Group 2 Alkaline Earth Metals, Group 7 Halide Reducing Power & Period 3 Oxides with Water
Inorganic chemistry at A-Level centres on three cornerstone periodic blocks: Group 2 (Alkaline Earth Metals), Group 17/7 (The Halogens), and Period 3 Elements and their Oxides (Na to S). Down Group 2, atomic radius and shielding increase, lowering ionisation energies so redox reactivity with water increases, while the solubilities of Group 2 hydroxides and sulfates follow exact opposite trends used in medical barium meals and flue-gas desulfurisation. Down Group 7, electronegativity and oxidising power of the halogens (X₂) decrease, whereas the reducing power of the halide ions (X⁻) increases dramatically, tested by reacting solid NaCl, NaBr, and NaI with concentrated sulfuric acid. Across Period 3, elements transition from giant metallic to giant covalent to simple molecular, mirrored by their oxides shifting from strongly alkaline ionic lattices (Na₂O, MgO) through amphoteric Al₂O₃ to strongly acidic covalent oxides (SiO₂, P₄O₁₀, SO₂, SO₃).
- Group 2 Trends: Reactivity, Hydroxide Solubility & Sulfate Solubility
- Down Group 2 (Mg → Ca → Sr → Ba), atomic radius and electron shielding increase, so first and second ionisation energies decrease and reactivity with water increases: Mg(s) reacts slowly with cold water (Mg + 2H₂O → Mg(OH)₂ + H₂) but vigorously with steam to form white MgO (Mg(s) + H₂O(g) → MgO(s) + H₂(g)); Ca, Sr, and Ba react with cold water with increasing vigour. Opposite Solubility Trends down Group 2: 1. Group 2 Hydroxides, M(OH)₂, become more soluble down the group (Mg(OH)₂ is sparingly soluble, pH ~9-10, used as 'Milk of Magnesia' antacid; Ba(OH)₂ is strongly soluble, pH ~13). 2. Group 2 Sulfates, MSO₄, become less soluble down the group (MgSO₄ is soluble; BaSO₄ is completely insoluble white precipitate).
- Uses of Group 2 Compounds (Ti Extraction, Flue Gas, Barium Meals & Testing SO₄²⁻)
- 1. Titanium extraction: TiO₂ is converted to TiCl₄, then reduced with Mg at 1000°C under Argon: TiCl₄ + 2Mg → Ti + 2MgCl₂. 2. Flue-gas desulfurisation: wet slurry of CaO or CaCO₃ reacts with acidic SO₂ gas from power stations: CaO(s) + SO₂(g) + 2H₂O(l) + ½O₂(g) → CaSO₄ 2H₂O(s) (gypsum plasterboard). 3. Barium meal: insoluble BaSO₄ is swallowed to absorb X-rays and image the gut without toxic Ba²⁺ ions being absorbed into the blood. 4. Testing for sulfate ions (SO₄²⁻): acidify first with dilute HCl or HNO₃ (to remove carbonate CO₃²⁻ impurities as CO₂ gas so white BaCO₃ cannot form a false positive), then add BaCl₂(aq) to observe a white precipitate: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
- Group 7: Oxidising Power of Halogens vs Reducing Power of Halides (with H₂SO₄)
- Down Group 7 (F₂ → Cl₂ → Br₂ → I₂), boiling point increases (more electrons → stronger van der Waals forces), electronegativity decreases, and oxidising power of X₂ decreases (Cl₂ displaces Br⁻ to orange Br₂ and I⁻ to brown I₂). Conversely, reducing power of halide ions (X⁻) increases down the group (Cl⁻ < Br⁻ < I⁻) because ionic radius and shielding increase so the outer electron is lost more easily. Reaction of solid NaX with concentrated H₂SO₄: 1. NaCl + H₂SO₄ → NaHSO₄ + HCl(g) (steamy white fumes; acid-base only, Cl⁻ is too weak a reducing agent to reduce S(+6)). 2. NaBr gives HBr(g) (acid-base) and reduces S(+6) to S(+4): 2H⁺ + 2Br⁻ + H₂SO₄ → Br₂(g/l) (orange-brown fumes) + SO₂(g) (choking gas) + 2H₂O. 3. NaI gives HI(g) and reduces S(+6) all the way to S(+4) in SO₂, S(0) in yellow solid sulfur S(s), and S(-2) in foul-smelling rotten-egg H₂S(g) (8H⁺ + 8I⁻ + H₂SO₄ → 4I₂(s) + H₂S(g) + 4H₂O, purple vapour/black solid I₂).
Group 2 Alkaline Earth Metals, Group 7 Halide Reducing Power & Period 3 Oxides with Water: Reaction Profile (ΔH & E_a) + Le Chatelier Yield Simulator
Toggle a catalyst on the enthalpy profile diagram or adjust temperature and pressure to see how dynamic equilibrium shifts.
Modelled on N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = −92 kJ/mol) (4 moles gas → 2 moles gas):
Higher pressure shifts equilibrium to the side with fewer gas moles (right, 4 → 2). Lower temperature favours the exothermic forward reaction.
Key Concepts & Exam Definitions (6)
- Group 2 Trends: Reactivity, Hydroxide Solubility & Sulfate Solubility
- Down Group 2 (Mg → Ca → Sr → Ba), atomic radius and electron shielding increase, so first and second ionisation energies decrease and reactivity with water increases: Mg(s) reacts slowly with cold water (Mg + 2H₂O → Mg(OH)₂ + H₂) but vigorously with steam to form white MgO (Mg(s) + H₂O(g) → MgO(s) + H₂(g)); Ca, Sr, and Ba react with cold water with increasing vigour. Opposite Solubility Trends down Group 2: 1. Group 2 Hydroxides, M(OH)₂, become more soluble down the group (Mg(OH)₂ is sparingly soluble, pH ~9-10, used as 'Milk of Magnesia' antacid; Ba(OH)₂ is strongly soluble, pH ~13). 2. Group 2 Sulfates, MSO₄, become less soluble down the group (MgSO₄ is soluble; BaSO₄ is completely insoluble white precipitate).
- Uses of Group 2 Compounds (Ti Extraction, Flue Gas, Barium Meals & Testing SO₄²⁻)
- 1. Titanium extraction: TiO₂ is converted to TiCl₄, then reduced with Mg at 1000°C under Argon: TiCl₄ + 2Mg → Ti + 2MgCl₂. 2. Flue-gas desulfurisation: wet slurry of CaO or CaCO₃ reacts with acidic SO₂ gas from power stations: CaO(s) + SO₂(g) + 2H₂O(l) + ½O₂(g) → CaSO₄ 2H₂O(s) (gypsum plasterboard). 3. Barium meal: insoluble BaSO₄ is swallowed to absorb X-rays and image the gut without toxic Ba²⁺ ions being absorbed into the blood. 4. Testing for sulfate ions (SO₄²⁻): acidify first with dilute HCl or HNO₃ (to remove carbonate CO₃²⁻ impurities as CO₂ gas so white BaCO₃ cannot form a false positive), then add BaCl₂(aq) to observe a white precipitate: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
- Group 7: Oxidising Power of Halogens vs Reducing Power of Halides (with H₂SO₄)
- Down Group 7 (F₂ → Cl₂ → Br₂ → I₂), boiling point increases (more electrons → stronger van der Waals forces), electronegativity decreases, and oxidising power of X₂ decreases (Cl₂ displaces Br⁻ to orange Br₂ and I⁻ to brown I₂). Conversely, reducing power of halide ions (X⁻) increases down the group (Cl⁻ < Br⁻ < I⁻) because ionic radius and shielding increase so the outer electron is lost more easily. Reaction of solid NaX with concentrated H₂SO₄: 1. NaCl + H₂SO₄ → NaHSO₄ + HCl(g) (steamy white fumes; acid-base only, Cl⁻ is too weak a reducing agent to reduce S(+6)). 2. NaBr gives HBr(g) (acid-base) and reduces S(+6) to S(+4): 2H⁺ + 2Br⁻ + H₂SO₄ → Br₂(g/l) (orange-brown fumes) + SO₂(g) (choking gas) + 2H₂O. 3. NaI gives HI(g) and reduces S(+6) all the way to S(+4) in SO₂, S(0) in yellow solid sulfur S(s), and S(-2) in foul-smelling rotten-egg H₂S(g) (8H⁺ + 8I⁻ + H₂SO₄ → 4I₂(s) + H₂S(g) + 4H₂O, purple vapour/black solid I₂).
- Testing for Halide Ions (AgNO₃ + Dilute/Concentrated NH₃) & Chlorine Disproportionation
- Acidify sample with dilute nitric acid (HNO₃, to remove CO₃²⁻/OH⁻ impurities; never use HCl), then add aqueous silver nitrate (AgNO₃): 1. Cl⁻ gives a white precipitate of AgCl(s), which dissolves in dilute aqueous NH₃ (forming [Ag(NH₃)₂]⁺). 2. Br⁻ gives a cream precipitate of AgBr(s), which is insoluble in dilute NH₃ but dissolves in concentrated NH₃. 3. I⁻ gives a yellow precipitate of AgI(s), which is insoluble in both dilute and concentrated NH₃. Disproportionation of Cl₂: 1. With cold water (water treatment): Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HClO(aq) (chlorate(I) acid kills bacteria; in bright sunlight: 2Cl₂ + 2H₂O → 4HCl + O₂). 2. With cold dilute NaOH (bleach manufacture): Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l) (Cl goes from 0 to -1 and +1).
- Period 3 Oxides: Bonding, Melting Points & Reactions with Water (pH)
- Across Period 3, elements react with oxygen to form oxides in their highest oxidation states (except S forms SO₂ and SO₃ with V₂O₅): 1. Ionic Oxides (Na₂O, MgO, Al₂O₃): high melting points due to strong electrostatic forces in giant ionic lattices. Na₂O(s) dissolves readily in water to form strongly alkaline NaOH(aq) (pH 13-14): Na₂O + H₂O → 2Na⁺ + 2OH⁻ (O²⁻ + H₂O → 2OH⁻). MgO(s) is only sparingly soluble in water (because the Mg²⁺/O²⁻ lattice is much stronger than Na⁺/O²⁻), forming weakly alkaline Mg(OH)₂(aq) (pH 9-10). Al₂O₃(s) has a very high lattice enthalpy with partial covalent character and is insoluble in water (pH 7). 2. Giant Covalent Oxide (SiO₂): macromolecular lattice, very high melting point, insoluble in water (pH 7). 3. Simple Molecular Covalent Oxides (P₄O₁₀(s), SO₂(g), SO₃(l/g)): low melting points (weak van der Waals/dipole forces), react vigorously with water to form strongly acidic solutions: P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq) (phosphoric(V) acid, pH 0-2); SO₂(g) + H₂O(l) → H₂SO₃(aq) (sulfuric(IV) acid, weak acid, pH 2-3); SO₃(g) + H₂O(l) → H₂SO₄(aq) (sulfuric(VI) acid, strong acid, pH 0-1).
- Acid-Base Reactions of Period 3 Oxides (Basic, Amphoteric Al₂O₃, Acidic)
- 1. Basic oxides (Na₂O, MgO) react with acids to form salt + water: MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l). 2. Amphoteric oxide (Al₂O₃) reacts with both acids and strong alkalis: With acid: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) (or Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O); With hot concentrated alkali: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2NaAl(OH)₄(aq). 3. Acidic oxides (SiO₂, P₄O₁₀, SO₂, SO₃) react with bases: SiO₂ reacts with hot concentrated NaOH: SiO₂(s) + 2NaOH(aq) → Na₂SiO₃(aq) + H₂O(l); P₄O₁₀ + 12NaOH → 4Na₃PO₄ + 6H₂O; SO₂ + 2NaOH → Na₂SO₃ + H₂O.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State the trends in solubility of the Group 2 hydroxides (M(OH)₂) and Group 2 sulfates (MSO₄) down the group from Mg to Ba.
Why must a solution be acidified with dilute HCl or HNO₃ before adding BaCl₂(aq) to test for sulfate ions (SO₄²⁻)?
Explain why iodide ions (I⁻) are much stronger reducing agents than chloride ions (Cl⁻), and state all the sulfur-containing products formed when solid NaI reacts with concentrated H₂SO₄.
Write the equations for the reaction of chlorine gas (Cl₂) with: (1) cold water, and (2) cold, dilute aqueous sodium hydroxide, and explain why both are disproportionation reactions.
Explain why Na₂O reacts with water to give a solution of pH 14, whereas MgO gives a solution of only pH 9-10, and Al₂O₃ gives pH 7.
Write balanced equations for the reactions of P₄O₁₀, SO₂, and SO₃ with water, and state the pH of each resulting solution.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Samples of solid sodium chloride (NaCl), sodium bromide (NaBr), and sodium iodide (NaI) were treated separately with concentrated sulfuric acid (H₂SO₄). Write balanced equations for the reactions of NaCl and NaBr with concentrated H₂SO₄, write the half-equations and overall equation for the reduction of H₂SO₄ to H₂S by NaI, and explain the trend in reducing ability of the halide ions. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Reaction with NaCl (acid-base reaction only, no redox; white steamy fumes of HCl): NaCl(s) + H₂SO₄(l) → NaHSO₄(s) + HCl(g) [1 mark]. 2. Reaction with NaBr (initial acid-base gives HBr, followed by redox reduction of S(+6) to S(+4) giving orange-brown Br₂ fumes and choking SO₂ gas): 2H⁺ + 2Br⁻ + H₂SO₄ → Br₂(g) + SO₂(g) + 2H₂O(l) [1 mark]. 3. Oxidation half-equation for iodide ions: 2I⁻ → I₂ + 2e⁻ [1 mark]. 4. Reduction half-equation for H₂SO₄ (S = +6) to H₂S (S = -2): H₂SO₄ + 8H⁺ + 8e⁻ → H₂S + 4H₂O [1 mark]. 5. Overall equation for NaI forming H₂S and black solid/purple vapour I₂: 8H⁺ + 8I⁻ + H₂SO₄ → 4I₂(s) + H₂S(g) + 4H₂O(l) [1 mark]. 6. Explanation of trend: Going down Group 17 from Cl⁻ to Br⁻ to I⁻, ionic radius and electron shielding by inner shells increase, weakening the electrostatic attraction between the nucleus and the outer-shell electron so the halide ion loses an electron more easily (stronger reducing agent) [1 mark].
- Q2
Explain the trend in melting points of the Period 3 elements from sodium (Na) to silicon (Si) to argon (Ar), making specific reference to structure and bonding. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Na, Mg, and Al have giant metallic structures; their melting points increase from Na to Mg to Al because metal ion charge increases (+1, +2, +3), ionic radius decreases, and the number of delocalised electrons per atom increases (1, 2, 3), resulting in stronger metallic bonding [1 mark]. 2. Silicon (Si) has the highest melting point in Period 3 because it has a giant covalent (macromolecular) lattice in which every Si atom is tetrahedrally bonded to four others by strong covalent bonds that require vast energy to break [1 mark]. 3. Phosphorus (P₄), sulfur (S₈), and chlorine (Cl₂) are simple molecular covalent substances with much lower melting points because only weak intermolecular van der Waals (induced dipole-dipole) forces are broken on melting [1 mark]. 4. Sulfur (S₈) has a higher melting point than phosphorus (P₄) and chlorine (Cl₂) because an S₈ molecule contains more atoms and electrons (Mr = 256) than P₄ (Mr = 124) or Cl₂ (Mr = 71), giving S₈ the strongest van der Waals forces [1 mark]. 5. Argon (Ar) has the lowest melting point because it exists as isolated monatomic atoms with the fewest electrons in a single particle and very weak van der Waals forces [1 mark].
- Q3
Aluminium oxide (Al₂O₃) is described as an amphoteric oxide, whereas silicon dioxide (SiO₂) is an acidic oxide, yet neither dissolves in water. Explain why Al₂O₃ has some covalent character, why neither oxide dissolves in water, and write balanced equations to show how Al₂O₃ demonstrates its amphoteric nature with hydrochloric acid and with sodium hydroxide. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. The Al³⁺ ion is very small and highly charged (+3), giving it a high charge density (polarising power) that distorts the electron cloud of the O²⁻ oxide ion, introducing partial covalent character into the ionic lattice [1 mark]. 2. Neither Al₂O₃ nor SiO₂ dissolves in water because the electrostatic forces in the Al₂O₃ ionic lattice and the giant network of Si-O covalent bonds in SiO₂ are far too strong to be overcome by hydration with water [1 mark]. 3. An amphoteric oxide reacts with both acids and bases to form salts [1 mark]. 4. Reaction of Al₂O₃ acting as a base with HCl: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) (or Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O) [1 mark]. 5. Reaction of Al₂O₃ acting as an acid with hot concentrated NaOH: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2NaAl(OH)₄(aq) (or Al₂O₃ + 2OH⁻ + 3H₂O → 2[Al(OH)₄]⁻) [1 mark].
- Q4
Explain how aqueous silver nitrate and aqueous ammonia are used together to distinguish between aqueous solutions of sodium chloride, sodium bromide, and sodium iodide, and discuss the benefits and risks of adding chlorine to drinking water. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Add dilute nitric acid followed by aqueous silver nitrate (AgNO₃): NaCl gives a white precipitate (AgCl), NaBr gives a cream precipitate (AgBr), and NaI gives a yellow precipitate (AgI) [1 mark]. 2. Because the three precipitate colours are difficult to tell apart cleanly by eye, add dilute aqueous ammonia (NH₃) first: only the white AgCl precipitate dissolves to form a colourless solution ([Ag(NH₃)₂]⁺) [1 mark]. 3. Next, add concentrated aqueous ammonia to the remaining two tubes: the cream AgBr precipitate dissolves to form a colourless solution, whereas the yellow AgI precipitate remains insoluble [1 mark]. 4. Benefit of chlorinating water: Chlorine reacts with water to form chlorate(I) acid (HClO), which is an oxidising agent that kills waterborne bacterial pathogens (preventing cholera and typhoid) [1 mark]. 5. Risk of chlorinating water: Chlorine is a toxic respiratory irritant gas, and it can react with dissolved organic matter from decaying vegetation to form chlorinated hydrocarbons (e.g. trichloromethane, CHCl₃) which are suspected carcinogens; however, the public health benefit of preventing fatal waterborne epidemics far outweighs the small risk [1 mark].
Common Exam Questions (4)
Why does white phosphorus have the formula P₄ and sulfur have the formula S₈ in Period 3 equations?
Under standard conditions, solid white phosphorus exists as tetrahedral P₄ molecules (4 phosphorus atoms), so its combustion equation is P₄(s) + 5O₂(g) → P₄O₁₀(s) (or 4P + 5O₂ → P₄O₁₀). Solid rhombic/monoclinic sulfur exists as crown-shaped S₈ rings (though S(s) + O₂(g) → SO₂(g) is accepted in exams).
Why does sulfur dioxide (SO₂) dissolve in water to give a higher pH (~pH 2-3) than sulfur trioxide (SO₃, ~pH 0-1)?
SO₂ reacts with water to form sulfuric(IV) acid (sulfurous acid, H₂SO₃), which is a weak acid that only partially dissociates in water (H₂SO₃ ⇌ H⁺ + HSO₃⁻). SO₃ reacts with water to form sulfuric(VI) acid (H₂SO₄), which is a strong acid that dissociates completely in its first step, producing a much higher concentration of H⁺(aq) ions.
How is barium chloride or magnesium hydroxide used medically, and how does this link to their Group 2 solubility trends?
Magnesium hydroxide, Mg(OH)₂, is at the top of Group 2 where hydroxides are sparingly soluble; therefore, a suspension of Mg(OH)₂ ('Milk of Magnesia') neutralises excess stomach acid (HCl) without releasing a dangerously high concentration of caustic OH⁻ ions. Conversely, barium sulfate, BaSO₄, is at the bottom of Group 2 where sulfates are completely insoluble, so patients can swallow a BaSO₄ X-ray 'barium meal' without absorbing toxic Ba²⁺ ions.
Why does adding concentrated H₂SO₄ to solid NaCl not produce chlorine gas (Cl₂)?
Chloride ions (Cl⁻) are at the top of the common halide series and have a small ionic radius and strong nuclear attraction for their outer electrons, making Cl⁻ too weak a reducing agent to reduce S(+6) in H₂SO₄. Therefore, only a non-redox acid-base proton transfer occurs (NaCl + H₂SO₄ → NaHSO₄ + HCl(g)).
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