OCR A Level Chemistry A: Group 2 (H432)

This study guide teaches section 3.1.2, Group 2, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. In the specification, Group 2 is the second section of Module 3, Periodic table and energy, with five lettered learning outcomes, 3.1.2(a) to (e), each taught below with worked examples. All of it is full A level content. Module 3 is examined in Periodic table, elements and physical chemistry (component 01) and in Unified chemistry (component 03), according to the specification. No PAG is listed against this section; PAG 4 uses a mixture of Group 2 salts, but the specification links it to 3.1.3(g), 3.1.4(a) and 5.3.2(a).

Find the other H432 topics through the OCR chemistry hub. Mark each outcome off as it becomes secure using the printable checklist. Where you are unsure, the free diagnostics help locate gaps.

Outcomes in 3.1.2

Outcome	What you must be able to do	Additional guidance in the specification
3.1.2(a)	Link the outer s² configuration to the loss of two electrons in redox reactions, forming 2+ ions	None
3.1.2(b)	Describe the relative reactivities of Mg to Ba through their redox reactions with (i) oxygen, (ii) water, (iii) dilute acids	Reactions with acids limited to those producing a salt and hydrogen
3.1.2(c)	Explain the trend in reactivity using first and second ionisation energies down the group	M3.1; a definition of second ionisation energy is not required, but you should be able to write an equation for the change (see also 3.1.1(c))
3.1.2(d)	Describe the action of water on Group 2 oxides, the approximate pH of the solutions and the trend of increasing alkalinity	None
3.1.2(e)	Describe uses of Group 2 compounds as bases, with equations	Examples (not limited to): Ca(OH)₂ to neutralise acid soils; Mg(OH)₂ and CaCO₃ as antacids

This section builds on oxidation numbers from 2.1.5 and ionisation energy from 3.1.1 (Periodicity). If oxidation numbers are shaky, revisit the Redox study guide first.

3.1.2(a): Two outer electrons, lost as a pair

Every Group 2 atom has two electrons in an outer s sub-shell.

Element	Electron configuration	Ion formed
Mg	[Ne] 3s²	Mg²⁺, [Ne]
Ca	[Ar] 4s²	Ca²⁺, [Ar]
Sr	[Kr] 5s²	Sr²⁺, [Kr]
Ba	[Xe] 6s²	Ba²⁺, [Xe]

In its reactions a Group 2 metal loses both s electrons. The ion left behind has the configuration of the noble gas before it. In oxidation-number terms, the metal goes from 0 in the element to +2 in the compound, so the metal is always oxidised. Something else in the reaction must gain those two electrons and be reduced: oxygen in a combustion, hydrogen in water or in an acid.

Worked example 1: redox in strontium burning

Strontium burns in oxygen: 2Sr(s) + O₂(g) → 2SrO(s). Show that this is a redox reaction and count the electrons transferred per mole of O₂.

Sr:  0 in Sr(s)   → +2 in SrO     each Sr atom loses 2 electrons: oxidised
O:   0 in O₂(g)   → −2 in SrO     each O atom gains 2 electrons: reduced
per mole of O₂: 2 mol Sr each lose 2 mol e⁻ = 4 mol e⁻ transferred
check: 2 mol O atoms each gain 2 mol e⁻ = 4 mol e⁻  ✓

Electrons lost must equal electrons gained.

3.1.2(b): Reactivity from magnesium to barium

Reactivity increases down the group. You show this by comparing three types of redox reaction. In each, the metal is oxidised from 0 to +2.

(i) With oxygen

All four metals burn in oxygen to give white solid oxides, MO:

2Mg(s) + O₂(g) → 2MgO(s)
2Ba(s) + O₂(g) → 2BaO(s)

Magnesium needs heating before it ignites, then burns with a very bright white light. The metals lower in the group react with oxygen more readily; barium tarnishes quickly in air, which is why it is stored under oil.

(ii) With water

• Magnesium reacts only very slowly with cold water, forming magnesium hydroxide and a few bubbles of hydrogen over a long time: Mg(s) + 2H₂O(l) → Mg(OH)₂(s) + H₂(g)
• Heated magnesium reacts quickly with steam, glowing white and forming the oxide: Mg(s) + H₂O(g) → MgO(s) + H₂(g)
• Calcium, strontium and barium react with cold water, forming the hydroxide and hydrogen, each faster than the one before: Sr(s) + 2H₂O(l) → Sr(OH)₂(aq) + H₂(g)

With calcium you see steady fizzing, the metal disappearing, the water warming and a cloudy white suspension forming, because calcium hydroxide is only slightly soluble. With barium the fizzing is much more vigorous.

In every case hydrogen goes from +1 in water to 0 in H₂, so hydrogen is reduced.

(iii) With dilute acids

Each metal reacts with a dilute acid to give a salt and hydrogen. The specification limits you to reactions of this type.

Ca(s) + 2HCl(aq) → CaCl₂(aq) + H₂(g)
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

You see fizzing as the metal dissolves. With the same acid at the same concentration, fizzing gets faster from Mg to Ba.

Worked example 2: calcium in water

0.802 g of calcium reacts completely with water. Write the equation with state symbols and calculate the volume of hydrogen, in cm³, at RTP. Use Ar(Ca) = 40.1 and 24.0 dm³ mol⁻¹ for the molar gas volume at RTP.

Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g)
n(Ca)   = 0.802 ÷ 40.1 = 0.0200 mol
n(H₂)   = n(Ca)        = 0.0200 mol      (1 : 1)
V(H₂)   = 0.0200 × 24.0 = 0.480 dm³ = 480 cm³

3.1.2(c): Why reactivity increases down the group

To react, a Group 2 atom must lose two electrons. That takes the first and second ionisation energies together. The first is the one you defined in 3.1.1(c). For the second, you need only the equation, which starts from a gaseous 1+ ion:

first:   Ca(g)  → Ca⁺(g)  + e⁻
second:  Ca⁺(g) → Ca²⁺(g) + e⁻

Both species are gaseous, and the electron leaves a 1+ ion.

The trend, in numbers

Standard values, in kJ mol⁻¹:

Element	First ionisation energy	Second ionisation energy	Sum
Mg	738	1451	2189
Ca	590	1145	1735
Sr	550	1064	1614
Ba	503	965	1468

Both ionisation energies fall down the group. The explanation uses the same three factors as 3.1.1:

1. Atomic radius increases. Going from Mg to Ba adds a shell of electrons at each step, which places the outer pair at a greater distance from the nucleus.
2. Shielding increases. More inner shells sit between the nucleus and the outer electrons.
3. Nuclear attraction on the outer electrons decreases. Nuclear charge does rise, but the extra distance and shielding outweigh it.

Less energy is needed to remove the two outer electrons, so they are lost more easily. The metal is oxidised more readily, so reactivity increases from Mg to Ba.

M3.1 means translating between forms: a table like this, a graph of ionisation energy against element, and a written trend.

Worked example 3: comparing calcium and barium

Use the table above to calculate the energy needed to form one mole of gaseous 2+ ions from gaseous atoms for Ca and for Ba. Explain what the difference shows.

Ca(g) → Ca²⁺(g) + 2e⁻    590 + 1145 = 1735 kJ mol⁻¹
Ba(g) → Ba²⁺(g) + 2e⁻    503 +  965 = 1468 kJ mol⁻¹
difference = 1735 − 1468 = 267 kJ mol⁻¹

Barium needs 267 kJ mol⁻¹ less. Its 6s electrons sit at a greater distance, behind more inner shells, so the nucleus holds them less tightly. Barium therefore reacts faster than calcium with water and with acids.

3.1.2(d): Group 2 oxides and water

Group 2 oxides are basic. They react with water to form hydroxides, which release hydroxide ions into solution:

CaO(s) + H₂O(l) → Ca(OH)₂(s)
Ca(OH)₂(s) + aq → Ca²⁺(aq) + 2OH⁻(aq)

This is not a redox reaction: every element keeps its oxidation number (Ca +2, O −2, H +1).

The hydroxides become more soluble down the group. More dissolved hydroxide means a higher OH⁻ concentration, so the solutions become more alkaline from Mg to Ba.

Oxide added to water	Approximate pH of the solution formed
MgO	about 10 (magnesium hydroxide is only sparingly soluble)
CaO	about 12
SrO	about 13
BaO	about 13 or above

These are approximate values for saturated solutions at room temperature. Learn the order and the reason.

Worked example 4: barium oxide in water

1.533 g of barium oxide is added to water and the solution made up to 250 cm³. Calculate the concentration of hydroxide ions. Use Mr(BaO) = 153.3.

BaO(s) + H₂O(l) → Ba(OH)₂(aq)
n(BaO)       = 1.533 ÷ 153.3 = 0.0100 mol
[Ba(OH)₂]    = 0.0100 ÷ 0.250 = 0.0400 mol dm⁻³
[OH⁻]        = 2 × 0.0400     = 0.0800 mol dm⁻³

Each Ba(OH)₂ releases two OH⁻ ions.

3.1.2(e): Group 2 compounds as bases

Because Group 2 hydroxides, oxides and carbonates neutralise acids, several are used to raise pH. The specification names two examples but says the list is not limited to them.

Agriculture. Calcium hydroxide (slaked lime) is spread on acid soils to raise the pH so that crops grow well:

Ca(OH)₂(s) + 2H⁺(aq) → Ca²⁺(aq) + 2H₂O(l)
Ca(OH)₂(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + 2H₂O(l)

Antacids. Excess hydrochloric acid in the stomach causes indigestion. Magnesium hydroxide and calcium carbonate neutralise it:

Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

The carbonate also releases carbon dioxide, which can cause wind.

Worked example 5: one antacid tablet

A tablet contains 0.600 g of CaCO₃ (Mr 100.1) and 0.150 g of Mg(OH)₂ (Mr 58.3). Calculate the volume of 0.100 mol dm⁻³ HCl it can neutralise, and the volume of CO₂ released at RTP.

n(CaCO₃)   = 0.600 ÷ 100.1 = 5.994 × 10⁻³ mol
n(Mg(OH)₂) = 0.150 ÷ 58.3  = 2.573 × 10⁻³ mol
n(HCl)     = 2 × (5.994 + 2.573) × 10⁻³ = 0.01713 mol
V(HCl)     = 0.01713 ÷ 0.100 = 0.171 dm³ = 171 cm³
n(CO₂)     = n(CaCO₃) = 5.994 × 10⁻³ mol
V(CO₂)     = 5.994 × 10⁻³ × 24.0 = 0.144 dm³ = 144 cm³

Only the carbonate releases gas.

Worked example 6: liming a vegetable bed

A soil test estimates that a 15.0 m² bed contains acid equivalent to 2.40 mol of H⁺ per m². Calculate the mass of Ca(OH)₂ (Mr 74.1) that would neutralise it.

n(H⁺)       = 2.40 × 15.0 = 36.0 mol
n(Ca(OH)₂)  = 36.0 ÷ 2    = 18.0 mol      (1 Ca(OH)₂ : 2 H⁺)
mass        = 18.0 × 74.1 = 1334 g = 1.33 kg

Common errors

• Writing the second ionisation equation from an atom, Ca(g) → Ca²⁺(g) + 2e⁻. That is the sum of the first two, not the second.
• Giving Mg(OH)₂ as the product of magnesium with steam. With steam the product is MgO.
• Saying ionisation energy decreases “because nuclear charge decreases”. It increases; distance and shielding win.
• Calling an oxide dissolving in water a redox reaction. No oxidation number changes.
• Forgetting that one Group 2 hydroxide or carbonate neutralises two H⁺ ions.
• Leaving out CO₂ from the carbonate antacid equation.

Where to go next

Condense this page with the Group 2 revision notes, then try the Group 2 practice questions. The mole methods used above are in the Amount of substance study guide, and s-orbital configurations are built up in Electron structure. Neutralisation is covered in Acids.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.1.2 Group 2, learning outcomes (a) to (e). Published by OCR.