Pearson Edexcel International GCSE Chemistry 4CH1: Group 1 (alkali metals) and Group 7 (halogens) – Study Guide

This guide teaches sub-topics (a) Group 1 (alkali metals) and (b) Group 7 (halogens) of Topic 2, Inorganic chemistry, in the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024). It covers points 2.1–2.8 in depth. The qualification is untiered. Points 2.4C and 2.8C carry a C reference (bold in the specification), so they are assessed on Paper 2 only, and this guide labels them Paper 2 only.

Links: 4CH1 course hub, printable checklist, Group 1 and Group 7 revision notes, the wider Reactivity and group trends guide, practice on Groups 1 and 7, acids, salts and tests, Reactivity practice and the free 4CH1 diagnostic.

What these sub-topics cover

Spec point	What you must be able to do	Paper
2.1	Use the similar reactions of lithium, sodium and potassium with water as evidence that they are one family	1 and 2
2.2	Use the differences in their reactions with air and water as evidence for the reactivity trend	1 and 2
2.3	Predict the properties of other alkali metals from the trends	1 and 2
2.4C	Explain the reactivity trend in Group 1 using electronic configurations	Paper 2 only
2.5	Know the colours, states at room temperature and physical trends of chlorine, bromine and iodine	1 and 2
2.6	Predict the properties of other halogens from the trends	1 and 2
2.7	Use halogen–halide displacement reactions as evidence for the reactivity trend	1 and 2
2.8C	Explain the reactivity trend in Group 7 using electronic configurations	Paper 2 only

The specification names lithium, sodium and potassium, and chlorine, bromine and iodine. Rubidium, caesium, fluorine and astatine appear only as predictions.

Group 1: why the alkali metals are a family (2.1)

A family of elements reacts in the same way and forms the same type of product. Lithium, sodium and potassium in water all show the same pattern:

• the metal floats (it is less dense than water)
• it fizzes, because hydrogen gas is given off
• it moves around on the surface
• it gets smaller and disappears as it reacts
• the water becomes alkaline: universal indicator turns blue or purple

Every one of them makes a metal hydroxide and hydrogen. The general word equation is:

alkali metal + water → metal hydroxide + hydrogen

The hydroxide dissolves, giving an alkaline solution. That is where the name “alkali metals” comes from. Each metal forms a 1+ ion (Li⁺, Na⁺, K⁺), so every hydroxide has the formula MOH and every equation has the same shape.

Worked example 1: one equation, three metals

Write balanced equations, with state symbols, for lithium and for potassium reacting with water, and say what the equations show about the group.

Step 1  Formulae: lithium hydroxide LiOH (Li⁺ and OH⁻), hydrogen H₂.
Step 2  Unbalanced: Li + H₂O → LiOH + H₂
Step 3  H₂ needs 2 H atoms from the water side, so double the
        water and the hydroxide: Li + 2H₂O → 2LiOH + H₂
Step 4  Now 2 Li on the right, so: 2Li + 2H₂O → 2LiOH + H₂
Step 5  State symbols:
        2Li(s) + 2H₂O(l) → 2LiOH(aq) + H₂(g)
Step 6  Potassium, same shape:
        2K(s)  + 2H₂O(l) → 2KOH(aq)  + H₂(g)

Only the symbol of the metal changes. Identical products and identical equations are the evidence the specification wants for 2.1.

Group 1: evidence for the reactivity trend (2.2)

The similarities show a family. The differences show a trend: reactivity increases down Group 1.

With water

Metal	What you see
Lithium	Fizzes steadily; moves slowly; stays solid
Sodium	Fizzes rapidly; melts into a silvery ball; moves quickly
Potassium	Reacts violently; melts; the hydrogen ignites with a lilac flame; may spark

Faster fizzing means faster hydrogen production, so a faster reaction.

With air

All three are stored under oil because they react with oxygen and water vapour in air. A freshly cut surface is shiny. It then tarnishes as a layer of oxide forms. Lithium tarnishes slowly, sodium faster, and potassium almost at once. When heated, all three burn to form oxides, for example:

4Li(s) + O₂(g) → 2Li₂O(s)
4Na(s) + O₂(g) → 2Na₂O(s)

Faster tarnishing down the group is the second piece of evidence for the trend.

Worked example 2: ranking from observations

Three unlabelled alkali metals, P, Q and R, are freshly cut. P stays shiny for several minutes. Q goes dull within seconds. R dulls within a minute. In water, Q produces a lilac flame. Place them in order of reactivity and identify each.

Step 1  Fastest tarnishing = most reactive: Q > R > P.
Step 2  Lilac flame in water is potassium, so Q = potassium.
Step 3  Down the group: Li < Na < K, so the slowest (P) is lithium
        and R is sodium.
Answer  Q (potassium) > R (sodium) > P (lithium).

Two lines of evidence (air and water) agree.

Group 1: predicting other alkali metals (2.3)

Rubidium and caesium sit below potassium. Predict them by continuing each trend.

Property	Li	Na	K	Rb	Cs
Melting point (°C)	181	98	63	39	28
Density (g/cm³)	0.53	0.97	0.89	1.53	1.93

Predictions for rubidium and caesium:

• more reactive than potassium, reacting explosively with water
• form RbOH and CsOH plus hydrogen, giving alkaline solutions
• form 1+ ions, so chlorides RbCl and CsCl and oxides Rb₂O and Cs₂O
• softer, with lower melting points

Worked example 3: predicting a melting point and a density effect

Using the melting points of lithium to rubidium, predict the melting point of caesium. Then predict whether caesium floats on water (density 1.00 g/cm³).

Step 1  Trend: melting point falls down the group.
Step 2  Size of each fall: 181 → 98 is 83; 98 → 63 is 35;
        63 → 39 is 24. The falls are getting smaller.
Step 3  So the next fall should be smaller than 24 °C:
        caesium about 25–35 °C. (Actual value: 28 °C.)
Step 4  Density: Li, Na and K are below 1.00 g/cm³ and float.
        Density rises overall down the group and rubidium is
        1.53 g/cm³, so caesium should be denser than water.
Answer  About 30 °C; caesium sinks.

State the direction of a trend before using it, and expect physical trends to be general rather than perfect (potassium is less dense than sodium).

Group 1: explaining the trend (2.4C, Paper 2 only)

An alkali metal reacts by losing its single outer electron to form a 1+ ion:

Na → Na⁺ + e⁻

The easier that electron is to lose, the more reactive the metal. Use the configurations from point 1.19:

Metal	Electronic configuration	Occupied shells
Lithium	2.1	2
Sodium	2.8.1	3
Potassium	2.8.8.1	4

Worked example 4: the full explanation chain

Explain why sodium is more reactive than lithium. (4 marks)

1  Sodium (2.8.1) has more occupied shells than lithium (2.1),
   so its outer electron is further from the nucleus.
2  There are more inner shells between the nucleus and that
   electron, so there is more shielding.
3  The attraction between the nucleus and the outer electron is
   weaker in sodium (even though sodium has more protons).
4  So sodium loses its outer electron more easily, and reacts
   more readily.

Each line is a separate scoring point. Stopping at “more shells” gives only the first.

Group 7: physical properties of the halogens (2.5)

The halogens exist as diatomic molecules: Cl₂, Br₂, I₂.

Halogen	Colour and state at room temperature	Melting point (°C)	Boiling point (°C)
Chlorine	Pale green (yellow-green) gas	−101.5	−34
Bromine	Red-brown liquid (gives off an orange-brown vapour)	−7	59
Iodine	Grey-black solid (gives a purple vapour when warmed)	114	184

Down the group:

• the colour gets darker
• the state changes gas → liquid → solid
• melting and boiling points increase

The halogens are simple molecular substances. Down the group, relative molecular mass rises, intermolecular forces get stronger and more energy is needed to separate the molecules (point 1.48).

Group 7: predicting other halogens (2.6)

Fluorine sits above chlorine and astatine below iodine.

Worked example 5: predicting fluorine

Use the data for chlorine, bromine and iodine to predict the state, colour and approximate boiling point of fluorine.

Step 1  Trend: boiling point rises down the group.
        Cl → Br rises 93 °C; Br → I rises 125 °C.
Step 2  Going UP from chlorine, the boiling point should fall
        by a similar large amount: well below −34 °C, roughly
        −120 °C or lower. (Actual value: −188 °C.)
Step 3  Boiling point far below room temperature → a gas.
Step 4  Colour gets darker down the group, so fluorine should be
        paler than chlorine: a pale yellow gas.
Step 5  Reactivity decreases down the group, so fluorine is the
        most reactive halogen.

The same method applied downwards predicts that astatine is a black solid, less reactive than iodine.

Group 7: displacement reactions as evidence (2.7)

A halide is a compound containing a halide ion: chloride Cl⁻, bromide Br⁻ or iodide I⁻. A more reactive halogen takes the electron from a less reactive halide ion, so it displaces the less reactive halogen from solution.

In solution, chlorine is very pale green, bromine is orange (yellow when dilute) and iodine is brown. The colour that appears tells you which halogen has been made.

Halogen added	Potassium chloride	Potassium bromide	Potassium iodide
Chlorine water	—	Turns orange (Br₂ formed)	Turns brown (I₂ formed)
Bromine water	No reaction	—	Turns brown (I₂ formed)
Iodine solution	No reaction	No reaction	—

Chlorine displaces both, bromine displaces only iodine, iodine displaces neither. So the order of reactivity is chlorine > bromine > iodine: reactivity decreases down Group 7.

Worked example 6: equations for chlorine and iodide

Chlorine water is added to potassium iodide solution. Give the observation, the balanced equation, and the ionic equation. Identify what is oxidised.

Observation  colourless solution turns brown
Equation     Cl₂(aq) + 2KI(aq) → 2KCl(aq) + I₂(aq)
Ionic        K⁺ appears unchanged on both sides (spectator ion):
             Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq)
Electrons    Cl₂ + 2e⁻ → 2Cl⁻   (chlorine gains electrons: reduced)
             2I⁻ → I₂ + 2e⁻     (iodide loses electrons: oxidised)

Chlorine is the oxidising agent. These redox terms come from point 2.20 in sub-topic (d).

Worked example 7: predicting a displacement

Predict whether iodine reacts with sodium astatide solution, NaAt.

Step 1  Astatine is below iodine, so iodine is more reactive.
Step 2  A more reactive halogen displaces a less reactive one.
Answer  Yes: I₂ + 2NaAt → 2NaI + At₂

State which halogen is more reactive before answering yes or no.

Group 7: explaining the trend (2.8C, Paper 2 only)

A halogen atom has seven outer electrons (chlorine is 2.8.7). It reacts by gaining one electron to complete its outer shell and form a 1− ion:

Cl₂ + 2e⁻ → 2Cl⁻

The more strongly an atom attracts an extra electron, the more reactive it is. Each halogen has one more occupied shell than the one above. So, down the group:

1. the outer shell is further from the nucleus
2. there is more shielding by inner shells
3. the attraction for an incoming electron is weaker
4. an electron is gained less easily, so reactivity decreases

This mirrors Group 1: distance and shielding make metals that lose electrons more reactive down the group, and halogens that gain electrons less reactive.

Common errors

• Saying a halogen “loses electrons” when it reacts. Halogens gain; halide ions lose.
• Writing “chlorine displaces chloride”. A halogen displaces a different, less reactive halogen.
• Describing solid iodine as purple. The solid is grey-black; the vapour is purple.
• Writing Cl for chlorine in equations. Halogens are diatomic: Cl₂.
• Writing the hydroxide as M(OH)₂. Group 1 ions are 1+, so it is MOH.

Where to go next

Condense this with the revision notes, then try the Groups 1 and 7 practice questions.

Official syllabus

Pearson Edexcel International GCSE in Chemistry (4CH1) specification, Issue 3, September 2024, published by Pearson Education Limited – Topic 2 Inorganic chemistry, sub-topics (a) Group 1 (alkali metals) and (b) Group 7 (halogens), points 2.1–2.8C.