Electron configuration and quantum numbers: the method

It is the end of September 2026: in Year 3 (the third year of Italian high school) at the liceo linguistico (language high school), science has just restarted with reactions and solutions, and in October electron configuration arrives — with the test typically falling between the end of October and November, so between five and eight weeks from now. With 2 hours a week and a programme that in Year 3 also has to cover inorganic nomenclature, Mendelian genetics and DNA, electron configuration is seen only once: there is no second round in January to put things right. At the science high school the same topic sits inside a broader block (atomic models, periodic table, bonds) and I have already talked about it in Chemistry in Year 3 in the human sciences: atom and bonds, which however stops at an introductory level. Here I go deeper, precisely on quantum numbers and on writing the configuration, because that is the point where most tests are decided.

From Bohr’s model to orbitals

Bohr’s model — electrons on circular orbits at a fixed distance from the nucleus — is probably the one in which you first met the atom, and it is useful for getting an idea, but it is not the one you need to write a correct electron configuration. The current model doesn’t talk about orbits but about orbitals: regions of space around the nucleus where the probability of finding an electron is high. An orbital is not a path, it is a “cloud” of probability with a precise shape, and each orbital can hold at most two electrons.

The four quantum numbers

Every electron in an atom is described by four quantum numbers, and no other electron in the same atom can have the same combination of the four (this is Pauli’s exclusion principle, which you will see shortly).

• Principal quantum number, $n$: indicates the energy level and is a positive integer, $n=1,2,3,\dots$. The higher it is, the further the electron is from the nucleus and the higher its energy.
• Secondary (or azimuthal) quantum number, $l$: indicates the type of orbital (its shape), and ranges from $0$ to $n-1$. $l=0$ corresponds to an $s$ orbital, $l=1$ to a $p$ orbital, $l=2$ to a $d$ orbital, $l=3$ to an $f$ orbital.
• Magnetic quantum number, $m_l$: indicates the orientation of the orbital in space, and ranges from $-l$ to $+l$, including zero. For $l=1$ ($p$ orbitals), $m_l$ can be $-1,0,+1$: three $p$ orbitals oriented along three different directions.
• Spin quantum number, $m_s$: indicates the direction of the electron’s rotation on itself, and can only be $+\frac{1}{2}$ or $-\frac{1}{2}$.

A combination that is not allowed — the most common mistake when an exercise asks you to check or build a set of quantum numbers — is one with $l\ge n$: for $n=2$, $l$ can only be $0$ or $1$, never $2$. Another combination that is not allowed is $m_l$ outside the range $[-l,+l]$.

Shape and number of orbitals

• The s subshell ($l=0$) has 1 orbital (spherical), so $2$ electrons at most.
• The p subshell ($l=1$) has 3 orbitals (double-lobed, along three axes), so $6$ electrons at most.
• The d subshell ($l=2$) has 5 orbitals, so $10$ electrons at most.
• The f subshell ($l=3$) has 7 orbitals, so $14$ electrons at most.

The method for writing the electron configuration

Step 1 — Pauli’s principle. At most two electrons can sit in the same orbital, and they must have opposite spin ($m_s=+\frac{1}{2}$ and $m_s=-\frac{1}{2}$). That is why each s, p, d, f subshell has 2, 6, 10, 14 places respectively and no other numbers.

Step 2 — the Aufbau principle (the filling order). Orbitals are filled in order of increasing energy, and energy increases with $n+l$: for equal $n+l$, the subshell with the lower $n$ wins (that is, it is filled first). In practice the order is:

The point almost everyone trips on: 4s is filled before 3d, even though the number “3” seems smaller than “4”. Applying $n+l$: for 4s, $n+l=4+0=4$; for 3d, $n+l=3+2=5$. 4s has the lower $n+l$, so lower energy, so it is filled first. A diagonal diagram (the arrows that cross the grid with rows for $n$ and columns for $l$, from top left to bottom right) is the quickest way of not having to remember it by heart: follow the diagonals one after another and you get exactly the order above.

Step 3 — Hund’s rule. When there are several orbitals of the same energy (for example the three $2p$ orbitals), electrons occupy them one at a time, all with the same spin, before starting to pair up. Filling one orbital with two electrons first and leaving others empty, when there are free ones at the same energy, is wrong.

Step 4 — box notation (orbitals) or compact form. The box notation shows each orbital as a little square with arrows (↑ and ↓) for the electrons; the compact notation writes only the subshell and the exponent, for example $1s^{2}2s^{2}2p^6$. For a written exam the second is almost always enough, but knowing how to draw the first helps you not get the filling order wrong in a $p$ or $d$ subshell.

Step 5 — abbreviated configuration. For elements with many electrons, you write the symbol of the noble gas preceding the element in square brackets, followed only by the subsequent electrons. For calcium ($Z=20$): instead of $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^2$, you write $[\mathrm{Ar}]4s^2$, because argon ($Z=18$) has exactly that configuration.

Valence electrons and position in the periodic table

Valence electrons are those of the last occupied energy level (the last $n$), plus the $d$ electrons in the process of filling when applicable. They are the ones that matter in chemical bonds, and they are also the key to reading the periodic table without learning it by heart:

• the period (row) in which an element sits is the maximum value of $n$ in its configuration;
• the group, for the representative elements, is linked to the number of electrons in the last s and p level;
• the block (s, p, d, f) tells you which subshell is being filled last: the first two groups are the s block, the last six main groups are the p block, transition metals are the d block, lanthanides and actinides are the f block.

Ions: 4s empties before 3d

When a transition metal forms a positive ion, the electrons don’t leave in the reverse order from the one in which they arrived: they are removed first from the subshell with the highest $n$, so from 4s before 3d, even though 4s had been filled first. It is the most frequent mistake on this topic: applying Aufbau “backwards” and removing electrons from 3d before 4s, ending up with an ion with the wrong configuration.

The exceptions: chromium and copper

Two elements don’t follow Aufbau to the letter: chromium ($Z=24$), which has configuration $[\mathrm{Ar}]4s^{1}3d^5$ instead of the “expected” $4s^{2}3d^4$, and copper ($Z=29$), which has $[\mathrm{Ar}]4s^{1}3d^{10}$ instead of $4s^{2}3d^9$. The reason is that a half-filled $d$ subshell (5 electrons) or completely filled one (10 electrons) is particularly stable, and it is energetically “worthwhile” to move an electron from 4s to 3d to reach that configuration. At this level it is enough to know them as exceptions to remember, without having to justify them in detail in a test.

The typical mistake, in brief

The quickest way to lose marks on this topic is to fill 3d before 4s (either when writing the configuration of a neutral atom, or — more often — by removing the wrong electrons to form an ion), or to miscount the total electrons when going from atom to ion (a $2+$ ion has two electrons fewer than the neutral atom, not two more). Always check, at the end, that the sum of the exponents in the configuration matches the number of electrons you expect.

Exercise 1 — configuration of iron and of the $\mathrm{F}{\mathrm{e}}^{3+}$ ion

Iron has $Z=26$, so 26 electrons to place in the neutral atom.

Following the filling order up to 26 electrons:

Check: $2+2+6+2+6+2+6=26$. Correct. In abbreviated form: $[\mathrm{Ar}]4s^{2}3d^6$.

For the $\mathrm{F}{\mathrm{e}}^{3+}$ ion you have to remove three electrons from the neutral atom, and as said above they are removed first from the subshell with the highest $n$: first the two electrons of 4s, then one of 3d.

Electron check: neutral atom 26, minus 3 (charge $3+$) = 23; in the configuration, $18$ (argon) $+5=23$. Correct.

Exercise 2 — quantum numbers of the last electron of bromine

Bromine has $Z=35$. The full electron configuration:

Check: $2+2+6+2+6+2+10+5=35$. Correct.

The last electron added sits in the $4p^5$ subshell: it is the fifth electron of that subshell.

• $n=4$ (the subshell is 4p).
• $l=1$ (p corresponds to $l=1$).
• For $l=1$, $m_l$ can be $-1,0,+1$: three orbitals. By Hund’s rule, the first three electrons occupy one orbital each ($m_l=-1,0,+1$) with parallel spin; the fourth and fifth electrons start to pair up, in the order $m_l=-1$, then $m_l=0$. The fifth electron pairs up in the orbital with $m_l=0$.
• $m_s=-\frac{1}{2}$, because in the $m_l=0$ orbital there was already an electron with $m_s=+\frac{1}{2}$ (the first of the three “unpaired” electrons), and by Pauli the second electron in that orbital must have opposite spin.

Set of numbers: $n=4,l=1,m_l=0,m_s=-\frac{1}{2}$.

Exercise 3 — recognising an element and its position from the configuration

An element has configuration $[\mathrm{Ne}]3s^{2}3p^4$. Which element is it, and what are its group and period?

Neon ($Z=10$) has 10 electrons. Adding $2+4=6$ electrons, the element has $Z=16$: it is sulphur.

The period is given by the maximum value of $n$ in the configuration: here $n=3$, so third period.

The group: sulphur has 6 valence electrons ($3s^{2}3p^4$, that is $2+4=6$), and is a p-block element with valence configuration $n{s}^{2}n{p}^4$: it belongs to group 16 (VIA in the traditional notation), the same as oxygen, with which it indeed shares several chemical properties.

How to practise after this post

There is not yet a dedicated chemistry practice test: you can find timed tests on other subjects at high school practice tests. For chemistry, the most effective way remains repeating configurations of different elements (with and without exceptions) until the diagonal diagram is no longer needed to remember the order.

The next step, once electron configuration has settled, is to understand how atoms are arranged in space in a molecule: I talk about it in Molecular geometry and VSEPR: the guide. If instead the October test is about the names of compounds, you will find the method in Chemical nomenclature: traditional and IUPAC without confusing them, and for calculations with moles in Mole and stoichiometry: the method in Year 3.

For other subjects and topics you will find everything in sciences. If after this post you still have doubts about electron configuration or quantum numbers, you can book a free 30-minute introductory call: it is the quickest way to find out where you really get stuck.