Q.Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

The key idea is that for a divalent ion in aqueous solution, the magnetic moment depends only on the number of unpaired electrons, as orbital angular momentum is quenched.

Step 1: Atomic number 25 corresponds to manganese (Mn). The ground state electron configuration is $[\text{Ar}]\,3d^5 4s^2$.

Step 2: For a divalent ion (Mn$^{2+}$), remove the two 4s electrons first. The configuration becomes $[\text{Ar}]\,3d^5$.

Step 3: In the $3d^5$ configuration, all five electrons occupy separate orbitals with parallel spins (Hund's rule). This gives $n = 5$ unpaired electrons. …

The key here is to connect atomic number to electronic configuration, then to the number of unpaired electrons in the aqueous ion. Magnetic moment in transition metal ions is almost always determined by the spin-only formula because orbital angular momentum is "quenched" by the surrounding water ligands.

Let’s walk through it.

1. Identify the element and its neutral configuration.

   Atomic number 25 is manganese (Mn). The ground state configuration of neutral Mn is:

   $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5$

   Or in condensed form: $[\text{Ar}] 4s^2 3d^5$.

2. Form the divalent ion (Mn²⁺).

   When a transition metal forms a cation, electrons are removed first from the 4s orbital (higher energy than 3d in the ion, despite being filled first in the neutral atom). So Mn²⁺ loses the two 4s electrons:

   $[\text{Ar}] 3d^5$.

3. Count unpaired electrons using Hund’s rule.

   The 3d subshell has five orbitals. With five electrons, Hund’s rule says each orbital gets one electron with parallel spins before any pairing occurs. So all five electrons are unpaired.

   $n = 5$.

4. Apply the spin-only magnetic moment formula.

   For a transition metal ion in solution, the orbital contribution is usually negligible due to interaction with water molecules (ligand field quenching). The magnetic moment is:

   $\mu = \sqrt{n(n+2)}$ BM

   Substitute $n = 5$:

   $\mu = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35}$.

5. Compute the numerical value. …

Method: Spin-Only Magnetic Moment Formula (for $3d$ transition metal ions in aqueous solution)

For first-row transition metal ions in aqueous solution, orbital angular momentum is quenched (due to ligand field effects), so the magnetic moment depends only on the number of unpaired electrons.

Steps

Step 1: Identify the ion and its electronic configuration

• Atomic number $Z = 25$ → element is Manganese (Mn).
• Divalent ion means loss of 2 electrons: $\text{Mn}^{2+}$.
• Ground state configuration of Mn: $[\text{Ar}]\, 3d^5 4s^2$.
• Remove $4s$ electrons first (as per Aufbau for ions): $\text{Mn}^{2+} = [\text{Ar}]\, 3d^5$.

Step 2: Determine number of unpaired electrons

• For $3d^5$ in a weak field (aqueous solution = high-spin), Hund's rule applies: All five $d$ orbitals are singly occupied before pairing.
• Unpaired electrons $n = 5$.

Step 3: Apply the spin-only formula

The magnetic moment $\mu$ (in Bohr magnetons, $\mu_B$) is:

$\mu = \sqrt{n(n+2)} \ \mu_B$ …

Common Mistakes in Magnetic Moment Calculation (Atomic Number 25, Divalent Ion)

Mistake 1: Incorrect Electronic Configuration

The error: Students write the neutral-atom configuration (Z=25) as $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5$ correctly, but then remove the two electrons from the 3d orbitals (because 3d filled last), giving a wrong $[Ar]\,4s^2 3d^3$ for the divalent ion.

Why it's wrong: For transition metal ions, the 4s orbital empties before the 3d orbital — even though 4s fills first in the neutral atom. The correct order of removal is: 4s electrons go first, then 3d.

Correct approach:

• Neutral Mn (Z=25): $[Ar] 4s^2 3d^5$
• Mn$^{2+}$: Remove two electrons from 4s → $[Ar] 3d^5$

How to avoid: Remember the mnemonic: "Last filled, first removed" for transition metal ions. Always write the neutral configuration, then strip the outermost (highest n) s-electrons first.

Mistake 2: Wrong Number of Unpaired Electrons

The error: Students count 3 unpaired electrons (thinking $3d^5$ means 5 electrons paired as 2+2+1) or 7 unpaired electrons (confusing with another element).

Why it's wrong: For $3d^5$, Hund's rule states that electrons occupy all five d-orbitals singly before pairing. So all 5 electrons are unpaired.

Correct count: 5 unpaired electrons

How to avoid: Draw the d-orbital box diagram:

↑   ↑   ↑   ↑   ↑
dxy dyz dxz dx²-y² dz²

Each arrow is one unpaired electron. Count them — 5 unpaired.

Mistake 3: Using Wrong Formula

The error: Using $\mu = \sqrt{n(n+2)}$ with $n$ = total number of d-electrons instead of the number of unpaired electrons.

Why it's wrong: In the formula, n = number of unpaired electrons. For Mn$^{2+}$ ($3d^5$) the two counts happen to coincide (all 5 d-electrons are unpaired), so the error stays hidden — but for an ion like Fe$^{2+}$ ($3d^6$, only 4 unpaired) the wrong count $n = 6$ gives $\sqrt{6 \times 8} = \sqrt{48} \approx 6.93$ BM instead of the correct $\sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM.

Correct formula: $\mu = \sqrt{n(n+2)}$ BM, where $n = 5$

Calculation:

How to avoid: Always write the formula with the definition: "n = number of unpaired electrons" before plugging in.

Mistake 4: Forgetting the Unit

The error: Writing the answer as just "5.92" without units.

Why it's wrong: Magnetic moment has a specific unit — Bohr Magneton (BM). …