What is the physical basis for Hund's first rule?

 

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According to Hund's first rule, all the orbitals are first singly occupied and then the other electrons start to fill the remaining orbital positions. This is quite intuitive because electron-electron repulsions would make an atom more unstable if the electrons start filling two at a time in a single orbital. However, the rule also states that electrons that singly fill the orbitals must be of the same spin. Why is it necessary for the electrons to have the same spin? quantum-chemistry electrons electronic-configuration spin shareimprove this question edited Sep 8 at 8:56 orthocresol♦ 15.4k33685 asked Sep 7 at 14:33 Aniansh 16518

1   It is a quantum mechanical effect that electrons with like spin repel each other less than electrons with unlike spin. bu.edu/quantum/notes/GeneralChemistry/… – orthocresol♦ Sep 7 at 14:47 @orthocresol Thanks for the reference! – Aniansh Sep 7 at 15:02 3   If you have only just learnt Hund's rules, you might not be able to understand it fully. The explanation is quite involved and you need to be familiar with some quantum mechanics. Just count it as one of those things that you postpone to a later date to find out about. (There's no shortage of such things in chemistry, sadly.) Exchange energy as Geoff described is half the explanation, I suppose. The origin of exchange energy lies in the QM. – orthocresol♦ Sep 7 at 15:04  @orthocresol I am finding it a little bit difficult but whatever I can make out of the notes helps to get a sense of what is going on. – Aniansh Sep 7 at 15:07 Related: chemistry.stackexchange.com/questions/14786 – orthocresol♦ Sep 24 at 3:38  add a comment

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The lowest energy state has parallel spins to maximize the exchange energy. As you say, there's a Coulomb repulsion between two electrons to put them in the same orbital. There's also a quantum mechanical effect. The exchange energy (which is favorable) increases with the number of possible exchanges between electrons with the same spin and energy. Going from the top state to the middle state, we remove the Coulomb repulsion between electrons in the same orbital. Going between the middle to the bottom (most stable state and that predicted by Hund's rule), we gain the exchange energy, because these two electrons are indistinguishable. shareimprove this answer answered Sep 7 at 14:50 Geoff Hutchison 15.1k33079

Can the exchange energy be understood as the energy released when electrons with different spins are made to have a single type of spin? And is it this energy that causes the stability in the atom? – Aniansh Sep 7 at 15:15 add a comment

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Warning: MathJaX heavy answer. Introduction The origin of Hund's first rule lies in "exchange energy", which is a way of saying that electrons with like spin repel each other less than electrons with unlike spin. To see why, let's consider an excited state of the helium atom: 1s12s1$1{\mathrm{s}}^{1}2{\mathrm{s}}^1$. (The ground state configuration 1s2$1{\mathrm{s}}^2$ is of no use to us because those electrons have to be paired with unlike spin.) In the excited state, the electrons can either have parallel spin (the triplet case) or paired spin (the singlet case). The total electronic wavefunction comprises a spatial part, which describes which orbitals the electrons are in (i.e. it specifies n,l,ml$n,l,m_l$), and a spin part, which describes the, well, spin (ms$m_s$). According to the Pauli exclusion principle, the total wavefunction must be antisymmetric with respect to interchange of the electron labels: Ψ(1,2)=−Ψ(2,1)$\mathrm{\Psi }(1,2)=-\mathrm{\Psi }(2,1)$. Spin wavefunctions The spin wavefunctions are simultaneous eigenfunctions of the S^2${\hat{S}}^2$ and S^z${\hat{S}}_z$ operators. The total spin quantum number, S$S$, can be found from the eigenvalue of the operator S^2${\hat{S}}^2$: S^2ψ=ℏ2S(S+1)ψ$${\hat{S}}^2\psi ={\hslash }^{2}S(S+1)\psi$$ For more detailed discussion, refer to any QM text on the coupling of angular momenta. There are three possible triplet spin wavefunctions, which are degenerate. ψspin,triplet(1,2)=⎧⎩⎨⎪⎪α(1)α(2)β(1)β(2)12√[α(1)β(2)+α(2)β(1)]$${\psi }_{\mathrm{s}\mathrm{p}\mathrm{i}\mathrm{n},\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}(1,2)=\{\begin{array}{l}\alpha (1)\alpha (2)\\ \beta (1)\beta (2)\\ \frac{1}{\sqrt{2}}[\alpha (1)\beta (2)+\alpha (2)\beta (1)]\end{array}$$ It turns out that S^2ψtriplet=2ℏ2ψtriplet${\hat{S}}^2{\psi }_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}=2{\hslash }^2{\psi }_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}$, which means that these have the quantum number S=1$S=1$. This corresponds to a multiplicity of M=2S+1=3$M=2S+1=3$, hence "triplet". These wavefunctions represent the case where the electrons have parallel spin. (Loosely speaking, each electron has a spin of 1/2$1/2$, and if these are aligned parallel the total spin is 1/2+1/2=1$1/2+1/2=1$.) The singlet state is ψspin,singlet(1,2)=12–√[α(1)β(2)−α(2)β(1)]$${\psi }_{\mathrm{s}\mathrm{p}\mathrm{i}\mathrm{n},\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{t}}(1,2)=\frac{1}{\sqrt{2}}[\alpha (1)\beta (2)-\alpha (2)\beta (1)]$$ and as you might guess, S^2ψsinglet=0${\hat{S}}^2{\psi }_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{t}}=0$, meaning that S=0$S=0$ and M=2S+1=1$M=2S+1=1$. This represents the case where the electrons are paired - one can think of it as the two 1/2$1/2$ electrons spins “cancelling each other out”. The important thing to notice is that the triplet spin wavefunctions are symmetric with respect to permutation of the labels, whereas the singlet spin wavefunction is antisymmetric. For example: ψsinglet(2,1)=12–√[α(2)β(1)−α(1)β(2)]=−12–√[α(1)β(2)−α(2)β(1)]=−ψsinglet(1,2)$$\begin{array}{rl}{\psi }_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{t}}(2,1)& =\frac{1}{\sqrt{2}}[\alpha (2)\beta (1)-\alpha (1)\beta (2)]\\ & =-\frac{1}{\sqrt{2}}[\alpha (1)\beta (2)-\alpha (2)\beta (1)]\\ & =-{\psi }_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{t}}(1,2)\end{array}$$ Spatial wavefunctions The only way of constructing these are the linear combinations below. ψspace,symmψspace,antisymm=12–√[1s(1)2s(2)+1s(2)2s(1)]=12–√[1s(1)2s(2)−1s(2)2s(1)]$$\begin{array}{rl}{\psi }_{\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e},\mathrm{s}\mathrm{y}\mathrm{m}\mathrm{m}}& =\frac{1}{\sqrt{2}}[1\mathrm{s}(1)2\mathrm{s}(2)+1\mathrm{s}(2)2\mathrm{s}(1)]\\ {\psi }_{\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e},\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{s}\mathrm{y}\mathrm{m}\mathrm{m}}& =\frac{1}{\sqrt{2}}[1\mathrm{s}(1)2\mathrm{s}(2)-1\mathrm{s}(2)2\mathrm{s}(1)]\end{array}$$ Remember that the atomic orbitals above are functions of the coordinates of the electron: 1s(1)=1s(r⃗ 1)=1s(r1,θ1,ϕ1)$1\mathrm{s}(1)=1\mathrm{s}({\overrightarrow{r}}_1)=1\mathrm{s}(r_1,{\theta }_1,{\varphi }_1)$. Now, the antisymmetric spatial wavefunction has an interesting property. Let's explore what happens when the electrons get very close together, i.e. r⃗ 1=r⃗ 2=r⃗ ${\overrightarrow{r}}_1={\overrightarrow{r}}_2=\overrightarrow{r}$. ψspace,antisymm=12–√[1s(r⃗ )2s(r⃗ )−1s(r⃗ )2s(r⃗ )]=0$$\begin{array}{rl}{\psi }_{\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e},\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{s}\mathrm{y}\mathrm{m}\mathrm{m}}& =\frac{1}{\sqrt{2}}[1\mathrm{s}(\overrightarrow{r})2\mathrm{s}(\overrightarrow{r})-1\mathrm{s}(\overrightarrow{r})2\mathrm{s}(\overrightarrow{r})]\\ & =0\end{array}$$ If the wavefunction is 0 when the two electrons are on top of each other, that means that there is no probability of this happening. The average distance between the two electrons ⟨|r⃗ 2−r⃗ 1|⟩$⟨\left|{\overrightarrow{r}}_2-{\overrightarrow{r}}_1\right|⟩$ will therefore be larger, and on average the electrons will repel each other less, since they are on average further away from each other. This is known as a Fermi hole. In fact, one can also show that the symmetric wavefunction has a larger probability amplitude when r⃗ 1=r⃗ 2${\overrightarrow{r}}_1={\overrightarrow{r}}_2$. This is known as a Fermi heap. The total wavefunction As I mentioned earlier, the Pauli exclusion principle dictates that the total wavefunction be antisymmetric. This means that the triplet spin wavefunctions, which are symmetric, must be paired with the antisymmetric spatial wavefunction. Likewise, the singlet spin wavefunction, which is antisymmetric, must be paired with the symmetric spatial wavefunction. SpinTriplet (S)Singlet (AS)SpatialASSTotalS × AS=ASAS × S=AS$$\begin{array}{ccc}\text{Spin}& \text{Spatial}& \text{Total}\\ \text{Triplet (S)}& \text{AS}& \text{S }\times \text{ AS}=\text{AS}\\ \text{Singlet (AS)}& \text{S}& \text{AS }\times \text{ S}=\text{AS}\end{array}$$ So, there is a decrease in electron-electron repulsion when the electrons are parallel - because this means they have a triplet spin wavefunction, and hence an antisymmetric spatial wavefunction, which has a Fermi hole. (image source: own work) Exchange integrals There is another, perhaps more mathematical, way of looking at it, and that is to calculate the expectation value of the electronic energy for both wavefunctions. For the helium atom, under the assumption of an infinitely heavy nucleus (essentially an atomic version of the Born-Oppenheimer approximation), the total Hamiltonian (in atomic units) is H^=−12(∇21+∇22)−2r1−2r2+1r12$$\hat{H}=-\frac{1}{2}({\mathrm{\nabla }}_1^2+{\mathrm{\nabla }}_2^2)-\frac{2}{r_1}-\frac{2}{r_2}+\frac{1}{r_{12}}$$ where ri$r_i$ is the distance of electron i$i$ from the nucleus, and r12$r_{12}$ is the distance between the two electrons. The expectation value of the energy is simply ⟨Ψ∣∣H^∣∣Ψ⟩$⟨\mathrm{\Psi }\left|\hat{H}\right|\mathrm{\Psi }⟩$ (assuming our wavefunction is already normalised). It turns out that the only important term in our present discussion is the electron-electron repulsion term 1/r12$1/r_{12}$. Let's look at the triplet spin wavefunction (of course paired with the antisymmetric spatial wavefunction) first. The operator doesn't depend on spin degrees of freedom, so we can separate those, and the inner product of the spin wavefunction with itself is just unity. ⟨Ψtriplet∣∣∣1r12∣∣∣Ψtriplet⟩=⟨ψspin,triplet|ψspin,triplet⟩⟨ψspace,antisymm∣∣∣1r12∣∣∣ψspace,antisymm⟩=12⟨1s(1)2s(2)−1s(2)2s(1)∣∣∣1r12∣∣∣1s(1)2s(2)−1s(2)2s(1)⟩=12{⟨1s(1)2s(2)∣∣∣1r12∣∣∣1s(1)2s(2)⟩−⟨1s(2)2s(1)∣∣∣1r12∣∣∣1s(1)2s(2)⟩−⟨1s(1)2s(2)∣∣∣1r12∣∣∣1s(2)2s(1)⟩+⟨1s(2)2s(1)∣∣∣1r12∣∣∣1s(2)2s(1)⟩}$$\begin{array}{rl}⟨{\mathrm{\Psi }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}\left|\frac{1}{r_{12}}\right|{\mathrm{\Psi }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}⟩& =⟨{\psi }_{\mathrm{s}\mathrm{p}\mathrm{i}\mathrm{n},\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}|{\psi }_{\mathrm{s}\mathrm{p}\mathrm{i}\mathrm{n},\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}⟩⟨{\psi }_{\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e},\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{s}\mathrm{y}\mathrm{m}\mathrm{m}}\left|\frac{1}{r_{12}}\right|{\psi }_{\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e},\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{s}\mathrm{y}\mathrm{m}\mathrm{m}}⟩\\ & =\end{array}$$$$\begin{array}{r}\frac{1}{2}⟨1\mathrm{s}(1)2\mathrm{s}(2)-1\mathrm{s}(2)2\mathrm{s}(1)\left|\frac{1}{r_{12}}\right|1\mathrm{s}(1)2\mathrm{s}(2)-1\mathrm{s}(2)2\mathrm{s}(1)⟩\\ & =\frac{1}{2}\{⟨1\mathrm{s}(1)2\mathrm{s}(2)\left|\frac{1}{r_{12}}\right|1\mathrm{s}(1)2\mathrm{s}(2)⟩\\ & -\end{array}$$$$\begin{array}{r}⟨1\mathrm{s}(2)2\mathrm{s}(1)\left|\frac{1}{r_{12}}\right|1\mathrm{s}(1)2\mathrm{s}(2)⟩\\ & -\end{array}$$$$\begin{array}{r}⟨1\mathrm{s}(1)2\mathrm{s}(2)\left|\frac{1}{r_{12}}\right|1\mathrm{s}(2)2\mathrm{s}(1)⟩\\ & +⟨1\mathrm{s}(2)2\mathrm{s}(1)\left|\frac{1}{r_{12}}\right|1\mathrm{s}(2)2\mathrm{s}(1)⟩\}\end{array}$$ Ugly, but that's what it is. We can simplify it however by noting that the first and the fourth brackets are the same (since the variables in the integral are just dummy variables, interchanging the electron labels does not affect the value of the integral). This integral is traditionally called a Coulomb integral, since it simply reflects the average Coulombic repulsion between two independent (uncorrelated) electrons, one in the 1s orbital and the second in the 2s orbital. It is given the symbol J1s2s$J_{1\mathrm{s}2\mathrm{s}}$ (the subscripts indicating the orbitals that the electrons are in). Because this is an expectation value for a potential energy arising from repulsion, it is necessarily a positive quantity. The second and third terms are actually also the same, again because the variables are just dummy variables and you can interchange the labels. This integral, however, does not have a physical interpretation; it is a purely quantum mechanical effect. It is called the exchange integral, and is given the symbol K1s2s$K_{1\mathrm{s}2\mathrm{s}}$. It's a bit harder to show, but the exchange integral is also a positive quantity (there is a question on it here, with some insight by resident QM master Wildmeow). However, its magnitude is not as large as J1s2s$J_{1\mathrm{s}2\mathrm{s}}$. Putting it all together (leaving out subscripts for clarity) we have ⟨Ψtriplet∣∣∣1r12∣∣∣Ψtriplet⟩=12(J−K−K+J)=J−K$$⟨{\mathrm{\Psi }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}\left|\frac{1}{r_{12}}\right|{\mathrm{\Psi }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{t}}⟩=\frac{1}{2}(J-K-K+J)=J-K$$ and if you were to perform the same calculation on the singlet spin wavefunction (not very difficult, just replace the minus signs with a plus!) you would get ⟨Ψsinglet∣∣∣1r12∣∣∣Ψsinglet⟩=12(J+K+K+J)=J+K$$⟨{\mathrm{\Psi }}_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{t}}\left|\frac{1}{r_{12}}\right|{\mathrm{\Psi }}_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{t}}⟩=\frac{1}{2}(J+K+K+J)=J+K$$ Since J>K>0$J>K>0$, the expectation value of the energy for the singlet state is higher. But I have more than two electrons. What now? Well, you just have to do a ton more ugly maths and perhaps a few hundred more integrals. But the idea is the same. For every pair of electrons with parallel spins, there is a corresponding exchange integral that reduces the expectation value of the energy. Therefore, in a 3d5$3{\mathrm{d}}^5$ manganese atom (we ignore the closed shells) there are 5!2!(5−2)!=10$$\frac{5!}{2!(5-2)!}=10$$ different pairs of d-d electrons with parallel spins. Therefore if you evaluate the expectation value of the energy, there will be a −10K3d3d$-10K_{3\mathrm{d}3\mathrm{d}}$ term, which is stabilising. This is precisely the exchange energy effect that Geoff describes in his answer. shareimprove this answer edited Sep 8 at 6:42 Martin - マーチン♦ 20.8k862121 answered Sep 8 at 5:26 orthocresol♦ 15.4k33685

"since the variables in the integral are just dummy variables" is this your way of saying that electrons are indistinguishable? – Martin - マーチン♦ Sep 8 at 6:25 1   @Martin I just meant that mathematically the integral is the same - in the sense that e.g. ∬3x2+4ydxdy=∬3y2+4xdydx$$\iint 3x^2+4y\mathrm{d}x\mathrm{d}y=\iint 3y^2+4x\mathrm{d}y\mathrm{d}x$$ (as long as the limits of integration are the same, which they are - 0 to infinity for both r1,r2$r_1,r_2$). If I am not wrong, this is a manifestation of indistinguishability, because it means you cannot assign a particular label 1 or 2 to a particular electron, which means r1$r_1$ and r2$r_2$ are interchangeable in the integral – orthocresol♦ Sep 8 at 6:54 Thanks for the answer, it is very informative. – Aniansh Sep 8 at 11:17 That is just beautiful and I love when the replies go the extra mile. Thank you. – Henrique Junior Sep 14 at 20:45 add a comment

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Some understanding can be gained by looking at the symmetry of the orbital parts of the wavefunctions involved. The total wavefunction for electrons must be anti-symmetric with respect to exchanging electrons, the Pauli Principle. The total wavefunction has spin and spatial parts, ψ=ψspaceψspin$\psi ={\psi }_{space}{\psi }_{spin}$. When electrons are paired with total spin = 0, (a singlet state) the spatial part of the wavefunction is symmetric and the spin part therefore antisymmetric. Its the other way round for the triplet state, spins parallel and total spin = 1, ψtriplet=ψspace(antisym)ψspin(sym)${\psi }_{triplet}={\psi }_{space}(antisym){\psi }_{spin}(sym)$. The effect is that the electrons occupy different regions of space in triplet states; they have a zero probability of being in the same place. In singlets the electrons crowd together; formally there is a non-zero probability of being in the same region of space. This is why the electrostatic repulsion, which gives energy e2/r1,2$e^2/r_{1,2}$ and is large when the electrons are near to one another, raises the energy of the singlet. Thus the degeneracy between singlet (spin paired) and triplet (spins 'parallel') is split. Therefore the energy difference between singlet and triplet ( and more generally paired and unpaired electrons) is a consequence of the wavefunction symmetry and so the value of the total spin. The picture shows the effect on the spatial part of wavefunctions illustrated with 2D 'particle in a box' wavefunctions. With respect to the coulomb and exchange integrals and energies, K and J respectively, both are positive, the total energy works out as K+J$K+J$ and K−J$K-J$ which shows the energy splitting. (For completeness both Coulomb and Exchange integrals are forms of electrostatic repulsion. If W is the potential energy, which may contain several terms positive and negative of the form e2/r$e^2/r$ , then the Coulomb integral with electrons 1 and 2 and orbitals a and b, is ∫ψ(1,a)ψ(2,b) W ψ(1,a)ψ(2,b)$\int \psi (1,a)\psi (2,b)W\psi (1,a)\psi (2,b)$ and exchange integral is ∫ψ(2,a)ψ(1,b) W ψ(1,a)ψ(2,b)$\int \psi (2,a)\psi (1,b)W\psi (1,a)\psi (2,b)$ where the electrons are 'exchanged') shareimprove this answer answered Sep 8 at 7:18 porphyrin 5,531423