Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
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In preparation: 0 of 7 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
S1.3.1 Emission spectrum
Emission spectrum
The pattern of electromagnetic radiation given out by a substance, spread out by wavelength (for example by a prism or diffraction grating). For atoms it is produced when electrons in excited states return to lower energy levels and the energy lost is emitted as photons.
Ground state and excited state
The ground state of an atom is its lowest-energy electron arrangement. When an atom absorbs energy (heat in a flame, or an electric discharge), an electron can move to a higher energy level; the atom is then in an excited state. An excited state is unstable, and the electron returns to a lower level, emitting the energy difference as a photon.
Photon
A discrete packet (quantum) of electromagnetic radiation. The energy of a photon is fixed by its frequency: the higher the frequency, the greater the energy. When an electron falls from a higher energy level to a lower one, one photon is emitted whose energy equals the difference in energy between the two levels.
Colour, wavelength, frequency and energy
Across the electromagnetic spectrum, wavelength and frequency are inversely related: as wavelength decreases, frequency increases, and photon energy increases with frequency. In the visible region, red light has the longest wavelength, lowest frequency and lowest photon energy; violet light has the shortest wavelength, highest frequency and highest photon energy. Infrared lies beyond red (lower energy) and ultraviolet beyond violet (higher energy).
Continuous spectrum
A spectrum containing all wavelengths over a range, with no gaps, such as the band of colours from red to violet produced when white light from a hot filament lamp or the Sun is passed through a prism.
Line spectrum
A spectrum containing only certain wavelengths. A line emission spectrum appears as separate bright coloured lines on a dark background. Each line corresponds to photons of one particular energy, emitted when electrons fall between two particular energy levels, so each element has its own characteristic line spectrum.
Students often think Light is given out when electrons are excited and jump up to higher energy levels. In fact No. An electron absorbs energy to move up to a higher level. A photon is emitted when an electron falls from a higher level to a lower one.
Students often think Atoms give out light when electrons are knocked out of them completely; the light is the energy released as the electrons escape. In fact No. Emission lines are produced by electrons moving between energy levels within the atom. The electron stays in the atom and falls to a lower level, emitting a photon.
S1.3.2 Discrete energy levels
Discrete energy levels
The electron in an atom can have only certain fixed energies, called energy levels. Because only these energies are allowed, only certain energy differences exist, so only photons of certain frequencies are emitted. The line emission spectrum of hydrogen is evidence for this: a continuous range of electron energies would produce a continuous spectrum.
Convergence of energy levels and spectral lines
The energy levels of the hydrogen atom get closer together as n increases (as their energy increases). As a result, within each set of lines in the emission spectrum the lines get closer together towards higher frequency (shorter wavelength) until they merge.
Hydrogen emission spectrum and energy transitions
Each line in the hydrogen emission spectrum is produced by an electron falling from a higher level to a lower one. Transitions ending at n = 1 involve the largest energy differences and produce lines in the ultraviolet region; transitions ending at n = 2 produce lines mainly in the visible region (the highest-energy lines of this set, near its convergence limit, lie in the near ultraviolet); transitions ending at n = 3 involve smaller energy differences and produce lines in the infrared region.
Students often think Each spectral line comes from a different electron in the atom, so the number of lines equals the number of electrons. In fact No. A hydrogen atom has only one electron. Each line corresponds to a different transition between two energy levels; a sample contains very many atoms, whose electrons make different transitions.
Students often think Energy levels are equally spaced, like the rungs of a ladder or the evenly drawn circles of a shell diagram, so each one-level fall releases the same energy. In fact No. The energy levels get closer together as their energy increases, which is why the lines in each set of the hydrogen spectrum converge at higher frequency.
S1.3.3 Main energy level and its maximum number of electrons
Main energy level and its maximum number of electrons
Main energy levels are numbered with an integer n = 1, 2, 3, … counting outwards from the nucleus. The main energy level n can hold a maximum of 2n² electrons: 2 for n = 1, 8 for n = 2, 18 for n = 3 and 32 for n = 4. This follows from the level containing n² orbitals, each holding a maximum of two electrons.
Students often think Electron shells fill 2, 8, 8, so the third energy level is full with 8 electrons. In fact No. The third main energy level can hold 2n² = 18 electrons: 2 in 3s, 6 in 3p and 10 in 3d.
Students often think Each orbital holds one electron, so the number of electrons a level or sublevel can hold equals its number of orbitals. In fact No. Each orbital holds up to two electrons of opposite spin, so a level with n² orbitals holds up to 2n² electrons.
S1.3.4 Sublevels
Sublevels
Each main energy level is divided into sublevels labelled s, p, d and f. Level n = 1 has only an s sublevel; n = 2 has s and p; n = 3 has s, p and d; n = 4 has s, p, d and f. In a many-electron atom the sublevels of one main level have successively higher energies in the order s < p < d < f.
Atomic orbital
A region of space around the nucleus in which there is a high probability of finding an electron. An orbital is not a path that the electron follows. Each orbital can hold a maximum of two electrons, which must have opposite spins.
Shape of an s orbital
An s orbital is spherical and centred on the nucleus, so it has no particular orientation. Each main energy level has one s orbital; the 2s orbital is larger than the 1s orbital.
Shapes and orientations of the p orbitals
A p orbital is dumbbell-shaped, with two lobes on opposite sides of the nucleus. Each p sublevel contains three p orbitals of equal energy, directed along the x, y and z axes at 90° to one another (px, py and pz).
Students often think Each sublevel (s, p, d, f) is a single orbital, so each sublevel holds just one pair of electrons. In fact No. A sublevel is a set of orbitals of equal energy: an s sublevel has one orbital, a p sublevel three, a d sublevel five and an f sublevel seven. Each orbital holds up to two electrons.
Students often think An orbital is a fixed path, like a planet’s orbit, that an electron travels along at a set distance from the nucleus. In fact No. An orbital is a region of space where there is a high probability of finding an electron. It is not a track or orbit.
S1.3.5 Number of orbitals in each sublevel
Number of orbitals in each sublevel
An s sublevel contains one orbital, a p sublevel three, a d sublevel five and an f sublevel seven. Because each orbital holds at most two electrons, the maximum numbers of electrons in s, p, d and f sublevels are 2, 6, 10 and 14.
Electron spin
A property of an electron that has two possible states, shown in orbital diagrams as an arrow pointing up or down. Two electrons that occupy the same orbital must have opposite spins.
Pauli exclusion principle
An orbital can hold a maximum of two electrons, and these must have opposite spins.
Hund’s rule
When electrons occupy a sublevel containing more than one orbital of equal energy (such as 2p or 3d), they occupy the orbitals singly, with parallel spins, before any orbital holds a pair.
Aufbau principle
In the ground state, electrons occupy the lowest-energy orbitals available first. The order of filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p: the 4s sublevel is filled before the 3d sublevel.
Full and condensed electron configurations
A full electron configuration lists every occupied sublevel with the number of electrons as a superscript, for example iron 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s². A condensed configuration replaces the electrons of the preceding noble gas with its symbol in square brackets (the noble gas core), for example iron [Ar] 3d⁶ 4s².
Orbital (arrow-in-box) diagram
A diagram in which each orbital is drawn as a box and each electron as an arrow whose direction shows its spin. Boxes are arranged in order of increasing energy and filled according to the Aufbau principle, the Pauli exclusion principle (at most two opposite arrows per box) and Hund’s rule (the boxes of one sublevel each receive one arrow, all pointing the same way, before any box receives a second).
Chromium and copper exceptions
Chromium and copper 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). These two configurations must be learned as exceptions for elements up to Z = 36.
Electron configurations of ions
A positive ion has fewer electrons than its atom and a negative ion more. For main-group ions, electrons are removed from, or added to, the outer sublevel (for example S²⁻ is 1s² 2s² 2p⁶ 3s² 3p⁶). When a transition element forms a positive ion, the 4s electrons are removed before the 3d electrons, even though 4s is filled first: Fe is [Ar] 3d⁶ 4s², Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵.
Students often think Hund’s rule is the rule that says which sublevel fills first, such as s before p or 4s before 3d. In fact No. The order of filling (lowest energy first) is the Aufbau principle. Hund’s rule deals only with how electrons occupy orbitals of equal energy within one sublevel.
Students often think Chromium and copper fill in the regular way, giving [Ar] 3d⁴ 4s² and [Ar] 3d⁹ 4s². In fact No. They are the exceptions for Z ≤ 36: chromium is [Ar] 3d⁵ 4s¹ and copper is [Ar] 3d¹⁰ 4s¹.
S1.3.6 Convergence limit HL
Convergence limit
The frequency (or wavelength) at which the lines of a set in an emission spectrum converge. For the set produced by transitions ending at n = 1 in hydrogen, it corresponds to an electron falling from outside the atom (n = ∞) into the ground state, so its energy equals the energy needed to remove the electron from the ground state: the ionization energy.
First ionization energy
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⁻. Units: kJ mol⁻¹.
Energy of a photon and the wave equation
The energy of one photon is E = hf, where h is the Planck constant (6.63 × 10⁻³⁴ J s) and f is the frequency in s⁻¹ (Hz). Frequency and wavelength are related by c = λf, where c = 3.00 × 10⁸ m s⁻¹ and λ is in metres, so E = hc/λ. Multiplying the energy per atom (in J) by the Avogadro constant, 6.02 × 10²³ mol⁻¹, gives the energy per mole, which is divided by 1000 to give kJ mol⁻¹.
Trend in first ionization energy across a period
First ionization energy generally increases across a period because nuclear charge increases while the electron removed is in the same main energy level with similar shielding, so it is more strongly attracted and lower in energy. There are two discontinuities: it falls from group 2 to group 13, because the electron removed is in a p sublevel of higher energy than the s sublevel; and it falls from group 15 to group 16, because the electron removed comes from a doubly occupied p orbital, where electron–electron repulsion raises its energy. Explanations are based on the energy of the electron removed, not on any 'special stability' of filled or half-filled sublevels.
Trend in first ionization energy down a group
First ionization energy decreases down a group because the electron removed is in a higher main energy level, further from the nucleus and more shielded by inner electrons. The increase in nuclear charge is outweighed by the greater distance and shielding, so the electron removed has a higher energy and less energy is needed to remove it.
Students often think The first ionization energy is the energy of the first line of the set, the transition between n = 2 and n = 1. In fact No. The lowest-frequency line of that set is the n = 2 → n = 1 transition. Ionization from the ground state corresponds to the convergence limit of the set, where the lines merge at the highest frequency (an electron from n = ∞ falling to n = 1).
Students often think The number obtained from E = hf × N_A can be written directly with the unit kJ mol⁻¹ asked for in the question. In fact No. E = hf gives joules; after multiplying by the Avogadro constant the energy is in J mol⁻¹, and it must be divided by 1000 to give kJ mol⁻¹.
S1.3.7 Successive ionization energies HL
Successive ionization energies
The energies needed to remove the first, second, third … electrons in turn from an atom: X(g) → X⁺(g) + e⁻, X⁺(g) → X²⁺(g) + e⁻, and so on. Each is larger than the one before because the electron is removed from an increasingly positive ion. A much larger increase (a 'big jump') occurs when the next electron must be removed from a lower main energy level, closer to the nucleus and less shielded.
Deducing the group from successive ionization energies
The number of electrons removed before the first big jump in successive ionization energies equals the number of electrons in the outer main energy level. For an s-block or p-block element, one electron before the jump places it in group 1, two in group 2, three in group 13, four in group 14, and so on up to group 18 (groups are numbered 1–18).
Students often think The number of outer electrons is the number of the ionization energy at which the large increase appears, so a jump at the fourth ionization energy means four outer electrons. In fact No. The number of outer electrons is the number of electrons removed BEFORE the big jump, one fewer than the position of the jump. For groups 1 and 2 this number is the group number; for groups 13–18 (helium aside) the group number is this number plus 10. If the jump appears at the fourth ionization energy, three electrons were removed before it, so the element is in group 13.
Students often think Any clear rise between successive ionization energies, such as the second being three times the first, marks the removal of an electron from a new inner level. In fact No. Successive ionization energies always rise, because each electron is removed from a more positive ion. The change of main level is shown by an increase much larger than the others, typically several times the neighbouring increases.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 When sodium compounds are heated in a flame, the flame turns yellow. Which process in the sodium atoms produces this yellow light?
Answer and reasoning
Excited electrons fall to lower energy levels, emitting photons. — Heat excites electrons to higher energy levels. When they fall back to lower levels, each transition emits a photon whose energy equals the difference between the two levels; the main sodium transitions give yellow light.
Electrons absorb energy and emit photons as they rise to higher levels. — Rising to a higher level requires energy to be absorbed, so no photon is emitted in that step. The light is emitted afterwards, when the excited electrons fall back to lower levels.
Electrons escape from the atoms entirely, releasing energy as photons. — Removing an electron from an atom is ionization, which requires energy rather than releasing light. The emission lines come from electrons that stay in the atom and fall between its energy levels.
The hot atoms glow with the yellow colour that sodium atoms possess. — Individual atoms do not have a colour. The yellow is the light emitted as photons of particular energies when excited electrons in sodium atoms fall to lower energy levels.
2 Consider the set of lines in the hydrogen emission spectrum produced by electrons falling to n = 2 from higher levels. Which statement about the lines in this set is correct?
Answer and reasoning
The lines are equally spaced in frequency across the set. — Equal spacing would require equally spaced energy levels. The levels get closer together at higher energy, so the lines crowd together towards higher frequency.
The lines get closer together towards higher frequency. — The energy levels get closer together as n increases, so the energy differences for falls to n = 2 from n = 3, 4, 5, … increase by ever smaller amounts. The lines therefore converge towards higher frequency.
The lines get closer together towards lower frequency. — This would be true only if the levels spread further apart at higher energy. The energy gaps shrink as n increases, so the lines converge at the high-frequency end of the set.
The lines are dark gaps across a continuous band of colour. — Dark lines on a continuous band form an absorption spectrum. An emission spectrum consists of bright lines, which in each set converge towards higher frequency.
3 What is the maximum total number of electrons that the first three main energy levels of an atom (n = 1, 2 and 3) can hold?
Answer and reasoning
18 — The '2, 8, 8' pattern describes the first 20 elements, not the capacity of the third level. The third level can hold 2 × 3² = 18 electrons (3s², 3p⁶, 3d¹⁰), so the total is 2 + 8 + 18 = 28.
28 — Level n holds a maximum of 2n² electrons: 2 × 1² = 2, 2 × 2² = 8 and 2 × 3² = 18, giving 2 + 8 + 18 = 28.
14 — 1 + 4 + 9 = 14 is the number of orbitals in the first three levels (n² for each). Each orbital holds two electrons of opposite spin, so the capacity is 2n² per level: 2 + 8 + 18 = 28.
12 — Giving each sublevel room for one pair (2, then 2 + 2, then 2 + 2 + 2) treats every sublevel as a single orbital. The p sublevel has three orbitals and the d sublevel five, so the capacities are 2, 8 and 18, totalling 28.
Working Maximum electrons in level n = 2n². n = 1: 2 × 1² = 2; n = 2: 2 × 2² = 8; n = 3: 2 × 3² = 18. Total = 2 + 8 + 18 = 28.
4 Which description of a 1s atomic orbital is correct?
Answer and reasoning
A circular path round the nucleus along which the electron travels at a fixed distance — An orbital is not an orbit. The electron does not follow a path; the orbital describes a region, here a sphere, where the electron is likely to be found.
A spherical region centred on the nucleus that contains the electron at every instant — The shape is right, but the drawn sphere encloses a high probability, not certainty. There is a small probability of finding the electron outside it.
A spherical region centred on the nucleus where the electron is most likely to be found — An s orbital is spherical and centred on the nucleus. An orbital is a region of space where there is a high probability of finding the electron, not a path or a container.
A spherical shell with its two electrons held at fixed points on opposite sides of the nucleus — Electrons are not held at fixed points; that is a shell-diagram drawing convention. The 1s orbital is a spherical region, centred on the nucleus, where there is a high probability of finding its electrons.
5 Which condensed electron configuration of a ground-state atom is correct?
Answer and reasoning
copper: [Ar] 3d⁹ 4s² — This is what the regular filling order predicts, but copper (Z = 29) is the other exception: it is [Ar] 3d¹⁰ 4s¹.
chromium: [Ar] 3d⁵ 4s¹ — Chromium (Z = 24) is one of the two exceptions to the simple filling order: it is [Ar] 3d⁵ 4s¹, not [Ar] 3d⁴ 4s².
potassium: [Ar] 3d¹ — After 3p the next electron enters 4s, not 3d, because 4s is lower in energy in potassium. Potassium (Z = 19) is [Ar] 4s¹.
iron: [Ar] 4s² 4p⁶ — The third level is not full at eight electrons; it can hold 18, including ten in 3d. After 4s the next electrons enter 3d, not 4p: iron (Z = 26) is [Ar] 3d⁶ 4s².
6 How does the first ionization energy change from lithium to sodium to potassium in group 1, and why? HL
Answer and reasoning
It decreases, as the outer electron is in a higher main level, further out and more shielded. — Down the group the electron removed is in a higher main energy level (n = 2, 3, 4), further from the nucleus and shielded by more inner electrons. This outweighs the rise in nuclear charge, so less energy is needed: Li 520, Na 496, K 419 kJ mol⁻¹.
It decreases, as the fixed pull of the nucleus is shared among a larger number of electrons. — The trend is right but the reason is not. A nucleus does not have a fixed amount of attraction to share out. The outer electron is held less strongly because it is further from the nucleus and more shielded.
It increases, as the nuclear charge rises from +3 to +19 and attracts the electron more strongly. — Nuclear charge does rise, but it is not the only factor. The outer electron is much further out and more shielded in potassium, so the first ionization energy falls from 520 to 419 kJ mol⁻¹.
It stays the same, as each atom must lose just one electron to reach a full outer shell. — Ionization energy measures how strongly the electron removed is attracted, not how many electrons stand between the atom and a full shell. The outer electron is further out and more shielded down the group, so the value falls.
7 The first four successive ionization energies of an element X are 578, 1817, 2745 and 11 577 kJ mol⁻¹. In which group of the periodic table is X? HL
Answer and reasoning
Group 14, as the large increase appears at the fourth ionization energy — The large value is the fourth, but the number of outer electrons is the number removed BEFORE the jump: three, not four. Three outer electrons (ns² np¹) place X in group 13.
Group 1, as the second value is already over three times larger than the first — Successive ionization energies always rise. The increase from 2745 to 11 577 is far larger than the earlier rises, so the jump comes after the third electron: group 13.
Group 13, as the large increase comes after the third electron is removed — The rises from 578 to 1817 to 2745 are steady, then the fourth value is about four times the third. Three electrons are removed before the big jump, so X has three outer electrons (ns² np¹): group 13. X is aluminium.
Group 3, as exactly three electrons are removed before the large increase occurs — Three outer electrons is correct, but in the 1–18 numbering a p-block element with three outer electrons is in group 13. Group 3 is a d-block group (scandium, yttrium).
8 Violet light and red light lie at opposite ends of the visible spectrum. How does violet light compare with red light?
Answer and reasoning
Violet light has the lower frequency, as it has the shorter wavelength. — Violet does have the shorter wavelength, but wavelength and frequency are inversely related, so the shorter wavelength of violet means a higher frequency, and therefore a greater photon energy, than red.
Violet light has the longer wavelength, as its photons carry more energy. — Violet photons do carry more energy, but more energetic photons have shorter wavelengths, not longer ones. Violet (about 400 nm) is shorter than red (about 700 nm).
Violet photons carry less energy, as red is the colour of hot, energetic things. — Red is not the high-energy end of the visible spectrum; it is the low-energy end, next to infrared. Violet, next to ultraviolet, has the shorter wavelength, higher frequency and greater photon energy.
Violet photons carry more energy, as violet light has the higher frequency. — Violet light has a shorter wavelength than red, so it has a higher frequency, and photon energy increases with frequency. Violet photons therefore carry more energy than red photons.
9 Which electron transition in a hydrogen atom emits the photon of highest frequency?
Answer and reasoning
n = 6 → n = 2 — This fall spans four levels, but the levels are not equally spaced: the gaps near n = 6 are tiny. The single gap from n = 2 to n = 1 alone is larger than the whole fall from n = 6 to n = 2, so n = 4 → n = 1 gives the higher frequency.
n = 1 → n = 5 — Moving from n = 1 up to n = 5 requires energy to be absorbed; no photon is emitted. Emission happens when an electron falls, and of the falls listed, n = 4 → n = 1 involves the largest energy difference.
n = 6 → n = 5 — The photon’s energy is the difference between the two levels, not the energy of the level the electron lands in. The gap between n = 6 and n = 5 is the smallest here, so this photon has the lowest frequency (it is in the infrared).
n = 4 → n = 1 — Photon energy, and therefore frequency, equals the energy difference between the two levels. Falls to n = 1 involve the largest gaps, because the levels are much further apart near n = 1. The n = 4 → n = 1 fall releases far more energy than 6 → 2 or 6 → 5.
10 A potassium atom (Z = 19) has the electron arrangement 2, 8, 8, 1: two electrons in the first main energy level, eight in the second, eight in the third and one in the fourth. Which statement about the third main energy level (n = 3) of this atom is correct?
Answer and reasoning
It is not full, because a level with n = 3 can hold up to 2 × 3² = 18 electrons. — The capacity of a main energy level is 2n², so the third level can hold 18 electrons (3s, 3p and 3d sublevels). Potassium’s third level holds only 8; its nineteenth electron enters the fourth level because 4s is lower in energy than 3d, not because n = 3 is full.
It is full, because every main energy level after the first holds a maximum of 8 electrons. — ‘2, 8, 8’ describes the filling of the first 20 elements, not the capacity of the third level. Capacity is 2n²: the third level can hold 18 electrons, and in the elements after calcium it fills up to 18 (3s² 3p⁶ 3d¹⁰).
It is full, because electrons only enter the fourth level once the third holds all it can. — Levels are not always filled completely before the next begins. The 4s sublevel is lower in energy than 3d, so the nineteenth electron enters n = 4 while n = 3 still has ten empty places (its capacity is 2 × 3² = 18).
It is not full, because it has n² = 9 orbitals and so has room for one more electron. — The third level does have n² = 9 orbitals, but each orbital holds two electrons of opposite spin, so the capacity is 2n² = 18, not 9. With 8 electrons present there is room for ten more, not one.
Working Maximum number of electrons in main energy level n = 2n². For n = 3: 2 × 3² = 18. The third level of potassium holds 8 of a possible 18, so it is not full; the nineteenth electron enters the fourth level (4s) because 4s is lower in energy than 3d, not because the third level is full.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
16 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 White light from a hot filament lamp gives a continuous spectrum when it is passed through a prism. Light from a hydrogen discharge tube passed through the same prism gives a line spectrum. How does the line spectrum differ from the continuous spectrum?
Answer and reasoning
It shows dark lines at certain wavelengths, across a continuous band of colour. — Dark lines across a continuous band make an absorption spectrum. The discharge tube emits light only at certain wavelengths, so its spectrum is bright lines on a dark background.
It shows only one bright line, since a hydrogen atom has only one electron. — The lines are not produced one per electron. Each line comes from a different transition between energy levels, and the many atoms in the tube make many different transitions, so hydrogen gives several lines.
It shows bright lines at only certain wavelengths, on a dark background. — A continuous spectrum contains every wavelength over a range. The hydrogen atoms emit only photons of certain energies, so the line emission spectrum consists of separate bright lines at particular wavelengths with darkness between them.
It shows bright lines at frequencies spaced at equal intervals across it. — The lines are not equally spaced in frequency, because the energy levels are not equally spaced. The levels get closer together at higher energy, so within each set the lines crowd together towards higher frequency.
2 In the hydrogen emission spectrum, transitions ending at n = 1 produce lines in the ultraviolet region. In which region are the lines produced by transitions ending at n = 3, and why?
Answer and reasoning
Infrared, because the energy gaps from higher levels down to n = 3 are smaller than those down to n = 2 — The levels get closer together as n increases, so every fall ending at n = 3 releases less energy than any fall ending at n = 2, including the smallest (n = 3 → n = 2, the red line of the visible set). Lower photon energy means lower frequency and longer wavelength than visible light: the infrared region.
Ultraviolet, because n = 3 is a higher energy level than n = 2, so its photons carry more energy — A photon’s energy is the difference between two levels, not the energy of the level the electron lands in. The gaps down to n = 3 are smaller than the gaps down to n = 2, so these lines are lower in energy than visible light: infrared.
Ultraviolet, because the levels above n = 3 are more widely spaced than the levels above n = 2 — The levels get closer together, not further apart, as n increases, even though the orbitals are further from the nucleus. The smaller gaps above n = 3 give lower-energy photons, in the infrared.
Visible, because the levels are equally spaced, so every fall releases the same size of energy step — The levels are not equally spaced; their gaps shrink as n increases. Falls to n = 3 release less energy than falls to n = 2, so their lines lie beyond the red end of the visible spectrum, in the infrared.
3 Which description of the shapes and orientations of the three 2p atomic orbitals is correct?
Answer and reasoning
Three dumbbell shapes along the x, y and z axes, at 90° to one another — Each p orbital is a dumbbell with two lobes on opposite sides of the nucleus. The three p orbitals of a sublevel point along the three perpendicular axes: px, py and pz.
Three dumbbell shapes lying in one plane, at 120° to one another — The shape is right but the orientation is not. The p orbitals are not arranged by spreading apart like electron domains; they lie along the x, y and z axes, at 90° to one another.
Three figure-of-eight paths that the electrons follow around the nucleus — Orbitals are not paths. Each p orbital is a dumbbell-shaped region where there is a high probability of finding an electron; the three lie along the x, y and z axes.
A single dumbbell shape that holds all six electrons of the 2p sublevel — The 2p sublevel is not one orbital. It contains three separate p orbitals, each holding up to two electrons, directed along the x, y and z axes.
4 In a potassium atom, the nineteenth electron occupies the 4s sublevel rather than the 3d sublevel. What is the reason?
Answer and reasoning
The 4s orbital is nearer the nucleus, and orbitals fill from the inside outward. — Filling follows energy, not distance. The 4s orbital actually extends further from the nucleus than 3d; it fills first because, in potassium, it is lower in energy.
Hund’s rule requires an s sublevel to be filled before any d sublevel. — Hund’s rule is about orbitals of equal energy within one sublevel: they are singly occupied, with parallel spins, before pairing. The filling order of sublevels comes from the Aufbau principle and the sublevel energies.
The third main energy level is already full, as it holds at most eight electrons. — The third level can hold 18 electrons, including ten in 3d; it is not full at eight. The 3d sublevel is simply higher in energy than 4s in potassium, and it fills later, from scandium onwards.
The 4s sublevel is lower in energy than the 3d sublevel in this atom. — Electrons occupy the lowest-energy orbitals available (the Aufbau principle). In potassium and calcium, the 4s sublevel is lower in energy than 3d, so it fills first: potassium is [Ar] 4s¹.
5 Which full electron configuration of a ground-state atom is correct?
Answer and reasoning
scandium: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4p¹ — The third level is not full at eight electrons; it can hold 18, including ten in 3d. After 4s the next electron enters 3d, not 4p: scandium (Z = 21) is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s².
cobalt: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ — The 4s sublevel is filled before 3d, so cobalt (Z = 27) keeps two electrons in 4s: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ 4s².
iron: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² — Iron (Z = 26) has 26 electrons. Filling in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d gives 18 electrons up to 3p⁶, then 4s², then the remaining six in 3d: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² ([Ar] 3d⁶ 4s²).
germanium: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹² 4s² — A 3d sublevel holds at most ten electrons (five orbitals × 2); 2n² = 18 is the capacity of the whole third level, shared by 3s, 3p and 3d. After 3d¹⁰ and 4s² the next electrons enter 4p: germanium (Z = 32) is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p².
6 An iron atom (Z = 26) has the condensed electron configuration [Ar] 3d⁶ 4s². How is an Fe²⁺ ion formed from this atom?
Answer and reasoning
Two of the six 3d electrons are removed from the atom. — Electrons are not removed in the reverse order of filling. Once 3d is occupied, the 4s electrons are removed first, so Fe²⁺ is [Ar] 3d⁶, not [Ar] 3d⁴ 4s².
One 4s electron and one 3d electron are each removed. — Ions do not rearrange their electrons to reach a half-filled 3d sublevel ([Ar] 3d⁵ 4s¹); the chromium pattern is an exception for that atom only. Both 4s electrons are removed, giving [Ar] 3d⁶.
Two electrons are added to the 3d sublevel of the atom. — A 2+ ion forms by losing two electrons, not gaining them; gaining two would give a 2− ion ([Ar] 3d⁸ 4s²). Fe²⁺ has 24 electrons, [Ar] 3d⁶.
Both of the 4s electrons are removed from the atom. — When a transition element forms a positive ion, the 4s electrons are removed before the 3d electrons. Removing both 4s electrons from [Ar] 3d⁶ 4s² gives Fe²⁺, [Ar] 3d⁶, with 24 electrons.
7 An orbital (arrow-in-box) diagram is drawn for a ground-state nitrogen atom, 1s² 2s² 2p³. The 1s box and the 2s box each contain two arrows pointing in opposite directions. Which description of the 2p part of the diagram is correct?
Answer and reasoning
Three boxes each hold one arrow, and all three arrows point the same way. — By Hund’s rule, the three electrons occupy the three 2p orbitals singly, with parallel spins, before any orbital holds a pair. Each 2p box therefore contains one arrow, all pointing in the same direction.
Three boxes: one holds two opposite arrows, one holds one, one is empty. — Electrons do not pair while orbitals of equal energy are empty. Hund’s rule places one arrow in each of the three 2p boxes, all pointing the same way.
Three boxes each hold one arrow, and the middle arrow is reversed. — Opposite spins are required only for two electrons in the same orbital. Singly occupied orbitals of one sublevel have parallel spins, so all three arrows point the same way.
One box holds all three arrows, as the 2p sublevel is only one orbital. — The 2p sublevel consists of three orbitals, drawn as three boxes, and no orbital can hold more than two electrons. The three electrons go one into each box, with parallel spins.
8 What is the maximum number of electrons that a 3d sublevel can hold, and why?
Answer and reasoning
20 electrons, as its ten orbitals can each hold two electrons of opposite spin — A d sublevel has five orbitals, not ten; ten is its electron capacity. Five orbitals × 2 electrons = 10.
18 electrons, as the n = 3 energy level holds a maximum of 2n² electrons — 2n² = 18 is the capacity of the whole third main level (3s² 3p⁶ 3d¹⁰), shared among three sublevels. The 3d sublevel alone holds 10.
10 electrons, as its five orbitals can each hold two electrons of opposite spin — A d sublevel contains five orbitals, and each orbital holds at most two electrons, which must have opposite spins (the Pauli exclusion principle): 5 × 2 = 10.
2 electrons, as the 3d sublevel is one orbital that holds a single pair — The 3d sublevel is not a single orbital; it contains five orbitals. Each holds up to two electrons, so the sublevel holds 10.
Working A d sublevel has 5 orbitals; each orbital holds a maximum of 2 electrons (opposite spins). Maximum = 5 × 2 = 10 electrons.
9 In the emission spectrum of hydrogen, the set of lines produced by electrons falling to n = 1 has its lowest-frequency line at a wavelength of 121.6 nm and converges to a limit at a wavelength of 91.18 nm. What is the first ionization energy of hydrogen? (h = 6.626 × 10⁻³⁴ J s; c = 2.998 × 10⁸ m s⁻¹; N_A = 6.022 × 10²³ mol⁻¹; E = hf; c = λf) HL
Answer and reasoning
983.8 kJ mol⁻¹ — Using 121.6 nm gives the energy of the n = 2 → n = 1 transition, which only excites the electron. Ionization corresponds to the convergence limit, 91.18 nm, which gives 1312 kJ mol⁻¹.
1312 kJ mol⁻¹ — The convergence limit corresponds to ionization from n = 1. E = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (91.18 × 10⁻⁹) = 2.179 × 10⁻¹⁸ J per atom; × 6.022 × 10²³ mol⁻¹ = 1.312 × 10⁶ J mol⁻¹ = 1312 kJ mol⁻¹.
1.312 × 10⁶ kJ mol⁻¹ — 1.312 × 10⁶ is the ionization energy in J mol⁻¹. Dividing by 1000 converts it to kJ mol⁻¹: 1312 kJ mol⁻¹.
2.179 × 10⁻²¹ kJ mol⁻¹ — 2.179 × 10⁻²¹ kJ (2.179 × 10⁻¹⁸ J) is the energy needed to ionize ONE atom, not one mole of atoms. Multiplying by N_A = 6.022 × 10²³ mol⁻¹ gives 1.312 × 10⁶ J mol⁻¹ = 1312 kJ mol⁻¹.
Working The convergence limit of the set ending at n = 1 corresponds to ionization from the ground state. f = c/λ = (2.998 × 10⁸ m s⁻¹) ÷ (91.18 × 10⁻⁹ m) = 3.288 × 10¹⁵ s⁻¹. E = hf = (6.626 × 10⁻³⁴ J s)(3.288 × 10¹⁵ s⁻¹) = 2.179 × 10⁻¹⁸ J per atom. Per mole: 2.179 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹ = 1.312 × 10⁶ J mol⁻¹ = 1312 kJ mol⁻¹ (4 significant figures, matching the data; this agrees with the accepted value, 1312 kJ mol⁻¹).
10 How does the first ionization energy generally change across period 3 from sodium to argon, and why? HL
Answer and reasoning
It decreases, as the fixed pull of the nucleus is shared among more and more electrons. — A nucleus does not have a fixed amount of pull to share among its electrons. The rising nuclear charge, with similar shielding, attracts the outer electrons more strongly, so the first ionization energy increases.
It decreases, as the atoms get larger because more electrons are added along the period. — Atoms get smaller across a period: the added electrons enter the same main level and the rising nuclear charge pulls them closer. The first ionization energy therefore increases.
It increases, as atoms closer to a full outer shell resist losing any of their outer electrons. — The trend is right but the reason is not. Atoms do not 'resist' losing electrons to keep a full shell; the outer electron is harder to remove because the nuclear charge increases while the shielding stays similar.
It increases, as the nuclear charge rises while the outer electrons have similar shielding. — Across the period the nuclear charge increases from +11 to +18, while the electron removed is in the third main level with similar shielding from the inner electrons. It is attracted more strongly, so the first ionization energy rises in general, from 496 to 1521 kJ mol⁻¹.
11 The first ionization energies (kJ mol⁻¹) of the period 2 elements are: Li 520, Be 900, B 801, C 1086, N 1402, O 1314, F 1681, Ne 2081. Which statement correctly explains both the fall from Be to B and the fall from N to O? HL
Answer and reasoning
Be’s filled 2s sublevel and N’s half-filled 2p sublevel have a special stability that makes them harder to ionize. — The explanations must be based on the energy of the electron removed, not on a 'special stability' of filled or half-filled sublevels. Boron’s electron is in the higher-energy 2p sublevel; oxygen’s is in a doubly occupied 2p orbital, raised in energy by repulsion.
B’s electron is in 2p, higher in energy than 2s; O’s is in a doubly occupied 2p orbital, where repulsion raises its energy. — Be is 1s² 2s² and B is 1s² 2s² 2p¹: boron’s electron is removed from the 2p sublevel, which is higher in energy than 2s. N is 2s² 2p³ with three singly occupied 2p orbitals; O is 2s² 2p⁴ with one 2p orbital holding a pair, and repulsion between the paired electrons raises the energy of the electron removed. In both cases less energy is needed.
B’s electron is in a doubly occupied 2p orbital, where repulsion raises its energy; O’s is in a higher sublevel than N’s. — The two explanations are the wrong way round. Boron has a single 2p electron, so pairing cannot apply; its electron is in 2p, above 2s. Nitrogen and oxygen both lose a 2p electron; oxygen’s comes from a doubly occupied orbital.
B and O atoms are larger than Be and N atoms, so the electron removed is further from the nucleus in each case. — Atoms get smaller across a period: boron is smaller than beryllium and oxygen smaller than nitrogen. The falls are caused by the higher energy of the 2p sublevel (Be → B) and by electron-pair repulsion in a 2p orbital (N → O).
12 The first four successive ionization energies of sodium are 496, 4562, 6910 and 9543 kJ mol⁻¹. Which explanation of the large increase from the first to the second value is correct? HL
Answer and reasoning
The Na⁺ ion has a stable noble-gas octet, and ions resist losing electrons from a complete outer shell. — Ions do not 'resist' losing a full shell. The second electron needs far more energy because it is in the second main level, closer to the nucleus and less shielded than the 3s electron.
The second electron is removed from the 2p sublevel, which is closer to the nucleus and less shielded. — Sodium is 1s² 2s² 2p⁶ 3s¹. The first electron comes from 3s; the second comes from 2p, in a lower main level, closer to the nucleus and shielded only by the 1s electrons. It is attracted much more strongly, so far more energy is needed.
After one electron is lost, the fixed pull of the nucleus is shared among 10 electrons, not 11. — The nucleus does not share a fixed amount of pull among its electrons. The big increase happens because the second electron is in a lower main level, closer to the nucleus and less shielded.
Losing the first electron raises the nuclear charge, so the next electron is held far more tightly. — The nuclear charge depends on the number of protons and stays +11 in Na⁺. The large increase arises because the second electron is removed from 2p, closer to the nucleus and less shielded.
13 Which of these condensed electron configurations of ground-state atoms is correct?
Answer and reasoning
chromium: [Ar] 3d⁴ 4s² — This is what the regular filling order predicts, but chromium (Z = 24) is the other exception: it is [Ar] 3d⁵ 4s¹.
scandium: [Ar] 3d³ — The 4s sublevel is filled before 3d, so scandium (Z = 21) has two electrons in 4s and one in 3d: [Ar] 3d¹ 4s².
copper: [Ar] 3d¹⁰ 4s¹ — Copper (Z = 29) is one of the two exceptions to the simple filling order for Z ≤ 36: it is [Ar] 3d¹⁰ 4s¹, not [Ar] 3d⁹ 4s². Filling 3d completely before 4s would also give [Ar] 3d¹⁰ 4s¹ for copper, but that filling order is wrong: it would make scandium [Ar] 3d³ instead of [Ar] 3d¹ 4s². Copper is an exception to the correct order, in which 4s fills before 3d.
manganese: [Ar] 4s² 4p⁵ — The third level is not full at eight electrons; it can hold 18, including ten in 3d. After 4s the next electrons enter 3d, not 4p: manganese (Z = 25) is [Ar] 3d⁵ 4s².
14 The diagram shows part of the emission spectrum of hydrogen plotted against frequency. The four lines P, Q, R and S lie in the visible region, and the lines converge towards the limit marked L. Which electron transition produces line P?
Answer and reasoning
n = 2 → n = 3 — An electron moving up from n = 2 to n = 3 absorbs energy; it does not emit a photon. Emission lines come from electrons falling to a lower level, so P, the lowest-frequency visible line, is the fall from n = 3 to n = 2.
n = 2 → n = 1 — A photon’s energy is the difference between the two levels, not the energy of the level the electron ends in. Falls to n = 1 are the largest and give ultraviolet lines, not the visible line P, which is the smallest fall into n = 2: from n = 3.
n = 6 → n = 2 — P has the lowest frequency, so it has the longest wavelength (red light, about 656 nm). A longer wavelength does not mean more energy: photon energy E = hf rises with frequency, so P carries the least energy of the four lines and comes from the smallest fall into n = 2, from n = 3. The fall from n = 6 to n = 2 is the largest of the four and gives S.
n = 3 → n = 2 — The visible lines of hydrogen are produced by electrons falling to n = 2. P is the line of lowest frequency, so it carries the least energy and comes from the smallest fall into n = 2, from n = 3. Larger falls (from n = 4, 5, 6 …) give the higher-frequency lines Q, R, S and the lines that crowd towards L.
Working Visible lines of hydrogen arise from transitions ending at n = 2. Photon energy E = hf, so the lowest-frequency line P is the smallest energy gap ending at n = 2, i.e. n = 3 → n = 2. Higher starting levels give higher-frequency lines that converge on L (n = ∞ → n = 2).
15 The diagram shows four arrow-in-box diagrams, W, X, Y and Z, each drawn for the 3d sublevel of a ground-state iron atom, [Ar] 3d⁶ 4s². Which diagram is correct?
Answer and reasoning
Diagram W — Filling three orbitals with pairs while two orbitals of the same energy stay empty breaks Hund’s rule. Electrons occupy each 3d orbital singly before any orbital holds a pair, so the ground state is ↑↓ ↑ ↑ ↑ ↑ (X).
Diagram X — By Hund’s rule the six 3d electrons first occupy the five orbitals singly with parallel spins; the sixth then pairs, with opposite spin, in one orbital (Pauli exclusion principle). X shows one paired box and four single up arrows.
Diagram Y — Opposite spins are required only for two electrons in the same orbital. Singly occupied orbitals of one sublevel hold electrons with parallel spins, so the four single arrows all point the same way (X), not alternately up and down.
Diagram Z — Two electrons in the same orbital must have opposite spins (Pauli exclusion principle); ↑↑ in one box is not allowed. Hund’s rule’s parallel spins apply to electrons in different orbitals, so the correct diagram is X: ↑↓ ↑ ↑ ↑ ↑.
16 The graph shows log₁₀ of the successive ionization energies of an element plotted against the number of electrons removed. In which group of the periodic table is the element? HL
Answer and reasoning
Group 14, because the first large jump comes after four electrons have been removed — The values rise steadily for the first four electrons and then jump sharply at the fifth, so four electrons are removed from the outer level before an inner level (n = 2) is reached. Four outer electrons (ns² np²) place the element in group 14; the second jump after the twelfth electron marks the n = 1 level, so the element is silicon.
Group 15, because the first large jump is seen at the fifth ionization energy — The fifth ionization energy is the first high value, but it is the number of electrons removed BEFORE the jump that counts: four. Four outer electrons mean group 14, not 15.
Group 12, because the largest jump on the graph comes after the twelfth electron — The jump after the twelfth electron marks the start of the n = 1 level, not the end of the outer electrons. It is the FIRST large jump, after the fourth electron, that shows how many outer electrons there are: four, so group 14.
Group 4, because four electrons are removed before the first large jump — Four outer electrons is the right reading of the graph, but in the 1–18 numbering a p-block element with four outer electrons (ns² np²) is in group 14. Group 4 is a d-block group (titanium, zirconium).
Working Read the graph: the points rise gently for electrons 1–4, then jump between the 4th and 5th values (the first large jump). Four electrons are removed before the jump, so the element has four outer electrons (ns² np²): group 14. The second jump between the 12th and 13th values marks the two n = 1 electrons, so the element has 14 electrons in total and is silicon.
That was your twenty minutes. Real practice on S1.3 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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