Inorganic Chemistry

Going down Group 2 from magnesium to barium, how do the solubilities of the hydroxides and the sulfates change?

• ABoth hydroxides and sulfates become more soluble
• BHydroxides become more soluble; sulfates become less soluble✓
• CHydroxides become less soluble; sulfates become more soluble
• DBoth hydroxides and sulfates become less soluble

Answer: Group 2 hydroxide solubility increases down the group (Mg(OH)₂ is sparingly soluble, Ba(OH)₂ is much more soluble), whereas sulfate solubility decreases down the group (MgSO₄ is soluble, BaSO₄ is insoluble — the basis of the sulfate test).

Why does barium sulfate, rather than barium chloride or nitrate, make the use of acidified barium chloride a reliable test for sulfate ions?

• ABaSO₄ is a soluble white solid
• BBaSO₄ is a soluble yellow solution
• CBaSO₄ decomposes on contact with acid
• DBaSO₄ is an insoluble white precipitate✓

Answer: Adding acidified BaCl₂ to a solution containing SO₄²⁻ gives a white precipitate of insoluble BaSO₄. The acid (e.g. dilute HCl or HNO₃) removes carbonate ions that would otherwise also precipitate, making the test specific for sulfate.

When chlorine reacts with cold dilute sodium hydroxide solution, the products include NaCl and NaClO. What does this tell you about the behaviour of chlorine in this reaction?

• AChlorine is only oxidised
• BChlorine is only reduced
• CChlorine undergoes disproportionation✓
• DChlorine is neither oxidised nor reduced

Answer: Cl₂ + 2NaOH → NaCl + NaClO + H₂O. Chlorine goes from 0 to −1 (in NaCl, reduced) and from 0 to +1 (in NaClO, oxidised) simultaneously. Being both oxidised and reduced in the same reaction is disproportionation.

A few drops of chlorine water are added to a colourless solution of potassium iodide. What is observed, and why?

• ANo change, because Cl₂ is a weaker oxidising agent than I₂
• BA white precipitate forms, because AgI is produced
• CThe solution turns pale green, because I⁻ reduces Cl₂ to Cl⁻ only
• DA brown colour forms, because Cl₂ oxidises I⁻ to I₂✓

Answer: Chlorine is a stronger oxidising agent than iodine, so it displaces iodine: Cl₂ + 2I⁻ → 2Cl⁻ + I₂. The liberated iodine gives a brown (yellow-brown) colour in solution.

Concentrated sulfuric acid is added to solid sodium halides. With which halide is the reducing power of the halide ion great enough to reduce sulfur from +6 in H₂SO₄ all the way to H₂S?

• ANaF
• BNaCl
• CNaBr
• DNaI✓

Answer: Reducing power increases down Group 7. Iodide is the strongest reducing agent of the common halides and can reduce S from +6 (H₂SO₄) to −2 (H₂S). Bromide only reduces it to SO₂ (+4); chloride and fluoride do not reduce sulfur at all (they give HCl/HF).

A solution gives a brick-red colour in a flame test and forms a white precipitate with dilute sulfuric acid that is insoluble. Which cation is present?

• ABa²⁺
• BSr²⁺
• CCa²⁺✓
• DMg²⁺

Answer: A brick-red flame colour is characteristic of calcium, Ca²⁺. Calcium also forms a white, sparingly soluble precipitate of CaSO₄ with sulfuric acid. (Ba²⁺ gives an apple-green flame; Sr²⁺ gives red/crimson; Mg²⁺ gives no characteristic colour.)

Across Period 3, the oxides change in character. Which sequence correctly describes the acid–base nature of Na₂O, Al₂O₃ and SO₃?

• AAcidic, amphoteric, basic
• BBasic, acidic, amphoteric
• CAmphoteric, basic, acidic
• DBasic, amphoteric, acidic✓

Answer: Period 3 oxides change from basic on the left (Na₂O, a metal oxide) through amphoteric (Al₂O₃) to acidic on the right (SO₃, a non-metal oxide). So the order is basic, amphoteric, acidic.

Which equation correctly represents the reaction of phosphorus(V) oxide with water?

• AP₄O₁₀ + 2H₂O → 4HPO₃
• BP₄O₁₀ + 6H₂O → 4H₃PO₄✓
• CP₄O₆ + 6H₂O → 4H₃PO₃
• DP₄O₁₀ + 6H₂O → 4H₃PO₃

Answer: Phosphorus(V) oxide reacts vigorously with water to give phosphoric(V) acid: P₄O₁₀ + 6H₂O → 4H₃PO₄. The oxidation state of phosphorus (+5) is unchanged, consistent with an acid–base (not redox) reaction.

Transition metals and their compounds are characteristically coloured. The colour of a transition-metal complex ion arises because:

• Aelectrons are promoted between split 3d energy levels, absorbing certain visible wavelengths✓
• Bthe 4s electrons are completely removed
• Cligands emit light of their own characteristic colour
• Dthe d sub-shell is completely full in all complexes

Answer: Ligands split the 3d orbitals into two energy levels. Electrons absorb visible-light photons of the right energy (ΔE) to be promoted to the higher level; the colour seen is the complementary colour of the light absorbed. A full or empty d sub-shell shows no such d–d transition (e.g. Zn²⁺ is colourless).

Which statement explains why scandium is often not regarded as a typical transition metal, whereas iron is?

• AScandium has no 3d electrons at all
• BScandium cannot form any ions
• CSc³⁺, its only common ion, has an empty 3d sub-shell, so it forms no coloured/partly-filled d-orbital ions✓
• DIron has a completely full 3d sub-shell in all its ions

Answer: A transition metal forms at least one stable ion with a partially filled d sub-shell. Scandium's only common ion is Sc³⁺ (3d⁰, empty), so it shows none of the characteristic transition-metal properties (variable oxidation states, coloured ions). Iron forms Fe²⁺ (3d⁶) and Fe³⁺ (3d⁵), both partly filled.

When excess concentrated hydrochloric acid is added to an aqueous solution of copper(II) sulfate, a colour change occurs. Which change and product are correct?

• APale blue [Cu(H₂O)₆]²⁺ changes to yellow/green [CuCl₄]²⁻✓
• BPale blue [Cu(H₂O)₆]²⁺ changes to deep blue [Cu(NH₃)₄(H₂O)₂]²⁺
• CColourless [CuCl₄]²⁻ changes to pale blue [Cu(H₂O)₆]²⁺
• DGreen [CuCl₄]²⁻ changes to a white precipitate of CuCl

Answer: Chloride ligands replace water in a ligand substitution: [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O. This changes the geometry (octahedral to tetrahedral) and the colour from pale blue to yellow/green. (The deep blue ammine forms with ammonia, not HCl.)