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?

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Accepted Answer

> [!TLDR]
> The number of unpaired electrons is found from the electronic configuration of each ion. $Mn^{3+}$ has 4 unpaired electrons, $Cr^{3+}$ has 3, $V^{3+}$ has 2, and $Ti^{3+}$ has 1. The most stable in aqueous solution is $Cr^{3+}$ because its half-filled $t_{2g}^3$ configuration gives extra stabilization energy.

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.

> [!TIP]
> A quick check: For a $d^n$ configuration in a free ion, the number of unpaired electrons is simply $n$ for $n \le 5$, because pairing only st

Material Type	$\chi_m$	Behaviour	Example
Diamagnetic	Small and negative ( $\approx -10^{-5}$ )	Weakly repelled by a magnet; $\vec{M}$ opposes $\vec{H}$	Water, copper, bismuth
Paramagnetic	Small and positive ( $\approx 10^{-5}$ to $10^{-3}$ )	Weakly attracted; $\vec{M}$ aligns with $\vec{H}$	Aluminium, oxygen gas
Ferromagnetic	Large and positive ( $\gg 1$ )	Strongly attracted; $\vec{M}$ can be huge and persists even after $\vec{H}$ is removed	Iron, nickel, cobalt

Ion	Neutral atom config	Ion config (after removing electrons)
$Mn^{3+}$	$[Ar]\,3d^5\,4s^2$	Remove 3 electrons → $[Ar]\,3d^4$
$Cr^{3+}$	$[Ar]\,3d^5\,4s^1$	Remove 3 electrons → $[Ar]\,3d^3$
$V^{3+}$	$[Ar]\,3d^3\,4s^2$	Remove 3 electrons → $[Ar]\,3d^2$
$Ti^{3+}$	$[Ar]\,3d^2\,4s^2$	Remove 3 electrons → $[Ar]\,3d^1$

Ion	$n$	$\mu$ (BM)
$Mn^{3+}$	4	$\sqrt{4 \times 6} = \sqrt{24} \approx 4.90$
$Cr^{3+}$	3	$\sqrt{3 \times 5} = \sqrt{15} \approx 3.87$
$V^{3+}$	2	$\sqrt{2 \times 4} = \sqrt{8} \approx 2.83$
$Ti^{3+}$	1	$\sqrt{1 \times 3} = \sqrt{3} \approx 1.73$

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 understanding — Magnetic Moment Calculation

From a Paperclip to a Magnet: The Intuition

The Precise Definition

How Magnetization Connects to the Magnetic Field

The Three Kinds of Magnetic Materials

Why this formula?

Magnetic Moment Calculation: Why the Formula Holds

1. What is Magnetic Moment?

2. The Fundamental Formula: Current Loop

The Setup

Why $\mu = I A$ ?

3. For a Single Moving Charge (Orbital Magnetic Moment)

The Setup

Why $\mu = \frac{e v r}{2}$ ?

4. For a Solenoid (Many Turns)

The Setup

Why $\mu = N I A$ ? …

Method: Electronic Configuration + Magnetic Moment Formula

Step 1 — Write the electronic configurations

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

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

Step 4 — Determine stability in aqueous solution …

🔍 Common Mistake #1: Wrong electronic configuration for ions

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

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

🔍 Common Mistake #4: Confusing stability with magnetic moment

Understanding the Configuration

Applying Hund's Rule

Q.Calculate the number of unpaired electrons in the following gaseous ions: Mn3+Mn^{3+}Mn3+, Cr3+Cr^{3+}Cr3+, V3+V^{3+}V3+ and Ti3+Ti^{3+}Ti3+. Which one of these is the most stable in aqueous solution?

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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?

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