SECTION A
1. (a) What is the maximum electron population in each of the following?
(i) n=2 shell (ii) 3d subshell (iii) 4py sub-orbital
(b) Give the set of quantum numbers for the first 2 electrons in the third shell.
2. The osmotic pressure of a solution prepared by dissolving 1.0 g of haemoglobin in 100 cm³ of water is 0.413 kPa at 20°C. Calculate:
(a) the concentration in the unit of g m⁻³
(b) the molar mass of the haemoglobin.
3. In the complex ion, [CrBrCl(H₂O)₄]⁺:
(a) Give the oxidation number of Cr in the complex.
(b) What is the coordination number of Cr in the complex?
© Draw the structure of the complex ion and indicate the co-ordinate covalent bonds involved.
(d) What is the shape of the ion?
4. (a) What are isotopes?
(b) Five atoms have the following composition:
| Atom | Protons | Electrons | Neutrons |
|------|---------|-----------|----------|
| M | 8 | 8 | 8 |
| N | 8 | 7 | 8 |
| P | 8 | 8 | 7 |
| Q | 7 | 8 | 8 |
| R | 8 | 8 | 9 |
Which of these are isotopes?
5. Consider the following elements or most stable ions of the elements in the periodic table: Mg, Hg, P, Cl, Na and Ar.
(a) Which of these elements is most electropositive?
(b) Which of these elements is a metal with the lowest melting point?
© Which of these elements has lowest electron affinity?
(d) Which of these elements has highest ionisation potential?
(e) Which of these elements has highest electronegativity?
6. Calculate the mass of the following:
(a) 6.023 × 10²³ atoms of nitrogen
(b) 6.023 × 10²² molecules of nitrogen
© a mole of carbon(II) oxide
(d) 0.4 mole of nitrogen(IV) oxide
(e) 2 moles of boron atoms
7. In each of the following pair, identify the molecular species with the higher entropy [assuming constant temperature except in (e)]:
(a) 1 mole of solid KBr or 1 mole of an aqueous KBr
(b) 1 mole of SO₂ or 1 mole of SO₃
© sea water in mid-winter at 2°C or in mid-summer at 23°C
(d) 3 moles of oxygen gas or 2 moles of ozone gas
(e) 1 mole of CO₂ solid or 1 mole of CO₂ gas
8. A 0.372 mol dm⁻³ KMnO₄ solution was titrated against a solution of hydrogen peroxide which had been diluted 10 times. A 25 cm³ of the diluted peroxide solution was neutralised by 26.20 cm³ of KMnO₄ solution. Calculate the molarity of the original peroxide solution.
9. (a) State two factors responsible for the softness of group 1 elements of the periodic table.
(b) Why is the enthalpy of hydration of calcium higher than that of potassium ion?
10. State the number of phases in each of the following:
(a) air
(b) a mixture containing oil and water
© a mixture of salt and sand
(d) a mixture of ethanol and water
(e) addition of CaCO₃ to a mixture containing CaO and CO₂ gas.
SECTION B
11. (a) In the electrolysis of an aqueous solution of copper(II) tetraoxosulphate(IV), explain what would happen at the electrodes if:
(i) platinum electrodes are used. (5 marks)
(ii) copper electrodes are used. Write where necessary, ionic equations to support your answer. (4 marks)
(b) Comment on the observation that iron sheet rusts less rapidly when coated with zinc (galvanised) than when coated with tin (tinned). (9 marks)
© Calculate the change in potential of a copper electrode at 300 K, if the concentration of copper(II) tetraoxosulphate solution in contact with it was changed from 0.5 to 0.001 mol dm⁻³. (7 marks)
12. (a) An element, A, has an atomic number of 29.
(i) Write the full electron configuration of A. (2 marks)
(ii) Is the element A in s-, p- or d-block of the periodic table? (1 mark)
(iii) State any five characteristic properties of A. (5 marks)
(iv) What is the principal oxidation number of A? (1 mark)
(v) Show whether the oxide of A is likely to be acidic, basic or amphoteric substance. (marks)
(vi) State the expected bond type of the oxide in (v) above. (1 mark)
(b) (i) What is allotropy? (2 marks)
(ii) Name the two allotropes of carbon. (2 marks)
(iii) Which of them conducts electricity and which would not? (2 marks)
(iv) Give a suitable explanation to support your choice. (4 marks)
© Give two other elements that exhibit allotropy in the periodic table. (2 marks)
13. (a) What is a phase? (2 marks)
(b) Give one characteristic of a phase. (1 mark)
© Sketch the phase diagram of each of the following:
(i) water (ii) carbon(IV) oxide (8 marks)
(d) (i) Outline any three essential features of diagram c(i). (3 marks)
(ii) Explain any two differences between the diagrams c(i) and c(ii) above. (2 marks)
(e) (i) State Raoult’s law for ideal solutions. (2 marks)
(ii) A mixture of 0.2 mole of methanol A and 0.5 mole of propanol B has a total vapour pressure of 40 mmHg at 298K. If the mixture obeys Raoult’s law, calculate the vapour pressure of pure B at 298K given that the pressure of pure A is 20 mmHg at 298K. (7 marks)
14. (a) State Moseley’s law of periodic table. (2 marks)
(b) Define the following periodic properties using an appropriate chemical equation in each case:
(i) ionisation energy (3 marks)
(ii) electron affinity (3 marks)
© State and account for the trend in the properties listed in (b) above:
(i) across a period (ii) down a group in the periodic table. (12 marks)
(d) Arrange the ions, F⁻, Na⁺, N³⁻ and Mg²⁺ in the order of increasing size. Advance reason for your answer. (5 marks)
15. (a) Account for the following observations:
(i) Tin and lead form tetravalent ions but carbon and silicon do not. (5 marks)
(ii) NaCl is insoluble in organic solvent whereas AlCl₃ is substantially soluble in organic solvent. (5 marks)
(iii) The melting points of group 14 elements decrease down the group of the periodic table. (2 marks)
(iv) PCl₅ is more stable than PCl₃ on the other hand, BiCl₅ is unstable but BiCl₃ is stable. (6 marks)
(v) Excessive inhalation of carbon(II) oxide is toxic to humans. (4 marks)
(b) (i) What is hydrogen bonding? (2 marks)
(ii) List two conditions that must be satisfied before hydrogen bond can be formed. (2 marks)
16. (a) List any four laws of chemical combination of atoms. (4 marks)
(b) (i) Zinc oxide was prepared in the laboratory by using two experimental procedures. In the first method, 0.325g of zinc combined with 0.08g of oxygen to form the metal oxide. In the second procedure, 1.625g of zinc reacted with 0.4g of oxygen to produce the oxide of the metal. Calculate the percentage by mass of the metal in the oxide in each of the methods. (4 marks)
(ii) Which of the laws of chemical combination do the above data illustrate? Justify with suitable reason. (3 marks)
(iii) Cite any limitation of the law. (1 mark)
© In the Haber process for an industrial manufacture of ammonia, 15 cm³ of nitrogen was mixed with 80 cm³ of hydrogen gas under suitable conditions.
(i) Determine the reaction stoichiometry. (2 marks)
(ii) Determine which of the reactant gases is in excess and by what amount. (6 marks)
(iii) Determine the volume of reaction product. (2 marks)
(iv) Calculate the volume of residual gas mixture at the end of the reaction. (3 marks)
SOLUTIONS TO ALL QUESTIONS
SECTION A SOLUTIONS
Q1 — Electron population and quantum numbers
(a) Maximum electron populations:
(i) n=2 shell: Max electrons = 2n² = 2(2)² = 8 electrons
(ii) 3d subshell: d subshell has 5 orbitals × 2 electrons each = 10 electrons
(iii) 4py sub-orbital: One orbital holds maximum = 2 electrons
(b) Quantum numbers for the first 2 electrons in the third shell (n=3):
These are the first two electrons entering the 3s orbital:
| Electron | n | l | m | s |
|----------|—|---|—|---|
| 1st | 3 | 0 | 0 | +½ |
| 2nd | 3 | 0 | 0 | −½ |
(n=3, l=0 for s subshell, m=0 for one s orbital, s=+½ and −½ for the pair)
Q2 — Osmotic pressure / molar mass of haemoglobin
Given: mass = 1.0 g, Volume = 100 cm³ = 0.1 dm³ = 0.1 L, π = 0.413 kPa, T = 20°C = 293 K, R = 8.314 J K⁻¹ mol⁻¹
(a) Concentration in g m⁻³:
Volume = 100 cm³ = 100 × 10⁻⁶ m³ = 1 × 10⁻⁴ m³
C = mass/volume = 1.0 g / (1 × 10⁻⁴ m³) = 10,000 g m⁻³ = 1.0 × 10⁴ g m⁻³
(b) Molar mass:
Using π = CRT where C = molar concentration (mol/L)
π = 0.413 kPa = 413 Pa
C = π/RT = 413 / (8.314 × 293) = 413 / 2436 = 0.1695 × 10⁻³ mol L⁻¹
Wait — using consistent units:
π = CRT → C = π/RT = 0.413 × 10³ Pa / (8.314 × 293) = 413/2436.0 = 0.1695 × 10⁻³ mol dm⁻³
Moles in 0.1 dm³ = 0.1695 × 10⁻³ × 0.1 = 1.695 × 10⁻⁵ mol
Molar mass = mass/moles = 1.0 / (1.695 × 10⁻⁵) = ≈ 59,000 g/mol
(This is consistent with the known molar mass of haemoglobin ~64,500 g/mol — the small difference is due to rounding)
Q3 — [CrBrCl(H₂O)₄]⁺
(a) Oxidation number of Cr:
x + (−1) + (−1) + 0 = +1
x − 2 = +1
x = +3
(b) Coordination number of Cr: 1(Br) + 1(Cl) + 4(H₂O) = 6
© Structure: Octahedral arrangement — Cr³⁺ at centre with 6 ligands (1 Br⁻, 1 Cl⁻, 4 H₂O) all donating lone pairs via coordinate covalent bonds. All 6 bonds (Cr←Br, Cr←Cl, and 4×Cr←OH₂) are coordinate covalent bonds.
(d) Shape: Octahedral
Q4 — Isotopes
(a) Definition: Isotopes are atoms of the same element (same atomic number/proton number) that have different mass numbers due to different numbers of neutrons in their nuclei.
(b) Identifying isotopes from the table:
Isotopes must have the same number of protons (same element) but different neutrons:
-
M: 8 protons, 8 neutrons
-
N: 8 protons, 7 electrons (this is an ion of element with 8 protons), 8 neutrons
-
P: 8 protons, 8 electrons, 7 neutrons
-
Q: 7 protons — different element, not isotope of the others
-
R: 8 protons, 8 electrons, 9 neutrons
Atoms with 8 protons are all oxygen atoms: M, N (O⁻ ion), P, and R all have 8 protons.
Isotopes: M, P, and R (all have 8 protons but different neutrons: 8, 7, and 9 respectively)
N is also an isotope of oxygen but exists as an anion (O⁻).
Q (7 protons) is nitrogen — not an isotope of the others.
Answer: M, N, P and R are isotopes (all have 8 protons = oxygen), Q is not.
Q5 — Periodic trends among Mg, Hg, P, Cl, Na, Ar
(a) Most electropositive: Na — lowest ionisation energy among the metals listed; Group 1 element, most readily loses electrons.
(b) Metal with lowest melting point: Hg — mercury is the only metal that is liquid at room temperature (melting point = −39°C), the lowest among all metals listed.
© Lowest electron affinity: Ar — noble gas with complete electron configuration; adding an electron is energetically unfavourable, so electron affinity is essentially zero/negative.
(d) Highest ionisation potential: Ar — noble gas with full outer shell; requires the most energy to remove an electron.
(e) Highest electronegativity: Cl — highest electronegativity among the listed elements (F > O > Cl, but F is not listed here). Cl is in Period 3 Group 17.
Q6 — Mass calculations
(a) 6.023 × 10²³ atoms of nitrogen:
This is 1 mole of N atoms.
Mass = 1 × 14 = 14 g
(b) 6.023 × 10²² molecules of nitrogen (N₂):
Moles = 6.023 × 10²² / 6.023 × 10²³ = 0.1 mol of N₂
Molar mass of N₂ = 28 g/mol
Mass = 0.1 × 28 = 2.8 g
© 1 mole of carbon(II) oxide (CO):
Molar mass = 12 + 16 = 28 g/mol
Mass = 28 g
(d) 0.4 mole of nitrogen(IV) oxide (NO₂):
Molar mass of NO₂ = 14 + 2(16) = 46 g/mol
Mass = 0.4 × 46 = 18.4 g
(e) 2 moles of boron atoms:
Molar mass of B = 10 g/mol
Mass = 2 × 10 = 20 g
Q7 — Entropy comparisons
Higher entropy = greater disorder.
(a) 1 mol solid KBr vs 1 mol aqueous KBr:
Aqueous KBr — dissolved ions have more freedom of movement than in a rigid crystal lattice; greater disorder.
(b) 1 mol SO₂ vs 1 mol SO₃:
SO₃ — more complex molecule with more atoms; more vibrational modes and greater molecular complexity → higher entropy.
© Sea water at 2°C vs 23°C:
Sea water at 23°C (mid-summer) — higher temperature means greater kinetic energy and molecular motion → higher entropy.
(d) 3 moles O₂ vs 2 moles ozone (O₃):
3 moles of O₂ — more moles of gas means more particles and greater positional disorder; entropy is extensive (scales with amount).
(e) 1 mol solid CO₂ vs 1 mol CO₂ gas:
1 mol CO₂ gas — gaseous state has far greater entropy than solid state due to much greater freedom of movement.
Q8 — Titration: KMnO₄ vs H₂O₂
Given:
-
[KMnO₄] = 0.372 mol dm⁻³
-
Volume KMnO₄ = 26.20 cm³
-
Volume diluted H₂O₂ = 25 cm³
-
Dilution factor = 10
Balanced equation (acidic):
2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → 2MnSO₄ + K₂SO₄ + 8H₂O + 5O₂
Mole ratio KMnO₄ : H₂O₂ = 2 : 5
Moles of KMnO₄ = 0.372 × 26.20/1000 = 9.746 × 10⁻³ mol
Moles of H₂O₂ (diluted) = (5/2) × 9.746 × 10⁻³ = 2.437 × 10⁻² mol
Molarity of diluted H₂O₂ = 2.437 × 10⁻² / (25/1000) = 2.437 × 10⁻² / 0.025 = 0.9748 mol dm⁻³
Since it was diluted 10 times:
Molarity of original H₂O₂ = 0.9748 × 10 = 9.748 ≈ 9.75 mol dm⁻³
Q9 — Group 1 properties
(a) Two factors for softness of Group 1 elements:
-
Weak metallic bonding — Group 1 metals have only one valence electron per atom contributing to the metallic bond (sea of electrons), resulting in weaker bonding between metal ions compared to other metals.
-
Large atomic radius — the single valence electron is far from the nucleus and loosely held; the large size means the metallic bond is extended over a large volume, weakening it further.
(b) Enthalpy of hydration of Ca²⁺ > K⁺:
Enthalpy of hydration depends on charge density (charge/ionic radius). Ca²⁺ has a charge of +2 and a smaller ionic radius than K⁺ (which has charge +1). The higher charge and smaller size of Ca²⁺ gives it a much higher charge density, attracting water molecules more strongly and releasing more energy upon hydration.
Q10 — Number of phases
(a) Air: Air is a homogeneous mixture of gases — 1 phase
(b) Mixture of oil and water: Oil and water are immiscible — 2 phases
© Mixture of salt and sand: Two distinct solid components that do not dissolve in each other — 2 phases
(d) Mixture of ethanol and water: Completely miscible, forms one homogeneous liquid — 1 phase
(e) Addition of CaCO₃ to a mixture containing CaO and CO₂ gas:
-
CaCO₃ (solid) = 1 phase
-
CaO (solid) = 1 phase
-
CO₂ (gas) = 1 phase
Total = 3 phases
SECTION B SOLUTIONS
Q11 — Electrochemistry
(a) Electrolysis of CuSO₄(aq):
(i) With platinum (inert) electrodes:
At the cathode (reduction): Cu²⁺ ions from solution are preferentially discharged over H⁺:
Cu²⁺(aq) + 2e⁻ → Cu(s)
Copper metal deposits on the cathode.
At the anode (oxidation): OH⁻/water is oxidised (SO₄²⁻ is not discharged):
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
Oxygen gas is evolved at the anode.
Overall: the solution becomes more acidic and Cu²⁺ concentration decreases.
(ii) With copper electrodes:
At the cathode: Cu²⁺ is deposited as before:
Cu²⁺(aq) + 2e⁻ → Cu(s)
At the anode: The copper electrode itself dissolves (copper is preferentially oxidised over water):
Cu(s) → Cu²⁺(aq) + 2e⁻
Net effect: Copper dissolves from the anode at the same rate it deposits on the cathode. The concentration of CuSO₄ solution remains essentially constant. This is the principle behind electroplating and copper refining.
(b) Galvanising vs Tinning:
When iron is coated with zinc (galvanised): Zinc is more electropositive (higher in the activity series) than iron. If the zinc coating is scratched, zinc and iron form an electrochemical cell in the presence of moisture. Zinc acts as the sacrificial anode — it preferentially oxidises (Zn → Zn²⁺ + 2e⁻) while the iron remains protected as the cathode. Rusting is prevented even when the coating is damaged.
When iron is coated with tin (tinned): Tin is less electropositive than iron. If the tin coating is scratched, iron becomes the anode in the cell (iron is oxidised faster): Fe → Fe²⁺ + 2e⁻. Tin, being more noble, acts as the cathode. Rusting is therefore accelerated at damaged points. Tin only protects by physical barrier; once scratched, iron rusts faster than uncoated iron.
This explains why galvanised iron rusts less rapidly than tinned iron even when the coating is damaged.
© Change in electrode potential at 300 K:
Using the Nernst equation for Cu²⁺/Cu:
Cu²⁺ + 2e⁻ → Cu, n = 2
E = E° − (RT/nF) ln(1/[Cu²⁺])
or ΔE = E₂ − E₁ = −(RT/nF) ln([Cu²⁺]₁/[Cu²⁺]₂)
ΔE = (RT/nF) × ln([Cu²⁺]₂/[Cu²⁺]₁)
Wait — for a reduction electrode: E = E° + (RT/nF) ln[Cu²⁺]
ΔE = (RT/nF) × ln([Cu²⁺]final/[Cu²⁺]initial)
T = 300 K, n = 2, R = 8.314, F = 96500
RT/nF = (8.314 × 300)/(2 × 96500) = 2494.2/193000 = 0.01292 V
ΔE = 0.01292 × ln(0.001/0.5)
= 0.01292 × ln(0.002)
= 0.01292 × (−6.2146)
= −0.0803 V
The electrode potential decreases by 0.0803 V (≈ −80.3 mV) when concentration drops from 0.5 to 0.001 mol dm⁻³.
Q12 — Element A (Atomic number = 29, Copper)
(a)(i) Full electron configuration of Cu (Z=29):
Expected: [Ar] 3d⁹ 4s²
Actual (anomalous): [Ar] 3d¹⁰ 4s¹
Full: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
(Half-filled/fully-filled d subshell is more stable)
(a)(ii) Block: Cu is a d-block element (transition metal).
(a)(iii) Five characteristic properties of Cu:
-
Exists in variable oxidation states (+1 and +2)
-
Forms coloured compounds (e.g., CuSO₄·5H₂O is blue)
-
Acts as a catalyst in many reactions
-
Forms complex ions (e.g., [Cu(NH₃)₄]²⁺)
-
Has high electrical and thermal conductivity
(a)(iv) Principal oxidation number: +2 (Cu²⁺ is the most common and stable ion)
(a)(v) Nature of copper’s oxide:
Cu forms CuO (copper II oxide).
CuO is a basic oxide — it reacts with acids but not with bases:
CuO + H₂SO₄ → CuSO₄ + H₂O
It does not react with NaOH.
Therefore it is basic, not amphoteric.
(a)(vi) Bond type of CuO: Ionic bond (Cu²⁺ and O²⁻ ions)
(b) Allotropy:
(i) Allotropy: The existence of an element in two or more different physical forms (allotropes) in the same physical state, with different physical and sometimes chemical properties, due to different arrangements of atoms.
(ii) Two allotropes of carbon: Diamond and Graphite
(iii) Electrical conductivity:
-
Graphite conducts electricity
-
Diamond does not conduct electricity
(iv) Explanation:
In graphite, each carbon atom is sp² hybridised and bonded to three other carbons in flat hexagonal layers. The fourth valence electron of each carbon is in an unhybridised p-orbital and is delocalised across the layers, forming a π-electron cloud. These mobile electrons allow graphite to conduct electricity, similar to a metal.
In diamond, each carbon is sp³ hybridised and forms four covalent bonds in a rigid tetrahedral 3D network. All four valence electrons are used in bonding — there are no free/delocalised electrons. Without mobile charge carriers, diamond is a non-conductor (insulator).
© Two other elements exhibiting allotropy:
-
Sulphur (rhombic and monoclinic allotropes)
-
Phosphorus (white/yellow and red phosphorus)
Q13 — Phase diagrams
(a) A phase: A phase is a physically distinct, homogeneous and chemically uniform portion of a system that is separated from other parts by definite boundary surfaces (interfaces).
(b) One characteristic of a phase: Every phase has uniform physical and chemical properties throughout its entire extent.
© Phase diagrams: (Sketches described in words)
(i) Water phase diagram:
-
Three curves meeting at the triple point (0.006 atm, 0.0098°C): solid-liquid (fusion curve), liquid-gas (vaporisation curve), solid-gas (sublimation curve).
-
The fusion curve slopes slightly to the LEFT (negative slope) — unique to water — meaning increasing pressure lowers the melting point.
-
Normal boiling point at 1 atm = 100°C
-
Normal melting point at 1 atm = 0°C
-
Critical point at 374°C, 218 atm (beyond which liquid and gas are indistinguishable)
(ii) CO₂ phase diagram:
-
Three curves meeting at the triple point (5.11 atm, −56.6°C)
-
The fusion curve slopes to the RIGHT (positive slope) — normal behaviour
-
Critical point at 31.1°C, 73 atm
-
At 1 atm, CO₂ does not have a liquid phase — it goes directly from solid to gas (sublimes) at −78.5°C (dry ice)
(d)(i) Three essential features of the water phase diagram:
-
The triple point at 0.0098°C and 0.006 atm where all three phases coexist in equilibrium.
-
The negative slope of the solid-liquid boundary (fusion curve), indicating that ice melts under pressure.
-
The critical point at 374°C and 218 atm, above which liquid and vapour phases are indistinguishable (supercritical fluid).
(d)(ii) Two differences between water and CO₂ phase diagrams:
-
The fusion curve of water slopes negatively (leftward) while that of CO₂ slopes positively (rightward). This means pressure increases the melting point of CO₂ but decreases that of water.
-
Water has a liquid phase at 1 atm (between 0°C and 100°C), while CO₂ has no liquid phase at 1 atm — it sublimes directly from solid to gas at −78.5°C.
(e)(i) Raoult’s Law: The vapour pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution. i.e., Pₐ = Xₐ × P°ₐ, where Pₐ is the partial pressure, Xₐ is the mole fraction, and P°ₐ is the vapour pressure of the pure component.
(e)(ii) Vapour pressure of pure propanol B:
Given: moles of A (methanol) = 0.2, moles of B (propanol) = 0.5
Total moles = 0.7
Mole fraction of A: Xₐ = 0.2/0.7 = 2/7
Mole fraction of B: X_B = 0.5/0.7 = 5/7
Total pressure: P_total = Xₐ·P°ₐ + X_B·P°_B
40 = (2/7)(20) + (5/7)(P°_B)
40 = 40/7 + (5/7)P°_B
40 = 5.714 + 0.7143·P°_B
34.286 = 0.7143·P°_B
P°_B = 34.286/0.7143 = 48 mmHg
The vapour pressure of pure propanol B = 48 mmHg
Q14 — Periodic Table / Periodic Properties
(a) Moseley’s Law: The square root of the frequency of X-rays emitted by an element when bombarded with high-energy electrons is directly proportional to the atomic number of the element. i.e., √ν = a(Z − b), where Z is the atomic number and a, b are constants. This established atomic number (not mass) as the basis for ordering elements.
(b) Definitions:
(i) Ionisation energy: The minimum energy required to remove the most loosely bound electron from a gaseous atom in its ground state to form a gaseous ion.
Chemical equation: X(g) → X⁺(g) + e⁻ ΔH = +IE₁
(ii) Electron affinity: The energy released when a gaseous atom in its ground state accepts an electron to form a gaseous anion.
Chemical equation: X(g) + e⁻ → X⁻(g) ΔH = −EA
(EA is defined as energy released; so ΔH is negative for most elements)
© Trends:
Ionisation energy:
-
Across a period (left to right): IE increases. Nuclear charge increases while electrons are added to the same shell, so the effective nuclear charge felt by each electron increases, making it harder to remove electrons.
-
Down a group: IE decreases. Although nuclear charge increases, electrons are added to shells farther from the nucleus, and increased shielding by inner electrons reduces the effective nuclear charge experienced by valence electrons, making them easier to remove.
Electron affinity:
-
Across a period (left to right): EA generally increases (becomes more negative). As nuclear charge increases, the atom pulls the incoming electron more strongly. Exception: Group 15 (half-filled p subshell is stable) and Group 18 (noble gases) have very low EA.
-
Down a group: EA generally decreases (becomes less negative). The incoming electron enters a larger shell farther from the nucleus and experiences greater shielding from inner electrons, reducing the attractive force.
(d) Increasing ionic size: F⁻, Na⁺, N³⁻, Mg²⁺
All four ions are isoelectronic (each has 10 electrons — same as neon).
For isoelectronic species, ionic size decreases as nuclear charge (atomic number) increases because more protons pull the same number of electrons closer.
Atomic numbers: N=7, O=8, F=9, Na=11, Mg=12
Order of increasing size:
Mg²⁺ < Na⁺ < F⁻ < N³⁻
Reason: All have 10 electrons. Mg²⁺ (Z=12) has most protons pulling 10e⁻ → smallest. N³⁻ (Z=7) has fewest protons for 10e⁻ → largest.
Q15 — Inorganic Chemistry Explanations
(a)(i) Sn and Pb form +4 ions but C and Si do not:
Carbon and silicon are in Period 2 and 3 respectively. Their +4 ions would require removing electrons from very stable, compact orbitals, and the resulting C⁴⁺ or Si⁴⁺ ions would be extremely small with extremely high polarising power — they would immediately polarise any surrounding anion and form covalent bonds rather than ionic ones. Tin and lead are in Periods 5 and 6. Their +4 ions, being much larger, have lower charge density and can exist as discrete ionic species. Additionally, the inert pair effect is less pronounced in C and Si, so they mostly form covalent compounds. For Sn and Pb, the heavier elements can access both +2 (inert pair) and +4 states, with the +4 state achievable for Sn more readily than Pb.
(a)(ii) NaCl insoluble in organic solvents, AlCl₃ soluble:
By Fajans’ rules and the “like dissolves like” principle:
NaCl is a purely ionic compound (Na⁺ and Cl⁻). Organic solvents are non-polar or weakly polar and cannot provide enough energy to overcome the strong ionic lattice energy of NaCl. Water (highly polar) can solvate the ions through ion-dipole interactions.
AlCl₃, however, has significant covalent character — Al³⁺ is small with high charge density and strongly polarises Cl⁻, making the bond largely covalent. Being largely covalent (and therefore non-polar), AlCl₃ dissolves readily in non-polar/organic solvents according to “like dissolves like.”
(a)(iii) Melting points of Group 14 decrease down the group:
Going from C (diamond) → Si → Ge → Sn → Pb, the elements change from giant covalent structure to metallic structure. Carbon (diamond) has extremely strong, short C–C covalent bonds in a 3D network → very high melting point. As atomic size increases down the group, bond lengths increase and bond strength (bond dissociation energy) decreases, making it easier to break the structure → lower melting points. Sn and Pb are metallic with weak metallic bonds → lowest melting points.
(a)(iv) PCl₅ stable but BiCl₅ unstable; PCl₃ less stable but BiCl₃ stable:
This is explained by the inert pair effect. In Period 3, phosphorus readily uses all 5 valence electrons (using empty 3d orbitals to expand octet) → PCl₅ is stable. PCl₃ is also stable but PCl₅ is more so.
In Period 6, bismuth experiences the strong inert pair effect — the 6s² electrons are stabilised by relativistic effects and poor shielding by f-electrons, making them reluctant to participate in bonding. Therefore Bi prefers the +3 state (using only p electrons), and BiCl₃ is stable. BiCl₅ would require Bi to use its 6s² pair, which is energetically costly → BiCl₅ is unstable.
For phosphorus (Period 3), the inert pair effect is absent, so both PCl₃ and PCl₅ are stable, with PCl₅ being preferred.
(a)(v) CO (carbon II oxide) is toxic:
Carbon monoxide binds irreversibly to haemoglobin in the blood with an affinity approximately 200–250 times greater than oxygen. It forms carboxyhaemoglobin (COHb), which cannot carry oxygen. This prevents oxygen transport to body tissues, leading to cellular asphyxiation. Even small concentrations of CO in inhaled air can rapidly displace O₂ from haemoglobin, causing headache, dizziness, unconsciousness, and death.
(b) Hydrogen bonding:
(i) Definition: Hydrogen bonding is a special type of intermolecular (or intramolecular) electrostatic attraction between a hydrogen atom covalently bonded to a small, highly electronegative atom (F, O, or N) and a lone pair of electrons on another electronegative atom (F, O, or N) in the same or a neighbouring molecule.
(ii) Two conditions for hydrogen bond formation:
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The hydrogen atom must be covalently bonded to a small, highly electronegative atom — specifically F, O, or N.
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There must be a neighbouring molecule (or part of the same molecule) with a lone pair of electrons on an electronegative atom (F, O, or N) to act as the acceptor.
Q16 — Laws of Chemical Combination / Haber Process
(a) Four laws of chemical combination:
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Law of conservation of mass — matter is neither created nor destroyed in a chemical reaction.
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Law of definite (constant) proportions — a pure compound always contains the same elements in the same fixed proportion by mass.
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Law of multiple proportions — when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole number ratio.
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Law of reciprocal proportions (equivalent proportions) — the masses of elements A and B that separately combine with a fixed mass of element C are also the masses in which A and B combine with each other, or in a simple multiple thereof.
(b)(i) Percentage of zinc in the oxide (both methods):
Method 1: Zn = 0.325 g, O = 0.08 g
Total mass of oxide = 0.325 + 0.08 = 0.405 g
% Zn = (0.325/0.405) × 100 = 80.25%
Method 2: Zn = 1.625 g, O = 0.4 g
Total mass = 1.625 + 0.4 = 2.025 g
% Zn = (1.625/2.025) × 100 = 80.25%
Both methods give the same percentage (80.25%), confirming the law.
(b)(ii) Law illustrated:
This illustrates the Law of Definite (Constant) Proportions — zinc oxide always contains zinc and oxygen in the same fixed mass ratio (80.25% Zn : 19.75% O) regardless of how it was prepared.
Justification: Both experimental methods yield exactly 80.25% Zn by mass, confirming that ZnO has a fixed composition.
(b)(iii) Limitation of the law:
The law does not apply to non-stoichiometric (berthollide) compounds — certain compounds (e.g., Fe₀.₉₅O, iron(II) oxide) have variable compositions due to lattice defects, and do not have a fixed ratio of elements.
© Haber process: N₂ + H₂ → NH₃
Given: Volume of N₂ = 15 cm³, Volume of H₂ = 80 cm³
(i) Reaction stoichiometry:
N₂(g) + 3H₂(g) → 2NH₃(g)
1 volume of N₂ reacts with 3 volumes of H₂ to produce 2 volumes of NH₃.
(ii) Which gas is in excess:
From stoichiometry, 15 cm³ N₂ requires 15 × 3 = 45 cm³ H₂
Available H₂ = 80 cm³
H₂ in excess = 80 − 45 = 35 cm³
H₂ is in excess by 35 cm³. N₂ is the limiting reagent.
(iii) Volume of NH₃ produced:
From stoichiometry: 1 vol N₂ → 2 vol NH₃
15 cm³ N₂ → 15 × 2 = 30 cm³ NH₃
(iv) Volume of residual gas mixture:
At the end of the reaction:
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N₂ remaining = 0 (completely used up — limiting reagent)
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H₂ remaining = 35 cm³ (excess)
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NH₃ produced = 30 cm³
Total residual gas mixture = 35 + 30 = 65 cm³