QUESTIONS
SECTION A
Atomic numbers: K=19, Al=13, Si=14, P=15, Zn=30, Cu=29, Co=27, S=16, V=23, Cr=24, Mg=12, Sn=50, Pb=82
Atomic masses: H=1, N=14, O=16, Mg=24, Cl=35.5, Fe=56, C=12
1a) Explain what happens when carbon(IV) oxide is passed into lime water for a long time. Write the necessary equation(s) to justify your answer.
2. Calculate the relative molecular mass of chloroform from the following data which was obtained by Victor Meyer's method: 0.220g of chloroform displaced 45cm³ of air at 20°C and 100.7 kNm⁻² pressure (saturated vapour pressure of water at 20°C = 2.32 kNm⁻²).
3. A metal, Y belongs to group 2 of the periodic table. On the basis of this information, answer the following questions:
a) Give the chemical formula of the possible oxide of Y.
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) A (atomic number = 25); ii) B (atomic number = 11); iii) C (atomic number = 16)
4b) The quantum numbers n, l, m, and s for the electrons in the valence shells of the atoms, Y and Z are given below:
| Y | n | l | m | s |
|---|---|---|---|---|
| 2 | 0 | 0 | −½ | |
| 2 | 0 | 0 | +½ |
| Z | n | l | m | s |
|---|---|---|---|---|
| 3 | 0 | 0 | −½ | |
| 3 | 0 | 0 | +½ | |
| 3 | 1 | 0 | −½ | |
| 3 | 1 | 0 | +½ |
Write out the electron configuration of the valence shell in each case.
5. Given that the volume of chlorine gas at saturated temperature and pressure is 2.24dm³, calculate the: (a) mole of chlorine gas; (b) mass of chlorine gas; (c) number of chlorine molecules; (d) number of chlorine atoms.
6a) Define the following terms: i) Enthalpy of solution; ii) Enthalpy of combustion.
6b) The heat of formation of carbon(IV) oxide is given by: C + O₂ → CO₂; ΔHf = −393 kJ/mol. What does this information indicate?
7. Outline five general features of group 1 elements of the periodic table.
8. The rate determining step of a reaction involves reaction between a mole each of P and Q to give a mole of R as a product.
a) Write the rate law for the reaction.
b) State four things you might do to increase the formation of the product, R.
9a) What is inert pair effect?
9b) Give the oxidation state of nitrogen in each of the following compounds: i) NH₃ ii) NH₂OH iii) N₂
10. A gas X diffuses four times as rapidly as another gas Y under the same condition of temperature and pressure. If the density of gas Y is 2.88×10³ gm⁻³, calculate the density of gas X.
SECTION B
11. Explain the following:
a) Tin and lead have variable oxidation states and form complex ions but are not regarded as transition metals. (2 marks)
b) An aqueous solution of KCl is colourless but that of CuCl₂ is green. (7 marks)
c) Although calcium chloride has a lower melting point than strontium chloride, it has a higher melting point than calcium iodide. (8 marks)
d) Although oxygen and sulphur belong to the same group of the periodic table, SF₆ is stable but OF₆ is not. (5 marks)
e) Occurrence of +3 oxidation state in scandium. (3 marks)
12a) Outline two similarities between electrochemical and electrolytic cells. (2 marks)
12b) In a tabular form, state four differences between electrochemical and electrolytic cells. (4 marks)
12c) Metal X and Y are both divalent electrodes in an electrochemical cell with E°X = −2.00V and E°Y = −4.25V 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 | Na | Mg | Al | Si | P | S | Cl |
|---|---|---|---|---|---|---|---|
| Atomic radius (nm) | 0.156 | 0.136 | 0.125 | 0.117 | 0.111 | 0.104 | 0.099 |
| Ionic radius (nm) | 0.095 | 0.065 | 0.050 | — | — | 0.184 | 0.181 |
| First ionisation energy (kJ/mol) | 492 | 743 | 579 | 791 | 1060 | 1003 | 1254 |
a) The atomic radius decreases from Na to Cl. (6 marks)
b) The ionic radii of the ions of S and Cl are greater than their respective atoms. (3 marks)
c) The ionic radii of the ions of Na, Mg and Al are less than their respective atoms. (2 marks)
d) The first ionisation energy of Al is less than that of Mg. (6 marks)
e) The first ionisation energy of S is less than that of P. (5 marks)
f) Which of the elements are s-block elements? (2 marks)
g) Which of the elements are d-block elements? (1 mark)
14a) Give the formula of: i) an example of cationic complex of chromium; ii) an example of anionic complex of chromium; iii) an example of neutral complex of chromium. (1 mark each)
14b) State three reasons why chromium is able to exist in the form of the above a(i)–(iii). (3 marks)
14c) The compound A, [CoCl(H₂O)₅]Cl₂·H₂O is isomeric with compound B, [Co(H₂O)₆]Cl₃
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 for any of the 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) Heisenberg's uncertainty principle; ii) Fajan's rule; iii) Nernst distribution law; iv) Hund's rule of maximum multiplicity; v) Hess' law of heat summation.
15b) A piece of impure iron scrap weighing 0.22g was dissolved in tetraoxosulphate(VI) acid, which oxidised the scrap to Fe²⁺. The resulting solution requires 36.50cm³ of 0.1 mole of potassium tetraoxomanganate(VII) for complete reaction. Calculate the percentage purity of the iron scrap. (5 marks)
15c) Give any two methods that can be employed to determine the end point of an acid-base titration. (5 marks)
16. At 200°C, the Kc for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ΔH = +124 kJ mol⁻¹, has a numerical value of 8×10⁻³.
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 PCl₅ is added ii) pressure is increased iii) temperature is increased? (2 marks)
e) Predict the effect on the equilibrium composition if: i) more PCl₅ is added (2 marks); ii) pressure is increased (2 marks); iii) temperature is increased (2 marks).
f) At 200°C and a total pressure of 1 atmosphere, the PCl₅ is found to be 50% 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 CO₂ is passed into lime water (Ca(OH)₂ solution) for a short time, a white precipitate of calcium carbonate forms, turning the solution milky:
CO₂ + Ca(OH)₂ → CaCO₃↓ + H₂O
When CO₂ is passed in for a long time (excess), the white precipitate dissolves, and the milkiness disappears, forming a clear solution of calcium hydrogencarbonate (bicarbonate), which is soluble in water:
CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂ (aq)
Question 2
Dry gas pressure = total pressure − water vapour pressure = 100.7 − 2.32 = 98.38 kNm⁻² = 98380 Pa
V = 45cm³ = 45×10⁻⁶ m³; T = 293K
n = PV/RT = (98380 × 45×10⁻⁶)/(8.314×293) = 4.4271/2436.0 = 0.001817 mol
M = mass/n = 0.220/0.001817 = ≈121 g/mol
(Compares well with actual M of CHCl₃ = 12+1+3(35.5) = 119.5 g/mol)
Question 3
a) YO
b) Ionic (electrovalent) bond
c) YO + H₂O → Y(OH)₂
d) Y(OH)₂ + 2HCl → YCl₂ + 2H₂O
Question 4
a)
i) A (Z=25, Mn): [Ar]3d⁵4s² → d-block
ii) B (Z=11, Na): [Ne]3s¹ → s-block
iii) C (Z=16, S): [Ne]3s²3p⁴ → p-block
b) Y: (2,0,0,−½) and (2,0,0,+½) → both electrons in the 2s orbital → Y valence configuration: 2s²
Z: (3,0,0,−½), (3,0,0,+½), (3,1,0,−½), (3,1,0,+½) → two electrons in 3s and two in 3p → Z valence configuration: 3s²3p²
Question 5
n(Cl₂) = V/Vm = 2.24/22.4 = 0.1 mol
mass = n×M = 0.1×71 = 7.1 g
molecules = n×NA = 0.1×6.023×10²³ = 6.023×10²² molecules
atoms = 2×molecules = 1.2046×10²³ atoms
Question 6
a) i) Enthalpy of solution: The enthalpy change when one mole of a solute is dissolved in a specified (usually large excess) amount of solvent, at constant pressure, to form a solution.
ii) Enthalpy of combustion: The enthalpy change when one mole of a substance is completely burnt in excess oxygen under standard conditions.
b) The negative sign indicates the reaction is exothermic; formation of one mole of CO₂ from its elements releases 393 kJ of heat, meaning CO₂ is more thermodynamically stable (lower in energy) than the separate elements C(s) and O₂(g).
Question 7
Five general features of Group 1 (alkali metals):
- One valence electron in the outermost s-orbital (ns¹ configuration).
- Highly reactive metals; reactivity increases down the group.
- Soft metals (can be cut with a knife) with low melting and boiling points, which decrease down the group.
- Readily lose one electron to form +1 cations, acting as strong reducing agents.
- React vigorously with water to liberate hydrogen gas and form strongly alkaline hydroxide solutions; their compounds are largely ionic and water-soluble.
Question 8
a) Rate = k[P][Q]
b) Four ways to increase formation of R:
- Increase the concentration of P and/or Q.
- Increase the temperature (raises rate constant k).
- Use of a suitable catalyst.
- Increase pressure (if P and Q are gases).
Question 9
a) Inert pair effect: The tendency of the outermost s-electron pair (ns²) of heavier p-block elements to resist participating in bonding (remaining "inert"/unshared), leading to the lower oxidation state (two less than the group oxidation state) being relatively more stable in heavier members of a group.
b)
i) NH₃: N = −3
ii) NH₂OH: N = −1
iii) N₂: N = 0
Question 10
By Graham's Law: rateX/rateY = √(ρY/ρX)
4 = √(2.88×10³/ρX)
16 = 2.88×10³/ρX
ρX = 2.88×10³/16 = 180 g/m³
Question 11
a) Tin and lead have variable oxidation states and form complex ions but are not regarded as transition metals because their d-orbitals are completely filled (d¹⁰) in all their common oxidation states (elemental and ionic forms); true transition metal character requires a partially-filled d-subshell in at least one stable oxidation state, which Sn and Pb lack.
b) KCl is colourless because K⁺ and Cl⁻ have complete, noble-gas-like electron configurations with no partially-filled d-orbitals, so no d-d electronic transitions can occur. CuCl₂ is green because Cu²⁺ has an incomplete 3d⁹ configuration; the ligand field splits the d-orbitals into different energy levels, and electrons undergo d-d transitions by absorbing specific wavelengths of visible light, with the complementary (unabsorbed) wavelengths transmitted, producing the observed green colour.
c) As the cation size increases down group 2 (Ca²⁺→Sr²⁺), the polarizing power of the cation decreases (Fajan's rules), reducing the covalent character of the M–Cl bond and increasing the purely ionic (electrostatic, more symmetric) character of the lattice, which raises the melting point — hence CaCl₂ (smaller, more polarizing Ca²⁺) has more covalent character and a lower melting point than SrCl₂.
Conversely, CaCl₂ has a higher melting point than CaI₂ because I⁻ is a much larger, more polarizable ion than Cl⁻; the small, highly polarizing Ca²⁺ distorts the electron cloud of the large I⁻ ion more readily than that of Cl⁻, introducing greater covalent character into CaI₂, weakening its ionic lattice and lowering its melting point relative to CaCl₂.
d) SF₆ is stable because sulphur (period 3) has energetically accessible 3d orbitals in its valence shell, allowing it to expand its octet and accommodate six bonding pairs (hexacoordination). Oxygen (period 2) has no accessible d-orbitals in its valence shell, so it cannot expand its octet beyond the normal limit, making OF₆ impossible to form.
e) Scandium's electron configuration is [Ar]3d¹4s²; losing all 3 valence electrons (2 from 4s, 1 from 3d) gives Sc³⁺ with a stable, fully-filled noble-gas configuration [Ar], which is energetically highly favourable — hence +3 is essentially the only stable oxidation state of scandium.
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 (ionic conduction) 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°X = −2.00V, E°Y = −4.25V
i) Metal X (less negative/higher reduction potential) serves as the cathode; Metal Y (more negative reduction potential) serves as the anode (Y more readily undergoes oxidation).
ii)
Anode (oxidation): Y → Y²⁺ + 2e⁻
Cathode (reduction): X²⁺ + 2e⁻ → X
iii) Overall: Y + X²⁺ → Y²⁺ + X
iv) E°cell = E°cathode − E°anode = −2.00 − (−4.25) = +2.25 V
v) ΔG = −nFE° = −(2)(96500)(2.25) = −434,250 J = −434.25 kJ/mol
vi) Since ΔG is negative, the reaction is spontaneous.
vii) ΔG° = −RT ln K
ln K = 434250/(8.314×298) = 434250/2477.6 = 175.3
log K = 175.3/2.303 = 76.1
K ≈ 10⁷⁶ (extremely large — the reaction proceeds essentially to completion)
Question 13
a) Across period 3 (Na→Cl), the number of protons (nuclear charge) increases while electrons are added to the same principal shell (n=3); increased effective nuclear charge, without corresponding increase in shielding, pulls the electron cloud in more strongly, causing atomic radius to progressively decrease.
b) S and Cl gain electrons to form S²⁻ and Cl⁻ 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) Na, Mg, Al lose their outer electron(s) to form cations, often losing the entire outer shell; the remaining electrons experience the same nuclear charge with less electron-electron repulsion and are pulled closer to the nucleus, making the cation smaller than the parent atom.
d) Mg's outer electron configuration (3s²) is a stable, fully-filled subshell, requiring extra energy to remove an electron from it. Al's outermost electron is in the 3p subshell (3s²3p¹), which is higher in energy, further from the nucleus, and shielded by the 3s² electrons, making it easier to remove — hence Al's first ionisation energy is lower than Mg's.
e) P has a half-filled 3p³ configuration, which is extra stable due to symmetrical electron distribution (exchange energy). S (3p⁴) has one 3p orbital doubly occupied; removing an electron from this paired orbital relieves extra electron-electron repulsion, making it easier to remove than from P's stable half-filled configuration — hence S's ionisation energy is lower than P's.
f) s-block elements: Na, Mg
g) d-block elements: None (all listed elements are s-block or p-block)
Question 14
a)
i) Cationic complex: [Cr(H₂O)₆]³⁺ (or [Cr(NH₃)₆]³⁺)
ii) Anionic complex: [CrCl₆]³⁻ (or [Cr(CN)₆]³⁻)
iii) Neutral complex: [Cr(NH₃)₃Cl₃] (or [Cr(CO)₆])
b) Three reasons:
- Chromium has several stable oxidation states (commonly +2, +3, +6), allowing different overall complex charges to form.
- Chromium has vacant d-orbitals available to accept lone electron pairs from 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. H₂O, NH₃) 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: Hydrate (solvate) isomerism
ii) In water:
Isomer A: [CoCl(H₂O)₅]Cl₂·H₂O → [CoCl(H₂O)₅]²⁺ + 2Cl⁻
Isomer B: [Co(H₂O)₆]Cl₃ → [Co(H₂O)₆]³⁺ + 3Cl⁻
iii) Test with AgNO₃ solution:
Isomer B (all Cl⁻ free/ionic): [Co(H₂O)₆]Cl₃ + 3AgNO₃ → Co(H₂O)₆₃ + 3AgCl↓ (white precipitate forms immediately, corresponding to all 3 chlorides)
Isomer A (one Cl coordinated): [CoCl(H₂O)₅]Cl₂ + 2AgNO₃ → CoCl(H₂O)₅₂ + 2AgCl↓ (only 2 moles of precipitate form per mole of complex, since 1 Cl is bound within the complex ion)
iv) Complex ion [CoCl(H₂O)₅]²⁺: Co + (−1) = +2 → Co = +3
Full electron configuration of Co³⁺ ([Ar]3d⁶): 1s²2s²2p⁶3s²3p⁶3d⁶
v) Coordination number of Co in isomer B: 6
vi) Structure: An octahedral complex ion, [Co(H₂O)₆]³⁺, with the six H₂O ligands symmetrically arranged at the six vertices around the central Co³⁺ ion, and 3 Cl⁻ as separate counter-ions outside the coordination sphere.
vii) With H₂O as a weak-field ligand, Co³⁺ (d⁶) adopts a high-spin configuration (t2g⁴eg²), giving 4 unpaired electrons, so isomer B is predicted to be paramagnetic.
Question 15
a)
i) Heisenberg's Uncertainty Principle: It is impossible to simultaneously determine, with complete precision, both the position and momentum of a subatomic particle such as an electron. Application: explains why electrons are described by probability distributions (orbitals) rather than fixed orbits.
ii) Fajan's Rule: Predicts the degree of covalent character in an ionic bond based on the cation's polarizing power and the anion's polarizability; covalency increases with small, highly-charged cations and large, easily-polarizable anions. Application: explains why AlCl₃ is significantly covalent while NaCl is essentially ionic.
iii) Nernst Distribution Law: At constant temperature, a solute distributes itself between two immiscible solvents such that the ratio of its concentrations in each solvent is a constant, provided the solute exists in the same molecular form in both. Application: forms the basis of solvent extraction techniques.
iv) Hund's Rule of Maximum Multiplicity: Electrons occupying degenerate (equal-energy) orbitals singly occupy each orbital first, with parallel spins, before any orbital is doubly filled. Application: explains the extra stability of half-filled and fully-filled subshells (e.g., Cr = [Ar]3d⁵4s¹ rather than [Ar]3d⁴4s²).
v) Hess' Law of Heat Summation: The total enthalpy change of a reaction is independent of the pathway taken, depending only on the initial and final states. Application: allows calculation of enthalpy changes that cannot be measured directly, by combining known enthalpy changes of related reactions.
b) Reaction: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
mol KMnO₄ = 0.1 × (36.50/1000) = 0.00365 mol
mol Fe²⁺ = 5 × 0.00365 = 0.01825 mol
mass Fe = 0.01825 × 56 = 1.022 g
% purity = (mass Fe/mass sample) × 100 = (1.022/0.22) × 100 = ≈464.5%
(Applying the standard method: moles KMnO₄ → moles Fe²⁺ via the 1:5 mole ratio → mass Fe → percentage purity, using the exact figures given. If the sample mass or titre in your source paper differs from what is shown here, substitute the correct value into the same method to obtain the percentage purity.)
c) Two methods to determine the end point of acid-base titration:
- Use of a suitable chemical indicator (e.g., phenolphthalein, methyl orange) that changes colour sharply at/near the equivalence point.
- Use of a pH meter, plotting a titration curve (pH vs volume of titrant) and identifying the point of steepest gradient as the equivalence point.
Question 16
a) Let n₀ = initial moles PCl₅, α = degree of dissociation, V = total volume:
At equilibrium: [PCl₅]=n₀(1−α)/V, [PCl₃]=[Cl₂]=n₀α/V
Kc = n₀α²/[V(1−α)]
b) Kc = [PCl₃][Cl₂]/[PCl₅], units = (mol/dm³)(mol/dm³)/(mol/dm³) = mol dm⁻³
c) i) For reverse reaction PCl₃+Cl₂⇌PCl₅: Kc(reverse) = 1/Kc(forward)
Kc(reverse) = V(1−α)/(n₀α²)
ii) Unit: dm³ mol⁻¹
d) (Kc depends only on temperature)
i) More PCl₅ 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 PCl₅ added: Equilibrium shifts right (forward), increasing PCl₃ and Cl₂ formed.
ii) Pressure increased: Forward reaction increases gas moles (1→2); equilibrium shifts left (backward), favouring PCl₅.
iii) Temperature increased: Since the reaction is endothermic, equilibrium shifts right (forward), favouring PCl₃ and Cl₂.
f) At α = 0.5, total moles at equilibrium = 1+α (basis n₀=1)
Mole fractions: x(PCl₅) = (1−α)/(1+α); x(PCl₃) = x(Cl₂) = α/(1+α)
Kp = [α/(1+α)·P]²/[(1−α)/(1+α)·P]
Kp = α²P/(1−α²)
At α=0.5, P=1atm:
Kp = (0.25)(1)/(1−0.25) = 0.25/0.75
Kp ≈ 0.333 atm