Q.Calculate the number of unpaired electrons in the following gaseous ions: $Mn^{3+}$, $Cr^{3+}$, $V^{3+}$ and $Ti^{3+}$. Which one of these is the most stable in aqueous solution?

Concept: Magnetic Moment Calculation – The number of unpaired electrons is found from the electronic configuration of the ion, and stability in aqueous solution relates to the standard reduction potential (or the tendency to resist further oxidation/reduction).

Step 1: Write the ground-state configurations of the neutral atoms and then the ions.

• Mn ($Z=25$): $[Ar]\,3d^5 4s^2$ → $Mn^{3+}$ loses $4s^2$ and one $3d$ electron → $[Ar]\,3d^4$ → 4 unpaired electrons.
• Cr ($Z=24$): $[Ar]\,3d^5 4s^1$ → $Cr^{3+}$ loses $4s^1$ and two $3d$ electrons → $[Ar]\,3d^3$ → 3 unpaired electrons.
• V ($Z=23$): $[Ar]\,3d^3 4s^2$ → $V^{3+}$ loses $4s^2$ and one $3d$ electron → $[Ar]\,3d^2$ → 2 unpaired electrons.
• Ti ($Z=22$): $[Ar]\,3d^2 4s^2$ → $Ti^{3+}$ loses $4s^2$ and one $3d$ electron → $[Ar]\,3d^1$ → 1 unpaired electron.

Step 2: Relate stability in aqueous solution. …

The key to solving this lies in understanding how electrons fill the d-orbitals in transition metal ions. For gaseous ions, we follow Hund’s rule — electrons occupy each orbital singly before pairing. The number of unpaired electrons directly tells us about the magnetic moment and, indirectly, about stability in solution.

Let’s work through each ion step by step.

1. Write the ground-state electronic configuration of the neutral atom.

   • Mn (Z = 25): $[Ar] 3d^5 4s^2$
   • Cr (Z = 24): $[Ar] 3d^5 4s^1$ (special case for half-filled stability)
   • V (Z = 23): $[Ar] 3d^3 4s^2$
   • Ti (Z = 22): $[Ar] 3d^2 4s^2$

2. Remove electrons to form the +3 ion.

   For transition metals, electrons are removed first from the 4s orbital (higher energy than 3d in ions), then from 3d.

   • $Mn^{3+}$: Remove 2 from 4s and 1 from 3d → $[Ar] 3d^4$
   • $Cr^{3+}$: Remove 1 from 4s and 2 from 3d → $[Ar] 3d^3$
   • $V^{3+}$: Remove 2 from 4s and 1 from 3d → $[Ar] 3d^2$
   • $Ti^{3+}$: Remove 2 from 4s and 1 from 3d → $[Ar] 3d^1$

3. Apply Hund’s rule to find unpaired electrons.

   For a gaseous ion (no ligand field), all five d-orbitals are degenerate. Electrons fill singly with parallel spins.

   • $3d^4$: Four electrons, each in a separate orbital → 4 unpaired electrons.
   • $3d^3$: Three electrons, each in a separate orbital → 3 unpaired electrons.
   • $3d^2$: Two electrons, each in a separate orbital → 2 unpaired electrons.
   • $3d^1$: One electron → 1 unpaired electron.

4. Now, which is most stable in aqueous solution? In water, the ions are surrounded by ligands (water molecules) that create a crystal field. For octahedral complexes (common for these +3 ions), the d-orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy). …

Method: Electronic Configuration + Magnetic Moment Formula

This problem uses the spin-only magnetic moment formula to find unpaired electrons, then judges stability in aqueous solution from the ion's electronic configuration and redox behaviour — not from the raw unpaired-electron count.

Step 1 — Write the electronic configurations

First, write the ground-state configurations of the neutral atoms, then remove electrons (from the 4s orbital first, then 3d).

Step 2 — Apply Hund’s rule to find unpaired electrons

Fill the 3d orbitals singly before pairing:

• $Mn^{3+}$ ($3d^4$): ↑ ↑ ↑ ↑ → 4 unpaired electrons
• $Cr^{3+}$ ($3d^3$): ↑ ↑ ↑ → 3 unpaired electrons
• $V^{3+}$ ($3d^2$): ↑ ↑ → 2 unpaired electrons
• $Ti^{3+}$ ($3d^1$): ↑ → 1 unpaired electron

Step 3 — Use the spin-only formula (optional verification)

The magnetic moment $\mu$ is given by:

where $n$ = number of unpaired electrons.

Step 4 — Determine stability in aqueous solution …

🔍 Common Mistake #1: Wrong electronic configuration for ions

The error: Students often write the configuration of the neutral atom and then remove electrons from the outermost shell without considering the energy order of orbitals.

Example: For $Mn^{3+}$, a student might write:

• Mn: $[Ar] 4s^2 3d^5$
• Remove 3 electrons from 4s → $[Ar] 4s^1 3d^3$ ✗

Why it’s wrong: In transition metal ions, electrons are removed from the 4s orbital first, even though 4s fills before 3d in the neutral atom. The correct removal order is: 4s before 3d.

How to avoid: Always write the neutral atom’s configuration, then remove from the outermost shell (highest n) — that’s 4s before 3d. For $Mn^{3+}$:

• Mn: $[Ar] 4s^2 3d^5$
• Remove 2 from 4s, then 1 from 3d → $[Ar] 3d^4$ ✓

🔍 Common Mistake #2: Forgetting Hund’s rule when filling d-orbitals

The error: After getting the correct $d^n$ configuration, students pair electrons prematurely.

Example: For $Cr^{3+}$ ($3d^3$), writing:

• $\uparrow\downarrow$ in one orbital, then one unpaired ✗

Why it’s wrong: Hund’s rule says electrons occupy all degenerate orbitals singly before pairing.

How to avoid: For $d^n$ (n ≤ 5), fill each of the 5 d-orbitals with one electron first before pairing. So:

• $d^1$: 1 unpaired
• $d^2$: 2 unpaired
• $d^3$: 3 unpaired
• $d^4$: 4 unpaired (high spin) — but note: $Cr^{3+}$ is $d^3$, so 3 unpaired ✓

🔍 Common Mistake #3: Using the wrong formula for magnetic moment

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

Example: For $Mn^{3+}$ ($d^4$) the total and unpaired counts coincide ($n = 4$, $\mu = \sqrt{24} \approx 4.9$ BM), so the error stays hidden — but for an ion like $Fe^{2+}$ ($d^6$, only 4 unpaired), plugging in the total ($n = 6$) gives $\sqrt{48} \approx 6.93$ BM instead of the correct $\sqrt{24} \approx 4.90$ BM.

How to avoid: First count unpaired electrons correctly (using Hund’s rule), then apply:

where $n$ = number of unpaired electrons only.

🔍 Common Mistake #4: Confusing stability with magnetic moment

The error: Assuming the ion with the highest magnetic moment is the most stable. …