The disproportionation of $$MnO_4^{2-}$$ in acidic medium resulted in the formation of two manganese compounds A and B. If the oxidation state of Mn in B is smaller than that of A, then the spin-only magnetic moment $$\mu$$ value of B in BM is _____ (Nearest integer)

In the manganate ion:

$$\mathrm{MnO_4^{2-}}$$

let the oxidation state of manganese be:

$$\mathrm{x}$$

Then:

$$\mathrm{x + 4(-2) = -2}$$

$$\mathrm{x - 8 = -2}$$

$$\mathrm{x = +6}$$

Thus, manganese in manganate ion has oxidation state:

$$\mathrm{+6}$$

In acidic medium, manganate ion undergoes disproportionation to form:

$$\mathrm{MnO_4^-}$$

and

$$\mathrm{MnO_2}$$

For permanganate ion:

$$\mathrm{MnO_4^-}$$

let oxidation state of manganese be:

$$\mathrm{y}$$

$$\mathrm{y + 4(-2) = -1}$$

$$\mathrm{y = +7}$$

For manganese dioxide:

$$\mathrm{MnO_2}$$

let oxidation state of manganese be:

$$\mathrm{z}$$

$$\mathrm{z + 2(-2) = 0}$$

$$\mathrm{z = +4}$$

Since oxidation state in compound $$\mathrm{B}$$ is smaller:

$$\mathrm{B = MnO_2}$$

In $$\mathrm{MnO_2}$$, manganese exists as:

$$\mathrm{Mn^{4+}}$$

Electronic configuration of neutral manganese:

$$\mathrm{[Ar]\ 3d^5 4s^2}$$

Electronic configuration of:

$$\mathrm{Mn^{4+}}$$

is:

$$\mathrm{3d^3}$$

Hence, number of unpaired electrons:

$$\mathrm{n = 3}$$

Spin-only magnetic moment:

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

$$\mathrm{\mu = \sqrt{3(3+2)}}$$

$$\mathrm{\mu = \sqrt{15}}$$

$$\mathrm{\mu \approx 3.87\ BM}$$

Rounding to nearest integer:

$$\mathrm{4}$$