## CHM 001 – General Chemistry
### QUESTION 1
**(a)(i)** State the Periodic Law. [2 marks]
The Periodic Law states that the physical and chemical properties of elements are a periodic function of their atomic numbers. When elements are arranged in order of increasing atomic number, elements with similar properties recur at regular intervals.
---
**(a)(ii)** Explain the organisation of the periodic table in terms of groups, periods, and blocks (s, p, d, f). [6 marks]
**Groups:**
Vertical columns numbered 1–18. Elements within the same group possess the same number of valence electrons, the same valence electron configuration, and consequently exhibit similar chemical properties and reactivity trends.
**Periods:**
Horizontal rows numbered 1–7. All elements in the same period have the same number of occupied electron shells (principal quantum number n). Across a period from left to right, atomic number increases by one at each step, and properties change progressively from metallic to non-metallic.
**Blocks:**
- **s-block:** Groups 1 and 2 (plus helium). The outermost electrons occupy s-orbitals. These are highly reactive metals (and helium, a noble gas).
- **p-block:** Groups 13–18. The outermost electrons occupy p-orbitals. Includes metals, metalloids, non-metals, and noble gases.
- **d-block:** Groups 3–12 (transition metals). Electrons fill d-orbitals. Characterised by multiple oxidation states, coloured compounds, and catalytic activity.
- **f-block:** Lanthanides (Period 6) and actinides (Period 7), placed below the main table. Electrons fill f-orbitals.
---
**(a)(iii)** Compare the properties of calcium (Ca) and bromine (Br) with respect to metallic character, electronegativity, and bonding type. [7 marks]
**Metallic character:**
Calcium (Group 2, Period 4) is a metal — it has a low ionisation energy, readily loses its two valence electrons, and exhibits typical metallic properties (lustrous, conducts electricity, malleable). Bromine (Group 17, Period 4) is a non-metal — it has a high ionisation energy, tends to gain electrons rather than lose them, and lacks metallic properties.
**Electronegativity:**
Calcium has a low electronegativity (Pauling value ≈ 1.0), reflecting its tendency to donate electrons. Bromine has a high electronegativity (Pauling value ≈ 2.96), reflecting its strong tendency to attract bonding electrons toward itself. Electronegativity increases across a period (left to right) and decreases down a group, explaining this contrast.
**Bonding type:**
Calcium forms predominantly **ionic bonds** — it donates electrons to electronegative non-metals to form Ca²⁺ cations (e.g., CaBr₂, CaCl₂). Bromine forms **covalent bonds** when combining with other non-metals (e.g., Br₂, HBr, CBr₄) due to similar electronegativities, but also forms ionic bonds as the bromide anion (Br⁻) in salts with metals such as calcium.
---
**(b)** Balance the following redox equation using the ion-electron method and identify the oxidising and reducing agents: [8 marks]
MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O
**Step 1 — Write the two half-equations:**
*Reduction half-reaction (MnO₄⁻ reduced to Mn²⁺):*
MnO₄⁻ → Mn²⁺
Balance O by adding H₂O; balance H by adding H⁺; balance charge by adding e⁻:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
*Oxidation half-reaction (Fe²⁺ oxidised to Fe³⁺):*
Fe²⁺ → Fe³⁺ + e⁻
**Step 2 — Equalise electrons transferred:**
Multiply the oxidation half-reaction by 5:
5Fe²⁺ → 5Fe³⁺ + 5e⁻
**Step 3 — Add the half-equations:**
MnO₄⁻ + 8H⁺ + 5e⁻ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ + 5e⁻
Cancel electrons:
**Balanced equation:**
**MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O**
- **Oxidising agent:** MnO₄⁻ (gains electrons; Mn reduced from +7 to +2)
- **Reducing agent:** Fe²⁺ (loses electrons; Fe oxidised from +2 to +3)
---
**(c)(i)** Calculate the number of moles of explosive that decomposed. [3 marks]
2C₃H₆N₆O₆(s) → 3N₂(g) + 6H₂O(g) + 6CO(g)
Molar mass of C₃H₆N₆O₆:
= (3×12) + (6×1) + (6×14) + (6×16)
= 36 + 6 + 84 + 96
= **222 g/mol**
Moles = mass / molar mass = 4.56 / 222
**= 0.02054 mol ≈ 0.0205 mol**
---
**(c)(ii)** Calculate the total volume of gases produced at STP. [4 marks]
From the equation, 2 mol explosive produces (3 + 6 + 6) = 15 mol gas.
Moles of gas from 0.0205 mol explosive:
= 0.0205 × (15/2) = 0.0205 × 7.5 = **0.1538 mol**
Volume at STP (1 mol gas = 22.4 L):
V = 0.1538 × 22.4
**V = 3.44 L**
---
### QUESTION 2
**(a)(i)** Distinguish between empirical formula and molecular formula. [4 marks]
**Empirical formula:** The simplest whole-number ratio of atoms of each element present in a compound. It does not necessarily reflect the actual number of atoms in a molecule. For example, glucose (C₆H₁₂O₆) has the empirical formula CH₂O.
**Molecular formula:** The actual number of atoms of each element present in one molecule of the compound. It is a whole-number multiple of the empirical formula. For glucose, the molecular formula is C₆H₁₂O₆, which is 6 × CH₂O.
---
**(a)(ii)** A hydrocarbon contains 85.7% C and 14.3% H by mass. Its relative molecular mass is 84. Determine its empirical and molecular formulae. [11 marks]
**Step 1 — Assume 100 g sample:**
- C: 85.7 g
- H: 14.3 g
**Step 2 — Convert to moles:**
- Moles of C = 85.7 / 12 = 7.142 mol
- Moles of H = 14.3 / 1 = 14.3 mol
**Step 3 — Find simplest ratio:**
Divide by smallest (7.142):
- C: 7.142 / 7.142 = 1
- H: 14.3 / 7.142 = 2.0
Ratio C : H = 1 : 2
**Empirical formula = CH₂**
**Step 4 — Find molecular formula:**
Empirical formula mass = 12 + 2 = 14 g/mol
n = Relative molecular mass / Empirical formula mass = 84 / 14 = **6**
**Molecular formula = C₆H₁₂** (cyclohexane)
---
**(b)** Using VSEPR theory, predict and explain the molecular shapes of CH₄, NH₃, and H₂O. [9 marks]
**VSEPR theory** states that electron pairs (both bonding and lone pairs) around a central atom repel one another and arrange themselves to minimise repulsion, determining the molecular geometry.
**CH₄ — Methane:**
Carbon has 4 bonding pairs and 0 lone pairs around it (4 electron domains). To minimise repulsion, the 4 electron pairs adopt a **tetrahedral arrangement**.
Molecular shape: **Tetrahedral**
Bond angle: **109.5°**
All electron domains are equivalent bonding pairs, so the geometry is perfectly symmetrical.
**NH₃ — Ammonia:**
Nitrogen has 3 bonding pairs and 1 lone pair (4 electron domains total). The electron geometry is tetrahedral, but the molecular shape is determined by the positions of atoms only.
Lone pair–bonding pair repulsion is greater than bonding pair–bonding pair repulsion, compressing the H–N–H angles.
Molecular shape: **Trigonal pyramidal**
Bond angle: **107°**
**H₂O — Water:**
Oxygen has 2 bonding pairs and 2 lone pairs (4 electron domains total). The electron geometry is tetrahedral, but with two lone pairs the molecular shape is determined by the two O–H bonds only. The two lone pairs exert even greater repulsion on the bonding pairs, further compressing the bond angle.
Molecular shape: **Bent (V-shaped)**
Bond angle: **104.5°**
---
**(c)** Define the following terms and give ONE example of each: [6 marks]
**(i) Absolute error:**
The absolute error is the magnitude of the difference between a measured (experimental) value and the accepted true value. It expresses the size of the error in the same units as the measurement.
*Example:* If the true mass is 10.0 g and the measured mass is 10.2 g, then:
Absolute error = |10.2 − 10.0| = **0.2 g**
**(ii) Relative error:**
The relative error is the ratio of the absolute error to the true (accepted) value, usually expressed as a decimal or percentage. It gives the error in proportion to the size of the measurement.
*Example:* Using the values above:
Relative error = 0.2 / 10.0 = 0.02 (or 2%)
**(iii) Standard deviation:**
Standard deviation (σ or s) is a statistical measure of the spread or dispersion of a set of data values about their mean. A small standard deviation indicates that data values cluster closely around the mean; a large standard deviation indicates greater variability.
Formula: s = √[Σ(xᵢ − x̄)² / (n − 1)]
*Example:* For repeated titration readings of 24.1, 24.3, 24.2, and 24.2 mL, the standard deviation quantifies how consistently the readings cluster around the mean of 24.2 mL.
---
## CHM 002 – Physical Chemistry
### QUESTION 3
**(a)(i)** Define a buffer solution. [2 marks]
A buffer solution is a solution that resists significant changes in pH upon the addition of small amounts of strong acid or strong base, or upon dilution. It typically consists of a weak acid and its conjugate base (acidic buffer), or a weak base and its conjugate acid (basic buffer), present in comparable concentrations.
---
**(a)(ii)** Explain the importance of buffer solutions in biological and industrial systems. [6 marks]
**Biological importance:**
1. **Blood pH regulation:** Human blood is maintained at pH 7.35–7.45 by the carbonic acid/bicarbonate buffer system (H₂CO₃/HCO₃⁻). Even small deviations cause acidosis or alkalosis, both of which are life-threatening. The buffer absorbs excess H⁺ or OH⁻ produced by metabolic processes.
2. **Intracellular and enzyme function:** Enzymes operate optimally within narrow pH ranges. Intracellular buffers (phosphate buffer system: H₂PO₄⁻/HPO₄²⁻) maintain the cytoplasmic pH necessary for normal enzyme activity, protein structure, and metabolic reactions.
3. **Digestive system:** Buffers in saliva (bicarbonate) help neutralise acids produced by oral bacteria, protecting teeth from erosion.
**Industrial importance:**
1. **Pharmaceutical manufacturing:** Buffer systems maintain the pH of drug formulations during production and storage, ensuring stability, efficacy, and safety of medicines (e.g., intravenous infusions, eye drops).
2. **Fermentation and biotechnology:** Industrial fermentation processes require stable pH environments for optimal microbial growth and product yield; buffer systems or controlled pH conditions are essential.
3. **Food preservation:** Buffers control pH in food products to inhibit microbial growth, preserve texture, flavour, and colour, and comply with food safety standards.
---
**(a)(iii)** Calculate the pH of a buffer containing 0.10 M CH₃COOH and 0.15 M CH₃COONa. (Kₐ = 1.8 × 10⁻⁵) [7 marks]
Using the Henderson–Hasselbalch equation:
pH = pKₐ + log([A⁻]/[HA])
**Step 1 — Calculate pKₐ:**
pKₐ = −log(1.8 × 10⁻⁵) = −log(1.8) − log(10⁻⁵) = −0.255 + 5 = **4.745**
**Step 2 — Substitute values:**
pH = 4.745 + log(0.15/0.10)
pH = 4.745 + log(1.5)
pH = 4.745 + 0.176
**pH = 4.92**
---
**(b)(i)** Define solubility product (Ksp). [2 marks]
The solubility product (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic salt in water at a given temperature. It equals the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the dissolution equation.
For AB(s) ⇌ A⁺(aq) + B⁻(aq): Ksp = [A⁺][B⁻]
---
**(b)(ii)** The solubility of AgCl in water at 25°C is 1.3 × 10⁻⁵ mol/L. Calculate the Ksp of AgCl. [6 marks]
Dissolution equation:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
If solubility = s = 1.3 × 10⁻⁵ mol/L, then:
[Ag⁺] = s = 1.3 × 10⁻⁵ mol/L
[Cl⁻] = s = 1.3 × 10⁻⁵ mol/L
Ksp = [Ag⁺][Cl⁻] = s²
= (1.3 × 10⁻⁵)²
**Ksp = 1.69 × 10⁻¹⁰**
---
**(b)(iii)** Will a precipitate form when equal volumes of 2.0 × 10⁻⁴ M AgNO₃ and 2.0 × 10⁻⁴ M NaCl are mixed? (Ksp of AgCl = 1.8 × 10⁻¹⁰) [7 marks]
**Step 1 — Find concentrations after mixing:**
When equal volumes are mixed, each concentration is halved:
[Ag⁺] = 2.0 × 10⁻⁴ / 2 = 1.0 × 10⁻⁴ M
[Cl⁻] = 2.0 × 10⁻⁴ / 2 = 1.0 × 10⁻⁴ M
**Step 2 — Calculate the ionic product (Q):**
Q = [Ag⁺][Cl⁻] = (1.0 × 10⁻⁴)(1.0 × 10⁻⁴) = 1.0 × 10⁻⁸
**Step 3 — Compare Q with Ksp:**
Q = 1.0 × 10⁻⁸ > Ksp = 1.8 × 10⁻¹⁰
Since Q > Ksp, the solution is supersaturated with respect to AgCl.
**A precipitate of AgCl will form.**
---
### QUESTION 4
**(a)(i)** Define an electrochemical cell. [2 marks]
An electrochemical cell is a device in which chemical energy is converted into electrical energy (galvanic/voltaic cell) or in which electrical energy is used to drive a non-spontaneous chemical reaction (electrolytic cell), both processes involving the transfer of electrons through an external circuit driven by oxidation–reduction (redox) reactions.
---
**(a)(ii)** Distinguish between galvanic and electrolytic cells with examples. [6 marks]
| Feature | Galvanic (Voltaic) Cell | Electrolytic Cell |
|---|---|---|
| Energy conversion | Chemical → Electrical | Electrical → Chemical |
| Reaction spontaneity | Spontaneous (ΔG < 0) | Non-spontaneous (ΔG > 0) |
| External power | Not required; generates EMF | Requires external power source |
| Anode sign | Negative (−) | Positive (+) |
| Cathode sign | Positive (+) | Negative (−) |
| Example | Daniell cell (Zn/Cu²⁺); dry cell battery | Electrolysis of molten NaCl; electroplating |
**Galvanic cell example — Daniell cell:**
Zinc anode oxidises spontaneously (Zn → Zn²⁺ + 2e⁻) and copper cathode reduces Cu²⁺ (Cu²⁺ + 2e⁻ → Cu), generating an EMF of approximately 1.10 V.
**Electrolytic cell example — Electrolysis of molten NaCl:**
An external current forces Na⁺ to be reduced at the cathode (Na⁺ + e⁻ → Na) and Cl⁻ to be oxidised at the anode (2Cl⁻ → Cl₂ + 2e⁻), producing sodium metal and chlorine gas — a reaction that would not occur spontaneously.
---
**(a)(iii)** Explain the electrochemical process involved in the rusting of iron. [7 marks]
Rusting is an electrochemical process that occurs when iron is exposed to oxygen and moisture (water). The surface of iron acts as a short-circuited electrochemical cell with anodic and cathodic regions:
**Anodic regions (oxidation):**
At areas of the iron surface with lower electrode potential (e.g., regions of stress, grain boundaries, or impurities), iron is oxidised:
Fe(s) → Fe²⁺(aq) + 2e⁻
**Cathodic regions (reduction):**
Electrons flow through the iron to adjacent cathodic regions where dissolved oxygen is reduced in the presence of water:
O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) (in acidic conditions)
or
O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq) (in neutral/alkaline conditions)
**Formation of rust:**
Fe²⁺ ions migrate through the electrolyte (moisture) and react with OH⁻ or O₂:
Fe²⁺ + 2OH⁻ → Fe(OH)₂
Fe(OH)₂ is further oxidised to Fe(OH)₃, which dehydrates to form hydrated iron(III) oxide — rust:
4Fe(OH)₂ + O₂ + 2H₂O → 4Fe(OH)₃
2Fe(OH)₃ → Fe₂O₃·3H₂O (rust)
The overall cell reaction is:
4Fe + 3O₂ + 2nH₂O → 2Fe₂O₃·nH₂O
Electrolytes in the moisture (e.g., dissolved salts, acids from atmospheric CO₂) accelerate rusting by increasing the conductivity of the electrolyte solution.
---
**(b)(i)** State Faraday's First Law of Electrolysis. [2 marks]
Faraday's First Law of Electrolysis states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte.
Mathematically: m ∝ Q, or m = (Q × M) / (n × F)
where m = mass deposited, Q = charge in coulombs, M = molar mass, n = number of electrons transferred per ion, F = Faraday constant (96,500 C/mol).
---
**(b)(ii)** Calculate the mass of copper deposited at the cathode when a current of 2.5 A is passed through CuSO₄ solution for 3 hours. (Cu = 64, F = 96,500 C/mol) [8 marks]
**Step 1 — Calculate total charge:**
Q = I × t = 2.5 × (3 × 3600) = 2.5 × 10,800 = **27,000 C**
**Step 2 — Calculate moles of electrons:**
Moles of e⁻ = Q / F = 27,000 / 96,500 = **0.2798 mol**
**Step 3 — Apply electrode reaction:**
Cu²⁺ + 2e⁻ → Cu
2 moles of electrons deposit 1 mole of copper:
Moles of Cu = 0.2798 / 2 = **0.1399 mol**
**Step 4 — Calculate mass:**
Mass = moles × molar mass = 0.1399 × 64
**Mass of copper deposited = 8.95 g**
---
**(c)(i)** Calculate the mass remaining after 24 days. [3 marks]
Half-life = 8 days; initial mass = 80 g
Number of half-lives in 24 days = 24 / 8 = **3**
Mass remaining = 80 × (1/2)³ = 80 × 1/8
**Mass remaining = 10 g**
---
**(c)(ii)** Calculate the time taken for the mass to reduce to 5 g. [3 marks]
Using: m = m₀ × (1/2)^(t/t½)
5 = 80 × (1/2)^(t/8)
(1/2)^(t/8) = 5/80 = 1/16 = (1/2)⁴
Therefore: t/8 = 4
**t = 32 days**
---
## CHM 003 – Inorganic and Organic Chemistry
### QUESTION 5
**(a)(i)** Define allotropy. [2 marks]
Allotropy is the existence of an element in two or more physically and structurally distinct forms (allotropes) in the same physical state, differing in the arrangement or bonding of their atoms and consequently in their physical and chemical properties. Example: carbon exists as diamond, graphite, and fullerene.
---
**(a)(ii)** Describe the structure and properties of TWO allotropes of carbon. [9 marks]
**Diamond:**
*Structure:* Each carbon atom is sp³ hybridised and forms four strong covalent bonds to four other carbon atoms arranged tetrahedrally. This creates an infinite, rigid, three-dimensional network covalent lattice with no free electrons. The C–C bond length is 0.154 nm.
*Properties:*
- Hardest known natural substance (Mohs hardness 10) due to the extensive 3D network of strong covalent bonds in all directions.
- Very high melting point (~3550°C) because an enormous amount of energy is required to break the covalent bonds.
- Electrical insulator — all four valence electrons are locked in covalent bonds; no free electrons for conduction.
- Transparent and highly refractive, making it valuable as a gemstone.
- Chemically inert under normal conditions.
**Graphite:**
*Structure:* Each carbon atom is sp² hybridised and forms three covalent bonds to three adjacent carbon atoms in the same plane, producing flat, extended hexagonal layers (graphene sheets). The fourth valence electron from each carbon is delocalised across the layer in a π-electron cloud. Adjacent layers are held together only by weak van der Waals (dispersion) forces. Layer spacing is 0.335 nm; C–C bond length within layers is 0.142 nm.
*Properties:*
- Soft and slippery — layers slide easily over one another because the interlayer van der Waals forces are weak, making graphite an excellent solid lubricant.
- Good electrical conductor — the delocalised π electrons are mobile within each layer, enabling electron flow parallel to the layers.
- High melting point — strong covalent bonds within layers require substantial energy to break.
- Opaque black-grey solid; lower density than diamond.
- Used in pencils, electrodes, lubricants, and as a moderator in nuclear reactors.
---
**(a)(iii)** Explain why diamond is hard while graphite is soft despite both being forms of carbon. [4 marks]
In **diamond**, every carbon atom is covalently bonded to four others in a rigid, continuous three-dimensional tetrahedral network. To deform or scratch diamond, strong covalent bonds must be broken in all three dimensions simultaneously — an extremely energy-demanding process. This makes diamond exceptionally hard.
In **graphite**, the carbon atoms within each layer are strongly bonded to three neighbours by covalent bonds, but adjacent layers are held to each other only by weak van der Waals (London dispersion) forces. These interlayer forces require very little energy to overcome, so the layers slide past one another with ease under shear stress. This accounts for graphite's characteristic softness and lubricating properties.
---
**(b)(i)** Define amphoteric oxides and give TWO examples. [4 marks]
Amphoteric oxides are metal oxides that react chemically with both acids and bases to form salt and water. They exhibit both acidic and basic behaviour depending on the reaction conditions.
**Examples:**
1. Aluminium oxide (Al₂O₃)
2. Zinc oxide (ZnO)
---
**(b)(ii)** Write balanced equations to show the amphoteric nature of aluminium oxide (Al₂O₃). [11 marks]
**Reaction with acid (behaving as a base):**
Al₂O₃ acts as a basic oxide and reacts with hydrochloric acid:
Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)
With sulphuric acid:
Al₂O₃(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂O(l)
**Reaction with base (behaving as an acid):**
Al₂O₃ acts as an acidic oxide and reacts with sodium hydroxide:
Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq)
Or in concentrated NaOH solution:
Al₂O₃(s) + 2NaOH(aq) → 2NaAlO₂(aq) + H₂O(l)
(forming sodium aluminate)
**Ionic equations:**
With acid: Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O
With base: Al₂O₃ + 2OH⁻ + 3H₂O → 2[Al(OH)₄]⁻
These equations demonstrate that Al₂O₃ neutralises both acids and bases — confirming its amphoteric character.
---
### QUESTION 6
**(a)(i)** What is the iodoform test? [2 marks]
The iodoform test is a chemical test used to identify compounds containing a methyl ketone group (CH₃CO–) or a secondary alcohol that can be oxidised to a methyl ketone (CH₃CH(OH)–). The compound is treated with iodine (I₂) in alkaline solution (NaOH). A **positive result** is indicated by the formation of a pale yellow crystalline precipitate of iodoform (triiodomethane, CHI₃), which also has a characteristic antiseptic odour.
---
**(a)(ii)** Write the structural formulae of THREE alcohols that give a positive iodoform test. [6 marks]
A positive iodoform test is given by ethanol and secondary alcohols of the form CH₃CH(OH)R (which are oxidised in situ to methyl ketones):
**1. Ethanol (CH₃CH₂OH):**
```
H H
| |
H — C — C — OH
| |
H H
```
CH₃CH₂OH
**2. Propan-2-ol (isopropanol, CH₃CH(OH)CH₃):**
```
H OH H
| | |
H — C — C — C — H
| | |
H H H
```
CH₃CH(OH)CH₃
**3. Butan-2-ol (CH₃CH(OH)CH₂CH₃):**
```
CH₃ — CH(OH) — CH₂ — CH₃
```
All three possess either the CH₃CH₂OH group (ethanol) or the CH₃CH(OH)– grouping required for a positive iodoform test.
---
**(a)(iii)** Write the equation for the oxidation of ethanol with acidified potassium dichromate. [7 marks]
**Stage 1 — Oxidation to ethanal (partial oxidation):**
CH₃CH₂OH + [O] → CH₃CHO + H₂O
Using K₂Cr₂O₇/H₂SO₄ (distil immediately to prevent further oxidation):
3CH₃CH₂OH + K₂Cr₂O₇ + 4H₂SO₄ → 3CH₃CHO + K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O
(Orange Cr₂O₇²⁻ reduced to green Cr³⁺)
**Stage 2 — Further oxidation to ethanoic acid (under reflux):**
CH₃CHO + [O] → CH₃COOH
**Overall oxidation of ethanol to ethanoic acid:**
3CH₃CH₂OH + 2K₂Cr₂O₇ + 8H₂SO₄ → 3CH₃COOH + 2K₂SO₄ + 2Cr₂(SO₄)₃ + 11H₂O
The colour change from orange (dichromate) to green (Cr³⁺) confirms oxidation has occurred.
---
**(b)(i)** Define geometrical isomerism. [2 marks]
Geometrical isomerism (also called cis-trans isomerism) is a type of stereoisomerism in which compounds have the same molecular formula and the same sequence of bonded atoms (constitutional structure), but differ in the spatial arrangement of groups or atoms about a bond or ring that restricts rotation. The two isomers are non-superimposable and cannot be interconverted without breaking a bond.
---
**(b)(ii)** Explain why geometrical isomerism occurs in alkenes but not in alkanes. [4 marks]
In **alkenes**, the carbon–carbon double bond (C=C) consists of a σ bond and a π bond. The π bond arises from sideways overlap of p orbitals above and below the σ bond framework. Rotation about the C=C bond would break the π overlap, requiring ~264 kJ/mol — far more energy than is available at room temperature. Consequently, the double bond is rigid and rotation is effectively prevented. This restricted rotation means that groups attached to the two sp² carbon atoms are locked in fixed spatial positions, giving rise to distinct cis and trans arrangements when each carbon bears two different substituents.
In **alkanes**, the carbon–carbon single bond is a σ bond only, formed by end-on overlap of orbitals. There is free rotation about the C–C single bond at room temperature because rotation does not disrupt the σ overlap. As a result, any spatial arrangement about a C–C single bond is rapidly and continuously interconverted by rotation, and no distinct fixed isomers can be isolated.
---
**(b)(iii)** Draw and name the geometrical isomers of but-2-ene. [9 marks]
But-2-ene: CH₃–CH=CH–CH₃ (double bond between C2 and C3)
Each doubly bonded carbon bears one CH₃ group and one H atom — two different groups on each carbon — fulfilling the condition for geometrical isomerism.
**cis-but-2-ene (Z-but-2-ene):**
Both CH₃ groups are on the **same side** of the double bond.
```
CH₃ CH₃
\ /
C = C
/ \
H H
```
Name: **cis-but-2-ene** (or (Z)-but-2-ene)
Boiling point: 3.7°C; has a small dipole moment.
**trans-but-2-ene (E-but-2-ene):**
The CH₃ groups are on **opposite sides** of the double bond.
```
CH₃ H
\ /
C = C
/ \
H CH₃
```
Name: **trans-but-2-ene** (or (E)-but-2-ene)
Boiling point: 0.9°C; dipole moment is approximately zero due to symmetry.
Both isomers have molecular formula C₄H₈ but differ in physical properties (boiling points, melting points, dipole moments) and cannot be interconverted without breaking the π bond.
---
**(c)(i)** Distinguish between cracking and reforming of petroleum. [4 marks]
**Cracking:**
Cracking is a thermal or catalytic process in which large, high-molecular-mass hydrocarbon molecules (long-chain alkanes from higher fractions of crude oil) are broken down into smaller, more useful molecules — including shorter-chain alkanes, alkenes (such as ethylene and propylene), and hydrogen. It involves C–C bond cleavage.
*Types:* Thermal cracking (high temperature, ~500°C, high pressure) and catalytic cracking (lower temperature with a zeolite catalyst, produces higher octane products).
**Reforming:**
Reforming (catalytic reforming) is a process in which straight-chain hydrocarbons (particularly naphtha fractions, C₆–C₁₀ alkanes) are structurally rearranged — without changing molecular formula — to form branched-chain alkanes, cycloalkanes, or aromatic compounds (benzene, toluene, xylene). It does not primarily break C–C bonds but rearranges them. A platinum-based catalyst is typically used at ~500°C and moderate pressure. It improves the octane rating of petrol.
**Key distinction:** Cracking reduces molecular size (breaks large molecules into smaller ones); reforming rearranges molecular structure without necessarily changing carbon chain length.
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**(c)(ii)** State the industrial importance of cracking. [2 marks]
1. **Increases petrol yield:** Cracking converts less economically valuable heavy fractions (gas oil, residual fuel oil) from crude oil distillation into high-demand lighter fractions, particularly high-octane petrol (gasoline), greatly increasing the yield of fuel from each barrel of crude oil.
2. **Produces petrochemical feedstocks:** Cracking generates short-chain alkenes — especially ethylene (ethene), propylene (propene), and butylene — which are essential raw materials for the manufacture of plastics (polyethylene, polypropylene), synthetic rubber, solvents, detergents, and numerous other organic chemicals.