Table of constants: R = 8.314 JK⁻¹mol⁻¹; Vm = 22.4 dm³; NA = 6.023×10²³ mol⁻¹; F = 96500 C mol⁻¹; 1 a.m.u = 931.5 MeV = 1.6023×10⁻¹³ J; RH = 109678 cm⁻¹; 1 atm = 760 mmHg = 1.013×10⁵ Nm⁻²
QUESTIONS
SECTION A
1. Explain what happens when sulphur(IV) oxide gas is passed into bromine water for a long time. Write the necessary equation(s) to justify your answer.
2. Calculate the relative molecular mass of a volatile liquid from the following data obtained by Victor Meyer's method: 0.150g of the liquid displaced 50cm³ of air at 25°C and 99.0 kNm⁻² pressure (saturated vapour pressure of water at 25°C = 3.17 kNm⁻²).
3. A metal, Z belongs to group 1 of the periodic table. On the basis of this information, answer the following:
a) Give the chemical formula of the possible oxide of Z.
b) Suggest the expected bond type in the oxide.
c) Write the chemical equation for the reaction of the metal oxide with water.
d) Write an equation for the effect of adding aqueous hydrogen chloride to the aqueous solution obtained in (c) above.
4a) Classify the following elements as s, p, d or f-block elements:
i) D (atomic number = 20); ii) E (atomic number = 26); iii) F (atomic number = 33)
4b) The quantum numbers n, l, m, and s for the electrons in the valence shells of atoms M and N are given below:
| M | n | l | m | s |
|---|---|---|---|---|
| 3 | 0 | 0 | −½ | |
| 3 | 0 | 0 | +½ |
| N | n | l | m | s |
|---|---|---|---|---|
| 2 | 1 | −1 | +½ | |
| 2 | 1 | 0 | +½ | |
| 2 | 1 | 1 | +½ |
Write out the electron configuration of the valence shell in each case.
5. Given that the volume of oxygen gas at standard temperature and pressure is 5.6 dm³, calculate the: (a) mole of oxygen gas; (b) mass of oxygen gas; (c) number of oxygen molecules; (d) number of oxygen atoms.
6a) Define the following terms: i) Enthalpy of hydration; ii) Enthalpy of neutralisation.
6b) The heat of formation of ammonia is given by: ½N₂ + 3/2H₂ → NH₃; ΔHf = −46 kJ/mol. What does this information indicate?
7. Outline five general features of group 17 (halogen) elements of the periodic table.
8. The rate determining step of a reaction involves reaction between two moles of A and one mole of B to give one mole of product C.
a) Write the rate law for the reaction.
b) State four things you might do to increase the formation of the product, C.
9a) What is lanthanide contraction?
9b) Give the oxidation state of sulphur in each of the following: i) H₂SO₄ ii) SO₃²⁻ iii) S₈
10. A gas M diffuses three times as rapidly as another gas L under the same conditions of temperature and pressure. If the density of gas L is 1.96×10³ gm⁻³, calculate the density of gas M.
SECTION B
11. Explain the following:
a) Zinc is a d-block element but is not regarded as a transition metal. (2 marks)
b) Anhydrous copper(II) sulphate is white, but hydrated copper(II) sulphate (CuSO₄·5H₂O) is blue. (7 marks)
c) Sodium chloride has a lower melting point than magnesium oxide. (8 marks)
d) Nitrogen forms NCl₃ but not NCl₅, whereas phosphorus forms both PCl₃ and PCl₅. (5 marks)
e) Occurrence of the +2 oxidation state (rather than +4) as the more stable state in lead. (3 marks)
12a) Outline two similarities between a Daniell cell and an electrolytic cell. (2 marks)
12b) In a tabular form, state four differences between electrochemical and electrolytic cells. (4 marks)
12c) Metal P and Q are both divalent electrodes in an electrochemical cell with E°P = +0.34V and E°Q = −0.76V at 298K.
i) With suitable reason, which of the metals will serve as the cathode and the anode? (2 marks)
ii) Write the reactions at the electrodes. (4 marks)
iii) Write the overall reaction in the cell. (1 mark)
iv) Determine the emf of the cell. (3 marks)
v) Calculate the value of ΔG for the reaction. (3 marks)
vi) Is the reaction spontaneous? Explain. (2 marks)
vii) Calculate the equilibrium constant, K, for the system. (4 marks)
13. Use the information in the following table to explain the following observations:
| Element | Li | Be | B | C | N | O | F |
|---|---|---|---|---|---|---|---|
| Atomic radius (nm) | 0.152 | 0.111 | 0.088 | 0.077 | 0.070 | 0.066 | 0.064 |
| Ionic radius (nm) | 0.060 | — | — | — | — | 0.140 | 0.133 |
| First ionisation energy (kJ/mol) | 520 | 899 | 801 | 1086 | 1402 | 1314 | 1681 |
a) The atomic radius decreases from Li to F. (6 marks)
b) The ionic radii of the ions of O and F are greater than their respective atoms. (3 marks)
c) The ionic radius of the ion of Li is less than its atom. (2 marks)
d) The first ionisation energy of B is less than that of Be. (6 marks)
e) The first ionisation energy of O is less than that of N. (5 marks)
f) Which of the elements are s-block elements? (2 marks)
g) Which of the elements are p-block elements? (1 mark)
14a) Give the formula of: i) an example of a cationic complex of nickel; ii) an example of an anionic complex of nickel; iii) an example of a neutral complex of nickel. (1 mark each)
14b) State three reasons why nickel is able to exist in the form of the above a(i)–(iii). (3 marks)
14c) The compound A, [Co(NH₃)₅Br]SO₄ is isomeric with compound B, [Co(NH₃)₅SO₄]Br
i) Name the type of isomerism. (1 mark)
ii) What ions would these two compounds produce when dissolved in polar medium like water? (4 marks)
iii) Using a balanced chemical equation, describe a simple chemical test to distinguish between the two isomers. (4 marks)
iv) Give the oxidation state of cobalt in isomer A and write its full electron configuration. (3 marks)
v) What is the coordination number of cobalt in isomer B? (1 mark)
vi) Draw the structure of isomer B. (3 marks)
vii) Predict the magnetic property of isomer B and give reason(s) for your answer. (3 marks)
15a) State the following laws or principles and give an application or consequence of each (3 marks each): i) Pauli's exclusion principle; ii) Le Chatelier's principle; iii) Raoult's law; iv) Aufbau principle; v) Kohlrausch's law of independent migration of ions.
15b) A sample of impure calcium carbonate weighing 0.50g was reacted with 50cm³ of 0.2M HCl (an excess). The excess acid required 15cm³ of 0.15M NaOH for complete neutralisation. Calculate the percentage purity of the calcium carbonate. (5 marks)
15c) Give any two methods (other than use of indicators or a pH meter) that can be employed to determine the end point of an acid-base titration. (5 marks)
16. At 25°C, the Kc for the reaction N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹, has a numerical value of 4.63×10⁻³.
a) Write an expression for Kc in terms of the degree of dissociation (α) and the total volume (V). (3 marks)
b) What is the unit of Kc? (2 marks)
c) i) What is the value of Kc for the reverse reaction at the same temperature in terms of α and V? (2 marks) ii) Write the unit for the reverse reaction. (1 mark)
d) What will be the effect on Kc if: i) more N₂O₄ is added ii) pressure is increased iii) temperature is increased? (2 marks)
e) Predict the effect on the equilibrium composition if: i) more N₂O₄ is added; (2 marks) ii) pressure is increased; (2 marks) iii) temperature is increased. (2 marks)
f) At 25°C and a total pressure of 2 atmospheres, N₂O₄ is found to be 20% dissociated. Write an expression for the Kp in terms of α and P. Calculate the equilibrium constant Kp at this temperature. (9 marks)
ANSWERS
Question 1
When SO₂ is passed into bromine water for a short time, the bromine water is decolourised as SO₂ acts as a reducing agent, being oxidised to sulphate ions while bromine is reduced:
SO₂ + Br₂ + 2H₂O → H₂SO₄ + 2HBr
Passed in for a long time (excess), all the bromine is consumed (decolourised completely) and the resulting solution simply becomes a dilute sulphuric acid/hydrobromic acid mixture — no further visible change occurs once the bromine colour is fully removed, since the reaction goes to completion in one step (unlike the CO₂/limewater case, there is no reversal on continued addition).
Question 2
Dry gas pressure = 99.0 − 3.17 = 95.83 kNm⁻² = 95830 Pa
V = 50cm³ = 50×10⁻⁶ m³; T = 298K
n = PV/RT = (95830×50×10⁻⁶)/(8.314×298) = 4.7915/2477.6 = 0.001935 mol
M = 0.150/0.001935 = ≈77.5 g/mol
Question 3
a) Z₂O
b) Ionic (electrovalent) bond
c) Z₂O + H₂O → 2ZOH
d) ZOH + HCl → ZCl + H₂O
Question 4
a)
i) D (Z=20, Ca): [Ar]4s² → s-block
ii) E (Z=26, Fe): [Ar]3d⁶4s² → d-block
iii) F (Z=33, As): [Ar]3d¹⁰4s²4p³ → p-block
b) M: (3,0,0,−½) and (3,0,0,+½) → both electrons in the 3s orbital → M valence configuration: 3s²
N: three electrons, one in each 2p orbital (m=−1,0,+1), all same spin (Hund's rule) → N valence configuration: 2p³
Question 5
n(O₂) = V/Vm = 5.6/22.4 = 0.25 mol
mass = n×M = 0.25×32 = 8 g
molecules = n×NA = 0.25×6.023×10²³ = 1.506×10²³ molecules
atoms = 2×molecules = 3.011×10²³ atoms
Question 6
a) i) Enthalpy of hydration: The enthalpy change when one mole of gaseous ions is dissolved in a large excess of water to form an infinitely dilute aqueous solution, at constant pressure.
ii) Enthalpy of neutralisation: The enthalpy change when one mole of water is formed from the reaction between an acid and a base under standard conditions.
b) The negative sign indicates the reaction is exothermic; formation of one mole of NH₃ from its elements releases 46 kJ of heat, meaning NH₃ is more thermodynamically stable (lower in energy) than the separate elements N₂(g) and H₂(g).
Question 7
Five general features of Group 17 (halogens):
- Seven valence electrons in the outermost shell (ns²np⁵ configuration).
- Highly reactive non-metals; reactivity decreases down the group.
- Exist as diatomic molecules (X₂) held together by covalent bonds.
- Readily gain one electron to form −1 anions, acting as strong oxidising agents (oxidising power decreases down the group).
- Physical states progress from gas (F₂, Cl₂) to liquid (Br₂) to solid (I₂) down the group, with melting and boiling points increasing due to stronger van der Waals forces.
Question 8
a) Rate = k[A]²[B]
b) Four ways to increase formation of C:
- Increase the concentration of A and/or B.
- Increase the temperature (raises rate constant k).
- Use of a suitable catalyst.
- Increase pressure (if A and B are gases).
Question 9
a) Lanthanide contraction: The steady, cumulative decrease in atomic and ionic radii of the lanthanide elements (Ce to Lu) across the series, caused by the poor shielding of the nuclear charge by the diffuse 4f electrons, resulting in a progressive increase in effective nuclear charge felt by the outer electrons.
b)
i) H₂SO₄: S = +6
ii) SO₃²⁻: S = +4
iii) S₈: S = 0
Question 10
By Graham's Law: rateM/rateL = √(ρL/ρM)
3 = √(1.96×10³/ρM)
9 = 1.96×10³/ρM
ρM = 1.96×10³/9 = 217.8 g/m³
Question 11
a) Zinc is a d-block element (electron configuration [Ar]3d¹⁰4s²) but not a transition metal because both its atom and its only common ion (Zn²⁺, [Ar]3d¹⁰) have a completely filled 3d subshell; transition metal character requires a partially-filled d-subshell in the atom or at least one common oxidation state.
b) Anhydrous CuSO₄ has no water molecules coordinated to Cu²⁺, so there are no ligands present to split the d-orbitals — hence no d-d electronic transition can occur, and the compound appears white. In hydrated CuSO₄·5H₂O, water molecules act as ligands, coordinating to Cu²⁺ and splitting its 3d orbitals into different energy levels; d-d electron transitions absorb specific wavelengths of visible light, with the unabsorbed (complementary) wavelengths transmitted, producing the observed blue colour.
c) NaCl is composed of singly-charged Na⁺ and Cl⁻ ions, giving a weaker electrostatic (ionic) lattice energy. MgO is composed of doubly-charged Mg²⁺ and O²⁻ ions, which are also smaller in size; the much higher ionic charges combined with smaller ionic radii produce a substantially stronger lattice energy (by Coulomb's law, F∝q₁q₂/r²) — hence MgO requires much more energy (higher temperature) to melt than NaCl.
d) Nitrogen (period 2) has no accessible d-orbitals in its valence shell, so it cannot expand its octet beyond 8 electrons, limiting it to a maximum of 3 covalent bonds (NCl₃) with its 5 valence electrons (one lone pair remaining). Phosphorus (period 3) has energetically accessible 3d orbitals, allowing it to expand its octet and form 5 bonds, giving both PCl₃ (using only 3p electrons) and PCl₅ (using excited-state d-orbital participation).
e) Lead's electron configuration is [Xe]4f¹⁴5d¹⁰6s²6p²; due to the inert pair effect, the 6s² electron pair is stabilised by poor shielding from the intervening 4f and 5d electrons and relativistic effects, making these electrons reluctant to participate in bonding. This makes the +2 oxidation state (using only the 6p electrons) more stable than the +4 state (using both 6s and 6p electrons) for lead, unlike its lighter group 14 congeners.
Question 12
a) Two similarities:
- Both involve redox reactions occurring at electrodes (oxidation at the anode, reduction at the cathode).
- Both consist of electrodes immersed in an electrolyte, with electron flow through an external circuit and ion flow through the electrolyte.
b) Four differences:
| Feature | Electrochemical (Galvanic) cell | Electrolytic cell |
|---|---|---|
| Energy conversion | Chemical energy → Electrical energy | Electrical energy → Chemical energy |
| Spontaneity | Spontaneous reaction (ΔG negative) | Non-spontaneous reaction, requires external EMF |
| Anode polarity | Anode is negative | Anode is positive |
| Cathode polarity | Cathode is positive | Cathode is negative |
c) E°P = +0.34V, E°Q = −0.76V
i) Metal Q (more negative reduction potential) serves as the anode (more readily oxidised); Metal P (higher/less negative reduction potential) serves as the cathode.
ii)
Anode (oxidation): Q → Q²⁺ + 2e⁻
Cathode (reduction): P²⁺ + 2e⁻ → P
iii) Overall: Q + P²⁺ → Q²⁺ + P
iv) E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V
v) ΔG = −nFE° = −(2)(96500)(1.10) = −212,300 J = −212.3 kJ/mol
vi) Since ΔG is negative, the reaction is spontaneous.
vii) ln K = 212300/(8.314×298) = 212300/2477.6 = 85.71
log K = 85.71/2.303 = 37.22
K ≈ 1.66×10³⁷
Question 13
a) Across period 2 (Li→F), the number of protons (nuclear charge) increases while electrons are added to the same principal shell (n=2); increased effective nuclear charge, without corresponding increase in shielding, pulls the electron cloud in more strongly, causing atomic radius to progressively decrease.
b) O and F gain electrons to form O²⁻ and F⁻ anions; the added electron(s) increase electron-electron repulsion while nuclear charge stays constant, causing the electron cloud to expand — hence the anion is larger than the parent atom.
c) Li loses its single outer electron to form Li⁺, losing the entire outer (n=2) shell; the remaining electron shell (n=1) experiences the same nuclear charge with far less electron-electron repulsion and shielding, pulling the remaining electrons much closer to the nucleus, making the cation significantly smaller than the parent atom.
d) Be's outer electron configuration (2s²) is a stable, fully-filled subshell, requiring extra energy to remove an electron from it. B's outermost electron is in the 2p subshell (2s²2p¹), which is higher in energy, further from the nucleus, and shielded by the 2s² electrons, making it easier to remove — hence B's first ionisation energy is lower than Be's.
e) N has a half-filled 2p³ configuration, which is extra stable due to symmetrical electron distribution (exchange energy). O (2p⁴) has one 2p orbital doubly occupied; removing an electron from this paired orbital relieves extra electron-electron repulsion, making it easier to remove than from N's stable half-filled configuration — hence O's ionisation energy is lower than N's.
f) s-block elements: Li, Be
g) p-block elements: B, C, N, O, F
Question 14
a)
i) Cationic complex: [Ni(NH₃)₆]²⁺
ii) Anionic complex: [NiCl₄]²⁻
iii) Neutral complex: [Ni(CO)₄]
b) Three reasons:
- Nickel has several stable oxidation states (commonly 0, +2), allowing different overall complex charges to form.
- Nickel has vacant/hybridisable orbitals available to accept lone electron pairs donated by ligands, forming coordinate bonds regardless of the ligand's charge type.
- The net charge of a complex is determined by the sum of the metal's oxidation state and the charges of its ligands; combining neutral ligands (e.g. NH₃, CO) and/or anionic ligands (e.g. Cl⁻, CN⁻) in different ratios allows overall cationic, anionic, or neutral complexes to be formed.
c) i) Type of isomerism: Ionisation isomerism
ii) In water:
Isomer A: [Co(NH₃)₅Br]SO₄ → [Co(NH₃)₅Br]²⁺ + SO₄²⁻
Isomer B: [Co(NH₃)₅SO₄]Br → [Co(NH₃)₅SO₄]⁺ + Br⁻
iii) Test with BaCl₂ solution:
Isomer A (free SO₄²⁻ ion): [Co(NH₃)₅Br]SO₄ + BaCl₂ → [Co(NH₃)₅Br]Cl₂ + BaSO₄↓ (white precipitate forms, since SO₄²⁻ is ionic/free)
Isomer B (SO₄ coordinated, Br⁻ free): no precipitate with BaCl₂, but with AgNO₃: [Co(NH₃)₅SO₄]Br + AgNO₃ → [Co(NH₃)₅SO₄]NO₃ + AgBr↓ (cream precipitate, since Br⁻ is the free ionic species)
iv) Complex ion [Co(NH₃)₅Br]²⁺: Co + 0(NH₃×5) + (−1)(Br) = +2 → Co = +3
Full electron configuration of Co³⁺ ([Ar]3d⁶): 1s²2s²2p⁶3s²3p⁶3d⁶
v) Coordination number of Co in isomer B (5 NH₃ + 1 SO₄ as monodentate ligand): 6
vi) Structure: An octahedral complex ion, [Co(NH₃)₅SO₄]⁺, with the five NH₃ ligands and one SO₄²⁻ ligand symmetrically arranged around the central Co³⁺ ion at the six vertices of an octahedron, with Br⁻ as a separate counter-ion outside the coordination sphere.
vii) With NH₃ as a strong-field ligand, Co³⁺ (d⁶) adopts a low-spin configuration (t2g⁶eg⁰), giving no unpaired electrons, so isomer B is predicted to be diamagnetic.
Question 15
a)
i) Pauli's Exclusion Principle: No two electrons in the same atom can have the same set of all four quantum numbers; consequently, an orbital can hold a maximum of two electrons, which must have opposite spins. Application: explains the maximum electron capacity of orbitals, subshells, and shells (2, 8, 18, 32...).
ii) Le Chatelier's Principle: When a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium shifts in the direction that tends to counteract the imposed change. Application: used to predict and optimise conditions for maximum yield in industrial processes, e.g., the Haber process.
iii) Raoult's Law: The 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. Application: used to calculate boiling point elevation and freezing point depression of solutions, and to explain colligative properties.
iv) Aufbau Principle: Electrons occupy the available orbitals of an atom starting from the lowest energy level before filling higher energy levels. Application: provides the basis for predicting the ground-state electron configuration of atoms.
v) Kohlrausch's Law of Independent Migration of Ions: At infinite dilution, each ion migrates independently of its co-ion, and the molar conductivity of an electrolyte is the sum of the individual contributions of its constituent ions. Application: allows calculation of the molar conductivity at infinite dilution of weak electrolytes, which cannot be measured directly by extrapolation.
b) Reaction: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
mol HCl (initial, excess) = 0.2 × (50/1000) = 0.01 mol
mol NaOH (used to neutralise excess HCl) = 0.15 × (15/1000) = 0.00225 mol = mol excess HCl
mol HCl reacted with CaCO₃ = 0.01 − 0.00225 = 0.00775 mol
mol CaCO₃ = 0.00775/2 = 0.003875 mol
mass CaCO₃ = 0.003875 × 100 = 0.3875 g
% purity = (0.3875/0.50) × 100 = 77.5%
c) Two additional methods to determine the end point of acid-base titration:
- Conductometric method — measuring the change in electrical conductivity of the solution as titrant is added; the end point corresponds to a sharp change (often a minimum, then rise) in the conductivity curve.
- Potentiometric method — measuring the electrode potential (using a suitable electrode pair) as titrant is added; the end point corresponds to the point of maximum rate of change of potential (steepest gradient) on the titration curve.
Question 16
a) Let n₀ = initial moles N₂O₄, α = degree of dissociation, V = total volume:
At equilibrium: [N₂O₄] = n₀(1−α)/V, [NO₂] = 2n₀α/V
Kc = 4n₀α²/[V(1−α)]
b) Kc = [NO₂]²/[N₂O₄], units = (mol/dm³)²/(mol/dm³) = mol dm⁻³
c) i) For reverse reaction 2NO₂⇌N₂O₄: Kc(reverse) = 1/Kc(forward)
Kc(reverse) = V(1−α)/(4n₀α²)
ii) Unit: dm³ mol⁻¹
d) (Kc depends only on temperature)
i) More N₂O₄ added: No effect (Kc unchanged)
ii) Pressure increased: No effect (Kc unchanged)
iii) Temperature increased: Since ΔH is positive (endothermic forward reaction), by Le Chatelier's principle, Kc increases
e)
i) More N₂O₄ added: Equilibrium shifts right (forward), increasing NO₂ formed.
ii) Pressure increased: Forward reaction increases gas moles (1→2); equilibrium shifts left (backward), favouring N₂O₄.
iii) Temperature increased: Since the reaction is endothermic, equilibrium shifts right (forward), favouring NO₂.
f) At α = 0.2, total moles at equilibrium = 1+α (basis n₀=1): N₂O₄ = 1−α = 0.8; NO₂ = 2α = 0.4; total = 1.2
Mole fractions: x(N₂O₄) = 0.8/1.2 = 0.667; x(NO₂) = 0.4/1.2 = 0.333
Kp = [x(NO₂)·P]²/[x(N₂O₄)·P]
Kp = 4α²P/(1−α²)
At α=0.2, P=2atm:
Kp = 4(0.04)(2)/(1−0.04) = 0.32/0.96
Kp ≈ 0.333 atm