Three teaching hours at both levels, outcomes 1.3.1 to 1.3.5, and three additional higher level hours, outcomes 1.3.6 and 1.3.7.
Guiding question: How can we model the energy states of electrons in atoms?
Structure 1.3 · Electron configurations
1Light, the electromagnetic spectrum and emission spectra 1.3.1 SL + HL Strong
Heat a little sodium chloride in a Bunsen flame and the flame turns orange-yellow; lithium gives crimson, copper blue-green. Pass the light from a hydrogen discharge tube through a prism and instead of a rainbow there are four sharp coloured lines on a dark background. Each element produces its own pattern of lines, so reliably that helium was discovered in the spectrum of the Sun before it was found on Earth. The lines are the first and most direct evidence for how electrons are arranged in atoms.
Wavelength, frequency and energy
Light is one region of the electromagnetic spectrum, which runs from gamma rays through X-rays, ultraviolet, visible light and infrared to microwaves and radio waves. All electromagnetic radiation travels through a vacuum at the same speed, c = 3.00 × 108 m s−1. Radiation can be described as a wave, with a wavelength λ (the distance between successive crests) and a frequency f (the number of waves passing a point per second), and also as a stream of packets of energy called photons. The two descriptions are linked by two equations given in the data booklet:
c = λf c in m s−1, λ in m, f in s−1 (Hz)
E = hf E = energy of one photon in J; h = 6.63 × 10−34 J s (the Planck constant)
Because c is constant, a shorter wavelength means a higher frequency, and because E is proportional to f, a higher frequency means more energy per photon. That single chain is all the qualitative relationship the syllabus asks for. Within the visible region, violet light (about 400 nm) has the shortest wavelength, highest frequency and most energetic photons; red light (about 700 nm) has the longest wavelength and the least energetic photons. Ultraviolet lies beyond violet and is more energetic still — which is why it can break bonds and damage skin — while infrared lies beyond red.
Continuous and line spectra
A hot solid such as a lamp filament emits a continuous spectrum: every wavelength is present and the colours merge into one another. A gaseous element excited by heat or an electric discharge emits a line spectrum: only certain wavelengths appear, as separate lines. Emission spectra are produced when atoms emit photons as electrons in excited states return to lower energy levels. The energy of each photon equals exactly the energy difference between the two levels, ΔE = hf. An atom has only certain allowed energy levels, so only certain energy differences — and therefore only certain frequencies — are possible. A line spectrum is thus evidence that the energy of an electron in an atom is quantized: restricted to fixed values.
In emission, an electron falls from a higher to a lower energy level and a photon is released. In absorption, an electron takes in a photon of exactly the right energy and is promoted to a higher level; the absorbed wavelengths appear as dark lines in an otherwise continuous spectrum. Each element's emission and absorption lines occur at the same wavelengths.
Question. When a chlorine emission spectrum is produced there is a strong line at 453 nm. Determine the energy of the photon emitted, in J. [2]
E = hc/λ = (6.63 × 10−34 × 3.00 × 108) ÷ (453 × 10−9) = 4.39 × 10−19 J
The step that trips people is the nanometre: 453 nm is 453 × 10−9 m. Check: visible photons carry energies of order 10−19 J, so an answer of 10−22 or 10−16 J signals a unit error, not chemistry.
Exam focus · what the published papers show
A multiple-choice item asks how emission spectra are formed and offers every combination of absorbed/emitted with promoted/returning. Only one is right: photons are emitted when promoted electrons return to a lower energy level. The word emission is the clue and it is in the question.
A report lists knowing that atomic emission spectra provide evidence for the electron energy levels in an atom among the things candidates handled well — and a separate item offers the line emission spectra of hydrogen produce four visible lines as a wrong answer to a question about evidence that matter is composed of atoms. Be precise: a line spectrum is evidence that energy levels are discrete, not by itself evidence for the existence of atoms.
2The hydrogen line emission spectrum 1.3.2 SL + HL Repeated
The line emission spectrum of hydrogen provides evidence for the existence of electrons in discrete energy levels, which converge at higher energies. The energy levels are numbered n = 1, 2, 3 … outwards from the nucleus; n = 1 is the ground state.
Hydrogen has one electron, so its spectrum is the simplest. The lines fall into groups, each group consisting of all the transitions that end on the same level. Transitions ending at n = 1 release the largest energies and lie in the ultraviolet. Transitions ending at n = 2 give the four lines in the visible region (656 nm red, 486 nm blue-green, 434 nm blue, 410 nm violet). Transitions ending at n = 3 release smaller energies and lie in the infrared. The names of the series are not assessed.
Within each group the lines get closer together towards higher frequency (shorter wavelength) and finally merge. This convergence of the lines mirrors the convergence of the energy levels themselves: the higher the level, the smaller the gap to the next. The point where the lines of a group merge is the convergence limit; for the group ending at n = 1 it corresponds to an electron being removed from the ground state altogether — ionization, which is developed at higher level in section 6.
Transitions to n = 2 from n = 3, 4, 5 … release energies that grow by smaller and smaller amounts, because the upper levels crowd together. So the lines crowd towards a limit. The convergence of the lines is a picture of the convergence of the levels. A published item asks which observation is the best evidence that energy levels are closer together further from the nucleus; the answer is that the lines in an emission spectrum converge at higher energy — not that ionization energies decrease down a group, which is true but concerns different atoms.
Electron configurations — the hydrogen emission spectrum
Click any transition. The energy gap, the wavelength and the position of the line are all computed from one equation: En = −13.6/n² eV.
Energy levels converge towards n = ∞. Click a downward arrow to emit a photon; the series a line belongs to is set by the level it lands on.
Where that photon lands. The visible window is 380–700 nm — only transitions to n = 2 (Balmer) fall inside it, which is why hydrogen's visible spectrum has just four lines.
The lines crowd together at short wavelength because the levels themselves converge. That convergence limit is how the ionisation energy of hydrogen is measured.
Exam focus · what the published papers show
Question. Which electron transition corresponds to the red line in the hydrogen line emission spectrum? An energy-level diagram shows transitions between n = 7, 4, 3, 2 and 1.
Two moves. Red is visible, and visible lines are transitions ending at n = 2 — so any transition ending at n = 1 (ultraviolet) or n = 3 (infrared) is out. Red is the least energetic visible colour, so it is the smallest gap ending at n = 2: n = 3 → n = 2.
A report says this item proved challenging: it needs both moves, and either alone gives a plausible wrong answer.
Asked to outline the model of electron configuration deduced from the hydrogen line emission spectrum, the scheme credits electrons in discrete/specific/certain/different shells/energy levels ✔ and energy levels converge/get closer together at higher energies ✔ — with the note Do not give marks for answers that refer to the lines in the spectrum. The question asks for the model the lines imply, not a description of the lines.
Asked for two limitations of a diagram of hydrogen energy levels, one scheme accepts, among others, does not represent sub-levels/orbitals, only applies to atoms with one electron/hydrogen and does not consider probability of finding electron at different positions — but refuses does not represent distance «from nucleus». The simple level model is the starting point that sublevels and orbitals refine.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
How are emission spectra formed?
Which are correct statements about the emission spectrum of hydrogen in the visible region?
I. The red line has a lower energy than the blue line.
II. The lines converge at longer wavelength.
III. The frequency of the blue line is greater than the frequency of the red line.
Which electron transition in the hydrogen atom emission spectrum emits radiation with the longest wavelength?
Which statements are correct for the emission spectrum of hydrogen?
I. The lines converge at higher frequencies.
II. Electron transitions to n = 2 are responsible for lines in the visible region.
III. Lines are produced when electrons move from lower to higher energy levels.
Which statement correctly describes the atomic emission spectrum of hydrogen?
The emission spectrum of an element can be used to identify it.
(a)(i) Draw the first four energy levels of a hydrogen atom on an energy axis, labelling n = 1, 2, 3 and 4. [1]
(ii) Draw the lines, on your diagram, that represent the electron transitions to n = 2 in the emission spectrum. [1]
Outline the model of electron configuration deduced from the hydrogen line emission spectrum (Bohr’s model). [2]
The diagram represents possible electron energy levels in a hydrogen atom, labelled n = 1 to n = 6 and n = ∞.
(i) All models have limitations. Suggest two limitations to this model of the electron energy levels. [2]
(ii) Draw an arrow, labelled X, to represent the electron transition for the ionization of a hydrogen atom in the ground state. [1]
(iii) Draw an arrow, labelled Z, to represent the lowest energy electron transition in the visible spectrum. [1]
Solutions and mark-scheme guidance · Set 1D
1D.1 D
Emission means photons given out, which happens when an excited electron falls back. B describes absorption; A and C mix the two.
1D.2 B
Red has the longest wavelength, lowest frequency and lowest energy of the visible lines, so I and III are correct. The lines converge at shorter wavelength (higher frequency), so II is false.
1D.3 D
Longest wavelength means smallest energy gap. n = 3 → 2 is far smaller than any transition to n = 1. B is an absorption, not an emission.
1D.4 A
I and II are correct. III describes absorption.
1D.5 B
A line spectrum, because the energy levels are discrete; converging at high frequency, because the levels get closer together as their energy rises.
1D.6 [2]
(i) Four levels showing convergence at higher energy ✔ — equally spaced levels do not score.
(ii) Arrows pointing down from n = 3 to n = 2 AND from n = 4 to n = 2 ✔. Upward arrows would be absorption.
1D.7 [2]
Electrons in discrete/specific/certain shells or energy levels ✔; the energy levels converge/get closer together at higher energies, or with distance from the nucleus ✔. Accept an appropriate diagram. Do not give marks for answers that refer to the lines in the spectrum.
1D.8 [4]
(i) Any two: does not represent sub-levels/orbitals; only applies to atoms with one electron; does not explain why only certain energy levels are allowed; treats the atom as isolated; does not consider the number of electrons a level can hold; does not consider the probability of finding the electron at different positions ✔✔. Do not accept “does not represent distance «from nucleus»”.
(ii) An upward arrow X from n = 1 to n = ∞ ✔.
(iii) An arrow between n = 3 and n = 2 ✔.
3Main energy levels and their capacity 1.3.3 SL + HL Repeated
The main energy level (shell) is given an integer number, n, and can hold a maximum of 2n2 electrons.
| n | Maximum electrons, 2n2 | Sublevels present | Orbitals |
|---|---|---|---|
| 1 | 2 | 1s | 1 |
| 2 | 8 | 2s, 2p | 1 + 3 = 4 |
| 3 | 18 | 3s, 3p, 3d | 1 + 3 + 5 = 9 |
| 4 | 32 | 4s, 4p, 4d, 4f | 1 + 3 + 5 + 7 = 16 |
The formula follows from what the next outcome adds. Level n contains n sublevels and n2 orbitals, and each orbital holds two electrons. The highest main level occupied in an atom's ground state is its period number (Structure 3.1): sodium, 2,8,1, has electrons in three main levels and is in period 3.
Multiple-choice items ask for the maximum number of electrons in a named level, for n = 3 and n = 4. The distractors catch two habits: 50 is the answer for n = 5, offered against n = 4; 8 and 18 belong to the levels either side. Compute 2n2 deliberately: square first, then double.
4Sublevels, orbitals and their shapes 1.3.4 SL + HL Strong
The emission spectra of atoms with more than one electron contain more lines than a model of single levels can explain. A more detailed model divides each main energy level into sublevels of successively higher energy, labelled s, p, d and f. Each sublevel consists of a fixed number of atomic orbitals: one s, three p, five d and seven f. An orbital is a region of space around the nucleus in which there is a high probability of finding an electron; it is a statement about probability, not a path like a planet's orbit.
| Sublevel | Orbitals | Maximum electrons | First appears in level |
|---|---|---|---|
| s | 1 | 2 | n = 1 |
| p | 3 | 6 | n = 2 |
| d | 5 | 10 | n = 3 |
| f | 7 | 14 | n = 4 |
You must recognize the shapes and orientation of an s orbital and the three p orbitals. An s orbital is spherical and centred on the nucleus; the 2s orbital is larger than 1s. A p orbital has two lobes on opposite sides of the nucleus, with a node — zero probability — at the nucleus. The three p orbitals of a sublevel (px, py, pz) are identical in shape and energy and lie along three mutually perpendicular axes.
Sublevels also explain the shape of the periodic table: the s-block is two groups wide because an s sublevel holds two electrons, the p-block six, the d-block ten and the f-block fourteen (Structure 3.1).
Exam focus · what the published papers show
Asked to sketch the occupied orbitals of boron, the scheme notes: Px,y or z can be used. M2 cannot be awarded if labels of orbital types are missing or incorrect. Node of p orbital must be at the origin. In another session the credit is for 1s AND 2s as spheres and one or more 2p orbital(s) as figure(s) of 8 shape(s) of any orientation.
A well-drawn but unlabelled sketch loses the mark; so does a p orbital with its node displaced from the nucleus.
One report records candidates who drew one lobe on the p orbital, instead of two. Another: one-third of candidates drew all three p-orbitals, but without the labels, these could be mistaken for d-orbitals. Draw exactly what the question names, and label it.
A report asks that candidates use appropriate terminology, for instance, distinguishing clearly between orbitals, energy sub-levels or subshells, and principal energy levels or shells. A level is n. A sublevel is s, p, d or f within a level. An orbital is one of the regions within a sublevel and holds at most two electrons.
5Writing electron configurations 1.3.5 SL + HL Strong
Each orbital has a defined energy for a given configuration, and electrons in the ground state occupy the lowest-energy arrangement available. Three rules decide that arrangement.
Aufbau principle — electrons occupy the lowest-energy sublevel available first.
Pauli exclusion principle — an orbital holds at most two electrons, and they must have opposite spins.
Hund’s rule — within a sublevel, electrons occupy orbitals singly, with parallel spins, before any orbital takes a pair.
Up to Z = 36 the order of filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. The only surprise is that 4s fills before 3d: in potassium and calcium the 4s sublevel is lower in energy than 3d. Reading the periodic table left to right, period by period, reproduces this order.
A full electron configuration lists every occupied sublevel with the number of electrons as a superscript: sulfur is 1s2 2s2 2p6 3s2 3p4. The condensed form replaces the inner electrons with the preceding noble gas in square brackets: [Ne] 3s2 3p4. An orbital diagram shows each orbital as a box and each electron as a half-arrow, with the boxes arranged to show relative energy; it is the representation that reveals unpaired electrons.
Chromium and copper
Two elements up to Z = 36 do not follow the simple filling order. Chromium is [Ar] 3d5 4s1, not [Ar] 3d4 4s2, and copper is [Ar] 3d10 4s1, not [Ar] 3d9 4s2. In each, a half-filled or completely filled 3d sublevel with one 4s electron is the lower-energy arrangement. These two exceptions must be learned.
Configurations of ions
For main-group ions, add or remove electrons from the outermost sublevel: O2− is 1s2 2s2 2p6, isoelectronic with neon; Ca2+ is [Ar]. For transition elements the 4s electrons are removed before the 3d. Once 3d is occupied, 4s is the higher-energy sublevel, so it empties first even though it filled first. Iron is [Ar] 3d6 4s2; Fe2+ is [Ar] 3d6 and Fe3+ is [Ar] 3d5.
Build an electron configuration
Fill the sub-levels in energy order, then read them back in shell order. The two things that catch people out — chromium and copper, and which electrons a transition metal loses first — are built into the model rather than listed as facts.
Orbital diagram. Boxes are orbitals, arrows are electrons. Within a sub-level the electrons occupy separate orbitals with parallel spins before any orbital takes a pair — Hund's rule, drawn rather than asserted.
The filling order. 4s fills before 3d because it is lower in energy when empty — but once occupied it lies above 3d, which is why it is also the first to be lost.
Exam focus · what the published papers show
Question. State the full electron configuration of an oxygen atom and the number of unpaired electrons in that atom. [2]
Z = 8: 1s takes 2, 2s takes 2, 2p takes the remaining 4: 1s2 2s2 2p4. In the three 2p boxes Hund’s rule places one electron in each before the fourth pairs up, leaving 2 unpaired. The average mark was 1.32 out of 2; the configuration itself was the part weaker candidates could not write.
The most-reported misconception in the topic. One report: Candidates struggled with writing the electron configuration of Co2+. Most candidates removed two of the 3d electrons of Co instead of the 4s electrons. Only about 30 % were correct. Another, later: Many candidates failed to recognize that 4s electrons are the ones removed.
Cobalt is [Ar] 3d7 4s2; Co2+ is [Ar] 3d7, not [Ar] 3d5 4s2. A published scheme prints exactly 1s2 2s2 2p6 3s2 3p6 3d7.
| Cr (24) | Expected [Ar] 3d4 4s2; actual [Ar] 3d5 4s1 — half-filled d sublevel |
| Cu (29) | Expected [Ar] 3d9 4s2; actual [Ar] 3d10 4s1 — filled d sublevel |
| Cr2+ | From [Ar] 3d5 4s1, remove the 4s electron and then one 3d: [Ar] 3d4. [Ar] 3d5 is Cr+. |
A report on the copper item confirms the trap: 56 % chose correctly, and the most commonly chosen distractor was 1s2 2s2 2p6 3s2 3p6 4s2 3d9 — the configuration you get by not knowing the exception.
One item gave calcium nitride and asked for the configurations of the species present. Reports: Quite a few students selected C over the correct answer D although the question asked for the electron configuration of the species in the compound not the neutral elements. In Ca3N2 the species are Ca2+, [Ar], and N3−, [Ne]. Read whether the question says atom, ion or species.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
What is the maximum number of electrons in energy level n = 4?
How many p-orbitals are occupied in a phosphorus atom?
Which is the electron configuration of a chromium atom in the ground state?
What is the condensed electron configuration of the Fe2+ ion?
Which species has the same electron configuration as argon?
Copper is widely used as an electrical conductor. Draw arrows in the boxes to represent the electronic configuration of copper in the 4s and 3d orbitals. [1]
Bromine can form the bromate(V) ion, BrO3−.
(a)(i) State the electron configuration of a bromine atom. [1]
(ii) Sketch the orbital diagram of the valence shell of a bromine atom (ground state) on an energy axis. Use boxes to represent orbitals and arrows to represent electrons. [1]
(a)(i) Annotate and label the ground state orbital diagram of boron, [He] + boxes, using arrows to represent electrons. [1]
(ii) Sketch the shapes of the occupied orbitals identified in part (a)(i), on x, y, z axes, and name each orbital type. [2]
State the condensed electron configurations for Cr and Cr3+. [2]
Solutions and mark-scheme guidance · Set 1E
1E.1 C
2n2 = 2 × 16 = 32 (4s, 4p, 4d and 4f: 2 + 6 + 10 + 14). 50 is the value for n = 5.
1E.2 D
P is 1s2 2s2 2p6 3s2 3p3. The three 2p orbitals are full and, by Hund’s rule, the three 3p electrons occupy all three 3p orbitals singly: 3 + 3 = 6 occupied p orbitals. The count is of orbitals, not sublevels.
1E.3 A
2s is paired with opposite spins (Pauli), which rules out C and D; the two 2p electrons occupy separate orbitals with parallel spins (Hund), which rules out B.
1E.4 D
Chromium is one of the two exceptions: 3d5 4s1. C is the configuration predicted by the simple filling order; A has only 23 electrons; B is Cr3+.
1E.5 A
Fe is [Ar] 3d6 4s2; the two 4s electrons are removed first. B removes 3d electrons instead — the misconception examiners report most often. D is the neutral atom.
1E.6 B
Ca2+ has 20 − 2 = 18 electrons, like argon. Br− is isoelectronic with krypton; Al3+ and Si4+ with neon.
1E.7 [1]
4s: one arrow; 3d: five boxes each containing a pair of opposite arrows ✔ — the 3d10 4s1 exception, drawn.
1E.8 [2]
(i) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 OR [Ar] 4s2 3d10 4p5 ✔ (accept 3d before 4s).
(ii) 4s box with a pair, and three 4p boxes, higher on the energy axis, containing two pairs and one single arrow ✔ (accept double-headed arrows).
1E.9 [3]
(i) Arrows AND identifies 2s and 2p sub-orbitals: 2s with a pair, one 2p box with one arrow ✔ (accept “hooks” for electrons).
(ii) s: a sphere centred on the origin, labelled s ✔; p: a figure-of-eight along one axis, labelled p ✔. M2 cannot be awarded if labels of orbital types are missing or incorrect; the node of the p orbital must be at the origin.
1E.10 [2]
Cr: [Ar] 4s1 3d5 ✔ (accept [Ar] 3d5 4s1). Cr3+: [Ar] 3d3 ✔ (accept [Ar] 3d3 4s0). The scheme gives [1 max] for two correct full electron configurations — the question asked for condensed ones.
Structure 1.3 · Additional higher level
6Ionization energy: trends, discontinuities and the convergence limit 1.3.6 HL only Strong
The first ionization energy is the minimum energy required to remove one mole of electrons from one mole of gaseous atoms in their ground state, forming one mole of gaseous 1+ ions.
X(g) → X+(g) + e− ΔH > 0, kJ mol−1
Three factors decide how strongly the outer electron is held: the nuclear charge, its distance from the nucleus (the energy level it occupies), and the shielding by inner electrons, which repel it and reduce the attraction it feels.
Trends
Down a group the first ionization energy decreases. The outer electron is in a higher main level, further from the nucleus and more shielded by extra inner shells; the increase in nuclear charge is outweighed. Across a period it increases in general. The nuclear charge rises by one with each element while electrons are added to the same main level, so shielding changes little and the outer electrons are held more strongly. The graph of first ionization energy against atomic number therefore has a peak at each noble gas and a minimum at each alkali metal.
The two discontinuities
The rise across a period is broken twice, and each break is evidence for sublevels.
- Be → B (Mg → Al). Boron’s outer electron is in a 2p orbital, beryllium’s in 2s. The 2p sublevel is higher in energy and its electron is shielded to some extent by the 2s electrons, so less energy is needed to remove it, despite boron’s extra proton.
- N → O (P → S). Nitrogen is 2p3, with one electron in each 2p orbital. Oxygen is 2p4, so two electrons share one orbital. The repulsion between the paired electrons makes one of them easier to remove, so oxygen’s first ionization energy is lower than nitrogen’s.
Ionization energy from the convergence limit
In the emission spectrum of hydrogen, the group of lines ending at n = 1 converges to a limit. Beyond it the spectrum is continuous, because the electron is no longer bound and can have any energy. The frequency of the convergence limit therefore corresponds to the energy needed to remove the electron from the ground state completely: the ionization energy. The energy of one photon at that frequency is the ionization energy of one atom; multiplying by the Avogadro constant gives the molar value.
IE (J mol−1) = h × flimit × NA and f = c ÷ λ
Question. The convergence limit of the group of lines ending at n = 1 in the hydrogen spectrum is at 91.2 nm. Calculate the first ionization energy of hydrogen in kJ mol−1.
| Frequency | f = c/λ = 3.00 × 108 ÷ 91.2 × 10−9 = 3.29 × 1015 s−1 |
| Energy per atom | E = hf = 6.63 × 10−34 × 3.29 × 1015 = 2.18 × 10−18 J |
| Per mole | 2.18 × 10−18 × 6.02 × 1023 = 1.31 × 106 J mol−1 = 1310 kJ mol−1 |
Check: the data-booklet value is 1312 kJ mol−1; the small difference comes from rounding the constants.
Exam focus · what the published papers show
| M1 | nuclear charge / number of protons increases for both |
| M2 | Li and Be outer electrons have the same subshell/shielding |
| M3 | electron in B lost from p-subshell whereas that in Be lost from s-subshell |
| M4 | outer electron in B/p-subshell experiences greater shielding / has higher energy |
And the note that decides how you argue: Do not accept explanations invoking distance of electrons from nucleus. Across a period distance barely changes; the answer is about which sublevel the electron comes from. The average mark was about 2 out of 4; a report notes that few mentioned that the 2p level is at a higher energy than 2s.
For a down-a-group comparison a scheme accepts electron removed from higher orbital/shell/energy level / further away from the nucleus but adds Do not accept increase in atomic radius on its own. A report gives the general instruction: Periodic trends (ionisation energy, etc.) should refer to the location of the valence electron, not just a physical description of the atom. Say where the electron is and what it feels.
The scheme: frequency/wavelength of the radiation at convergence limit is proportional to the ionization energy, with Accept highest frequency/shortest wavelength. A report calls the question probably the most difficult question on the paper, with only about a fifth of candidates connecting ionization energy to the convergence limit — and notes that even very strong students described what happened in the ionization of an electron, but never described or named the limit of convergence.
One report, on a thallium version: candidates failed to use Avogadro's number (to calculate per atom) and also did not convert the ionisation energy of Tl from kJ to J. The next session, on a phosphorus version: candidates did not convert the ionisation energy value in the data book to joules.
Forgetting the Avogadro constant leaves an energy per mole, wrong by a factor of 6 × 1023; forgetting kJ → J makes the energy 1000 times too small. Write the units on every line.
| Element | IE / kJ mol−1 | E per atom / J | f / s−1 | λ / m |
|---|---|---|---|---|
| thallium | 589 | 9.78 × 10−19 | 1.48 × 1015 | 2.03 × 10−7 |
| iron | 762 | 1.27 × 10−18 | 1.91 × 1015 | 1.57 × 10−7 |
| phosphorus | 1012 | 1.68 × 10−18 | 2.54 × 1015 | not asked |
Thallium in full: E = 589 × 103 ÷ 6.02 × 1023 = 9.78 × 10−19 J; f = E ÷ h = 1.48 × 1015 s−1; λ = c ÷ f = 2.03 × 10−7 m (203 nm). Check: 203 nm is ultraviolet — ionization needs more energy than any visible transition. The phosphorus question stops at the frequency and is worth 2 marks rather than 3.
7Successive ionization energies 1.3.7 HL only Strong
Electrons can be removed from an atom one at a time. The second ionization energy is X+(g) → X2+(g) + e−, the third X2+(g) → X3+(g) + e−, and so on. Each is larger than the one before, because each electron is removed from an increasingly positive ion: the same nuclear charge now holds fewer electrons, which repel one another less. But the increase is not smooth. When all the electrons of the outer main level have gone, the next must come from a level closer to the nucleus and much less shielded, and the ionization energy jumps sharply.
Counting the electrons removed before the first large jump gives the number of valence electrons, and therefore the group. Magnesium’s jump comes between the second and third ionization energies: two valence electrons, group 2, and a stable Mg2+ ion. A logarithmic scale is used because the values span more than two orders of magnitude; on a linear scale the first few would be indistinguishable from zero. The pattern of jumps is evidence for main energy levels, just as the dips in first ionization energy are evidence for sublevels.
Exam focus · what the published papers show
Question. Successive ionization energies of element E, in kJ mol−1: 1000, 2295, 3375, 4565, 6950, 8490, 27 107. Identify the group in which E is located, giving a reason. [1]
Take the ratios, not the differences. The consecutive ratios are 2.30, 1.47, 1.35, 1.52, 1.22 and 3.19. The largest by a clear margin is between the sixth and seventh electrons: six come from the outer level.
Group 16. The published answer is 16 AND 7th IE much higher — one mark requiring both the group and the reason. A report notes candidates performed well in identifying the jump in ionisation energy but sometimes misinterpreted the data and how it related to the group number.
In the data above the first ratio is 2.30 — larger than the next four. That is normal: the second electron is always removed from a positive ion. The test is not “where is there a big jump” but “where is the biggest jump”. Stopping at the first large increase would give group 1 instead of group 16.
Question. First three ionization energies, kJ mol−1: X 900, 1757, 14 849; Y 1086, 2350, 4620. Which pair of elements are X and Y?
For X the third value is more than eight times the second: two valence electrons — beryllium. For Y the values rise smoothly with no jump in the first three: more than three valence electrons — carbon. A report records that the vast majority of students correctly used the successive ionisation energy data to identify the pair of elements.
Comparing iron and beryllium, the scheme credits: IE values of Fe gradually increase AND IE values of Be show a sudden rise; Be always loses 2 electrons / forms Be2+; and further IEs of Fe are close to second IE, so the oxidation state/number of electrons Fe loses can vary. A report records that the vast majority argued from orbitals rather than following the instruction to use successive ionization energies — correct chemistry, but not the evidence the question named.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
The elements argon, potassium and calcium are consecutive in the periodic table. Which gives the correct order of increasing first ionization energies?
X, Y and Z represent the successive elements Ne, Na and Mg, but not necessarily in that order. First ionization energies / kJ mol−1: X 2081; Y 496; Z 738. What is the order of increasing atomic number?
Which statement explains why the second ionization energy of aluminium is higher than the first ionization energy of magnesium?
Which element is in group 13?
| Ionization energy / kJ mol−1 | ||||
|---|---|---|---|---|
| 1st | 2nd | 3rd | 4th | |
| A. | 789 | 1580 | 3230 | 4360 |
| B. | 578 | 1820 | 2750 | 11 600 |
| C. | 738 | 1450 | 7730 | 10 500 |
| D. | 496 | 4560 | 6910 | 9540 |
A period 3 element, M, forms an oxide of the type M2O. Which represents the first four successive ionization energies of M?
| First | Second | Third | Fourth | |
|---|---|---|---|---|
| A. | 496 | 4563 | 6913 | 9544 |
| B. | 738 | 1451 | 7733 | 10541 |
| C. | 578 | 1817 | 2745 | 11578 |
| D. | 787 | 1577 | 3232 | 4356 |
Successive ionization energies of an element X are 740, 1450, 7730 and 10 540 kJ mol−1. What energy, in kJ mol−1, is required for element X to reach its most stable oxidation state in ionic compounds?
(d)(i) Explain the convergence of lines in a hydrogen emission spectrum. [1]
(ii) State what can be determined from the frequency of the convergence limit. [1]
The electron configuration of copper makes it a useful metal. (a) Determine the frequency of a photon that will cause the first ionization of copper. Use sections 1, 2 and 8 of the data booklet. [2]
Explain why the first ionization energy of sulfur is lower than that of phosphorus. [2]
Explain the general increase in trend in the first ionization energies of the period 3 elements, Na to Ar. [2]
Sketch a graph to show the relative values of the successive ionization energies of boron, for electrons 1 to 5. [2]
Plot the relative values of the first four ionization energies of sodium. [1]
Solutions and mark-scheme guidance · Set 1F
1F.1 D
K begins period 4: its outer electron is in 4s, far from the nucleus and well shielded — lowest. Ca has one more proton in the same level — higher. Ar, a noble gas with its outer electron in 3p, is highest.
1F.2 A
The noble gas Ne has the highest value (X = 2081); Na, the next element, the lowest (Y = 496); Mg follows (Z = 738). So Ne < Na < Mg is X < Y < Z.
1F.3 D
The second IE of Al removes an electron from Al+, 1s2 2s2 2p6 3s2; the first IE of Mg removes it from Mg, with the same configuration. The same electrons, the same shielding, and one extra proton in Al.
1F.4 B
Group 13 means three valence electrons, so the big jump comes between the third and fourth: B (2750 → 11 600). C jumps after two (group 2), D after one (group 1), and A has no jump in the first four (group 14).
1F.5 A
M2O contains M+, so M is in group 1 and the jump comes after the first electron. The item links Structure 1.3 to the formula of an ionic compound (Structure 2.1).
1F.6 C
The jump after the second electron shows X forms X2+. Reaching it requires both the first and second ionization energies: 740 + 1450 = 2190 kJ mol−1. B is the second alone.
1F.7 [2]
(i) Energy levels are closer together at high energy / high frequency / short wavelength ✔.
(ii) Ionization energy ✔.
1F.8 [2]
E = 745 000 J mol−1 ÷ 6.02 × 1023 mol−1 = 1.24 × 10−18 J ✔; ν = E ÷ h = 1.24 × 10−18 ÷ 6.63 × 10−34 = 1.87 × 1015 s−1 ✔. Award [2] for correct final answer. Award [1] for 1.12 × 1039 Hz — the value obtained by forgetting to divide by the Avogadro constant.
1F.9 [2]
P has three unpaired electrons in the 3p sublevel AND S has one full 3p orbital (P: [Ne]3s23px13py13pz1; S: [Ne]3s23px23py13pz1) ✔; repulsion between the paired electrons in sulfur makes one easier to remove ✔. Accept orbital diagrams for M1. This scheme also accepts “removing electron from S gives more stable half-filled sub-level” for M2; the repulsion argument is the one that works in every session.
1F.10 [2]
Increasing number of protons / nuclear charge ✔; atomic radius decreases OR same number of shells / electrons occupy the same shell OR similar shielding by inner electrons ✔. A general rise across a period is SL content (Structure 3.1); the discontinuities are the higher level extension.
1F.11 [2]
IE1 < IE2 < IE3 < IE4 < IE5 ✔; the largest increase between the third and fourth ionization energies ✔ — boron has three valence electrons (2s2 2p1), and the fourth comes from 1s.
1F.12 [1]
A rising plot in which the difference between the first two points is much larger than between points 2 and 3, and 3 and 4 ✔ — one valence electron, so the jump comes after the first.
Review · Structure 1.3
8Misconceptions, the examiner’s view, and the question types
- “Emission lines are produced when electrons jump up.” Why it is wrong: jumping up absorbs energy. Correct model: emission is an electron falling to a lower level and releasing a photon of energy ΔE = hf. Consequence: the distractor in every emission item.
- “Longer wavelength means more energy.” Longer wavelength means lower frequency and less energy per photon.
- “The lines converge because the electrons slow down.” They converge because the energy levels converge.
- “An orbital is the path an electron follows.” It is a region of high probability of finding the electron.
- “Transition metals lose their 3d electrons first because they were added last.” The 4s electrons are lost first.
- “Level, sublevel and orbital mean the same.” n; s/p/d/f; one box holding two electrons.
- HL “Boron’s electron is easier to remove because it is further from the nucleus.” Schemes refuse distance: the reason is the higher-energy 2p sublevel.
- HL “The first large increase in successive IEs is the jump.” Compare all the ratios; the second IE always rises noticeably.
- HL “The ionization energy from a spectrum is hf.” hf is per atom; multiply by NA and convert J to kJ.
Evidence base: the IB Diploma chemistry subject reports quoted in the exam-focus sections of this page.
Answered well: knowing that emission spectra are evidence for energy levels; writing configurations of main-group atoms; identifying a pair of elements from their successive ionization energies (the vast majority of students); spotting the jump in successive ionization data.
Found difficult: (1) configurations of transition-metal ions, where most candidates removed 3d rather than 4s electrons (about 30 % correct for Co2+); (2) the chromium and copper exceptions (56 % correct, with the Aufbau-order configuration the favourite distractor); (3) choosing the transition that gives a named visible line; (4) sketching and labelling orbitals, including drawing one lobe for a p orbital and unlabelled p sets; (5) using level, sublevel and orbital precisely; (6) at HL, explaining discontinuities in terms of sublevels and electron repulsion rather than distance, naming the convergence limit (about a fifth of candidates), and carrying out the kJ → J and ÷ NA conversions.
What successful answers did: named the sublevel from which the electron is removed and the factor that changes; used ratios to find the largest ionization-energy jump; labelled every sketch; wrote units on every line of a calculation; and used the evidence the question specified.
Seven question types cover most of this topic.
| If the question asks… | …then |
|---|---|
| Which transition / which line? | Region first (n = 1 UV, n = 2 visible, n = 3 IR), then rank by energy gap. |
| Photon energy or wavelength | E = hf, c = λf; nm × 10−9 = m; check the order of magnitude. |
| Maximum electrons in a level | 2n2: square, then double. |
| Configuration of an atom or ion | Aufbau order; Cr and Cu exceptions; remove 4s before 3d. |
| Sketch an orbital | Sphere for s; two lobes with the node at the origin, along an axis, for p; label. |
| Explain an IE trend or dip HL | Nuclear charge, shielding, sublevel; paired-electron repulsion for N → O. No “distance” across a period. |
| Group from successive IEs HL | Largest ratio between consecutive values; electrons before it = group (13–18: add 10). |
9Quick check
- State the maximum number of electrons in the n = 2 and n = 5 levels.
- A red line and a violet line appear in the visible hydrogen spectrum. Which corresponds to the larger energy transition, and why?
- Write the full and condensed configurations of a sulfur atom.
- Give the configuration of Fe3+ and the number of unpaired electrons.
- Explain why copper is not [Ar] 3d9 4s2.
- State two features a published scheme requires of a p-orbital sketch.
- HL Calculate the energy per atom, in J, and the frequency corresponding to an ionization energy of 1000 kJ mol−1.
- HL Explain why the first ionization energy of boron is lower than that of beryllium without using the word radius.
- HL Successive ionization energies jump sharply between the third and fourth. Deduce the group.
- HL A calculated convergence-limit wavelength is 2.0 × 10−4 m. What has probably gone wrong?
| 1 | 8 and 50 (2n2). |
| 2 | Violet: shorter wavelength, higher frequency, higher photon energy. Both end at n = 2; violet starts from a higher level. |
| 3 | 1s2 2s2 2p6 3s2 3p4; [Ne] 3s2 3p4. |
| 4 | [Ar] 3d5 — two 4s electrons then one 3d removed; five unpaired, one in each 3d orbital. |
| 5 | A filled 3d sublevel with one 4s electron, [Ar] 3d10 4s1, is the lower-energy arrangement. |
| 6 | Labelled; node at the origin (the p orbital along an axis). An unlabelled sketch loses the mark. |
| 7 | 1.00 × 106 ÷ 6.02 × 1023 = 1.66 × 10−18 J; f = 1.66 × 10−18 ÷ 6.63 × 10−34 = 2.51 × 1015 s−1. |
| 8 | Boron’s outer electron is removed from the 2p sublevel, which is higher in energy (and partly shielded by 2s), while beryllium’s comes from 2s; this outweighs boron’s extra proton. |
| 9 | Three valence electrons: group 13. |
| 10 | 200 μm is infrared — the energy used was about 1000 times too small: the kJ → J conversion was missed. |
10Summary and knowledge organiser
Essential knowledge
- c = λf and E = hf: shorter wavelength → higher frequency → more energy per photon; violet > red; UV > visible > IR.
- A line emission spectrum arises from electrons falling between discrete energy levels; each line has ΔE = hf. A continuous spectrum contains all wavelengths.
- Hydrogen: transitions to n = 1 (UV), n = 2 (visible), n = 3 (IR); lines converge at higher energy because the levels converge.
- Level n holds 2n2 electrons; sublevels s, p, d, f have 1, 3, 5, 7 orbitals; each orbital holds two electrons of opposite spin.
- s orbitals are spherical; p orbitals have two lobes along x, y or z with a node at the nucleus.
- Aufbau, Pauli and Hund give configurations to Z = 36; Cr 3d54s1, Cu 3d104s1; transition-metal ions lose 4s first.
- HL First IE rises across a period (nuclear charge) with dips at groups 13 and 16 (p sublevel; paired-electron repulsion) and falls down a group (level, shielding).
- HL IE = h × flimit × NA; successive IE jumps give the number of valence electrons and the group.
Examination checklist
- Emission: electron falls, photon emitted. Absorption: electron promoted.
- Convert nm to m; check the answer’s order of magnitude.
- Write condensed configurations with the noble-gas core when asked; full when asked for “full”.
- Label every orbital sketch.
- HL Use sublevels and repulsion, not distance, for discontinuities; name the convergence limit; ÷ NA and kJ → J.
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| Spectra 1.3.1 | c = λf; E = hf; line vs continuous. | Emission = falling electron. Short λ = high E. |
| Hydrogen 1.3.2 | n=1 UV, n=2 visible, n=3 IR; convergence at high energy. | Red = 3 → 2. Lines converge because levels do. |
| Levels 1.3.3 | Capacity 2n2: 2, 8, 18, 32. | 50 is n = 5. |
| Sublevels 1.3.4 | s1, p3, d5, f7 orbitals; s sphere, p dumbbell. | Level ≠ sublevel ≠ orbital; label sketches. |
| Configurations 1.3.5 | 1s 2s 2p 3s 3p 4s 3d 4p; Aufbau, Pauli, Hund. | Cr, Cu exceptions; 4s lost before 3d. |
| HL IE trends 1.3.6 | Dips Be→B, N→O; IE = hfNA at the convergence limit. | No “distance” across a period; ÷ NA, kJ → J. |
| HL Successive IE 1.3.7 | Biggest jump → number of valence electrons → group. | Use ratios; the first increase is not the jump. |