Course 2.3Lesson 4 of 4

Filling the orbitals: electron configurations

Electrons fill the orbitals by three rules. The Aufbau principle fills the lowest energy first. Pauli's exclusion principle allows at most two electrons in an orbital, with opposite spins. Hund's rule puts one electron in each orbital of a subshell before any pair. Chromium and copper are exceptions, because half-filled and full d subshells are extra stable.

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Three rules

To write the electron configuration of an atom, we place its electrons in the orbitals, one by one, by three rules.

1. The Aufbau principle (from the German word for building up). Electrons go first into the orbital with the lowest energy. The order is the one that we found in the last lesson:

A chart of the subshells 1s, 2s, 2p, 3s, 3p, 3d, up to 7p, in rows. Numbered orange arrows show the order of filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, and so on.

2. Pauli's exclusion principle. No two electrons in an atom can have the same set of all four quantum numbers. Two electrons in the same orbital already share $n$ , $l$ and $m_l$ . So their spins must be different. An orbital holds at most two electrons, with opposite spins. We say that they are paired.

3. Hund's rule. In a subshell with several orbitals of the same energy, the electrons go one in each orbital first, with parallel spins. They pair up only when every orbital has one. The reason is that electrons repel each other, so they stay apart if they can, and parallel spins lower the energy a little more.

How we write a configuration

We write each occupied subshell with the number of electrons as a small superscript. For example, $1s^2\ 2s^2\ 2p^3$ is nitrogen. For a heavier atom, we use the previous noble gas as a short form, for example $[\text{Ne}]\,3s^2 3p^5$ for chlorine.

Atom	$Z$	Configuration	Unpaired electrons
Carbon	$6$	$1s^2\ 2s^2\ 2p^2$	$2$
Nitrogen	$7$	$1s^2\ 2s^2\ 2p^3$	$3$
Oxygen	$8$	$1s^2\ 2s^2\ 2p^4$	$2$
Sodium	$11$	$[\text{Ne}]\,3s^1$	$1$
Chlorine	$17$	$[\text{Ne}]\,3s^2 3p^5$	$1$
Potassium	$19$	$[\text{Ar}]\,4s^1$	$1$
Scandium	$21$	$[\text{Ar}]\,3d^1 4s^2$	$1$
Iron	$26$	$[\text{Ar}]\,3d^6 4s^2$	$4$

Box diagrams show the spins. For oxygen, the four electrons in $2p$ go as one pair and two singles, and not as two pairs:

Orbital box diagrams. Nitrogen, 2p3: three boxes with one up arrow each. Oxygen, 2p4: the first box has an up and a down arrow, and the other two have one up arrow each. Chromium, 4s1 3d5: one up arrow in the 4s box, and one up arrow in each of the five 3d boxes. Copper, 4s1 3d10: one up arrow in 4s, and a pair of arrows in each of the five 3d boxes.

Two exceptions: chromium and copper

By the rules, chromium ( $Z = 24$ ) should be $[\text{Ar}]\,3d^4 4s^2$ , and copper ( $Z = 29$ ) should be $[\text{Ar}]\,3d^9 4s^2$ . The real configurations are

\displaystyle \text{Cr}: [\text{Ar}]\,3d^5 4s^1 \qquad \text{Cu}: [\text{Ar}]\,3d^{10} 4s^1

An electron moves from $4s$ to $3d$ . A half-filled ( $d^5$ ) or completely filled ( $d^{10}$ ) subshell is specially stable, for two reasons. The electrons are arranged symmetrically, and electrons with the same spin in different orbitals can exchange places, which lowers the energy (the exchange energy). The $4s$ and $3d$ orbitals are close in energy, so the move costs very little, and the gain is larger. Molybdenum and silver behave in the same way in the next row.

Ions

Anions gain electrons in the next empty orbitals. $\text{Cl}^-$ is $[\text{Ne}]\,3s^2 3p^6$ , which is the configuration of argon. $\text{O}^{2-}$ is $1s^2\ 2s^2\ 2p^6$ , the configuration of neon.
Cations lose electrons from the outermost shell first, the one with the highest $n$ , and not from the orbital that filled last. Iron is $[\text{Ar}]\,3d^6 4s^2$ . $\text{Fe}^{2+}$ loses the two $4s$ electrons, so it is $[\text{Ar}]\,3d^6$ . $\text{Fe}^{3+}$ loses one more, from $3d$ , and is $[\text{Ar}]\,3d^5$ , which is half-filled. This is why iron(III) is a stable ion.

Unpaired electrons and magnetism

An atom or ion with unpaired electrons is paramagnetic: a magnet attracts it. If all the electrons are paired, it is diamagnetic, and it is weakly pushed away. The strength of the magnetism can be found from the number of unpaired electrons $n$ . The spin-only magnetic moment is

\displaystyle \mu = \sqrt{n(n + 2)}\ \text{BM}

where BM is the Bohr magneton, the unit. For $\text{Fe}^{3+}$ , $n = 5$ , so $\mu = \sqrt{5 \times 7} = 5.92\ \text{BM}$ . For $\text{Zn}^{2+}$ ( $3d^{10}$ ), $n = 0$ , and $\mu = 0$ .

You should find that the moment grows a little less than in proportion to the number of unpaired electrons.

Practise with new numbers

Try it in Python

This program builds the configuration of an atom with the $n + l$ order.

Remember these

Aufbau: lowest energy first. Pauli: at most two electrons in an orbital, with opposite spins. Hund: singly in the orbitals of a subshell first.
Write a configuration with superscripts, and use the noble gas short form.
Cr is $[\text{Ar}]\,3d^5 4s^1$ and Cu is $[\text{Ar}]\,3d^{10} 4s^1$ , because half-filled and full subshells are extra stable.
A cation loses its outermost electrons first: $\text{Fe}^{2+}$ is $[\text{Ar}]\,3d^6$ .
Unpaired electrons give paramagnetism, with $\mu = \sqrt{n(n+2)}\ \text{BM}$ .