(A) The electronic configuration of a divalent metal ion with atomic number $25$ $(Mn^{2 })$ is $[Ar] 3d^5 4s^0$.The number of unpaired electrons $(n)

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(A) The electronic configuration of a divalent metal ion with atomic number $25$ $(Mn^{2 })$ is $[Ar] 3d^5 4s^0$.The number of unpaired electrons $(n)$ is $5$.The spin-only magnetic moment $(\mu)$ is calculated using the formula:$\mu = \sqrt{n(n 2)} \ BM$Substituting $n = 5$:$\mu = \sqrt{5(5 2)} = \sqrt{35} \ BM \approx 5.92 \ BM$.

Class 12 Chemistry Coordination Compounds Isomerism and Magnetic properties

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What is the spin-only magnetic moment value $(BM)$ of a divalent metal ion with atomic number $25$ in its aqueous solution?

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A
$5.92$
B
$5.0$
C
$0$
D
$5.26$

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What is the spin-only magnetic moment value $(BM)$ of a divalent metal ion with atomic number $25$ in its aqueous solution?

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