C. Structure and Properties of Matter · Structure and Properties of MatterWriting Electron Configurations: Aufbau, Pauli and Hund
C2.2 C3.2About 75 minutes
Learning goals- I can use the aufbau principle, the Pauli exclusion principle and Hund's rule to write electron configurations.
- I can write full and condensed configurations and orbital box diagrams for atoms and ions.
- I can explain the exceptions for chromium and copper and remove 4s electrons first for transition-metal ions.
- I can predict whether an atom or ion is paramagnetic or diamagnetic.
Words to know
electron configuration
A list of the orbitals occupied by an atom's electrons, such as 1s² 2s² 2p³.
aufbau principle
Electrons fill the lowest-energy available orbitals first.
Hund's rule
In orbitals of equal energy, electrons occupy separate orbitals with parallel spins before pairing.
valence electrons
Electrons in the highest occupied main energy level (for main-group elements), which take part in bonding.
isoelectronic
Having the same number of electrons and the same configuration, such as Na⁺, Ne and F⁻.
paramagnetic / diamagnetic
Attracted into a magnetic field because of unpaired electrons / weakly repelled because all electrons are paired.
Filling the address book
You now know the "addresses" (orbitals) available to electrons. An electron configuration lists which orbitals an atom's electrons actually occupy in the ground state. For example, sodium is 1s² 2s² 2p⁶ 3s¹.
In each term such as 2p⁶, the 2 is n, the p is the sublevel, and the superscript 6 is the number of electrons in that sublevel.
Remember from Grade 11: you arranged the first 20 elements' electrons in shells of 2, 8, 8, 2. Electron configurations add the sublevel detail that explains why the fourth row of the periodic table suddenly gets 10 extra elements (the transition metals).
Three rules
- Aufbau principle (German for "building up"): electrons fill the lowest-energy orbitals first.
- Pauli exclusion principle: an orbital holds at most two electrons, with opposite spins.
- Hund's rule: when electrons fill orbitals of equal energy (such as the three 2p orbitals), they go into separate orbitals with parallel spins first, and pair up only when every orbital in the sublevel has one electron. (Electrons repel each other, so they spread out.)
The filling order
In a multi-electron atom, sublevels in the same level have different energies: s < p < d < f. As a result the levels overlap, and 4s is lower in energy than 3d for the atoms where it is being filled. The order is:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p
A handy rule: sublevels fill in order of increasing n + l; when two have the same n + l, the one with lower n fills first (3d has n + l = 5, the same as 4p, but 3d has lower n, so 3d comes first).
The diagonal (aufbau) diagram gives the order in which sublevels fill. Each arrow runs from upper right to lower left; read the arrows in order from the top.Three ways to write a configuration
- Full configuration: every sublevel. Phosphorus (15 electrons): 1s² 2s² 2p⁶ 3s² 3p³.
- Condensed (noble-gas core) configuration: replace the core with the previous noble gas in square brackets. Phosphorus: [Ne] 3s² 3p³. The electrons after the core in the highest n level are the valence electrons (5 for phosphorus).
- Orbital box diagram: one box per orbital, arrows for electrons. This is the only form that shows unpaired electrons.
Orbital box diagrams. Each box is one orbital; arrows are electrons with opposite spins. In a sublevel, electrons spread out one per orbital, with parallel spins, before any pairing.Ions
- Anions: add electrons in the normal filling order. Cl (17 e⁻): [Ne] 3s² 3p⁵, so Cl⁻ (18 e⁻) is [Ne] 3s² 3p⁶, the same as [Ar]. Atoms and ions with the same configuration are isoelectronic.
- Cations of main-group elements: remove the outermost electrons. Na⁺ is 1s² 2s² 2p⁶, isoelectronic with neon.
- Transition-metal cations: remove the 4s electrons before the 3d. Once 3d starts to fill, 4s electrons are the outermost (they are farther from the nucleus on average). Iron is [Ar] 4s² 3d⁶; Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵.
Two famous exceptions
Chromium is [Ar] 4s¹ 3d⁵, not [Ar] 4s² 3d⁴. Copper is [Ar] 4s¹ 3d¹⁰, not [Ar] 4s² 3d⁹. A half-filled (d⁵) or completely filled (d¹⁰) d sublevel is especially stable, and 4s and 3d are so close in energy that moving one electron pays off. Elements below them (molybdenum, silver, gold) show similar patterns. Use the regular rules for everything else in this course.
Magnetism: a test of the rules
A substance with unpaired electrons is paramagnetic (pulled into a magnetic field). If all electrons are paired, it is diamagnetic. Measurements agree with the predictions: Zn ([Ar] 4s² 3d¹⁰, all paired) is diamagnetic, while Fe³⁺ ions (3d⁵, five unpaired electrons) are strongly paramagnetic.
Worked exampleExample 1: Full and condensed configurations
Write the full and condensed electron configurations of sulfur (Z = 16) and state its number of valence electrons.
- Count 16 electrons into the sublevels in order: 1s² (2), 2s² (4), 2p⁶ (10), 3s² (12), 3p⁴ (16).
- Full: 1s² 2s² 2p⁶ 3s² 3p⁴.
- The previous noble gas is neon (10 electrons): [Ne] 3s² 3p⁴.
- Valence electrons are those in level 3: 2 + 4 = 6.
1s² 2s² 2p⁶ 3s² 3p⁴ = [Ne] 3s² 3p⁴, with 6 valence electrons (group 16).
Worked exampleExample 2: A transition metal and its ion
Write the condensed configuration of iron (Z = 26) and of Fe³⁺. Draw the 3d box diagram for Fe³⁺ and give the number of unpaired electrons.
- Argon has 18 electrons, so 8 remain: 4s² (20), then 3d⁶ (26). Fe = [Ar] 4s² 3d⁶.
- Fe³⁺ loses 3 electrons: the two 4s electrons first, then one 3d. Fe³⁺ = [Ar] 3d⁵.
- Hund's rule: five 3d electrons go one per orbital: [↑][↑][↑][↑][↑].
Fe: [Ar] 4s² 3d⁶; Fe³⁺: [Ar] 3d⁵, with 5 unpaired electrons (strongly paramagnetic).
Worked exampleExample 3: Spot the error
Each configuration below is wrong. Explain and correct it. (a) Ca: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² (b) N (box diagram): 2p [↑↓][↑][ ] (c) Cu: [Ar] 4s² 3d⁹
- (a) 4s fills before 3d. Ca is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² = [Ar] 4s².
- (b) Hund's rule: the three 2p electrons must occupy separate orbitals with parallel spins: [↑][↑][↑].
- (c) Copper is an exception: [Ar] 4s¹ 3d¹⁰, with a full d sublevel.
See each step for the corrected configuration.
Checks that save marks
- Check your total: the superscripts must add up to the number of electrons (Z for a neutral atom).
- Many textbooks write configurations in level order (Fe = [Ar] 3d⁶ 4s²); others in filling order ([Ar] 4s² 3d⁶). Both are accepted; just be consistent.
- For a transition-metal ion, always remove 4s before 3d.
Just for funElectrons follow Hund's rule like passengers on an almost empty city bus: everyone takes an empty double seat first, and only when every seat has someone in it does anyone sit beside a stranger.
Hands-on activityThe bus-seat model of Hund's rule
You need: a muffin tin or 7 small cups or bowls, 14 coins or buttons, and a pencil and paper. Grown-up note: keep small objects away from young children.
- Line up 5 cups for a d sublevel (each cup is one orbital; each holds two coins: one heads up for ↑, one tails up for ↓).
- Fill the cups for d¹ through d¹⁰, one coin at a time, following Hund's rule: one heads-up coin in each cup before any cup gets a second (tails-up) coin.
- For each step, record the number of unpaired coins in a table (d¹, d², ... d¹⁰).
- Make a quick bar graph of unpaired electrons against the number of d electrons.
What to look for: the number of unpaired electrons rises 1, 2, 3, 4, 5 and then falls 4, 3, 2, 1, 0. The peak is at d⁵, the half-filled sublevel. Transition-metal ions with configurations near d⁵ (such as Mn²⁺ and Fe³⁺) are the most strongly paramagnetic, and Zn²⁺ (d¹⁰) is diamagnetic. Repeat with 3 cups for a p sublevel: the peak is at p³ (nitrogen).
Your turn
1Write the full configuration of silicon (Z = 14).
2Write the condensed configuration of bromine (Z = 35).
[Ar] 4s² 3d¹⁰ 4p⁵ (or [Ar] 3d¹⁰ 4s² 4p⁵). It has 7 valence electrons (4s² 4p⁵).
3Write the condensed configuration of titanium (Z = 22).
4Draw the 2p box diagram for fluorine and give the number of unpaired electrons.
2p⁵: [↑↓][↑↓][↑], 1 unpaired electron.
5Write the configuration of S²⁻. Which noble gas is it isoelectronic with?
S has 16 e⁻; S²⁻ has 18: 1s² 2s² 2p⁶ 3s² 3p⁶, isoelectronic with argon.
6Write the configuration of Zn²⁺ and say whether it is paramagnetic or diamagnetic.
Zn = [Ar] 4s² 3d¹⁰; remove the two 4s electrons: [Ar] 3d¹⁰. All electrons paired: diamagnetic.
7Which element has the configuration [Kr] 5s² 4d¹⁰ 5p²?
Count: 36 + 2 + 10 + 2 = 50: tin (Sn).
More practice: warm-up
1Write the configuration of lithium.
2Write the configuration of neon.
3Which sublevel fills right after 3p?
4How many electrons fit in a d sublevel?
5Write the condensed configuration of potassium.
More practice: on your own
1Write the condensed configuration of manganese (Z = 25) and its number of unpaired electrons.
[Ar] 4s² 3d⁵; the five 3d electrons are all unpaired (Hund's rule): 5 unpaired.
2Write the configuration of Cu⁺ and Cu²⁺.
Cu = [Ar] 4s¹ 3d¹⁰. Cu⁺ loses the 4s electron: [Ar] 3d¹⁰. Cu²⁺ also loses one 3d: [Ar] 3d⁹.
3Which of these is diamagnetic: O, Mg, N, Fe²⁺? Explain.
Mg ([Ne] 3s²): all electrons are paired. O has 2 unpaired, N has 3, and Fe²⁺ (3d⁶) has 4.
4Give the condensed configuration of iodine (Z = 53).
[Kr] has 36; 17 remain: [Kr] 5s² 4d¹⁰ 5p⁵.
5Write the full set of four quantum numbers for the last electron added to aluminium (1s² 2s² 2p⁶ 3s² 3p¹), taking the first electron in a sublevel to have mₗ = −l and mₛ = +½.
The last electron is the first 3p electron: n = 3, l = 1, mₗ = −1, mₛ = +½ (other mₗ choices are acceptable if a different convention is used).
Challenge
1Predict the configuration of silver (Z = 47), using the copper exception as a guide.
Like copper (same group): [Kr] 5s¹ 4d¹⁰, with a full 4d sublevel. (The regular rules would give [Kr] 5s² 4d⁹.)
2An ion X²⁺ has the configuration [Ar] 3d⁷. Identify X and give its neutral configuration.
X has 2 more electrons: 18 + 7 + 2 = 27, so X is cobalt: [Ar] 4s² 3d⁷. (The 2+ ion loses the 4s electrons.)
3Use the n + l rule to predict which sublevel fills first, 6s or 4f, and which fills first, 4f or 5d.
6s: n + l = 6 + 0 = 6; 4f: 4 + 3 = 7; 5d: 5 + 2 = 7. So 6s fills before 4f. 4f and 5d tie at 7, and the lower n fills first: 4f before 5d.
Check your understandingQuiz
1. What is the electron configuration of chlorine (Z = 17)?
2. Which rule says electrons occupy equal-energy orbitals singly before pairing?
3. What is the configuration of Fe²⁺?
4. Which sublevel is filled immediately after 4s?
5. How many unpaired electrons does a ground-state oxygen atom have?
6. Which particle is isoelectronic with neon?
For parents and teachersBig idea (C2.2, C3.2): three rules (aufbau, Pauli, Hund) predict how electrons fill orbitals, and the resulting configurations explain magnetism, ion charges and the layout of the periodic table.
How to help: give your teen an element number and ask for the condensed configuration, then check the total of the superscripts. Ask "how many unpaired electrons?" to test Hund's rule.
Common difficulties:
- Filling 3d before 4s, or removing 3d electrons before 4s when forming transition-metal ions.
- Pairing electrons too early in box diagrams.
- Superscripts that do not add up to the electron count, especially for ions.
- Over-applying the Cr and Cu exceptions to other elements.