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 10 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
S3.1.1 Period
Period
A horizontal row of the periodic table. Atomic number increases by one from each element to the next along a period. The elements of a period have the same number of occupied main energy levels, and the period number is the number of the outer energy level that is occupied (for example, every period 3 element has its outer electrons in the third energy level).
Group
A vertical column of the periodic table. Groups are numbered from 1 to 18. The elements of a group have the same number of valence electrons and the same outer-level electron arrangement (for example ns¹ in group 1 and ns² np⁵ in group 17), which is why they have similar chemical properties.
Blocks (s, p, d and f)
The periodic table divides into four blocks named after the sublevel that is being filled, in the aufbau order, across that part of the table (the sublevel to which the last electron is added). The s-block is groups 1 and 2 (plus helium), the p-block is groups 13 to 18, the d-block is groups 3 to 12, and the f-block is the two rows (lanthanoids and actinoids) usually printed below the main table. For example, gallium, [Ar] 3d¹⁰ 4s² 4p¹, is a p-block element because its last electron enters the 4p sublevel.
Metals, metalloids and non-metals
Metals occupy the left and centre of the periodic table: the s-block (except hydrogen and helium), the d-block, the f-block and the lower left of the p-block (for example aluminium, tin and lead). Non-metals occupy the upper right of the p-block, together with hydrogen. The metalloids lie along the stepped diagonal line that separates the two; the elements commonly classified as metalloids are boron, silicon, germanium, arsenic, antimony and tellurium. Metalloids have properties intermediate between those of metals and non-metals; for example silicon is a shiny solid but a semiconductor, not a good electrical conductor.
Students often think Any element that touches the stepped metal/non-metal line is a metalloid. In fact No. Aluminium is a metal: it is shiny, malleable and a good electrical conductor, and forms an amphoteric oxide. The metalloids are boron, silicon, germanium, arsenic, antimony and tellurium.
Students often think An element can be classified as a metal or metalloid from one visible property, such as a shiny surface. In fact No. Iodine is a non-metal: it is a molecular solid of I₂ molecules that does not conduct electricity and forms an acidic oxide. Its lustre alone does not make it metallic.
S3.1.2 Valence electrons
Valence electrons
The electrons in the outer (highest occupied) energy level of an atom of a main-group element, which take part in bonding. With groups numbered 1 to 18, atoms of groups 1 and 2 have 1 and 2 valence electrons, and atoms of groups 13 to 18 have (group number − 10) valence electrons, for example 6 in group 16 (helium, in group 18, has 2).
Deducing electron configuration from position
The period gives the outer energy level and the block gives the sublevel being filled, so an atom's configuration can be read from its position (up to Z = 36). Selenium is in period 4, group 16: the core is argon, and across period 4 the 4s, then 3d, then 4p sublevels fill, giving [Ar] 3d¹⁰ 4s² 4p⁴. Conversely, a configuration ending 4s² 4p⁴ places an element in period 4 and the fourth column of the p-block, group 16.
Alkali metals
The elements of group 1 below hydrogen: lithium, sodium, potassium, rubidium, caesium and francium. Their atoms have one valence electron (ns¹). They are reactive metals that react with water to form hydrogen and an alkaline solution of the metal hydroxide. Hydrogen is placed at the top of group 1 because of its 1s¹ configuration but is a non-metal and not an alkali metal.
Halogens
The elements of group 17: fluorine, chlorine, bromine, iodine and astatine. Their atoms have seven valence electrons (ns² np⁵). They are non-metals that exist as diatomic molecules (F₂, Cl₂, Br₂, I₂) and form halide ions (F⁻, Cl⁻, Br⁻, I⁻).
Noble gases
The elements of group 18: helium, neon, argon, krypton, xenon and radon. Their atoms have a full outer energy level (1s² for helium, ns² np⁶ for the rest). They are monatomic gases with very low reactivity.
Transition elements
The metals of the central d-block of the periodic table, whose atoms are filling d sublevels. In period 4 the d-block runs from scandium to zinc, filling the 3d sublevel; scandium to copper are the transition elements, because zinc's 3d sublevel is full. (At HL the definition is made precise in terms of an incomplete d sublevel.)
Students often think The group number is always the number of valence electrons, so a group 16 atom has 16 and the halogens, with seven, are in group 7. In fact For the s- and p-block groups, only in groups 1 and 2. For groups 13 to 18 the number of valence electrons is the group number minus 10 (helium has 2).
Students often think Periods and groups are interchangeable, so the period number can give the valence electrons and the group the energy level. In fact No. It is the other way round: the period number is the outer energy level occupied, and the group identifies the number of valence electrons.
S3.1.3 Periodicity
Periodicity
The repeating pattern of physical and chemical properties across periods and down groups, which arises because the arrangement of outer electrons repeats in each period. The properties explained in terms of periodicity include atomic radius, ionic radius, ionization energy, electron affinity and electronegativity.
Nuclear charge and shielding
Nuclear charge is the positive charge of the nucleus, set by the number of protons. Shielding is the reduction in the attraction between the nucleus and an outer electron caused by repulsion from electrons in inner energy levels. Across a period, nuclear charge increases while the added electrons go into the same outer level, so shielding changes little and outer electrons are held more strongly. Down a group, a new energy level is added each period, so the outer electrons are further from the nucleus and more shielded, which outweighs the increase in nuclear charge.
Atomic radius
Half the distance between the nuclei of two bonded atoms of the same element, usually given in picometres (pm). It decreases across a period (nuclear charge increases while electrons are added to the same energy level) and increases down a group (an extra occupied energy level each period).
Ionic radius
The radius of an ion in an ionic lattice, usually given in pm. A positive ion is smaller than its atom (it has lost its outer level, or has fewer electrons for the same nuclear charge); a negative ion is larger than its atom (extra electrons increase electron–electron repulsion while the nuclear charge is unchanged). In an isoelectronic series such as O²⁻, F⁻, Na⁺, Mg²⁺ (all 1s² 2s² 2p⁶), radius decreases as nuclear charge increases.
First ionization energy
The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions: X(g) → X⁺(g) + e⁻. Units kJ mol⁻¹. It generally increases across a period and decreases down a group; for example sodium 496 kJ mol⁻¹, rubidium 403 kJ mol⁻¹.
Electron affinity
The energy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous 1− ions: X(g) + e⁻ → X⁻(g). Units kJ mol⁻¹. For most elements the first electron affinity is negative (energy is released). It generally becomes more negative across a period and less negative down a group, with irregularities; for example chlorine (−349 kJ mol⁻¹) has a more negative value than fluorine (−328 kJ mol⁻¹).
Electronegativity
The ability of an atom to attract a shared pair of electrons in a covalent bond towards itself. It is measured on the Pauling scale, which has no units. It increases across a period and decreases down a group, so fluorine is the most electronegative element. Noble gases such as neon form no bonds in ordinary conditions and are not given Pauling values. Electronegativity is a property of bonded atoms; electron affinity is an energy change for isolated gaseous atoms.
Students often think Atoms are driven to gain or lose electrons to reach a stable full outer shell, and this stability explains their size, ionization energies and reactions. In fact No. Atoms do not 'want' full shells. Properties such as size, ionization energy and reactivity are explained by the electrostatic attraction between the nucleus and the electrons, using nuclear charge, distance and shielding.
Students often think Atoms with more electrons (or more particles) are larger, because the extra electrons need more space. In fact No. Atomic radius decreases across a period. The added electrons go into the same energy level, while the nuclear charge increases, so the electrons are pulled closer.
S3.1.4 Metallic and non-metallic character
Metallic and non-metallic character
Metallic character is the tendency of an element to lose electrons and form positive ions; non-metallic character is the tendency to gain electrons. Down group 1, the outer electron is further from the nucleus and more shielded, so it is lost more easily and metallic character (reactivity) increases. Down group 17, an incoming electron is attracted less strongly, so non-metallic character (oxidizing ability, reactivity) decreases.
Reaction of group 1 metals with water
Group 1 metals react with water to form hydrogen gas and an alkaline solution of the metal hydroxide, for example 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g). The metal floats, fizzes and moves about; the reaction becomes more vigorous down the group (lithium fizzes steadily, potassium ignites the hydrogen) because the outer electron is lost more easily.
Halogen displacement reactions
A halogen oxidizes the halide ions of any halogen below it in group 17, taking their electrons: for example Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq), and Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq). A halogen does not react with the halide ions of a halogen above it, so iodine does not react with chloride or bromide ions. Halide ions themselves are not oxidizing agents.
Students often think A metal reacting with water forms the metal oxide and hydrogen. In fact No. The products are the metal hydroxide, which dissolves to give an alkaline solution, and hydrogen gas: 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g).
Students often think If a reaction produces hydrogen, the solution becomes acidic, because hydrogen is associated with acids. In fact No. The solution is alkaline, because it contains hydroxide ions from the sodium hydroxide formed. Hydrogen gas is not an acid and does not make a solution acidic.
S3.1.5 Basic, amphoteric and acidic oxides
Basic, amphoteric and acidic oxides
Oxide character changes continuously across a period from basic metal oxides, through amphoteric oxides, to acidic non-metal oxides. Group 1 and group 2 metal oxides react with water to form alkaline hydroxide solutions: Na₂O(s) + H₂O(l) → 2NaOH(aq); CaO(s) + H₂O(l) → Ca(OH)₂(aq). Non-metal oxides react with water to form acids: CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq); SO₂(g) + H₂O(l) → H₂SO₃(aq) (sulfurous acid); SO₃(g) + H₂O(l) → H₂SO₄(aq) (sulfuric acid). An amphoteric oxide, such as aluminium oxide, reacts with both acids and bases.
Acid rain
Rain with a pH below about 5.6, the pH of unpolluted rain in equilibrium with atmospheric carbon dioxide. It is caused by gaseous non-metal oxides, mainly sulfur oxides from burning sulfur-containing fossil fuels and nitrogen oxides from vehicle engines, which react with water in the atmosphere to form acids such as sulfurous, sulfuric and nitric acids. Sulfur dioxide itself forms sulfurous acid; sulfuric acid forms after the sulfur dioxide is oxidized to sulfur trioxide.
Ocean acidification
The decrease in the pH of seawater caused by the oceans absorbing some of the increased atmospheric carbon dioxide: CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq), and carbonic acid releases H⁺ ions. The average pH of surface seawater has fallen from about 8.2 to about 8.1 since pre-industrial times; seawater has become less alkaline but remains above pH 7.
Students often think Any sulfur oxide dissolving in water gives sulfuric acid, the acid in acid rain. In fact No. Sulfur dioxide forms sulfurous acid: SO₂(g) + H₂O(l) → H₂SO₃(aq). Sulfuric acid, H₂SO₄, forms from sulfur trioxide: SO₃(g) + H₂O(l) → H₂SO₄(aq).
Students often think All oxides behave like metal oxides: they react with water to form hydroxides, which are bases. In fact No. Metal oxides form hydroxides and alkaline solutions, but non-metal oxides such as CO₂ and SO₂ react with water to form acids.
S3.1.6 Oxidation state
Oxidation state
A number assigned to an atom to show the number of electrons transferred in forming a bond: it is the charge the atom would have if the compound were composed of ions. It is written with the sign first, for example +2 or −1. The terms 'oxidation state' and 'oxidation number' are used interchangeably.
Deducing oxidation states
The oxidation state of an atom in an uncombined element is zero, because its atoms are identical and no electrons are transferred between them. In a compound the oxidation states sum to zero, and in an ion they sum to the charge of the ion. Oxygen is usually −2, but −1 in peroxides such as H₂O₂ and Na₂O₂; hydrogen is usually +1, but −1 in metal hydrides such as NaH and CaH₂. For example, in NO₂⁻, x + 2(−2) = −1, so nitrogen is +3.
Names of oxyanions
Systematic names show the oxidation state of the central atom in Roman numerals, but generic names persist and are acceptable: NO₃⁻ nitrate (nitrogen +5), NO₂⁻ nitrite (nitrogen +3), SO₄²⁻ sulfate (sulfur +6), SO₃²⁻ sulfite (sulfur +4). The '-ate' ion has the central atom in the higher oxidation state and the '-ite' ion the lower; '-ide' (as in sulfide, S²⁻) names a simple anion with no oxygen.
Students often think An oxidation state is the real charge on an ion, so it applies only to ions, and in a polyatomic ion the ion's charge is the oxidation state. In fact No. An oxidation state is the charge an atom would have if the compound were composed of ions. It is assigned to atoms in covalent molecules and in polyatomic ions too; for example nitrogen is +3 in NO₂⁻, even though the whole ion carries a 1− charge.
Students often think The oxidation states in any species add up to zero. In fact No. In an ion the oxidation states add up to the charge on the ion. In NO₂⁻ they add up to −1, so nitrogen is +3, not +4.
S3.1.7 Discontinuities in first ionization energy as evidence for sublevels HL
Discontinuities in first ionization energy as evidence for sublevels
Across period 3, first ionization energy generally increases but falls from magnesium (738 kJ mol⁻¹) to aluminium (578 kJ mol⁻¹) and from phosphorus (1012 kJ mol⁻¹) to sulfur (1000 kJ mol⁻¹). Aluminium's electron is removed from the 3p sublevel, which is higher in energy than the 3s sublevel from which magnesium's is removed. Sulfur's electron is removed from a 3p orbital that already holds another electron; repulsion between the paired electrons raises its energy. In each case less energy is needed to remove a higher-energy electron. If all the electrons of an energy level had the same energy, the increase would be smooth, so the discontinuities are evidence for sublevels. The explanation rests on the energy of the electron removed, not on 'special stability' of filled or half-filled sublevels.
Students often think Filled and half-filled sublevels have a special stability, which explains ionization-energy discontinuities and the configurations of ions. In fact No. The IB explains the fall by the energy of the electron removed: in sulfur the electron comes from a 3p orbital that already holds another electron, and the repulsion between the paired electrons raises its energy, so less energy is needed to remove it. The 'special stability' of filled and half-filled sublevels is not accepted as an explanation.
Students often think Either explanation (a new sublevel, or paired electrons) can be used for either discontinuity. In fact No. It is the other way round. Mg → Al: aluminium's electron is removed from the higher-energy 3p sublevel rather than 3s. P → S: both are removed from 3p, but sulfur's comes from an orbital holding two electrons, whose repulsion raises its energy.
S3.1.8 Transition element (incomplete d sublevel) HL
Transition element (incomplete d sublevel)
An element whose atoms, or common ions, have an incomplete d sublevel. In the first row, scandium to copper are transition elements; zinc, [Ar] 3d¹⁰ 4s², whose only ion Zn²⁺ is [Ar] 3d¹⁰, has a complete 3d sublevel and is a d-block element but not a transition element.
Characteristic properties of transition elements
Properties that arise from the incomplete d sublevel: variable oxidation states (for example iron +2 and +3, manganese +2 to +7); high melting points; magnetic properties; catalytic activity of the elements and their compounds (for example iron in the Haber process, manganese(IV) oxide in the decomposition of hydrogen peroxide), the catalyst being regenerated at the end of the reaction; formation of coloured compounds; and formation of complex ions, in which ligands (species with a lone pair, such as H₂O, NH₃, Cl⁻ and CN⁻) form coordination bonds to a central metal ion, for example [Cu(H₂O)₆]²⁺.
Students often think Every compound of a transition (d-block) element is coloured, whatever its oxidation state. In fact No. Colour from electron promotion between split d orbitals needs a partly filled d sublevel, so compounds of Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) are white or colourless.
Students often think A catalyst takes part in the reaction and is gradually used up, like a reactant. In fact No. A catalyst takes part in the reaction but is regenerated, so it is not used up. A transition element catalyst may change oxidation state during the reaction and then return to its original state.
S3.1.9 Electron configurations of first-row transition element ions HL
Electron configurations of first-row transition element ions
When a first-row transition element forms a positive ion, electrons are removed from the 4s sublevel before the 3d sublevel, because once the 3d sublevel is occupied the 4s electrons are the higher in energy. So Fe, [Ar] 3d⁶ 4s², forms Fe²⁺, [Ar] 3d⁶, and Fe³⁺, [Ar] 3d⁵; Cu, [Ar] 3d¹⁰ 4s¹, forms Cu²⁺, [Ar] 3d⁹.
Variable oxidation states and successive ionization energies
The 4s and 3d electrons of a transition element are close in energy, so its successive ionization energies are close in value with no large jump until the 3d and 4s electrons are used up. The energy needed to form a higher oxidation state can therefore be recovered from bonding or lattice formation, and several oxidation states are stable. In a group 2 metal, by contrast, the large jump after the second ionization energy limits it to the +2 state.
Students often think Electrons are removed in the reverse order of filling (last in, first out), so the 3d electrons, filled after 4s, are removed first. In fact No. The 4s electrons are removed first. Fe, [Ar] 3d⁶ 4s², forms Fe²⁺, [Ar] 3d⁶.
Students often think A positive charge means extra particles have been added, so a 2+ ion has gained two electrons. In fact No. A positive ion forms by losing electrons: Fe²⁺ has two electrons fewer than an Fe atom (24 instead of 26).
S3.1.10 Colour of transition element complexes HL
Colour of transition element complexes
In a complex ion the ligands split the d sublevel of the central metal ion into orbitals of slightly different energies. When the d sublevel is partly filled, an electron can be promoted from a lower to a higher split d orbital by absorbing visible light whose energy matches the gap. The light that is not absorbed is transmitted or reflected, and this is the colour seen. Ions with an empty (d⁰) or full (d¹⁰) d sublevel, such as Sc³⁺ and Zn²⁺, cannot undergo this promotion and form colourless compounds.
Complementary colour and the colour wheel
On the colour wheel, colours opposite each other are complementary: red is opposite green, orange opposite blue and yellow opposite violet. The colour observed is complementary to the colour absorbed, so a complex that absorbs orange light appears blue. Wavelength λ and frequency f of light are related by c = λf, where c = 3.00 × 10⁸ m s⁻¹, so f = c/λ with λ in metres (1 nm = 10⁻⁹ m).
Students often think The colour we see is the colour of light the substance absorbs. In fact No. It looks blue because it absorbs light of the complementary colour, orange, and the remaining light, which appears blue, reaches the eye.
Students often think Coloured compounds give out light of the colour we see, as in a flame test. In fact No. A complex is coloured because it absorbs some wavelengths of white light as electrons are promoted to higher split d orbitals; the colour seen is the light not absorbed.
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 Which element is classified as a metalloid?
Answer and reasoning
Silicon — Silicon lies on the stepped line between metals and non-metals and has intermediate properties: it is a shiny solid but a semiconductor, and its oxide is acidic. The metalloids are boron, silicon, germanium, arsenic, antimony and tellurium.
Aluminium — A student who counts every element touching the stepped line as a metalloid picks aluminium. Aluminium borders the line on the metal side and is a typical metal: shiny, malleable and a good electrical conductor.
Iodine — A student who classifies by appearance picks iodine because its crystals are shiny. Iodine is a non-metal in group 17: a molecular solid of I₂ that does not conduct electricity.
Scandium — A student who reads 'transition' as 'between metal and non-metal' picks this d-block element. Scandium, like every transition element, is a metal; the metalloids are in the p-block along the stepped line.
Krypton — A student who merges the two right-hand groups picks krypton. Krypton is a noble gas in group 18, with a full outer level; the halogens are the adjacent group 17.
Iodine — The halogens are the group 17 elements: fluorine, chlorine, bromine, iodine and astatine. Iodine atoms have seven valence electrons (5s² 5p⁵) and the element exists as I₂ molecules.
Manganese — A student who carries over the old 'group VII' label into the 1–18 numbering looks in group 7 and finds manganese. Group 7 is a d-block group; the halogens, with seven valence electrons, are group 17.
Hydrogen — A student who groups hydrogen with the halogens, because it forms H₂ and H⁻ as they form X₂ and X⁻, picks hydrogen. Hydrogen is not in group 17; the halogens are fluorine to astatine.
3 Which element has the highest electronegativity on the Pauling scale?
Answer and reasoning
Neon, Ne — A student who extends the trend to the end of the period picks neon. Electronegativity refers to attracting a bonding pair, and neon forms no covalent bonds in ordinary conditions, so it is not given a Pauling value.
Chlorine, Cl — A student who treats electronegativity as electron affinity picks chlorine, which has the most negative first electron affinity (−349 kJ mol⁻¹). Electronegativity is a different property, and fluorine's is the highest.
Iodine, I — A student who thinks more protons always means a stronger pull picks iodine. Down group 17 the bonding pair is further from the nucleus and more shielded, so electronegativity decreases; iodine is the least electronegative of these halogens.
Fluorine, F — Electronegativity increases across a period and decreases down a group, so fluorine, at the top of group 17, is the most electronegative element (3.98 on the Pauling scale). Its small atoms attract a bonding pair strongly.
4 A small piece of sodium is added to a large volume of water containing universal indicator. Which statement describes what happens?
Answer and reasoning
Hydrogen and sodium oxide form, and the oxide stays as a solid — A student who expects the metal oxide as product picks this. Sodium reacts with cold water to form sodium hydroxide, which dissolves; even sodium oxide would react with water to give sodium hydroxide.
Hydrogen gas is released, so the solution turns acidic — A student who links hydrogen with acids picks this. Hydrogen gas is not an acid; the reaction leaves hydroxide ions in solution, so the solution becomes alkaline.
Hydrogen and sodium hydroxide form, so the solution turns alkaline — 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g). The sodium fizzes as hydrogen is released, and the hydroxide ions formed turn the indicator to an alkaline colour.
The sodium dissolves unchanged, and the fizzing is air escaping — A student who confuses reacting with dissolving picks this. Sodium reacts: the gas is hydrogen (it pops with a lighted splint), and the solution contains sodium hydroxide, not sodium metal.
5 Which statement about ocean acidification is correct?
Answer and reasoning
Dissolved CO₂ forms carbonic acid, lowering the pH, but seawater stays alkaline — CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq), and carbonic acid releases H⁺ ions. The average surface pH has fallen from about 8.2 to about 8.1 since pre-industrial times: seawater is less alkaline, not acidic.
Seawater has now become acidic, with a pH below 7, because of absorbed CO₂ — A student who reads 'acidification' as 'becoming acidic' picks this. The pH is falling, but at about 8.1 seawater is still above 7.
Acid rain falling on the sea is the main cause of the fall in seawater pH — A student who merges the two environmental problems picks this. The main cause is the absorption of atmospheric CO₂; acid rain from sulfur and nitrogen oxides has only a small effect on the whole ocean.
Warmer seawater dissolves more CO₂, which is the main reason the pH is falling — A student who assumes gases dissolve better when warm picks this. Gas solubility decreases as temperature rises; the pH falls because the concentration of CO₂ in the atmosphere has risen.
6 Every atom in an uncombined element, such as O₂ or Cl₂, has an oxidation state of zero. Why?
Answer and reasoning
It has no overall charge, so each of its atoms must also be zero — A student who equates each atom's oxidation state with the overall charge picks this. Water is neutral, yet hydrogen is +1 and oxygen −2; elements are zero because their atoms are identical.
Oxidation states are given only to ions, and it contains none — A student who thinks oxidation states are real ionic charges picks this. Oxidation states are assigned to atoms in covalent molecules too, such as carbon (+4) in CO₂.
Its atoms are all identical, so no electrons are transferred between them — An oxidation state shows the electrons transferred in forming a bond. Between identical atoms there is no difference in attraction for electrons, so no atom gains or loses electrons relative to another, and each is zero.
Its atoms already have full outer shells by sharing, so they need no transfer — A student who explains chemistry by atoms reaching stable full shells picks this. Chlorine also has a full outer shell by sharing in HCl, yet there it is −1; the oxidation state of an element is zero because identical atoms do not transfer electrons to one another.
7 The first ionization energies of magnesium and aluminium are 738 kJ mol⁻¹ and 578 kJ mol⁻¹. Which explanation of the lower value for aluminium is correct? HL
Answer and reasoning
Aluminium's electron is removed from 3p, higher in energy than magnesium's 3s — Magnesium is [Ne] 3s² and aluminium [Ne] 3s² 3p¹. Aluminium's electron is removed from the 3p sublevel, which is higher in energy than the 3s sublevel, so less energy is needed. This drop is evidence that the third energy level is divided into sublevels.
Magnesium's filled 3s sublevel has a special stability that resists ionization — A student using the 'special stability' of filled sublevels picks this. The IB explanation is based on the energy of the electron removed: aluminium's 3p electron is higher in energy than magnesium's 3s electron.
Aluminium's electron is removed from a paired orbital, where repulsion aids removal — A student who applies the paired-electron explanation to the wrong discontinuity picks this. Aluminium's single 3p electron is unpaired; electron-pair repulsion explains the fall from phosphorus to sulfur, not from magnesium to aluminium.
Aluminium atoms are larger than magnesium atoms, so the outer electron is further away — A student who assumes more electrons make a bigger atom picks this. Aluminium atoms are smaller than magnesium atoms; the drop in ionization energy comes from the higher energy of the 3p sublevel.
8 Which statement describes a characteristic property of the transition elements? HL
Answer and reasoning
Many of them and their compounds act as catalysts — Catalytic activity is a characteristic property: for example iron in the Haber process and manganese(IV) oxide in the decomposition of hydrogen peroxide. The catalyst is regenerated at the end of the reaction.
Their catalysts are consumed, so they must be added in excess — A student who thinks catalysts are used up like reactants picks this. A transition element catalyst may change oxidation state during the reaction but is regenerated, so it is not consumed and only a small amount is needed.
All of their compounds are coloured, even those of Sc³⁺ — A student who treats colour as universal for the d-block picks this. Sc³⁺ is 3d⁰, so it has no d electrons to promote between split d orbitals, and scandium(III) compounds are colourless.
Each shows a single oxidation state, as group 2 metals do — A student who thinks each element has one fixed oxidation state picks this. Variable oxidation state is a characteristic property of transition elements, for example iron +2 and +3.
9 Manganese forms compounds in oxidation states from +2 to +7. Which statement explains why transition elements show variable oxidation states? HL
Answer and reasoning
They lose different numbers of electrons, each ion reaching a stable noble-gas configuration — A student who explains ion formation by reaching a noble-gas configuration picks this. Most transition element ions, such as Mn²⁺ ([Ar] 3d⁵), do not have noble-gas configurations; variable states arise from closely spaced ionization energies.
Their 3d electrons, filled last, are the outermost and are removed first in any number — A student who removes electrons in reverse filling order picks this. The 4s electrons are removed first; what allows several oxidation states is that the 4s and 3d ionization energies are close in value.
Each oxidation state leaves a half-filled or filled d sublevel, which has special stability — A student relying on 'special stability' picks this. Most oxidation states do not leave half-filled or filled d sublevels (Mn³⁺ is 3d⁴, Mn⁴⁺ is 3d³); the explanation is that successive ionization energies are close in value.
Their 4s and 3d electrons are close in energy, so successive ionization energies are close in value — Because the 4s and 3d electrons have similar energies, successive ionization energies rise gradually, with no large jump until these electrons are used up. The energy needed for a higher oxidation state can be recovered from bonding, so several states are stable.
10 A solution of a transition element complex looks blue in white light. On the colour wheel, blue is opposite orange. Which statement explains the blue colour? HL
Answer and reasoning
It absorbs blue light as electrons are promoted between split d orbitals — A student who thinks the colour seen is the colour absorbed picks this. A complex that absorbed blue light would appear orange; the colour observed is complementary to the colour absorbed.
Its electrons fall back between split d orbitals, emitting blue light — A student who carries over the idea of flame tests picks this. A solution at room temperature is seen by the light it transmits; the complex absorbs orange light and does not emit blue.
It absorbs orange light as electrons are promoted between split d orbitals — Promotion of an electron between the split d orbitals absorbs light of the complementary colour, orange. The remaining light, which appears blue, is transmitted to the eye.
It absorbs no light at all; blue is the natural colour of the metal ion itself — A student who treats colour as a fixed property of the ion picks this. The colour results from absorption of visible light, and changing the ligands changes the colour of the same metal ion.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
21 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Gallium (Z = 31) has the electron configuration [Ar] 3d¹⁰ 4s² 4p¹. Which statement classifies gallium correctly?
Answer and reasoning
A p-block metalloid, as it borders the stepped metal/non-metal line — A student who counts every element touching the stepped line as a metalloid picks this. Gallium borders the line next to germanium, but on the metal side, and has metallic properties; the metalloids are boron, silicon, germanium, arsenic, antimony and tellurium.
A p-block non-metal, as metals are in the s- and d-blocks — A student who thinks the p-block holds only non-metals sees that gallium is a p-block element and concludes it is a non-metal. The metal/non-metal boundary runs diagonally through the p-block, and its lower-left elements, including aluminium and gallium, are metals.
A d-block element, as each atom holds a filled 3d sublevel — A student who assigns the block from any d electrons present picks this. The 3d sublevel in gallium is full and is an inner sublevel; the sublevel being filled is 4p, so gallium is in the p-block.
A p-block metal, as its last electron enters the 4p sublevel — The block is named after the sublevel being filled in the aufbau order. After zinc the 3d sublevel is full, and gallium's last electron enters 4p, so gallium is the first p-block element of period 4 (group 13). It lies below and to the left of the stepped line, and is a metal: shiny and an electrical conductor.
2 Selenium is in period 4 and group 16 of the periodic table. Which statement about a selenium atom is correct?
Answer and reasoning
It has sixteen valence electrons, in the fourth energy level — A student who takes the group number (1–18) as the number of valence electrons gets sixteen. With the 1–18 numbering, a p-block atom has (group number − 10) valence electrons, so selenium has six.
It has four valence electrons, located in the sixth energy level — A student who swaps the roles of period and group takes the period (4) as the number of valence electrons and reads group 16 as the old group VI, placing them in the sixth level. The period number (4) gives the outer energy level, and the group gives the valence electrons (16 − 10 = 6).
It has six valence electrons, in the fourth energy level — Period 4 means the outer electrons are in the fourth energy level; group 16 means 16 − 10 = 6 valence electrons. The configuration is [Ar] 3d¹⁰ 4s² 4p⁴, with 4s² 4p⁴ in the outer level.
It has only two valence electrons, in the fourth energy level — A student who confuses valence electrons with valency counts the two electrons selenium needs to reach eight. Selenium has six valence electrons (4s² 4p⁴); its valency of 2 is a separate idea.
3 How, and why, does atomic radius change across period 3 from sodium to chlorine?
Answer and reasoning
It increases, as each atom has more electrons, which need more space — A student who thinks more electrons means a bigger atom picks this. Atomic radius decreases across the period; the extra electrons enter the same energy level while the nuclear charge rises, so they are pulled in.
It increases, as the nucleus's attraction is shared out among more electrons — A student who pictures the nuclear attraction as a fixed amount shared out among the electrons picks this. The attraction on each electron depends on nuclear charge and distance and is not divided up; radius decreases across the period.
It decreases, as nuclear charge rises while electrons enter the same level — Across period 3 the nuclear charge increases from +11 to +17, while the added electrons go into the third energy level, where they shield each other very little. The outer electrons are pulled closer, so atomic radius decreases.
It decreases, as atoms closer to a full outer shell are more stable and compact — A student who uses the stability of a full shell as a cause picks this. The direction is right but the reason is not: size is explained by the increasing nuclear charge acting on electrons in the same energy level, not by a drive towards a full shell.
4 The ions Na⁺, F⁻, Mg²⁺ and O²⁻ all have the electron configuration 1s² 2s² 2p⁶. Which lists them in order of decreasing ionic radius?
Answer and reasoning
O²⁻ > F⁻ > Na⁺ > Mg²⁺ — All four ions have the same 10 electrons in the same arrangement, so size depends on nuclear charge: 8, 9, 11 and 12 protons. The greater the nuclear charge acting on the same electrons, the more strongly they are pulled in, so radius decreases from O²⁻ to Mg²⁺.
O²⁻ = F⁻ = Na⁺ = Mg²⁺ — A student who thinks the electrons alone decide size expects equal radii. Isoelectronic ions differ in nuclear charge, and a greater nuclear charge pulls the same 10 electrons closer, so the radii differ.
Mg²⁺ > Na⁺ > F⁻ > O²⁻ — A student who links size with the mass of the nucleus ranks the ions by nuclear mass (Mg 24, Na 23, F 19, O 16). The nucleus is a tiny part of the ion's volume; more protons acting on the same electrons make the ion smaller, so this order is reversed.
Na⁺ > Mg²⁺ > O²⁻ > F⁻ — A student who assumes ions keep the size order of their parent atoms uses atomic radius (Na > Mg > O > F). Na⁺ and Mg²⁺ have lost their third energy level and are much smaller than their atoms, while O²⁻ and F⁻ are larger than theirs, so the anions are the larger ions here.
5 How does the first ionization energy of rubidium (Z = 37) compare with that of sodium (Z = 11)?
Answer and reasoning
Higher, as rubidium's nucleus carries a much larger positive charge — A student who considers nuclear charge alone picks this. Down a group the increase in distance and shielding outweighs the larger nuclear charge, so the first ionization energy decreases.
Lower, as rubidium's outer electron is further out and more shielded — Rubidium's outer electron is in the fifth energy level, sodium's in the third. The greater distance and the extra shielding by inner electrons outweigh the larger nuclear charge, so less energy is needed (Rb 403 kJ mol⁻¹, Na 496 kJ mol⁻¹).
Equal, as each atom loses one electron to gain a noble-gas configuration — A student who explains ionization by reaching a full shell expects equal values. Both atoms do form 1+ ions, but the energy needed depends on how strongly the outer electron is attracted, which is less for rubidium.
Lower, as rubidium's nuclear attraction is shared among more electrons — A student who pictures a fixed amount of nuclear attraction shared out picks this. The direction is right but the reason is not: the attraction on the outer electron is weaker because of its greater distance and shielding, not because it is divided among the electrons.
6 Bromine water is added separately to aqueous potassium chloride and to aqueous potassium iodide. Which observation is expected?
Answer and reasoning
A reaction occurs with the potassium chloride only — A student who thinks reactivity increases down every group expects bromine to displace chlorine. Down group 17 oxidizing ability decreases, so bromine reacts with iodide but not chloride ions.
There is no reaction with either of the two solutions — A student who thinks halide ions are unreactive because they have full outer shells picks this. Iodide ions are oxidized by bromine, which attracts electrons more strongly than iodine does.
A reaction occurs with both of the salt solutions — A student who thinks all halogens behave identically expects bromine to react with both. A halogen oxidizes only the halide ions of halogens below it, so bromine reacts with iodide only.
A reaction occurs with the potassium iodide only — Bromine oxidizes iodide ions, which are below it in group 17: Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq), and the solution darkens to brown. Bromine cannot oxidize chloride ions, because chlorine attracts electrons more strongly than bromine.
7 Sulfur dioxide released by burning fossil fuels contributes to acid rain. What is formed when sulfur dioxide itself reacts with water, before any oxidation takes place?
Answer and reasoning
Sulfuric acid, which makes the solution acidic — A student who assumes every sulfur oxide gives sulfuric acid picks this. Sulfuric acid, H₂SO₄ (sulfur +6), forms from sulfur trioxide; SO₂ must first be oxidized to SO₃.
Sulfurous acid, which makes the solution acidic — SO₂(g) + H₂O(l) → H₂SO₃(aq). Sulfur is +4 in both SO₂ and sulfurous acid, and the acid releases H⁺ ions, lowering the pH.
Sulfur hydroxide, which makes the solution alkaline — A student who applies 'oxide + water → hydroxide' to every oxide picks this. That pattern holds for metal oxides; sulfur dioxide is a non-metal oxide and forms an acid.
No new substance, so the solution remains neutral — A student who thinks a gas dissolves only physically picks this. Sulfur dioxide reacts with water to form sulfurous acid, which is why the solution becomes acidic.
8 What is the oxidation state of nitrogen in the nitrite ion?
Answer and reasoning
+4 — A student who makes the oxidation states sum to zero gets x + 2(−2) = 0, so x = +4. In an ion they must sum to the ion's charge, −1, giving +3.
−1 — A student who treats the oxidation state as the actual charge on the ion gives nitrogen the ion's charge, −1. Oxidation states are assigned atom by atom; nitrogen in NO₂⁻ is +3.
+5 — A student who swaps the endings uses nitrate, NO₃⁻: x + 3(−2) = −1, so x = +5. Nitrite is NO₂⁻, with nitrogen +3; +5 is nitrogen in nitrate.
+3 — Nitrite is NO₂⁻. With oxygen −2 and the sum equal to the ion's charge: x + 2(−2) = −1, so x = +3. This matches the rule that the '-ite' ion has the lower oxidation state.
Working The nitrite ion is NO₂⁻. Oxygen is −2 and the oxidation states sum to the charge on the ion: x + 2(−2) = −1, so x = +3.
9 What are the oxidation states of hydrogen in sodium hydride, NaH, and of oxygen in sodium peroxide, Na₂O₂?
Answer and reasoning
H: +1; O: −1 — A student who thinks hydrogen is always +1 picks this. In a metal hydride hydrogen is bonded to a less electronegative metal and is −1; with sodium +1, NaH is neutral only if hydrogen is −1.
H: −1; O: −1 — In NaH, sodium is +1, so hydrogen is −1 (a metal hydride contains H⁻). In Na₂O₂, 2(+1) + 2x = 0, so oxygen is −1 (the peroxide ion O₂²⁻).
H: −1; O: −2 — A student who treats the peroxide as an oxide gives oxygen −2. Then Na₂O₂ would sum to 2(+1) + 2(−2) = −2, not zero; oxygen in a peroxide is −1.
H: +1; O: −2 — A student who applies a fixed table of values (H always +1, O always −2) picks this. Neither compound then sums to zero; hydrides and peroxides are the standard exceptions, with hydrogen and oxygen both −1.
Working NaH: Na is +1 and the compound is neutral, so H = 0 − (+1) = −1. Na₂O₂: 2(+1) + 2x = 0, so x = −1.
10 The first ionization energies of phosphorus and sulfur are 1012 kJ mol⁻¹ and 1000 kJ mol⁻¹. Which explanation of the lower value for sulfur is correct? HL
Answer and reasoning
Phosphorus's half-filled 3p sublevel has a special stability that resists ionization — A student using the 'special stability' of half-filled sublevels picks this. The IB guide requires the explanation to be based on the energy of the electron removed: sulfur's paired 3p electron is raised in energy by repulsion.
Sulfur's electron is removed from a sublevel of higher energy than phosphorus's — A student who applies the new-sublevel explanation to the wrong discontinuity picks this. Both electrons are removed from the 3p sublevel; the difference is that sulfur's comes from a doubly occupied orbital.
Sulfur loses an electron from a doubly occupied 3p orbital, where repulsion raises its energy — Phosphorus is [Ne] 3s² 3p³ with three unpaired 3p electrons; sulfur is [Ne] 3s² 3p⁴, with one 3p orbital doubly occupied. Repulsion between the paired electrons raises the energy of the electron removed, so slightly less energy is needed.
Sulfur atoms are larger than phosphorus atoms, so their outer electron is further away — A student who assumes more electrons make a bigger atom picks this. Sulfur atoms are smaller than phosphorus atoms; the fall is caused by electron-pair repulsion in the 3p orbital.
11 Iron has the electron configuration [Ar] 3d⁶ 4s². Which statement about the formation of the Fe³⁺ ion is correct? HL
Answer and reasoning
Three electrons are removed from the 3d sublevel, giving [Ar] 3d³ 4s² — A student who removes electrons in reverse filling order (last in, first out) picks this. The 4s electrons are removed before any 3d electron, so Fe³⁺ is [Ar] 3d⁵, not [Ar] 3d³ 4s².
Both 4s electrons and then one 3d electron are removed, giving [Ar] 3d⁵ — Once the 3d sublevel is occupied, the 4s electrons are the higher in energy, so they are removed first (Fe²⁺ is [Ar] 3d⁶); the third electron then comes from 3d. Fe³⁺ is [Ar] 3d⁵ (23 electrons).
Three electrons are added to the 3d sublevel, giving [Ar] 3d⁹ 4s² — A student who thinks a positive ion gains electrons picks this. A 3+ ion has lost three electrons: Fe → Fe³⁺ + 3e⁻, leaving 23 electrons, [Ar] 3d⁵.
One 3d electron is removed before the 4s electrons, as 3d⁵ is specially stable — A student who thinks the special stability of a half-filled 3d⁵ sublevel decides the order picks this. The 4s electrons are removed first because they are higher in energy once 3d is occupied; the configuration [Ar] 3d⁵ follows from that order, not from a special stability.
12 A solution of a transition element complex appears red. On the colour wheel, red is opposite green. Take the wavelength of green light as 530 nm and of red light as 700 nm, and c = 3.00 × 10⁸ m s⁻¹. What is the frequency of the light absorbed most strongly? HL
Answer and reasoning
4.29 × 10¹⁴ Hz — A student who takes the observed colour as the absorbed one uses red light: 3.00 × 10⁸/(7.00 × 10⁻⁷) = 4.29 × 10¹⁴ Hz. The complex absorbs green, the complement of red.
5.66 × 10¹¹ Hz — A student who converts nanometres with 10⁻⁶ gets 3.00 × 10⁸/(530 × 10⁻⁶) = 5.66 × 10¹¹ Hz. 1 nm = 10⁻⁹ m, so λ = 5.30 × 10⁻⁷ m.
1.8 × 10⁻¹⁵ Hz — A student who rearranges c = λf as f = λ/c gets (5.30 × 10⁻⁷)/(3.00 × 10⁸) = 1.8 × 10⁻¹⁵ (1.77 × 10⁻¹⁵). That quantity has units of seconds, not hertz, so it cannot be a frequency; f = c/λ.
5.66 × 10¹⁴ Hz — Red is observed, so the complementary colour, green, is absorbed. f = c/λ = (3.00 × 10⁸ m s⁻¹)/(5.30 × 10⁻⁷ m) = 5.66 × 10¹⁴ Hz.
Working The colour absorbed is complementary to the colour observed: red is seen, so green light is absorbed, λ = 530 nm = 530 × 10⁻⁹ m = 5.30 × 10⁻⁷ m. f = c/λ = (3.00 × 10⁸ m s⁻¹)/(5.30 × 10⁻⁷ m) = 5.66 × 10¹⁴ s⁻¹ = 5.66 × 10¹⁴ Hz.
13 The first electron affinity of chlorine is −349 kJ mol⁻¹. How does the first electron affinity of bromine compare with this value, and why?
Answer and reasoning
More negative, as bromine's nucleus carries a larger positive charge — A student who considers nuclear charge alone expects bromine, with 35 protons, to attract the added electron more strongly. Down a group the added electron is further out and more shielded, and these effects outweigh the larger nuclear charge; bromine's value is −325 kJ mol⁻¹.
Less negative, as the added electron enters a more distant, more shielded level — Bromine's added electron enters the 4p sublevel, one energy level further from the nucleus than chlorine's 3p, and is shielded by an extra inner level. This outweighs the larger nuclear charge, so the electron is attracted less strongly and less energy is released: Br(g) + e⁻ → Br⁻(g) has a first electron affinity of −325 kJ mol⁻¹.
The same, as each atom gains just one electron to complete a stable octet — A student who explains electron gain by completing an octet expects equal values, since both atoms gain one electron to reach eight. The energy released depends on how strongly the nucleus attracts the added electron, which is less for bromine (−325 kJ mol⁻¹).
Less negative, as bromine's nuclear attraction is shared out among more electrons — A student who pictures a fixed amount of nuclear attraction shared out picks this. The direction is right but the reason is not: the added electron is attracted less strongly because it is further from the nucleus and more shielded, not because the attraction is divided among the electrons.
14 Small pieces of lithium and of potassium are added separately to cold water. How does the reaction of potassium compare with that of lithium, and why?
Answer and reasoning
More vigorous, as its outer electron is further out and more shielded — Both metals react to form the hydroxide and hydrogen, for example 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g). Potassium's outer electron is in the fourth energy level and lithium's in the second, so potassium's is further from the nucleus and shielded by more inner electrons. It is lost more easily, so potassium reacts more vigorously (the hydrogen often ignites): metallic character increases down group 1.
Less vigorous, as its larger nuclear charge holds its outer electron more tightly — A student who considers nuclear charge alone expects potassium (19 protons) to hold its outer electron more tightly than lithium (3 protons). Down a group the greater distance and shielding outweigh the larger nuclear charge, so potassium loses its electron more easily and reacts more vigorously.
Equally vigorous, as both are group 1 metals with one outer electron — A student who treats elements of one group as identical expects the same reaction. Group 1 metals react similarly, forming the hydroxide and hydrogen, but reactivity increases down the group: potassium reacts far more vigorously than lithium.
More vigorous, as its nuclear attraction is shared among more electrons — A student who pictures a fixed amount of nuclear attraction shared out picks this. The direction is right but the reason is not: potassium's outer electron is held less strongly because it is further from the nucleus and more shielded, not because the attraction is divided among more electrons.
15 Which statement about aluminium oxide, Al₂O₃, is correct?
Answer and reasoning
It reacts with acids only, as every metal oxide is basic — A student who has learned that every metal oxide is basic picks this. Aluminium oxide does react with acids, but it also reacts with bases such as sodium hydroxide: it is amphoteric, part of the continuum from basic to acidic oxides.
It reacts with bases only, as it is a p-block non-metal oxide — A student who thinks the p-block holds only non-metals treats aluminium oxide as a non-metal oxide, and so as acidic. Aluminium is a metal, and its oxide reacts with acids as well as with bases.
It reacts with neither acids nor bases, being neutral — A student who reads 'amphoteric', between basic and acidic, as 'neutral' picks this. An amphoteric oxide reacts with both acids and bases; aluminium oxide's insolubility in water does not make it unreactive towards them.
It reacts with both acids and bases, since it is amphoteric — Aluminium oxide is amphoteric: it reacts with acids, for example Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l), and also with bases such as sodium hydroxide. In period 3 it lies between the basic oxides of sodium and magnesium and the acidic oxides of the non-metals.
16 Calcium oxide is added to water containing universal indicator. Which statement describes what happens?
Answer and reasoning
Calcium hydroxide forms, so the solution turns alkaline — CaO(s) + H₂O(l) → Ca(OH)₂(aq). Calcium oxide is a basic metal oxide: the oxide ion takes H⁺ from water, and the hydroxide ions formed turn the indicator to an alkaline colour. Calcium hydroxide is only slightly soluble, but enough dissolves to make the solution alkaline.
The mixture fizzes, as hydrogen gas is released from the water — A student who applies the pattern for a metal reacting with water (hydroxide + hydrogen) expects hydrogen. Calcium oxide contains no metal atoms to reduce the water: CaO(s) + H₂O(l) → Ca(OH)₂(aq), and all the hydrogen atoms of the water end up in the calcium hydroxide, so no hydrogen gas forms.
The oxide dissolves without reacting, so the solution stays neutral — A student who thinks dissolving is always a physical change picks this. Calcium oxide reacts with water to form calcium hydroxide; the reaction releases heat, and the hydroxide ions make the solution alkaline.
An acid forms, as with every oxide, so the solution turns acidic — A student who generalizes the acid rain reactions to every oxide picks this. Non-metal oxides such as CO₂ and SO₂ form acids with water; calcium oxide is a basic metal oxide and forms an alkaline solution.
Working CaO(s) + H₂O(l) → Ca(OH)₂(aq); in solution Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq), so the solution is alkaline. The equation balances with no H₂ as a product: 2 H atoms on each side.
17 Copper, [Ar] 3d¹⁰ 4s¹, is a d-block element. Is copper a transition element? HL
Answer and reasoning
No, as its atom, [Ar] 3d¹⁰ 4s¹, has a full 3d sublevel — A student who applies the definition to the atom alone sees a full 3d sublevel and excludes copper. The definition also covers common ions: Cu²⁺, [Ar] 3d⁹, has an incomplete 3d sublevel, so copper is a transition element. Zinc is excluded because its ion Zn²⁺, [Ar] 3d¹⁰, is full as well.
Yes, as every metal in the d-block is a transition element — A student who uses 'transition element' for the whole d-block reaches 'Yes' for the wrong reason. Position in the d-block is not enough: zinc is a d-block metal but not a transition element. Copper qualifies only because its ion Cu²⁺, [Ar] 3d⁹, has an incomplete 3d sublevel.
Yes, as its ion Cu²⁺, [Ar] 3d⁹, has an incomplete 3d sublevel — A transition element has an incomplete d sublevel in its atom or in a common ion. Copper's atom has a full 3d sublevel, but copper loses its 4s electron and one 3d electron to form Cu²⁺, [Ar] 3d⁹, its most common ion. That ion has an incomplete 3d sublevel, so copper is a transition element. Consistent with this, copper(II) compounds are coloured and copper shows the oxidation states +1 and +2.
No, as its only ion, Cu⁺, [Ar] 3d¹⁰, has a full 3d sublevel — A student who thinks each element has one fixed oxidation state reads +1 from copper's single 4s electron and considers only Cu⁺. Copper also forms Cu²⁺, [Ar] 3d⁹, which is its most common ion and has an incomplete 3d sublevel, so copper is a transition element.
18 An atom has the electron configuration [Ar] 3d¹⁰ 4s² 4p³. Which statement gives the group of the element correctly?
Answer and reasoning
Group 5, as its outer level holds five valence electrons — A student who takes the group number as the number of valence electrons gets 5. With groups numbered 1 to 18, a p-block atom has (group number − 10) valence electrons, so five valence electrons means group 15. Group 5 is a d-block group (vanadium).
Group 15, as its 4p³ electrons place it in the third p-block column — The configuration ends in 4p³, so the element is in period 4 and in the third column of the p-block. With groups numbered 1 to 18, the p-block is groups 13 to 18, so it is group 15 (arsenic, Z = 33); it has 2 + 3 = 5 valence electrons, and 5 + 10 = 15.
Group 12, as its full 3d sublevel places it in the d-block — A student who assigns the block from any d electrons present places the element in the d-block, in zinc's column. The full 3d sublevel is an inner sublevel; the last electron enters 4p, so the element is in the p-block, group 15.
Group 4, as its outer electrons occupy the fourth energy level — A student who swaps the roles of period and group takes the outer energy level, 4, as the group. The outer energy level gives the period (period 4); the number of valence electrons gives the group (5 + 10 = 15).
19 The graph shows the first ionization energies of the elements with atomic numbers 3 to 18. Which statement explains why the value at the point labelled X is lower than that of the element before it? HL
Answer and reasoning
The element before X has a half-filled p sublevel, whose special stability makes its electron harder to remove — A student who explains the dip by the 'special stability' of nitrogen's half-filled 2p sublevel picks this. The IB explanation compares the energies of the electrons removed: oxygen's electron comes from a 2p orbital that already holds another electron, and the repulsion between the pair raises its energy.
Its electron is removed from the p sublevel, which is higher in energy than the s sublevel of the element before X — A student who applies the new-sublevel explanation to the wrong dip picks this. That explanation fits the dip at Z = 5 (boron, 2p¹, after beryllium, 2s²). Nitrogen's outer electron is also in 2p, so the fall at oxygen comes from the pairing of electrons in one 2p orbital, not from a change of sublevel.
Its nucleus must attract one more electron, so the attraction available for each electron is reduced — A student who pictures the nuclear attraction as a fixed amount shared among the electrons picks this. If that were so, the graph would fall at every step, but it rises from Z = 5 to 7 and from 8 to 10. The attraction on an electron depends on nuclear charge and distance; the dip at oxygen comes from repulsion between the two electrons sharing one 2p orbital.
Its electron is removed from a p orbital that already holds another electron, whose repulsion raises its energy — Reading Z = 8 off the axis, X is oxygen, [He] 2s² 2p⁴; the element before it is nitrogen, [He] 2s² 2p³. Nitrogen's three 2p electrons occupy separate orbitals, but oxygen's fourth 2p electron must pair with one of them. Repulsion between the two electrons in that orbital raises the energy of the electron removed, so less energy is needed than for nitrogen. The dip is evidence that the 2p sublevel consists of separate orbitals that fill singly before pairing.
20 A student shakes a small sample of each of six period 3 oxides with water and measures the pH of the mixture. The bar chart shows the readings. Two of the oxides give the same reading. Which statement about those two oxides is correct?
Answer and reasoning
Both are insoluble in water, which hides that one is amphoteric and the other acidic — The two bars at 7 are Al₂O₃ and SiO₂. Neither oxide dissolves in or reacts with water to a measurable extent, so the water stays at pH 7 and the test cannot show their character. Their reactions with acids and bases do: aluminium oxide reacts with both hydrochloric acid and sodium hydroxide, so it is amphoteric, while silicon dioxide reacts only with bases (hot concentrated sodium hydroxide), so it is acidic. Across period 3 the oxides run from basic (Na₂O, MgO) through amphoteric (Al₂O₃) to acidic (SiO₂, P₄O₁₀, SO₂).
Both are amphoteric, as their elements lie on the boundary between metals and non-metals — A student who maps the metal → metalloid → non-metal sequence directly onto basic → amphoteric → acidic picks this. Aluminium oxide is amphoteric, but silicon dioxide is an acidic oxide: it reacts with hot concentrated sodium hydroxide and not with acids. The pH reading of 7 shows only that neither oxide reacts with water.
Both are neutral oxides, which react with neither acids nor bases and so leave water at pH 7 — A student who reads a neutral pH in water as 'reacts with nothing' picks this. Aluminium oxide reacts with both acids and bases, and silicon dioxide reacts with bases. The reading of 7 arises because the oxides are insoluble in water, not because they are unreactive towards acids and bases.
Both dissolve in water without reacting, so no acid or alkali forms to alter the pH — A student who treats dissolving as a physical process that leaves water neutral picks this. Aluminium oxide and silicon dioxide do not dissolve in water at all; that insolubility is why the pH stays at 7. The oxides that do enter solution here react: Na₂O forms NaOH and SO₂ forms H₂SO₃, which is why their readings are far from 7.
21 The diagram shows a blank outline of periods 1 to 4 of the periodic table, with the s-, d- and p-blocks labelled and four cells lettered. Which cell is occupied by an element whose atoms have six valence electrons in the third energy level?
Answer and reasoning
The cell labelled X — A student who takes the group number to equal the number of valence electrons counts six columns from the left and picks X. Column 6 lies in the d-block, whose atoms have two 4s valence electrons and partly filled 3d sublevels. For a p-block element the group number is the number of valence electrons plus 10, so six valence electrons place the element in column 16.
The cell labelled Y — Three occupied energy levels means period 3, the third row. Six valence electrons means 3s² 3p⁴, so the atom's last electrons are in the p-block, in its fourth column: counting across, that is column 16 (the group number is 10 more than the number of valence electrons for p-block elements). Cell Y, period 3 and column 16, is sulfur.
The cell labelled W — A student who reads 'six valence electrons' as 'six electrons needed to complete the outer level' looks for an atom with two valence electrons and picks W, in column 2 of period 3 (magnesium). Valence electrons are the electrons in the outer energy level; an atom with six of them is in column 16 of the p-block.
The cell labelled Z — A student who counts only the electrons of the last-written sublevel as valence electrons looks for a period 3 atom with 3p⁶ and picks Z, in column 18 (argon). The valence electrons are all the electrons in the outer energy level, 3s and 3p together, so argon has eight and the atom with six is in column 16.
That was your twenty minutes. Real practice on S3.1 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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