IB DP Chemistry Structure 1.3: Electron configurations

This study guide teaches Structure 1.3, Electron configurations, as set out in the IB Chemistry guide for the Diploma Programme (published February 2023 for first assessment in 2025). It works through understandings Structure 1.3.1 to 1.3.7. Understandings 1.3.1 to 1.3.5 are taught at both SL and HL; 1.3.6 and 1.3.7 sit in the additional higher level (AHL) part of the guide, so they are HL only. The page applies to the course as first assessed in 2025.

The Structure 1 summary page gives context. Links to the notes, questions, hub and checklist are at the end.

What this unit covers

Guide reference	Skill you need	Level
Structure 1.3.1	Link colour, wavelength, frequency and energy; tell continuous from line spectra	SL and HL
Structure 1.3.2	Describe the hydrogen spectrum and its drops to levels 1, 2 and 3	SL and HL
Structure 1.3.3	Use 2n² for the capacity of a main level	SL and HL
Structure 1.3.4	Name s, p, d, f sublevels; recognise s and p orbital shapes	SL and HL
Structure 1.3.5	Write full and condensed configurations and orbital diagrams up to Z = 36, including Cr and Cu	SL and HL
Structure 1.3.6	Explain first IE trends; calculate first IE from a convergence limit	HL only
Structure 1.3.7	Deduce a group from successive ionization energies	HL only

Light, colour and emission spectra (1.3.1)

Electromagnetic radiation travels as photons. Three linked quantities describe each photon:

• Frequency (f) and energy (E) rise together. A photon with double the frequency carries double the energy.
• Wavelength (λ) runs the other way. Shorter wavelength means higher frequency and higher energy.

In order of increasing energy: radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays. Within the visible region (roughly 400 nm to 700 nm), red has the longest wavelength and lowest energy; violet has the shortest wavelength and highest energy.

How an emission spectrum forms

Heat a gaseous element or pass an electric discharge through it, and electrons absorb energy and move up to higher levels. These excited states are unstable. When an electron drops back to a lower level, the atom gives out one photon whose energy equals the gap between the two levels. A large gap gives a high-frequency, short-wavelength photon.

Continuous versus line spectra

• A continuous spectrum contains every wavelength across a range, with no gaps. White light through a prism gives one: the colours merge smoothly from red to violet.
• A line spectrum contains only certain wavelengths. It appears as separate bright lines on a dark background.

Each element has its own set of energy-level gaps, so its own pattern of lines. A line spectrum shows that electron energies in an atom take only fixed values, and it identifies the element.

Worked example 1: ranking three photons

Rank by photon energy: yellow light from a sodium lamp (589 nm), infrared from a heat lamp, ultraviolet from a sunbed. Wavelength falls from infrared to yellow to ultraviolet, so frequency and energy rise in that order: infrared < yellow < ultraviolet.

The hydrogen emission spectrum (1.3.2)

Every hydrogen line comes from its single electron dropping from a higher level to a lower one.

 n = ∞  ----------------------  (electron free of the atom)
 n = 5  ----------------------
 n = 4  ----------------------

 n = 3  ----------------------


 n = 2  ----------------------
                                 gaps shrink as n rises



 n = 1  ----------------------  (ground state)

Three families of lines matter:

• Drops to n = 1 release the most energy: ultraviolet lines.
• Drops to n = 2 release less: visible lines.
• Drops to n = 3 release less again: infrared lines.

Within each family, the lines get closer together as frequency rises, because the levels themselves crowd together at high n. The lines finally merge at a convergence limit. This is evidence for discrete levels whose spacing shrinks as energy increases. You do not need the names of the series.

Worked example 2: comparing two visible lines

Use these values for hydrogen energy levels, in kJ mol⁻¹: n = 2, −328; n = 3, −145.8; n = 4, −82.0.

(a) Which drop gives the higher-energy visible line, n = 3 → 2 or n = 4 → 2? (b) Which of these two lines is closer to the violet end?

(a) n = 3 -> 2: -145.8 - (-328) = 182.2 kJ mol^-1
    n = 4 -> 2: -82.0 - (-328)  = 246.0 kJ mol^-1
    The n = 4 -> 2 drop releases more energy.
(b) More energy means higher frequency and shorter wavelength,
    so the n = 4 -> 2 line lies nearer the violet end.

The n = 3 → 2 line is the red line of the hydrogen spectrum, at about 656 nm.

Main energy levels (1.3.3)

Each main energy level has an integer number n, counting outwards from the nucleus. Level n holds at most 2n² electrons.

n	Maximum electrons, 2n²
1	2
2	8
3	18
4	32

Worked example 3

Show that level 4 holds 32 electrons: 2n² = 2 × 4² = 32. Check by sublevel: 4s (2) + 4p (6) + 4d (10) + 4f (14) = 32.

Sublevels and orbital shapes (1.3.4)

A more detailed model splits each main level into sublevels, labelled s, p, d and f. Within one main level their energy increases in that order: s < p < d < f. Level 1 has only 1s; level 2 has 2s and 2p; level 3 has 3s, 3p and 3d; level 4 has 4s, 4p, 4d and 4f.

An orbital is a region around the nucleus in which an electron is very likely to be found.

• An s orbital is a sphere with the nucleus at its centre.
• A p sublevel has three dumbbell-shaped p orbitals, each with two lobes either side of the nucleus. They lie along the x, y and z axes at right angles to each other (px, py, pz) and have equal energy.

Orbitals and the three filling rules (1.3.5)

Each sublevel contains a fixed number of orbitals, and each orbital holds two electrons of opposite spin.

Sublevel	Maximum electrons	Orbitals
s	2	1
p	6	3
d	10	5
f	14	7

Three rules decide where electrons go:

1. Aufbau principle: electrons enter the lowest-energy orbital available. Up to Z = 36: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. Note that 4s fills before 3d.
2. Pauli exclusion principle: an orbital holds at most two electrons, and they must have opposite spins.
3. Hund’s rule: in a sublevel with several orbitals of equal energy, each orbital takes one electron, all spinning the same way, before any orbital takes a second.

In an orbital diagram (arrow-in-box diagram), each box is an orbital and each arrow is an electron. Boxes are drawn higher up the page for higher-energy sublevels.

Worked example 4: sulfur, Z = 16

Full:       1s² 2s² 2p⁶ 3s² 3p⁴
Condensed:  [Ne] 3s² 3p⁴

Orbital diagram (highest energy at the top):
3p  [↑↓][↑ ][↑ ]
3s  [↑↓]
2p  [↑↓][↑↓][↑↓]
2s  [↑↓]
1s  [↑↓]

The fourth 3p electron must pair up, because each 3p orbital already holds one. Hund’s rule leaves two unpaired electrons.

Worked example 5: manganese and the Mn²⁺ ion, Z = 25

Mn atom:   1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²    = [Ar] 3d⁵ 4s²
Mn²⁺ ion:  remove 2 electrons from 4s first
           1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵        = [Ar] 3d⁵

3d  [↑ ][↑ ][↑ ][↑ ][↑ ]   five unpaired electrons

Once 3d is occupied, 4s lies higher in energy, so a first-row transition element loses its 4s electrons first. [Ar] 3d³ 4s² for Mn²⁺ is wrong.

Chromium and copper

Two atoms below Z = 36 break the simple filling order:

• Chromium, Z = 24: [Ar] 3d⁵ 4s¹, not [Ar] 3d⁴ 4s².
• Copper, Z = 29: [Ar] 3d¹⁰ 4s¹, not [Ar] 3d⁹ 4s².

The 4s and 3d sublevels are close in energy. Moving one electron from 4s into 3d gives a half-filled 3d⁵ set (Cr) or a filled 3d¹⁰ set (Cu), which lowers the overall energy. Chromium ends up with six unpaired electrons.

Worked example 6: the bromide ion, Z = 35

Br atom:  [Ar] 3d¹⁰ 4s² 4p⁵      (35 electrons)
Br⁻ ion:  add 1 electron to 4p
          1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶   (36 electrons)

Br⁻ has the same configuration as krypton. For a negative ion, add electrons to the next free place in the Aufbau order. For a positive ion, remove electrons from the highest main level first.

HL only: the convergence limit and first ionization energy (1.3.6)

At the convergence limit, the “upper level” is n = ∞: the electron is no longer bound to the atom. The photon energy at that limit equals the energy needed to remove the electron from the ground state. For lines that end on the lowest level, it is the first ionization energy: per mole, the energy taken to strip one electron from every gaseous atom in its ground state.

To get a molar value from spectral data, use E = hf (or E = hc/λ when you are given a wavelength) for one atom, then multiply by the Avogadro constant.

Worked example 7: sodium from a wavelength

The lines in the emission spectrum of sodium that end on its ground state converge at 241 nm. Constants: Planck constant h = 6.63 × 10⁻³⁴ J s; speed of light c = 3.00 × 10⁸ m s⁻¹; Avogadro constant L = 6.02 × 10²³ mol⁻¹.

λ = 241 nm = 241 × 10⁻⁹ m
E (one atom) = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (241 × 10⁻⁹)
             = 8.25 × 10⁻¹⁹ J
E (one mole) = 8.25 × 10⁻¹⁹ × 6.02 × 10²³ = 4.97 × 10⁵ J mol⁻¹
First IE     = 497 kJ mol⁻¹

Trends across a period

Period 2 first ionization energies, in kJ mol⁻¹ (use these values):

Li	Be	B	C	N	O	F	Ne
520	900	801	1086	1402	1314	1681	2081

• General rise: nuclear charge increases by one each step, while the electron removed stays in the second main level with similar shielding, so it is held more tightly.
• Dip from Be to B: boron’s electron comes from 2p, a sublevel of higher energy than 2s, so less energy is needed.
• Dip from N to O: in oxygen, one 2p orbital holds a pair. Repulsion between the paired electrons makes one of them easier to remove than any of nitrogen’s three unpaired 2p electrons.

Trend down a group

First IE falls down a group. The outer electron is in a higher main level, further from the nucleus and more shielded, which outweighs the extra nuclear charge.

HL only: successive ionization energies (1.3.7)

Each successive ionization energy is larger than the last, because each electron leaves an ion that is already more positive. A large jump shows that the next electron comes from a main level nearer the nucleus. Count how many electrons come off before that jump. That count equals the outer electrons, and so fixes the group. A log scale makes the jumps easier to see.

Worked example 8: silicon

Successive ionization energies of silicon, in kJ mol⁻¹ (use these values): 786, 1577, 3232, 4356, 16 091.

Increases: +791, +1655, +1124, +11 735
The largest increase is between the 4th and 5th values.
4 electrons are removed before the jump -> 4 outer electrons
Silicon is in group 14 (configuration [Ne] 3s² 3p²).

Common errors

• Writing [Ar] 3d⁴ 4s² for chromium or [Ar] 3d⁹ 4s² for copper.
• Removing 3d electrons before 4s electrons when forming a transition-element ion.
• Pairing electrons in a p or d orbital before every orbital in that sublevel holds one, or drawing paired arrows that point the same way.
• Saying the hydrogen lines converge at low energy, or linking long wavelength with high energy.
• (HL) Leaving λ in nm, or not multiplying by L.
• (HL) Explaining the N to O dip by “half-filled stability” alone, without naming repulsion between paired electrons.
• (HL) Giving the group as the position of the jump, not the number of electrons removed before it.

Next steps

• Structure 1.3 revision notes and practice questions
• The full-course syllabus guide, course hub and printable checklist
• Free diagnostics to find your gaps

Official syllabus

Chemistry guide, Diploma Programme, International Baccalaureate Organization: first assessment 2025, published February 2023. Sub-topic Structure 1.3, Electron configurations (understandings 1.3.1 to 1.3.5 for SL and HL; 1.3.6 and 1.3.7 additional higher level).