The spin only magnetic moment of $$[Mn(H_2O)_6]^{2+}$$ complexes is ______ B.M. (Nearest integer)
(Given atomic number of Mn = 25)

$$[Mn(H_2O)_6]^{2+}$$ complex. Atomic number of Mn = 25.

Mn: $$[Ar] 3d^5 4s^2$$. Mn²⁺ loses 2 electrons (from 4s): $$[Ar] 3d^5$$.

$$H_2O$$ is a weak field ligand (low in the spectrochemical series), so it does not cause pairing of electrons. The complex is high spin.

In a high-spin $$d^5$$ octahedral complex, all five 3d orbitals have one electron each (following Hund's rule): $$t_{2g}^3 e_g^2$$.

Number of unpaired electrons = 5.

The spin-only magnetic moment formula is:

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

where $$n$$ is the number of unpaired electrons.

$$\mu = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35} = 5.92 \approx 6$$ B.M.

The answer is 6 (nearest integer).