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Educerie · IB Diploma · Chemistry
Structure 3 Classification of matter · S3.1 The periodic table: Classification of elements
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Place metals, metalloids, non-metals and the s, p, d and f blocks on the table | SL, HL | Paper 1A: "Which element is a metalloid?" (1 mark) |
| Deduce a configuration up to Z = 36 from the element's position, and the position from the configuration | SL, HL | "Deduce the full electron configuration of the element in period 4, group 15" (1–2 marks) |
| Name the alkali metals, halogens, transition elements and noble gases; number groups 1 to 18 | SL, HL | Recall inside a longer question |
| Explain the periodicity of atomic radius, ionic radius, ionization energy, electron affinity and electronegativity | SL, HL | "Explain why atomic radius decreases across period 3" (2 marks) |
| Describe and explain group 1 metals with water and halogens with halide ions | SL, HL | Observations, an equation with state symbols, a reason (3–4 marks) |
| Deduce equations for the oxides of groups 1 and 2, carbon and sulfur with water; link to acid rain and ocean acidification | SL, HL | "Write an equation for sulfur trioxide reacting with water" (1 mark) |
| Deduce oxidation states, and explain why an element's is zero | SL, HL | Paper 1A, and inside redox questions |
| Explain how dips in first IE across a period are evidence for sublevels | HL only | "Explain why the first IE of oxygen is lower than that of nitrogen" (2 marks) |
| Recognize the characteristic properties of transition elements | HL only | "State two characteristic properties" (2 marks) |
| Deduce configurations of first-row transition element ions; link close successive IEs to variable oxidation states | HL only | "Deduce the electron configuration of Fe³⁺" (1 mark) |
| Explain the colour of complexes; use the colour wheel with c = λf | HL only | "Explain why [Cu(H₂O)₆]²⁺ is coloured" (3 marks); a frequency (2 marks) |
Before you start
You need electron configurations from S1.3: the filling order 1s 2s 2p 3s 3p 4s 3d 4p and the two exceptions, chromium and copper. You need ionic and covalent bonding from Structure 2, because the oxides in section 6 behave as they do because of how they are bonded. For HL, you need successive ionization energies from S1.3.
1The idea in one paragraph
The periodic table lists the elements in order of atomic number and starts a new row each time a new electron shell starts. So an element's row tells you how many shells its atoms have, and its column tells you how many electrons sit in the outer shell. Almost every property that matters depends on two things: how strongly the nucleus pulls on those outer electrons, and how far away they are. So properties change regularly across each row and steadily down each column. That repeating pattern is periodicity, and it lets you predict an element you have never met from its neighbours.
2The map: periods, groups and blocks
A period is a horizontal row; there are seven. A group is a vertical column, numbered 1 to 18 from left to right, counting the transition elements. Figure 1 shows both, and the blocks that cut across them.
Each block is named after the sublevel the last electron goes into. Groups 1 and 2, and helium, are the s-block; groups 13 to 18 are the p-block; groups 3 to 12 are the d-block; the two rows printed underneath are the f-block. The widths are the sublevel capacities: s holds 2, p holds 6, d holds 10, f holds 14.
Four families have names you must know. Alkali metals are group 1, except hydrogen, which has one outer electron but is not a metal. Halogens are group 17. Noble gases are group 18. Transition elements are in the d-block; HL section 9 gives the exact definition.
The table also sorts by character, as Figure 2 shows. Metals fill the left and centre, non-metals the top right, and along the staircase between them are the metalloids: boron, silicon, germanium, arsenic, antimony and tellurium. Their properties are in between: silicon looks metallic and conducts a little, but its oxide behaves like a non-metal oxide.
When Mendeleev arranged the known elements by their properties in the 1860s, nobody knew about electrons. He left gaps where his pattern demanded an element nobody had found, and predicted its properties from its neighbours. Gallium and germanium were later discovered with properties close to his predictions. That is the guide's nature-of-science point: a good classification tells you where to look next.
3Reading an electron configuration off the table
The period number is the outer energy level that holds electrons. Groups 1 and 2 have 1 and 2 valence electrons; groups 13 to 18 have the group number minus 10.
Valence electrons are the electrons in the outer energy level, the ones that take part in reactions. Every element in a group has the same number, which is why a group reacts alike.
To write a full configuration, read along the rows from hydrogen to your element, writing each block as you cross it. The one trap is period 4: you cross 4s, then 3d, then 4p. The d sublevel is always one energy level behind the period.
Worked example 1. Selenium is in period 4, group 16.
Worked example 2, backwards. An element is 1s² 2s² 2p⁶ 3s² 3p³.
Worked example 3, the d-block. Vanadium is the third d-block element of period 4, so it is [Ar] 3d³ 4s². Two elements break the pattern and you must know them: chromium is [Ar] 3d⁵ 4s¹ and copper is [Ar] 3d¹⁰ 4s¹. The guide stops at krypton, Z = 36, so these are the only exceptions you meet.
4Periodicity: every trend has the same two causes
Learn the two causes first. Every one of the five properties follows from them.
Across a period, the nuclear charge rises but the shielding barely changes. Each step right adds one proton to the nucleus and one electron to the same outer level. Electrons in the same level shield each other poorly and the inner electrons are unchanged, so the outer electrons feel a steadily stronger pull, often called the effective nuclear charge.
Down a group, a shell is added each time. The nuclear charge rises too, but the extra inner shell shields it. The outer electrons are further away and more shielded, so they are held less tightly.
Atomic radius is half the distance between the nuclei of two bonded atoms of the element. Across a period it decreases: the stronger pull draws the same shells in closer. Down a group it increases: more shells. Figure 3 shows both; period 3 lies above period 2 at every group because it has one more shell.
Ionic radius. A positive ion, a cation, is smaller than its atom: sodium loses its only 3s electron and with it the whole third shell. A negative ion, an anion, is larger than its atom: the extra electron adds repulsion in the outer shell while the nuclear charge is unchanged, so the electron cloud spreads out. Figure 4(a) draws both to scale.
Across period 3, Na⁺, Mg²⁺ and Al³⁺ all have neon's configuration and shrink as the charge rises. P³⁻, S²⁻ and Cl⁻ all have argon's, one shell more, so they are much larger, and they too shrink left to right. The rule, drawn in Figure 4(b): in an isoelectronic series (ions with the same number of electrons), more protons means a smaller ion. Down a group, ions of the same charge get larger, as atoms do.
First ionization energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms: X(g) → X⁺(g) + e⁻. Across a period it generally increases, because the outer electron is held more tightly. Down a group it decreases, because the outer electron is further away and more shielded. Figure 5 is periodicity in one picture: a peak at every noble gas, a collapse at every alkali metal where a new shell starts.
The rise across a period has two small dips, at groups 13 and 16. At SL, just notice them; at HL they are evidence for sublevels (section 8).
Electron affinity is the energy change when one mole of electrons is added to one mole of gaseous atoms: X(g) + e⁻ → X⁻(g). For most elements energy is released, so the value is negative. Across a period it generally becomes more negative, as the incoming electron feels a stronger pull. Down a group it generally becomes less negative, as the electron enters a shell further out. The trend is untidy. The best-known exception: chlorine (−349 kJ mol⁻¹) releases more than fluorine (−328 kJ mol⁻¹), because fluorine's small 2p sublevel is so crowded that the new electron is repelled by those already there.
Electronegativity is the ability of an atom to attract the shared pair of electrons in a covalent bond, measured on the unitless Pauling scale. It increases across a period (a smaller atom with a stronger pull holds a bonding pair more tightly) and decreases down a group. Fluorine is the most electronegative element at 4.0.
Put the five together and one picture remains. Up and to the right, atoms are smaller and hold electrons more tightly. Down and to the left, the reverse, which is why the metals are there: a metal is an element whose atoms lose electrons easily.
5Down a group: the alkali metals and the halogens
The guide picks two groups whose reactivity runs in opposite directions.
Group 1 with water. Every alkali metal gives a hydroxide and hydrogen:
The reaction gets more violent down the group. Lithium floats and fizzes steadily. Sodium melts into a silvery ball and skates across the surface. Potassium does the same and the hydrogen catches fire with a lilac flame. Each leaves an alkaline solution that turns universal indicator blue or purple.
Why? The metal atom loses its single outer electron: M → M⁺ + e⁻. Down the group the atoms get larger, the outer electron is further from the nucleus and more shielded, so it is lost more easily. That is increasing metallic character down group 1.
Group 17 with halide ions. A halogen takes the electron from a halide ion. Chlorine turns colourless potassium bromide solution orange-yellow as bromine forms:
This displacement reaction only goes one way: a halogen displaces the halide of a halogen below it, never above. Figure 6 shows all nine combinations. Iodine in water is brown, which tells it apart from bromine.
Why? The halogen atom gains an electron, so reactivity depends on how strongly it attracts one. Higher up the group the atom is smaller, the incoming electron gets closer to the nucleus with less shielding, and the attraction is stronger. Chlorine is a stronger oxidising agent than bromine, and bromine than iodine: decreasing non-metallic character down group 17.
Both explanations turn on the same fact, that atoms get bigger down a group. Group 1 gets more reactive and group 17 less because one loses electrons and the other gains them.
Rubidium and caesium react explosively, and fluorine is too toxic to handle in school. That answers the guide's question about simulations: a simulation or video shows the whole trend safely, and the trend then predicts what you have not tested, such as astatine.
6From metal to non-metal: the oxides
Metallic and non-metallic character are the two ends of a continuum. The clearest place to see it is the oxides across period 3, in Figure 7.
Metal oxides are basic. Sodium oxide and magnesium oxide are ionic. Their oxide ion, O²⁻, takes a proton from water and leaves hydroxide ions. An oxide that gives an alkaline solution, or reacts with an acid to form a salt, is a basic oxide.
Non-metal oxides are acidic. The oxides of carbon, sulfur, phosphorus and nitrogen are covalent molecules that react with water to give acids.
In between is amphoteric. Aluminium oxide is ionic with some covalent character. It is insoluble in water but reacts with acids and with bases, so it is amphoteric:
Silicon dioxide is insoluble too, but reacts only with bases, so it is acidic. "Insoluble" does not mean "neutral".
Bonding explains the pattern, the guide's link to Structure 2. An ionic oxide supplies O²⁻, which is a base. A molecular oxide has no oxide ion; water attacks the electron-poor non-metal atom and the product releases H⁺ ions.
Two environmental problems come straight from these equations. Acid rain: rain is naturally about pH 5.6 because of dissolved CO₂. Burning sulfur-containing fuels releases SO₂, and the heat inside engines makes nitrogen oxides; these dissolve to give stronger acids, for example 2NO₂(g) + H₂O(l) → HNO₃(aq) + HNO₂(aq). Acid rain erodes limestone, leaches metal ions from soils and harms life in lakes. Ocean acidification: as atmospheric CO₂ rises, more dissolves in seawater. The carbonic acid releases H⁺, which reacts with carbonate ions (H⁺ + CO₃²⁻ → HCO₃⁻). The ocean's pH falls, and the carbonate that corals and shellfish need to build calcium carbonate is used up.
7Oxidation states
An oxidation state is a number given to an atom to show how many electrons it has lost or gained, fully or in part, in bonding. It is the charge the atom would carry if the compound were entirely ionic. It is written sign first: +2, −1. "Oxidation number" means the same and is accepted.
Oxidation states in a neutral compound add up to zero; in an ion they add up to the charge on the ion.
Apply these in order:
- An atom in an element (Na, O₂, Cl₂, S₈) is 0.
- Group 1 metals are +1, group 2 are +2, aluminium is +3 in compounds.
- Fluorine is always −1.
- Hydrogen is +1, except in metal hydrides such as NaH, where it is −1.
- Oxygen is −2, except in peroxides such as H₂O₂, where it is −1.
- Everything else is whatever makes the total come out right.
Why an element is zero. In Cl₂ the two atoms are identical, so they attract the shared pair exactly equally; neither has gained or lost any share of an electron. The same holds in a lump of sodium: an element has no partner of different electronegativity.
Roman numerals in a name are oxidation states: iron(II) chloride is FeCl₂, iron(III) chloride FeCl₃. Oxyanions can be named this way, but the older names are accepted. Learn four: nitrate NO₃⁻ (N +5), nitrite NO₂⁻ (N +3), sulfate SO₄²⁻ (S +6), sulfite SO₃²⁻ (S +4). The "-ate" ion has the higher oxidation state. Reactivity 3.2 uses exactly these numbers to find what is oxidised and reduced.
8HLIonization energy dips as evidence for sublevels
SL students can skip to section 12.
If electrons simply sat in shells, first IE would rise smoothly across a period. Figure 8 shows it does not: IE falls from beryllium to boron, and from nitrogen to oxygen. Period 3 repeats both dips, Mg to Al and P to S. The dips are evidence that each shell is split into sublevels of different energy.
Beryllium to boron. Beryllium's electron is removed from 2s, boron's from 2p. A 2p electron is higher in energy and partly shielded by the 2s pair, so less energy is needed, even though boron has one more proton.
Nitrogen to oxygen. Nitrogen's three 2p electrons are each alone in an orbital. Oxygen's fourth 2p electron shares an orbital, and the repulsion between the pair raises its energy, so it is easier to remove.
The guide is precise, and so are mark schemes: explain the dip by the energy of the electron removed. "Half-filled sublevels are especially stable" describes the pattern without explaining it and does not earn the mark.
9HLTransition elements
A transition element has an incomplete d sublevel in its atom, or forms ions with an incomplete d sublevel. In period 4 that is titanium to copper for certain, with an argument at each edge.
- Zinc is not one: the atom is [Ar] 3d¹⁰ 4s² and Zn²⁺ is [Ar] 3d¹⁰. The d sublevel is full in both.
- Scandium is the guide's nature-of-science debate. The atom is [Ar] 3d¹ 4s², so it meets the definition. But its only ion, Sc³⁺, is [Ar] 3d⁰, so it shows none of the typical properties: one oxidation state, white compounds.
The incomplete d sublevel gives six characteristic properties:
- Variable oxidation states: manganese is +2 in Mn²⁺, +4 in MnO₂ and +7 in MnO₄⁻ (section 10).
- High melting points: 3d and 4s electrons are both delocalised into the metallic bonding. Iron melts at 1538 °C, calcium at 842 °C.
- Magnetic properties from unpaired d electrons. The types of magnetism are not assessed.
- Catalytic properties: iron in the Haber process, vanadium(V) oxide in the Contact process, manganese(IV) oxide for decomposing hydrogen peroxide, nickel for adding hydrogen to alkenes. Variable oxidation states let them pass electrons to and from reactants.
- Coloured compounds: copper(II) compounds are blue or green, iron(III) yellow-brown (section 11).
- Complex ions: a ligand is a molecule or ion that donates a lone pair to a central metal ion, forming a coordination bond. Six water molecules around Cu²⁺ make [Cu(H₂O)₆]²⁺.
10HLVariable oxidation states and the configurations of ions
Why does titanium form Ti²⁺, Ti³⁺ and Ti⁴⁺ while calcium only forms Ca²⁺? Figure 9 answers with ionization energies.
Calcium's first two electrons come from 4s. The third would come from the argon core and costs over four times as much as the second, so calcium stops at +2. Titanium's first four come from 4s and 3d, which are close in energy, so its successive IEs climb gently. The energy released when the ion forms a lattice or is hydrated can pay for two, three or four of them, depending on the compound. Successive ionization energies close in value are why transition elements have variable oxidation states.
To write a transition element ion, write the atom, then remove electrons from 4s first, then 3d.
4s fills before 3d because it is lower in energy in the empty atom, but once 3d holds electrons, the 4s electrons are outermost and highest in energy, so they leave first. Fe²⁺ written as [Ar] 3d⁴ 4s² scores zero.
11HLWhy transition element complexes are coloured
In an isolated ion the five d orbitals have the same energy. When ligands surround the ion, their lone pairs repel the d electrons unequally: orbitals pointing at the ligands are raised more than the others. The d sublevel splits into two sets with a gap ΔE, as Figure 10(a) shows.
For first-row transition elements that gap matches the energy of a visible photon. When white light passes through, an electron in the lower set absorbs the photon whose energy equals ΔE and is promoted to the upper set. That colour is removed. What reaches your eye looks like the complementary colour, opposite the absorbed one on the colour wheel in Figure 10(b). The data booklet prints its own wheel with wavelength ranges; use that one in the exam.
This needs a partly filled d sublevel. With 3d⁰ (Sc³⁺, Ti⁴⁺) there is no electron to promote; with 3d¹⁰ (Zn²⁺, Cu⁺) there is no space above. Both give white compounds.
Worked example. A complex absorbs most strongly at 600 nm. Deduce its colour and the frequency absorbed.
ΔE, and so the colour, depends on the metal, its oxidation state and the ligand, which is why [Cu(H₂O)₆]²⁺ changes colour when other ligands replace the water. And because a solution absorbs more light the more concentrated it is, colorimetry can measure the concentration of a coloured ion.
12Where marks are lost
Saying "more shells" for a trend across a period. Across a period the number of shells is constant. The cause is rising nuclear charge with similar shielding.
Using "stable" instead of a cause. For ionization energy and its dips, say where the electron is (distance, shielding, sublevel, pairing) and so how much energy it needs.
Mixing up electronegativity and electron affinity. Electronegativity concerns a shared pair in a bond and has no units. Electron affinity is an energy change for a gaseous atom gaining an electron, in kJ mol⁻¹.
Getting halogen reactivity backwards. Reactivity rises down group 1 but falls down group 17. Say which process happens, losing or gaining, before you explain.
Calling an insoluble oxide neutral. Al₂O₃ and SiO₂ leave water at pH 7 because they do not dissolve. Their reactions with acids and bases show they are amphoteric and acidic.
Writing oxidation states number first. An oxidation state is +2; an ionic charge is 2+.
Removing 3d before 4s (HL). Fe³⁺ is [Ar] 3d⁵.
Naming the colour absorbed as the colour seen (HL). A complex that looks blue absorbs orange.
13Draw it right
- A trend sketch has both axes labelled with quantity and unit: "first ionization energy / kJ mol⁻¹" against "atomic number".
- A first IE sketch across a period shows the general rise, the peak at the noble gas and the drop to the next alkali metal; at HL, the dips at groups 13 and 16.
- Radii are compared on the same scale, or the comparison shows nothing.
- Full configurations go in filling order with superscripts; condensed ones start from the previous noble gas in square brackets.
- Every equation is balanced with state symbols; displacement answers also give the ionic equation.
- Orbital boxes (HL): one box per orbital, electrons as arrows, paired only once every box in the sublevel has one.
- A colour-wheel answer (HL) names the colour absorbed and the colour observed separately.
14Try it
Marks in brackets. Answers and marker's notes are at the end.
Q1. Which statement explains why atomic radius decreases from sodium to chlorine? 1 mark
A. The number of occupied energy levels decreases.
B. The nuclear charge increases while shielding by inner electrons stays about the same.
C. The number of neutrons increases.
D. The atoms change from metallic to covalent bonding.
Q2. Element X is in period 4, group 15. Deduce the full electron configuration of X and state its block. 2 marks
Q3. A small piece of potassium is added to water containing universal indicator. Describe two observations, write an equation with state symbols, and explain why potassium reacts more vigorously than sodium. 4 marks
Q4. A student adds a little of each oxide to water, shakes it, and measures the pH. Her results (invented for this question):
| Oxide | Na₂O | MgO | Al₂O₃ | SiO₂ | SO₂ |
|---|---|---|---|---|---|
| pH of mixture | 13 | 10 | 7 | 7 | 3 |
(a) Write equations for the reactions of Na₂O and SO₂ with water. 2 marks
(b) The student concludes that Al₂O₃ and SiO₂ are neutral oxides. Evaluate this conclusion. 2 marks
(c) Explain how the result for SO₂ relates to acid rain. 2 marks
Q5. Deduce the oxidation state of: (a) Cl in ClO₃⁻, (b) H in CaH₂, (c) O in BaO₂. (d) Explain why the oxidation state of nitrogen in N₂ is zero. 4 marks
Q6 (HL).
(a) Explain why the first ionization energy of sulfur is lower than that of phosphorus. 2 marks
(b) Deduce the electron configuration of Fe³⁺, and explain why Fe³⁺ compounds are coloured but Zn²⁺ compounds are white. 4 marks
(c) A complex ion absorbs most strongly at 510 nm. Deduce the colour of its solution and calculate the frequency absorbed. 2 marks
15In one breath
The table orders elements by atomic number in 7 periods and 18 groups, split into s, p, d and f blocks by the sublevel filled last; metals are left and centre, non-metals top right, metalloids on the staircase. The period is the outer level and the group gives the valence electrons, so position and configuration translate both ways. Across a period nuclear charge rises with little extra shielding, so atoms shrink and ionization energy, electron affinity and electronegativity rise; down a group a shell is added and all reverse. Cations are smaller than their atoms, anions larger, and in an isoelectronic series more protons means smaller. Group 1 gets more reactive down the group because the outer electron is lost more easily; group 17 gets less reactive because an electron is gained less strongly. Oxides run from basic and ionic through amphoteric Al₂O₃ to acidic and covalent, which is where acid rain and ocean acidification come from. Oxidation states add to zero or the ion's charge; elements are 0. HL: the dips at groups 13 and 16 come from the higher energy of a p electron and of a paired electron; transition elements have incomplete d sublevels, close successive IEs and so variable oxidation states, lose 4s before 3d, and are coloured because ligands split the d orbitals and an electron absorbs the photon matching the gap, leaving the complementary colour.
Answers
Q1. B. Each step across period 3 adds a proton, and the extra electron joins the same outer level, adding little shielding, so the outer electrons are pulled in more strongly. B only. A is false: three levels are occupied throughout. C is irrelevant: neutrons are uncharged. D describes the elements, not the cause.
Q2. Group 15 in period 4 means five valence electrons, 4s² 4p³, after crossing the d-block. X is arsenic: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p³, in the p-block. 1 for the full configuration with 3d¹⁰ and 33 electrons in total, 1 for p-block. Leaving out 3d¹⁰ scores 0 for the first mark.
Q3. Any two of: floats; melts into a ball; moves rapidly on the surface; fizzes; the gas burns with a lilac flame; the indicator turns blue or purple. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g). Potassium atoms are larger, so the outer electron is further from the nucleus and more shielded; it is lost more easily, so potassium reacts faster. 1 for two observations, 1 for the balanced equation with state symbols, 1 for a larger atom with the electron further out, 1 for more shielding or lower ionization energy so the electron is lost more easily. "Because it is lower in the group" restates the question and scores 0.
Q4. (a) Na₂O(s) + H₂O(l) → 2NaOH(aq); SO₂(g) + H₂O(l) → H₂SO₃(aq). (b) Not supported. pH 7 shows only that the oxides did not dissolve. Al₂O₃ reacts with acids and with bases, so it is amphoteric; SiO₂ reacts with bases, so it is acidic. She would need to test each with acid and with alkali. (c) SO₂ reacts with water to form an acid, hence pH 3. SO₂ from burning sulfur-containing fuels dissolves in rainwater the same way, taking rain below its natural pH of about 5.6. (a) 1 for each equation with state symbols. (b) 1 for pH 7 reflecting insolubility, 1 for the correct character of Al₂O₃ or SiO₂ from its reactions. (c) 1 for SO₂ forming an acid in water, 1 for fuel combustion as the source and rain below pH 5.6. "SO₂ is acid rain" scores 0.
Q5. (a) Cl + 3(−2) = −1, so Cl is +5. (b) Ca is +2, so each H is −1, as in any metal hydride. (c) Ba is +2, so each O is −1, as in any peroxide. (d) The two identical atoms attract the shared electrons equally, so no electrons are transferred, even partly; each is 0. 1 each for (a), (b), (c) with the sign first, 1 for equal sharing between identical atoms in (d). "5+" loses the mark in (a).
Q6 (HL). (a) In phosphorus, 3s² 3p³, each 3p electron is alone in an orbital. In sulfur, 3s² 3p⁴, the electron removed shares a 3p orbital; repulsion between the pair raises its energy, so less energy is needed to remove it. (b) Fe is [Ar] 3d⁶ 4s²; removing both 4s electrons and one 3d gives Fe³⁺ [Ar] 3d⁵. Ligands split the d orbitals into two sets. With a partly filled d sublevel, an electron can absorb visible light of energy ΔE and be promoted, and the complementary colour is seen. Zn²⁺ is [Ar] 3d¹⁰, so no promotion is possible and no visible light is absorbed. (c) 510 nm is green, so the solution appears red. f = 3.00 × 10⁸ ÷ 5.10 × 10⁻⁷ = 5.88 × 10¹⁴ s⁻¹. (a) 1 for the paired 3p electron in S, 1 for repulsion making it easier to remove; "half-filled is stable" alone scores 0. (b) 1 for 3d⁵ with 4s removed, 1 for splitting by ligands, 1 for promotion absorbing visible light with the complementary colour seen, 1 for Zn²⁺ 3d¹⁰ with no possible transition. (c) 1 for red, 1 for the frequency with its unit; "green" scores 0 for the colour.
Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section S3.1 The periodic table: Classification of elements. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.