JEE MAIN - Chemistry (2024 - 4th April Morning Shift - No. 30)
Consider the following reaction
$$\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{A}+\mathrm{H}_2 \mathrm{O} \text {. }$$
Product '$$\mathrm{A}$$' in neutral or acidic medium disproportionate to give products '$$\mathrm{B}$$' and '$$\mathrm{C}$$' along with water. The sum of spin-only magnetic moment values of $$\mathrm{B}$$ and $$\mathrm{C}$$ is ________ BM. (nearest integer) (Given atomic number of $$\mathrm{Mn}$$ is 25)
Answer
4
Explanation
$$\mathrm{A}$$ is $$\mathrm{K}_2 \mathrm{MnO}_4$$
$$\mathrm{B}$$ and $$\mathrm{C}$$ are $$\mathrm{KMnO}_4$$ and $$\mathrm{MnO}_2$$
$$\mathrm{KMnO}_4(\mu=0)$$
$$\mathrm{MnO}_2(\mathrm{Mn}^{4+})(\mu=3.87)$$
Sum $$=3.87=4$$ (Nearest integer)
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