What is the spin-only magnetic moment value (B.M.) of a divalent metal ion with atomic number 25, in its aqueous…

🧠 Translate Atomic Number to d-Count Fast

Atomic number 25 → manganese. Divalent → $\mathrm{Mn^{2+}}$. Configuration: $[\mathrm{Ar}]\,3\mathrm{d^5}$. In aqueous solution ($\mathrm{H_2O}$ ligand), $\mathrm{Mn^{2+}}$ is high-spin → all 5 d-electrons unpaired. Plug into the spin-only formula: $\mu = \sqrt{n(n+2)} = \sqrt{5 \cdot 7} = \sqrt{35} \approx 5.92$ BM.

🗺️ Two-Step Calculation

Step 1 — d-count. Mn ($Z=25$): $[\mathrm{Ar}]\,3\mathrm{d^5}\,4\mathrm{s^2}$. Lose 2 electrons (4s²) → $\mathrm{Mn^{2+}} = [\mathrm{Ar}]\,3\mathrm{d^5}$.

Step 2 — unpaired electrons. d⁵ in aqueous solution (= $[\mathrm{Mn(H_2O)_6}]^{2+}$, weak/moderate field) → high-spin $\mathrm{t_{2g}^3\,e_g^2}$ → 5 unpaired.

Step 3 — magnetic moment. $\mu = \sqrt{n(n+2)} = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM.

⚡ The "$\mathrm{Mn^{2+}}$ Constant"

$\mathrm{Mn^{2+}}$ has the highest spin-only moment routinely encountered: 5.92 BM (= maximum 5 unpaired electrons in 3d). Any time the question gives "$Z = 25$, divalent" or just says "$\mathrm{Mn^{2+}}$ in aqueous solution", this number is the answer. Memorise it as a constant.

⚠️ The "Ground-State Atom" Slip

Don't mistakenly use Mn's ground-state atom configuration ($\mathrm{d^5\,s^2}$) and count 7 unpaired. The 4s electrons pair up trivially and contribute nothing; only 3d unpaired electrons count. Only 5 unpaired, not 7.

$\boxed{\text{Answer: (1) 5.92 BM}}$