</p> <p>μ = $\sqrt {5\left( {5 + 2} \right)}$ = $\sqrt {35}$ = 5.9 Bohr Magnetons</p> <p>The nearest integer to 5.9 is 6.</p> <p>Hence, the spin only magnetic moment of $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is 6 Bohr magnetons.</p>