In transition metal complexes, the total magnetic moment ($\mu_{eff}$) arises from two components: the spin of the electrons and their orbital motion around the nucleus. While many simple systems can be described using the "spin-only" formula, the orbital contribution often plays a significant role in determining the actual magnetic behavior.
1. The Total Magnetic Moment
The theoretical magnetic moment, taking both spin ($S$) and orbital ($L$) angular momentum into account, is given by the following expression:
2. Quenching of Orbital Contribution
In many first-row transition metal complexes (3d series), the experimental magnetic moment is very close to the spin-only value:
(where $n$ is the number of unpaired electrons)
This occurs because the electric field of the ligands (the Crystal Field) restricts the orbital motion of the electrons. This phenomenon is known as quenching. For an orbital contribution to exist, there must be a way to transform one orbital into another of equivalent energy by rotation.
3. Conditions for Orbital Contribution
For a coordination compound to exhibit a significant orbital contribution, it must meet specific symmetry requirements:
- The ground state must be orbitally degenerate (e.g., $T_{1g}$ or $T_{2g}$ states).
- The orbitals involved in the degeneracy must be capable of being interconverted by rotation about an axis.
- In an octahedral field, this specifically means the $t_{2g}$ set ($d_{xy}, d_{yz}, d_{zx}$) can contribute, while the $e_g$ set ($d_{x^2-y^2}, d_{z^2}$) cannot.
4. High-Spin vs. Low-Spin Examples
| Configuration | Ground State | Orbital Contribution? |
|---|---|---|
| $d^1$ (Octahedral) | $t_{2g}^1$ | Yes (Triply degenerate) |
| $d^3$ (Octahedral) | $t_{2g}^3$ | No (Half-filled shell) |
| $d^7$ (High Spin Oct.) | $t_{2g}^5 e_g^2$ | Yes |
| $d^9$ (Octahedral) | $t_{2g}^6 e_g^3$ | No (Degeneracy in $e_g$ only) |
5. Spin-Orbit Coupling
In cases where the orbital contribution is partially quenched, the magnetic moment is further modified by spin-orbit coupling. The effective moment can be approximated by:
Where:
- $\lambda$ is the spin-orbit coupling constant.
- $\Delta$ is the crystal field splitting energy.
- $\alpha$ is a constant depending on the ground term.
Note: If the shell is less than half-full, $\lambda$ is positive (reducing the moment); if more than half-full, $\lambda$ is negative (increasing the moment).
Practice Questions for Entrance Exams
- (A) [Fe(H2O)6]2+ (High Spin)
- (B) [Mn(H2O)6]2+ (High Spin)
- (C) [Ni(H2O)6]2+
- (D) [Co(H2O)6]2+ (High Spin)
Reason: Co2+ in high spin is a d7 system ($t_{2g}^5 e_g^2$). The $t_{2g}$ level is unsymmetrically filled, leading to a T1g ground state which allows for a large orbital contribution.
- (A) 2.83 B.M.
- (B) Higher than 2.83 B.M.
- (C) Lower than 2.83 B.M.
- (D) Exactly 1.73 B.M.
Reason:A $d^8$ tetrahedral complex has a $e^4 t_2^4$ configuration. Because the $t_2$ orbitals are unsymmetrically filled, there is an orbital contribution that adds to the spin-only value ($\sqrt{2(2+2)} = 2.83$).
$\mu_{s.o.} = 2.83$ B.M., $\lambda = -315$ cm-1, and $\Delta_o = 8500$ cm-1.
Using the formula: $\mu_{eff} = \mu_{s.o.} (1 - \frac{4\lambda}{\Delta_o})$
$\mu_{eff} = 2.83 \times (1 - \frac{4 \times -315}{8500})$
$\mu_{eff} = 2.83 \times (1 + 0.148) \approx \mathbf{3.25}$ B.M.
Reason (R): $[Mn(CN)_6]^{3-}$ is a low-spin complex with fewer unpaired electrons, while $[MnCl_6]^{3-}$ is a high-spin complex.
- (A) Both A and R are true, and R is the correct explanation of A.
- (B) Both A and R are true, but R is not the correct explanation of A.
- (C) A is true, but R is false.
- (D) A is false, but R is true.
Reason:(A).Context: $Mn^{3+}$ is $d^4$. $CN^-$ (strong field) creates a low-spin state ($t_{2g}^4$, $n=2$), while $Cl^-$ (weak field) creates a high-spin state ($t_{2g}^3 e_g^1$, $n=4$). Note that the low-spin $t_{2g}^4$ state also has an orbital contribution, but the reduction in unpaired electrons dominates the total moment.