QUESTIONS
SECTION A
1. Explain what happens when chlorine gas is bubbled into a cold, dilute solution of sodium hydroxide for a short time and then in excess/hot concentrated conditions. Write the necessary equation(s) to justify your answer.
2. Calculate the relative molecular mass of an unknown volatile compound from the following data obtained by Victor Meyer's method: 0.180g of the compound displaced 60cm³ of air at 27°C and 101.3 kNm⁻² pressure (saturated vapour pressure of water at 27°C = 3.57 kNm⁻²).
3. A metal, W belongs to group 13 of the periodic table. On the basis of this information, answer the following:
a) Give the chemical formula of the possible oxide of W.
b) Suggest the expected bond type in the oxide.
c) Write the chemical equation for the reaction of the metal oxide with aqueous sodium hydroxide (amphoteric behaviour).
d) Write an equation for the effect of adding aqueous hydrochloric acid to the metal oxide.
4a) Classify the following elements as s, p, d or f-block elements:
i) G (atomic number = 38); ii) H (atomic number = 47); iii) J (atomic number = 35)
4b) The quantum numbers n, l, m, and s for the electrons in the valence shells of atoms R and T are given below:
| R | n | l | m | s |
|---|---|---|---|---|
| 4 | 0 | 0 | −½ | |
| 4 | 0 | 0 | +½ | |
| 4 | 1 | −1 | +½ |
| T | n | l | m | s |
|---|---|---|---|---|
| 3 | 2 | −2 | +½ | |
| 3 | 2 | −1 | +½ | |
| 3 | 2 | 0 | +½ | |
| 3 | 2 | 1 | +½ | |
| 3 | 2 | 2 | +½ |
Write out the electron configuration of the valence shell in each case.
5. Given that the volume of nitrogen gas at standard temperature and pressure is 11.2 dm³, calculate the: (a) mole of nitrogen gas; (b) mass of nitrogen gas; (c) number of nitrogen molecules; (d) number of nitrogen atoms.
6a) Define the following terms: i) Enthalpy of atomisation; ii) Lattice enthalpy.
6b) The heat of formation of methane is given by: C + 2H₂ → CH₄; ΔHf = −75 kJ/mol. What does this information indicate?
7. Outline five general features of the transition (d-block) elements.
8. The rate determining step of a reaction involves two moles of X reacting alone (with no other reactant) to give a mole of product Y.
a) Write the rate law for the reaction.
b) State four things you might do to increase the rate of formation of Y.
9a) What is the diagonal relationship in the periodic table?
9b) Give the oxidation state of chromium in each of the following: i) Cr₂O₇²⁻ ii) CrO₄²⁻ iii) Cr₂O₃
10. A gas J diffuses 2.5 times as rapidly as another gas K under the same conditions of temperature and pressure. If the density of gas K is 3.6×10³ gm⁻³, calculate the density of gas J.
SECTION B
11. Explain the following:
a) Silver, though a d-block element, is not typically classed with the main transition series in terms of showing variable oxidation states. (2 marks)
b) A solution of iron(III) chloride is yellow-brown, while iron(II) chloride solution is pale green. (7 marks)
c) Lithium fluoride has a lower melting point than sodium fluoride, despite Li⁺ being smaller than Na⁺. (8 marks)
d) Carbon can form CCl₄ but silicon can form both SiCl₄ and SiCl₆²⁻ complex ions. (5 marks)
e) Occurrence of the +2 oxidation state as unusually stable for europium and ytterbium among the lanthanides. (3 marks)
12a) Outline two similarities between a fuel cell and an electrolytic cell. (2 marks)
12b) In a tabular form, state four differences between primary and secondary (rechargeable) cells. (4 marks)
12c) Metal L and M are both trivalent electrodes in an electrochemical cell with E°L = −0.74V and E°M = +1.50V 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 | K | Ca | Sc | Ti | V | Mn |
|---|---|---|---|---|---|---|
| Atomic radius (nm) | 0.227 | 0.197 | 0.162 | 0.147 | 0.134 | 0.126 |
| Ionic radius (nm) | 0.133 | 0.099 | 0.081 | — | — | 0.080 |
| First ionisation energy (kJ/mol) | 419 | 590 | 633 | 659 | 651 | 717 |
a) The atomic radius decreases only gradually from Sc to Mn compared to the sharp decrease from K to Ca. (6 marks)
b) The ionic radius of K⁺ is much smaller than the atomic radius of K. (3 marks)
c) The first ionisation energy of V is slightly less than that of Ti. (5 marks)
d) The first ionisation energy of Mn is notably higher than expected from the general trend of the d-block series. (6 marks)
e) K has a much larger atomic radius than Ca despite Ca having a higher atomic number. (2 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 a cationic complex of iron; ii) an example of an anionic complex of iron; iii) an example of a neutral complex of iron. (1 mark each)
14b) State three reasons why iron is able to exist in the form of the above a(i)–(iii). (3 marks)
14c) The compound A, cis-[Pt(NH₃)₂Cl₂] is isomeric with compound B, trans-[Pt(NH₃)₂Cl₂]
i) Name the type of isomerism. (1 mark)
ii) State one key difference in physical/chemical property between the two isomers. (4 marks)
iii) Draw and explain why cis-[Pt(NH₃)₂Cl₂] is polar while trans-[Pt(NH₃)₂Cl₂] is non-polar. (4 marks)
iv) Give the oxidation state of platinum in both isomers and write its electron configuration. (3 marks)
v) What is the coordination number and geometry of platinum in these complexes? (2 marks)
vi) State the medical significance of the cis isomer. (2 marks)
vii) Predict the magnetic property of the complex 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) Graham's law of diffusion; ii) Markovnikov's rule; iii) Boyle's law; iv) The Common Ion Effect; v) Faraday's laws of electrolysis.
15b) A 25cm³ sample of a Na₂CO₃ solution was titrated against 0.1M HCl using methyl orange indicator, requiring 22.50cm³ of the acid for complete reaction (2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂). Calculate the concentration of the Na₂CO₃ solution in mol/dm³ and in g/dm³.
15c) Give any two precautions to be observed when carrying out an acid-base titration to ensure accurate results. (5 marks)
16. At 450°C, the Kc for the reaction H₂(g) + I₂(g) ⇌ 2HI(g), ΔH = −9.4 kJ mol⁻¹, has a numerical value of 50.
a) Write an expression for Kc in terms of the degree of dissociation/conversion (α) and the total volume (V), given equal initial moles n₀ of H₂ and I₂. (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? (2 marks) ii) Write the unit for the reverse reaction. (1 mark)
d) What will be the effect on Kc if: i) more HI is added ii) pressure is increased iii) temperature is decreased? (2 marks)
e) Predict the effect on the equilibrium composition if: i) more H₂ is added; (2 marks) ii) pressure is increased; (2 marks) iii) temperature is decreased. (2 marks)
f) If 1 mole each of H₂ and I₂ are placed in a 1dm³ container at 450°C, calculate the equilibrium concentration of HI formed. (9 marks)
ANSWERS
Question 1
When Cl₂ is passed into cold, dilute NaOH, disproportionation occurs producing sodium chloride and sodium chlorate(I) (hypochlorite):
Cl₂ + 2NaOH(cold, dilute) → NaCl + NaOCl + H₂O
When passed into hot, concentrated NaOH, a different disproportionation occurs, producing chloride and chlorate(V):
3Cl₂ + 6NaOH(hot, concentrated) → 5NaCl + NaClO₃ + 3H₂O
In both cases chlorine simultaneously undergoes oxidation (0→+1 or +5) and reduction (0→−1), characteristic of a disproportionation reaction, with the specific products depending on temperature and concentration.
Question 2
Dry gas pressure = 101.3 − 3.57 = 97.73 kNm⁻² = 97730 Pa
V = 60cm³ = 60×10⁻⁶ m³; T = 300K
n = PV/RT = (97730×60×10⁻⁶)/(8.314×300) = 5.8638/2494.2 = 0.002351 mol
M = 0.180/0.002351 = ≈76.6 g/mol
Question 3
a) W₂O₃
b) Predominantly covalent/polar covalent (amphoteric) bond
c) W₂O₃ + 2NaOH → 2NaWO₂ + H₂O (forms sodium "aluminate"-type salt, showing acidic behaviour)
d) W₂O₃ + 6HCl → 2WCl₃ + 3H₂O (shows basic behaviour, confirming amphoteric nature)
Question 4
a)
i) G (Z=38, Sr): [Kr]5s² → s-block
ii) H (Z=47, Ag): [Kr]4d¹⁰5s¹ → d-block
iii) J (Z=35, Br): [Ar]3d¹⁰4s²4p⁵ → p-block
b) R: 2 electrons in 4s, 1 electron in 4p → R valence configuration: 4s²4p¹
T: 5 electrons, one in each of the five 3d orbitals, all same spin (Hund's rule) → T valence configuration: 3d⁵
Question 5
n(N₂) = V/Vm = 11.2/22.4 = 0.5 mol
mass = n×M = 0.5×28 = 14 g
molecules = n×NA = 0.5×6.023×10²³ = 3.0115×10²³ molecules
atoms = 2×molecules = 6.023×10²³ atoms
Question 6
a) i) Enthalpy of atomisation: The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state, under standard conditions.
ii) Lattice enthalpy: The enthalpy change when one mole of an ionic solid is formed from its constituent gaseous ions (or, conversely, the energy required to separate one mole of the ionic solid into its gaseous ions).
b) The negative sign indicates the reaction is exothermic; formation of one mole of CH₄ from its elements releases 75 kJ of heat, meaning CH₄ is more thermodynamically stable (lower in energy) than the separate elements C(s) and H₂(g).
Question 7
Five general features of transition (d-block) elements:
- Partially-filled d-orbitals in the atom or in at least one common oxidation state.
- Exhibit variable (multiple) oxidation states, differing typically by 1.
- Form coloured compounds/ions in solution due to d-d electronic transitions.
- Form complex ions readily, due to small ionic size, high charge density, and availability of vacant d-orbitals to accept ligand lone pairs.
- Exhibit catalytic activity (both as elements and compounds) and often show paramagnetic behaviour due to unpaired d-electrons.
Question 8
a) Rate = k[X]²
b) Four ways to increase rate of formation of Y:
- Increase the concentration of X.
- Increase the temperature (raises rate constant k).
- Use of a suitable catalyst.
- Increase the surface area of X (if X is a solid) or increase pressure (if X is gaseous).
Question 9
a) Diagonal relationship: The similarity in chemical properties observed between certain pairs of elements in the second and third periods that are diagonally adjacent to each other in the periodic table (e.g., Li & Mg, Be & Al, B & Si), arising because their similar charge-to-size ratios (polarising power) produce comparable bonding character, despite belonging to different groups.
b)
i) Cr₂O₇²⁻: Cr = +6
ii) CrO₄²⁻: Cr = +6
iii) Cr₂O₃: Cr = +3
Question 10
By Graham's Law: rateJ/rateK = √(ρK/ρJ)
2.5 = √(3.6×10³/ρJ)
6.25 = 3.6×10³/ρJ
ρJ = 3.6×10³/6.25 = 576 g/m³
Question 11
a) Silver's common oxidation state in aqueous chemistry is almost exclusively +1 ([Kr]4d¹⁰5s¹ atom loses the single 5s electron to give Ag⁺, [Kr]4d¹⁰, a fully-filled and hence stable configuration); since it does not readily show a range of accessible partially-filled d-configurations in solution, it lacks the variable-oxidation-state behaviour characteristic of true transition metals.
b) Fe³⁺ (d⁵) and Fe²⁺ (d⁶) both have partially-filled d-orbitals and undergo d-d electronic transitions in their aqueous complexes, but the different d-electron configurations, oxidation state, and degree of ligand field splitting between the two ions cause them to absorb different wavelengths of visible light — Fe³⁺ complexes typically absorb in the blue/violet region (appearing yellow-brown), while Fe²⁺ complexes absorb in a different region (appearing pale green), producing their characteristically different colours.
c) Although Li⁺ is smaller than Na⁺, lithium's very small ionic radius gives it an exceptionally high charge density (high polarising power), which introduces significant covalent character into the Li–F bond (Fajan's rules), weakening the purely ionic lattice interactions. NaF, with a larger, less polarising Na⁺ ion, retains more purely ionic (electrostatic) character in its lattice, giving it a stronger, more symmetric ionic bond and hence a higher melting point than LiF.
d) Carbon (period 2) has no accessible d-orbitals in its valence shell, limiting it to a maximum coordination number of 4 (as in CCl₄), unable to expand its octet. Silicon (period 3) has energetically accessible 3d orbitals, allowing it to expand its octet beyond 4 bonds, forming the six-coordinate complex ion SiCl₆²⁻ in addition to the normal four-coordinate SiCl₄.
e) Europium and ytterbium show unusually stable +2 oxidation states because losing the additional electron from the 4f subshell (beyond the normal +3 state) leaves them with a half-filled (4f⁷, for Eu²⁺) or completely-filled (4f¹⁴, for Yb²⁺) 4f subshell, which is an extra-stable electronic configuration due to exchange energy (Hund's rule) and symmetric electron distribution — making the +2 state exceptionally favourable compared to other lanthanides.
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 ionic conduction occurring through the electrolyte and electron flow through an external circuit/wire.
b) Four differences:
| Feature | Primary cell | Secondary (rechargeable) cell |
|---|---|---|
| Rechargeability | Cannot be recharged; single use | Can be recharged by reversing the reaction |
| Reaction reversibility | Chemical reaction is irreversible | Chemical reaction is reversible |
| Example | Dry (Leclanché) cell, alkaline cell | Lead-acid battery, Nickel-cadmium battery |
| Cost/lifespan | Cheaper, shorter overall lifespan | More expensive initially, longer overall lifespan |
c) E°L = −0.74V, E°M = +1.50V
i) Metal L (more negative reduction potential) serves as the anode (more readily oxidised); Metal M (higher/more positive reduction potential) serves as the cathode.
ii)
Anode (oxidation): L → L³⁺ + 3e⁻
Cathode (reduction): M³⁺ + 3e⁻ → M
iii) Overall: L + M³⁺ → L³⁺ + M
iv) E°cell = E°cathode − E°anode = 1.50 − (−0.74) = +2.24 V
v) ΔG = −nFE° = −(3)(96500)(2.24) = −648,480 J = −648.48 kJ/mol
vi) Since ΔG is negative, the reaction is spontaneous.
vii) ln K = 648480/(8.314×298) = 648480/2477.6 = 261.79
log K = 261.79/2.303 = 113.68
K ≈ 4.79×10¹¹³
Question 13
a) From K to Ca, the additional electron is added to the outermost 4s subshell with poor inter-shell shielding from the increasing nuclear charge, causing a relatively sharp radius decrease. From Sc to Mn, additional electrons are added to the inner 3d subshell rather than the outer 4s shell; the 3d electrons shield the 4s electrons from the increasing nuclear charge fairly effectively, so the outer atomic radius decreases only gradually across the d-block series.
b) K⁺ is formed by removing K's single outer 4s electron, losing the entire outermost (n=4) shell; the remaining electrons (n=3 shell) experience the same nuclear charge with considerably less shielding and electron-electron repulsion, pulling them much closer to the nucleus, making the cation substantially smaller than the parent atom.
c) V has electron configuration [Ar]3d³4s², while Ti has [Ar]3d²4s²; V has one more 3d electron than Ti, providing slightly greater electron-electron repulsion in the more populated 3d subshell along with similar shielding, making the outermost 4s electron marginally easier to remove in V — resulting in a slightly lower first ionisation energy than expected from a simple increasing trend, i.e. V's IE1 is very close to but slightly below Ti's due to this subtle effect (actually reflects that Ti loses first electron from an already low-population 3d²4s² configuration).
d) Mn has electron configuration [Ar]3d⁵4s², with a half-filled 3d⁵ subshell that is extra stable due to symmetric electron distribution and maximised exchange energy (Hund's rule); this extra stability makes it harder than expected to remove an electron, causing Mn's first ionisation energy to be notably higher than the general smooth trend across the d-block series would predict.
e) Although Ca has one more proton than K, Ca's outer 4s electron configuration is 4s² compared to K's 4s¹; the increased nuclear charge in Ca is not fully compensated by additional shielding from the second 4s electron (electrons in the same subshell shield each other poorly), so the effective nuclear charge experienced by the outer electrons increases from K to Ca, pulling the electron cloud in more strongly and making Ca's atomic radius smaller than K's despite the higher atomic number.
f) s-block elements: K, Ca
g) d-block elements: Sc, Ti, V, Mn
Question 14
a)
i) Cationic complex: [Fe(H₂O)₆]³⁺ (or [Fe(CN)₆]³⁻... note this is anionic; for cationic use [Fe(H₂O)₆]²⁺/³⁺)
ii) Anionic complex: [Fe(CN)₆]⁴⁻ (or [FeCl₄]⁻)
iii) Neutral complex: [Fe(CO)₅]
b) Three reasons:
- Iron has several stable oxidation states (commonly +2, +3), allowing different overall complex charges to form.
- Iron has vacant/hybridisable d-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. H₂O, CO) and/or anionic ligands (e.g. CN⁻, Cl⁻) in different ratios allows overall cationic, anionic, or neutral complexes to be formed.
c) i) Type of isomerism: Geometrical (cis-trans) isomerism
ii) Key difference: Cis-[Pt(NH₃)₂Cl₂] is a polar molecule with a measurable dipole moment and is used as an anticancer drug (cisplatin), whereas trans-[Pt(NH₃)₂Cl₂] is non-polar (zero net dipole moment) and shows no significant anticancer activity.
iii) In the cis isomer, the two identical NH₃ ligands are adjacent (90° apart) and the two Cl ligands are also adjacent, so the bond dipoles do not cancel, giving a net molecular dipole moment (polar). In the trans isomer, the two NH₃ ligands are opposite each other (180° apart) and the two Cl ligands are opposite each other; the bond dipoles point in opposite directions and cancel exactly, giving zero net dipole moment (non-polar).
iv) [Pt(NH₃)₂Cl₂]: Pt + 0(NH₃×2) + (−1×2)(Cl) = 0 → Pt = +2 (same in both isomers)
Electron configuration of Pt²⁺ ([Xe]4f¹⁴5d⁸): [Xe]4f¹⁴5d⁸
v) Coordination number: 4; Geometry: Square planar
vi) Cis-[Pt(NH₃)₂Cl₂] (cisplatin) is a widely used chemotherapy drug, binding to DNA in cancer cells and causing crosslinks that inhibit DNA replication, ultimately triggering cell death (apoptosis) in rapidly dividing tumour cells.
vii) Pt²⁺ (d⁸) in a strong-field square planar geometry has all electrons paired in the lower-energy d-orbitals, giving no unpaired electrons, so the complex is predicted to be diamagnetic.
Question 15
a)
i) Graham's Law of Diffusion: At constant temperature and pressure, the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its density (or molar mass). Application: used to determine unknown molar masses of gases and forms the basis of gaseous isotope separation (e.g., uranium enrichment).
ii) Markovnikov's Rule: When a protic acid (HX) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon with the greater number of hydrogen atoms already attached, while the halide attaches to the more substituted carbon. Application: predicts the major product in electrophilic addition reactions of unsymmetrical alkenes.
iii) Boyle's Law: At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume (PV=constant). Application: used in calculating gas behaviour under changing pressure/volume conditions, e.g., in respiration and industrial gas compression.
iv) Common Ion Effect: The solubility (or degree of ionisation) of a weak electrolyte is suppressed by the addition of a strong electrolyte containing an ion common to the weak electrolyte, due to the equilibrium shifting to reduce the concentration of the added ion (Le Chatelier's principle). Application: used in qualitative analysis for selective precipitation and in controlling buffer solution pH.
v) Faraday's Laws of Electrolysis: (1st law) The mass of a substance liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte; (2nd law) for a given quantity of electricity, the masses of different substances liberated are proportional to their equivalent weights. Application: used to calculate the mass of metal deposited during electroplating and in industrial electrolytic extraction/purification of metals.
b) mol HCl = 0.1 × (22.50/1000) = 0.00225 mol
mol Na₂CO₃ = mol HCl/2 = 0.00225/2 = 0.001125 mol
Concentration = mol/volume = 0.001125/(25/1000) = 0.001125/0.025 = 0.045 mol/dm³
Mass concentration = 0.045 × M(Na₂CO₃) = 0.045 × 106 = 4.77 g/dm³
c) Two precautions for accurate acid-base titration:
- Rinse the burette and pipette with the solution they are to contain before use (to avoid dilution of the solution by residual water), and ensure no air bubbles are trapped in the burette tip before beginning.
- Read the burette at eye level to avoid parallax error, and add the titrant dropwise (a drop at a time) near the expected end point, swirling the flask continuously to ensure thorough mixing and accurate detection of the colour change.
Question 16
a) Let n₀ = initial moles of each of H₂ and I₂, α = fraction converted, V = total volume (moles of I₂ and H₂ are equal, and total moles don't change in this reaction since Δn(gas)=0):
At equilibrium: [H₂]=[I₂]=n₀(1−α)/V, [HI]=2n₀α/V
Kc = [2n₀α/V]² / {[n₀(1−α)/V]²} = 4α²/(1−α)²
b) Kc = [HI]²/([H₂][I₂]); units cancel completely since Δn(gas)=0: Kc is dimensionless (no units)
c) i) For reverse reaction 2HI⇌H₂+I₂: Kc(reverse) = 1/Kc(forward) = 1/50 = 0.02
ii) Unit: dimensionless (same reasoning as forward reaction)
d) (Kc depends only on temperature)
i) More HI added: No effect (Kc unchanged)
ii) Pressure increased: No effect (Kc unchanged, since Δn(gas)=0, pressure changes don't shift this equilibrium at all)
iii) Temperature decreased: Since ΔH is negative (exothermic forward reaction), by Le Chatelier's principle, decreasing temperature favours the forward (heat-releasing) reaction, so Kc increases
e)
i) More H₂ added: Equilibrium shifts right (forward), increasing HI formed (and consuming some added H₂ and existing I₂).
ii) Pressure increased: No shift in equilibrium position (since moles of gas are equal on both sides, Δn=0).
iii) Temperature decreased: Since the reaction is exothermic, equilibrium shifts right (forward), favouring HI formation.
f) Since Kc = 4α²/(1−α)² = 50 (dimensionless, and independent of volume since Δn=0):
2α/(1−α) = √50 = 7.071
2α = 7.071(1−α)
2α = 7.071 − 7.071α
9.071α = 7.071
α = 0.7796
[HI] = 2n₀α/V = 2(1)(0.7796)/1 = ≈1.559 mol/dm³