The spin only magnetic moment ( $$\mu$$) value (B.M.) of the compound with strongest oxidising power among $$Mn_{2}O_{3},$$ TiO and VO is ____ B.M. (Nearest integer).

We need to find the spin-only magnetic moment of the compound with the strongest oxidising power among $$Mn_2O_3$$, TiO, and VO.

In $$Mn_2O_3$$, Mn is in the +3 oxidation state as shown by $$2x + 3(-2) = 0 \Rightarrow x = +3$$, while in TiO and VO, Ti and V are both in the +2 oxidation state.

$$Mn^{3+}$$ in $$Mn_2O_3$$ is the strongest oxidising agent because it has the highest reduction potential among the three. It readily accepts an electron to become the stable $$Mn^{2+}$$ (d$$^5$$, half-filled), whereas $$Ti^{2+}$$ and $$V^{2+}$$ act as better reducing agents.

The electronic configuration of Mn is [Ar] 3d$$^5$$ 4s$$^2$$, so for $$Mn^{3+}$$ it becomes [Ar] 3d$$^4$$.

Because the oxide ion ($$O^{2-}$$) is a weak-field ligand, $$Mn^{3+}$$ in $$Mn_2O_3$$ remains high-spin. A high-spin d$$^4$$ configuration has all four electrons unpaired by Hund’s rule, giving $$n = 4$$ unpaired electrons.

The spin-only magnetic moment is given by the formula

$$ \mu = \sqrt{n(n+2)} \text{ BM} $$

where $$n$$ is the number of unpaired electrons. Substituting $$n = 4$$ yields

$$ \mu = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6} \approx 4.90 \text{ BM} $$

Nearest integer: 5 BM.

The answer is 5.