Noesis
Level 2 — RecallInorganic Qualitative Analysis

Inorganic Qualitative Analysis

30 marksprintable — key stays hidden on paper

Level: 2 (Recall / Standard Textbook) Time: 30 minutes Total Marks: 30

Answer all questions. Write balanced equations where asked.

Q1. State the group reagent used for the following cation groups in systematic analysis: (a) Group I, (b) Group II, (c) Group III, (d) Group IV. (4 marks)

Q2. A white precipitate forms when dilute HCl is added to a salt solution. The precipitate is soluble in hot water and gives a yellow precipitate with KI. Identify the cation and give the group. (3 marks)

Q3. Give the confirmatory test (reagent + observation) for the chloride ion. Write the balanced ionic equation for the confirmatory precipitation. (3 marks)

Q4. State the flame test colours for: (a) Na⁺, (b) K⁺, (c) Ca²⁺, (d) Ba²⁺, (e) Cu²⁺. (5 marks)

Q5. Explain the brown ring test for the nitrate ion. Give the reagents and the formula of the coloured complex responsible for the brown ring. (3 marks)

Q6. Name the cation group precipitated by H₂S in the presence of dilute HCl, and explain briefly why dilute acid (not neutral/alkaline) medium is used. (3 marks)

Q7. In the borax bead test, a salt gives a blue bead in both oxidising and reducing flames. Identify the metal ion. Write the general reaction of borax on heating that forms the bead. (3 marks)

Q8. Distinguish between Br⁻ and I⁻ using chlorine water and carbon tetrachloride (layer test). State the colour of the CCl₄ layer in each case. (3 marks)

Q9. In the charcoal cavity test, a substance gives a grey metallic bead with no incrustation. Name a possible metal. Separately, name a metal that gives a yellow incrustation (hot) that turns white (cold). (3 marks)

Answer keyMark scheme & solutions

Q1. (4 marks) — 1 mark each

  • (a) Group I: dilute HCl (precipitates as chlorides).
  • (b) Group II: H₂S in presence of dilute HCl (precipitates as sulphides).
  • (c) Group III: NH₄Cl + NH₄OH (precipitates as hydroxides).
  • (d) Group IV: H₂S in presence of NH₄OH (ammoniacal, precipitates as sulphides). Why: Group reagents adjust conditions (sulphide ion concentration via [H⁺]) to selectively precipitate cations by their solubility products.

Q2. (3 marks)

  • Cation: Pb²⁺ (lead) — 1
  • Group: Group I — 1
  • Reasoning: PbCl₂ white ppt, soluble in hot water; Pb²⁺ + 2I⁻ → PbI₂ (yellow ppt) — 1

Q3. (3 marks)

  • Reagent/observation: Add dilute HNO₃ then AgNO₃ → white precipitate soluble in NH₄OH (1)
  • Equation: Ag⁺ + Cl⁻ → AgCl↓ (white) (1)
  • Solubility confirmation: AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻ (1)

Q4. (5 marks) — 1 mark each

  • (a) Na⁺: golden yellow
  • (b) K⁺: lilac / violet (pink through blue glass)
  • (c) Ca²⁺: brick red
  • (d) Ba²⁺: apple green / yellowish green
  • (e) Cu²⁺: blue / green (bluish-green)

Q5. (3 marks)

  • Reagents: salt + FeSO₄ solution, then conc. H₂SO₄ added slowly along the side (1)
  • Observation: brown ring at the junction of the two liquids (1)
  • Complex: [Fe(H₂O)₅NO]²⁺ (nitrosoferrous sulphate / nitrosyl-iron complex, brown) (1) Reaction basis: NO₃⁻ + 3Fe²⁺ + 4H⁺ → 3Fe³⁺ + NO + 2H₂O; NO + Fe²⁺ → complex.

Q6. (3 marks)

  • Group: Group II (Cu²⁺, Pb²⁺, Cd²⁺, Hg²⁺, Bi³⁺, As, Sb, Sn) (1)
  • Reason: In dilute HCl the H⁺ suppresses ionisation of H₂S (common-ion effect), keeping [S²⁻] low (2)
  • So only cations with very low Ksp sulphides (Group II) precipitate, while Group IV sulphides (higher Ksp) do not.

Q7. (3 marks)

  • Metal ion: Cu²⁺ — gives blue bead in oxidising flame (CuO/cupric metaborate); in reducing flame it may appear colourless (hot) / red opaque (Cu₂O), often taken as blue → Cu. (Accept Cu²⁺.) (1)
  • Borax reaction on heating: Na₂B₄O₇·10H₂O → Na₂B₄O₇ → 2NaBO₂ + B₂O₃ (transparent bead) (1)
  • Bead formation with metal oxide: B₂O₃ + CuO → Cu(BO₂)₂ (metaborate, coloured) (1)

Q8. (3 marks)

  • Add chlorine water + CCl₄ and shake (1)
  • Br⁻: liberated Br₂ → CCl₄ layer orange/brown; 2Br⁻ + Cl₂ → Br₂ + 2Cl⁻ (1)
  • I⁻: liberated I₂ → CCl₄ layer violet/purple; 2I⁻ + Cl₂ → I₂ + 2Cl⁻ (1)

Q9. (3 marks)

  • Grey metallic bead, no incrustation: Ag (or Cu — a red bead; accept metals whose oxides are non-volatile). Ag gives silvery-white/grey malleable bead (1.5)
  • Yellow (hot) → white (cold) incrustation: Zn (ZnO) (1.5)
[
  {"claim":"Pb + 2 I(reduced charge) forms PbI2 with 1:2 Pb:I stoichiometry balances charge",
   "code":"Pb_charge=2; I_count=2; I_charge=-1; result=(Pb_charge + I_count*I_charge)==0"},
  {"claim":"Brown ring test redox: NO3- + 3Fe2+ + 4H+ -> 3Fe3+ + NO + 2H2O is charge balanced",
   "code":"left=(-1)+3*(2)+4*(1); right=3*(3)+0+0; result=(left==right)"},
  {"claim":"2Br- + Cl2 -> Br2 + 2Cl- balances charge and atoms",
   "code":"Br_left=2; Cl_left=2; Br_right=2; Cl_right=2; charge_left=2*(-1); charge_right=2*(-1); result=(Br_left==Br_right and Cl_left==Cl_right and charge_left==charge_right)"},
  {"claim":"Borax decomposition Na2B4O7 -> 2NaBO2 + B2O3 balances Na, B, O",
   "code":"Na=(2, 2*1); B=(4, 2*1+2); O=(7, 2*2+3); result=(Na[0]==Na[1] and B[0]==B[1] and O[0]==O[1])"}
]