Calculated value of magnetic moment (BM) of an aqueous solution of $Mn^{2+}$ is .........
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5.92
Electronic configuration of Mn²⁺
Atomic number of Mn = 25
Ground state configuration of Mn: Mn = [Ar] 3d⁵ 4s²
When manganese forms Mn²⁺: Two electrons are removed from the 4s orbital.
Mn²⁺ = [Ar] 3d⁵
Thus Mn²⁺ has: Number of unpaired electrons (n) = 5
Using the spin-only magnetic moment formula:
$\mu = \sqrt{n(n+2)} \text{ BM}$
where μ = magnetic moment in Bohr Magneton (BM)
n = number of unpaired electrons
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$