OxfordAQA International GCSE Chemistry: Group properties (9202)

This study guide teaches group properties for OxfordAQA International GCSE Chemistry (9202). It follows the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, and covers every statement in section 3.7.1 (statements a to f). The qualification is untiered, so all of it applies to every candidate. The specification allows content from any part of the course on Paper 1 or Paper 2, so expect this unit on either paper.

Quick links: course hub · printable checklist · revision notes for this unit · practice questions for this unit · free diagnostics

What this unit covers

Statement	What you must be able to do
3.7.1 a	Describe the alkali metals: low density, +1 ions in white ionic compounds that give colourless solutions, hydrogen released with water, alkaline hydroxide solutions
3.7.1 b	State that Group 1 reactivity increases down the group
3.7.1 c	Describe halogens reacting with metals to give ionic compounds containing 1− halide ions
3.7.1 d	State that, down Group 7, reactivity falls while melting and boiling points rise
3.7.1 e	Explain and predict halogen displacement reactions in aqueous solution
3.7.1 f	Explain both group trends using the energy level of the outer electrons

Section 6.2 of the specification attaches no required practical to 3.7.1. You still need practical knowledge, such as what you would see when an alkali metal meets water or when two halogen solutions are mixed with salts.

This unit builds on electron arrangements. If “one outer electron” or “seven outer electrons” is not automatic for you yet, revisit the atomic structure and periodic table guide and the ionic, covalent and metallic bonding guide first.

3.7.1 a: what the alkali metals have in common

Group 1 runs lithium, sodium, potassium, rubidium, caesium, francium. Every atom has a single electron in its highest occupied energy level. That shared structure produces shared chemistry.

Low density. For metals, these are unusually light. Lithium, sodium and potassium are all less dense than water, so a piece placed on water floats while it reacts. Rubidium and caesium are denser than water.

Ionic compounds with +1 ions. When an alkali metal reacts with a non-metal, each metal atom gives away its one outer electron and becomes an M⁺ ion. The products are ionic. Typical examples:

2Na + Cl₂ → 2NaCl        (Na⁺ and Cl⁻)
4K + O₂  → 2K₂O          (K⁺ and O²⁻)
2Li + S  → Li₂S          (Li⁺ and S²⁻)

Because the metal ion is always 1+, the formula only depends on the charge of the non-metal ion: one M⁺ per Cl⁻, two M⁺ per O²⁻ or S²⁻. The compounds are white solids that dissolve in water to give colourless solutions. If an exam question shows a coloured solution, the colour is not coming from the Group 1 ion.

Reaction with water. Every alkali metal reacts with water to release hydrogen and leave a metal hydroxide in solution:

metal + water → metal hydroxide + hydrogen
2M + 2H₂O → 2MOH + H₂        (M = any Group 1 metal)

The gas can be identified because a lighted splint held near it gives a pop.

Alkaline solutions. The hydroxides (LiOH, NaOH, KOH and so on) dissolve in water. The solution contains OH⁻ ions, so it is alkaline: pH above 7. Universal indicator turns blue or purple. This is why the group is named “alkali metals”.

Worked example 1: a Group 1 salt from first principles

1.40 g of lithium reacts completely with bromine. Write the equation and calculate the mass of lithium bromide made. (Ar: Li = 7, Br = 80)

Step 1  Ions: Li⁺ and Br⁻, so the formula is LiBr.
Step 2  Bromine is diatomic: 2Li + Br₂ → 2LiBr
Step 3  Moles of Li = 1.40 ÷ 7 = 0.20 mol
Step 4  Ratio Li : LiBr = 2 : 2 = 1 : 1, so 0.20 mol LiBr
Step 5  Mr of LiBr = 7 + 80 = 87
Step 6  Mass = 0.20 × 87 = 17.4 g

Answer: 17.4 g of lithium bromide, a white solid that would dissolve to give a colourless solution.

Worked example 2: hydrogen and alkali from one reaction

0.92 g of sodium reacts with water. The solution is then made up to 200 cm³. Find the hydrogen volume at RTP and the sodium hydroxide concentration. (Ar: Na = 23; molar gas volume 24 dm³ at RTP)

Step 1  2Na + 2H₂O → 2NaOH + H₂
Step 2  Moles of Na = 0.92 ÷ 23 = 0.040 mol
Step 3  Moles of H₂ = 0.040 ÷ 2 = 0.020 mol
Step 4  Volume of H₂ = 0.020 × 24 = 0.48 dm³ (480 cm³)
Step 5  Moles of NaOH = 0.040 mol (1 : 1 with Na)
Step 6  200 cm³ = 0.200 dm³
Step 7  Concentration = 0.040 ÷ 0.200 = 0.20 mol/dm³

Answers: 480 cm³ of hydrogen; 0.20 mol/dm³ sodium hydroxide. The quantitative chemistry guide covers the mole and gas-volume methods used here.

3.7.1 b: reactivity increases down Group 1

The further down Group 1 an element is, the more reactive it is. You can see it in the water reaction:

Metal	What you see on cold water
Lithium	Floats; steady stream of bubbles; shrinks gradually
Sodium	Floats; turns into a shiny molten sphere; skims rapidly across the surface; bubbles vigorously
Potassium	Floats; reacts so fast that the hydrogen ignites with a lilac flame

Each metal makes the same type of products. Only the speed and violence change. So for a metal lower in the group, predict the same products (hydroxide and hydrogen) but a faster, more violent reaction. Rubidium and caesium are not normally reacted with water in school laboratories because the reactions are too violent.

3.7.1 c: halogens react with metals to form halides

Group 7 contains fluorine, chlorine, bromine, iodine and astatine. Each atom has seven outer electrons. When a halogen reacts with a metal, each halogen atom gains one electron to become a halide ion with a 1− charge: F⁻, Cl⁻, Br⁻, I⁻. The metal forms a positive ion, so the product is ionic.

To write a formula, balance charges so the total is zero.

Metal ion	Halide ion	Formula	Name
Ca²⁺	I⁻	CaI₂	calcium iodide
Al³⁺	Cl⁻	AlCl₃	aluminium chloride
Fe³⁺	Br⁻	FeBr₃	iron(III) bromide
K⁺	F⁻	KF	potassium fluoride

Then balance the equation, remembering that the halogen element is a diatomic molecule:

2Al + 3Cl₂ → 2AlCl₃
Ca + I₂ → CaI₂

Note the naming change: the element is chlorine, but the ion in a compound is chloride.

3.7.1 d: trends down Group 7

The further down Group 7 an element is:

• the less reactive it is;
• the higher its melting point and boiling point.

At room temperature the states change down the group:

Halogen	Appearance at room temperature
Fluorine	Pale yellow gas
Chlorine	Pale yellow-green gas
Bromine	Red-brown liquid that gives off orange-brown vapour
Iodine	Grey-black solid that forms a purple vapour on warming

Gas, then liquid, then solid is the boiling-point trend made visible. The halogens exist as small diatomic molecules. Larger molecules further down the group have stronger forces between the molecules, so more energy is needed to separate them. This links to the bonding, structure and properties guide.

3.7.1 e: halogen displacement reactions

If a halogen is more reactive than the halogen present as halide ions in a dissolved salt, it takes that halide’s place. The reactivity order is:

fluorine > chlorine > bromine > iodine

So chlorine water added to a bromide solution gives bromine, and the colourless solution turns yellow-orange. Chlorine or bromine added to an iodide solution gives iodine, and the solution turns brown. Iodine added to a chloride or bromide solution causes no reaction, because iodine is less reactive than both.

Use this grid to predict (✓ = reaction, ✗ = no reaction):

Halogen added	chloride solution	bromide solution	iodide solution
Chlorine	–	✓	✓
Bromine	✗	–	✓
Iodine	✗	✗	–

Fluorine is not used in aqueous tests because it reacts with water itself, so displacement questions normally use chlorine, bromine and iodine.

Worked example 3: predicting and calculating a displacement

Bromine water is added to calcium iodide solution containing 2.94 g of CaI₂. Explain whether a reaction happens, write the equation, and calculate the maximum mass of iodine produced. (Ar: Ca = 40, I = 127)

Step 1  Bromine is above iodine in Group 7, so it is more
        reactive and displaces iodine.
Step 2  Br₂ + CaI₂ → CaBr₂ + I₂
Step 3  Mr of CaI₂ = 40 + (2 × 127) = 294
Step 4  Moles of CaI₂ = 2.94 ÷ 294 = 0.010 mol
Step 5  Ratio CaI₂ : I₂ = 1 : 1, so 0.010 mol I₂
Step 6  Mr of I₂ = 254; mass = 0.010 × 254 = 2.54 g

Answer: yes, it reacts; maximum mass of iodine = 2.54 g. You would see the solution turn brown. The ionic equation removes the calcium ions, which do not change: Br₂ + 2I⁻ → 2Br⁻ + I₂.

3.7.1 f: explaining the trends with energy levels

The specification explains both trends with one idea: the higher the energy level of the outer electrons, the more easily electrons are lost and the less easily electrons are gained. You may answer in terms of energy levels or shells.

Going down any group, each element has one more occupied energy level. The outer electrons are therefore further from the nucleus and more shielded by inner electrons. The attraction between the nucleus and the outer electrons is weaker.

Group 1 (metals lose an electron). A weaker attraction means the single outer electron is lost more easily. Reacting means losing that electron, so reactivity increases down the group.

Group 7 (non-metals gain an electron). The incoming electron would go into the outer energy level. Further down, that level is further from the nucleus and more shielded, so the attraction for an extra electron is weaker. The electron is gained less easily, so reactivity decreases down the group.

Worked example 4: writing the explanation

Explain why rubidium reacts faster than lithium, whereas iodine reacts more slowly than chlorine.

A full answer links four points in order:

1. Rubidium has five occupied energy levels and lithium has two, so rubidium’s single outer electron sits in a much higher energy level.
2. That electron lies further out, with more inner electrons screening it, so the nucleus holds it less tightly.
3. Rubidium therefore gives up its outer electron more readily and reacts faster.
4. Iodine’s outer energy level is higher than chlorine’s, so an extra electron arriving there feels a weaker pull from the nucleus. Iodine takes on an electron less readily, so it reacts more slowly.

The same structural change (an extra energy level) causes opposite trends, because metals react by losing electrons and halogens react by gaining them.

Common errors

• Writing Cl or Br as single atoms in equations. The elements are Cl₂, Br₂ and I₂.
• Saying the alkali metal “dissolves” in water. It reacts; the product dissolves.
• Calling the gas from Group 1 and water “oxygen”. It is hydrogen.
• Mixing up chlorine (element) and chloride (ion in a compound).
• Saying iodine displaces chlorine. A less reactive halogen never displaces a more reactive one.
• Explaining Group 7 reactivity as “harder to lose electrons”. Halogens react by gaining electrons.
• Using “bigger atom” without the reason. Link to distance from the nucleus, shielding and weaker attraction.
• Stating that Group 1 compounds are coloured. They are white and give colourless solutions.

Where next

Condense what you have learned with the group properties revision notes, then test yourself with the group properties practice questions. For other units, use the OxfordAQA IGCSE Chemistry hub, the printable checklist and the free 10-minute diagnostics.

Official syllabus

OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, published by OxfordAQA. This guide covers section 3.7.1 Group properties (statements a to f).