Q.Which of the following ions show higher spin only magnetic moment value?

Concept: Magnetic Moment Calculation – The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. Higher $n$ gives a higher $\mu$.

Step 1: Determine the electronic configuration and number of unpaired electrons for each ion (considering weak field / high-spin case for $d^4$–$d^7$).

• $Ti^{3+}$: $[Ar]\,3d^1$ → $n = 1$
• $Mn^{2+}$: $[Ar]\,3d^5$ → $n = 5$
• $Fe^{2+}$: $[Ar]\,3d^6$ → $n = 4$ (high-spin)
• $Co^{3+}$: $[Ar]\,3d^6$ → $n = 4$ (high-spin, same as $Fe^{2+}$) …

The spin-only magnetic moment formula is a direct consequence of the electron's spin angular momentum. For transition metal ions, the orbital contribution is often quenched in complexes, so the observed magnetic moment is primarily due to unpaired electrons. The formula $\mu = \sqrt{n(n+2)}\,\mu_B$ (where $\mu_B$ is the Bohr magneton) tells us that the moment grows faster than linearly with $n$ — each additional unpaired electron adds more than just a constant increment.

To compare the ions, we need to determine the number of unpaired electrons in each. This requires writing their electronic configurations, remembering that for transition metal ions, electrons are removed first from the 4s orbital (which is higher in energy than 3d for neutral atoms, but after ionization, the 3d becomes lower).

1. $Ti^{3+}$: Titanium atomic number 22. Neutral Ti: $[Ar]\,3d^2\,4s^2$. Removing three electrons: first two from 4s, then one from 3d. So $Ti^{3+}$: $[Ar]\,3d^1$. One unpaired electron. $\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73\,\mu_B$.

2. $Mn^{2+}$: Manganese atomic number 25. Neutral Mn: $[Ar]\,3d^5\,4s^2$. Removing two electrons: both from 4s. So $Mn^{2+}$: $[Ar]\,3d^5$. By Hund's rule, all five d-orbitals are singly occupied before pairing occurs. Five unpaired electrons. $\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92\,\mu_B$.

3. $Fe^{2+}$: Iron atomic number 26. Neutral Fe: $[Ar]\,3d^6\,4s^2$. Removing two electrons: both from 4s. So $Fe^{2+}$: $[Ar]\,3d^6$. For a high-spin configuration (which is typical for aqueous or common complexes), the six electrons fill as: five orbitals each with one electron, and the sixth pairs in one orbital. That gives four unpaired electrons. $\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90\,\mu_B$.

Method: Spin-Only Magnetic Moment Calculation (Using Number of Unpaired Electrons)

This method uses the formula:

where:

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

Steps

Step 1 — Write the electronic configuration of each ion (ground state)

• Ti³⁺: Atomic number 22 → Ti: $[Ar]\,3d^2\,4s^2$

  Remove 3 electrons → Ti³⁺: $[Ar]\,3d^1$

  → n = 1

• Mn²⁺: Atomic number 25 → Mn: $[Ar]\,3d^5\,4s^2$

  Remove 2 electrons → Mn²⁺: $[Ar]\,3d^5$

  → n = 5 (half-filled, all parallel spins)

• Fe²⁺: Atomic number 26 → Fe: $[Ar]\,3d^6\,4s^2$

  Remove 2 electrons → Fe²⁺: $[Ar]\,3d^6$

  → n = 4 (Hund’s rule: four unpaired, one paired)

• Co³⁺: Atomic number 27 → Co: $[Ar]\,3d^7\,4s^2$

  Remove 3 electrons → Co³⁺: $[Ar]\,3d^6$

  → n = 4 (same d⁶ configuration as Fe²⁺)

Step 2 — Calculate $\mu$ for each

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

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

Mistake 1: Forgetting to find the correct electronic configuration first

The error: Students jump straight to counting unpaired electrons without writing the configuration of the ion (not the neutral atom). For example, they might use the configuration of neutral Ti instead of $Ti^{3+}$.

How to avoid: Always write the configuration of the neutral atom first, then remove electrons from the outermost shell (typically 4s before 3d for transition metals).

• Correct approach for $Ti^{3+}$:

  • Neutral Ti: $[Ar] 4s^2 3d^2$
  • Remove 3 electrons: $[Ar] 3d^1$
  • Unpaired electrons = 1

• Correct approach for $Mn^{2+}$:

  • Neutral Mn: $[Ar] 4s^2 3d^5$
  • Remove 2 electrons: $[Ar] 3d^5$
  • Unpaired electrons = 5

Mistake 2: Using the wrong formula or forgetting the units

The error: Students use $\mu = \sqrt{n(n+2)}$ but forget that $n$ is the number of unpaired electrons, not the total electrons. Some also write the answer in Bohr Magneton (BM) but forget to state the unit.

How to avoid: Memorise the formula clearly:

Where $n$ = number of unpaired electrons. Always write the unit BM (Bohr Magneton) after the value.

Mistake 3: Assuming higher atomic number means higher magnetic moment

The error: Students think $Co^{3+}$ (atomic number 27) will have a higher moment than $Ti^{3+}$ (atomic number 22) just because it is heavier.

How to avoid: Magnetic moment depends only on the number of unpaired electrons, not on atomic mass or atomic number. Compare $n$ values:

Key insight: $Mn^{2+}$ has the maximum unpaired electrons (5) in the 3d series because it is exactly half-filled ($d^5$).

Mistake 4: Ignoring the possibility of low-spin vs high-spin

The error: For $Co^{3+}$ ($d^6$), students automatically assume 4 unpaired electrons (high spin) without checking if the question specifies a ligand or complex type. …