SECTION A

1. State the type of bonding in each of the following:

(a) CaF₂(s) (b) N₂(g) © Na(s) (d) O₂(g) (e) MgCl₂(s)

2. (a) Suggest the IUPAC names of two oxides of nitrogen that are neutral to moist litmus paper.

(b) Define the following terms: (i) Molecularity of a reaction (ii) Unit cell of a crystal (iii) Lattice defect

3. (a) Define an element.

(b) How many atoms are there in each of the following:

(i) 3 moles of scandium (ii) 32 g of hydrogen sulphide (iii) 13 g of silicon?

4. In the complex ion, [Zn(OH)₄]²⁻:

(a) Give the oxidation number of Zn in the complex.

(b) What is the coordination number of Zn in the complex ion?

(d) What is the shape of the ion?

5. A 1.32 g of magnesium was dissolved in dilute hydrochloric acid and the solution was heated in a stream of hydrogen chloride. If 5.26 g of an anhydrous metal chloride remained, what is the simplest formula of the metal chloride formed?

6. Consider the following elements in the periodic table: F, He, Cs, Cl, Mg and K. Which of them:

(a) has the lowest electronegativity value?

(b) has the highest electron affinity value?

(d) has the largest atomic size?

(e) has the highest ionisation potential?

7. (a) Explain briefly how change in temperature can affect solubility of a solute in a given solvent.

(b) If the solubility of copper(II) tetraoxosulphate(VI) pentahydrate (CuSO₄·5H₂O) at 30°C is 25 g per 100 g of water, what is the maximum mass of crystals that will be obtained from 10 g of the solution?

8. The isotope ⁴²₁₉K undergoes β-decay to form ⁴²₂₀Ca. If after 62.0 hours, 96.88% of ⁴²₂₀Ca was found to have undergone transformation, what is the half-life of the isotope?

9. State the reason for each of the following statements about alkali metals:

(a) They are univalent.

(b) They have poor complexing tendency.

(d) They have the lowest first ionisation enthalpy values in their respective periods.

(e) They are largely ionic in nature.

10. (a) Define rate of a chemical reaction.

(b) A drop of water of volume 0.05 cm³ from a pipette contains 3.0 × 10⁴ moles of hydrogen ion. If the rate of disappearance of the hydrogen ion is 1.00 × 10⁻⁷ dm³ mol⁻¹ s⁻¹, how long would it take for the hydrogen ion in the drop to disappear?

SECTION B

11. (a) Define the following terms (8 marks):

(i) Standard heat of formation of a substance

(ii) Standard heat of combustion

(iii) Standard heat of sublimation

(iv) Standard heat of atomisation

(b) State Hess’s law of heat summation. (2 marks)

Keto form (A): CH₃–CO–CH₂–CO–CH₃

Enol form (B): CH₃–C(OH)=CH–CO–CH₃

Given bond energy values: C–H = 435 kJ/mol; C–C = 368 kJ/mol; C=C = 610 kJ/mol; C–O = 357 kJ/mol; C=O = 748 kJ/mol; O–H = 462 kJ/mol

(i) Calculate the enthalpy change from the keto form (A) to the enol form (B). (7 marks)

(ii) Assuming the entropy change for the conversion of (A) to (B) is zero, calculate the equilibrium constant for the keto–enol equilibrium at 27°C. (8 marks)

12. (a) Write the electron configuration of the valence shell of the following group of elements in the periodic table: (i) group 2; (ii) group 13; (iii) group 15. (3 marks)

(b) Using an example of any member of group 2:

(i) Write the formula of an oxide of a group 2 element. (1 mark)

(ii) What is the bond type of the oxide in (i) above? (1 mark)

(iii) Write the chemical equation for the reaction of the oxide with water. (2 marks)

(iv) Give an equation of the reaction between HCl and the product formed in (iii) above. (2 marks)

(i) Whereas both zinc and copper have the same d electron configuration, zinc or its ion is a non-transition element but copper or its ion is regarded as a transition element. (4 marks)

(ii) The maximum covalency of carbon is four but that of silicon is six. (5 marks)

(iii) The melting point of AlF₃ is greater than that of AlCl₃. (5 marks)

(d) What do you understand by the term “catenation”? (2 marks)

13. (a) State five assumptions of kinetic theory of gases. (5 marks)

(b) Deduce the following gas laws from kinetic theory equation of an ideal gas (10 marks):

(i) Avogadro’s law (ii) Graham’s law of diffusion

14. (a) Explain the following concepts of acids and bases giving appropriate examples in each case. (9 marks):

(i) Arrhenius (ii) Lewis (iii) Brønsted–Lowry

(b) Classify the reactants in the following reactions as acid or base and identify the acid–base conjugate pairs:

(i) CH₃COO⁻ + HCN ⇌ CH₃COOH + CN⁻ (2 marks)

(ii) H₂PO₄⁻ + NH₃ ⇌ HPO₄²⁻ + NH₄⁺ (2 marks)

(iii) HClO + CH₃NH₂ ⇌ CH₃NH₃⁺ + ClO⁻ (2 marks)

(ii) Give two physical methods to show that HCl dissociates more in aqueous solution than methanoic acid. (4 marks)

15. (a) Give 4 postulates of Bohr atomic theory. (4 marks)

(b) State 2 each of the successes and limitations of Bohr atomic theory. (4 marks)

(d) Calculate the ionisation enthalpy of hydrogen atom. Given that the frequency of convergence limit of the Lyman series of a hydrogen atom is 3.29 × 10¹⁵ Hz. (4 marks)

(f) Explain the importance of n, l, m, and s quantum numbers in the orbital arrangement of electrons in atoms. (8 marks)

16. (a) Compound A, [FeBr(H₂O)₅]SO₄ is isomeric with Compound B, [FeSO₄(H₂O)₅]Br.

(i) What type of isomerism exists between compounds A and B? (1 mark)

(ii) What ions would these isomers yield in aqueous solution? (4 marks)

(iii) Using simple laboratory tests, explain how you can differentiate between the two isomers. (6 marks)

(iv) State the oxidation state, electron configuration and coordination number of Fe in compound A. (4 marks)

(b) Draw the structure of [FeBr(H₂O)₅]²⁺ and indicate all the coordinate covalent bonds involved. (2 marks)

(i) cationic complex of cobalt (ii) anionic complex of cobalt (iii) neutral complex of cobalt

(d) Give two examples of double salts. (2 marks)

SOLUTIONS

Q1 — Types of Bonding

(a) CaF₂(s) — Ionic bonding (Ca²⁺ and F⁻ ions held by electrostatic attraction)

(b) N₂(g) — Covalent bonding (triple bond between two N atoms; both non-metals)

(d) O₂(g) — Covalent bonding (double bond between two O atoms)

(e) MgCl₂(s) — Ionic bonding (Mg²⁺ and Cl⁻ ions)

(a) Two neutral oxides of nitrogen:

Dinitrogen oxide, N₂O (nitrous oxide)
Nitrogen(II) oxide, NO (nitric oxide)

(These are neutral — they do not react with moist litmus paper unlike NO₂ or N₂O₅)

(b) Definitions:

(i) Molecularity of a reaction — the number of molecules (or ions/atoms) that actually collide and participate in a single elementary step of a reaction.

(ii) Unit cell of a crystal — the smallest repeating structural unit of a crystalline solid that, when stacked in three dimensions, produces the entire crystal lattice.

(iii) Lattice defect — an irregularity or imperfection in the regular arrangement of atoms/ions in a crystal lattice (e.g., Schottky defect, Frenkel defect).

(a) Definition of an element: A pure substance that cannot be broken down into simpler substances by ordinary chemical means, and whose atoms all have the same atomic number.

(b) Number of atoms:

(i) 3 moles of scandium:

Number of atoms = 3 × 6.023 × 10²³ = 1.807 × 10²⁴ atoms

(ii) 32 g of hydrogen sulphide (H₂S): Molar mass of H₂S = 2(1) + 32 = 34 g/mol

Moles = 32/34 = 0.941 mol

Each molecule has 3 atoms (2H + 1S)

Number of atoms = 0.941 × 3 × 6.023 × 10²³ = 1.70 × 10²⁴ atoms

(iii) 13 g of silicon: Molar mass of Si = 28 g/mol

Moles = 13/28 = 0.464 mol

Number of atoms = 0.464 × 6.023 × 10²³ = 2.80 × 10²³ atoms

Q4 — [Zn(OH)₄]²⁻

(a) Oxidation number of Zn:

Let oxidation state of Zn = x

x + 4(−1) = −2

x − 4 = −2

x = +2

(b) Coordination number of Zn: 4 (four OH⁻ ligands bonded to Zn)

© Structure: Tetrahedral arrangement — four OH⁻ ligands each donate a lone pair to the central Zn²⁺ ion via coordinate (dative) covalent bonds:


[OH]

|

HO — Zn — OH (all four bonds are coordinate covalent bonds; OH⁻ is the donor)

|

[OH]

All four Zn–O bonds are coordinate covalent bonds (OH⁻ donates lone pairs to Zn²⁺).

(d) Shape: Tetrahedral (4 bonding pairs, no lone pairs on Zn in this complex)

Q5 — Formula of metal chloride

Given: 1.32 g Mg dissolved in HCl → anhydrous metal chloride = 5.26 g

Mass of Cl in the chloride = 5.26 − 1.32 = 3.94 g

Moles of Mg = 1.32/24 = 0.055 mol

Moles of Cl = 3.94/35.5 = 0.111 mol

Ratio Mg : Cl = 0.055 : 0.111 = 1 : 2

Simplest formula = MgCl₂

Q6 — Periodic trends among F, He, Cs, Cl, Mg, K

(a) Lowest electronegativity — Cs (furthest left and down in periodic table; large atomic radius, low nuclear pull on bonding electrons)

(b) Highest electron affinity — Cl (F has anomalously lower EA than Cl due to small size and inter-electron repulsion; Cl has highest EA among common elements)

(d) Largest atomic size — Cs (lowest in Group 1; most electron shells, weakest effective nuclear charge per shell)

(e) Highest ionisation potential — He (noble gas; full 1s² shell, highest IE of all elements)

Q7 — Solubility

(a) Effect of temperature on solubility:

For most solid solutes in liquid solvents, increasing temperature increases solubility because the dissolving process is usually endothermic — added heat energy disrupts the lattice and favours dissolution. However, for a few substances (e.g., Ce₂(SO₄)₃), solubility decreases with increasing temperature because their dissolution is exothermic. For gases dissolved in liquids, solubility always decreases with increasing temperature (increased kinetic energy allows gas molecules to escape).

(b) Mass of CuSO₄·5H₂O crystals from 10 g solution:

Solubility = 25 g CuSO₄·5H₂O per 100 g water

So in 125 g solution → 25 g CuSO₄·5H₂O

In 10 g solution → (25/125) × 10 = 2 g of CuSO₄·5H₂O

Maximum mass of crystals = 2.0 g

Q8 — Half-life of ⁴²K

Given: After 62.0 hours, 96.88% of ⁴²Ca formed (i.e., 96.88% of ⁴²K has decayed)

So fraction of ⁴²K remaining = 100 − 96.88 = 3.12% = 0.0312

Using: N/N₀ = (1/2)ⁿ where n = number of half-lives

0.0312 = (1/2)ⁿ

(1/2)ⁿ = 0.0312

n × ln(0.5) = ln(0.0312)

n = ln(0.0312)/ln(0.5) = (−3.467)/(−0.693) = 5.0

So n = 5 half-lives in 62.0 hours

t₁/₂ = 62.0/5 = 12.4 hours

Q9 — Alkali metals

(a) Univalent: Their outermost shell has only one electron (ns¹). Losing this one electron achieves a stable noble gas configuration, making +1 the only stable oxidation state.

(b) Poor complexing tendency: Their ions are large with low charge density (+1 charge over a large radius). They therefore have weak electrostatic attraction for ligands and cannot polarise them effectively.

(d) Lowest first ionisation enthalpy in their periods: Their valence electron is in the outermost shell, well-shielded from the nucleus by all inner electrons, with the lowest effective nuclear charge of all elements in that period.

(e) Largely ionic in nature: The large difference in electronegativity between alkali metals and non-metals (especially halogens and oxygen) means electron transfer rather than sharing occurs, forming ionic bonds.

Q10 — Reaction rate

(a) Definition: The rate of a chemical reaction is the change in concentration of a reactant or product per unit time.

(b) Time for hydrogen ion to disappear:

Volume of drop = 0.05 cm³ = 0.05 × 10⁻³ dm³ = 5 × 10⁻⁵ dm³

Moles of H⁺ = 3.0 × 10⁴ mol (as given)

Wait — re-reading: the drop contains 3.0 × 10⁴ moles seems unreasonable for 0.05 cm³. This is likely 3.0 × 10⁻⁴ mol (a typographical issue in the paper).

Concentration of H⁺ = moles/volume = (3.0 × 10⁻⁴)/(5 × 10⁻⁵) = 6.0 mol dm⁻³

Rate of disappearance = 1.00 × 10⁻⁷ mol dm⁻³ s⁻¹

Time = concentration/rate = 6.0/(1.00 × 10⁻⁷) = 6.0 × 10⁷ seconds

SECTION B SOLUTIONS

Q11 — Thermochemistry

(a) Definitions:

(i) Standard heat of formation — the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states at 298 K and 1 atm pressure.

(ii) Standard heat of combustion — the enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions (298 K, 1 atm).

(iii) Standard heat of sublimation — the enthalpy change when one mole of a solid is converted directly into gaseous state at standard conditions without passing through the liquid phase.

(iv) Standard heat of atomisation — the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state.

(b) Hess’s Law: The total enthalpy change of a chemical reaction is independent of the pathway taken — it depends only on the initial and final states.

©(i) Enthalpy change: Keto (A) → Enol (B)

Keto (A): CH₃–CO–CH₂–CO–CH₃

Bonds broken in A (going to atoms, then recounted in B):

The conversion: one C–H bond is broken, one C=O becomes C–OH (so C=O broken, O–H formed), and a C–C single bond becomes C=C.

Bonds broken in A → B:

1 C–H: +435 kJ/mol
1 C=O: +748 kJ/mol
1 C–C: +368 kJ/mol

Total broken = 435 + 748 + 368 = +1551 kJ/mol

Bonds formed in B:

1 O–H: −462 kJ/mol
1 C=C: −610 kJ/mol
1 C–O (single, now C–OH): −357 kJ/mol

Total formed = −(462 + 610 + 357) = −1429 kJ/mol

ΔH = Energy broken − Energy formed = 1551 − 1429 = +122 kJ/mol

The keto → enol conversion is endothermic by 122 kJ/mol.

©(ii) Equilibrium constant at 27°C (300 K):

Given: ΔS = 0, T = 300 K

ΔG = ΔH − TΔS = 122,000 − 300(0) = +122,000 J/mol

ΔG° = −RT ln K

122,000 = −(8.314)(300) ln K

ln K = −122,000/2494.2 = −48.91

K = e⁻⁴⁸·⁹¹ ≈ 6.4 × 10⁻²²

This extremely small K confirms the keto form is overwhelmingly favoured at equilibrium.

Q12 — Periodic Table / Group Chemistry

(a) Valence shell electron configurations:

(i) Group 2: ns²

(ii) Group 13: ns² np¹

(iii) Group 15: ns² np³

(b) Using Magnesium (Mg) as example of Group 2:

(i) Formula of oxide: MgO

(ii) Bond type: Ionic bond (Mg²⁺ and O²⁻)

(iii) Reaction with water:

MgO + H₂O → Mg(OH)₂

(iv) Reaction of Mg(OH)₂ with HCl:

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

© Accounts:

(i) Zn vs Cu as transition elements:

Transition elements are defined as those that form at least one stable ion with a partially filled d-subshell.

Cu (Z=29): [Ar] 3d¹⁰ 4s¹. Cu²⁺ ion: [Ar] 3d⁹ — partially filled d-subshell → transition element.
Zn (Z=30): [Ar] 3d¹⁰ 4s². Zn²⁺ ion: [Ar] 3d¹⁰ — completely filled d-subshell → non-transition element.

Even though both atoms have the same d¹⁰ configuration in their ground state, only Cu forms an ion (Cu²⁺) with an incomplete d shell.

(ii) Carbon max covalency 4, Silicon max covalency 6:

Carbon is in Period 2 and only has 2s and 2p orbitals in its valence shell — a total of 4 orbitals available for bonding, giving a maximum covalency of 4. Silicon is in Period 3 and has 3s, 3p, and 3d orbitals available. The empty 3d orbitals allow Si to expand its octet and accommodate up to 6 bond pairs (sp³d² hybridisation), giving a maximum covalency of 6.

(iii) AlF₃ melting point > AlCl₃:

AlF₃ is predominantly ionic due to the high electronegativity and small size of F, which polarises the Al–F bond minimally, keeping it strongly ionic. Ionic compounds have high melting points due to strong electrostatic lattice forces. AlCl₃, however, is largely covalent — Cl is larger and less electronegative than F, leading to greater covalent character (Fajans’ rules). Covalent AlCl₃ exists as Al₂Cl₆ dimers with weak van der Waals forces between molecules, hence a much lower melting point.

(d) Catenation: The ability of atoms of an element to form long chains or rings by bonding with other atoms of the same element. Carbon exhibits the most extensive catenation due to the strength and stability of the C–C bond, giving rise to the vast diversity of organic compounds.

Q13 — Kinetic Theory of Gases

(a) Five assumptions of kinetic theory:

Gases consist of a very large number of tiny particles (molecules) in constant, rapid, random motion.
The volume of gas molecules is negligible compared to the total volume occupied by the gas.
There are no intermolecular forces of attraction or repulsion between gas molecules (except during collisions).
Collisions between gas molecules and with the walls of the container are perfectly elastic (no net loss of kinetic energy).
The average kinetic energy of the molecules is directly proportional to the absolute temperature of the gas.

(b) Deductions:

(i) Avogadro’s Law:

From kinetic theory: PV = ⅓Nmv² (where N = number of molecules, m = mass of one molecule, v² = mean square speed)

Also, average KE = ½mv² = (3/2)kT

So PV = NkT

For two different gases at same P and T occupying same V:

N₁kT = N₂kT → N₁ = N₂

Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules — Avogadro’s Law.

(ii) Graham’s Law of Diffusion:

Rate of diffusion r ∝ average speed of molecules.

From kinetic theory: ½mv²(rms) = (3/2)kT → v(rms) = √(3kT/m) = √(3RT/M) where M = molar mass.

For two gases at same T and P:

r₁/r₂ = v₁/v₂ = √(M₂/M₁)

Rate of diffusion is inversely proportional to the square root of molar mass — Graham’s Law.

Q14 — Acids and Bases

(a) Concepts:

(i) Arrhenius theory: An acid is a substance that produces H⁺ (proton) ions in aqueous solution; a base produces OH⁻ ions. Example: HCl → H⁺ + Cl⁻ (acid); NaOH → Na⁺ + OH⁻ (base). Limitation: applies only to aqueous solutions.

(ii) Lewis theory: An acid is an electron-pair acceptor; a base is an electron-pair donor. Example: BF₃ (acid) + NH₃ (base) → BF₃·NH₃. This is the broadest definition, covering non-aqueous and non-proton systems.

(iii) Brønsted–Lowry theory: An acid is a proton (H⁺) donor; a base is a proton acceptor. Example: CH₃COOH (acid) + H₂O (base) ⇌ CH₃COO⁻ + H₃O⁺. Applies to non-aqueous systems and explains amphiprotic behaviour.

(b) Conjugate pairs:

(i) CH₃COO⁻ + HCN ⇌ CH₃COOH + CN⁻

HCN = acid (proton donor); CN⁻ = its conjugate base
CH₃COO⁻ = base (proton acceptor); CH₃COOH = its conjugate acid

(ii) H₂PO₄⁻ + NH₃ ⇌ HPO₄²⁻ + NH₄⁺

H₂PO₄⁻ = acid; HPO₄²⁻ = conjugate base
NH₃ = base; NH₄⁺ = conjugate acid

(iii) HClO + CH₃NH₂ ⇌ CH₃NH₃⁺ + ClO⁻

**HClO# 2023 IJMB CHEMISTRY PAPER I — FULL TRANSCRIPTION & SOLUTIONS

TRANSCRIPTION OF ALL QUESTIONS

SUBJECT: Chemistry Paper I

DATE: Saturday 22nd July, 2023 | TIME: 3 Hours

Instructions: Answer ALL questions in Section A and any FOUR (4) from Section B. Section A = 5 marks each; Section B = 25 marks each. No scientific programmable calculator.