IB DP Chemistry Structure 1.3: Electron configurations – Revision Notes

These are recall notes for Structure 1.3, Electron configurations, from the IB Chemistry guide for the Diploma Programme (published February 2023; first assessment 2025). Understandings 1.3.1 to 1.3.7 are all here. Sections 1.3.1 to 1.3.5 are SL and HL; 1.3.6 and 1.3.7 are additional higher level, so HL only. Stuck on a point? Go back to the study guide, which gives the full reasoning and longer examples.

Also useful: practice questions, Structure 1 overview notes, the nuclear atom notes for Structure 1.2, the course hub page, a printable tick-list and free diagnostics to find gaps.

Key definitions

Term	Meaning
Photon	A packet of electromagnetic energy; its energy rises with frequency
Excited state	Any arrangement where an electron sits above its lowest available level
Continuous spectrum	All wavelengths in a range, no gaps
Line spectrum	Only certain wavelengths, seen as separate lines
Main energy level	Shell numbered n = 1, 2, 3 …; holds up to 2n² electrons
Sublevel	s, p, d or f division of a main level
Orbital	Region around the nucleus where an electron is likely to be; holds two electrons of opposite spin
Convergence limit	Point where spectral lines merge at high frequency (HL: corresponds to ionization)
First ionization energy	(HL) Energy per mole to take one electron from each gaseous ground-state atom

1.3.1 Spectra in four lines

• Higher frequency = higher energy = shorter wavelength.
• Increasing energy: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.
• Visible light: red is lowest energy, violet is highest.
• Electrons fall from excited states to lower levels and each fall emits one photon. Gap size fixes the photon’s frequency, so a fixed set of gaps gives a fixed set of lines.

Distinction: a rainbow or white light through a prism is continuous. A gas discharge tube gives a line spectrum unique to the element.

1.3.2 Hydrogen emission spectrum

Electron falls to	Energy released	Region
n = 1	Largest	Ultraviolet
n = 2	Middle	Visible
n = 3	Smallest of the three	Infrared

• Lines in each family crowd together towards higher frequency, because the levels get closer together as n increases.
• The merging point is the convergence limit.
• The names of the series are not needed.

1.3.3 to 1.3.5 Levels, sublevels and orbitals

n	Sublevels	Orbitals (n²)	Max electrons (2n²)
1	1s	1	2
2	2s, 2p	4	8
3	3s, 3p, 3d	9	18
4	4s, 4p, 4d, 4f	16	32

Per sublevel: s has 1 orbital, p has 3, d has 5, f has 7.

Shapes: s is a sphere centred on the nucleus. Each p orbital is a dumbbell; the three (px, py, pz) lie along the x, y and z axes at 90° to each other.

The three rules

Rule	In one line
Aufbau principle	Fill the lowest-energy orbital first: 1s 2s 2p 3s 3p 4s 3d 4p
Pauli exclusion principle	Two electrons at most per orbital, with opposite spins
Hund’s rule	Spread out singly with parallel spins across equal-energy orbitals before pairing

Method in steps: configuration of an atom

1. Read off Z; a neutral atom has Z electrons.
2. Fill sublevels in the Aufbau order until all electrons are placed.
3. Check for Cr (Z = 24) and Cu (Z = 29): move one 4s electron into 3d.
4. For a condensed form, replace the inner part with the previous noble gas in square brackets.
5. Check that the superscripts add up to Z.

Method in steps: configuration of an ion

1. Write the atom first.
2. Positive ion: remove electrons from the highest main level first. For first-row transition elements that means 4s before 3d.
3. Negative ion: add electrons to the next free orbital in the Aufbau order.
4. Check the total equals Z minus the charge.

Small worked reminders

Ni (Z = 28):    [Ar] 3d⁸ 4s²
Ni²⁺:           [Ar] 3d⁸            (4s emptied first)
3d  [↑↓][↑↓][↑↓][↑ ][↑ ]            2 unpaired electrons

Cu⁺ (Z = 29):   [Ar] 3d¹⁰           (the single 4s electron goes)
S²⁻ (Z = 16):   1s² 2s² 2p⁶ 3s² 3p⁶ (same as argon)

Cr and Cu exceptions

Atom	Expected	Actual	Reason
Cr, Z = 24	[Ar] 3d⁴ 4s²	[Ar] 3d⁵ 4s¹	Half-filled 3d gives lower overall energy
Cu, Z = 29	[Ar] 3d⁹ 4s²	[Ar] 3d¹⁰ 4s¹	Filled 3d gives lower overall energy

1.3.6 HL only: convergence limit and first IE

Formulas: E = hf and c = λf, so E = hc/λ. Molar energy = E × L, where L is the Avogadro constant.

Method in steps

1. Convert the wavelength to metres if needed (1 nm = 10⁻⁹ m).
2. Find the energy of one photon with E = hf or E = hc/λ.
3. Multiply by L to get J mol⁻¹.
4. Divide by 1000 for kJ mol⁻¹ and give 3 significant figures.

Worked reminder. The convergence limit for lithium’s ground-state lines is at a frequency of 1.30 × 10¹⁵ Hz. Taking L = 6.02 × 10²³ mol⁻¹ and h = 6.63 × 10⁻³⁴ J s:

E = 6.63 × 10⁻³⁴ × 1.30 × 10¹⁵ = 8.62 × 10⁻¹⁹ J
First IE = 8.62 × 10⁻¹⁹ × 6.02 × 10²³ = 5.19 × 10⁵ J mol⁻¹ = 519 kJ mol⁻¹

First IE trends

Pattern	Cause
Rises across a period	Nuclear charge increases; electron removed from the same main level with similar shielding
Falls down a group	Outer electron in a higher main level, further away and more shielded
Drop from group 2 to group 13	Electron removed from a p sublevel, higher in energy than s
Drop from group 15 to group 16	Electron removed from a doubly occupied p orbital; pair repulsion

1.3.7 HL only: successive ionization energies

• Every value is larger than the one before: the ion left behind is more positive each time.
• Look for the biggest increase, not the biggest value.
• Electrons removed before that jump = outer electrons = group (add 10 for groups 13 to 18).

Worked reminder. Boron, kJ mol⁻¹ (use these values): 801, 2427, 3660, 25 026. Increases: +1626, +1233, +21 366. The jump comes after the third value, so boron has three outer electrons and is in group 13.

Must-know distinctions

• Frequency vs wavelength: energy rises with frequency and falls with wavelength.
• Main level vs sublevel vs orbital: n = 3 is a main level; 3p is a sublevel; 3px is one orbital.
• Filling order vs loss order: 4s fills before 3d but empties before 3d.
• Full vs condensed: full lists every sublevel from 1s; condensed starts from a noble gas core.
• Convergence: the lines converge at high frequency, not low.

Where marks are usually lost

• Writing [Ar] 4s² 3d⁴ or [Ar] 3d⁴ 4s² for chromium, or the 3d⁹ form for copper.
• Removing 3d electrons before the 4s pair for ions such as Ni²⁺ or Co²⁺.
• Drawing two electrons in one box with the same spin, breaking the Pauli principle.
• Pairing up in a p or d sublevel before every orbital holds one electron.
• Saying an s orbital is circular rather than spherical, or showing p orbitals not at 90°.
• Mixing up which hydrogen lines are ultraviolet (drops to n = 1) and which are visible (drops to n = 2).
• (HL) Using nm directly in E = hc/λ, or giving the energy of one atom when a molar value is asked for.
• (HL) Explaining the group 15 to 16 drop as “extra stability” with no mention of paired-electron repulsion.

Quick self-test

1. Which photon has more energy: 450 nm or 650 nm light?
2. State one difference between a continuous spectrum and a line spectrum.
3. In which region are hydrogen lines produced by electrons falling to n = 3?
4. What is the maximum number of electrons in the third main energy level?
5. How many orbitals are in a d sublevel?
6. Give the condensed configuration of Ni²⁺.
7. Write the full electron configuration of copper.
8. Which 3+ ion with Z ≤ 36 has the configuration [Ar] 3d³?
9. How many unpaired electrons are in an Fe³⁺ ion?
10. (HL) Explain why the first IE of oxygen is lower than that of nitrogen.
11. (HL) The first IE of caesium is 376 kJ mol⁻¹. Calculate the frequency at the convergence limit. Take L = 6.02 × 10²³ mol⁻¹ and h = 6.63 × 10⁻³⁴ J s.
12. (HL) Successive IEs of an element (kJ mol⁻¹): 419, 3052, 4420, 5877. Deduce its group.

Answers

1. 450 nm: shorter wavelength, higher frequency, higher energy.
2. A continuous spectrum contains all wavelengths in a range; a line spectrum contains only specific wavelengths.
3. Infrared.
4. 18 (2 × 3²).
5. Five.
6. [Ar] 3d⁸.
7. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.
8. Cr³⁺ (chromium, Z = 24: 24 − 3 = 21 electrons).
9. Five. Fe³⁺ is [Ar] 3d⁵, one electron in each 3d orbital.
10. Oxygen’s fourth 2p electron shares an orbital with another electron. Repulsion between the pair makes it easier to remove.
11. Energy per atom = 376 000 ÷ (6.02 × 10²³) = 6.246 × 10⁻¹⁹ J; f = E ÷ h = 9.42 × 10¹⁴ Hz.
12. The large jump is between the first and second values, so one outer electron: group 1.

Next steps

Work through the Structure 1.3 practice questions, then go back to the study guide for any topic you missed. Tick off progress on the printable checklist.

Official syllabus

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