QJEEM23Chd2d2: JEE Main 2023 Chemistry - Class 12 JEEM23Chd2d2 Solution: The spin only magnetic moment of complexes is _________ B.M.…

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.