How can the ground state electron configuration for a d3 complex correspond to an F term symbol?

Question

The ground-state term symbol from a $\mathrm{d^3}$ Tanabe-Sugano diagram is $\mathrm{^4F}$. My question is how the total orbital quantum number $\Lambda=3$, or $\mathrm{F}$ term, arises.

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For a $\mathrm{d^3}$ metal, I'd expect the following ground state d-electron configuration:

where notably the $\mathrm{t_{2g}}$ orbitals correspond to $\mathrm d_{xy}$, $\mathrm d_{xz}$, and $\mathrm d_{yz}$ suborbitals.

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From the spherical harmonics, $\mathrm d_{xy}$ results from a linear combination of:

$$\begin{align} Y_2^{-2}(\theta,\varphi) &= \frac{1}{4}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{-2\mathrm{i}\varphi}\cdot \sin^2\theta \\ Y_2^{+2}(\theta,\varphi) &= \frac{1}{4}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{2\mathrm{i}\varphi}\cdot \sin^2\theta \end{align}$$

i.e. $m_l=\pm2$ equations.

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$\mathrm d_{xz}$ and $\mathrm d_{yz}$ result from:

$$\begin{align} Y_2^{-1}(\theta,\varphi) &= \frac{1}{2}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{-\mathrm{i}\varphi} \cdot \sin\theta \cdot \cos\theta \\ Y_2^{1}(\theta,\varphi) &= -\frac{1}{2}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{\mathrm{i}\varphi} \cdot \sin\theta \cdot \cos\theta \end{align}$$

i.e. $m_l=\pm1$ equations.

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From this assumption–which I may very well have presented incorrectly – the $\mathrm{t_{2g}}$ orbitals would therefore correspond to $m_l$ values of $\pm1$ and either $+2$ or $-2$. This would mean the greatest $\Lambda$ from one electron in each $\mathrm{t_{2g}}$ orbital (a $\mathrm d^3$ ground-state electron configuration), would be $\mathrm d_{xy} + \mathrm d_{xz} + \mathrm d_{yz} = \pm2+1-1=\pm2$, or a $\mathrm{D}$ term symbol.

Would anyone be able to correct me where my logic went wrong? I have a feeling I'm not allowed to restrict the $\mathrm{t_{2g}}$ orbitals to those $m_l$ values, but why would that not be allowed, when those are the equations that derive the $\mathrm{t_{2g}}$ d-orbitals?

Thank you!

Answer 1

Free ion

The ground-state term symbol is only $\mathrm{^4F}$ in the case of a free ion. If you take a closer look at the Tanabe-Sugano diagram, the $\mathrm{^4F}$ term only appears at the far left-hand side of the diagram, where $\Delta = 0$. $\Delta$ refers to the ligand-field splitting parameter, and $\Delta = 0$ indicates that there is no ligand field, i.e. a free ion.

The quantum number $L$ (total orbital angular momentum) of the ground state can be obtained by coupling the individual orbital angular momenta of the d electrons using a Clebsch-Gordan series. The way to do this is described in most physical chemistry textbooks under the Russell-Saunders coupling scheme. For example in Atkins 10th ed. it is in page 386 under the chapter on "Atomic structure and spectra".

(Note that the symbol $\Lambda$ is used for diatomic molecules, not atoms.)

$L$ is said to be a "good" quantum number, in that the operator $\hat{L}^2$ (almost - this neglects spin-orbit coupling) commutes with the Hamiltonian $\hat{H}$. Quantum mechanically this means that the $\hat{H}$ and $\hat{L}^2$ (almost) share a set of eigenstates, and therefore for every state of the Hamiltonian (which correspond to the electronic configurations we are familiar with) one can (almost) calculate the corresponding value of $L$.

$$\hat{L}^2|\psi\rangle = L(L+1)\hbar^2|\psi\rangle$$

Octahedral complex

The ground state term symbol of a $(\mathrm{t_{2g}})^3$ ion is $\mathrm{^4\!A_2}$, not $\mathrm{^4F}$!

The $\mathrm{t_{2g}}$ set comprises the $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ and $\mathrm{d}_{xy}$ orbitals. These three d orbitals are what we call "real" spherical harmonics, which are linear combinations of the complex spherical harmonics which you have quoted. As such it is not possible to assign $m_l$ values like you have to the $\mathrm{t_{2g}}$ orbitals.

It's not correct to say that $\mathrm{d}_{xy}$ can have "either $m_l = +2$ or $-2$". That would mean that $\mathrm{d}_{xy}$, at any one point in time, is either equal to $Y_2^{+2}$ or equal to $Y_2^{-2}$, which makes no sense. It is not flip-flopping between the two spherical harmonics, it is its own thing: a linear combination of the two spherical harmonics, or a superposition, if you prefer that word. Furthermore, the spherical harmonics only have significance under spherical symmetry, where they act as simultaneous eigenstates of $\hat{H}$, $\hat{L}^2$, and $\hat{L}_z$. In octahedral symmetry, the spherical harmonics aren't of any significance at all and trying to "resolve" the $\mathrm{t_{2g}}$ orbitals into their components as spherical harmonics is physically unmeaningful (it helps with the maths, but that's all).

Under octahedral symmetry, the total orbital angular momentum $L$ is no longer a good quantum number (i.e. $\hat{L}$ no longer commutes with the Hamiltonian) and therefore the term symbol doesn't say anything about it!

Answer 2

The ground state of the free ion is $^4F$ but is $^4A_2 (t_{2g}^3)$ in a cubic field such as in an octahedral complex with $O_h$ symmetry. This term is shown on the abscissa of the Tanabe-Sugano plot. Thus, even though there is an energy difference between the free ion and when it is in an octahedral field, this is not shown in the plot. The lines representing the higher energy states measure the increase in energy from the ground state

The way to calculate the term symbol for the free ion is explained in detail in many textbooks and in my answer to How does one find the ground-state term symbol for a configuration that is exactly half-filled? .

Why the ground state term symbol is $^4A_2$ in an octahedral complex does need some explanation, however. In $O_h$ (and $T_d$) point groups the highest dimension of an irreducible representation is three-fold; Mulliken symbol T. As a result the states with orbital degeneracies greater than this e.g. $D, F, G..$ etc. must split into new terms of degeneracy no greater than three.
The calculation for S, P, D, F, G etc. terms is outlined below with an example of F terms.

The effect that the symmetry imposed by the ligands has on the d-orbitals means that these have to be rotated, inverted or reflected according to the operations of the point group. This does not change the energy as it corresponds only to a change in the direction of the axes. Operating this way leads to a reducible representation that is then analysed to obtain its makeup as irreducible representations (irreps).

In $O_h$ the symmetry operations are $E, C_3, C_2, C_4, i, S_4,S_6, \sigma_h, \sigma_d$. The equations to use for rotation are shown in the notes below. Applying these operations produces the following reducible representation for an F term with orbital angular momentum $L=3$. $$ \begin{array}{c|cccccccccc} O_h & E & 8C_3 & 6C_2 & 6C_4 & 3C_2 & i & 6S_4 & 8S_6 & 3\sigma_h & 6\sigma_d \\ \hline \chi = & 7 & 1 & -1 & -1 &-1 & 7 & -1 & 1 & -1 & -1 \ \end{array}$$ Using the tabular method (see my answer to this question Understanding group theory easily and quickly ) produces the irreps $A_{2g}+T_{1g}+T_{2g}$. Thus an $F$ state splits into a non-degenerate $A_{2g}$ ground state and two triply degenerate states of higher energy. The splitting of other terms (S, D, G etc) is determined in a similar manner.

As the d orbitals are inherently ‘gerade’ or g this subscript is usually dropped from terms in the Tanabe-Sugano plots. Unless spin-orbit coupling is exceptionally strong the spin of the final states is the same as that of the free ion.

The next table shows some free ion and $O_h$ terms. $$ \begin{array}{clcr} \text{Free ion} ~~& ~~ O_h\\ \hline S & A_{1g}\\ P & T_{1g}\\ D & E_g + T_{2g}\\ F & A_{2g} + T_{1g} + T_{2g}\\ G & A_{1g} + E_g + T_{1g} + T_{2g}\\ H & E_g + 2T_{1g} + T_{2g} \end{array}$$

Using spherical harmonics to produce the energy splitting, which means calculating the potential energy and wavefunctions, is significantly harder and is only sketched.(See Balhausen, 'Introduction to Ligand Field Theory' for all the gory details.)

We suppose that the potential is caused by 6 charges around the central ion, and choose to use the sum of spherical harmonics $Y_l^m$ to form the potential since these are solutions to a problem of full spherical symmetry. The general potential for i electrons is thus $V=\sum_i\sum_l\sum_m Y_l^m(\theta_i\phi_i)R_{nl}(r_i)$ where the R is the radial function which can be dropped from now on as a common factor. The specific potential must transform as the totally symmetric representation of the point group of the molecule, ($A_{1g}$ in $O_h$) because the Hamiltonian must remain totally symmetric under all symmetry operations. It turns out that only terms in $l=0, 2, 4$ can contribute to the potential. The $l=0$ term is the largest but as it is spherically symmetric it has little effect on electronic properties as it only shifts energy levels. The $l=2$ harmonics produce irreps only of $E_g$ and $T_{2g}$, so are not suitable as the totally symmetric representation is absent, but the $l=4$ harmonics produce irreps of $A_{1g},~ E_g,~ T_{1g}$ and $T_{2g}$ which means that there is a linear transform of $Y_4^m$ that transforms as $A_{ig}$. If the $C_4$ axis is taken as the axis to be quantised then the potential $V_4$ of $A_{1g} $ symmetry (excluding that from $l=0$) is proportional to a linear combination of the harmonics $V_4 \approx Y_4^0+b( Y_4^{+4}+Y_4^{-4})$, b is a constant. (These are the only harmonics that satisfy $\hat C_4 V_4=V_4$.)

To find the wavefunctions we use the fact that the d orbitals transform as $E_g$ and $T_{2g}$ in $O_h$. These can be combined to produce the familiar ‘real’ d orbitals shown in textbooks, $ d_{z^2}, d_{x^2-y^2} $ etc, by quantising along the $C_4$ axis.

The energy splitting $e_g-t_{2g}$ of a single electron in a d orbital e.g. $\ce{Ti^{3+}}$, is conventionally set to $\Delta = 10Dq$ and is positive. The energy of each level calculated as $E_{eg} = \epsilon_0 +\int \phi^*(e_g)V\phi(e_g)d\tau$ and $E_{t2g}=\epsilon_0 +\int \phi^*(t_{2g})V\phi(t_{2g})d\tau=-4Dq$ where $\epsilon_0$ is the spherically symmetric part of the potential. The energy gap is then $10Dq=E_{eg}-E_{t2g}$ and when all energy levels are filled with 10 electrons (an S state) then $0=4E_{eg}+6E_{t2g}$ from which $E_{eg}= 6Dq$ and $E_{t2g}=-4Dq$.

As the electron density of the $e_g$ orbitals are directed towards the ligands these are of higher energy than the $t_{2g}$.

Notes:

For quantum number k these relationships can be used with any point group because all point groups are subgroups of a sphere’s symmetry. Remember that $C_n$, is rotation by $2\pi/n$ radians.

$$ \chi(E)= 2k+1\\ \chi(C(x)) = \frac{\sin((k+1/2)x)}{\sin(x/2)}\\ \chi(i) = \pm (2k+1)\\ \chi(S(x)) = \frac{\sin((k+1/2)(x+\pi))}{\sin((x+\pi)/2)}\\ \chi(\sigma)=\pm \sin((k+1/2)\pi) $$

The + sign is used with gerade, - with ungerade.