JEE MAIN - Chemistry (2022 - 28th July Morning Shift - No. 20)
The disproportionation of $$\mathrm{MnO}_{4}^{2-}$$ in acidic medium resulted in the formation of two manganese compounds $$\mathrm{A}$$ and $$\mathrm{B}$$. If the oxidation state of $$\mathrm{Mn}$$ in $$\mathrm{B}$$ is smaller than that of A, then the spin-only magnetic moment $$(\mu)$$ value of B in BM is __________. (Nearest integer)
Answer
4
Explanation
$3 \overset{+6}{\mathrm{MnO}_{4}^{-2}}+4 \mathrm{H}^{+} \rightarrow \overset{+4}{\mathrm{MnO}_{2}}+\overset{+7}{\mathrm{MnO}_{4}^{-}}$
$\mathrm{Mn} \rightarrow 4 s^{2} 3 d^{5}$
$\mathrm{Mn}^{+4} \rightarrow 3 d^{3}$
$$ \mathrm{n}=3 $$
$$ \begin{aligned} \mu &=\sqrt{n(n+2)} \\\\ &=\sqrt{3(5)} \\\\ &=\sqrt{15} \\\\ &=3.87 \approx 4 \text { B.M. } \end{aligned} $$
$\mathrm{Mn} \rightarrow 4 s^{2} 3 d^{5}$
$\mathrm{Mn}^{+4} \rightarrow 3 d^{3}$
$$ \mathrm{n}=3 $$
$$ \begin{aligned} \mu &=\sqrt{n(n+2)} \\\\ &=\sqrt{3(5)} \\\\ &=\sqrt{15} \\\\ &=3.87 \approx 4 \text { B.M. } \end{aligned} $$
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