### SECTION A
**1.** State the type of bonding in each of the following:
(a) CsF(s); (b) N₂(g); (c) Na(s); (d) O₃(g); (e) MgCl₂(s).
**2.** (a) Suggest the IUPAC names of two oxides of nitrogen that are neutral to moist litmus paper.
(b) Define the following terms:
(i) Molecularity of a reaction; (ii) Unit cell of a crystal; (iii) Lattice defect.
**3.** (a) Define an element.
(b) How many atoms are there in each of the following:
(i) 3 moles of scandium? (ii) 32 g of hydrogen sulphide? (iii) 13 g of silicon?
**4.** In the complex ion, [Zn(OH)₄]²⁻
(a) Give the oxidation number of Zn in the complex.
(b) What is the coordination number of Zn in the complex ion?
(c) Draw the structure of the complex ion and indicate any coordinate covalent bond(s) involved. (d) What is the shape of the ion?
**5.** A 1.32 g of magnesium was dissolved in diluted hydrochloric acid and the solution was heated in a stream of hydrogen chloride. If 5.26 g of an anhydrous metal chloride remained, what is the simplest formula of the metal chloride formed?
**6.** Consider the following elements of the elements in the periodic table: F, He, Cs, Cl, Mg and K. Which of them:
(a) has the lowest electronegativity value? (b) has the highest electron affinity value?
(c) exist as monoatomic species at room temperature? (d) has the largest atomic size?
(e) has the highest ionisation potential?
**7.** (a) Explain briefly how change in temperature can affect solubility of a solute in a given solvent.
(b) If the solubility of copper(II) tetraoxosulphate(VI) pentahydrate (CuSO₄.5H₂O) at 30°C is 25g per 100g of water, what is the maximum mass of crystals that will be obtained from 10g of the solution?
**8.** The isotope ⁴²₁₉K undergoes β-decay to form ⁴²₂₀Ca. If after 62.0 hours, 96.88% of ⁴²₂₀Ca was found to have undergone transformation, what is the half-life of the isotope?
**9.** State the reason for each of the following statement about alkali metals:
(a) They are univalent.
(b) They have poor complexing tendency.
(c) They are strong reducing agents.
(d) They have the lowest first ionisation enthalpy values in their respective periods.
(e) They are largely ionic in nature.
---
**10.** (a) Define rate of a chemical reaction.
(b) A drop of water of volume 0.05 cm³ from a pipette contains 3.0 × 10⁻⁶ mole of hydrogen ion. If the rate of disappearance of the hydrogen ion is 1.00 × 10⁻⁹ dm³ mol⁻¹ s⁻¹, how long would it take for the hydrogen ion in the drop to disappear?
---
### SECTION B
**11.** (a) Define the following terms: (8 marks)
(i) Standard heat of formation of a substance; (ii) Standard heat of combustion;
(iii) Standard heat of sublimation; (iv) Standard heat of atomisation.
(b) State Hess' law of heat summation. (2 marks)
(c) A tautomeric keto (A) - enol (B) equilibrium can be represented as follows:
*(Structural diagram showing keto form A: H-C(-H)(-H)-C(=O)-H and enol form B: H-C(=C(-OH)(-H))-H)*
Given the following bond energy values:
C-H = 435 kJmol⁻¹; C-C = 368 kJmol⁻¹; C=C = 610 kJmol⁻¹; C-O = 357 kJmol⁻¹
C=O = 748 kJmol⁻¹; O-H = 462 kJmol⁻¹
(i) Calculate the enthalpy change from the keto form (A) to the enol form (B). (7 marks)
(ii) Assuming the entropy change for the conversion of (A) to (B) is zero, calculate the equilibrium constant for the keto – enol equilibrium at 27°C. (8 marks)
**12.** (a) Write the electron configuration of the valence shell of the following group of elements in the periodic table: (i) group 2; (ii) group 13; (iii) group 15. (3 marks)
(b) Using an example of any member of group 2,
(i) write the formula of an oxide of a group 2 element; (1 mark)
(ii) what is the bond type of the oxide in (i) above? (1 mark)
(iii) Write the chemical equation for the reaction of the oxide with water. (2 marks)
(iv) Give an equation of the reaction between HCl and the product formed in (iii) above. (2 marks)
(c) Account for each of the following observations:
(i) Whereas both zinc and copper have the same *d* electron configuration, zinc or its ion is a non-transition element but copper or its ion is regarded as transition element. (4 marks)
(ii) The maximum covalency of carbon is four but that of silicon is six. (5 marks)
(iii) The melting point of AlF₃ is greater than that of AlCl₃. (5 marks)
(d) What do you understand by the term "catenation". (2 marks)
**13.** (a) State five assumptions of kinetic theory of gases. (5 marks)
(b) Deduce the following gas laws from kinetic theory equation of an ideal gas: (10 marks)
(i) Avogadro's law; (ii) Graham's law of diffusion.
---
## PAGE 4 — 2023 IJMBE CHEMISTRY I contd.
(c) The van der Waal's equation for a mole of real gas is
(P + a/V²)(V-b) = RT
Compare this equation with the equation, PV = RT for a mole of an ideal gas and explain why the volume and pressure corrections are necessary. (7 marks)
(d) Calculate the pressure of one mole of oxygen gas at 200K when placed in 1 dm³ container assuming ideal behaviour. (3 marks)
**14.** (a) Explain the following concepts of acids and bases giving appropriate examples in each case: (9 marks)
(i) Arrhenius; (ii) Lewis; (iii) Brønsted-Lowry.
(b) Classify the reactants in the following reactions as acid or base and identify the acid-base conjugate pairs:
(i) CH₃COO⁻ + HCN ⇌ CH₃COOH + CN⁻ (2 marks)
(ii) H₂PO₄⁻ + NH₃ ⇌ HPO₄²⁻ + NH₄⁺ (2 marks)
(iii) HClO + CH₃NH₂ ⇌ CH₃NH₃⁺ + ClO⁻ (2 marks)
(c) (i) Arrange the following equimolar solutions in order of increasing pH:
NH₄Cl, KOH, HCl, KCl, HCOOH, and HCOONa. Give reasons for your order of arrangement. (6 marks)
(ii) Give two physical methods to show that HCl dissociates more in aqueous solution than methanoic acid. (4 marks)
**15.** (a) Give 4 postulates of Bohr atomic theory. (4 marks)
(b) State 2 each of the successes and limitations of Bohr atomic theory. (4 marks)
(c) Explain the experimental evidence for the small size of the nucleus of an atom (detail of the experiment is not required). (5 marks)
(d) Calculate the ionisation enthalpy of hydrogen atom. Given that the frequency of convergence limit of the Lyman series of a hydrogen atom is 3.29 × 10¹⁵ Hz. (4 marks)
(f) Explain the importance of *n*, *l*, *m*, and *s* quantum numbers in the orbital arrangement of electrons in atoms. (8 marks)
**16.** (a) Compound A, [FeBr(H₂O)₅]SO₄ is isomeric with another compound B, [FeSO₄(H₂O)₅]Br
(i) What type of isomerism exists between the compounds, A and B? (1 mark)
(ii) What ions would these isomers yield in aqueous solution? (4 marks)
(iii) Using simple laboratory tests, explain how you can differentiate between the two isomers? (6 marks)
(iv) State the oxidation state, electron configuration and coordination number of Fe in compound A? (4 marks)
(b) Draw the structure of [FeBr(H₂O)₅]²⁺ and indicate all the coordinate covalent bonds involved. (2 marks)
(c) Give the formula of an example of each of the following: (6 marks)
(i) cationic complex of cobalt; (ii) anionic complex of cobalt; (iii) neutral complex of cobalt.
(d) Give two examples of double salts. (2 marks)
---
COMPLETE SOLUTIONS
## SECTION A
### Question 1: State the type of bonding in each of the following:
**(a) CsF(s)** — **Ionic bonding.** Caesium (a metal) transfers an electron to fluorine (a non-metal), forming Cs⁺ and F⁻ ions held together by strong electrostatic attraction.
**(b) N₂(g)** — **Covalent bonding (triple covalent bond).** Two nitrogen atoms share three pairs of electrons to achieve a stable octet configuration: N≡N.
**(c) Na(s)** — **Metallic bonding.** Sodium atoms release their valence electrons into a "sea" of delocalized electrons, creating a lattice of positive ions held together by electrostatic attraction to the mobile electrons.
**(d) O₃(g)** — **Covalent bonding** (with resonance). Oxygen atoms share electrons covalently. Ozone has a resonance structure with one double bond and one coordinate (dative) covalent bond, giving a bond order of 1.5.
**(e) MgCl₂(s)** — **Ionic bonding.** Magnesium transfers two electrons (one to each chlorine atom), forming Mg²⁺ and 2Cl⁻ ions held by electrostatic forces.
---
### Question 2:
**(a) Two IUPAC names of oxides of nitrogen neutral to moist litmus paper:**
- **Dinitrogen oxide (N₂O)** — nitrous oxide; neutral oxide
- **Nitrogen(II) oxide (NO)** — neutral oxide
Both N₂O and NO are neutral oxides — they neither turn blue litmus red nor red litmus blue when in contact with moist litmus paper.
**(b) Definitions:**
**(i) Molecularity of a reaction:**
Molecularity is the **total number of reacting species (atoms, ions, or molecules) that collide simultaneously in an elementary step** of a chemical reaction. It is always a whole number (1, 2, or 3) and applies only to elementary reactions, not overall reactions.
**(ii) Unit cell of a crystal:**
A unit cell is the **smallest repeating structural unit of a crystalline solid** that, when repeated in three dimensions, generates the entire crystal lattice. It defines the geometry and symmetry of the crystal structure.
**(iii) Lattice defect:**
A lattice defect (or crystal defect) is an **irregularity or imperfection in the regular, repeating arrangement of atoms, ions, or molecules** in a crystal lattice. Examples include Schottky defects (missing ions) and Frenkel defects (displaced ions).
---
### Question 3:
**(a) Definition of an element:**
An element is a **pure substance that cannot be broken down into simpler substances by ordinary chemical means.** It consists of only one type of atom, characterized by a unique atomic number (number of protons in the nucleus).
**(b) Number of atoms in each:**
Using: Number of atoms = number of moles × Nₐ
Nₐ = 6.023 × 10²³ mol⁻¹
**(i) 3 moles of scandium:**
Number of atoms = 3 × 6.023 × 10²³
**= 1.807 × 10²⁴ atoms**
**(ii) 32 g of hydrogen sulphide (H₂S):**
Molar mass of H₂S = 2(1) + 32 = 34 g/mol
Moles of H₂S = 32/34 = 0.941 mol
Each H₂S molecule contains 3 atoms (2H + 1S)
Number of atoms = 0.941 × 3 × 6.023 × 10²³
= 0.941 × 1.807 × 10²⁴
**= 1.700 × 10²⁴ atoms**
**(iii) 13 g of silicon (Si):**
Molar mass of Si = 28 g/mol
Moles of Si = 13/28 = 0.4643 mol
Number of atoms = 0.4643 × 6.023 × 10²³
**= 2.796 × 10²³ atoms**
---
### Question 4: Complex ion [Zn(OH)₄]²⁻
**(a) Oxidation number of Zn:**
Let oxidation number of Zn = x
Each OH⁻ = −1; four OH⁻ = −4
Overall charge = −2
x + (−4) = −2
x = −2 + 4
**x = +2**
The oxidation number of Zn is **+2.**
**(b) Coordination number of Zn:**
The coordination number is the number of ligands directly bonded to the central metal ion.
There are **4 OH⁻ ligands** surrounding Zn.
**Coordination number = 4**
**(c) Structure of [Zn(OH)₄]²⁻:**
```
[OH]⁻
|
[OH]⁻ — Zn²⁺ — [OH]⁻
|
[OH]⁻
```
Each OH⁻ donates a lone pair of electrons to Zn²⁺ through **coordinate (dative) covalent bonds** — all four Zn–O bonds are coordinate covalent bonds where oxygen donates the electron pair.
**(d) Shape of the ion:**
With 4 ligands and no lone pairs on Zn, the shape is **tetrahedral** (bond angle ≈ 109.5°).
---
### Question 5: Empirical formula of metal chloride
**Given:**
- Mass of Mg = 1.32 g
- Mass of anhydrous metal chloride = 5.26 g
- Mass of Cl = 5.26 − 1.32 = **3.94 g**
**Molar masses:** Mg = 24 g/mol; Cl = 35.5 g/mol
**Moles:**
- Moles of Mg = 1.32/24 = **0.055 mol**
- Moles of Cl = 3.94/35.5 = **0.111 mol**
**Ratio:**
Mg : Cl = 0.055 : 0.111 = 1 : 2.018 ≈ **1 : 2**
**Simplest formula = MgCl₂**
---
### Question 6: Elements — F, He, Cs, Cl, Mg, K
**(a) Lowest electronegativity value:**
**Cs (Caesium)** — it is the largest atom in the list, furthest from the nucleus effect on bonding electrons; electronegativity decreases down a group and across from right to left.
**(b) Highest electron affinity value:**
**Cl (Chlorine)** — chlorine has the highest electron affinity among these elements. (Fluorine has a slightly lower electron affinity than chlorine due to its small size causing electron-electron repulsion.)
**(c) Exist as monoatomic species at room temperature:**
**He (Helium)** — it is a noble gas and exists as individual atoms (monoatomic) at room temperature.
**(d) Largest atomic size:**
**Cs (Caesium)** — it is in Period 6, Group 1, making it the largest atom in this list as atomic size increases down a group and decreases across a period.
**(e) Highest ionisation potential:**
**He (Helium)** — noble gases have completely filled electron shells, making them extremely stable and requiring the most energy to remove an electron.
---
### Question 7:
**(a) Effect of temperature on solubility:**
For **most solid solutes** dissolving in liquid solvents, solubility **increases with increasing temperature.** This is because dissolving is generally an endothermic process — increasing temperature provides more energy to overcome lattice energy, allowing more solute to dissolve. However, for some salts (e.g., Ce₂(SO₄)₃), solubility **decreases** with increasing temperature (exothermic dissolution). For **gases dissolved in liquids**, solubility always **decreases** with increasing temperature, as higher temperatures give gas molecules enough kinetic energy to escape from the solution.
**(b) Maximum mass of CuSO₄·5H₂O crystals from 10 g of solution:**
**Given:** Solubility = 25 g CuSO₄·5H₂O per 100 g water at 30°C
This means in 125 g of solution: 25 g CuSO₄·5H₂O is dissolved in 100 g water.
**In 10 g of solution:**
Mass of CuSO₄·5H₂O in 10 g solution = (25/125) × 10 = **2.0 g**
Since the solution is saturated, cooling will precipitate dissolved salt. However since we are asked for maximum crystals obtainable from 10 g solution (assuming complete crystallization upon cooling to a temperature where solubility → 0):
**Maximum mass of crystals = 2.0 g**
---
### Question 8: Half-life of ⁴²₁₉K
**Given:**
- 96.88% of ⁴²₁₉K has transformed to ⁴²₂₀Ca after 62.0 hours
- Therefore, fraction of ⁴²₁₉K remaining = 100 − 96.88 = **3.12% = 0.0312**
**Using:** N/N₀ = (1/2)ⁿ where n = number of half-lives
0.0312 = (1/2)ⁿ
Taking log of both sides:
log(0.0312) = n × log(0.5)
−1.506 = n × (−0.3010)
n = 1.506/0.3010 = **5.003 ≈ 5 half-lives**
**Half-life = Total time / n = 62.0/5 = 12.4 hours**
**∴ Half-life of ⁴²₁₉K = 12.4 hours**
---
### Question 9: Reasons for properties of alkali metals
**(a) They are univalent:**
Alkali metals have **one electron in their outermost shell (ns¹).** They lose this single valence electron easily to form a +1 ion, which gives them a noble gas configuration. Since they have only one electron to lose, they exhibit only a +1 oxidation state — hence they are univalent.
**(b) They have poor complexing tendency:**
Alkali metals have **large ionic radii and low charge density (+1).** Complex formation requires a central metal ion with high charge density and available empty orbitals of suitable energy to accept lone pairs from ligands. Because alkali metal ions have low charge/size ratio and their orbitals are too diffuse and high in energy for effective overlap with ligand orbitals, they form complexes poorly.
**(c) They are strong reducing agents:**
Alkali metals have **very low ionisation energies** — they lose their single valence electron very easily. This makes them excellent electron donors (reducing agents). The ease with which they release electrons to other species makes them the strongest reducing agents among metals.
**(d) They have the lowest first ionisation enthalpy values in their respective periods:**
In any period, alkali metals are the **largest atoms** (leftmost in the period) with the **least nuclear charge effectively experienced by the valence electron** (large atomic radius + effective shielding). This means very little energy is required to remove the single outermost electron — hence they have the lowest first ionisation enthalpies in their periods.
**(e) They are largely ionic in nature:**
The large difference in **electronegativity** between alkali metals and most non-metals (especially oxygen, halogens) means that when they bond, electron transfer is essentially complete, forming ionic compounds. Their low ionisation energies make electron donation energetically favorable, resulting in predominantly ionic bonding in their compounds.
---
### Question 10:
**(a) Definition of rate of a chemical reaction:**
The rate of a chemical reaction is defined as the **change in concentration of a reactant or product per unit time.** It is expressed as:
Rate = −Δ[Reactant]/Δt = +Δ[Product]/Δt
Units: mol dm⁻³ s⁻¹
**(b) Time for hydrogen ion to disappear:**
**Given:**
- Volume of drop = 0.05 cm³ = 0.05 × 10⁻³ dm³ = 5 × 10⁻⁵ dm³
- Moles of H⁺ = 3.0 × 10⁻⁶ mol
- Rate of disappearance = 1.00 × 10⁻⁹ mol dm⁻³ s⁻¹
**Concentration of H⁺:**
[H⁺] = moles/volume = (3.0 × 10⁻⁶)/(5 × 10⁻⁵)
= **0.06 mol dm⁻³** = 6 × 10⁻² mol dm⁻³
**Time = Concentration/Rate:**
t = [H⁺]/Rate = (6 × 10⁻²)/(1.00 × 10⁻⁹)
**t = 6 × 10⁷ seconds**
---
## SECTION B
---
### Question 11:
**(a) Definitions:**
**(i) Standard heat of formation:**
The standard heat of formation (ΔH°f) is the **enthalpy change when one mole of a compound is formed from its constituent elements** in their standard states under standard conditions (298 K, 1 atm). The standard heat of formation of an element in its standard state is zero.
**(ii) Standard heat of combustion:**
The standard heat of combustion (ΔH°c) is the **enthalpy change when one mole of a substance is completely burned in excess oxygen** under standard conditions (298 K, 1 atm), with all products in their standard states.
**(iii) Standard heat of sublimation:**
The standard heat of sublimation (ΔH°sub) is the **enthalpy change when one mole of a solid substance is converted directly into gaseous state** without passing through the liquid phase, under standard conditions.
**(iv) Standard heat of atomisation:**
The standard heat of atomisation (ΔH°at) is the **enthalpy change when one mole of gaseous atoms is formed from an element** in its standard state under standard conditions. For diatomic molecules, it is half the bond dissociation energy.
---
**(b) Hess' Law of Heat Summation:**
Hess' Law states that **the total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place**, provided the initial and final states are the same. In other words, the total heat change is the same whether the reaction occurs in one step or in several steps.
---
**(c) Keto-Enol Tautomerism:**
**Structure Analysis:**
**Keto form (A):** CH₃−C(=O)−H (ethanal/acetaldehyde equivalent shown)
Bonds present: C-H (×4), C-C (×1), C=O (×1)
**Enol form (B):** CH₂=C(−OH)−H
Bonds present: C-H (×3), C=C (×1), C-O (×1), O-H (×1)
**Bond energy values:**
- C-H = 435 kJmol⁻¹
- C-C = 368 kJmol⁻¹
- C=C = 610 kJmol⁻¹
- C-O = 357 kJmol⁻¹
- C=O = 748 kJmol⁻¹
- O-H = 462 kJmol⁻¹
**(i) Enthalpy change from keto (A) to enol (B):**
ΔH = Energy absorbed (bonds broken) − Energy released (bonds formed)
**Bonds broken in keto form A:**
- 4 × C-H = 4 × 435 = 1740 kJmol⁻¹
- 1 × C-C = 368 kJmol⁻¹
- 1 × C=O = 748 kJmol⁻¹
- **Total broken = 2856 kJmol⁻¹**
**Bonds formed in enol form B:**
- 3 × C-H = 3 × 435 = 1305 kJmol⁻¹
- 1 × C=C = 610 kJmol⁻¹
- 1 × C-O = 357 kJmol⁻¹
- 1 × O-H = 462 kJmol⁻¹
- **Total formed = 2734 kJmol⁻¹**
**ΔH = 2856 − 2734 = +122 kJmol⁻¹**
The enthalpy change from keto to enol form is **+122 kJmol⁻¹** (endothermic).
---
**(ii) Equilibrium constant at 27°C (ΔS = 0):**
Using: ΔG = ΔH − TΔS
Since ΔS = 0:
ΔG = ΔH = +122 kJmol⁻¹ = +122,000 Jmol⁻¹
T = 27 + 273 = 300 K
R = 8.314 JK⁻¹mol⁻¹
Using: ΔG = −RT ln K
122,000 = −8.314 × 300 × ln K
122,000 = −2494.2 × ln K
ln K = −122,000/2494.2 = −48.91
**K = e⁻⁴⁸·⁹¹ = 6.56 × 10⁻²²**
The very small value of K confirms that the **keto form strongly predominates** at equilibrium at 27°C.
---
### Question 12:
**(a) Valence shell electron configurations:**
**(i) Group 2:** ns² (e.g., Mg: 3s²)
**(ii) Group 13:** ns²np¹ (e.g., Al: 3s²3p¹)
**(iii) Group 15:** ns²np³ (e.g., N: 2s²2p³)
---
**(b) Using Magnesium (Mg) as example of Group 2:**
**(i) Formula of oxide of Mg:**
**MgO** (magnesium oxide)
**(ii) Bond type:**
**Ionic bonding** — Mg²⁺ and O²⁻ ions held by electrostatic attraction.
**(iii) Chemical equation for reaction of MgO with water:**
MgO(s) + H₂O(l) → Mg(OH)₂(aq)
**(iv) Equation for reaction between HCl and Mg(OH)₂:**
Mg(OH)₂(aq) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)
---
**(c) Accounts for observations:**
**(i) Zinc is non-transition but copper is transition element:**
Both Zn and Cu have the electronic configuration with completely filled or partially filled d-orbitals. However:
- **Cu** (Z=29): [Ar] 3d¹⁰4s¹ — but Cu²⁺ ion has configuration [Ar] 3d⁹, which is **partially filled d-orbitals.** A transition element is defined as one whose atom or at least one of its common ions has an incompletely filled d-subshell. Cu²⁺ satisfies this criterion.
- **Zn** (Z=30): [Ar] 3d¹⁰4s² — Zn²⁺ ion has configuration [Ar] 3d¹⁰, which is a **completely filled d-subshell.** Since neither Zn nor its ion has an incomplete d-subshell, it does NOT qualify as a transition element.
**(ii) Maximum covalency of carbon is four but silicon is six:**
Carbon is in Period 2 and has no available **d-orbitals** in its valence shell (only 2s and 2p orbitals are available, giving maximum 4 bonds). Silicon is in Period 3 and has **available 3d orbitals** of relatively low energy that can participate in bonding. These empty 3d orbitals allow silicon to expand its valence shell beyond 4 and accommodate up to 6 bonding pairs, giving a maximum covalency of 6 (e.g., SiF₆²⁻).
**(iii) Melting point of AlF₃ > AlCl₃:**
AlF₃ has a **predominantly ionic structure** with strong electrostatic interactions between Al³⁺ and F⁻ ions, giving it a high melting point. AlCl₃, on the other hand, has significant **covalent character** because Cl⁻ is larger and more polarizable than F⁻; the highly charged small Al³⁺ polarizes the Cl⁻ electron cloud extensively (Fajans' rules), resulting in a layer structure with weaker van der Waals forces between layers. Therefore AlCl₃ has a much lower melting point than AlF₃.
**(d) Catenation:**
Catenation is the **ability of atoms of the same element to form long chains or rings by bonding with each other** through covalent bonds. Carbon exhibits the greatest catenation of all elements, forming chains, branched chains, and rings of varying lengths — this is the basis of organic chemistry. Silicon also shows catenation but to a much lesser extent.
---
### Question 13:
**(a) Five assumptions of kinetic theory of gases:**
1. A gas consists of a large number of **tiny particles (molecules or atoms)** that are in constant, rapid, random motion.
2. The **volume of the gas molecules themselves is negligible** compared to the total volume occupied by the gas.
3. There are **no intermolecular forces of attraction or repulsion** between gas molecules (except during collisions).
4. Collisions between gas molecules and between molecules and the container walls are **perfectly elastic** — no net loss of kinetic energy.
5. The **average kinetic energy of gas molecules is directly proportional to the absolute temperature** of the gas: KE = ½mv² ∝ T.
---
**(b) Deduction of gas laws from kinetic theory:**
The kinetic theory equation for an ideal gas is:
**PV = ⅓Nmv²** or equivalently **PV = NkT = nRT**
Where: P = pressure, V = volume, N = number of molecules, m = mass of one molecule, v² = mean square speed, k = Boltzmann constant, T = absolute temperature.
**(i) Avogadro's Law:**
Avogadro's law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
**Derivation:**
From kinetic theory: PV = ⅓Nmv²
Also, average KE = ½mv² = (3/2)kT
Therefore: PV = NkT
For two different gases (1 and 2) at the same P, V, and T:
- Gas 1: PV = N₁kT
- Gas 2: PV = N₂kT
Since P, V, and T are the same:
**N₁kT = N₂kT → N₁ = N₂**
Therefore equal volumes of different gases at the same T and P contain equal numbers of molecules — **Avogadro's Law is proved.**
**(ii) Graham's Law of Diffusion:**
Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass at constant temperature and pressure.
**Derivation:**
From kinetic theory, average KE = ½mv² = (3/2)kT
Therefore: v² = 3kT/m → v = √(3kT/m)
The rate of diffusion r ∝ v (root mean square speed)
For two gases at same T:
r₁/r₂ = v₁/v₂ = √(3kT/m₁)/√(3kT/m₂) = √(m₂/m₁)
Since m ∝ M (molar mass):
**r₁/r₂ = √(M₂/M₁)**
This means r ∝ 1/√M — **Graham's Law of Diffusion is proved.**
---
**(c) Van der Waals equation vs Ideal gas equation:**
**Ideal gas:** PV = RT (for one mole)
**Van der Waals:** (P + a/V²)(V − b) = RT
**Volume correction (b):**
In the ideal gas equation, the volume V is assumed to be the free space available for molecules to move. But real gas molecules have **finite volume.** The term **b** (the co-volume) corrects for this by representing the volume actually occupied by the molecules themselves. The effective free volume is therefore (V − b) rather than V.
**Pressure correction (a/V²):**
In an ideal gas, there are no intermolecular forces. In real gases, there are **attractive forces** between molecules. Molecules near the wall of the container experience a net inward pull from surrounding molecules, reducing the force with which they hit the wall. The actual pressure is therefore less than ideal. The term **a/V²** corrects for this by adding the "internal pressure" back to give the effective pressure.
At **high temperatures and low pressures**, real gases approximate ideal behavior because: molecules move fast enough to overcome attractive forces, and intermolecular distances are large enough to make molecular volume negligible.
---
**(d) Pressure of one mole of O₂ at 200 K in 1 dm³:**
Using ideal gas equation: PV = nRT
- n = 1 mol
- R = 8.314 JK⁻¹mol⁻¹ = 8.314 × 10⁻³ kJ = 8.314 Pa·m³·K⁻¹·mol⁻¹
- T = 200 K
- V = 1 dm³ = 1 × 10⁻³ m³
P = nRT/V = (1 × 8.314 × 200)/(1 × 10⁻³)
P = 1662.8/10⁻³
**P = 1,662,800 Pa = 1.663 × 10⁶ Pa ≈ 1.663 × 10⁶ Nm⁻²**
Or in atm: P = 1,662,800/101,325 = **16.41 atm**
---
### Question 14:
**(a) Concepts of acids and bases:**
**(i) Arrhenius Theory:**
An **Arrhenius acid** is a substance that produces hydrogen ions (H⁺) when dissolved in water. An **Arrhenius base** is a substance that produces hydroxide ions (OH⁻) in water.
- Example acid: HCl → H⁺ + Cl⁻
- Example base: NaOH → Na⁺ + OH⁻
- **Limitation:** Only applies to aqueous solutions.
**(ii) Lewis Theory:**
A **Lewis acid** is an electron pair **acceptor.** A **Lewis base** is an electron pair **donor.**
- Example Lewis acid: BF₃ (accepts lone pair from NH₃)
- Example Lewis base: NH₃ (donates lone pair to BF₃)
- BF₃ + :NH₃ → F₃B←NH₃
- **Advantage:** Broadest definition — includes reactions without H⁺ or OH⁻.
**(iii) Brønsted-Lowry Theory:**
A **Brønsted-Lowry acid** is a **proton donor.** A **Brønsted-Lowry base** is a **proton acceptor.**
- Example: HCl + H₂O → H₃O⁺ + Cl⁻
(HCl = acid/proton donor; H₂O = base/proton acceptor)
- **Advantage:** Not limited to aqueous solutions; introduces the concept of conjugate pairs.
---
**(b) Acid-base conjugate pairs:**
**(i) CH₃COO⁻ + HCN ⇌ CH₃COOH + CN⁻**
- **HCN** = Brønsted-Lowry **acid** (proton donor)
- **CH₃COO⁻** = Brønsted-Lowry **base** (proton acceptor)
- Conjugate pairs: HCN/CN⁻ and CH₃COOH/CH₃COO⁻
**(ii) H₂PO₄⁻ + NH₃ ⇌ HPO₄²⁻ + NH₄⁺**
- **H₂PO₄⁻** = **acid** (donates proton)
- **NH₃** = **base** (accepts proton)
- Conjugate pairs: H₂PO₄⁻/HPO₄²⁻ and NH₄⁺/NH₃
**(iii) HClO + CH₃NH₂ ⇌ CH₃NH₃⁺ + ClO⁻**
- **HClO** = **acid** (proton donor)
- **CH₃NH₂** = **base** (proton acceptor)
- Conjugate pairs: HClO/ClO⁻ and CH₃NH₃⁺/CH₃NH₂
---
**(c)(i) Order of increasing pH for equimolar solutions:**
**HCl < HCOOH < NH₄Cl < KCl < HCOONa < KOH**
**Reasons:**
- **HCl** — strong acid, fully dissociates, lowest pH (~1)
- **HCOOH** — weak acid, partially dissociates, low but higher pH than HCl
- **NH₄Cl** — salt of weak base + strong acid, slightly acidic (pH slightly below 7)
- **KCl** — salt of strong acid + strong base, neutral (pH = 7)
- **HCOONa** — salt of weak acid + strong base, slightly basic (pH slightly above 7)
- **KOH** — strong base, fully dissociates, highest pH (~13)
**(c)(ii) Physical methods to show HCl dissociates more than methanoic acid:**
1. **Electrical conductivity:** A solution of HCl conducts electricity much better than an equimolar solution of methanoic acid (HCOOH). Since electrical conductivity depends on the number of ions in solution, the greater conductivity of HCl indicates greater dissociation and more ions present.
2. **Depression of freezing point / Elevation of boiling point:** An equimolar solution of HCl shows a greater depression of freezing point (or elevation of boiling point) than methanoic acid, because these colligative properties depend on the total number of solute particles. HCl produces more ions per molecule dissolved, giving a greater effect.
---
### Question 15:
**(a) Four postulates of Bohr atomic theory:**
1. Electrons revolve around the nucleus in **fixed circular orbits (shells)** called stationary states or energy levels, without radiating energy.
2. Each orbit corresponds to a **definite fixed energy level.** The energy of an electron is quantized — it can only have specific energy values.
3. Electrons can **jump from one energy level to another.** When an electron jumps from a higher to a lower energy level, it emits a photon of energy equal to the difference: ΔE = hν = E₂ − E₁. When it absorbs energy, it jumps to a higher level.
4. The **angular momentum of an electron** in any permitted orbit is a whole number multiple of h/2π: mvr = nh/2π, where n is the principal quantum number.
---
**(b) Successes and limitations of Bohr theory:**
**Successes:**
1. Successfully explained the **line spectrum of hydrogen** and hydrogen-like ions (He⁺, Li²⁺) — predicting the wavelengths of spectral lines with great accuracy.
2. Introduced the concept of **quantized energy levels**, providing the foundation for quantum mechanics and explaining why atoms are stable.
**Limitations:**
1. **Failed to explain the spectra of multi-electron atoms** (helium, lithium, etc.) — only works for one-electron systems.
2. **Failed to explain the fine structure** of spectral lines (splitting of lines in magnetic fields — Zeeman effect, and electric fields — Stark effect).
---
**(c) Experimental evidence for the small size of the nucleus:**
The **Rutherford gold foil (α-particle scattering) experiment** provides the evidence. When alpha particles were directed at a thin gold foil:
- The **vast majority passed straight through** with little or no deflection — indicating that most of the atom is empty space.
- A **small fraction were deflected at large angles**, some even bouncing back — indicating they encountered a region of concentrated positive charge.
- The probability of large-angle deflection was extremely small — consistent with the positive charge being concentrated in an **extremely small, dense region** (the nucleus).
The calculations from the scattering data showed the nucleus to be approximately **10⁻¹⁵ m** in diameter, compared to the atom's diameter of ~10⁻¹⁰ m — demonstrating the nucleus is about 100,000 times smaller than the atom.
---
**(d) Ionisation enthalpy of hydrogen:**
**Given:** Frequency of convergence limit of Lyman series = 3.29 × 10¹⁵ Hz
Using: E = hν
where h = 6.626 × 10⁻³⁴ Js
E = 6.626 × 10⁻³⁴ × 3.29 × 10¹⁵
E = 2.180 × 10⁻¹⁸ J per atom
**Per mole:**
E = 2.180 × 10⁻¹⁸ × 6.023 × 10²³
E = 1.313 × 10⁶ J mol⁻¹
**E = 1313 kJmol⁻¹**
**∴ Ionisation enthalpy of hydrogen = 1313 kJmol⁻¹**
---
**(f) Importance of quantum numbers n, l, m, and s:**
**1. Principal quantum number (n):**
- Determines the **main energy level or shell** of an electron
- n = 1, 2, 3, 4... (positive integers)
- Determines the **size and energy** of the orbital — higher n means larger orbital, higher energy, and greater distance from nucleus
- Maximum electrons in shell n = 2n²
**2. Azimuthal/Angular momentum quantum number (l):**
- Determines the **shape of the orbital** (subshell)
- l = 0 to (n−1); l=0 (s), l=1 (p), l=2 (d), l=3 (f)
- Also determines **orbital angular momentum:** L = √[l(l+1)]·h/2π
- Describes the **subshell** within a main shell
**3. Magnetic quantum number (m or mₗ):**
- Determines the **orientation of the orbital** in space relative to a magnetic field
- mₗ = −l to +l (including 0); gives (2l+1) possible values
- For l=1 (p): mₗ = −1, 0, +1 → three p orbitals (pₓ, p_y, p_z)
- Explains **splitting of spectral lines in a magnetic field** (Zeeman effect)
**4. Spin quantum number (s or mₛ):**
- Describes the **intrinsic angular momentum (spin)** of an electron
- mₛ = +½ (spin up, ↑) or −½ (spin down, ↓)
- By the **Pauli exclusion principle**, no two electrons in an atom can have the same set of all four quantum numbers — this limits each orbital to a maximum of **two electrons** with opposite spins
- Explains the **fine structure** of spectral lines and magnetic properties of atoms
---
### Question 16:
**(a) [FeBr(H₂O)₅]SO₄ (A) and [FeSO₄(H₂O)₅]Br (B):**
**(i) Type of isomerism:**
**Ionisation isomerism** — the two compounds have the same molecular formula but differ in which ligand is inside the coordination sphere and which is outside (as a counter ion), resulting in different ions in solution.
**(ii) Ions yielded in aqueous solution:**
**Compound A: [FeBr(H₂O)₅]SO₄**
→ [FeBr(H₂O)₅]²⁺ + SO₄²⁻
Ions: the complex cation **[FeBr(H₂O)₅]²⁺** and **SO₄²⁻**
**Compound B: [FeSO₄(H₂O)₅]Br**
→ [FeSO₄(H₂O)₅]⁺ + Br⁻
Ions: the complex cation **[FeSO₄(H₂O)₅]⁺** and **Br⁻**
**(iii) Differentiation using laboratory tests:**
**Test 1 — Test for free Br⁻ (to identify Compound B):**
Add **silver nitrate solution (AgNO₃)** to solutions of both compounds:
- Compound B will give a **pale yellow precipitate** of AgBr (insoluble), confirming free Br⁻ ions in solution.
- Compound A will give **no precipitate** with AgNO₃ (Br⁻ is inside the coordination sphere and not free).
**Test 2 — Test for free SO₄²⁻ (to identify Compound A):**
Add **barium chloride solution (BaCl₂)** to solutions of both compounds:
- Compound A will give a **white precipitate** of BaSO₄ (insoluble in dilute HCl), confirming free SO₄²⁻ ions.
- Compound B will give **no white precipitate** with BaCl₂ (SO₄²⁻ is coordinated inside the sphere).
**(iv) Oxidation state, electron configuration and coordination number of Fe in Compound A:**
**Compound A: [FeBr(H₂O)₅]SO₄**
Overall charge on complex ion = +2 (to balance SO₄²⁻)
Let Fe oxidation state = x:
x + (−1) + 0(×5) = +2
x − 1 = +2
**x = +3 → Fe is in +3 oxidation state**
**Electron configuration of Fe³⁺:**
Fe (Z=26): [Ar] 3d⁶4s²
Fe³⁺ (loses 3 electrons): **[Ar] 3d⁵** (1s²2s²2p⁶3s²3p⁶3d⁵)
**Coordination number:**
Ligands = 1 Br⁻ + 5 H₂O = **6**
**Coordination number = 6**
---
**(b) Structure of [FeBr(H₂O)₅]²⁺:**
The complex has an **octahedral geometry** with Fe³⁺ at the center, one Br⁻ and five H₂O ligands surrounding it. All six Fe←ligand bonds are **coordinate covalent (dative) bonds** — each ligand donates a lone pair of electrons to the empty d-orbitals of Fe³⁺.
```
H₂O
|
H₂O — Fe³⁺ — H₂O
/ | \
H₂O Br⁻ H₂O
```
(Octahedral arrangement — all bonds are coordinate covalent bonds →)
---
**(c) Formulae of cobalt complexes:**
**(i) Cationic complex of cobalt:**
**[Co(NH₃)₆]Cl₃** — hexaamminecobalt(III) chloride
(The complex cation is [Co(NH₃)₆]³⁺)
**(ii) Anionic complex of cobalt:**
**K₃[Co(CN)₆]** — potassium hexacyanocobaltate(III)
(The complex anion is [Co(CN)₆]³⁻)
**(iii) Neutral complex of cobalt:**
**[Co(NH₃)₃(NO₂)₃]** — triamminetrinitrocobalt(III)
(Overall charge = 0; no counter ions needed)
---
**(d) Two examples of double salts:**
1. **Mohr's salt** — FeSO₄·(NH₄)₂SO₄·6H₂O (iron(II) ammonium sulphate hexahydrate)
2. **Potash alum** — KAl(SO₄)₂·12H₂O (potassium aluminium sulphate dodecahydrate)
*(Note: Double salts differ from complex salts — they dissociate completely in solution to give all constituent ions, unlike coordination compounds which retain the complex ion in solution.)*