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I read that the electronic configuration of uranium is [Rn] 5f³ 6d¹ 7s² . Given that the subshells fill in the order 5f --> 6d, why is the 5f subshell only partially filled? Why do electrons fill the 5f subshell partially then proceed to fill the 6d subshell?

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    $\begingroup$ Nobody really knows. $\endgroup$ Commented May 2, 2023 at 17:50
  • $\begingroup$ @OscarLanzi maybe when we will get a chance to simulate atoms using quantum computer we will realize why it is what it is. $\endgroup$
    – Volpina
    Commented May 2, 2023 at 23:58
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    $\begingroup$ @Volpina that will depend on how well developed is the theory with which we program the computers. $\endgroup$ Commented May 3, 2023 at 1:30
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    $\begingroup$ @OscarLanzi IBM's quantum computer is already at 500+ qubits so if each qubit represents 1 electron then we already are very capable of simulating such atoms but I am not a expert in quantum computing so I could be wrong. $\endgroup$
    – Volpina
    Commented May 3, 2023 at 6:44
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    $\begingroup$ @Volpina We aren't either. We need to effectively translate that information into what we, as Chemistry folks, could understand. $\endgroup$ Commented May 3, 2023 at 9:50

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Nobody really knows. Our theory of how electrons are ordered in the ground states of atoms is a case of good news and bad news. And, as in human endeavors, the bad news ends up getting the headlines.

First, the good news. Electrons do occupy shells (numbered $n$ in typical nonenclature) in perfect order, just as Bohr imagined -- when we look only at each individual level of angular momentum ($l$). Among $s$ orbitals ($l=0$) we have $1s$ at the lowest level, then $2s$, then $3s$ and so on. With one unit of angular momentum, which we don't have until the second shell, the energies are $2p<3p<4p<...$. And so on with $d$ orbitals ($l=2$), $f$ orbitals ($l=3$), etc.

The bad news creeps in when we try to consider variation in $n$ as well as $l$; that is, when we try to consider more than one of the angular-monentum-based tiers of shells at the same time. We know that with one electron, the energy level depends on just the shell number $n$; thus both $2p$ as well as $2s$ will be at higher energy than $1s$, and $4s$ has to sit above $3p$ and $3d$ as well as $3s$. But with multiple electrons, the shielding effects of electron-electron repulsions causes higher angular momentum states of each shell to rise above lower angular momentum states; we have $3d>3p>3s$ as well as $4s>3s$, but these relationships may allow, let us say, $3d>4s$ even though $3<4$. We get neutral iron atoms with the $4s$ orbital filled while $3d$ orbitals are left short, or worse yet a situation where the clashing orbital levels are balanced in such a way that filled and empty orbitals are most stably distributed between both (chromium = $\ce{[Ar]}3d^54s^1$).

We might explain the configurations of chromium and copper in terms of exchange energy or spherical symmetry, but such arguments cannot begin to cope with the fifth-period elements, let alone subsequent periods when $f$ orbitals come into play. By the time we get to uranium we are guessing as much as knowing, even with computer outputs whose underlying programs are also based on educated guesses.

There is one way to simplify things — a little. If we add extra protons (and any needed neutrons for nuclear stability) to the nucleus and thus form a positive ion, we tend to push the electrons back towards the shell-by-shell order that the energy levels of a single electron would obey. This is because the extra nuclear charge breaks through the electron-electron repulsive shielding. Even a small ionic charge can have a large effect on valence electrons, whose effective nuclear charge is also relatively small in the neutral atoms of most metals. The case of $24$ electrons is described in both Jan's and my own answers to this question: adding just two extra positive charges to the nucleus converts the $\ce{[Ar]}3d^54s^1$ of $\ce{Cr^0}$ to $\ce{[Ar]}3d^6$ of $\ce{Fe^{2+}}$. When we put in the full electronic configuration of the argon core, we find that the extra protons in the ferrous ion have pushed all the electrons into shell-by-shell order.

Similarly, with $88$ electrons the configuration of $\ce{Ra^0}$ is $\ce{[Rn]}7s^2$ whereas that of $\ce{U^{4+}}$ (a common oxidation state of uranium) is $\ce{[Rn]}5f^2$. As with the ferrous ion, the valence electrons have dropped into a lower shell under the influence of the ionic charge on the uranium. However, with the heavier atoms we still have lower angular momentum states ($s,p$) occupied up to the sixth shell even as higher angular momentum states fail to fill the fifth shell. The radon core, containing electrons in the low-angular momentum $6s$ and $6p$ subshell but none in the higher angular-momentum $5g$, is itself not in shell-by-shell order so the idea that it holds up in seventh-period metal ions is also sonewhat of a guess. So the shell-by-shell ordering remains imperfect even for chemically accessible ions in this case and generally beyond the fourth period.

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Uranium has the electronic configuration $\ce{[Rn] 5f^3 6d^1 7s^2}$. You might alternately ask as to why it doesn't have the configuration $\ce{[Rn] 5f^4 6d^0 7s^2}$ if it should follow Aufbau principle.

But remember one more fact that d orbitals are slightly better shielders than f orbitals. Now the problem with $\ce{[Rn] 5f^4 6d^0 7s^2}$ is that there are no d electrons which causes the effective nuclear charge $(Z_{eff})$ to increase on the $\ce {7s^2}$ electrons. But this increased $Z_{eff}$ would tend to draw $\ce{7s^2}$ electrons toward the nucleus and thus decreasing the stability of a large orbital of $\ce{7s^2}$ and the electrons in it.

On the other hand, $\ce{[Rn] 5f^3 6d^1 7s^2}$ contains one d orbital which will, with it's slightly better screening, would decrease $Z_{eff}$ on $\ce{7s^2}$ electrons and thus keeping it distant from the nucleus and hence stabilizing it.

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I am sure you are familiar with the rules for assigning electron orbitals, I will briefly descibe them here:

Electrons fill orbitals in a way to minimize the energy of the atom. Therefore, the electrons in an atom fill the principal energy levels in order of increasing energy (the electrons are getting farther from the nucleus). The order of levels filled looks like this:

Afbau build up

Pauli Exclusion Principle

The Pauli exclusion principle states that no two electrons can have the same four quantum numbers. The first three (n, l, and ml) may be the same, but the fourth quantum number must be different. A single orbital can hold a maximum of two electrons, which must have opposing spins; otherwise they would have the same four quantum numbers, which is forbidden.

Hund's Rule

When assigning electrons in orbitals, each electron will first fill all the orbitals with similar energy (also referred to as degenerate) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. This explains the behaviour of Chromium: Z:24 [Ar] 3d54s1 (note here the the one electron in 4s orbital while the d orbitals are occupied with single electrons of one spin direction)

Exceptions

Although the Aufbau rule accurately predicts the electron configuration of most elements, there are notable exceptions among the transition metals and heavier elements. The reason these exceptions occur is that some elements are more stable with fewer electrons in some subshells and more electrons in others and a notable example is uranium, for it to acquire maximum stability is usually having this ground state: Uranium: Z:92 [Rn] 7s2 5f3 6d1

References

  1. Rules for Assigning Electron Orbitals
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  • $\begingroup$ Ah so uranium is an exception to this rule. What is it about this specific configuration that makes it so stable? $\endgroup$ Commented Mar 17, 2017 at 15:30
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    $\begingroup$ Its not only Uranium, read again I have also mentioned Chromium. There are other elements for example Copper, Niobium, Palladium, Silver, Thorium etc which deviate from this trend. The reason like described is partly based on the combination of the rules. Remember that in a ground state of an element the electron configuration has its lowest energy. The lower the energy the more the stability. In some cases this kind of stability can only be obtained when there are fewer electrons in a particular orbital say Uranium configuration. $\endgroup$ Commented Mar 17, 2017 at 15:35

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