## Transition Metals

The transition metals are more complicated than lighter elements.

Why?

Because they’re the first whose electron wavefunctions are described by quadratic functions of $x,y,$ and $z$ — not just linear or constant. These are called ‘d orbitals’, and they look sort of like this:

More precisely: the wavefunctions of electrons in atoms depend on the distance $r$ from the nucleus and also the angles $\theta, \phi.$ The angular dependence is described by ‘spherical harmonics’, certain functions on the sphere. These are gotten by taking certain polynomials in $x,y,z$ and restricting them to the unit sphere. Chemists have their own jargon for this:

• constant polynomial: s orbital

• linear polynomial: p orbital

• cubic polynomial: f orbital

and so on.

To be even more precise, a spherical harmonic is an
eigenfunction of the Laplacian on the sphere. Any such function is the restriction to the sphere of some homogeneous polynomial in $x,y,z$ whose Laplacian in 3d space is zero. This polynomial can be constant, linear, etc.

The dimension of the space of spherical harmonics goes like 1, 3, 5, 7,… as we increase the degree of the polynomial starting from 0:

• constant: $1$

• linear: $x, y, z$

• quadratic: $xy, xz, yz, x^2 - y^2, x^2 - z^2$

etcetera. So, we get one s orbital, three p orbitals, five d orbitals and so on. Here I’ve arbitrarily chosen a basis of the space of quadratic polynomials with vanishing Laplacian, and I’m not claiming this matches the d orbitals in the pictures!

The transition metals are the first to use the d orbitals. This is why they’re so different than lighter elements.

Although there are 5 d orbitals, an electron occupying such an orbital can have spin up or down. This is why there are 10 transition metals per row!

This chart doesn’t show the last row of highly radioactive transition metals, just the ones you’re likely to see:

Look: 10 per row, all because there’s a 5d space of quadratic polynomials in $x,y,z$ with vanishing Laplacian. Math becomes matter.

Can we understand why the first transition element, scandium, has 21 electrons? Yes, if we’re willing to use the ‘Madelung rules’ explained last time. Let me review them rapidly here.

You’ll notice this chart has axes called $n$ and $\ell.$

As I just explained, the angular dependence of an orbital is determined by a homogeneous polynomial with vanishing Laplacian. In the above chart, the degree of this polynomial is called $\ell.$ The space of such polynomials has dimension $2\ell + 1.$

But an orbital has an additional radial dependence, described using a number called $n.$ The math, which I won’t go into, requires that $0 \le \ell \le n.$ That gives the above chart its roughly triangular appearance.

The letters s, p, d, f are just chemistry jargon for $\ell = 0,1,2,3.$

Thanks to spin and the Pauli exclusion principle, we can pack at most $2(2\ell + 1)$ electrons into the orbitals with a given choice of $n$ and $\ell.$ This bunch of orbitals is called a ‘subshell’.

The Madelung rules say the order in which subshells get filled:

1. Electrons are assigned to subshells in order of increasing values of $n + \ell$.
2. For subshells with the same value of $n + \ell$, electrons are assigned first to the subshell with lower $n.$

So let’s see what happens. Only when we hit $\ell = 2$ will we get transition metals!

$\boxed{n + \ell = 1}$

$n = 1, \ell = 0$

This is called the 1s subshell, and we can put 2 electrons in here. First we get hydrogen with 1 electron, then helium with 2. At this point all the $n = 1$ subshells are full, so the ‘1st shell’ is complete, and helium is called a ‘noble gas’.

$\boxed{n + \ell = 2}$

$n = 2, \ell = 0$

This is called the 2s subshell, and we can put 2 more electrons in here. We get lithium with 3 electrons, and then beryllium with 4.

$\boxed{n + \ell = 3}$

$n = 2, \ell = 1$

This is called the 2p subshell, and we can put 6 more electrons in here. We get:

◦ boron with 5 electrons,
◦ carbon with 6,
◦ nitrogen with 7,
◦ oxygen with 8,
◦ fluorine with 9,
◦ neon with 10.

At this point all the $n = 2$ subshells are full, so the 2nd shell is complete and neon is another noble gas.

$n = 3, \ell = 0$

This is is called the 3s subshell, and we can put 2 more electrons in here. We get sodium with 11 electrons, and magnesium with 12.

$\boxed{n + \ell = 4}$

$n = 3, \ell = 1$

This is called the 4p subshell, and we can put 6 more electrons in here. We get:

◦ aluminum with 13 electrons,
◦ silicon with 14,
◦ phosphorus with 15,
◦ sulfur with 16,
◦ chlorine with 17,
◦ argon with 18.

At this point all the $n = 3$ subshells are full, so the 3rd shell is complete and argon is another noble gas.

$n = 4, \ell = 0$

This is called the 4s subshell, and we can put 2 more electrons in here. We get potassium with 19 electrons and calcium with 20.

$\boxed{n + \ell = 5}$

$n = 3, \ell = 2$

This is called the 3d subshell, and we can put 10 electrons in here. Since now we’ve finally hit $\ell = 2,$ and thus a d subshell, these are transition metals! We get:

◦ scandium with 21 electrons,
◦ titanium with 22,
◦ chromium with 24,
◦ manganese with 25,
◦ iron with 26,
◦ cobalt with 27,
◦ nickel with 28,
◦ copper with 29,
◦ zinc with 30.

And the story continues—but at least we’ve seen why the first batch of transition elements starts where it does!

### The scandal of scandium

For a strong attack on the Madelung rules, see:

• Eric Scerri, The problem with the Aufbau principle for finding electronic configurations, 24 June 2012.

But it’s important to realize that he’s attacking a version of the Madelung rules that is different, and stronger than the version stated above. My version only concerned atoms, not ions. The stronger version claims that you can use the Madelung rules not only to determine the ground state of an atom, but also those of the positive ions obtained by taking that atom and removing some electrons!

This stronger version breaks down if you consider scandium with one electron removed. As we’ve just seen, scandium has the electrons as in argon together with three more: two in the 4s orbital and one in the 3d orbital. This conforms to the Madelung rules.

But when you ionize scandium and remove one electron, it’s not the 3d electron that leaves—it’s one of the 4s electrons! This breaks the stronger version of the Madelung rules.

The weaker version of the Madelung rules also breaks down, but later in the transition metals. The first problem is with chromium, the second is with copper:

By the Madelung rules, chromium should have 2 electrons in the 4s shell and 4 in the 3d shell. But in fact it has just 1 in the 4s and 5 in the 3d.

The second is with copper. By the Madelung rules, this should have 2 electrons in the 4s shell and 9 in the 3d. But in fact it has just 1 in the 4s and 10 in the 3d.

There are also other breakdowns in heavier transition metals, listed here:

• Wikipedia, Aufbau principle: exceptions in the d block.

These subtleties can only be understood by digging a lot deeper into how the electrons in an atom interact with each other. That’s above my pay grade right now. If you know a good place to learn more about this, let me know! I’m only interested in atoms here, not molecules.

### Oxidation states of transition metals

Transition metals get some of their special properties because the electrons in the d subshell are easily removed. For example, this is why the transition metals conduct electricity.

Also, when reacting chemically with other elements, they lose different numbers of electrons. The different possibilities are called ‘oxidation states’.

For example, scandium has all the electrons of argon (Ar) plus two in an s orbital and one in a d orbital. It can easily lose 3 electrons, giving an oxidation state called Sc3+. Titanium has one more electron, so it can lose 4 and form Ti4+. And so on:

This accounts for the most obvious pattern in the chart below: the diagonal lines sloping up.

The red dots are common oxidation states, while the white dots are rarer oxidation states. For example iron (Fe) can lose 2 electrons, 3 electrons, 4 electrons (more rarely), 5 electrons, or 6 electrons (more rarely).

The diagonal lines sloping up come from the simple fact that as we move through a group of transition metals, there are more and more electrons in the d subshell, so more can be easily be removed. But everything is complicated by the fact that electrons interact! So the trend doesn’t go on forever: manganese gives up 8 electrons but iron doesn’t easily give up 8, only at most 6. And there’s much more going on, too.

Note also that the two charts above don’t actually agree: the chart in color includes more rare oxidation states.

### References

• Wikipedia, Transition metals.

Oxidation states of transition metals, Chemistry LibreTexts.

The colored chart of oxidation states in this post is from Wikicommons,
made by Felix Wan, corrected to include the two most common oxidation
states of ruthenium. The black-and-white chart is from the Chemistry
Libretexts
webpage.

### 24 Responses to Transition Metals

1. Tying in to John’s recent posts on music (and the Father of Galileo Galilei, hence also a connection to physics), I have to admit that “transition metal” is a new one for me! :-)

https://en.wikipedia.org/wiki/Heavy_metal_genres

2. jackjohnson says:

This is wonderful. It is hard to find an accessible discussion of this anywhere, and it is one of the most beautiful applications of Schur-Weyl duality in representation theory. It’s a great shame that AAIK there is no account of this anywhere else for serious students of chemistry or mathematicians. Kudos props and thanks!

• John Baez says:

Thanks very much! I seem to recall that Shlomo Sternberg’s Group Theory and Physics gets into this material. But I always found that book a bit hard to read—I’m not sure why—so I don’t actually know if this in there.

3. ericscerri says:

Dear John, if I may?

Thank you for citing my blog on the configuration of Sc and the aufbau. You state that “he (meaning me Scerri) is attacking a different and stronger version of the aufbau than stated above”.

Regardless of what is stated above I dont think there is one unequivocal version of the aufbau as you seem to be claiming to have stated.

Secondly, I believe that obtaining the configuration of ions does not depend of the aufbau but just on the common sense notion that one needs to reverse the presumed order of filling of orbitals in the atom in question.

You appear to rescue your own version of the aufbau that you restrict to just atoms by ignoring the empirical evidence on the order of filling.

As I and Eugen Schwarz, whom I cited in my blog have stressed, in a number of articles, the 3d orbitals are in fact filled ‘before’ or preferentially to the 4s orbital in transition metals. This may disagree with what is commonly stated in chemistry and physics textbooks but I assure you it is supported by the spectroscopic evidence on these atoms. Moreover this alternative way of teaching the 4s 3d conundrum is beginning to find it’s way into textbooks such as Atkins and Jones, Chemical Principles (The Quest for Knowledge) as well as Oxtoby, Gillis and Campion, both of which books are used regularly to teach thousands of undergraduate students here in the UCLA chemistry department.

If you are interested in seeing articles on this issue rather than just the blog by me which you cited please visit my website at
http://www.ericscerri.com and look at my publications list or request particular papers you see listed.

Better still, for a discussion of the 4s-3d conundrum see my YouTube workshop on this subject, https://www.youtube.com/watch?v=5cfKot3nBFA

• John Baez says:

Hi!

You appear to rescue your own version of the aufbau that you restrict to just atoms by ignoring the empirical evidence on the order of filling.

I’m not trying to ‘rescue’ anything, just clarify that there are (at least) two statements of the Madelung rules: the one I stated in my blog articles, which makes claims only about neutral atoms and fails first at chromium, and the stronger one you discuss, which additionally makes claims about positive ions and fails already at scandium.

• ericscerri says:

Thank you but the failure of the Madelung rule at Sc has nothing to do with the ion.

The order of filling in the Sc neutral atom is in fact 3d before 4s as Schwarz and I have argued in a number of articles. I also discuss this in my books on the periodic table including.

Eric Scerri, The Periodic Table, Its Story and Its Significance, OUP, 2020.

and my “A Very Short Introduction to the Periodic Table”, OUP, 2019.

http://www.ericscerri.com

• John Baez says:

Can you explain how we operationally determine the “order of filling” if we only consider an unionized atom of scandium? The only operational way that I know to determine a kind of “order of filling” is to ionize the atom, removing one electron, and see which orbital is now empty.

Of course, I’m not a chemist so there could be tricks I don’t know.

Anyway, the version of the Madelung rules that I stated in my blog says nothing about the order of filling of orbitals in a single unionized atom. It merely says which orbitals will be filled in that atom. So this whole question is orthogonal to what I was discussing in my blog—except for the section “The scandal of scandium”, which was about your thoughts.

• Raphael says:

The problem does not arise at the the antisymmetrized product level of approximation, since at this approximation you still have one-electron wavefunctions that are eigenfunctions of a related model Hamiltonian. The problem arizes if the simple antisymmetrized products are not suitable anymore. And that is pretty quickly relevant epspecially for transition elements. At multideterminant approximations the orbitals can not be assigned energy eigenvalues anymore, and they do not form a orthonormal basis of 1-electron wave functions.

4. ericscerri says:

You also briefly mention the anomalous electronic configurations found in Cr and Cu and incidentally about 20 atoms altogether in the periodic table.

You wonder whether anybody can provide you with a more detailed explanation of these violations of the Aufbau in what you call the weaker sense.

I have written an article on this subject which appears in an edited collection on the Philosophy of Chemistry. Since this is not generally accessible I would be happy to send copies on E-mail request.

The reference is E.R. Scerri, The Changing Views of a Philosopher of Chemistry on the Question of Reduction, in E.R. Scerri, G. Fisher eds. Essays in the Philosophy of Chemistry, Oxford University Press, New York, 2016.

• John Baez says:

Thanks, I’d be happy to get a copy of your article, so I will email you! I’ll also delete the email address you kindly provided in your comment, since publicly visible email addresses tend to attract spam.

5. Crust says:

Typo: Should $z^2-x^2-y^2$ be $z^2 -x^2$ (when you’re giving a basis for quadratic eigenfunctuons of the Laplacian)?

• John Baez says:

Thanks! You’re right, $z^2 - x^2 - z^2$ doesn’t have vanishing Laplacian. I’ve changed it to $x^2 - z^2.$ Since all I’m doing here is describing some randomly chosen basis of quadratic functions of $x,y,z$ with vanishing Laplacian, I could also have chosen $z^2 - \frac{1}{2} (x^2 + y^2).$ I’m not particularly trying to choose a basis that matches those shown in the pictures.

6. Crust says:

So why does the $d_{z^2}$ orbital look different from the other d orbitals and why is it indexed by a polynomial on which the 3D Laplacian doesn’t vanish?

• John Baez says:

I don’t know what people mean by $d_{z^2}.$ There are various ways to mess with the function $z^2$ and get a homogeneous quadratic polynomial in $x,y,z$ on which the Laplacian vanishes. For example, you can subtract off a suitable multiple of $x^2 + y^2 + z^2,$ and then multiply by any nonzero constant. One possible result from doing this is $z^2 - \frac{1}{2}(x^2 + y^2).$ This has rotational symmetry around the $z$ axis, like the picture of $d_{z^2}$ does. But I’m not claiming this is the official definition of $d_{z^2}.$

In my blog article here, I didn’t really care what basis we pick for the space of homogeneous quadratic polynomials in $x,y,z$ on which the Laplacian vanishes. I just wanted to exhibit some basis, to show that this space is 5-dimensional. (Even so, I screwed up until you corrected me above.)

If we want a ‘physically well-motivated’ choice of basis, we could take the basis of eigenvectors of orbital angular momentum around the $z$ axis, called $L_z.$ This has eigenvalues 2, 1, 0, -1 and -2 on the space of homogeneous quadratic polynomials in $x,y,z$ on which the Laplacian vanishes. But its eigenvectors are, in general, complex functions. $z^2 - \frac{1}{2}(x^2 + y^2)$ is real, and it’s an eigenvector with eigenvalue 0 because it has rotational symmetry around the $z$ axis.

• ericscerri says:

Here is a simple explanation of why the $d_{z^2}$ orbital differs in appearance from the other four d orbitals which I teach to undergrad chemists at UCLA.

﻿There are potentially 6 equivalent d orbitals which would be labeled, $d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}, d_{z^2 - x^2}, d_{z^2 - y^2}$.

But there can only be 5, due to quantum mechanics and the fact that if $l = 2$, $m_l$ can only be -2, -1, 0, +1, +2

We therefore take a linear combination of the last two listed above, by convention, to give $d_{2z^2 - x^2 - y^2}$.

This orbital is usually abbreviated to just $d_{z^2}$.

Its shape is the result of combining the shapes of $d_{z^2 - x^2}$ and $d_{z^2 - y^2}$. For diagrams see the inorganic chemistry textbook by Huheey, Keiter and Keiter.

http://www.ericscerri.com

Hi, Just a couple of corrections. The elements from Al – Ar are 3p, not 4p, and the metals from Sc – Zn are 3d, not 5d.

• John Baez says:

Thanks! I’ll have to see where I made those mistakes and fix them!

8. Raphael says:

If you happen to continue this excellent series on chemistry from a mathematicians perspective, I would love to read about your digest of the famous Jahn-Teller effect. I bet you’ll find it fascinating, too.

9. Wolfgang says:

As a chemist by training it’s extremely interesting to see an exposition of chemistry through the eyes of a theoretical physicist, in particular, when the point of view is not one of ‘pure quantum mechanics’ (Schroedinger equation calculations), but including algebra, group theory (by the topic of symmetry), etc. This is rarely, in my point of view, treated by chemists this way. Either they treat it purely empirically, as facts to learn, or purely theoretically, as the mentioned Schroedinger equation calculations, but the middle ground in between, being more conceptual, seems very nice as well.

Two questions:

I wonder if the author is aware of the work of Jan C. A. Boeyens, which was a theoretical chemist, and if so, what he thinks of them? I only got a glimpse of ‘Number Theory and the Periodicity of Matter’ and it was as fascinating as appearing on the fringe of exact science. Bordering as much on number theory as on numerology it appeared to me. Wikipedia has it as ‘As an emeritus he wrote books challenging the current scientific consensus about the adequacy of quantum mechanics in which he presented a way to establish more accurate modern physics and chemistry without using higher mathematics by using elementary number theory.’

As a chemistry student I also wondered about the odd shapes of the orbitals. I did not see a plot of them like here, which nicely puts them into a triangle, with analogous orbitals above each other. It looks much more compelling this way. For instance, I wondered about the nice symmetry of the p orbitals, the same shape in the three possible axis orientations, and the breaking of this pattern with the d orbitals. For me the x^2 – y^2 always was an abomination and the z^2 even more so, being unique in shape among the orbitals. For me it would be nicer for all d orbitals to have again the same shape only in different orientations. My fellow students didn’t seem to care at all. I once read in some book that this might be even possible by choosing another basis for the construction of the orbital pictures, but I do not understand enough of how this can be, since then it would appear that the pictures of the shapes of the orbitals are non-unique, which would be horrific, since so many chemical discussions about chemical reactivity are based on the textbook shapes.

• John Baez says:

Hi! I haven’t heard of the work of Jan C. A. Boeyens, and from your description it’s hard to tell if it’s worth taking seriously. Do you think it is?

Everyone with a lively sense of curiosity wonders why the usually chosen set of d orbitals looks more ad hoc and asymmetrical than the s and p orbitals. We’ve had a good conversation about this in the comments on this blog article above yours!

The short answer is that the set of d orbitals is a choice, not a divine commandment: in choosing them one is trying to choose a convenient basis of the 5-dimensional space of quadratic polynomials in x, y, and z whose Laplacian is zero. The most convenient choices do not have all 5 basis elements being just rotated versions of each other. There are bases of the space of d orbitals where all the basis elements are rotated versions of each other, but these are less convenient in calculations.

• Wolfgang says:

Hi. Well, I would say I doubt his work by 51% :). I think as a theoretical chemist he had a thorough understanding of quantum mechanics and would not write total nonsense in this regard, and certain plots he makes about trends in atomic and nucleon numbers, discussing element stability, relating them with the golden section and primes and the like, seem valid, too, because they are just empirical observations. When he starts to interpret them, however, he seems to go too far into some esoteric realm, emphasizing the role of prime numbers too much. I have to admit though that my critique is more based on some intuitive fishy feeling than the ability to really tell the true from the false, because in many cases one could say, it could be his explanation, which is true, but no one can prove or disprove it, so no one really knows.

Yes, the conversation above is about the same topic. I am happy that this seems to be a common point of interest for many people. In the chemistry textbooks I know this is not discussed.

Yes, that’s what I read. And then this confuses me. If one opens a textbook about chemistry one can find quite a lot of orbital diagrams explaining some chemical reactivity and spatial arrangements of atoms by the means of showing how certain orbital combinations can overlap and thus how certain shapes can interact. This is especially true for inorganic complex chemistry, where there is a metal atom interacting and bonding with some molecules in its periphery. But I always only find discussions with the standard set of orbital shapes. But shouldn’t it be sometimes the case that this is not the best choice? If the shapes are different, doesn’t this mean the probability of finding an electron is different, too? For me it looks like that such a free choice of technically equivalent constructions leads to different geometrical outcomes in the interpretation but this should not be the case (and possibly cannot, because of some clever argument).

• John Baez says:

But I always only find discussions with the standard set of orbital shapes. But shouldn’t it be sometimes the case that this is not the best choice?

I’m not a chemist. Sometimes it might not be the best choice. What’s really good about it is that these 5 orbitals are the eigenstates of orbital angular momentum along a specific axis. Conventionally that’s chosen to be the z axis, but if it were some other axis you’d just rotate all these orbitals, and I don’t think that’s the kind of different choice you’re talking about. If you were more concerned about some other observable—not angular momentum along a given axis—then you’d want to choose orbitals that are eigenstates of that other observable. And these could look really different.

What would that other observable be? Well, energy comes to mind. For example, if you have an atom with some electrons in d orbitals interacting with some other nearby atom, you’ll care about energy eigenstates, and they’ll depend on the details of the interaction.

If the shapes are different, doesn’t this mean the probability of finding an electron is different, too?

There’s a certain amount of subtlety in the pictures: are we drawing the absolute value of the wavefunction, whose square is the probability, or just the real part of the wavefunction, which has no significant physical meaning? But anyway: there are choices of 5 basis orbitals where the probabilities of finding electrons are different than in the conventional orbitals (and not just rotated versions, either). And these could be useful, as above.

• Wolfgang says:

I see. Yes, this is an important part I did not think of, that you will get different results for different choice of observable. In this way I am not even sure, if the pictures I have in mind were not already referring to the energy observable. Could be. It would make a lot of sense in terms of chemical reactivity. On the other hand, if the z direction is singled out for orbital angular momentum it also makes a lot of sense that the orbital corresponding to this axis is special, too. I think the main problem arises from the fact that many chemists in their education become used to these pictures as showing THE orbitals, forgetting about the details of what is depicted and how it is depicted. Thanks for clarifying this a little.

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