Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the mangenese product formed from the above reaction is ______ B.M. (Nearest Integer)

When acidified potassium permanganate ($$KMnO_4$$) oxidises oxalic acid ($$H_2C_2O_4$$), the balanced reaction is:

$$2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \to 2MnSO_4 + K_2SO_4 + 10CO_2 + 8H_2O$$

In this reaction, manganese goes from +7 oxidation state (in $$KMnO_4$$) to +2 oxidation state (in $$MnSO_4$$).

The manganese product is $$Mn^{2+}$$.

Electronic configuration of $$Mn^{2+}$$:

Mn has atomic number 25. Its electronic configuration is [Ar] 3d$$^5$$ 4s$$^2$$.

$$Mn^{2+}$$ loses 2 electrons from 4s: [Ar] 3d$$^5$$

With 5 electrons in the 3d subshell, each orbital gets one electron (by Hund's rule):

$$d_{xy}^1 \; d_{xz}^1 \; d_{yz}^1 \; d_{x^2-y^2}^1 \; d_{z^2}^1$$

Number of unpaired electrons ($$n$$) = 5

The spin-only magnetic moment formula is:

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

Substituting $$n = 5$$:

$$\mu = \sqrt{5(5+2)}$$

$$= \sqrt{5 \times 7}$$

$$= \sqrt{35}$$

$$= 5.916 \text{ B.M.}$$

Rounding to the nearest integer:

$$\mu \approx$$ 6 B.M.