## CHM 001: GENERAL CHEMISTRY
**3(a)**
(i) What is an energy level?
(ii) List two differences between ground state and excited state. **[2 marks]**
**(b)** Manganate ion, MnO₄⁻ reacts with Nitrogen (IV) oxide gas in medium to give Mn²⁺ and NO₃⁺, in the determination of NO (pollutant) in a sample of air collected from an agrochemical industry, 10.5 cm³ of 4.5×10⁻⁴ M KMnO₄ solution (in acid) require for equivalence.
(i) Write a balanced ionic equation for each half reaction above, hence the net ionic equation. **[2 marks]**
(ii) Calculate the number of moles of NO₂ gas in the air sample. **[1 mark]**
**(c)** Calculate the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from the energy level n = 3 to energy level n = 2?
(Given R_H = 169678 cm⁻¹) **[2 marks]**
**(d)** Define electronegativity and justify its trend in the periodic table. **[2 marks]**
**(e)** Write the chemical formula of ammonium chloride and name the types of bonds in it. **[1 mark]**
**(f)** Define the term precision.
**(g)** Four replicate measurements were obtained from the determination of zinc in water sample as follows: 0.520 mg, 0.519 mg, and 0.522 mg. Determine the mean absolute error of the measurement if the acceptable true value is 0.521 mg. **[3 marks]**
---
## (Organic Chemistry Section — Image 1)
**(b)** Explain the following terms with their examples:
(i) Electron donating substituent
(ii) Electron withdrawing substituent **[2 marks]**
**(c)** Write an equation to show how and under what condition Propanoic acid can be converted to:
(i) CH₃CH(Cl)COOH
(ii) CH₃CH₂CH₂OH
(iii) CH₃CH₂COOC₂H₅
(iv) CH₃CH₂COCl **[2 marks]**
**(d)** A compound J is an alcohol with molecular formula C₄H₁₀O. On oxidation gave K (C₄H₈O). K reacts with 2,4-dinitrophenylhydrazine to form a yellow crystal solid and forms a silver mirror with ammoniacal silver nitrate.
(i) Deduce the structures of J and K. **[1 mark]**
(ii) Write an equation to show how K reacts with 2,4-dinitrophenylhydrazine and name the product formed. **[2 marks]**
**(a)** Draw the structure of the following compounds:
(i) 2, 2, 9–trimethyldeca-3, 7-diene
(ii) Ethylbenzene
(iii) Heptan-1, 3-triol **[1½ marks]**
**(b)(i)** Define petrochemicals. **[1 mark]**
**(b)(ii)** List three applications of petrochemicals. **[1½ marks]**
**(c)** Consider the reaction scheme below:
CH₃CH₂OH → (III) → CH₃CH₂COOH → (I, CH₃CH₂OH/H⁺) → U → (II) → CH₃CH₂COO... (V)
W ← (III) ← CH₃CH₂COOH
(I) Provide the name and structure of compound U.
---
## CHM 002: PHYSICAL CHEMISTRY
**(b)** With examples, briefly explain the following bonding types:
(i) Hydrogen bonding
(ii) Co-ordinate covalent bonding
(iii) Electrovalent bonding **[3 marks]**
**(c)(i)** Balance the redox reaction below in an acidic medium:
Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ **[3 marks]**
**(c)(ii)** Identify the reducing and oxidizing agents in the balanced equation from b(i). **[1 mark]**
**3(a)(i)** State Raoult's law.
**(a)(ii)** When 5.59 g of an organic compound was dissolved in 250 g of water, the boiling point of the solution was 100.13°C at 1 atm. Assuming the solute is non-volatile, calculate the molecular mass of the solute. If the empirical formula of this compound was found to be C₂H₄O, determine the molecular formula of the compound. [K_b = 0.512] **[4 marks]**
**(b)** Define buffer solution and give an example of buffer solutions.
**(c)** The pH of a 0.1 M solution of a monoprotic acid is 4.2. What is the value of Kₐ? **[3 marks]**
**(d)** The standard reduction potential of Cu²⁺/Cu and Ag⁺/Ag electrodes are +0.337 V and +0.799 V respectively.
- Using cell notation, construct a galvanic cell using these electrodes so that its standard e.m.f. is positive.
- For what concentration of Ag⁺ will the e.m.f. of the cell at 25°C be zero if the concentration of Cu²⁺ is 0.01 M? **[3 marks]**
**(e)** State the Le Chatelier's principle.
# Complete Solutions
-
## CHM 001: GENERAL CHEMISTRY
---
### 3(a)(i) What is an energy level?
An energy level (or shell) is a fixed region around the nucleus of an atom where electrons are found, each having a specific, quantized amount of energy. Electrons can only exist at these discrete energy states, not between them.
---
### 3(a)(ii) Two differences between Ground State and Excited State
| Ground State | Excited State |
|---|---|
| Lowest energy state of an electron | Higher energy state after absorbing energy |
| Most stable configuration | Unstable; electron returns to ground state quickly |
---
### 3(b)(i) Balanced Ionic Equations
**Half reactions:**
**Oxidation (NO₂ → NO₃⁺):**
$$\text{NO}_2 + \text{H}_2\text{O} \rightarrow \text{NO}_3^- + 3\text{H}^+ + e^-$$
**Reduction (MnO₄⁻ → Mn²⁺):**
$$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
**Net ionic equation** (multiply oxidation ×5, reduction ×1):
$$5\text{NO}_2 + \text{MnO}_4^- + \text{H}_2\text{O} + 3\text{H}^+ \rightarrow 5\text{NO}_3^- + \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
Simplified:
$$\boxed{5\text{NO}_2 + \text{MnO}_4^- \rightarrow 5\text{NO}_3^- + \text{Mn}^{2+} + \text{products balanced with H}^+/\text{H}_2\text{O}}$$
### 3(b)(ii) Moles of NO₂
$$n(\text{KMnO}_4) = C \times V = 4.5\times10^{-4} \times \frac{10.5}{1000} = 4.725\times10^{-6} \text{ mol}$$
From the equation, ratio of NO₂ : MnO₄⁻ = 5 : 1
$$n(\text{NO}_2) = 5 \times 4.725\times10^{-6} = \boxed{2.3625\times10^{-5} \text{ mol}}$$
---
### 3(c) Wavelength of light emitted (n=3 → n=2)
Using the Rydberg formula:
$$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
$$\frac{1}{\lambda} = 169678\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 169678\left(\frac{1}{4} - \frac{1}{9}\right)$$
$$= 169678 \times \frac{5}{36} = 169678 \times 0.13889 = 23566.4 \text{ cm}^{-1}$$
$$\lambda = \frac{1}{23566.4} = 4.244\times10^{-5} \text{ cm} = \boxed{424.4 \text{ nm}}$$
(This is in the visible red-orange region — Balmer series)
---
### 3(d) Electronegativity
**Definition:** Electronegativity is the ability of an atom in a molecule to attract shared electrons (bonding pair) towards itself.
**Periodic Trend:**
- **Across a period (left → right):** Electronegativity **increases** because nuclear charge increases while atomic radius decreases, so the nucleus attracts bonding electrons more strongly.
- **Down a group (top → bottom):** Electronegativity **decreases** because atomic radius increases and electron shielding increases, reducing nuclear attraction on bonding electrons.
---
### 3(e) Ammonium Chloride
**Formula:** NH₄Cl
**Bonds present:**
- **Covalent bonds** — between N and three H atoms (N–H)
- **Co-ordinate (dative) covalent bond** — between N and the fourth H⁺ (N donates lone pair)
- **Electrovalent (ionic) bond** — between NH₄⁺ and Cl⁻
---
### 3(f) Precision
**Definition:** Precision is the closeness of agreement between repeated measurements of the same quantity under the same conditions. It indicates reproducibility, regardless of whether the values are close to the true value.
---
### 3(g) Mean Absolute Error
Measurements: 0.520, 0.519, 0.522 mg (note: only 3 values are clearly readable)
**Mean:**
$$\bar{x} = \frac{0.520 + 0.519 + 0.522}{3} = \frac{1.561}{3} = 0.5203 \text{ mg}$$
**Absolute errors** (|measured − true value|, true = 0.521):
| Measurement | |x − true| |
|---|---|
| 0.520 | 0.001 |
| 0.519 | 0.002 |
| 0.522 | 0.001 |
$$\text{Mean Absolute Error} = \frac{0.001 + 0.002 + 0.001}{3} = \frac{0.004}{3} = \boxed{0.00133 \text{ mg}}$$
---
---
## ORGANIC CHEMISTRY
---
### (b) Electron Donating vs Withdrawing Substituents
**(i) Electron Donating Substituent (EDG):**
A group that releases electron density into an aromatic ring or molecule, increasing electron density and stabilizing positive charge. They activate the ring toward electrophilic substitution.
**Example:** –OH, –NH₂, –OCH₃, –CH₃
**(ii) Electron Withdrawing Substituent (EWG):**
A group that pulls electron density away from the ring or molecule, decreasing electron density and destabilizing it toward electrophilic substitution (deactivating).
**Example:** –NO₂, –COOH, –CHO, –CN, –SO₃H
---
### (c) Conversion of Propanoic Acid (CH₃CH₂COOH)
**(i) → CH₃CH(Cl)COOH** (α-chloro propanoic acid)
$$\text{CH}_3\text{CH}_2\text{COOH} + \text{Cl}_2 \xrightarrow{\text{red P / heat}} \text{CH}_3\text{CH(Cl)COOH} + \text{HCl}$$
**Condition:** Red phosphorus (Hell-Volhard-Zelinsky reaction)
**(ii) → CH₃CH₂CH₂OH** (propan-1-ol)
$$\text{CH}_3\text{CH}_2\text{COOH} + 4[\text{H}] \xrightarrow{\text{LiAlH}_4 / \text{dry ether}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{H}_2\text{O}$$
**Condition:** LiAlH₄ (lithium aluminium hydride) in dry ether
**(iii) → CH₃CH₂COOC₂H₅** (ethyl propanoate)
$$\text{CH}_3\text{CH}_2\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, \Delta} \text{CH}_3\text{CH}_2\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$
**Condition:** Concentrated H₂SO₄ catalyst, heat (Fischer esterification)
**(iv) → CH₃CH₂COCl** (propanoyl chloride)
$$\text{CH}_3\text{CH}_2\text{COOH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{CH}_2\text{COCl} + \text{POCl}_3 + \text{HCl}$$
**Condition:** PCl₅ or SOCl₂
---
### (d) Compound J and K
**Given:** J = C₄H₁₀O (alcohol), oxidized to K = C₄H₈O. K gives yellow precipitate with 2,4-DNPH (∴ K is a carbonyl) AND gives silver mirror test (∴ K is an **aldehyde**).
**K must be:** Butanal — **CH₃CH₂CH₂CHO**
**J must be:** Butan-1-ol — **CH₃CH₂CH₂CH₂OH**
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{[\text{O}], K_2\text{Cr}_2\text{O}_7/\text{H}^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$$
---
**(d)(ii) Reaction of K with 2,4-DNPH:**
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} + \text{2,4-(NO}_2)_2\text{C}_6\text{H}_3\text{NHNH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH=N-NH-C}_6\text{H}_3(\text{NO}_2)_2 + \text{H}_2\text{O}$$
**Product:** Butanal 2,4-dinitrophenylhydrazone (a yellow crystalline solid)
---
### Petrochemicals
**(b)(i) Definition:** Petrochemicals are chemical compounds derived from petroleum (crude oil) or natural gas through refining and chemical processing.
**(b)(ii) Three Applications:**
1. Manufacture of plastics and polymers (e.g. polyethylene, PVC)
2. Production of fertilizers (e.g. ammonia, urea)
3. Manufacture of synthetic fibres (e.g. nylon, polyester)
---
### Reaction Scheme — Compound U
The scheme shows:
$$\text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\text{CH}_3\text{CH}_2\text{OH}/\text{H}^+} U$$
**U = Ethyl propanoate: CH₃CH₂COOC₂H₅**
This is an esterification reaction between propanoic acid and ethanol under acid catalysis.
---
---
## CHM 002: PHYSICAL CHEMISTRY
---
### (b) Types of Bonding
**(i) Hydrogen Bonding:**
A strong electrostatic intermolecular attraction between a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) and a lone pair on another electronegative atom.
**Example:** Between water molecules: O–H···O
**(ii) Co-ordinate (Dative) Covalent Bonding:**
A type of covalent bond where both electrons in the shared pair are donated by the same atom (the donor).
**Example:** Formation of NH₄⁺ — nitrogen donates its lone pair to H⁺
**(iii) Electrovalent (Ionic) Bonding:**
The electrostatic attraction between oppositely charged ions formed by complete transfer of electrons from a metal to a non-metal.
**Example:** NaCl — Na transfers one electron to Cl, forming Na⁺ and Cl⁻
---
### (c)(i) Balancing the Redox Reaction in Acidic Medium
$$\text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+}$$
**Reduction half (Cr):**
$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$
**Oxidation half (Fe):**
$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$ (×6)
$$6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-$$
**Balanced equation:**
$$\boxed{\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}}$$
---
### (c)(ii) Oxidizing and Reducing Agents
- **Oxidizing agent** (gains electrons, gets reduced): **Cr₂O₇²⁻** (dichromate ion)
- **Reducing agent** (loses electrons, gets oxidized): **Fe²⁺** (iron(II) ion)
---
### 3(a)(i) Raoult's Law
**Statement:** The partial vapour pressure of a solvent above a solution is equal to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
$$P_A = X_A \cdot P_A^°$$
Where P_A = partial vapour pressure, X_A = mole fraction of solvent, P°_A = vapour pressure of pure solvent.
---
### 3(a)(ii) Molecular Mass and Molecular Formula
**Given:** mass of solute = 5.59 g, mass of solvent = 250 g, ΔT_b = 100.13 − 100 = 0.13°C, K_b = 0.512
$$\Delta T_b = K_b \times m = K_b \times \frac{w \times 1000}{M \times W}$$
$$0.13 = 0.512 \times \frac{5.59 \times 1000}{M \times 250}$$
$$M = \frac{0.512 \times 5590}{0.13 \times 250} = \frac{2862.08}{32.5} = \boxed{88.07 \approx 88 \text{ g/mol}}$$
**Molecular formula from empirical formula C₂H₄O:**
Empirical mass = 2(12) + 4(1) + 16 = **44 g/mol**
$$n = \frac{88}{44} = 2$$
$$\boxed{\text{Molecular formula} = \text{C}_4\text{H}_8\text{O}_2}$$
---
### (b) Buffer Solution
**Definition:** A buffer solution is a solution that resists significant changes in pH when small amounts of acid or base are added to it.
**Examples:**
- Acidic buffer: CH₃COOH / CH₃COONa (acetic acid / sodium acetate)
- Basic buffer: NH₃ / NH₄Cl (ammonia / ammonium chloride)
---
### (c) Ka from pH
**Given:** pH = 4.2, [HA] = 0.1 M, monoprotic acid
$$[\text{H}^+] = 10^{-4.2} = 6.31\times10^{-5} \text{ M}$$
For a weak acid: HA ⇌ H⁺ + A⁻
$$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{(6.31\times10^{-5})^2}{0.1 - 6.31\times10^{-5}} \approx \frac{(6.31\times10^{-5})^2}{0.1}$$
$$K_a = \frac{3.98\times10^{-9}}{0.1} = \boxed{3.98\times10^{-8}}$$
---
### (d) Galvanic Cell with Cu²⁺/Cu and Ag⁺/Ag
**Standard potentials:**
- Cu²⁺/Cu: E° = +0.337 V
- Ag⁺/Ag: E° = +0.799 V
For positive e.m.f., Ag⁺/Ag is the **cathode** (reduction), Cu/Cu²⁺ is the **anode** (oxidation).
**Cell notation:**
$$\boxed{\text{Cu}(s) | \text{Cu}^{2+}(aq) || \text{Ag}^+(aq) | \text{Ag}(s)}$$
**Standard EMF:**
$$E°_{cell} = 0.799 - 0.337 = +0.462 \text{ V}$$
**Cell reaction:**
$$\text{Cu} + 2\text{Ag}^+ \rightarrow \text{Cu}^{2+} + 2\text{Ag}$$
**When EMF = 0, using Nernst equation** (n = 2, T = 25°C):
$$0 = 0.462 - \frac{0.0592}{2}\log\frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2}$$
$$\log\frac{0.01}{[\text{Ag}^+]^2} = \frac{0.462 \times 2}{0.0592} = 15.608$$
$$\frac{0.01}{[\text{Ag}^+]^2} = 10^{15.608} = 4.06\times10^{15}$$
$$[\text{Ag}^+]^2 = \frac{0.01}{4.06\times10^{15}} = 2.46\times10^{-18}$$
$$\boxed{[\text{Ag}^+] = 1.57\times10^{-9} \text{ M}}$$
---
### (e) Le Chatelier's Principle
**Statement:** If a system at equilibrium is subjected to a change in concentration, temperature, pressure, or volume, the system will shift its equilibrium position in the direction that tends to counteract or minimize that change, until a new equilibrium is established.