Q.Although $Cr^{3+}$ and $Co^{2+}$ ions have same number of unpaired electrons but the magnetic moment of $Cr^{3+}$ is 3.87 B.M. and that of $Co^{2+}$ is 4.87 B.M. Why?

The key idea is that magnetic moment depends not only on the number of unpaired electrons but also on the orbital contribution, which varies with the electronic configuration and the ligand field.

Reasoning:

1. Number of unpaired electrons: Both $Cr^{3+}$ ($d^3$) and $Co^{2+}$ ($d^7$) have three unpaired electrons in their ground states.

   • $Cr^{3+}$: $t_{2g}^3$ (all unpaired).
   • $Co^{2+}$ (high-spin): $t_{2g}^5 e_g^2$ — three unpaired electrons.

2. Spin-only formula: The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ B.M. For $n = 3$, this gives $\sqrt{3 \times 5} = \sqrt{15} \approx 3.87$ B.M.

3. Orbital contribution:

   • $Cr^{3+}$ has a $^4F$ ground term with $L = 3$, but in an octahedral field the orbital angular momentum is largely quenched (the $t_{2g}^3$ configuration has no orbital degeneracy in a cubic field). Hence $\mu$ matches the spin-only value (3.87 B.M.). …

The key here is that magnetic moment is not simply a function of the number of unpaired electrons — it also depends on whether the orbital angular momentum contributes to the total moment. For transition metal ions in complexes, the crystal field can "quench" the orbital contribution, but not always.

Let’s break this down.

1. The spin-only formula For an ion with $n$ unpaired electrons, the spin-only magnetic moment is:

For $n = 3$, this gives $\sqrt{3 \times 5} = \sqrt{15} \approx 3.87$ B.M.

So $Cr^{3+}$ matches the spin-only value exactly. That tells us immediately that its orbital angular momentum is completely quenched.

2. Why is $Cr^{3+}$ spin-only?

   $Cr^{3+}$ has the electronic configuration $[Ar]\,3d^3$. In an octahedral crystal field, the three electrons occupy the $t_{2g}$ orbitals ($d_{xy}, d_{yz}, d_{zx}$) with parallel spins (Hund’s rule).

   The $t_{2g}$ set is triply degenerate, but in a perfect octahedral field, the orbital angular momentum is quenched because the orbitals are real (not complex) and the ground term is $^4A_{2g}$ — an orbital singlet. No orbital degeneracy means no orbital contribution.

3. Now $Co^{2+}$: same $n$, different story

   $Co^{2+}$ has the configuration $[Ar]\,3d^7$. In an octahedral field, this is $t_{2g}^5 e_g^2$. The ground term is $^4T_{1g}(F)$ — an orbital triplet.

   The $T_{1g}$ term has orbital angular momentum that is not fully quenched. The magnetic moment is therefore higher than the spin-only value. The experimental value of 4.87 B.M. is close to what is expected when spin-orbit coupling mixes in some orbital contribution.

4. Quantifying the difference For $Co^{2+}$ in an octahedral field, the effective magnetic moment is given by:

where $\lambda$ is the spin-orbit coupling constant (negative for $d^7$), $\Delta$ is the crystal field splitting, and $\alpha$ is a factor depending on the term. The negative $\lambda$ and the mixing cause $\mu_{\text{eff}}$ to be larger than $\mu_{s.o.}$. …

Method: Spin-Only Magnetic Moment Calculation

This problem is solved using the spin-only formula for magnetic moment, combined with the electronic configuration of each ion.

Step 1 – Recall the spin-only formula

The magnetic moment $\mu$ (in Bohr Magnetons, B.M.) for a transition metal ion is given by:

where $n$ = number of unpaired electrons.

This formula assumes that orbital angular momentum is quenched (i.e., not contributing), which is often true for first-row transition metals in complexes.

Step 2 – Find the electronic configuration of each ion

• Cr³⁺:

  Atomic number of Cr = 24

  Ground state: $[Ar]\,3d^5\,4s^1$

  After losing 3 electrons (2 from 4s and 1 from 3d):

  $[Ar]\,3d^3$

  Number of unpaired electrons = 3

• Co²⁺:

  Atomic number of Co = 27

  Ground state: $[Ar]\,3d^7\,4s^2$

  After losing 2 electrons (from 4s):

  $[Ar]\,3d^7$

  In a weak field (high spin) environment, the 3d⁷ configuration gives:

  Number of unpaired electrons = 3

Step 3 – Apply the formula for each

• For Cr³⁺ ($n = 3$):

• For Co²⁺ ($n = 3$):

Step 4 – Identify the reason for the discrepancy

The spin-only formula gives 3.87 B.M. for both, but the observed value for Co²⁺ is 4.87 B.M.

Why? …

✗ Mistake 1: Assuming same number of unpaired electrons means same magnetic moment

Why it's wrong:

Magnetic moment ($\mu$) depends not just on the number of unpaired electrons ($n$), but also on the orbital contribution (via the spin–orbit coupling). The formula $\mu = \sqrt{n(n+2)}$ assumes only spin contribution — this works for most first-row transition metals, but not always.

How to avoid:

Always check if the ion is in a weak field or strong field environment, and whether orbital angular momentum is quenched or not. For $Co^{2+}$ in an octahedral field, there is an orbital contribution that increases $\mu$ beyond the spin-only value.

✗ Mistake 2: Forgetting the electronic configuration details

Why it's wrong:

Students often write $Cr^{3+}$ as $[Ar] 3d^3$ and $Co^{2+}$ as $[Ar] 3d^7$ — correct so far. But then they assume both have 3 unpaired electrons and stop there. They miss that the arrangement of electrons in $t_{2g}$ and $e_g$ orbitals affects orbital angular momentum.

How to avoid:

Draw the d-orbital splitting diagram for each ion in an octahedral field:

• $Cr^{3+}$ ($d^3$): All three electrons go into $t_{2g}$ (one each) — all unpaired, but no orbital degeneracy → no orbital contribution.
• $Co^{2+}$ ($d^7$): High-spin configuration: $t_{2g}^5 e_g^2$ — this has orbital degeneracy in the $t_{2g}$ set → orbital angular momentum is not fully quenched.

✗ Mistake 3: Using the spin-only formula blindly

Why it's wrong:

For $Cr^{3+}$: $n = 3$, so $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ B.M. — matches.

For $Co^{2+}$: $n = 3$, same formula gives $3.87$ B.M., but the actual value is 4.87 B.M. — the formula fails here.

How to avoid:

Remember: The spin-only formula works well when orbital angular momentum is quenched (common for $d^3$, $d^5$, $d^8$ high-spin, $d^6$ low-spin). For $d^7$ high-spin (like $Co^{2+}$), orbital contribution is significant — use the formula $\mu = \sqrt{4S(S+1) + L(L+1)}$ or simply memorise the experimental trend.

✗ Mistake 4: Confusing high-spin vs low-spin

Why it's wrong: …