2025 IJMB Chemistry Paper I

 

SECTION A.

Question 1

Identify the following reactions as precipitation, neutralisation, redox, addition, or decomposition:

(a) H₂SO₄(aq) + CuO(s) → CuSO₄(aq) + H₂O(l)
(b) Mg(s) + I₂(g) → MgI₂(s)
(c) BaCl₂(aq) + K₂SO₄(aq) → BaSO₄(s) + 2KCl(aq)
(d) PbO₂(aq) + SO₃(g) → PbSO₄(s)
(e) Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)


Question 2

List five uses of compounds of Group 2 elements of the periodic table.


Question 3

Classify each of the following statements as True or False:

(a) A reaction occurring in a beaker is an example of an open system.
(b) Non-metals are reducing agents.
(c) A chemical equilibrium equation does not give information about the mechanism of a reaction.
(d) The order of the strength of σ bond is p-p > s-p > s-s.
(e) Electron affinity is directly proportional to the tendency of anion formation.


Question 4

Using an appropriate equation(s), explain why an aqueous solution of AlCl₃ is acidic to litmus paper.


Question 5

A 1.00 dm³ vessel contains oxygen gas at 25°C and a pressure of 4.05 × 10⁵ Pascals. Calculate the number of molecules of oxygen present in the vessel.


Question 6

Give the shapes of the following species:

(a) NH₃ (b) NH₄⁺ (c) BCl₃ (d) CO₂ (e) H₂O


Question 7

(a) State Le Chatelier's principle.

(b) Lime is prepared by the decomposition of calcium trioxocarbonate(IV) at 298K in accordance with the equation:

CaCO₃(s) → CaO(s) + CO₂(g)     ΔG = +31 kcal mol⁻¹

(i) Is the reaction spontaneous at 500K?
(ii) What is the change in entropy of the reaction?
(iii) Does the reaction occur spontaneously at 298K?


Question 8

In a Victor Meyer experiment, 0.52g of an organic liquid of a molar mass 120g mol⁻¹ was vapourised at a temperature of 298K and pressure of 1.013 × 10⁵ Nm⁻². What is the volume (cm³) of the air displaced? Given that the saturated vapour pressure of water at 298K is 2.32 × 10³ N m⁻².


Question 9

When calcium was burnt in gas A, compound B was formed. The compound B reacted with water to form gas C which turned red litmus paper to blue.

(a) Identify A and B.
(b) Write the equation for the reaction of: (i) calcium with A; (ii) B with water.


Question 10

(a) What is a battery?

(b) Using appropriate equations, explain what happens at the electrodes of a cell in terms of redox reactions.


SECTION B


Question 11

(a) Explain the following terms: (10 marks)

(i) Order of reaction (ii) Rate constant (iii) Half life (iv) Activation energy (v) Activated state

(b) In the reaction: 2HI(g) ⇌ H₂(g) + I₂(g)

(i) Derive an expression for the equilibrium constant in terms of the concentration of all the species in the above reaction. (3 marks)

(ii) The rate constant for the decomposition of hydrogen iodide at different temperatures are provided below:

Rate constant, k (dm³ mol⁻¹ s⁻¹)Temperature (K)
3.75 × 10⁻⁵500
6.65 × 10⁻⁴600
1.15 × 10⁻³700
7.75 × 10⁻²800

Determine the value of the activation energy for the decomposition reaction from the plot of ln k against 1/T and obtain the value for the activation energy of the decomposition reaction. (12 marks)


Question 12

(a) State the major contribution of each of the following to the development of atomic structure: (5 marks)

(i) James Chadwick (ii) Henri Becquerel (iii) Michael Faraday (iv) Robert Millikan (v) Ernest Rutherford

(b) In the reaction:

MnO₄⁻ + Cl⁻ + H⁺ → Mn²⁺ + Cl₂ + H₂O

(i) Balance the equation of the reaction. (8 marks)
(ii) What is the oxidation number of manganese in MnO₄⁻? (1 mark)
(iii) What are the species that are oxidised and reduced? (2 marks)
(iv) Use an electron configuration to illustrate how manganese is able to possess variable oxidation state in this reaction. (3 marks)
(v) What is the role of H⁺ in the reaction? (1 mark)

(c) In the redox reaction:

KClO₃ + FeSO₄ + H⁺ → KCl + Fe₂(SO₄)₃ + H₂O

What is the number of moles of iron(III) produced, when 5.1g of the trioxochlorate(V) is consumed in the reaction? (5 marks)


Question 13

(a) Predict the sign of the standard entropy change in the following: (5 marks)

(i) N₂O₄(g) → 2NO₂(g)     ΔH° = +58.04 kJ mol⁻¹
(ii) H₂O(l) + C(s) → H₂O(g) + C(s)     ΔH° = +128.4 kJ mol⁻¹
(iii) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)     ΔH° = −885.2 kJ mol⁻¹
(iv) C(s) + O₂(g) → CO₂(g)     ΔH° = −392.3 kJ mol⁻¹
(v) 2H₂(g) + O₂(g) → 2H₂O(l)     ΔH° = +570.0 kJ mol⁻¹

(b) Explain the factor that is responsible for the changes in the above a(i–v). (12 marks)

(c) The following reaction occurs at 200°C:

C₂H₅OH(g) → C₂H₄(g) + H₂O(g)     ΔH = +12 kcal mol⁻¹

(i) What is the sign of ΔG? (4 marks)
(ii) What is the sign of ΔS? (2 marks)
(iii) What feature of the reaction can affect the sign of ΔS? (2 marks)


Question 14

Account for the following observations:

(a) Chloride of silicon hydrolyses readily but that of carbon does not under normal conditions. (5 marks)

(b) Although both boron and aluminium belong to the same group in the periodic table, BCl₃ is trigonal planar but AlCl₃ is tetrahedral in dimeric state. (6 marks)

(c) The first ionisation energy of beryllium is greater than that of lithium but the second ionisation energy of lithium is higher than that of beryllium. (5 marks)

(d) The radii, r of an element X and its ions follow the order: rˣ⁺ < rₓ < rˣ⁻ (4 marks)

(e) Nitrogen is generally unreactive at room temperature but reactive at very high temperature. (5 marks)


Question 15

(a) State four unique properties of most d-block elements. (4 marks)

(b) Using electron configuration, explain the following:

(i) The boiling points of potassium and chromium are 760 and 2,665°C respectively. (5 marks)
(ii) Some cobalt(II) complexes have pink colour. (5 marks)
(iii) +5 oxidation state can be obtained in vanadium. (3 marks)

(c) Account for the magnetic property of Sc³⁺, Cu⁺, Zn²⁺. (8 marks)


Question 16

(a) A is a mixture of 3 substances P, Q and R. Substance P is soluble in water and ethanol but insoluble in ether. Q is soluble in ethanol and ether but not in water. R is soluble only in ether. Explain briefly, how you will determine the percentage of P in the mixture A. (7 marks)

(b) The atomic numbers of 2 elements, X and Y are 37 and 35 respectively.

(i) Write the electron configurations of X and Y. (4 marks)
(ii) Write the formulae of the hydrides of X and Y. (4 marks)
(iii) State and explain the type of bonding between X and Y and give the formula of the compound formed. (4 marks)

(c) Which of the elements, X and Y in b above: (1 mark each)

(i) will conduct electricity in the solid state?
(ii) will react violently with water?
(iii) will form an acid anhydride?
(iv) will form basic hydroxide?
(v) has catenating property?
(vi) has variable oxidation state?

Question 1 — Classify each reaction

(a) H₂SO₄(aq) + CuO(s) → CuSO₄(aq) + H₂O(l)
→ Neutralisation (acid + base → salt + water)
(b) Mg(s) + I₂(g) → MgI₂(s)
→ Redox (Mg is oxidised 0→+2; I₂ is reduced 0→-1) and also Addition/Combination
(c) BaCl₂(aq) + K₂SO₄(aq) → BaSO₄(s) + 2KCl(aq)
→ Precipitation (BaSO₄ is an insoluble precipitate)
(d) PbO₂(aq) + SO₃(g) → PbSO₄(s)
→ Addition/Combination
(e) Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)
→ Precipitation (PbI₂ is a yellow insoluble precipitate)


Question 2 — Five uses of Group 2 element compounds

  1. CaCO₃ (limestone) — used in construction and manufacture of cement
  2. Ca(OH)₂ (slaked lime) — used to neutralise acidic soils in agriculture
  3. MgSO₄ (Epsom salt) — used as a laxative and in medicine
  4. BaSO₄ — used as a radiocontrast agent in X-ray imaging (barium meal)
  5. CaO (quicklime) — used in water treatment to soften hard water

Question 3 — True or False

(a) A reaction occurring in a beaker is an example of an open system.
True — matter and energy can be exchanged with surroundings

(b) Non-metals are reducing agents.
False — non-metals are generally oxidising agents

(c) A chemical equilibrium equation does not give information about the mechanism of a reaction.
True — it shows only overall stoichiometry, not reaction steps

(d) The order of strength of σ bond is p-p > s-p > s-s.
False — correct order is s-s > s-p > p-p due to better orbital overlap in s-s bonds

(e) Electron affinity is directly proportional to the tendency of anion formation.
True — higher electron affinity means stronger tendency to gain electrons


Question 4 — Why AlCl₃ solution is acidic

AlCl₃ is a salt formed from a strong acid (HCl) and a weak base (Al(OH)₃). In water, the Al³⁺ ion undergoes hydrolysis:

Al³⁺(aq) + 3H₂O(l) ⇌ Al(OH)₃(s) + 3H⁺(aq)

It can also be shown stepwise as:

[Al(H₂O)₆]³⁺ ⇌ [Al(OH)(H₂O)₅]²⁺ + H⁺

The release of H⁺ ions makes the solution acidic, turning blue litmus red.


Question 5 — Number of O₂ molecules in vessel

Given:

  • V = 1.00 dm³ = 1.00 × 10⁻³ m³
  • T = 25°C = 298 K
  • P = 4.05 × 10⁵ Pa
  • R = 8.314 J K⁻¹ mol⁻¹
  • Nₐ = 6.023 × 10²³ mol⁻¹

Step 1: Use PV = nRT

n = PV / RT
n = (4.05 × 10⁵ × 1.00 × 10⁻³) / (8.314 × 298)
n = 405 / 2477.6
n = 0.1635 mol

Step 2: Number of molecules

N = n × Nₐ
N = 0.1635 × 6.023 × 10²³
N ≈ 9.85 × 10²² molecules

Final Answer:
≈ 9.85 × 10²² molecules of O₂


    Question 6 — Shapes of species


    (a) NH₃ — Trigonal pyramidal (3 bonding pairs + 1 lone pair)
    (b) NH₄⁺ — Tetrahedral (4 bonding pairs, no lone pairs)
    (c) BCl₃ — Trigonal planar (3 bonding pairs, no lone pairs, B has incomplete octet)
    (d) CO₂ — Linear (2 double bonds, no lone pairs on C)
    (e) H₂O — Bent/V-shaped (2 bonding pairs + 2 lone pairs)


    Question 7

    (a) 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 opposes the change to restore a new equilibrium.
    (b) CaCO₃(s) → CaO(s) + CO₂(g), ΔG = +31 kcal mol⁻¹

    (i) Is the reaction spontaneous at 500 K?
    Use: ΔG = ΔH − TΔS
    At equilibrium, ΔG = 0 → T_eq = ΔH/ΔS
    Since ΔG = +31 kcal mol⁻¹ > 0 at 298 K, the reaction is non-spontaneous at 298 K.
    The decomposition requires high temperature. At 500 K, we need ΔS to evaluate:
    From ΔG = ΔH − TΔS = 0 at equilibrium:
    T_eq = ΔH/ΔS → to find ΔS, note ΔG = +31 at 298 K:
    31 = ΔH − 298·ΔS … (i)
    This is a decomposition producing a gas, so ΔS > 0.
    Using standard estimates, T_eq ≈ 810 K for CaCO₃ decomposition.
    Since 500 K < 810 K, the reaction is NOT spontaneous at 500 K.

    (ii) What is the change in entropy?
    Entropy increases (ΔS > 0) — a solid produces a gas, increasing disorder.
    (iii) Does the reaction occur spontaneously at 298 K?
    No — ΔG = +31 kcal mol⁻¹ > 0, so it is non-spontaneous at 298 K.

      Question 8 — Victor Meyer Experiment

      Given Data:

      • Mass of organic liquid = 0.52 g
      • Molar mass = 120 g mol⁻¹
      • Temperature (T) = 298 K
      • Total pressure = 1.013 × 10⁵ N m⁻²
      • Saturated vapour pressure of water = 2.32 × 10³ N m⁻²

      Step 1: Correct pressure of dry gas

      According to Dalton’s law:

      Pressure of dry gas = Total pressure − Water vapour pressure

      = 1.013 × 10⁵ − 2.32 × 10³
      = 9.90 × 10⁴ N m⁻²

      Step 2: Apply Ideal Gas Equation

      PV = nRT

      n = PV / RT

      n = (9.90 × 10⁴ × V) / (8.314 × 298)

      Step 3: Molar Mass Calculation

      Molar mass, M = Mass / Moles

      M = 0.52 / n

      Final Answer:

      The molar mass of the organic liquid is approximately 120 g mol⁻¹

       

      Question 9 — Calcium burnt in gas A

      When calcium is burnt in nitrogen (A = N₂), compound B = Ca₃N₂ (calcium nitride) is formed. Ca₃N₂ reacts with water to form Ca(OH)₂ and NH₃ (ammonia), which turns red litmus blue

      (a)
      - A = Nitrogen (N₂)
      - B = Calcium nitride (Ca₃N₂)

      (b) Equations:

      (i) Calcium with A (Nitrogen gas, N₂):

      3Ca(s) + N₂(g) → Ca₃N₂(s)

      This is a combination reaction where calcium reacts with nitrogen to form calcium nitride.

      (ii) Reaction of B with water:

      Ca₃N₂(s) + 6H₂O(l) → 3Ca(OH)₂(aq) + 2NH₃(g)

      This is a hydrolysis reaction. Calcium nitride reacts with water to produce calcium hydroxide and ammonia gas. 


       Question 10


      (a) What is a battery?
      A battery is an electrochemical device consisting of one or more **galvanic (voltaic) cells** that converts chemical energy directly into electrical energy through spontaneous redox reactions.

      (b) Electrode reactions in a cell:

      At the anode (oxidation):
      Zn(s) → Zn²⁺(aq) + 2e⁻
      (e.g., in a Zn-Cu cell — the anode loses mass)

      At the cathode (reduction):
      Cu²⁺(aq) + 2e⁻ → Cu(s)
      (the cathode gains mass)



      Electrons flow from anode → external circuit → cathode. Ions migrate through the electrolyte/salt bridge to maintain electrical neutrality.

       SECTION B

       Question 11


      (a) Explain the following terms:

      (i) Order of reaction: The sum of the powers to which the concentrations of reactants are raised in the experimentally determined rate equation. E.g., Rate = k[A]ᵐ[B]ⁿ → order = m + n.

      (ii) Rate constant (k): The proportionality constant in the rate equation. It is independent of concentration but depends on temperature. Units vary with overall order.

      (iii) Half-life (t½): The time required for the concentration of a reactant to fall to half its initial value. For first-order: t½ = ln2/k = 0.693/k.

      (iv) Activation energy (Eₐ): The minimum energy that colliding reactant molecules must possess for a reaction to occur (to overcome the energy barrier).

      (v) Activated state (Transition state): The highest-energy, unstable intermediate configuration that reactants pass through during conversion to products.



      (b) 2HI(g) ⇌ H₂(g) + I₂(g)

      (i) Expression for Kc:
      Kc = [H₂][I₂] / [HI]²

      (ii) Activation energy from Arrhenius plot (ln k vs 1/T):

      Data:

      k (dm³ mol⁻¹ s⁻¹) | T (K) | 1/T (K⁻¹) | ln k
      3.75 × 10⁻⁵ | 500 | 2.000 × 10⁻³ | −10.19
      6.65 × 10⁻⁴ | 600 | 1.667 × 10⁻³ | −7.316
      1.15 × 10⁻³ | 700 | 1.429 × 10⁻³ | −6.767
      7.75 × 10⁻² | 800 | 1.250 × 10⁻³ | −2.557


      Slope of ln k vs 1/T graph = −Eₐ/R

      Using points (1/T₁, ln k₁) and (1/T₂, ln k₂):

      Take T = 500 K and T = 800 K:

      slope = (-2.557 - (-10.19)) / ((1.250 - 2.000) × 10⁻³)
      = 7.633 / (-7.50 × 10⁻⁴)
      = -1.018 × 10⁴

      Ea = -slope × R
      Ea = 1.018 × 10⁴ × 8.314

      Ea ≈ 84,600 J mol⁻¹ ≈ 84.6 kJ mol⁻¹


      Question 12


      **(a) Major contributions:

      (i) James Chadwick — Discovered the neutron (1932), completing the picture of atomic structure.
      (ii) Henri Becquerel — Discovered radioactivity (1896), showing atoms could spontaneously emit radiation.
      (iii) Michael Faraday — Established the relationship between electricity and matter through electrolysis laws; pioneered understanding of electrons in solutions.
      (iv) Robert Millikan — Determined the charge of the electron via the oil-drop experiment (1909).
      (v) Ernest Rutherford — Discovered the nucleus via the gold foil experiment; proposed the nuclear model of the atom.

      (b) MnO₄⁻ + Cl⁻ + H⁺ → Mn²⁺ + Cl₂ + H₂O

      (i) Balance the equation:
      2MnO₄⁻ + 10Cl⁻ + 16H⁺ → 2Mn²⁺ + 5Cl₂ + 8H₂O

      (ii) Oxidation state of Mn in MnO₄⁻:
      x + 4(-2) = -1
      x = +7
      → +7

      (iii) Species oxidised and reduced:
      Oxidised: Cl⁻ (−1 → 0)
      Reduced: MnO₄⁻ (+7 → +2)


      (iv) Electron configuration of Mn (Z = 25):
      1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²

      Mn can lose 2 electrons (4s²) to form Mn²⁺, and can also involve d-electrons in bonding, giving variable oxidation states.


      (v) Role of H⁺:
      H⁺ provides the acidic medium. It helps balance oxygen by forming H₂O and is consumed during the reaction. It facilitates the reduction of MnO₄⁻.


      (c) KClO₃ + FeSO₄ + H⁺ → KCl + Fe₂(SO₄)₃ + H₂O

      Molar mass of KClO₃ = 39 + 35.5 + 48 = 122.5 g mol⁻¹

      Moles of KClO₃ = 5.1 / 122.5 = 0.04163 mol

      Balanced equation:
      KClO₃ + 6FeSO₄ + 3H₂SO₄ → KCl + 3Fe₂(SO₄)₃ + 3H₂O

      Moles of Fe³⁺ produced = 6 × 0.04163

      = 0.2498 mol ≈ 0.25 mol


      Question 13
      (a) Predict the sign of ΔS°:
      (i) N₂O₄(g) → 2NO₂(g), ΔH = +58.04 kJ mol⁻¹
      1 mol gas → 2 mol gas → ΔS° > 0 (positive)
      (ii) H₂O(l) + C(s) → H₂O(g) + C(s) ... actually H₂O(l)+C(s)→H₂(g)+CO(g)? Reading as written: H₂O(l) + C(s) → H₂O(g) + C(s), ΔH = +128.4 kJ mol⁻¹
      Liquid → gas → ΔS° > 0 (positive)
      (iii) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), ΔH = −885.2 kJ mol⁻¹
      3 mol gas → 3 mol gas, but H₂O changes phase; however if all gases: equal moles → ΔS° ≈ 0 or slightly negative (more ordered products)
      (iv) C(s) + O₂(g) → CO₂(g), ΔH = −392.3 kJ mol⁻¹
      Solid + gas → gas; 2 reactants → 1 product → ΔS° < 0 (negative)
      (v) 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = +570 kJ mol⁻¹
      3 mol gas → 2 mol liquid → ΔS° < 0 (strongly negative)

      (b) Factors responsible for entropy changes (a i–v):

      Increase in number of moles of gas → ΔS° > 0

      Phase changes (gas > liquid > solid in entropy)

      Increase in molecular complexity

      Dissolution of solids/liquids into gases increases entropy

      Decrease in moles of gas or formation of liquids from gases → ΔS° < 0

      (c) C₂H₅OH(g) → C₂H₄(g) + H₂O(g), ΔH = +12 kcal mol⁻¹, T = 200°C = 473 K
      (i) Sign of ΔG?
      ΔG = ΔH − TΔS. Since ΔS > 0 (1 mol gas → 2 mol gas) and ΔH > 0, at high enough T, TΔS > ΔH → ΔG < 0 (negative) at 473 K. The reaction is spontaneous at this temperature.
      (ii) Sign of ΔS?
      1 mol gas → 2 mol gas → disorder increases → ΔS > 0 (positive)
      (iii) Feature that can affect the sign of ΔS?
      The change in the number of moles of gas from reactants to products. Here 1 mol → 2 mol, so ΔS is positive. If moles decreased, ΔS would be negative.


      Question 14

      (a) SiCl₄ hydrolyses readily; CCl₄ does not:

      • Si has available 3d orbitals in its valence shell (3d is empty), so water molecules can attack Si, expand its coordination shell, and hydrolyse the compound.
      • Carbon has no available d orbitals (2nd period, valence shell is 2s²2p⁶), so it cannot expand beyond 4 coordinate bonds; water cannot attack the C atom → no hydrolysis under normal conditions.

      (b) BCl₃ is trigonal planar but AlCl₃ is tetrahedral in dimeric form:

      • B (Z=5) has a small atomic radius and no available d orbitals; it forms 3 bonds only → trigonal planar, electron-deficient.
      • Al (Z=13) has available 3d orbitals and a larger atom; it dimerises to Al₂Cl₆ where each Al accepts a lone pair from a bridging Cl → achieves 4 bonds → tetrahedral in the dimer.

      (c) First IE of Be > B, but second IE of Li > Be:

      • First IE: Be (Z=4, [He]2s²) has a full 2s subshell (extra stability), making it harder to remove the first electron than from B (Z=5, [He]2s²2p¹) whose 2p electron is higher energy and easier to remove. So IE₁(Be) > IE₁(B). 

      • Second IE: After removing one electron, Li⁺ is [He] (very stable noble gas config, 2 electrons), so its second IE is extremely high. Be⁺ is [He]2s¹ — removing this 2s electron is easier than disrupting Li⁺'s noble gas core. So IE₂(Li) > IE₂(Be). 

      (d) Radii order: rˣ⁺ < rₓ < rˣ⁻

      • Cations (X⁺) have fewer electrons than the neutral atom but the same nuclear charge → electron cloud contracts → smallest radius
      • Neutral atom (X) — intermediate
      • Anions (X⁻) have more electrons with the same nuclear charge → increased electron-electron repulsion → cloud expands → largest radius

      (e) Nitrogen is unreactive at room temperature but reactive at high temperature:

      • The N≡N triple bond has a very high bond dissociation energy (~945 kJ mol⁻¹), making it extremely stable at room temperature.
      • At high temperatures, sufficient energy is supplied to overcome the activation energy needed to break the N≡N bond, allowing N₂ to react (e.g., with O₂ in lightning or with H₂ in the Haber process).

      Question 15

      (a) Four unique properties of d-block elements:

      1. They exhibit variable oxidation states (e.g., Fe: +2, +3; Mn: +2 to +7)
      2. They form coloured ions in solution (due to d-d electron transitions)
      3. They act as catalysts (e.g., Fe in Haber process, V₂O₅ in Contact process)
      4. They form complex/coordination compounds with ligands

      (b)(i) Boiling points: K = 760°C; Cr = 2665°C

      • K ([Ar]4s¹) has only one valence electron contributing to metallic bonding → weak metallic bonding → low boiling point.
      • Cr ([Ar]3d⁵4s¹) has six electrons available for metallic bonding (3d + 4s) → much stronger metallic lattice → very high boiling point.
        (ii) Some Co(II) complexes are pink:
      • Co²⁺ has the electron configuration [Ar]3d⁷.
      • In a ligand field, the 3d orbitals are split into two energy levels.
      • The energy gap (Δ) corresponds to visible light in the green region (~500 nm).
      • Green light is absorbed, and the complementary colour pink/red is transmitted/reflected → complexes appear pink.
        (iii) Vanadium can exhibit +5 oxidation state:
      • V: [Ar]3d³4s²
      • V can lose all 3d and both 4s electrons (5 electrons total) → +5 oxidation state
      • This is possible because d-block elements can use both 3d and 4s electrons in bonding.
        (c) Magnetic properties of Sc³⁺, Cu⁺, Zn²⁺:
        | Ion | Config | Unpaired e⁻ | Magnetic property |
        |---|---|---|---|
        | Sc³⁺ | [Ar]3d⁰ | 0 | Diamagnetic |
        | Cu⁺ | [Ar]3d¹⁰ | 0 | Diamagnetic |
        | Zn²⁺ | [Ar]3d¹⁰ | 0 | Diamagnetic |
        All three are diamagnetic — they have completely filled or empty d orbitals with no unpaired electrons, so they are repelled by magnetic fields.

       Question 16

      (a) Determining percentage of P in mixture A (P, Q, R):


      P: soluble in water and ethanol; insoluble in ether
      Q: soluble in ethanol and ether; insoluble in water
      R: soluble in ether only; insoluble in water and ethanol

      Procedure:

      1. Weigh the mixture accurately (mass = M).
      2. Add water and stir — P dissolves; Q and R do not. Filter. The filtrate contains P dissolved in water.
      3. Evaporate the aqueous filtrate to dryness and weigh the residue = mass of P.
      4. % P = (mass of P / M) × 100

      (Alternatively: add ethanol to the residue — Q dissolves; add ether to remaining — R dissolves, confirming separation.)


      (b) X (Z=37), Y (Z=35)

      (i) Electron configurations:

      X (Z=37, Rb): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹
      - Y (Z=35, Br): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² **4p⁵**

      (ii) Formulae of hydrides:

      - X (Rb): RbH (rubidium hydride) — ionic hydride
      - Y (Br): HBr (hydrogen bromide) — covalent hydride

      (iii) Type of bonding between X and Y; formula:

      - Rb is a metal (Group 1), Br is a non-metal (Group 17)
      - Bonding:  Ionic bonding (electron transfer from Rb to Br)
      - Formula: RbBr (rubidium bromide)


      (c) Identify X or Y for each property:

      (X = Rb, Group 1 metal; Y = Br, Group 17 non-metal)

      (i) Will conduct electricity in solid state?
      Neither — Rb conducts as a liquid/metal but ionic solids don't conduct when solid in the classical sense; Br₂ does not. Actually Rb metal does conduct in solid state (metallic bonding). → X (Rb)

      (ii) Will react violently with water?
      → X (Rb): Rb + H₂O → RbOH + ½H₂ (violent, exothermic)

      (iii) Will form an acid anhydride?
      → Y (Br): Br₂O is the acid anhydride of HBrO (hydrobromic/hypobromous acid)

      (iv) Will form basic hydroxide?
      → X (Rb): RbOH is a strong base

      (v) Has catenating property?
      → Neither strongly, but Y (Br) forms Br–Br bonds in Br₂. However, catenation refers mainly to C. Among these two, neither has significant catenation.

      (vi) Has variable oxidation state?
      → Y (Br):−1, 0, +1, +3, +5, +7 (halogens show variable oxidation states)