1. Identify the reacting species

Acidified manganate means the manganate ion $\mathrm{MnO_4^{2-}}$, where manganese is in oxidation state $+6$.

In acidic medium, manganate undergoes disproportionation:

$3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$

So the products contain manganese in:

• higher oxidation state: $\mathrm{MnO_4^-}$ with Mn in $+7$
• lower oxidation state: $\mathrm{MnO_2}$ with Mn in $+4$

We need the spin-only magnetic moment of the product having manganese in the higher oxidation state, i.e. $\mathrm{MnO_4^-}$.

2. Find electronic configuration of Mn in $+7$ oxidation state

Atomic number of Mn = 25

Ground-state configuration of Mn:

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

For $\mathrm{Mn^{7+}}$, remove 7 electrons:

• 2 from $4s$
• 5 from $3d$

Hence,

$\mathrm{Mn^{7+}}: [\mathrm{Ar}]\,3d^0$

So number of unpaired electrons,

$n = 0$

3. Calculate spin-only magnetic moment

Formula:

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

Substitute $n=0$:

$\mu = \sqrt{0(0+2)} = 0\ \text{B.M.}$

4. Nearest integer

$0$

5. Compare with stored answer

Stored correct answer = 0, which matches the derived answer.