DP Chemistry HL S3.1 The Periodic Table: Classification of Elements

DP Chemistry HL S3.1 The Periodic Table: Classification of Elements - Full notes

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DP Chemistry HL | S3.1 The Periodic Table: Classification of Elements

DP CHEMISTRY · STRUCTURE

S3.1

The Periodic Table: Classification of Elements

Full notes for HL · IB-style

Contents

1	The structure of the periodic table
2	Periodic trends: radius, ionization energy, electron affinity, electronegativity
3	Metallic and non-metallic character
4	Oxides across period 3
5	Group 1: the alkali metals
6	Group 17: the halogens
7	Oxidation states
8	Transition elements: properties and variable oxidation states (HL only)
9	Complex ions and colour (HL only)
10	The history of the periodic table
11	Periods, groups and blocks
12	Effective nuclear charge and shielding
13	Atomic radius: data and explanation
14	Ionization energy trends in detail
15	Electron affinity
16	Electronegativity trends
17	Metallic and non-metallic character
18	Group 1: the alkali metals
19	Group 17: the halogens
20	Oxides across period 3
21	Oxidation states
22	Mixed review: worked problems
23	Additional exam-style practice
24	Periodicity in physical properties
25	Common misconceptions and links
26	Exam-style question with markscheme
27	Check your understanding
28	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe the periodic table in terms of periods, groups and s, p, d, f blocks
explain trends in atomic radius, ionic radius, ionization energy, electron affinity and electronegativity
describe the change from metallic to non-metallic character
explain the acid-base character of period 3 oxides and write equations
describe the reactions of group 1 metals with water and halogens with halides
deduce oxidation states and name compounds using oxidation states
explain the characteristic properties of transition elements
explain the formation of complex ions and the origin of colour

Sections marked HL only (Sections 8 and 9) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How does the periodic table help us predict the properties of elements? What causes the patterns in properties across periods and down groups?

How this topic is examined

S3.1 is examined through explanations of periodic trends, equations for reactions of oxides, group 1 and group 17 elements, and oxidation states. HL questions add transition element properties and complex ion colour.

Explain trends using nuclear charge, distance of outer electrons from the nucleus, and shielding.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 The structure of the periodic table

Elements arranged by increasing atomic number.

Think about it

Mendeleev left gaps in his periodic table for elements nobody had discovered. How could he be confident they existed?

The periodic table arranges elements in order of increasing atomic number. Horizontal rows are periods; the period number equals the number of occupied electron shells. Vertical columns are groups, numbered 1 to 18; elements in the same group have the same number of valence electrons and similar chemical properties.

The table is divided into blocks according to the sub-level being filled: the s-block (groups 1 and 2), the p-block (groups 13 to 18), the d-block (groups 3 to 12, transition elements), and the f-block (lanthanoids and actinoids). For s- and p-block elements, the number of valence electrons equals the group number (groups 1 and 2) or the group number minus 10 (groups 13 to 18).

Dmitri Mendeleev (1869) arranged elements by atomic mass and similar properties, leaving gaps for undiscovered elements and predicting their properties. The discovery of gallium and germanium with properties close to his predictions strongly supported the periodic law.

WORKED EXAMPLE 1 Locating an element

The configuration of an element is [Ne] 3s2 3p4. Deduce its period, group, block and identity.

Solution

Highest shell n = 3: period 3A1

Six valence electrons in s and p: group 16, p-blockA1

The element is sulfurA1

Quick check

1. What does the period number show?

2. Which block contains the transition elements?

Answers: 1. number of occupied electron shells 2. d-block

2 Periodic trends: radius, ionization energy, electron affinity, electronegativity

Nuclear charge, distance and shielding explain everything.

Three factors explain most trends. Nuclear charge increases across a period (one more proton each time). Shielding by inner electrons stays about the same across a period but increases down a group. Distance of outer electrons from the nucleus increases down a group as shells are added. The combined effect of nuclear charge and shielding is the effective nuclear charge felt by outer electrons.

Atomic radius decreases across a period (greater effective nuclear charge pulls the same shell closer) and increases down a group (more shells).

Ionic radius: cations are smaller than their atoms (they lose the outer shell); anions are larger than their atoms (extra electrons increase repulsion). Across period 3, Na+, Mg2+, Al3+ get smaller (more protons for the same electrons); then P3−, S2−, Cl− are much larger and also decrease.

First ionization energy generally increases across a period and decreases down a group (S1.3).

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 it is exothermic. Halogens have the most exothermic values.

Electronegativity increases across a period and decreases down a group. Fluorine is the most electronegative element.

Property	Across a period	Down a group	Main reason
Atomic radius	decreases	increases	nuclear charge; number of shells
First ionization energy	increases	decreases	attraction to outer electrons
Electronegativity	increases	decreases	attraction for bonding pair
Metallic character	decreases	increases	ease of losing electrons

Figure 1. Atomic radius decreases across period 3.

Common mistake

Saying atoms get smaller across a period because they have fewer electrons. They have more electrons, but the increasing nuclear charge pulls the same shell closer.

Quick check

1. Why does atomic radius increase down a group?

2. Why is Cl− larger than Cl?

Answers: 1. more electron shells 2. extra electron increases repulsion; same nuclear charge

3 Metallic and non-metallic character

Metals on the left, non-metals on the right.

Metallic character (the tendency to lose electrons and form cations) decreases across a period and increases down a group. Metals are found on the left and in the centre; non-metals on the upper right. Metalloids (such as boron, silicon, germanium, arsenic) lie along the boundary and have intermediate properties: silicon is a semiconductor.

Across period 3: sodium, magnesium and aluminium are metals; silicon is a metalloid; phosphorus, sulfur, chlorine and argon are non-metals.

4 Oxides across period 3

Basic on the left, acidic on the right.

The acid-base character of oxides changes across period 3. Metal oxides are basic: they react with acids, and soluble ones form alkaline solutions with water. Non-metal oxides are acidic: they react with water to form acids, and react with bases. Aluminium oxide is amphoteric: it reacts with both acids and bases.

Oxide	Bonding	Character	Reaction with water
Na2O	ionic	basic	Na2O + H2O → 2NaOH
MgO	ionic	basic	MgO + H2O → Mg(OH)2
Al2O3	ionic (some covalent character)	amphoteric	insoluble
SiO2	giant covalent	acidic	insoluble
P4O10	covalent	acidic	P4O10 + 6H2O → 4H3PO4
SO3	covalent	acidic	SO3 + H2O → H2SO4

Non-metal oxides released by human activity cause acid deposition: sulfur dioxide from burning fossil fuels forms sulfurous and sulfuric acid; nitrogen oxides from engines form nitric acid. Effects include damage to forests, lakes and limestone buildings. Carbon dioxide dissolves in seawater to form carbonic acid, causing ocean acidification.

Quick check

1. Why is Al2O3 described as amphoteric?

2. Write the equation for SO3 with water.

Answers: 1. reacts with both acids and bases 2. SO3 + H2O → H2SO4

5 Group 1: the alkali metals

Reactive metals that get more reactive down the group.

Group 1 metals (lithium, sodium, potassium, rubidium, caesium) are soft, have low densities and low melting points, and are very reactive, so they are stored under oil. They have one valence electron, which they lose easily to form 1+ ions.

They react with water to form a hydroxide and hydrogen: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g). Lithium fizzes steadily; sodium melts into a ball and moves on the surface; potassium ignites with a lilac flame. Reactivity increases down the group, because the outer electron is further from the nucleus and more shielded, so it is lost more easily (lower ionization energy).

6 Group 17: the halogens

Reactive non-metals that get less reactive down the group.

The halogens exist as diatomic molecules: F2 (pale yellow gas), Cl2 (green gas), Br2 (red-brown liquid), I2 (grey solid, purple vapour). Melting and boiling points increase down the group because molecules have more electrons and stronger London forces.

Halogens gain one electron to form halide ions (1−). Reactivity decreases down the group, because the incoming electron is further from the nucleus and more shielded, so it is attracted less strongly.

Displacement reactions: a more reactive halogen displaces a less reactive halide from solution. Chlorine displaces bromide and iodide: Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(aq) (solution turns orange-yellow). Bromine displaces iodide (solution turns brown). Iodine cannot displace chloride or bromide.

Halide ions with silver nitrate form precipitates: AgCl white, AgBr cream, AgI yellow. Ag+(aq) + Cl−(aq) → AgCl(s).

	Cl−	Br−	I−
Cl2	no reaction	Br2 formed (orange)	I2 formed (brown)
Br2	no reaction	no reaction	I2 formed (brown)
I2	no reaction	no reaction	no reaction

Quick check

1. Why does reactivity increase down group 1?

2. Write an equation for chlorine with potassium iodide.

Answers: 1. outer electron further and more shielded, lost more easily 2. Cl2 + 2KI → 2KCl + I2

7 Oxidation states

Bookkeeping for electrons.

The oxidation state is the charge an atom would have if all its bonds were ionic. Rules: the oxidation state of an uncombined element is 0; of a simple ion equals its charge; in compounds, the sum of oxidation states equals zero (or the charge of a polyatomic ion); fluorine is always −1; oxygen is usually −2 (except in peroxides, −1, and with fluorine); hydrogen is usually +1 (except in metal hydrides, −1); group 1 metals are +1 and group 2 are +2.

Oxidation states are used in names, written as Roman numerals: iron(III) chloride, FeCl3; manganese(IV) oxide, MnO2; potassium manganate(VII), KMnO4.

WORKED EXAMPLE 2 Deducing oxidation states

Deduce the oxidation state of Cr in K2Cr2O7 and of S in SO42−.

Solution

K2Cr2O7: 2(+1) + 2x + 7(−2) = 0, so 2x = 12, x = +6A1

SO42−: x + 4(−2) = −2, x = +6A1

Quick check

1. What is the oxidation state of Mn in KMnO4?

2. Name FeO using oxidation states.

Answers: 1. +7 2. iron(II) oxide

8 Transition elements: properties and variable oxidation states (HL only)

Incomplete d sub-levels give distinctive properties.

HL only

This section is Additional Higher Level content. SL students do not need it.

A transition element is an element that has an incomplete d sub-level in the atom or in one of its common ions. Scandium to copper are transition elements; zinc is not, because Zn and Zn2+ both have a full 3d10 sub-level.

Characteristic properties: variable oxidation states; formation of complex ions with ligands; coloured compounds; catalytic activity (iron in the Haber process, V2O5 in the Contact process, MnO2 decomposing H2O2); and magnetic properties (paramagnetism from unpaired electrons).

Variable oxidation states arise because the 4s and 3d sub-levels are close in energy, so different numbers of electrons can be lost without a large jump in ionization energy. All first-row transition elements show +2 (loss of 4s electrons). Manganese shows every state from +2 to +7.

Element	Common oxidation states	Examples
Fe	+2, +3	FeSO4, FeCl3
Cu	+1, +2	Cu2O, CuSO4
Cr	+3, +6	Cr2O3, K2Cr2O7
Mn	+2, +4, +7	MnCl2, MnO2, KMnO4

9 Complex ions and colour (HL only)

Ligands split the d orbitals.

HL only

This section is Additional Higher Level content. SL students do not need it.

A complex ion consists of a central metal ion surrounded by ligands: molecules or ions with a lone pair that form coordination bonds to the metal. Common ligands are H2O, NH3, Cl−, CN−. The coordination number is the number of coordination bonds. Examples: [Cu(H2O)6]2+ (octahedral, blue), [CuCl4]2− (tetrahedral, yellow-green).

Colour: 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, and the d orbitals split into two sets of different energy. When white light passes through the solution, an electron absorbs a photon of a particular frequency and is promoted to the higher set. The complementary colour of the absorbed light is seen.

The size of the splitting, and so the colour, depends on: the identity of the metal, its oxidation state (higher charge gives larger splitting), the ligand (the spectrochemical series: I− < Cl− < H2O < NH3 < CN−), and the geometry of the complex. Ions with no d electrons (Sc3+) or a full d sub-level (Zn2+, Cu+) are colourless, because no d-d transition is possible.

Figure 2. Origin of colour in transition metal complexes.

Quick check

1. Why is zinc not a transition element?

2. Why is Sc3+ colourless?

3. Define ligand.

Answers: 1. Zn and Zn2+ have full 3d10 2. no d electrons, so no d-d transitions 3. species with a lone pair that forms a coordination bond to a metal ion

10 The history of the periodic table

Mendeleev's predictions.

Mendeleev (1869) arranged elements by atomic mass and chemical properties, leaving gaps for undiscovered elements and predicting their properties. Gallium and germanium were later discovered with properties close to his predictions. Moseley (1913) showed that atomic number, not mass, is the true basis of ordering.

WORKED EXAMPLE 3 Evaluating Mendeleev (standard)

Give two reasons why Mendeleev's table was accepted.

Solution

He left gaps and correctly predicted properties of undiscovered elementsR1

He grouped elements with similar properties together, sometimes swapping order by massR1

Quick check

1. Why was Moseley's work important?

Answers: 1. showed elements are ordered by atomic number (proton number)

11 Periods, groups and blocks

Reading the table.

Periods are horizontal rows (number of occupied shells). Groups are vertical columns (similar outer electron configurations). Elements are classified as metals, metalloids and non-metals; the s, p, d and f blocks show the sub-level being filled.

WORKED EXAMPLE 4 Position from configuration (standard)

An element has configuration 1s22s22p63s23p5. State its period, group and name.

Solution

Period 3, group 17A1

ChlorineA1

Quick check

1. State the group and period of an element with configuration [Ar]4s2.

2. Which block contains the halogens?

Answers: 1. group 2, period 4 (calcium) 2. p-block

12 Effective nuclear charge and shielding

The key to all trends.

The effective nuclear charge felt by the outer electrons is roughly the nuclear charge minus the shielding by inner electrons. Across a period, nuclear charge increases while shielding stays similar, so the effective nuclear charge increases. Down a group, extra shells increase distance and shielding.

WORKED EXAMPLE 5 Effective charge (standard)

Estimate the effective nuclear charge on the outer electron of sodium and of chlorine (simple model).

Solution

Na: 11 − 10 = +1A1

Cl: 17 − 10 = +7A1

Quick check

1. Use this model to explain why chlorine atoms are smaller than sodium atoms.

Answers: 1. greater effective charge pulls outer electrons closer

13 Atomic radius: data and explanation

Decreases across, increases down.

Atomic radius decreases across a period because the effective nuclear charge increases with the same number of shells. It increases down a group because of additional electron shells.

Element	Na	Mg	Al	Si	P	S	Cl
Radius / pm	160	140	124	114	109	104	100

WORKED EXAMPLE 6 Explaining (exam level)

Explain why potassium atoms are larger than sodium atoms.

Solution

K has an extra electron shellR1

Outer electron further from the nucleus despite greater nuclear charge (more shielding)R1

Quick check

1. Which is larger: Mg or S?

Answers: 1. Mg

14 Ionization energy trends in detail

Across and down.

First ionization energy generally increases across a period and decreases down a group. Small dips occur: from group 2 to 13 (the electron removed is in a higher-energy p sub-level) and from group 15 to 16 (pair repulsion in the p sub-level).

Figure 3. First ionization energies of period 3 elements.

WORKED EXAMPLE 7 Dip at aluminium (exam level)

Explain why aluminium has a lower first ionization energy than magnesium.

Solution

Al's outer electron is in 3p, Mg's in 3sR1

3p is higher in energy and slightly more shielded, so easier to removeR1

Quick check

1. Explain why sulfur's first ionization energy is lower than phosphorus's.

Answers: 1. S has a paired 3p electron; repulsion makes it easier to remove

15 Electron affinity

Energy change on gaining an electron.

The first electron affinity is the energy change when one mole of electrons is added to one mole of gaseous atoms: X(g) + e− → X−(g). It is usually exothermic (negative). Halogens have the most negative values. Second electron affinities are endothermic, because an electron is added to a negative ion.

WORKED EXAMPLE 8 Second electron affinity (standard)

Explain why the second electron affinity of oxygen is endothermic.

Solution

Adding an electron to O−R1

Repulsion between the negative ion and the electron must be overcomeR1

Quick check

1. Write the equation for the first electron affinity of chlorine.

Answers: 1. Cl(g) + e− → Cl−(g)

16 Electronegativity trends

Attracting shared electrons.

Electronegativity (Pauling scale) increases across a period and decreases down a group, for the same reasons as ionization energy. Fluorine is the most electronegative element (4.0). Noble gases are usually not assigned values.

WORKED EXAMPLE 9 Comparing (standard)

Explain why fluorine is more electronegative than chlorine.

Solution

F's outer electrons are closer to the nucleusR1

Less shielding: stronger attraction for bonding electronsR1

Quick check

1. Order by electronegativity: O, N, F, C.

Answers: 1. C < N < O < F

17 Metallic and non-metallic character

Reactivity trends.

Metallic character (tendency to lose electrons) increases down a group and decreases across a period. Non-metallic character (tendency to gain electrons) shows the opposite trends. This explains the reactivity of group 1 increasing down the group and group 17 decreasing down the group.

WORKED EXAMPLE 10 Group reactivity (exam level)

Explain why caesium is more reactive than lithium.

Solution

Cs outer electron much further from the nucleus and more shieldedR1

Lower ionization energy: lost more easilyR1

Quick check

1. Why is fluorine the most reactive halogen?

Answers: 1. smallest atom; strongest attraction for an extra electron

18 Group 1: the alkali metals

Reactions with water.

Group 1 metals react with water to form hydroxides and hydrogen: 2M(s) + 2H2O(l) → 2MOH(aq) + H2(g). Reactivity increases down the group: lithium fizzes steadily, sodium melts into a ball, potassium ignites with a lilac flame.

WORKED EXAMPLE 11 Equation (basic)

Write the equation for potassium and water, with state symbols.

Solution

2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)A2

Quick check

1. Predict the pH of the solution formed.

2. Why are alkali metals stored under oil?

Answers: 1. above 7 (alkaline) 2. to prevent reaction with air and moisture

19 Group 17: the halogens

Displacement reactions.

A more reactive halogen displaces a less reactive halide from solution: Cl2(aq) + 2Br−(aq) → 2Cl−(aq) + Br2(aq). Colour changes identify the products (bromine orange, iodine brown in water, violet in hexane).

	Cl−	Br−	I−
Cl2	–	orange (Br2)	brown (I2)
Br2	no reaction	–	brown (I2)
I2	no reaction	no reaction	–

WORKED EXAMPLE 12 Displacement (standard)

Write the ionic equation for bromine water reacting with potassium iodide solution.

Solution

Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(aq)A2

Quick check

1. Explain why iodine does not react with sodium chloride solution.

Answers: 1. iodine is less reactive (weaker oxidizing agent) than chlorine

20 Oxides across period 3

Basic to acidic.

Period 3 oxides change from basic (Na2O, MgO) through amphoteric (Al2O3) to acidic (P4O10, SO3). Na2O + H2O → 2NaOH; SO3 + H2O → H2SO4. Bonding changes from ionic to covalent across the period.

WORKED EXAMPLE 13 Acid–base character (standard)

Write equations for MgO and P4O10 reacting with water.

Solution

MgO(s) + H2O(l) → Mg(OH)2(aq) (slightly soluble)A1

P4O10(s) + 6H2O(l) → 4H3PO4(aq)A1

In the real world

Acid rain forms when sulfur and nitrogen oxides from fuel combustion dissolve in rainwater, forming sulfuric and nitric acids.

Quick check

1. Describe the amphoteric behaviour of Al2O3.

Answers: 1. reacts with both acids and bases

21 Oxidation states

Bookkeeping for electrons.

Oxidation states are assigned using rules: elements 0; monatomic ions equal their charge; O usually −2 (−1 in peroxides); H usually +1 (−1 in metal hydrides); F always −1; the sum equals the overall charge. Roman numerals in names give oxidation states: iron(III), manganese(VII).

WORKED EXAMPLE 14 Assigning (standard)

Find the oxidation states of Mn in KMnO4 and Cr in Cr2O72−.

Solution

Mn: +1 + x − 8 = 0: +7A1

Cr: 2x − 14 = −2: +6A1

Quick check

1. Find the oxidation state of S in H2SO4 and in Na2S2O3.

2. Find the oxidation state of N in NH4+.

Answers: 1. +6; +2 2. −3

22 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 15 Mixed problem (exam level)

Element E is in period 3. Its oxide reacts with both HCl and NaOH. (a) Identify E. (b) State the type of oxide. (c) Write the equation for the oxide reacting with HCl.

Solution

(a) aluminiumA1

(b) amphotericA1

Quick check

1. Predict the formula and nature of the oxide of selenium in its +6 state.

Answers: 1. SeO3; acidic

23 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 16 Paper 1 question (standard)

Which has the largest radius? A. Na+ B. Mg2+ C. F− D. O2−

Solution

Isoelectronic; fewest protons: DA1

WORKED EXAMPLE 17 Paper 2 question (exam level)

(a) Explain why first ionization energy decreases down group 1. (b) Predict and explain the reactivity of rubidium with water.

Solution

(a) outer electron further from nucleus, more shieldingR1 R1

(b) very vigorous, more than potassium: lower ionization energyR1

24 Periodicity in physical properties

Melting points across period 3.

Melting points rise from Na to Si (metallic bonding strengthens, then Si is giant covalent), then fall sharply to P4, S8, Cl2 and Ar (simple molecular with London forces). S8 melts higher than P4 because it has more electrons.

Figure 4. Melting points across period 3.

WORKED EXAMPLE 18 Silicon (standard)

Explain why silicon has the highest melting point in period 3.

Solution

Giant covalent structureR1

Many strong covalent bonds to breakR1

25 Common misconceptions and links

Students often think…	Actually…
atoms get bigger across a period	they get smaller
halogens get more reactive down the group	less reactive
all metal oxides are basic	Al2O3 is amphoteric
colour is the light absorbed	colour seen is complementary to the light absorbed

How S3.1 connects to the rest of the course

Topic	Connection
S1.3 Electron configurations	blocks and ionization energy
S2.2 Covalent model	electronegativity
R3.1 Proton transfer	acidic and basic oxides
R3.2 Electron transfer	oxidation states and redox

26 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Figure 5. Atomic radii across period 3.

(a) Explain the trend in atomic radius across period 3. [2]

(b) Write an equation for the reaction of sodium oxide with water and state the pH of the solution. [2]

(d) Deduce the oxidation state of N in HNO3. [1]

(e) Predict and explain the result of adding bromine water to potassium chloride solution. [2]

(f) Explain why [Cu(H2O)6]2+ is coloured. [3]

Markscheme

Part	Answer	Marks
(a)	nuclear charge increases; electrons in same shell	R1
	stronger attraction pulls electrons closer	R1
(b)	Na2O + H2O → 2NaOH	A1
	pH about 13 to 14 (alkaline)	A1
(c)	atoms larger; electron added further from nucleus, more shielded	R1
	weaker attraction for incoming electron	R1
(d)	+5	A1
(e)	no reaction	A1
	bromine less reactive than chlorine	R1
(f)	ligands split d orbitals	R1
	d electron absorbs visible light, promoted	R1
	complementary colour (blue) seen	R1

Examiner's comments. In (a), weaker answers did not mention that electrons are added to the same shell.

27 Check your understanding

Section A: multiple choice

1. Atomic radius across a period

A. decreases B. increases C. stays constant D. increases then decreases

2. The most electronegative element is

A. oxygen B. chlorine C. fluorine D. caesium

3. SO3 is

A. basic B. acidic C. amphoteric D. neutral

4. Group 1 reactivity

A. decreases down the group B. is constant C. depends on colour D. increases down the group

5. AgBr is

A. white B. a cream precipitate C. yellow D. soluble

6. Oxidation state of Cr in Cr2O72−

A. +6 B. +7 C. +3 D. +12

7. Which ion is colourless?

A. Cu2+ B. Fe3+ C. Ni2+ D. Zn2+

8. Variable oxidation states arise because

A. d orbitals are empty B. ligands are present C. 4s and 3d are close in energy D. atoms are small

Section B: structured

9. Explain the trend in first ionization energy down group 1. [3]

10. Describe the trend in acid-base character of period 3 oxides. [3]

11. Outline the reactions of halogens with halide ions. [3]

12. State four characteristic properties of transition elements. [4]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Decreases: outer electron further away, more shielding, weaker attraction.

10. Na, Mg oxides basic; Al2O3 amphoteric; Si, P, S oxides acidic.

11. More reactive halogen displaces less reactive halide; Cl2 + 2Br− → 2Cl− + Br2; colour changes.

12. Variable oxidation states, complex ions, coloured compounds, catalysis (or magnetism).

28 Summary and topic map

Idea	Remember
Structure	periods, groups, s/p/d/f blocks
Trends	radius down across; IE and electronegativity up across
Character	metallic decreases across, increases down
Oxides	basic to amphoteric to acidic
Group 1	react with water; more reactive down
Group 17	displacement; less reactive down
Oxidation states	rules; Roman numerals
HL	transition elements; complex ions; d-orbital splitting and colour

Before the exam, make sure you…

•use nuclear charge, distance and shielding
•write equations for oxides with water
•predict displacement reactions
•calculate oxidation states algebraically

The Periodic Table on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 6. Topic map for The Periodic Table.

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