OxfordAQA A-Level Chemistry: Group 2, the alkaline earth metals (9620)

This study guide teaches section 3.2.2.1 Group 2, the alkaline earth metals of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. The whole section is International AS content, so none of it is labelled International A-level only. The specification places 3.2.1 to 3.2.3 in Unit 1 (Inorganic 1 and Physical 1), and Unit 5 (Practical and synoptic) can draw on all content.

Condense what you learn here with the Group 2 revision notes, then test yourself with the Group 2 practice questions. The OxfordAQA A-Level Chemistry hub lists every topic, and the printable checklist lets you tick off each outcome. To find your weakest outcomes first, try a free 10-minute diagnostic.

Background: atomic structure (shells and shielding), bonding and structure (metallic bonding), periodicity and redox.

What this section covers

Spec 3.2.2.1	What you must be able to do	Stage
Physical trends	Explain why atomic radius rises and first ionisation energy falls from Mg to Ba	International AS
Physical trends	Account for the melting points using structure and bonding	International AS
Reactions	Describe the reactions of Mg–Ba with water	International AS
Titanium	Show how magnesium reduces TiCl₄ to titanium metal	International AS
Hydroxides	Relative solubilities Mg–Ba; Mg(OH)₂ is sparingly soluble	International AS
Uses	Medicine: Mg(OH)₂. Farming: Ca(OH)₂. Flue-gas cleaning: CaO or CaCO₃ capturing SO₂	International AS
Sulfates	Relative solubilities Mg–Ba; BaSO₄ is insoluble	International AS
Sulfate test	Use acidified BaCl₂; explain why BaCl₂ is used and why it is acidified	International AS
Medicine	The use of BaSO₄ in medicine	International AS
Required practical 3	Test-tube reactions to identify Group 2 cations (and anions including SO₄²⁻)	International AS

Required practical 3 is printed after the halogens section (3.2.3). Its Group 2 part is covered here; the rest is in the halogens study guide.

Why the group behaves as a family

Mg, Ca, Sr and Ba each have two outer electrons in an s sub-shell (3s² to 6s²), so they are s-block elements. They react by losing both and forming M²⁺ ions (oxidation state +2). Down the group, the outer electrons sit in a higher shell and are held less tightly; most of this topic follows from that.

Atomic radius

Atomic radius increases from Mg to Ba. Each step down adds a whole new occupied shell, so the outer electrons are further out. Nuclear charge rises too, but the new inner shell shields the outer electrons from most of that extra charge.

First ionisation energy

First ionisation energy measures how much energy it takes to strip one electron from every atom in one mole of gaseous atoms, leaving one mole of 1+ ions. For strontium:

Sr(g) → Sr⁺(g) + e⁻

First ionisation energy decreases from Mg to Ba. Three points build the explanation:

• the outer electron is in a higher shell, so it is further from the nucleus;
• more inner shells sit between it and the nucleus, so shielding is greater;
• together these weaken the attraction to the nucleus by more than the larger nuclear charge strengthens it, so less energy is needed.

Worked example 1: calcium against magnesium

Explain why calcium has a lower first ionisation energy than magnesium.

1. Calcium’s outer electron is in the 4s sub-shell; magnesium’s is in 3s. Calcium’s is further from its nucleus.
2. Calcium has three full inner shells; magnesium has two. Calcium’s outer electron is more shielded.
3. Calcium has more protons, but distance and shielding outweigh this. Its outer electron is attracted less strongly, so less energy removes it.

Melting point

Each element is a giant lattice of M²⁺ ions held together by metallic bonding: the electrostatic attraction between the positive ions and the sea of delocalised electrons.

From calcium to barium the melting point falls. Every ion still carries a 2+ charge, but the ions grow larger, so the delocalised electrons are further from the centre of each ion. The attraction is weaker and less energy breaks up the lattice.

Magnesium breaks the pattern: it melts at a lower temperature than calcium. The usual explanation is that magnesium packs its atoms in a different crystal arrangement. Treat magnesium as the exception and apply the model to Ca, Sr and Ba.

Worked example 2: strontium against barium

Explain why strontium has a higher melting point than barium.

• Both have giant metallic lattices.
• Both form 2+ ions and contribute two delocalised electrons per atom.
• The Sr²⁺ ion is smaller than Ba²⁺, so the delocalised electrons are closer to the ion’s nucleus.
• The metallic bonding in strontium is stronger, so more energy is needed to overcome it.

Reactions with water

Each metal is oxidised from 0 to +2, and hydrogen in water is reduced from +1 to 0. The reactions get more vigorous down the group, because the two outer electrons are lost more easily as ionisation energies fall.

Metal	What you see	Product with water
Mg	almost no reaction with cold water; burns with a bright white flame in steam	Mg(OH)₂ (cold, very slow); MgO (steam)
Ca	steady fizzing; the metal dissolves; the liquid turns cloudy white	Ca(OH)₂
Sr	faster fizzing than calcium	Sr(OH)₂
Ba	the most vigorous of the four	Ba(OH)₂

With liquid water the general equation is:

M(s) + 2H₂O(l) → M(OH)₂(aq) + H₂(g)

The cloudiness with calcium is undissolved Ca(OH)₂. With steam, magnesium forms the oxide instead:

Mg(s) + H₂O(g) → MgO(s) + H₂(g)

The reaction of magnesium with water appears again at International A2 in Period 3 elements and their oxides.

Worked example 3: the solution left after strontium reacts

1.314 g of strontium reacts completely with water and the solution is made up to 500 cm³. Calculate the concentration of hydroxide ions. (Aᵣ Sr = 87.6)

n(Sr)       = 1.314 ÷ 87.6          = 0.0150 mol
n(Sr(OH)₂)  = n(Sr)                 = 0.0150 mol
[Sr(OH)₂]   = 0.0150 ÷ 0.500        = 0.0300 mol dm⁻³
[OH⁻]       = 2 × 0.0300            = 0.0600 mol dm⁻³

Each Sr(OH)₂ releases two OH⁻ ions, hence the doubling.

Extracting titanium with magnesium

Titanium is obtained from titanium(IV) chloride by heating it with magnesium:

TiCl₄ + 2Mg → Ti + 2MgCl₂

Titanium goes from +4 to 0, so it is reduced. Magnesium goes from 0 to +2, so it is oxidised and acts as the reducing agent.

Worked example 4: percentage yield

A batch of 250 kg of TiCl₄ gives 58.5 kg of titanium. Calculate the percentage yield. (Aᵣ Ti = 47.9, Cl = 35.5)

Mr of TiCl₄       = 47.9 + 142.0          = 189.9
amount of TiCl₄   = 250 × 10³ ÷ 189.9     = 1316 mol
theoretical Ti    = 1316 × 47.9           = 63.1 × 10³ g = 63.1 kg
percentage yield  = 58.5 ÷ 63.06 × 100    = 92.8%

The ratio of TiCl₄ to Ti is 1 : 1, so the moles carry straight across.

Hydroxides: solubility and uses

Going from Mg to Ba, each hydroxide is more soluble than the one above it:

Mg(OH)₂ < Ca(OH)₂ < Sr(OH)₂ < Ba(OH)₂
sparingly soluble  →  most soluble

A saturated solution of barium hydroxide therefore holds far more OH⁻ ions than a saturated solution of magnesium hydroxide, and is more strongly alkaline.

Mg(OH)₂ in medicine. A suspension of magnesium hydroxide (milk of magnesia) is taken as an antacid. It neutralises excess hydrochloric acid in the stomach:

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

Being sparingly soluble, it releases little free hydroxide at once, so the stomach never becomes strongly alkaline.

Ca(OH)₂ in agriculture. Farmers spread calcium hydroxide on acidic soil to raise its pH, which helps crops grow:

Ca(OH)₂ + 2H⁺ → Ca²⁺ + 2H₂O

Worked example 5: one dose of antacid

A 10.0 cm³ dose of milk of magnesia contains 0.800 g of Mg(OH)₂. What volume of 0.150 mol dm⁻³ HCl can it neutralise? (Mr Mg(OH)₂ = 58.3)

n(Mg(OH)₂) = 0.800 ÷ 58.3          = 0.01372 mol
n(HCl)     = 2 × 0.01372           = 0.02744 mol
volume     = 0.02744 ÷ 0.150       = 0.183 dm³ = 183 cm³

Removing SO₂ from flue gases

Power stations that burn sulfur-containing fuel release sulfur dioxide, an acidic gas. Passing the flue gases through calcium oxide or calcium carbonate removes it, because both are bases:

CaCO₃ + SO₂ → CaSO₃ + CO₂      (limestone)
CaO + SO₂ → CaSO₃              (lime)

CaCO₃ releases one mole of CO₂ per mole of SO₂ captured; CaO does not.

Worked example 6: impure limestone

A plant removes 0.800 tonnes of SO₂ a day with powdered limestone that is 90.0% CaCO₃ by mass. What mass of limestone is used each day? (Mr SO₂ = 64.1, CaCO₃ = 100.1)

n(SO₂)          = 0.800 × 10⁶ ÷ 64.1     = 1.248 × 10⁴ mol
mass CaCO₃      = 1.248 × 10⁴ × 100.1    = 1.249 × 10⁶ g
mass limestone  = 1.249 × 10⁶ ÷ 0.900    = 1.39 × 10⁶ g = 1.39 tonnes

Divide by the purity fraction, because the impure rock must contain the required CaCO₃.

Sulfates: solubility, the sulfate test and BaSO₄ in medicine

Solubility of the sulfates decreases down the group: MgSO₄ dissolves readily, CaSO₄ and SrSO₄ much less, and the specification describes BaSO₄ as insoluble. This runs the opposite way to the hydroxides.

The test for sulfate ions. Add dilute hydrochloric acid to the sample, then a few drops of barium chloride solution. A white precipitate shows sulfate ions:

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

• Why barium chloride? BaSO₄ does not dissolve, so added barium ions pull sulfate out of solution as a visible solid even when little sulfate is present.
• Why acidified? Carbonate ions would also give a white precipitate (BaCO₃) with barium ions. Added acid turns carbonate into carbon dioxide gas, 2H⁺(aq) + CO₃²⁻(aq) → CO₂(g) + H₂O(l), so it cannot interfere. Sulfuric acid must never be used, since it supplies sulfate ions itself.

Worked example 7: a misleading result

A student adds barium chloride, without acid, to a washing soda (sodium carbonate) solution and sees a white precipitate. Does this prove sulfate is present?

No. The solution contains carbonate ions, so the solid could be BaCO₃, BaSO₄ or both. After adding dilute HCl the carbonate is gone; any precipitate that still forms must be BaSO₄.

BaSO₄ in medicine. Patients swallow a “barium meal”, a suspension of barium sulfate. It absorbs X-rays, so the gut lining shows clearly on an X-ray image. Dissolved barium ions are toxic, but BaSO₄ is insoluble, so they do not pass into the blood. Magnesium sulfate, by contrast, is soluble and is also used in medicine (as Epsom salts).

Required practical 3: telling Group 2 ions apart

Add sodium hydroxide solution to one portion of the unknown and dilute sulfuric acid to a fresh portion. The two solubility trends predict the results.

Ion	With NaOH(aq)	With dilute H₂SO₄
Mg²⁺	white precipitate	none
Ca²⁺	slight white precipitate	slight white precipitate
Sr²⁺	very little	white precipitate
Ba²⁺	none	dense white precipitate

Use solutions of similar concentration, or the amounts of precipitate cannot be compared. Limewater, the test reagent for carbonate in the same practical, is itself saturated Ca(OH)₂ solution: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O.

Common errors

• Writing “nuclear charge decreases down the group”. It increases; distance and shielding win.
• Describing metallic bonding as attraction “between atoms” rather than between ions and delocalised electrons.
• Giving Mg(OH)₂ as the product with steam (it is MgO).
• Forgetting to double [Sr(OH)₂] or [Ba(OH)₂] to get [OH⁻].
• Saying the sulfate-test acid “helps the precipitate form”. It removes carbonate.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.2.2.1 Group 2, the alkaline earth metals, with the Group 2 part of Required practical 3.