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Educerie · IB Diploma · Chemistry
Structure 1 Models of the particulate nature of matter · S1.3 Electron configurations
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Relate colour, wavelength, frequency and energy across the electromagnetic spectrum, qualitatively | SL, HL | Paper 1A: "Which has the highest energy per photon?" |
| Distinguish a continuous spectrum from a line spectrum | SL, HL | "Distinguish between a continuous spectrum and a line spectrum" (1 to 2 marks) |
| Describe the hydrogen emission spectrum and link its lines to transitions ending on n = 1, 2 and 3 | SL, HL | "Explain how the lines in the visible spectrum of hydrogen arise, and why they converge" (3 marks) |
| Deduce the maximum number of electrons in a main energy level from 2n² | SL, HL | Paper 1A, 1 mark |
| Recognise the shape of an s orbital and the three p orbitals | SL, HL | "Sketch the shape of a p orbital" (1 mark) |
| Apply the Aufbau principle, Hund's rule and the Pauli exclusion principle to write full and condensed configurations and orbital diagrams for atoms and ions up to Z = 36, including Cr and Cu | SL, HL | "Deduce the electron configuration of Fe³⁺" (1 mark); orbital diagram (2 marks) |
| Explain the trends and the two discontinuities in first ionization energy across a period and down a group | HL only | "Explain why the first ionization energy of sulfur is lower than that of phosphorus" (2 marks) |
| Calculate a first ionization energy from the wavelength or frequency of a convergence limit | HL only | Paper 2, 2 to 3 marks, using E = hf and c = λf from the data booklet |
| Deduce the group of an element from successive ionization energy data | HL only | Paper 1B or Paper 2, 2 marks |
Before you start
You need S1.2: an atom has a positive nucleus with Z protons and, if neutral, Z electrons around it. For the HL calculation you need standard form on a calculator and the data booklet, which gives the Planck constant h, the speed of light c, the Avogadro constant, the equations E = hf and c = λf, a chart of the electromagnetic spectrum and a table of first ionization energies.
1The idea in one paragraph
Electrons in an atom can have only certain fixed energies, called energy levels, like the rungs of a ladder. When an electron drops from a higher rung to a lower one, the atom gives out a single packet of light, a photon, whose energy is exactly the size of the drop. Because only certain drops are possible, an atom gives out only certain colours of light, and that pattern of lines is the evidence that energy levels exist. Each main level is split into sublevels (s, p, d, f), each sublevel into orbitals, and each orbital holds at most two electrons. Three rules tell you how electrons fill these spaces, and the result, the electron configuration, explains most of the chemistry in the rest of the course.
2Light: wavelength, frequency, energy
Light is electromagnetic radiation, and visible light is a narrow band of a much wider electromagnetic spectrum. Figure 1 lays it out; the data booklet has a similar chart.
Wavelength, λ, is the distance between successive peaks of the wave (visible light is quoted in nanometres, 1 nm = 10⁻⁹ m). Frequency, f, is the number of waves passing a point each second, in hertz (s⁻¹). All electromagnetic radiation travels at the same speed, so shorter wavelength means higher frequency; and a photon's energy is proportional to its frequency, so higher frequency means more energy. Violet light therefore has a shorter wavelength, higher frequency and more energy per photon than red, and ultraviolet has more still. From low to high energy the visible colours run red, orange, yellow, green, blue, violet. The equations behind this, c = λf and E = hf, are in the data booklet; at SL you need only the relationships.
3Continuous spectra and line spectra
Pass white light from a lamp through a prism and it spreads into a continuous spectrum: every wavelength across the visible range is present, and the colours merge into one another with no gaps.
Now pass an electric current through hydrogen gas in a discharge tube, so the gas glows, and look at that light through a prism. You do not get a rainbow. You get a line spectrum: a few sharp, bright lines of particular colours on a dark background. Figure 2 sets the two side by side.
A continuous spectrum contains all wavelengths in a range. A line spectrum contains only certain separate wavelengths.
A line emission spectrum is produced when atoms are given energy (by heating or by an electric discharge), so that electrons move up to higher energy levels, called excited states. The excited electrons then fall back to lower levels, and each fall emits a photon whose energy equals the energy difference between the two levels. Only certain differences exist, so only certain frequencies of light appear.
Every element has its own energy levels and so its own line spectrum, as individual as a fingerprint. That is how a spectrum identifies the elements in a sample, or in a star: helium was first identified from lines in the Sun's spectrum that matched no element known on Earth.
4The hydrogen spectrum and energy levels
Hydrogen has one electron, so its spectrum is the simplest, and it is the one the guide asks you to describe. Figure 3 shows its energy levels, numbered by the principal quantum number n, with n = 1 the lowest.
Each line is one downward transition. An electron falling from n = 3 to n = 2 emits a photon with exactly the energy of that gap: the red line at 656 nm.
Where the electron lands decides the region of the spectrum. Transitions ending on n = 1 are the largest drops, so they emit high-energy ultraviolet photons. Transitions ending on n = 2 emit visible light: the four lines of Figure 2 come from n = 3, 4, 5 and 6 falling to n = 2. Transitions ending on n = 3 are smaller still and emit infrared. You do not need the names of these series.
A bigger drop means a higher-energy line. Within the visible set, 3 → 2 is the smallest drop and gives the red line (lowest frequency); 6 → 2 is a bigger drop and gives the violet line (highest frequency).
The levels get closer together as n increases. The gap between n = 1 and n = 2 is large; the gap between n = 5 and n = 6 is tiny. The levels crowd together until they merge at n = ∞, the level at which the electron is no longer held by the atom at all.
That last point explains the most distinctive feature of the spectrum. Because the upper levels crowd together, the photons from 5 → 2, 6 → 2, 7 → 2 and so on differ in energy by less and less, so the lines they produce get closer and closer together at higher frequency until they merge. This is convergence, and Figure 4 plots it on a frequency scale.
The lines in an emission spectrum converge at higher frequency (higher energy, shorter wavelength) because the energy levels converge at higher energy.
An answer that says only "the lines get closer" describes the spectrum. The mark for explaining it is the reason: the energy levels themselves get closer together.
5Main energy levels, sublevels and orbitals
The hydrogen spectrum shows that electrons occupy main energy levels, n = 1, 2, 3 and so on. A more detailed model splits each main level into sublevels, labelled s, p, d and f, of successively higher energy within that level. Each sublevel is made of orbitals.
An orbital is a region of space in which there is a high probability of finding an electron. Each orbital holds at most two electrons, of opposite spin.
An orbital is not a path the electron follows. It is where the electron is likely to be found. "Spin" is a property of the electron with two possible values, drawn as an up arrow and a down arrow.
| Sublevel | Orbitals in it | Maximum electrons |
|---|---|---|
| s | 1 | 2 |
| p | 3 | 6 |
| d | 5 | 10 |
| f | 7 | 14 |
Each main level contains a fixed set of sublevels: n = 1 has only 1s; n = 2 has 2s and 2p; n = 3 has 3s, 3p and 3d; n = 4 has 4s, 4p, 4d and 4f. Adding up their capacities gives the maximum for each main level.
Maximum number of electrons in main level n = 2n²
Main level n also contains n² orbitals: one in n = 1, four in n = 2, nine in n = 3.
Shapes. You must recognise two shapes, drawn in Figure 5. An s orbital is a sphere centred on the nucleus, with no direction. A p orbital is a dumbbell, two lobes on opposite sides of the nucleus. There are three p orbitals in each p sublevel, identical in shape and energy but pointing along three axes at right angles: pₓ, py and pz (written px, py, pz).
6Writing electron configurations
An electron configuration lists the occupied sublevels in order, with the number of electrons in each as a superscript: 1s² 2s² 2p⁶ means two electrons in 1s, two in 2s and six in 2p. Three rules decide it.
The Aufbau principle. Electrons fill the lowest-energy orbitals available first. The order of sublevel energies is shown in Figure 6. The one surprise, for atoms up to Z = 36, is that 4s fills before 3d, because an empty 4s orbital is lower in energy than an empty 3d one.
Filling order: 1s 2s 2p 3s 3p 4s 3d 4p
The Pauli exclusion principle. An orbital holds at most two electrons, of opposite spins: one up arrow and one down arrow in a box, never two up.
Hund's rule. When electrons go into a sublevel with several orbitals of equal energy (p or d), they occupy the orbitals singly first, with parallel spins, and only pair up once every orbital has one. Electrons repel each other, so they spread out when they can.
Figure 7 puts all three rules into orbital diagrams (arrow-in-box diagrams), where each box is an orbital and the boxes of one sublevel sit side by side.
Nitrogen's three 2p electrons sit one in each p orbital, all spins parallel. Oxygen has one more, and it has nowhere to go but into an orbital that is already half full, so it pairs up. That pairing matters in HL section 7.
Full and condensed configurations. A full configuration lists everything. A condensed configuration replaces the inner electrons with the symbol of the previous noble gas in square brackets.
Writing 3d before 4s (as above) groups the sublevels by main level; writing 4s² 3d⁶ follows the filling order. Both are accepted. The count is what matters: the superscripts must add up to Z.
Two exceptions: chromium and copper. By the Aufbau order, chromium (Z = 24) should be [Ar] 3d⁴ 4s² and copper (Z = 29) should be [Ar] 3d⁹ 4s². In fact:
Cr: [Ar] 3d⁵ 4s¹ · Cu: [Ar] 3d¹⁰ 4s¹
In each, one 4s electron moves into 3d, giving a half-filled (Cr) or full (Cu) 3d sublevel. The 3d and 4s energies are so close that these arrangements give the lower-energy atom. These are the only two exceptions you need up to Z = 36.
Ions. For a negative ion, add electrons to the next available orbitals: S²⁻ is [Ne] 3s² 3p⁶. For a positive ion, remove electrons from the highest main level first. For the first-row transition elements, that means 4s electrons are removed before 3d, even though 4s filled first. Once 3d is occupied, the 4s electrons are the outermost and the easiest to remove.
Configurations and the periodic table (Structure 3.1 builds on this). The highest main level occupied is the period: sulfur's outer electrons are in n = 3, and sulfur is in period 3. The sublevel filled last gives the block: s for groups 1 and 2, p for groups 13 to 18, d for the transition elements.
7HLFirst ionization energy: the convergence limit and the trends
SL students can skip to section 9.
First ionization energy is the minimum energy needed to remove one mole of electrons from one mole of gaseous atoms in their ground state:
X(g) → X⁺(g) + e⁻
Learn the equation with its state symbols.
Ionization and the convergence limit. In Figure 3, ionization is an electron moving from the ground state to n = ∞. The lines that end on the ground state converge on a limit, and that convergence limit is the photon emitted by a fall from n = ∞ to the ground state, so its energy is the ionization energy of one atom. Multiply by the Avogadro constant for one mole.
Worked example. The lines of hydrogen that end on n = 1 converge at a wavelength of 91.2 nm. Calculate the first ionization energy of hydrogen.
The data booklet gives 1312 kJ mol⁻¹. Convert nm to m first, never forget the Avogadro constant, and divide by 1000 for kJ. Given a frequency, skip the first line.
The trends. Figure 8 plots the first ionization energies of the first twenty elements, from the data booklet.
Every explanation in this section comes from the same three factors: the nuclear charge, the distance of the outer electron from the nucleus, and the shielding of the outer electron from the nucleus by the electrons in inner levels.
Across a period, first IE generally increases. Nuclear charge rises by one each time, but each new electron enters the same main level, so shielding by inner electrons hardly changes. The outer electrons are pulled more strongly, sit closer to the nucleus, and are harder to remove.
Down a group, first IE decreases. Each step down adds a main level. The outer electron is further from the nucleus and is shielded by more inner electrons. The nuclear charge rises too, but the extra distance and shielding outweigh it, so the outer electron is held less strongly. Compare Li (520), Na (496) and K (419 kJ mol⁻¹).
Two discontinuities in each period. They are marked in Figure 8, and they are evidence for sublevels.
- From group 2 to group 13 (Be → B, Mg → Al). Boron's outer electron is the first in a 2p orbital, which is higher in energy than 2s and slightly further from the nucleus, and is partly shielded by the 2s electrons. It is removed more easily than one of beryllium's 2s electrons, even though boron has the larger nuclear charge.
- From group 15 to group 16 (N → O, P → S). Nitrogen's three 2p electrons are in separate orbitals. Oxygen's fourth 2p electron must pair with another in the same orbital, as Figure 7 showed. The two electrons in that orbital repel each other, so one of them is easier to remove, and oxygen's first IE is lower than nitrogen's.
Ionization energies are often plotted on a log scale (a Nature of science link in the guide): successive values span more than a hundredfold range, and a linear scale squashes the small ones flat. Section 8 uses one.
8HLSuccessive ionization energies
An atom can lose its electrons one at a time. The second ionization energy is the energy to remove one mole of electrons from one mole of gaseous 1+ ions, X⁺(g) → X²⁺(g) + e⁻, and so on. Figure 9 plots all thirteen for aluminium on a log scale.
Two patterns carry all the information.
Every successive ionization energy is larger than the one before. Each electron is being removed from an ion with the same nuclear charge but fewer electrons, so there is less repulsion between electrons, the ion is smaller, and the remaining electrons are held more tightly.
Big jumps mark a change of main level. Between the third and fourth ionization energies of aluminium, the value leaps from 2745 to 11 577 kJ mol⁻¹. The first three electrons came from n = 3; the fourth must come from n = 2, which is much closer to the nucleus and far less shielded. There is another leap between the 11th and 12th, where removal starts from n = 1. So the electrons fall into groups of 3, 8 and 2, which is aluminium's configuration by main level, 2, 8, 3, read backwards.
The number of electrons removed before the first big jump is the number of outer electrons, which for the s and p blocks gives the group (1 or 2 directly; 3 to 8 outer electrons means groups 13 to 18).
Worked example. The first five ionization energies of an element, in kJ mol⁻¹, are 590, 1145, 4912, 6491 and 8153. Deduce its group.
Two electrons are removed relatively easily; the third needs over four times as much as the second because it comes from a lower main level, closer to the nucleus and less shielded. Two outer electrons: group 2. Compare the ratios, not the differences, since later values are larger in every case.
9Where marks are lost
Saying the lines converge "because the electrons get closer". The lines converge because the energy levels converge at higher energy. Name the levels.
Getting the energy direction of the spectrum backwards. Shorter wavelength means higher frequency and higher energy. Violet is more energetic than red; the visible hydrogen lines converge towards the violet end.
Filling 3d before 4s, or emptying 3d before 4s. 4s fills first and, in positive ions, 4s empties first. Fe²⁺ is [Ar] 3d⁶, never [Ar] 3d⁴ 4s².
Writing Cr and Cu by the rule. Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹.
Pairing electrons too early in an orbital diagram. Hund's rule: one in each orbital of a sublevel, spins parallel, before any pairing.
HL: leaving out the Avogadro constant or the nm-to-m conversion. E = hf gives the energy for one atom in joules. The question wants kJ per mole.
HL: explaining the N → O dip with "a half-filled sublevel is stable". The mark scheme wants the mechanism: the paired electrons in one 2p orbital repel, so one is easier to remove.
HL: dropping the state symbols from the ionization equation. X(g) → X⁺(g) + e⁻: the atom and the ion are both gaseous.
10Draw it right
- Orbital diagrams: one box per orbital, with p and d sublevels drawn as 3 and 5 boxes side by side. Label each sublevel. Arrows in a pair point opposite ways; single electrons in a sublevel all point the same way.
- Energy level diagram: horizontal lines that get closer together going up, with n = ∞ at the top. Emission transitions are downward arrows that end on a level, never floating between levels.
- s orbital: a sphere (a circle in two dimensions) centred on the nucleus. p orbital: two lobes on opposite sides of the nucleus along one axis, with the nucleus between them; label px, py or pz by the axis.
- Line spectrum: separate vertical lines on a dark or blank background, getting closer together towards the high-frequency (short-wavelength) end.
- HL: label IE graphs with units, kJ mol⁻¹, and say "log scale" when you use one.
11Try it
Marks in brackets. Answers and marker's notes are at the end. Use the data booklet where you need constants or data.
Q1. Which electron transition in a hydrogen atom emits the photon of highest energy? 1 mark
A. n = 2 → n = 1 B. n = 3 → n = 2 C. n = 6 → n = 2 D. n = 4 → n = 3
Q2. (a) Write the full electron configuration of phosphorus. 1 mark
(b) Write condensed electron configurations for manganese and for the Mn²⁺ ion. 2 marks
(c) Draw an orbital diagram for the 3d sublevel of Co²⁺ and state the number of unpaired electrons. 2 marks
Q3. (a) Distinguish between a continuous spectrum and a line spectrum. 1 mark
(b) The four visible lines of hydrogen are at 410, 434, 486 and 656 nm. Identify the electron transition that produces the line at 486 nm. 1 mark
(c) Explain why the lines in the visible emission spectrum of hydrogen converge at higher frequency. 2 marks
Q4. State the maximum number of electrons in the fourth main energy level, and the number of orbitals in the third. 2 marks
Q5 (HL). In the emission spectrum of an element, the lines that end on the ground state converge at a frequency of 1.24 × 10¹⁵ s⁻¹. Calculate the first ionization energy of the element in kJ mol⁻¹, and use the data booklet to suggest which element it is. 3 marks
Q6 (HL). The first seven ionization energies of an element, in kJ mol⁻¹, are 1000, 2252, 3357, 4556, 7004, 8496 and 27 107.
(a) Deduce the group of the element, giving your reasoning. 2 marks
(b) The element is sulfur. Explain why the first ionization energy of sulfur is lower than that of phosphorus. 2 marks
12In one breath
Shorter wavelength means higher frequency and more energy per photon. A continuous spectrum has every wavelength; a line spectrum has only certain ones, because each fall of an excited electron to a lower level emits a photon of exactly that energy difference. In hydrogen, drops to n = 1 give ultraviolet, to n = 2 visible, to n = 3 infrared, and the lines converge at higher frequency because the energy levels converge at higher energy. Main level n holds 2n² electrons in s, p, d, f sublevels of 1, 3, 5, 7 orbitals; an orbital is a region of high probability for an electron, holding two of opposite spin; s is a sphere, p a dumbbell. Fill by Aufbau (1s 2s 2p 3s 3p 4s 3d 4p), Pauli (two per orbital, opposite spins) and Hund (singly, parallel, before pairing); Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹; positive ions lose 4s before 3d. HL: first IE is X(g) → X⁺(g) + e⁻, equal per atom to hf at the convergence limit, times the Avogadro constant; it rises across a period (more nuclear charge, same shielding) and falls down a group (more distance and shielding), with dips at group 13 (new p sublevel) and group 16 (paired-electron repulsion); successive IEs always rise, and the big jump shows how many outer electrons there are, and so the group.
Answers
Q1. A. The n = 2 to n = 1 gap is larger than any gap between higher levels. A only.
Q2. (a) 1s² 2s² 2p⁶ 3s² 3p³ [1] (b) Mn: [Ar] 3d⁵ 4s² (or [Ar] 4s² 3d⁵). Mn²⁺: [Ar] 3d⁵. 1 for each. [Ar 3d³ 4s² for the ion scores 0, because 4s electrons are removed first.] (c) Co is [Ar] 3d⁷ 4s², so Co²⁺ is [Ar] 3d⁷. Five 3d boxes: ↑↓ ↑↓ ↑ ↑ ↑. Three unpaired electrons. 1 for five boxes with two pairs and three single parallel electrons, 1 for three unpaired. A diagram that pairs all seven as far as possible (↑↓ ↑↓ ↑↓ ↑ _) breaks Hund's rule and scores 0 for the first mark.
Q3. (a) A continuous spectrum contains all wavelengths (colours) in a range with no gaps; a line spectrum contains only certain separate wavelengths, seen as distinct lines. both halves needed. (b) n = 4 → n = 2. The visible lines all end on n = 2; 656 nm is the smallest drop (3 → 2), and 486 nm is the next line up in energy. n = 4 → n = 2 only. (c) The lines are produced by electrons falling to n = 2 from higher levels. The energy levels get closer together as their energy increases (they converge), so the differences in energy between successive transitions get smaller, and the lines get closer together until they merge at higher frequency. 1 for energy levels converging at higher energy, 1 for smaller energy differences between successive transitions. "The lines get closer together" alone scores 0.
Q4. Fourth level: 2 × 4² = 32 electrons. Third level: 3² = 9 orbitals (one 3s, three 3p, five 3d). 1 each.
Q5 (HL).
The closest first ionization energy in the data booklet is sodium's, 496 kJ mol⁻¹, so the element is probably sodium. M1 for E = hf per atom, M1 for multiplying by the Avogadro constant, A1 for 495 kJ mol⁻¹ with sodium. An answer of 8.22 × 10⁻¹⁹ with no conversion scores M1 only.
Q6 (HL). (a) The large jump is between the sixth (8496) and the seventh (27 107) ionization energies: a rise of about 18 600 kJ mol⁻¹, more than a threefold increase, against no more than about 2 500 kJ mol⁻¹ for any earlier step. Six electrons come from the outer main level before one must come from a lower level, closer to the nucleus and less shielded. Six outer electrons: group 16. 1 for locating the jump between the 6th and 7th with a reason (new main level, closer, less shielded), 1 for group 16. (b) Phosphorus has three 3p electrons, one in each 3p orbital. Sulfur has four, so two of its 3p electrons are paired in the same orbital. Repulsion between the paired electrons makes one of them easier to remove, so sulfur's first ionization energy is lower (1000 against 1012 kJ mol⁻¹). 1 for sulfur having a pair of electrons in one 3p orbital, 1 for repulsion between them making removal easier. "Sulfur has a larger nuclear charge" argues the wrong way and scores 0.
Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section S1.3 Electron configurations. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.