Edexcel International A Level Chemistry: The elements of Groups 1 and 2 (YCH11)

This guide teaches Topic 8B, The elements of Groups 1 and 2, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, and covers outcomes 8.11 to 8.23, including Core Practical 3 (8.21) and Core Practical 4 (8.23). The topic belongs to Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols, one of the IAS units.

For quick recall there are revision notes; for exam-style work, try the practice questions. The course hub links to every Edexcel IAL Chemistry page. Use the printable checklist to track outcomes, and take a free 10-minute diagnostic to find your gaps. Titration maths builds on formulae, equations and amount of substance.

Data used on this page (use these values): Sr 87.6, O 16.0, N 14.0, Cl 35.5, H 1.0. Molar volume of any gas (room temperature and pressure): 24.0 dm³ mol⁻¹.

What Topic 8B covers

Spec ref	You need to be able to
8.11	Explain the trend in ionisation energy down Groups 1 and 2
8.12	Explain the trend in reactivity, Li to K and Mg to Ba
8.13	Know reactions of the metals with oxygen, chlorine and water
8.14	Know reactions of the oxides with water and dilute acid, and of the hydroxides with dilute acid
8.15	Know the solubility trends of Group 2 hydroxides and sulfates
8.16	Explain thermal stability of nitrates and carbonates using cation size and charge
8.17	Explain flame colours by electron transitions, and know the colours
8.18	Know procedures for thermal decomposition patterns and flame tests
8.19	Identify carbonate, hydrogencarbonate, sulfate and ammonium ions
8.20	Calculate concentrations in mol dm⁻³ and g dm⁻³, including titrations
8.21	Core Practical 3: find the concentration of hydrochloric acid
8.22	Minimise measurement uncertainty in volumetric analysis and estimate overall uncertainty
8.23	Core Practical 4: make a standard solution from a solid acid and use it to find the concentration of sodium hydroxide

8.11 and 8.12 Ionisation energy and reactivity

Use these first ionisation energies (kJ mol⁻¹): Li 520, Na 496, K 419; Mg 738, Ca 590, Sr 550, Ba 503.

Going down either group:

1. Each element has one more occupied shell, so the atomic radius increases.
2. Inner shells shield the outer electron from the nucleus.
3. Nuclear charge rises, but the larger distance and extra shielding outweigh it.
4. The outer electron is held less tightly, so removing it takes less energy.

Group 1 metals react by losing one electron (M → M⁺ + e⁻) and Group 2 metals two (M → M²⁺ + 2e⁻). These are removed more easily down the group, so reactivity increases from Li to K and Mg to Ba. In redox terms the metals are reducing agents, and their oxidation number rises from 0 to +1 or +2. For Group 2 the first and second ionisation energies both fall down the group.

8.13 Reactions with oxygen, chlorine and water

Reaction	Group 1 (Li to K)	Group 2 (Mg to Ba)
Oxygen	4Li + O₂ → 2Li₂O (oxide). Sodium in plenty of oxygen gives mainly the peroxide, 2Na + O₂ → Na₂O₂. Potassium gives the superoxide, K + O₂ → KO₂	2M + O₂ → 2MO. Magnesium burns with a brilliant white flame
Chlorine	2M + Cl₂ → 2MCl, white solids	M + Cl₂ → MCl₂, white solids
Water	2M + 2H₂O → 2MOH + H₂. Li fizzes steadily; Na melts into a ball and darts about; K also ignites the hydrogen with a lilac flame	M + 2H₂O → M(OH)₂ + H₂. Mg reacts very slowly with cold water but glows with steam: Mg + H₂O → MgO + H₂. Ca fizzes steadily, leaving a cloudy white suspension; Sr and Ba react faster

Oxygen has oxidation number −2 in an oxide, −1 in a peroxide and −½ in a superoxide.

8.14 Oxides and hydroxides

The oxides are basic. With water they give hydroxides, which is why the solutions are alkaline:

• Li₂O + H₂O → 2LiOH and CaO + H₂O → Ca(OH)₂ (the second is strongly exothermic).
• Magnesium oxide is only slightly soluble, so the solution is only weakly alkaline.

With dilute acid, oxides and hydroxides both give a salt and water:

• Na₂O + 2HCl → 2NaCl + H₂O; BaO + 2HNO₃ → Ba(NO₃)₂ + H₂O
• KOH + HNO₃ → KNO₃ + H₂O; Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
• Ionic forms: O²⁻ + 2H⁺ → H₂O and OH⁻ + H⁺ → H₂O

With dilute sulfuric acid, the calcium, strontium and barium compounds can stop reacting early, because a coating of poorly soluble sulfate forms on the solid.

8.15 Solubility of Group 2 hydroxides and sulfates

Compound	Trend down the group (Mg → Ba)
Hydroxides	Solubility increases: Mg(OH)₂ is sparingly soluble, Ba(OH)₂ dissolves most
Sulfates	Solubility decreases: MgSO₄ is soluble, BaSO₄ is insoluble

The trends run in opposite directions. That is why barium ions are used to test for sulfate (8.19), and why aqueous sodium hydroxide gives a white precipitate with Mg²⁺(aq) but little or none with dilute Ba²⁺(aq).

8.16 Thermal stability of carbonates and nitrates

Salt	Group 1 (Na, K)	Lithium	Group 2
Carbonate	No decomposition at Bunsen temperatures	Li₂CO₃ → Li₂O + CO₂	MCO₃ → MO + CO₂
Nitrate	2MNO₃ → 2MNO₂ + O₂	4LiNO₃ → 2Li₂O + 4NO₂ + O₂	2M(NO₃)₂ → 2MO + 4NO₂ + O₂

The explanation in four links:

1. The carbonate and nitrate ions are large, with a spread-out electron cloud.
2. A cation that is small and highly charged (high charge density) polarises that cloud, pulling electron density towards itself.
3. This weakens the bonds inside the anion, so less heat is needed to break it into CO₂, or NO₂ and O₂.
4. Down a group the charge stays the same but the cation gets larger, so it polarises less and the salt is more stable.

Group 1 salts are more stable than Group 2 salts because M⁺ carries a smaller charge and is larger than the M²⁺ ion in the same period. Li⁺ is very small, so lithium salts decompose the way Group 2 salts do.

Worked example 1. 4.23 g of strontium nitrate is heated until it has fully decomposed. Calculate the total volume of gas at room temperature and pressure, and the mass of solid left.

Mr Sr(NO₃)₂ = 87.6 + 2(14.0) + 6(16.0) = 211.6
n = 4.23 / 211.6 = 0.0200 mol
2Sr(NO₃)₂ → 2SrO + 4NO₂ + O₂
n(NO₂) = 0.0400 mol; n(O₂) = 0.0100 mol; total gas = 0.0500 mol
volume = 0.0500 × 24.0 = 1.20 dm³
mass SrO = 0.0200 × 103.6 = 2.07 g

8.17 Flame colours

Heat from the flame promotes electrons to higher energy levels. When they fall back to lower levels they release the energy as light. Each energy gap has a fixed size, so each element emits particular frequencies and gives its own colour.

Ion	Flame colour
Li⁺	red (crimson)
Na⁺	yellow-orange
K⁺	lilac
Ca²⁺	brick red (orange-red)
Sr²⁺	crimson red
Ba²⁺	pale (apple) green

Magnesium compounds give no flame colour, because the energy released lies outside the visible range.

8.18 Practical procedures

Flame test. Clean a nichrome wire by dipping it in concentrated hydrochloric acid and heating it until the flame shows no colour. Dip it in the acid again, then into the solid, and hold it in the hottest part of a non-luminous Bunsen flame.

Thermal decomposition patterns. Heat equal amounts (in moles) of each salt in identical boiling tubes over the same flame.

• Carbonates: pass the gas into a fixed volume of limewater and time how long it takes to turn cloudy. Carbon dioxide turns limewater milky.
• Nitrates: time how long it takes for brown gas to appear, or test for oxygen with a glowing splint, which relights. Nitrogen dioxide is brown and acidic; it turns damp blue litmus red. Use a fume cupboard, since NO₂ is toxic.
• Take the delivery tube out of the limewater before you stop heating, so liquid is not sucked back.

8.19 Tests for ions

Ion	Test	Positive result and equation
CO₃²⁻ / HCO₃⁻	Add a dilute acid and pass any gas into limewater	Effervescence; the limewater goes cloudy (milky). CO₃²⁻ + 2H⁺ → CO₂ + H₂O; HCO₃⁻ + H⁺ → CO₂ + H₂O; Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
SO₄²⁻	Acidify with dilute HCl, then add BaCl₂(aq)	A white precipitate forms. Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
NH₄⁺	Warm with sodium hydroxide solution	Gas turns damp red litmus blue and gives white smoke with HCl fumes. NH₄⁺ + OH⁻ → NH₃ + H₂O; NH₃ + HCl → NH₄Cl

The acid in the sulfate test removes carbonate ions, which would otherwise give a white barium carbonate precipitate.

8.20 and 8.21 Titrations and Core Practical 3

• c (mol dm⁻³) = n ÷ V, with V in dm³. Mass concentration (g dm⁻³) = c × M.
• Methyl orange: red in acid, yellow in alkali, orange at the end point.
• Phenolphthalein: colourless in acid, pink in alkali.

Core Practical 3 outline. Pipette 25.0 cm³ of the hydrochloric acid into a conical flask with two or three drops of indicator. Fill a burette with sodium hydroxide of known concentration. Run in alkali, swirling, then add it dropwise near the end point. Repeat until you have concordant titres (typically within 0.10 cm³) and average only those.

Worked example 2. 25.0 cm³ of hydrochloric acid needs 0.150 mol dm⁻³ NaOH. Titres: rough 21.20; then 20.80, 20.55, 20.65 cm³.

Concordant: 20.55 and 20.65 → mean = 20.60 cm³ (20.80 is 0.15 from 20.65)
n(NaOH) = 0.150 × 20.60/1000 = 3.09 × 10⁻³ mol = n(HCl)
c(HCl) = 3.09 × 10⁻³ / 0.0250 = 0.124 mol dm⁻³
mass concentration = 0.1236 × 36.5 = 4.51 g dm⁻³

8.22 and 8.23 Uncertainty and Core Practical 4

Making a standard solution. Weigh the solid acid by difference. Dissolve it in a small volume of distilled water in a beaker. Transfer it to a volumetric flask, rinsing the beaker and glass rod into the flask. Fill to the mark so that, viewed at eye level, the bottom of the meniscus touches the graduation line. Stopper and invert several times.

Reducing uncertainty: use larger masses and larger titres; rinse the burette and pipette with the solution they will hold; remove the funnel and fill the burette jet; read meniscuses at eye level against a white tile.

Percentage uncertainty = (uncertainty ÷ value measured) × 100. A titre uses two burette readings, so double the reading uncertainty. Add the percentage uncertainties to estimate the overall percentage uncertainty.

Worked example 3. 2.95 g of butanedioic acid (HOOCCH₂CH₂COOH, Mr 118.1, diprotic) is made up to 250 cm³. 25.0 cm³ portions need a mean of 23.65 cm³ of NaOH. Use these uncertainties: balance ±0.005 g per reading, flask ±0.2 cm³, pipette ±0.04 cm³, burette ±0.05 cm³ per reading.

n(acid) in flask = 2.95 / 118.1 = 0.02498 mol; in 25.0 cm³ = 2.498 × 10⁻³ mol
n(NaOH) = 2 × 2.498 × 10⁻³ = 4.996 × 10⁻³ mol (1 : 2)
c(NaOH) = 4.996 × 10⁻³ / 0.02365 = 0.211 mol dm⁻³
% uncertainties: mass 0.01/2.95 = 0.34%; flask 0.08%; pipette 0.16%; titre 0.10/23.65 = 0.42%
total ≈ 1.0%, so c = 0.211 ± 0.002 mol dm⁻³

The mass and the titre dominate, so increasing either would improve the result most.

Common errors

• Saying the nuclear charge falls down a group. It rises; distance and shielding win.
• Writing Na₂O for sodium burnt in plenty of oxygen without mentioning the peroxide, or forgetting that potassium gives KO₂.
• Reversing the solubility trends: hydroxides become more soluble, sulfates less soluble.
• Explaining thermal stability by “the metal gets bigger” without linking cation size to polarisation of the anion.
• Giving sodium nitrate the Group 2 equation. Sodium and potassium nitrates give the nitrite and oxygen only.
• Averaging every titre, including the rough one or one that is not concordant.
• Using one burette reading’s uncertainty for a whole titre.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Scope: Topic 8B, outcomes 8.11 to 8.23.