Inorganic Chemistry Problem Solving

This final lesson brings together everything you have learned about inorganic chemistry and focuses on the types of problem-solving questions that appear in Edexcel A-Level exams. You will practise identifying unknown salts, predicting products, working through ligand substitution reactions, calculating oxidation states, and interpreting data.

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Identifying Unknown Salts: Systematic Approach

Exam questions frequently describe a series of tests performed on an unknown compound and ask you to identify it. The approach is systematic:

1. Flame test — identifies the metal cation (Li⁺, Na⁺, K⁺, Ca²⁺, Ba²⁺, Cu²⁺)
2. NaOH test — identifies transition metal cations by precipitate colour and behaviour in excess
3. Anion tests — identifies the anion (CO₃²⁻ with acid + limewater, SO₄²⁻ with BaCl₂, halides with AgNO₃)

Worked Example 1: White Solid

A white solid gives the following results:

• Flame test: pale green
• Dissolves in water to give a colourless solution
• Adding dilute HNO₃ then AgNO₃ gives a white precipitate that dissolves in dilute NH₃

Analysis:

• Pale green flame → Ba²⁺
• White precipitate with AgNO₃ that dissolves in dilute NH₃ → Cl⁻
• The compound is barium chloride (BaCl₂)

Worked Example 2: Coloured Solid

An unknown green solid dissolves in water to give a green solution. The following tests are performed:

• Flame test: no distinctive colour
• NaOH added: green precipitate; dissolves in excess NaOH to give a dark green solution
• Dilute HCl added to a fresh portion: effervescence; gas turns limewater milky

Analysis:

• Green precipitate + dissolves in excess NaOH → Cr³⁺ (amphoteric)
• Gas turns limewater milky → CO₂ → anion is CO₃²⁻
• The compound is chromium(III) carbonate, Cr₂(CO₃)₃ (though this is unstable; the exam may present it hypothetically)

Worked Example 3: Identifying from Multiple Observations

A student has four unlabelled bottles of white solids: NaCl, BaSO₄, CaCO₃, and KBr. Design tests to identify each.

Test	NaCl	BaSO₄	CaCO₃	KBr
Dissolve in water?	Yes	No (insoluble)	No (slightly)	Yes
Add dil HCl to solid	No reaction	No reaction	Effervescence (CO₂)	No reaction
Flame test on original	Yellow (Na⁺)	Pale green (Ba²⁺)	Brick red (Ca²⁺)	Lilac (K⁺)
AgNO₃ test on solution	White ppt (AgCl)	N/A	N/A	Cream ppt (AgBr)

Steps:

1. Add dilute HCl to each — CaCO₃ effervesces (CO₂ identified)
2. Add water — BaSO₄ does not dissolve
3. Flame test on BaSO₄ confirms pale green (Ba²⁺)
4. Dissolve NaCl and KBr in water; flame test distinguishes Na (yellow) from K (lilac through cobalt glass)
5. Confirm with AgNO₃: NaCl → white AgCl; KBr → cream AgBr

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Predicting Products of Transition Metal Reactions

Addition of NaOH

Adding NaOH to transition metal solutions produces hydroxide precipitates:

Reaction	Precipitate Colour	In Excess NaOH?
Cu²⁺ + 2OH⁻ → Cu(OH)₂	Blue	Insoluble
Fe²⁺ + 2OH⁻ → Fe(OH)₂	Green	Insoluble (darkens)
Fe³⁺ + 3OH⁻ → Fe(OH)₃	Red-brown	Insoluble
Cr³⁺ + 3OH⁻ → Cr(OH)₃	Green	Dissolves (amphoteric)
Al³⁺ + 3OH⁻ → Al(OH)₃	White	Dissolves (amphoteric)

Addition of NH₃ (in excess)

The Cu²⁺ reaction sequence is a classic:

1. [Cu(H₂O)₆]²⁺ — pale blue solution
2. Add a small amount of NH₃: Cu(OH)₂ precipitates — blue precipitate
3. Add excess NH₃: Cu(OH)₂ dissolves: Cu(OH)₂ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2OH⁻ — deep blue solution

This colour change from pale blue → blue precipitate → deep blue is one of the most frequently tested sequences in A-Level chemistry.

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Ligand Substitution Reactions

Ligand substitution occurs when one ligand replaces another in a complex ion.

Key Reactions to Know

Starting Complex	Reagent	Product	Colour Change
[Cu(H₂O)₆]²⁺ (blue)	Conc HCl / Cl⁻	[CuCl₄]²⁻ (yellow-green)	Blue → yellow-green
[Cu(H₂O)₆]²⁺ (blue)	Excess NH₃	[Cu(NH₃)₄(H₂O)₂]²⁺ (deep blue)	Blue → deep blue
[Co(H₂O)₆]²⁺ (pink)	Conc HCl / Cl⁻	[CoCl₄]²⁻ (blue)	Pink → blue
[Co(H₂O)₆]²⁺ (pink)	Excess NH₃	[Co(NH₃)₆]²⁺ (straw/yellow)	Pink → straw
[Cr(H₂O)₆]³⁺ (green)	Excess NH₃	[Cr(NH₃)₆]³⁺ (purple)	Green → purple
[Fe(H₂O)₆]³⁺ (yellow)	SCN⁻	[Fe(H₂O)₅(SCN)]²⁺ (blood red)	Yellow → blood red

Coordination Number Changes

When H₂O is replaced by Cl⁻, the coordination number often drops from 6 to 4 because Cl⁻ is much larger than H₂O. The geometry changes from octahedral to tetrahedral.

When H₂O is replaced by NH₃, the coordination number usually stays at 6 because H₂O and NH₃ are similar in size. The geometry remains octahedral.

flowchart TD
    A["[M(H₂O)₆]²⁺<br>Octahedral, CN=6"] --> B{"Incoming ligand?"}
    B -->|"Cl⁻ (large)"| C["[MCl₄]²⁻<br>Tetrahedral, CN=4<br>Different colour"]
    B -->|"NH₃ (similar size to H₂O)"| D["[M(NH₃)₆]²⁺<br>Octahedral, CN=6<br>Different colour"]
    B -->|"en (bidentate)"| E["[M(en)₃]²⁺<br>Octahedral, CN=6<br>More stable (chelate)"]

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Calculating Oxidation States in Complex Ions

To find the oxidation state of the central metal:

1. Identify the charges on all ligands (neutral ligands = 0, Cl⁻ = −1, CN⁻ = −1, etc.)
2. Use the overall charge of the complex to set up an equation

Worked Examples

Complex	Ligand Charges	Equation	Metal Ox. State
[Fe(CN)₆]⁴⁻	6 × (−1) = −6	Fe + (−6) = −4	Fe = +2
[Cr(NH₃)₄Cl₂]⁺	4 × 0 + 2 × (−1) = −2	Cr + (−2) = +1	Cr = +3
MnO₄⁻	4 × (−2) = −8	Mn + (−8) = −1	Mn = +7
[V(H₂O)₆]³⁺	6 × 0 = 0	V + 0 = +3	V = +3
[CoCl₄]²⁻	4 × (−1) = −4	Co + (−4) = −2	Co = +2
[Cu(NH₃)₄(H₂O)₂]²⁺	4 × 0 + 2 × 0 = 0	Cu + 0 = +2	Cu = +2

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Exam-Style Data Analysis

Example: Interpreting Colour Changes

A student adds concentrated HCl dropwise to a blue solution of CuSO₄:

Volume of HCl Added	Colour	Species Present
0 cm³	Blue	[Cu(H₂O)₆]²⁺
5 cm³	Blue-green	Mixture
10 cm³	Green	Approximately equal mixture
20 cm³	Yellow-green	Mostly [CuCl₄]²⁻

The blue [Cu(H₂O)₆]²⁺ is gradually converted to yellow-green [CuCl₄]²⁻. The intermediate green colour is a mixture of both. This is a ligand substitution with coordination number change (6 → 4) and geometry change (octahedral → tetrahedral).

Example: Interpreting Redox Titrations

In a titration of Fe²⁺ with MnO₄⁻:

MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

The end point is when the purple MnO₄⁻ is no longer decolourised — a faint permanent pink colour persists. This is a self-indicating reaction (no separate indicator needed).

Worked calculation: A student titrates 25.00 cm³ of FeSO₄ solution with 0.0200 mol dm⁻³ KMnO₄ and requires 23.50 cm³ for the end point.

1. Moles of MnO₄⁻ = 0.0200 × 23.50/1000 = 4.70 × 10⁻⁴ mol
2. From equation: 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺
3. Moles of Fe²⁺ = 5 × 4.70 × 10⁻⁴ = 2.35 × 10⁻³ mol
4. [Fe²⁺] = 2.35 × 10⁻³ / (25.00/1000) = 0.0940 mol dm⁻³

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Cobalt Chloride Equilibrium

[Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l)

• Pink (octahedral) ⇌ Blue (tetrahedral)
• Forward reaction is endothermic

This equilibrium is the basis of cobalt chloride indicator paper:

• Dry (blue): Water is removed, shifting equilibrium to the right
• Wet (pink): Water is added, shifting equilibrium to the left
• Heated (blue): Endothermic forward reaction is favoured at higher temperature

This is an excellent Le Chatelier's principle example combining both concentration and temperature effects.

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Thiocyanate Test for Iron(III)

Adding potassium thiocyanate (KSCN) to a solution containing Fe³⁺ produces a distinctive blood red colour:

[Fe(H₂O)₆]³⁺(aq) + SCN⁻(aq) ⇌ [Fe(H₂O)₅(SCN)]²⁺(aq) + H₂O(l)

This is a very sensitive and specific test for Fe³⁺. It does not give a positive result with Fe²⁺.

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Common Exam Pitfalls: Master List

1. Confusing Fe²⁺ and Cr³⁺: Both give green precipitates with NaOH. Use excess NaOH (Cr dissolves; Fe does not) and check for darkening (Fe only)
2. Forgetting that 4s electrons are lost first: Fe²⁺ is [Ar] 3d⁶, not [Ar] 3d⁴ 4s²
3. Confusing displacement and disproportionation: Displacement = more reactive replaces less reactive. Disproportionation = same element both oxidised and reduced
4. Using the wrong acid for anion tests: HNO₃ for halide tests (to avoid introducing Cl⁻), HCl for sulfate tests (to remove carbonates)
5. Catalysts do not change equilibrium position: They increase the rate of both forward and reverse reactions equally
6. Writing "bond between metal and carbonate breaks" instead of "bonds within the carbonate ion are weakened by polarisation"
7. Saying "d-block element" when asked to define a transition metal: Must say "forms at least one stable ion with an incomplete d sub-shell"
8. Forgetting coordination number changes: When Cl⁻ replaces H₂O, CN often drops from 6 to 4 (Cl⁻ is larger)
9. Not explaining colour in terms of ΔE: Always mention d-orbital splitting, d-d transitions, and complementary colour
10. Confusing Ba²⁺ (pale green flame) and Cu²⁺ (blue-green flame): Check solution colour to distinguish

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Summary

Inorganic problem solving requires you to integrate knowledge across all topics: Group 2 trends, halogen chemistry, transition metal properties, complex ions, colour, catalysis, and qualitative analysis. Use a systematic approach: identify the cation (flame test + NaOH), identify the anion (specific tests), then combine. For transition metal questions, always consider oxidation states, ligand substitution, coordination number changes, and colour. Show all working in calculations and always link explanations to underlying chemical principles.

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A-Level Deep Dive: Inorganic Problem Solving

Spec mapping

This synoptic lesson integrates Edexcel 9CH0 specification Topics 4 and 15 with bonding (Topic 2), redox (Topic 8), energetics (Topic 13), and equilibria (Topic 11), as required for the high-AO3 questions on Paper 3 (9CH0/03: General and Practical Principles of Chemistry) that pull together inorganic content from across the specification (refer to the official specification document for exact wording). Paper 3 multi-step problems typically present unknown compounds, complex reaction sequences, or thermodynamic challenges that demand systematic application of CP4, CP10, CP12 and CP16 protocols. Examined principally in Paper 3 with crossover into Paper 1 multi-mark synoptic items.

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Worked example with full mark scheme

Question (12 marks): A pale blue crystalline solid X dissolves in water to give a pale blue solution Y. The following observations are made on Y:

(i) Adding NaOH(aq) gives a pale blue gelatinous precipitate that does not redissolve in excess NaOH but dissolves in excess concentrated NH₃ to give a deep royal blue solution.

(ii) Adding excess concentrated HCl turns the solution yellow-green; on dilution with water, the original pale blue colour is restored.

(iii) Acidified BaCl₂(aq) gives a white precipitate insoluble in dilute HCl.

(iv) On heating dry crystals of X, water vapour is evolved and the residue is white.

Identify X. For each observation, write balanced equations and explain the underlying chemistry, including geometry and ligand-field considerations where relevant. Suggest one further test that would confirm your identification.

Solution with mark scheme:

Identification: X = CuSO₄·5H₂O (copper(II) sulfate pentahydrate, "blue vitriol").

M1 (AO2) — pale blue solution + Cu²⁺ test outcomes consistent with CuSO₄.

M1 (AO2) — sulfate confirmed by white BaSO₄ insoluble in HCl.

M1 (AO3) — water of crystallisation explains the dehydration on heating: CuSO₄·5H₂O → CuSO₄ (white anhydrous) + 5H₂O.

Observation (i) explanation:

Y contains [Cu(H₂O)₆]²⁺. Adding OH⁻ deprotonates two of the water ligands and produces a precipitate:

[Cu(H₂O)₆]²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(H₂O)₄ + 2H₂O(l)

usually written as Cu(OH)₂(s).

B1 — equation balanced.

In excess NaOH it remains insoluble (Cu(OH)₂ is not amphoteric).

In excess conc. NH₃, ligand exchange:

[Cu(OH)₂(H₂O)₄] + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2OH⁻ + 2H₂O

M1 (AO2) — equation balanced; product deep blue square-bipyramidal Jahn-Teller-distorted complex (4 NH₃ equatorial, 2 H₂O axial).

The deeper blue is because NH₃ is higher in the spectrochemical series than H₂O → larger Δ_oct → absorption shifts to shorter wavelength → complementary colour darker/more intense.

A1 (AO3) — Δ_oct argument with spectrochemical series.

Observation (ii) explanation:

[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O

B1 — equation balanced.

Excess Cl⁻ drives equilibrium to right (Le Chatelier); product is tetrahedral 4-coordinate. Cl⁻ is much larger than H₂O (~0.18 nm vs ~0.14 nm) — only four can fit. Δ_t = ⁴⁄₉ Δ_oct, so the d-d absorption shifts to longer wavelength (lower energy); complementary colour shifts from blue (high Δ absorbed) to yellow-green (lower Δ absorbed).

M1 (AO2) — geometry change explained.

A1 (AO3) — colour shift quantified.

On dilution, [Cl⁻] falls; equilibrium shifts back to [Cu(H₂O)₆]²⁺; pale blue restored.

Observation (iii) explanation:

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) (white)

B1 — equation balanced.

BaSO₄ is insoluble in dilute HCl (Ksp ~ 1.1 × 10⁻¹⁰); the acid does not protonate SO₄²⁻ enough to dissolve. This is the diagnostic test for sulfate.

Observation (iv) explanation:

CuSO₄·5H₂O(s) → CuSO₄(s) + 5H₂O(g)

B1 — equation balanced.

Removal of water ligands changes the coordination environment of Cu²⁺ from [Cu(H₂O)₆]²⁺ to a more covalent solid. Without the water ligands, the Cu²⁺ no longer experiences the same ligand-field splitting; the d–d absorption shifts and intensity drops. The white anhydrous CuSO₄ regains blue colour on rehydration — this is the classic test for water.