The spin-only magnetic moment value of the compound with strongest oxidizing ability among $$MnF_4$$, $$MnF_3$$ and $$MnF_2$$ is ______ B.M. (nearest integer)

We need to find the spin-only magnetic moment of the compound with the strongest oxidizing ability among $$MnF_4$$, $$MnF_3$$, and $$MnF_2$$.

Since fluorine is always -1, in $$MnF_2$$ manganese is in the +2 oxidation state ($$3d^5$$ configuration), in $$MnF_3$$ it is in the +3 oxidation state ($$3d^4$$ configuration), and in $$MnF_4$$ it is in the +4 oxidation state ($$3d^3$$ configuration).

Among these, $$MnF_3$$ is the strongest oxidizing agent because $$Mn^{3+}$$ ($$3d^4$$) has a very strong tendency to gain one electron and form $$Mn^{2+}$$ ($$3d^5$$), which has the exceptionally stable half‐filled $$d$$‐subshell configuration. The large $$Mn^{3+}/Mn^{2+}$$ reduction potential ($$E^\circ = +1.51$$ V) confirms this.

Manganese (Z = 25) in its ground state is $$[Ar]\,3d^5\,4s^2$$.
Removing two electrons from 4s and one from 3d gives $$Mn^{3+} = [Ar]\,3d^4$$.

Because fluoride ($$F^-$$) is a weak field ligand, it does not cause pairing of electrons, so the $$3d^4$$ configuration remains high spin:
$$\uparrow\;|\;\uparrow\;|\;\uparrow\;|\;\uparrow\;|\;\_$$
Thus the number of unpaired electrons is $$n = 4$$.

The spin‐only magnetic moment is given by $$\mu_s = \sqrt{n(n+2)}\text{ B.M.}$$
Substituting $$n=4$$ gives $$\mu_s = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6} = 4.899\text{ B.M.}$$
Rounding to the nearest integer yields $$\mu_s \approx 5\text{ B.M.}$$

Therefore, the spin‐only magnetic moment of the strongest oxidizing compound is $$\boxed{5}$$ B.M.