Q.Amongst the following ions which one has the highest magnetic moment value?

Concept: Magnetic Moment Calculation

The magnetic moment $\mu$ depends on the number of unpaired electrons $n$:

Step 1 – Find $n$ for each ion

• $[Cr(H_2O)_6]^{3+}$: Cr in +3 state → $[Ar]3d^3$ → $n = 3$
• $[Fe(H_2O)_6]^{2+}$: Fe in +2 state → $[Ar]3d^6$ (high-spin, weak field ligand H₂O) → $n = 4$
• $[Zn(H_2O)_6]^{2+}$: Zn in +2 state → $[Ar]3d^{10}$ → $n = 0$

Step 2 – Compute $\mu$ …

The magnetic moment of a transition metal complex is a direct window into its electronic structure. For first-row transition metals, the spin-only formula $\mu = \sqrt{n(n+2)}$ BM (where $n$ is the number of unpaired electrons) works beautifully because orbital contributions are usually quenched by the ligand field. So the question reduces to: which ion has the most unpaired electrons?

Let’s examine each complex one by one.

1. $[Cr(H_2O)_6]^{3+}$

   Chromium in its +3 oxidation state: atomic number 24, so Cr$^{3+}$ has $24 - 3 = 21$ electrons. The configuration is $[Ar]\,3d^3$. Water is a weak field ligand, so no pairing occurs — the three $d$ electrons occupy three different $t_{2g}$ orbitals (Hund’s rule). Unpaired electrons: $n = 3$.

   Magnetic moment: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM.

2. $[Fe(H_2O)_6]^{2+}$

   Iron in +2 state: atomic number 26, so Fe$^{2+}$ has $26 - 2 = 24$ electrons. Configuration: $[Ar]\,3d^6$. Water is again weak field, so this is a high-spin complex. The six electrons fill the $t_{2g}$ set (three orbitals, each with one electron first, then one pairs) and then two go into $e_g$ orbitals. The $t_{2g}$ set has 4 electrons (one orbital doubly occupied, two singly), and $e_g$ has 2 electrons (one each). Total unpaired: $4$ (two in $t_{2g}$ and two in $e_g$).

   Magnetic moment: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM.

3. $[Zn(H_2O)_6]^{2+}$ …

Method: Spin-Only Magnetic Moment Formula

We use the spin-only formula for first-row transition metal complexes (where orbital contribution is often quenched):

Where:

• $\mu$ = magnetic moment in Bohr magnetons (B.M.)
• $n$ = number of unpaired electrons

Steps

Step 1: Find the oxidation state and electron configuration of the metal ion.

• (i) $[Cr(H_2O)_6]^{3+}$

  Cr atomic number = 24

  Cr$^{3+}$: remove 3 electrons → $[Ar]\,3d^3$

  Unpaired electrons: 3

• (ii) $[Fe(H_2O)_6]^{2+}$

  Fe atomic number = 26

  Fe$^{2+}$: remove 2 electrons → $[Ar]\,3d^6$

  $H_2O$ is a weak field ligand → high spin

  Unpaired electrons: 4

• (iii) $[Zn(H_2O)_6]^{2+}$

  Zn atomic number = 30

  Zn$^{2+}$: remove 2 electrons → $[Ar]\,3d^{10}$

  Unpaired electrons: 0

--- …

Here are the most common mistakes students make when solving this magnetic moment problem, along with how to avoid each.

Mistake 1: Forgetting to find the oxidation state of the metal

The Error:

Students often directly write the electronic configuration of the neutral atom (e.g., Cr: $3d^5 4s^1$) and count unpaired electrons without adjusting for the charge on the complex.

Why it’s wrong:

The magnetic moment depends on the number of unpaired electrons in the metal ion, not the neutral atom. The ligands ($H_2O$) are neutral, so the charge on the complex equals the charge on the metal ion.

How to avoid:

Always determine the oxidation state of the metal first.

• For $[Cr(H_2O)_6]^{3+}$: Cr is in +3 state. Cr (24): $[Ar] 3d^5 4s^1$ → Cr$^{3+}$: $[Ar] 3d^3$ → 3 unpaired electrons.
• For $[Fe(H_2O)_6]^{2+}$: Fe is in +2 state. Fe (26): $[Ar] 3d^6 4s^2$ → Fe$^{2+}$: $[Ar] 3d^6$ → 4 unpaired electrons (high spin, as $H_2O$ is a weak field ligand).
• For $[Zn(H_2O)_6]^{2+}$: Zn is in +2 state. Zn (30): $[Ar] 3d^{10} 4s^2$ → Zn$^{2+}$: $[Ar] 3d^{10}$ → 0 unpaired electrons.

Mistake 2: Ignoring the ligand field strength (high spin vs. low spin)

The Error:

Students assume all $d^4$, $d^5$, $d^6$, $d^7$ configurations have the same number of unpaired electrons, forgetting that strong-field ligands can cause pairing.

Why it’s wrong:

$H_2O$ is a weak field ligand (for first-row transition metals). It does not cause pairing — the complex remains high spin. But if the ligand were strong (e.g., CN$^-$, CO), the electron configuration would change.

How to avoid:

Memorize the spectrochemical series and know that $H_2O$ is weak field. For $[Fe(H_2O)_6]^{2+}$ ($d^6$):

• Weak field (high spin): $t_{2g}^4 e_g^2$ → 4 unpaired electrons.
• Strong field (low spin): $t_{2g}^6 e_g^0$ → 0 unpaired electrons.

Since $H_2O$ is weak, the answer is 4 unpaired electrons.

Mistake 3: Using the wrong formula for magnetic moment

The Error:

Students sometimes use $\mu = \sqrt{n(n+2)}$ incorrectly — either forgetting the square root or using $n$ as total electrons instead of unpaired electrons.

Why it’s wrong:

The formula is $\mu = \sqrt{n(n+2)}$ Bohr magnetons, where $n$ = number of unpaired electrons. Using total electrons gives a meaningless number.

How to avoid:

Always compute $n$ first, then plug into $\mu = \sqrt{n(n+2)}$.

• Cr$^{3+}$: $n = 3$ → $\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87$ BM
• Fe$^{2+}$: $n = 4$ → $\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90$ BM …