My textbook (Chapter: The d- and f- Block Elements) makes an interesting assertion, however, without any reason to back it up.

Paramagnetism arises from the presence of unpaired electons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e.,

$$\mu = \sqrt{n (n + 2)}$$

where $n$ is the number of unpaired electrons and $\mu$ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of $1.73$ Bohr magnetons (BM).

Now from what I've learnt from my physics classes last week, the magnetic moment of an electron is calculated using the formula:

$$\frac{m}{L} = \frac{e}{2 m_e}$$

Magnetic moment of revolving electron,

$$m = \frac{e}{2 m_e}L$$

In vector form,

$$\vec{m} = - \frac{e}{2m_e}\vec{L}$$

Where $m$ is the mass of the electron, $e$ is the magnitude of charge associated with the electron and $L$ is the angular momentum of the electron.

Now as I understand it, the angular momentum of an electron in an atom is the resultant of the electron's spin angular momentum and its orbital angular momentum.

But (as you can clearly see) my textbook exhibits an unexplained enthusiasm for only the spin angular momentum. It dismisses the contribution of the orbital angular momentum, as inconsequential, when determining the magnetic moments of electrons in the 3d elements.

As for the 'reason' provided by my textbook:

For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and is hence of no significance.

Well, I fail to see how that even counts as a 'reason'. I did ask my teacher about this, but she expressed her reluctance to go into the details; she's of the opinion that my particular query is “not worth clarifying” since it has no significance from the examination point of view (“Just give them what's in the textbook Aaron, nothing more and certainly not anything less”).

Subsequent internet searches have not yield any satisfying explanation (Perhaps I didn't use the right keywords while searching?).

So I'll just break up my question, point-wise, and list them below:

  1. Is the orbital angular really of no significance when calculating the magnetic dipole moment for 3d elements? If it is significant, is there any 'educational' purpose behind that omission in my textbook, or is it just a mistake?

  2. (As stated in the book) What causes the 'quenching'? How is this 'quenching' effect caused?

  3. Do the transition elements of the other periods also have "insignificant" orbital angular momentum contributions? Is it for the same reason as the 3d transition elements.

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    $\begingroup$ It is generally quenched for symmetry reasons ( lower symmetry, less orbital contribution) $\endgroup$
    – Greg
    Jan 27, 2017 at 16:17

1 Answer 1


Simplistically speaking orbital angular momentum is present when some conditions are satisfied:

  • A set of orbitals are degenerate;
  • These orbitals can be "interconverted" by rotation about a certain axis;
  • The set of orbitals is not empty, half-filled, or fully filled.

Physically speaking, the "rotation" of an electron from one orbital to the next generates orbital angular momentum. The rotation must not have an energy cost, which necessitates degeneracy. It also means that the set of orbitals must have at least one electron, as well as one empty space (a so-called hole) for the electron to "rotate" into. The hole must have the same spin as the electron, which means that a half-filled subshell doesn't allow for orbital angular momentum (since all the electrons have parallel spins).

This is most easily illustrated with the three 2p orbitals in an atom. Rotation of a $\mathrm{p}_x$ orbital by 90 degrees around the $y$-axis, for example, turns it into a $\mathrm{p}_z$ orbital.

Any atom with a $\mathrm{p}^1$, $\mathrm{p}^2$, $\mathrm{p}^4$, or $\mathrm{p}^5$ configuration therefore possesses nonzero angular momentum. This is also reflected in their term symbols where $L = 1$. This is said to be "one unit of angular momentum". An atom with a half-filled shell (for example, $\ce{N}$) or a closed-shell $\mathrm{p}^6$ configuration (for example, $\ce{Ne}$) does not have any orbital angular momentum.

The five 3d orbitals in an atom also satisfy these requirements. It is less intuitive to see this as compared to the 2p orbitals, because we have taken specific linear combinations of the complex 3d orbitals and converted them into real orbitals. Nevertheless one can use quantum mechanics to prove that the 3d orbitals possess angular momentum. The maths can be found in Figgis and Hitchman's Ligand Field Theory and Its Applications (it is very early in the book, but I do not have it with me now and cannot remember offhand where it is). As such, for free 3d metal atoms or ions, the orbital angular momentum has to be incorporated into the calculation of the magnetic moment. Exceptions are, of course, $\mathrm{d^0}$, $\mathrm{d^5}$, and $\mathrm{d^{10}}$ ions (e.g. $\ce{Sc^3+}$, $\ce{Mn^2+}$, and $\ce{Zn^2+}$).

That is applicable for free ions. However, 3d metals do not commonly exist as free ions, but exist in coordination complexes with a ligand field. For simplicity we'll consider a perfect octahedral ligand field, which causes d orbitals to split into a $\mathrm{t_{2g}}$ ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$) and $\mathrm{e_{g}}$ ($\mathrm{d}_{z^2}$, $\mathrm{d}_{x^2-y^2}$) set. This means that the five d orbitals are no longer all degenerate, breaking the first condition stipulated above.

At a simplistic level, many textbooks will claim that this leads to a "quenching" of orbital angular momentum, which just means that there is no longer any orbital angular momentum, and as such, the spin-only formula applies. [Note that there are some second-order effects (quantum mechanical mixing of excited states into the ground state) that can lead to deviations from the spin-only formula.]

This is true to a certain extent. However, the five 3d orbitals are not split into individually nondegenerate orbitals; the three $\mathrm{t_{2g}}$ orbitals, for example, are still degenerate. On top of that, one can verify pictorially that it is possible to rotate the $\mathrm{t_{2g}}$ orbitals into one another. In this aspect the $\mathrm{t_{2g}}$ orbitals behave quite similarly to the three $\mathrm{p}$ orbitals in a free atom; this is called the "$\mathrm{t_2}$-$\mathrm{p}$ isomorphism". As such, the $\mathrm{t_{2g}}$ orbitals still have orbital angular momentum associated with them.

Any complex with unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals (i.e. $\mathrm{t_{2g}}^1$, $\mathrm{t_{2g}}^2$, $\mathrm{t_{2g}}^4$, $\mathrm{t_{2g}}^5$) therefore possesses orbital angular momentum and exhibit great deviations from the spin-only formula. These complexes all have a $\mathrm{T_{1g}}$ or $\mathrm{T_{2g}}$ ground state term symbol, so it is commonly said that "$\mathrm{T}$ terms possess one unit of orbital angular momentum". Spin-orbit coupling has to be invoked to obtain a better agreement between the calculated and measured magnetic moments.

As an example of how large the deviations can be, consider the compound $\ce{(NH4)2Co(SO4)2.2H2O}$. This features octahedral Co(II) which adopts a high-spin configuration $(\mathrm{t_{2g}})^5(\mathrm{e_g})^2$. From the spin-only formula, we would expect $\mu_\mathrm{so} = \sqrt{3(3+2} = 3.87$. However, the unsymmetrical population of the $\mathrm{t_{2g}}$ orbitals leads to the generation of orbital angular momentum, and at $300~\mathrm{K}$ the measured effective magnetic moment is actually $5.1$. (Data from here.)

A more detailed discussion can be found in any of the standard inorganic chemistry textbooks. Figgis and Hitchman's book (mentioned earlier) is more advanced and has a very comprehensive treatment of the subject.

  • 1
    $\begingroup$ Thank you for this amazing answer! I just wonder why doesn't JT distortion break the symmetry of e.g. $t_{2g}^1$ set, whenever it occurs? $\endgroup$ Feb 12, 2017 at 0:54
  • 3
    $\begingroup$ It does occur, and $\mathrm{d}^1$ complexes are susceptible to JT distortions. However, usually, it is the unsymmetrical occupancy of $\mathrm{e_g}$ (for example, high-spin d4 Cr(II) and d9 Cu(II)) that lead to large JT distortions. The $\mathrm{e_g}$ orbitals point directly towards the ligands, such that there is a greater change in the energy (i.e. greater stabilisation) upon geometrical distortion $\endgroup$ Feb 12, 2017 at 1:02
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    $\begingroup$ whereas for the $\mathrm{t_{2g}}$ set which point in between the ligands there is not much difference in the energies upon geometrical distortion, so the corresponding distortion is smaller (think of it as not having any incentive to distort). So, it's a decent approximation to treat the complex as having an undistorted geometry. However I am fairly sure that JT distortions will come into play at some point in time when calculating magnetic moments, and I think it is also in Figgis/Hitchman, but I do not have the book with me right now (and I did not read it - it is a complicated matter!). $\endgroup$ Feb 12, 2017 at 1:04
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    $\begingroup$ I try to understand that from a more physical perspective: If the electronic Hamiltonian does not commute with the angular momentum operator $\hat{L}_z$ then the eigenstates are in general no eigenfunctions of $\hat{L}_z$. If I now take some linear triatomic molecule with non-zero $m_z$ (like $\hat{L}_z|0> = m_z \hbar |0>$) and bend it slightly, by an angle $\epsilon$ what will happen? I suppose I still can calculate the expectation value $<0(\epsilon)|\hat{L}_z|0(\epsilon)>$. But what exactly will get? I suppose some real value ($\hat{L}_z$ is hermititan after all) close to $m\hbar$.Any idea? $\endgroup$ Nov 9, 2019 at 11:19
  • 1
    $\begingroup$ I suppose in most cases of non-symmetric molecules this expectation value is close to (or =) zero. Since in the symmetrized case $m$ should be usually $=0$ in the first instance (for ground states), right? $\endgroup$ Nov 9, 2019 at 11:28

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