Calculated value of magnetic moment (BM) of an aqueous solution of $Mn^{2+}$ is .........

5.92

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

$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$