SL & HL Structure · Models of the particulate nature of matter BETA S1.3

Electron configurations

IB syllabus topic Structure 1.3 — Electron configurations

Electron configurations describe how electrons are arranged in the energy levels, sublevels and atomic orbitals of an atom; this revision note covers emission spectra, the order of sublevels, the Aufbau, Hund and Pauli rules, the shapes of s and p orbitals, and (HL) how ionization energy data, the convergence limit and successive ionization energies provide evidence for these arrangements.

Part of IB Chemistry (2025-2030 syllabus) — Standard and Higher Level.

Items marked HL are Higher Level only — SL students can skip them.

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Key points

  • Electrons occupy quantized energy levels (shells); when an excited electron falls to a lower level it emits a photon of fixed energy, giving a line emission spectrum rather than a continuous one.
  • Each main energy level n contains sublevels labelled s, p, d, f; the s sublevel holds 2 electrons, p holds 6, d holds 10 and f holds 14.
  • Electron configurations are built using three rules: Aufbau (fill lowest-energy sublevels first), Pauli exclusion (max 2 electrons per orbital with opposite spins) and Hund's rule (singly fill degenerate orbitals before pairing).
  • Because 4s is lower in energy than 3d, the 4s sublevel fills before 3d; chromium and copper are anomalies (4s1 3d5 and 4s1 3d10) due to extra stability of half-filled and full d sublevels.
  • An s orbital is spherical; the three p orbitals are dumbbell-shaped and point along the x, y and z axes; each orbital holds a maximum of 2 electrons.
  • HL(HL) First ionization energy is the energy to remove one mole of electrons from one mole of gaseous atoms; emission-spectrum lines converging to a limit (the convergence limit) allow the ionization energy to be calculated.
  • HL(HL) Successive ionization energies of one element always increase, and large jumps reveal when an electron is removed from a new, inner main energy level, confirming the shell structure and group of the element.

Subtopic by subtopic

Emission spectra

When atoms are given energy (by heating or an electric discharge), electrons are promoted to higher energy levels. They are unstable there and quickly fall back to lower levels, releasing the energy difference as a photon of light.

Because energy levels are quantized (only certain values are allowed), only photons of certain fixed energies can be emitted. This produces a line emission spectrum - sharp coloured lines on a black background - rather than a continuous rainbow.

The photon energy is fixed by the gap between the two levels:

E=hf=hcλE = hf = \frac{hc}{\lambda}

so a larger energy gap gives a higher frequency (shorter wavelength, towards the blue/violet end).

Key features of the hydrogen spectrum:

  • Lines appear in series (e.g. the visible Balmer series, where electrons fall to n=2n=2).
  • Within each series the lines get closer together (converge) at higher frequency.
  • The convergence happens because the energy levels themselves get closer together as nn increases.

The existence of discrete lines is the central piece of evidence that electrons occupy discrete energy levels, not a continuous range of energies. The booklet provides c=fλc = f\lambda and E=hfE = hf for these calculations.

Energy levels and sublevels

Each main energy level (shell), numbered by the principal quantum number n=1,2,3,...n = 1, 2, 3, ..., is divided into sublevels (subshells) labelled s, p, d, f.

The number of sublevels equals nn:

  • n=1n = 1: 1s only
  • n=2n = 2: 2s, 2p
  • n=3n = 3: 3s, 3p, 3d
  • n=4n = 4: 4s, 4p, 4d, 4f

Each sublevel holds a fixed maximum number of electrons:

  • s holds 2 (1 orbital)
  • p holds 6 (3 orbitals)
  • d holds 10 (5 orbitals)
  • f holds 14 (7 orbitals)

The sublevels are not filled strictly by nn. Their energies overlap, and the order of filling is:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p ...1s\ 2s\ 2p\ 3s\ 3p\ 4s\ 3d\ 4p\ 5s\ 4d\ 5p\ ...

The key crossover is that 4s is lower in energy than 3d, so 4s fills first. This explains why the d-block (transition metals) starts in period 4. Sublevels within the same level rise in energy in the order s < p < d < f.

Electron configurations (Aufbau, Hund, Pauli)

An electron configuration shows how electrons are distributed among sublevels, e.g. iron is 1s2 2s2 2p6 3s2 3p6 4s2 3d61s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6. Three rules govern the filling:

  • Aufbau principle: fill the lowest-energy sublevel available first (follow the order 1s,2s,2p,3s,3p,4s,3d...1s, 2s, 2p, 3s, 3p, 4s, 3d...).
  • Pauli exclusion principle: an orbital holds a maximum of 2 electrons, and they must have opposite spins (drawn as ↑↓\uparrow\downarrow).
  • Hund's rule: within a set of equal-energy (degenerate) orbitals, electrons occupy them singly with parallel spins first, then pair up. This minimizes repulsion.

Condensed notation uses the previous noble gas, e.g. [Ar] 4s2 3d6[\text{Ar}]\,4s^2\,3d^6 for iron.

Forming ions: when transition metals form cations, electrons are removed from 4s before 3d (e.g. Fe2+\text{Fe}^{2+} is [Ar] 3d6[\text{Ar}]\,3d^6).

Two anomalies to learn:

  • Chromium: [Ar] 4s1 3d5[\text{Ar}]\,4s^1\,3d^5 (not 4s2 3d44s^2\,3d^4)
  • Copper: [Ar] 4s1 3d10[\text{Ar}]\,4s^1\,3d^{10} (not 4s2 3d94s^2\,3d^9)

A half-filled (3d53d^5) or fully filled (3d103d^{10}) d sublevel gives extra stability.

Atomic orbitals (s and p)

An atomic orbital is a region where an electron is most likely to be found. Each orbital holds at most 2 electrons with opposite spins.

s orbitals are spherical and centred on the nucleus. There is one s orbital in each level (1s, 2s, 3s...). Higher s orbitals are larger.

p orbitals are dumbbell-shaped (two lobes either side of the nucleus). There are three p orbitals in any p sublevel, identical in shape and energy but pointing along the three perpendicular axes:

  • pxp_x along the x-axis
  • pyp_y along the y-axis
  • pzp_z along the z-axis

Because the three p orbitals are degenerate (equal energy), Hund's rule applies when filling them: they each take one electron before any pairs up.

Useful summary of capacities:

  • s sublevel: 1 orbital, 2 e−^-
  • p sublevel: 3 orbitals, 6 e−^-
  • d sublevel: 5 orbitals, 10 e−^-

For IB you only need to know the shapes of s and p orbitals; you do not need the shapes of d or f orbitals. An orbital diagram uses a box (or line) per orbital and arrows for electron spins.

Ionization energy and the convergence limitHL

The first ionization energy is the energy to remove one mole of electrons from one mole of gaseous atoms:

X(g)→X+(g)+e−\text{X(g)} \rightarrow \text{X}^+\text{(g)} + e^-

It is always endothermic (positive). Ionization energy gives direct evidence for electron arrangement.

Link to emission spectra: in the hydrogen spectrum, the lines of a series get closer together at higher frequency and eventually converge to a limit. This convergence limit corresponds to the electron jumping from n=1n=1 all the way to n=∞n=\infty - that is, leaving the atom completely. So the energy of the photon at the convergence limit equals the ionization energy of one atom.

To find the ionization energy per mole:

  • Read off the frequency at the convergence limit, flimitf_{\text{limit}}.
  • Energy of one photon: E=hflimitE = hf_{\text{limit}}.
  • Multiply by the Avogadro constant: Emol=NA hflimitE_{\text{mol}} = N_A\,hf_{\text{limit}}.
  • Convert from J mol−1\text{J mol}^{-1} to kJ mol−1\text{kJ mol}^{-1}.

Trends to know: ionization energy generally increases across a period (rising nuclear charge) and decreases down a group (electrons further from the nucleus, more shielding). Small dips (e.g. from group 2 to 13, and group 15 to 16) give evidence for sublevels and orbital pairing.

Successive ionization energiesHL

Successive ionization energies are the energies to remove the 1st, 2nd, 3rd... electron from the same atom, one after another:

X(g)→X+→X2+→X3+...\text{X(g)} \rightarrow \text{X}^+ \rightarrow \text{X}^{2+} \rightarrow \text{X}^{3+} ...

They always increase for a given element, because:

  • Each electron is pulled from an increasingly positive ion, so it is held more strongly.
  • The remaining electrons feel less repulsion and are drawn closer.

The most useful feature is the large jumps. A sudden big rise occurs when an electron must be removed from a new, inner main energy level that is closer to the nucleus and less shielded.

Reading the data:

  • Plot log⁡(ionization energy)\log(\text{ionization energy}) against electron number to make the jumps clear.
  • Count the electrons removed before the first big jump - this equals the number of electrons in the outer shell, i.e. the group number.
  • The pattern of jumps confirms the shell structure (2, 8, 8...).

For example, sodium shows a huge jump after the 1st electron, confirming 1 outer electron (group 1). Magnesium jumps after the 2nd (group 2). This data is strong evidence for the existence of discrete energy levels.

Formulae

c=fλc = f\lambda

Given in the data booklet. Links the speed of light c=3.00×108 m s−1c = 3.00\times10^8\ \text{m s}^{-1} to the frequency ff (in Hz\text{Hz}) and wavelength λ\lambda (in m\text{m}) of a spectral line. Use to convert a measured wavelength into a frequency.

E=hfE = hf

Given in the data booklet. The energy EE (in J\text{J}) of one photon emitted or absorbed, where h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s} is the Planck constant and ff is the frequency. Larger frequency (shorter wavelength) means a more energetic photon.

E=hcλE = \dfrac{hc}{\lambda}

Constructed by the student by combining the two booklet equations c=fλc = f\lambda and E=hfE = hf. Use to get photon energy directly from wavelength λ\lambda.

Emol=NA hflimitE_{\text{mol}} = N_A \, h f_{\text{limit}}HL

(HL) Not a booklet equation; constructed to find the ionization energy per mole from the convergence-limit frequency flimitf_{\text{limit}}. Multiply the single-photon energy hfhf by the Avogadro constant NA=6.02×1023 mol−1N_A = 6.02\times10^{23}\ \text{mol}^{-1}, then convert from J mol−1\text{J mol}^{-1} to kJ mol−1\text{kJ mol}^{-1}.

Definitions

Emission spectrum
The pattern of discrete coloured lines produced when atoms emit light; each line corresponds to a photon released as an electron falls from a higher to a lower energy level.
Energy level (principal quantum number, n)
A quantized shell that an electron can occupy; levels are numbered n = 1, 2, 3..., with higher n further from the nucleus and higher in energy.
Sublevel (subshell)
A subdivision of a main energy level, labelled s, p, d or f, each made up of one or more atomic orbitals of the same energy.
Atomic orbital
A region of space around the nucleus where there is a high probability of finding an electron; each orbital holds a maximum of two electrons with opposite spins.
Aufbau principle
Electrons fill the lowest available energy sublevels first before occupying higher ones.
First ionization energyHL
The energy required to remove one mole of electrons from one mole of gaseous atoms in their ground state: X(g)→X+(g)+e−\text{X(g)} \rightarrow \text{X}^+\text{(g)} + e^-.
Convergence limitHL
(HL) The high-frequency point in an emission spectrum where successive lines merge together; it corresponds to the electron escaping the atom completely and is used to find the ionization energy.

Worked examples

1

(SL) A line in the emission spectrum of hydrogen has a wavelength of 4.34×10−74.34\times10^{-7} m. Calculate the frequency of this light and the energy of one photon. Use c=3.00×108c = 3.00\times10^8 m s−1^{-1} and h=6.63×10−34h = 6.63\times10^{-34} J s.

  1. 1
    Rearrange the booklet equation c=fλc = f\lambda to get f=c/λf = c/\lambda.
  2. 2
    f=(3.00×108)/(4.34×10−7)=6.91×1014f = (3.00\times10^8) / (4.34\times10^{-7}) = 6.91\times10^{14} Hz.
  3. 3
    Use the booklet equation E=hfE = hf for the photon energy.
  4. 4
    E=(6.63×10−34)(6.91×1014)=4.58×10−19E = (6.63\times10^{-34})(6.91\times10^{14}) = 4.58\times10^{-19} J.

Answer: Frequency f=6.91×1014 Hzf = 6.91\times10^{14}\ \text{Hz} and photon energy E=4.58×10−19 JE = 4.58\times10^{-19}\ \text{J}. This line is in the visible (Balmer) series of hydrogen.

2

(SL) Write the full electron configuration of a chromium atom (Z = 24) and explain why it is not [Ar] 4s2 3d4[\text{Ar}]\,4s^2\,3d^4.

  1. 1
    Chromium has 24 electrons. The Aufbau order is 1s, 2s, 2p, 3s, 3p, 4s, 3d.
  2. 2
    Filling normally to argon (18 electrons): 1s2 2s2 2p6 3s2 3p61s^2\,2s^2\,2p^6\,3s^2\,3p^6.
  3. 3
    Six electrons remain. The expected arrangement would be 4s2 3d44s^2\,3d^4.
  4. 4
    Chromium is an anomaly: one 4s electron moves into 3d to give a half-filled d sublevel.
  5. 5
    A half-filled 3d53d^5 set (one electron in each of the five d orbitals) is extra stable.

Answer: 1s2 2s2 2p6 3s2 3p6 4s1 3d51s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1\,3d^5, i.e. [Ar] 4s1 3d5[\text{Ar}]\,4s^1\,3d^5. The configuration 4s1 3d54s^1\,3d^5 is adopted because a half-filled d sublevel (3d53d^5) is more stable than 4s2 3d44s^2\,3d^4.

3

(HL) The convergence limit in the emission spectrum of hydrogen (transitions to n=1n=1) occurs at a frequency of 3.29×10153.29\times10^{15} Hz. Calculate the first ionization energy of hydrogen in kJ mol−1\text{kJ mol}^{-1}. Use h=6.63×10−34h = 6.63\times10^{-34} J s and NA=6.02×1023N_A = 6.02\times10^{23} mol−1^{-1}.HL

  1. 1
    The convergence limit corresponds to the electron escaping the atom, so its photon energy equals the ionization energy of one atom.
  2. 2
    Energy of one photon: E=hf=(6.63×10−34)(3.29×1015)=2.18×10−18E = hf = (6.63\times10^{-34})(3.29\times10^{15}) = 2.18\times10^{-18} J.
  3. 3
    Energy per mole: multiply by the Avogadro constant: Emol=(2.18×10−18)(6.02×1023)=1.31×106E_{\text{mol}} = (2.18\times10^{-18})(6.02\times10^{23}) = 1.31\times10^{6} J mol−1^{-1}.
  4. 4
    Convert to kJ mol−1^{-1}: divide by 1000: 1.31×106/1000=1.31×1031.31\times10^{6} / 1000 = 1.31\times10^{3} kJ mol−1^{-1}.

Answer: First ionization energy ≈1.31×103 kJ mol−1\approx 1.31\times10^{3}\ \text{kJ mol}^{-1} (about 1310 kJ mol−11310\ \text{kJ mol}^{-1}), in close agreement with the accepted value for hydrogen.

Common mistakes

  • ×Worrying about the order of 4s and 3d when writing a configuration: IB accepts either order (e.g. [Ar] 3d6 4s2[\text{Ar}]\,3d^6\,4s^2 or [Ar] 4s2 3d6[\text{Ar}]\,4s^2\,3d^6). What does matter is that 4s electrons are removed before 3d when forming ions, so Fe2+^{2+} is [Ar] 3d6[\text{Ar}]\,3d^6, not [Ar] 4s2 3d4[\text{Ar}]\,4s^2\,3d^4.
  • ×Forgetting Hund's rule in orbital diagrams: pairing electrons in one p orbital before the other p orbitals each have one. Fill degenerate orbitals singly (parallel spins) first.
  • ×Confusing energy levels (shells, n) with sublevels (s, p, d). The number of sublevels equals n; the number of electrons per sublevel is 2, 6, 10, 14 for s, p, d, f.
  • ×(HL) Forgetting to multiply the single-photon energy by the Avogadro constant when finding ionization energy per mole, or leaving the answer in J mol−1\text{J mol}^{-1} instead of converting to kJ mol−1\text{kJ mol}^{-1}.

Exam tips

  • ✓Learn the filling order and the two anomalies (Cr is 4s1 3d54s^1\,3d^5, Cu is 4s1 3d104s^1\,3d^{10}); these are very common exam targets.
  • ✓For emission spectra, remember the lines converge at the high-frequency (high-energy, short-wavelength) end, and that converging lines mean energy levels getting closer together as n increases.
  • ✓(HL) For successive ionization energies, count the electrons removed before the first big jump to find the group number; a log plot makes the jumps easy to spot.
  • ✓(HL) The convergence limit gives the ionization energy: use E=hfE = hf for one atom, then multiply by NAN_A and divide by 1000 for kJ mol−1\text{kJ mol}^{-1}. Keep the speed of light, h and NAN_A to 3 significant figures from the booklet.

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