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Number One
(1)
===========================
Number Two
(2)
===========================
Number Three
(3)
===========================
Number Four
(4ai)
Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
Last electron enters the d-subshell.
Therefore, A is a d-block element
(4aii)
Electronic configuration = 1s² 2s² 2p⁶ 3s¹
Last electron enters the s-subshell.
Therefore, B is an s-block element.
(4aiii)
Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁴
Last electron enters the p-subshell.
Therefore, C is a p-block element.
(4b)
=For Y=
Given quantum numbers:
n = 2, l = 0, m = 0, s = -½
n = 2, l = 0, m = 0, s = +½
Since l = 0, the electrons are in the 2s orbital.
Therefore, the valence shell configuration is:
Y = 2s²
=For Z=
Given quantum numbers:
n = 3, l = 0, m = 0, s = -½
n = 3, l = 0, m = 0, s = +½
n = 3, l = 1, m = 0, s = -½
n = 3, l = 1, m = 0, s = +½
Since l = 0 corresponds to 3s and l = 1 corresponds to 3p,
there are two electrons in 3s and two electrons in 3p.
Therefore, the valence shell configuration is:
Z = 3s² 3p²
===========================
Number Five
(5)
===========================
Number Six
(6ai)
Enthalpy of solution is the heat change that occurs when one mole of a substance is completely dissolved in a large amount of solvent to form a dilute solution under standard conditions.
(6aii)
Enthalpy of combustion is the heat change evolved when one mole of a substance is completely burned in excess oxygen under standard conditions.
(6b)
Given:
C(s) + O₂(g) → CO₂(g) ΔHf = −393 kJ mol⁻¹
The equation indicates that when one mole of carbon reacts completely with one mole of oxygen to form one mole of carbon(IV) oxide, 393 kJ of heat is released. The negative sign (−393 kJ mol⁻¹) shows that the reaction is exothermic, meaning heat is given out to the surroundings.
Therefore, the standard enthalpy of formation of carbon(IV) oxide is −393 kJ mol⁻¹.
===========================
Number Seven
(7)
(i) They have one valence electron: Each Group 1 element has only one electron in its outermost shell, which they readily lose during chemical reactions.
(ii) They are highly reactive metals: They react vigorously with water, oxygen, and other non-metals. Their reactivity increases down the group.
(iii) They form positive ions (+1): Since they lose one electron easily, they form univalent cations with a charge of +1.
(iv) They are soft and have low densities: Group 1 metals are soft enough to be cut with a knife and are generally less dense than most other metals.
(v) They form alkaline hydroxides: When they react with water, they produce soluble hydroxides that are strongly alkaline, which is why they are called alkali metals.
===========================
Number Eight
(8)
Since the rate-determining step involves the reaction of one mole of P with one mole of Q to form one mole of R, the rate is directly proportional to the concentrations of P and Q.
(8b)
(PICK ANY FOUR)
(i) Increase the concentration of reactant P: A higher concentration of P increases the frequency of effective collisions between P and Q, thereby increasing the rate of formation of R.
(ii) Increase the concentration of reactant Q: When more Q particles are present, the chances of collision with P particles increase, leading to greater production of R.
(iii) Increase the temperature of the reaction mixture: Higher temperature increases the kinetic energy of the reacting particles, resulting in more frequent and effective collisions that form R.
(iv) Add a suitable catalyst: A catalyst provides an alternative reaction pathway with lower activation energy, thereby increasing the rate at which R is formed.
(v) Increase the pressure (for gaseous reactants): Higher pressure brings gas molecules closer together, increasing the frequency of collisions and enhancing the formation of R.
(vi) Increase the surface area of solid reactants (if any): Breaking solids into smaller pieces exposes more surface area for reaction, leading to faster formation of R.
(vii) Continuously remove the product R as it is formed: Removing R shifts the equilibrium towards the product side, causing more reactants to convert into R.
(viii) Ensure thorough mixing or stirring of the reactants: Proper mixing brings reactant particles into closer contact, increasing the chances of effective collisions and product formation.
===========================
Number Nine
(9a)
The inert pair effect is the tendency of the outermost s-electrons (ns² electrons) in heavier elements to remain unshared or inactive in chemical bonding. As a result, these electrons do not participate readily in bond formation, leading to the stability of lower oxidation states.
(9bi)
NH₃
Let the oxidation state of N = x
x + 3(+1) = 0
x + 3 = 0
x = -3
∴ Oxidation state of nitrogen = –3
(9bii)
NH₂OH
Let the oxidation state of N = x
x + 3(+1) + (-2) = 0
x + 1 = 0
x = -1
∴ Oxidation state of nitrogen = –1
(9biii)
N₂
Since N₂ is an element in its free state, its oxidation state is 0.
∴ Oxidation state of nitrogen = 0
===========================
Number Ten
(10)
===========================
Number Eleven
(11a)
Tin (Sn) and lead (Pb) are not regarded as transition metals because they do not possess partially filled d-orbitals in their atoms or in any of their common oxidation states. Although they exhibit variable oxidation states and can form complex ions, they belong to the p-block of the periodic table rather than the d-block. Therefore, they do not satisfy the definition of transition elements.
(11b)
Potassium chloride (KCl) forms K⁺ ions which have a completely filled electronic configuration and no partially filled d-orbitals. Consequently, no d-d electronic transitions can occur, and the solution remains colourless. Copper(II) chloride (CuCl₂), on the other hand, contains Cu²⁺ ions with partially filled d-orbitals. Electrons in these orbitals absorb certain wavelengths of visible light and undergo d-d transitions. The remaining transmitted light appears green, giving the solution its characteristic colour.
(11c)
Calcium chloride (CaCl₂) has a lower melting point than strontium chloride (SrCl₂) because strontium ions are larger and the crystal lattice formed is more stable due to stronger overall ionic interactions, requiring more energy to break. However, calcium chloride has a higher melting point than calcium iodide (CaI₂) because chloride ions are smaller than iodide ions. The smaller chloride ions allow stronger electrostatic attraction between Ca²⁺ and Cl⁻ ions, resulting in a higher lattice energy and therefore a higher melting point. The larger iodide ions reduce the lattice attraction, leading to a lower melting point.
(11d)
Sulphur can expand its valence shell and accommodate more than eight electrons because it has vacant 3d orbitals available for bonding. Hence, sulphur can form six covalent bonds with fluorine atoms, producing stable SF₆. Oxygen, however, belongs to the second period and has no vacant d-orbitals. It cannot expand its octet and therefore cannot accommodate six fluorine atoms around it. As a result, OF₆ does not exist as a stable compound.
(11e)
Scandium has the electronic configuration:
Sc = [Ar] 3d¹ 4s²
During chemical reactions, scandium loses its two 4s electrons and one 3d electron to attain the stable noble gas configuration of argon. Sc → Sc³⁺ + 3e⁻
Therefore, the +3 oxidation state is the most stable and common oxidation state of scandium.
===========================
Number Twelve
(12a)
(PICK ANY TWO)
(i) Both involve oxidation and reduction (redox) reactions.
(ii) Both consist of an anode, a cathode and an electrolyte.
(iii) In both cells, oxidation occurs at the anode.
(iv) In both cells, reduction occurs at the cathode.
(v) Both involve the transfer of electrons through an external circuit.
(vi) Both involve the movement of ions through an electrolyte.
(vii) Both obey Faraday’s laws of electrolysis and electrochemical principles.
(viii) Both can be represented using electrode reactions and overall cell reactions.
(12b)
In a Tabular form
=Electrochemical Cell=
(PICK ANY FOUR)
(i) Converts chemical energy into electrical energy.
(ii) Operates through a spontaneous chemical reaction.
(iii) Produces electrical energy.
(iv) Does not require an external source of electricity.
(v) Has a negatively charged anode.
(vi) Has a positively charged cathode.
(vii) Commonly used in batteries and power generation.
(viii) Usually consists of two half-cells connected by a salt bridge.
=Electrolytic Cell=
(PICK ANY FOUR)
(i) Converts electrical energy into chemical energy.
(ii) Operates through a non-spontaneous chemical reaction.
(iii) Consumes electrical energy.
(iv) Requires an external source of electricity.
(v) Has a positively charged anode.
(vi) Has a negatively charged cathode.
(vii) Commonly used in electroplating, electrorefining and extraction of metals.
(viii) Usually consists of a single container and does not require a salt bridge.
(12c)
Given:
E°ₓ = −2.00 V
E°ᵧ = −4.25 V
Temperature = 298 K
n = 2
F = 96,500 C mol⁻¹
R = 8.314 J mol⁻¹ K⁻¹
(12ci)
The electrode with the higher (less negative) reduction potential acts as the cathode.
E°ₓ = −2.00 V
E°ᵧ = −4.25 V
Since −2.00 V > −4.25 V,
Cathode = X
Anode = Y
Reason: X is more readily reduced while Y is more readily oxidized.
(12cii)
Cathode (Reduction):
X²⁺(aq) + 2e⁻ → X(s)
Anode (Oxidation):
Y(s) → Y²⁺(aq) + 2e⁻
(12ciii)
X²⁺(aq) + Y(s) → X(s) + Y²⁺(aq)
(12civ)
E°cell = E°cathode − E°anode
= (−2.00) − (−4.25)
= −2.00 + 4.25
= +2.25 V
Therefore, E°cell = +2.25 V
(12cv)
Formula:
ΔG = −nFE°cell
Substituting values:
ΔG = −(2)(96,500)(2.25)
ΔG = −434,250 J mol⁻¹
ΔG = −434.25 kJ mol⁻¹
Therefore, ΔG = −434.25 kJ mol⁻¹
(12cvi)
Yes, the reaction is spontaneous.
Reason:
E°cell = +2.25 V (positive)
ΔG = −434.25 kJ mol⁻¹ (negative)
Since E°cell is positive and ΔG is negative, the reaction occurs spontaneously.
(12cvii)
Formula:
lnK = nFE°cell / RT
Substituting values:
lnK = (2 × 96,500 × 2.25) / (8.314 × 298)
lnK = 434,250 / 2,477.57
lnK = 175.28
K = e¹⁷⁵·²⁸
K ≈ 1.0 × 10⁷⁶
Therefore, K ≈ 1.0 × 10⁷⁶
===========================
Number Fifteen
(15ai)
Statement: It is impossible to determine simultaneously and accurately both the exact position and momentum (or velocity) of an electron.
Application/Consequence: It led to the development of the quantum mechanical model of the atom and explains why electrons are described by orbitals rather than fixed paths.
(15aii)
Statement: The covalent character of an ionic compound increases with increasing polarizing power of the cation and increasing polarizability of the anion.
Application/Consequence: It is used to predict whether a bond will be predominantly ionic or covalent. For example, AlCl₃ is more covalent than NaCl.
(15aiii)
Statement: At constant temperature, a solute distributes itself between two immiscible solvents such that the ratio of its concentrations in the two solvents remains constant.
Application/Consequence: It forms the basis of solvent extraction used in the separation and purification of substances.
(15aiv)
Statement: Electrons occupy degenerate orbitals singly with parallel spins before pairing occurs.
Application/Consequence: It is used in writing electronic configurations and determining the magnetic properties of atoms.
(15av)
Statement: The total enthalpy change of a reaction is independent of the route by which the reaction occurs, provided the initial and final states are the same.
Application/Consequence: It is used to calculate enthalpy changes that cannot be measured directly, such as heats of formation and bond energies.
(15b)
Volume of KMnO₄ = 36.50 cm³ = 0.03650 dm³
Concentration of KMnO₄ = 0.1 mol dm⁻³
Moles of KMnO₄
= CV
= 0.1 × 0.03650
= 0.00365 mol
From:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
1 mole KMnO₄ reacts with 5 moles Fe²⁺
Moles of Fe²⁺
= 5 × 0.00365
= 0.01825 mol
Mass of pure Fe
= 0.01825 × 56
= 1.022 g
Percentage purity
= (1.022 / 2.20) × 100
= 46.45%
Percentage purity = 46.45%
(15c)
(i) Indicator method
(ii) pH meter method
===========================
Number Sixteen
===========================
COMPLETED.
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