Question

The disproportionation of MnO42- 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 (μ) value of B in BM is _______. (Nearest integer)

Answer 1

$\begin{array}{l}\overset{+6}{3Mn}O_4^{-2}+4H^+\rightarrow \overset{+4}{Mn}O_2 + \overset{+7}{Mn}O_4^-\end{array}$

Mn → 4 s 23 d 5

Mn+4 → 3 d 3

n = 3

$\begin{array}{l}\mu = \sqrt{n(n+2)}\end{array}$

$\begin{array}{l}=\sqrt{3(5)}\end{array}$

$\begin{array}{l}=\sqrt{15} = 3.87 \approx 4 ~\text{B.M.}\end{array}$

So, the spin-only magnetic moment (μ) value of B in BM is 4.