Consider the following reaction (MnO _ 2 + KOH+O _ 2 arrow A+H _ 2 O ). Product ' (A ) ' in neutral or acidic medium disproportionate to give products ' (B ) ' and ' (C ) ' along with water. The sum of spin-only

Correct answer: 0 MnO_2+KOH+O_2 arrow K_2 MnO_4+H_2 O (A) aligned & K_2 MnO_4 Neutral/acidic solution KMnO_4+MnO_2 & Mn^ +4 :-[Ar] 3 ~d^3 & n=3, μ= 3(3+2) =3.87 B.M. aligned Nearest integer is (4)

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Chemistry
Chapter: d and f Block Elements

JEE Main 2024 · 4 April · Morning Shift · Chemistry · d and f Block Elements

Consider the following reaction (MnO _ 2 + KOH+O _ 2 arrow A+H _ 2...

Consider the following reaction (\mathrm{MnO}}_{2}+ \mathrm{KOH}+\mathrm{O}

}_{2}

\rightarrow \mathrm{A}+\mathrm{H}}_{2}

\mathrm{O}

). 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 (\qquad) BM. (nearest integer) (Given atomic number of (\mathrm{Mn}

) is 25)

Correct answer

Step-by-step solution

\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{H}_2 \mathrm{O} (A)

\begin{aligned} & \mathrm{K}_2 \mathrm{MnO}_4 \xrightarrow{\text{ Neutral/acidic solution }} \mathrm{KMnO}_4+\mathrm{MnO}_2 \\ & \mathrm{Mn}^{+4}:-[\mathrm{Ar}] 3 \mathrm{~d}^3 \\ & \mathrm{n}=3, \mu=\sqrt{3(3+2)}=3.87 \text{ B.M. } \end{aligned}

<br />Nearest integer is (4)

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