2023 JUPEB chemistry

CHM 001: GENERAL CHEMISTRY

1. (a) 2.21g of calcium is reacted with dilute hydrochloric acid, to give 6.15g of anhydrous metal chloride. Find the empirical formula of the metal chloride. [3 Marks]

(b) What are the oxidation numbers of nitrogen in the following compounds?

(i) Dinitrogen oxide
(ii) Sodium nitrite
(iii) Dinitrogen trioxide (N₂O₃) [3 Marks]

(i) explain disproportionation of chlorine. [2 Marks]
(ii) find the oxidation number of oxygen in sodium peroxide (Na₂O₂) and caesium superoxide (CsO₂). [2 Marks]

[Total = 10 Marks]

2. (a)

(i) State the Periodic law. [1 Mark]
(ii) List any TWO atomic properties and state their trends in the periodic table. [2 Marks]

(b)

(i) Define the term Standard solution. Mention ONE application of standard solution. [1 Mark]
(ii) Differentiate between empirical formula and molecular formula. Name ONE other chemical formula that describes atoms in a molecule of a compound. [2 Marks]
(iii) An organic compound J is known to contain carbon, hydrogen and oxygen only. When burnt completely in excess oxygen, carbon dioxide and water are given out as the only product. It is found that 0.46 g of J gives 0.88 g carbon dioxide and 0.54 g water. Find the empirical formula of compound J. (C = 12.0, O = 16.0, H = 1.0)

CHM 002: PHYSICAL CHEMISTRY

3. (a)

(i) State Hess’ law of enthalpy summation. [1 Mark]
(ii) Hydrogen sulphide, H₂S, is a poisonous gas with the odour of rotten eggs. The reaction for the formation of H₂S from the elements is:

H₂(g) + ⅛S(rhombic) → H₂S(g)

Use Hess’ law to obtain the enthalpy change for this reaction from the following enthalpy changes:

H₂S(g) + ³⁄₂O₂(g) → H₂O(g) + SO₂(g) ΔH = −518 kJ mol⁻¹

H₂(g) + ½O₂(g) → H₂O(g) ΔH = −242 kJ mol⁻¹

S(rhombic) + O₂(g) → SO₂(g) ΔH = −297 kJ mol⁻¹ [2½ Marks]

(b) Complete the following reactions:

(i) ²³Na + ⁴He → ²⁶Mg + ?
(ii) ⁶⁴Cu → ⁶⁴e + ?
(iii) ¹¹B + ⁴He → ¹⁴N + ?
(iv) ¹⁰⁸Pd + ? → ¹⁰⁸Ag + |p [2 Marks]
(v) Give THREE general applications of radioactive elements. [1½ Marks]

[Total = 10 Marks]

4. (a) What do you understand by the term “Standard enthalpy of combustion”? [1 Mark]

(b) A sample of butane gas, measured at 25°C and 98 kPa having a volume of 200 cm³, was completely burnt in air. The heat produced raised the temperature of 230g of water by 14.4°C.

(i) Assuming no heat losses occurred during this experiment, calculate the mass of butane used.
(ii) Determine the amount of heat released if the specific heat capacity of water is 4.2 Jkg⁻¹K⁻¹.
(iii) Calculate the standard enthalpy of combustion of butane.

[R = 8.31 J/K/mol, C = 12, H = 1] [4½ Marks]

© Given that 0.153g of dichloromethane displaced 45 cm³ of air at 20°C and 100.7 kNm⁻² pressure. If at 20°C the saturated vapour pressure of water is 2.319 kNm⁻², calculate the relative molecular mass of the dichloromethane.

(Avogadro’s constant = 6.02 × 10²³ mol⁻¹, R = 8.314 J/K/mol) [4½ Marks]

[Total = 10 Marks]

CHM 003: INORGANIC CHEMISTRY

5. (a) Define the following:

(i) A ligand
(ii) Coordination number [2 Marks]

(b) Given the complex ion [Cu(NH₃)₄]²⁺:

(i) What is the coordination number of the complex ion?
(ii) Give the IUPAC name of the complex ion.
(iii) What is the oxidation state of the central metal ion?
(iv) What is the shape of the complex ion? [3 Marks]

(d) Explain why the alkaline earth metals are harder and have higher melting point than the alkali metals? [3 Marks]

[Total = 10 Marks]

6. (a)

(i) Why do halogens form interhalogen compounds? [½ Mark]
(ii) Give FOUR examples of interhalogen compounds. [2 Marks]

(b) Suggest reasons for the following:

(i) Ionisation energy of Manganese (Mn) is greater than Iron (Fe) despite the trend that ionisation energy increases across the period on the periodic table.
(ii) Compounds of Sc³⁺ and Zn²⁺ are mostly white while those of Se²⁺ and Cu²⁺ are coloured.
(iii) Ionization energy of gallium (Ga) is slightly higher than that of aluminium (Al). [4½ Marks]

CHM 004: ORGANIC CHEMISTRY

7. (a) Draw the structures of THREE isomers of C₃H₆O. [1½ Marks]

(b) The chlorination of methane can be achieved through a free-radical reaction. Write a balanced equation for each stage. [3 Marks]

(i) bromine (ii) hydrogen bromide (iii) water (iv) Hydrogen (with catalyst) [2½ Marks]

(d)

(i) Given the reaction scheme below:

CH₂=CHCH₂CH₂COOH →(H₂/Ni) X →(CH₃COCl/AlCl₃) Y

Write the chemical formula and name of X and Y.

(ii) Give the name of the reactions producing X and Y respectively. [3 Marks]

[Total = 10 Marks]

8. (a)

(i) Define the term hybridization.
(ii) Give the name of the following compound and the hybridization pattern of first and last carbon atom: CH₂=CHCH₂CH₂COOH [2 Marks]

(b) Aldehydes and ketones are an important class of organic compounds.

(i) Using butanal and 2-butanone as examples, what type of isomerism is exhibited by these two classes of organic compounds?
(ii) Give the products formed when butanal and 2-butanone are reacted with:
i. Na₂Cr₂O₇/H₂SO₄
ii. NaBH₄, H₃O⁺ [2½ Marks]

(i) Define the term polymerisation.
(ii) Give TWO differences between condensation polymerisation and addition polymerisation.
(iii) Using chemical equations ONLY, describe the polymerisation reactions for:
a condensation polymer
an addition polymer [3½ Marks]

(d) Using appropriate examples, describe a chemical test to distinguish between a reducing sugar and a non-reducing sugar. [3 Marks]

Answers

CHM 001: GENERAL CHEMISTRY

Question 1

(a) Empirical formula of metal chloride

Ca reacts with HCl → metal chloride.

Moles of Ca = 2.21/40 = 0.05525 mol

Let metal chloride = MCl_n. Mass of Cl = 6.15 − 2.21 = 3.94 g

Moles of Cl = 3.94/35.5 = 0.1109 mol

Ratio Cl : Ca = 0.1109/0.05525 = 2.008 ≈ 2

∴ Empirical formula = CaCl₂

(b) Oxidation numbers of nitrogen

| Compound | Calculation | O.N. of N |

|—|---|—|

| (i) Dinitrogen oxide (N₂O) | 2x + (−2) = 0 → x = +1 | +1 |

| (ii) Sodium nitrite (NaNO₂) | (+1) + x + 2(−2) = 0 → x = +3 | +3 |

| (iii) N₂O₃ | 2x + 3(−2) = 0 → x = +3 | +3 |

Balanced equation: Cl₂ + H₂O ⇌ HCl + HOCl

(i) Disproportionation:

Cl₂ undergoes disproportionation because the same element (Cl) is simultaneously oxidised and reduced. In Cl₂, Cl is 0. In HCl, Cl = −1 (reduced). In HOCl, Cl = +1 (oxidised). One Cl atom gains electrons while the other loses electrons.

(ii) Oxidation numbers of oxygen:

Na₂O₂ (sodium peroxide): 2(+1) + 2x = 0 → x = −1
CsO₂ (caesium superoxide): (+1) + 2x = 0 → x = −½

Question 2

(a)

(i) Periodic Law: The physical and chemical properties of elements are a periodic function of their atomic numbers (when elements are arranged in order of increasing atomic number, properties repeat at regular intervals).

(ii) Two atomic properties and trends:

| Property | Trend |

|—|---|

| Atomic radius | Decreases across a period (left→right); increases down a group |

| Ionisation energy | Increases across a period; decreases down a group |

(b)(i) Standard solution: A solution of accurately known concentration. Application: used as a titrant in volumetric analysis (titrations) to determine unknown concentrations.

(ii)

Empirical formula: Shows the simplest whole-number ratio of atoms of each element in a compound.
Molecular formula: Shows the actual number of atoms of each element in one molecule.
Another chemical formula: Structural formula (shows how atoms are bonded/arranged).

(iii) Empirical formula of compound J:

From 0.46 g of J:

CO₂ = 0.88 g → moles C = 0.88/44 = 0.02 mol → mass C = 0.02 × 12 = 0.24 g
H₂O = 0.54 g → moles H₂O = 0.54/18 = 0.03 mol → moles H = 0.06 → mass H = 0.06 g
Mass O = 0.46 − 0.24 − 0.06 = 0.16 g → moles O = 0.16/16 = 0.01 mol

Ratio C : H : O = 0.02 : 0.06 : 0.01 = 2 : 6 : 1

∴ Empirical formula = C₂H₆O

CHM 002: PHYSICAL CHEMISTRY

Question 3

(a)(i) Hess’ Law: The total enthalpy change for a reaction is independent of the route taken, provided initial and final states are the same.

(ii) Finding ΔH for: H₂(g) + ⅛S(rhombic) → H₂S(g)

Given:

Eq.1: H₂S(g) + 3/2O₂(g) → H₂O(g) + SO₂(g) ΔH₁ = −518 kJ/mol
Eq.2: H₂(g) + ½O₂(g) → H₂O(g) ΔH₂ = −242 kJ/mol
Eq.3: S(rhombic) + O₂(g) → SO₂(g) ΔH₃ = −297 kJ/mol

Target: H₂ + S → H₂S

Use: Eq.2 + Eq.3 − Eq.1

ΔH = (−242) + (−297) − (−518)

ΔH = −539 + 518 = −21 kJ/mol

∴ ΔH(formation of H₂S) = −21 kJ mol⁻¹

(b) Complete nuclear reactions:

(i) ²³Na + ⁴He → ²⁶Mg + ¹H (proton)

(ii) ⁶⁴Cu → ⁶⁴e + ⁶⁴Zn (beta decay; Cu→Zn)

(iii) ¹¹B + ⁴He → ¹⁴N + ¹n (neutron)

(iv) ¹⁰⁸Pd + ¹n → ¹⁰⁸Ag + p (neutron capture)

(v) THREE applications of radioactive elements:

Medical diagnosis and treatment — e.g., radiotherapy for cancer using Co-60
Carbon dating — C-14 used to determine age of ancient materials
Nuclear power generation — U-235 used as fuel in nuclear reactors

© Disintegrations per year from 1g of Uranium-236

Half-life t½ = 4.5 × 10⁹ years

Decay constant: λ = 0.693/t½ = 0.693/(4.5 × 10⁹) = 1.54 × 10⁻¹⁰ yr⁻¹

Moles of U = 1/236 = 4.237 × 10⁻³ mol

Number of atoms N = 4.237 × 10⁻³ × 6.02 × 10²³ = 2.55 × 10²¹ atoms

Activity A = λN = 1.54 × 10⁻¹⁰ × 2.55 × 10²¹

A = 3.93 × 10¹¹ disintegrations per year

Question 4

(a) Standard enthalpy of combustion: The enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions (298 K, 100 kPa), with all reactants and products in their standard states.

(b)

(i) Mass of butane used:

Using ideal gas law: PV = nRT

P = 98 kPa = 98,000 Pa; V = 200 cm³ = 200 × 10⁻⁶ m³; T = 25°C = 298 K

n = PV/RT = (98000 × 200 × 10⁻⁶)/(8.31 × 298)

n = 19.6/2476.38 = 7.92 × 10⁻³ mol

Molar mass of butane (C₄H₁₀) = 4(12) + 10(1) = 58 g/mol

Mass = 7.92 × 10⁻³ × 58 = 0.459 g ≈ 0.46 g

(ii) Heat released:

Q = mcΔT = 0.230 × 4200 × 14.4

Q = 13,910.4 J = 13.91 kJ

(iii) Standard enthalpy of combustion:

Moles of butane = 0.459/58 = 7.92 × 10⁻³ mol

ΔH°c = −Q/n = −13.91/(7.92 × 10⁻³)

ΔH°c = −1756 kJ/mol

(Literature value ≈ −2877 kJ/mol; discrepancy due to heat losses despite the assumption)

© Relative molecular mass of dichloromethane

Mass of vapour = 0.153 g

V = 45 cm³ = 45 × 10⁻⁶ m³

T = 20°C = 293 K

Corrected pressure = total − vapour pressure of water:

P = 100.7 − 2.319 = 98.381 kNm⁻² = 98,381 Pa

Using PV = nRT:

n = PV/RT = (98381 × 45 × 10⁻⁶)/(8.314 × 293)

n = 4.427/2.436 = 1.817 × 10⁻³ mol

Mr = mass/moles = 0.153/1.817 × 10⁻³ = 84.2 g/mol

∴ Relative molecular mass of dichloromethane = 84.2 (theoretical = 85, ✓)

CHM 003: INORGANIC CHEMISTRY

Question 5

(a) Definitions:

(i) Ligand: An ion or molecule that donates a lone pair of electrons to a central metal ion to form a coordinate (dative) bond in a complex.

(ii) Coordination number: The total number of donor atoms (ligands) bonded directly to the central metal ion in a complex.

(b) [Cu(NH₃)₄]²⁺

(i) Coordination number = 4 (four NH₃ ligands attached)

(ii) IUPAC name: Tetraamminecopper(II) ion

(iii) Oxidation state of Cu:

x + 4(0) = +2 → x = +2

(iv) Shape: Square planar

© Why I₂ is solid, Br₂ liquid, Cl₂ gas:

All are non-polar molecules held together by van der Waals (London dispersion) forces. The strength of these forces increases with increasing molecular mass and electron cloud size:

Cl₂ (Mr = 71) — weakest forces → gas at room temperature
Br₂ (Mr = 160) — intermediate forces → liquid
I₂ (Mr = 254) — strongest forces → solid

(d) Alkaline earth metals harder with higher melting points than alkali metals:

Alkaline earth metals (Group 2) have two valence electrons compared to one in alkali metals (Group 1).
They form smaller, more highly charged ions with shorter metallic bond lengths.
The metallic bonding is stronger due to higher charge density and greater electron delocalization.
This results in stronger electrostatic attraction between metal ions and delocalized electrons → greater hardness and higher melting points.

Question 6

(a)(i) Halogens form interhalogen compounds because different halogen atoms have similar electronegativities and atomic sizes (though slightly different), allowing them to bond covalently. The smaller halogen bonds to the larger one. Also, they all need one electron to complete their octet, facilitating bonding.

(ii) Four examples of interhalogen compounds:

ClF (chlorine monofluoride)
BrF₃ (bromine trifluoride)
ICl (iodine monochloride)
IF₅ (iodine pentafluoride)

(b) Reasons:

(i) IE of Mn > Fe despite left-to-right trend:

Mn has the configuration [Ar] 3d⁵4s². The 3d subshell is exactly half-filled, which is extra stable (exchange energy stabilisation). Removing an electron disrupts this stability, so more energy is needed. Fe is [Ar] 3d⁶4s² — one electron is paired in 3d, making it easier to remove → lower IE than Mn.

(ii) Sc³⁺ and Zn²⁺ are white; Se²⁺ and Cu²⁺ are coloured:

Colour in transition metal compounds arises from d-d electron transitions (absorption of visible light).

Sc³⁺: [Ar] — no d electrons → no d-d transition → white/colourless
Zn²⁺: [Ar] 3d¹⁰ — completely filled d → no d-d transition → white
Se²⁺ and Cu²⁺ have partially filled d orbitals → d-d transitions occur → coloured

(iii) IE of Ga slightly higher than Al:

Ga ([Ar] 3d¹⁰4s²4p¹) has a filled 3d subshell which provides poor shielding of the nucleus. Despite being below Al in the group, the effective nuclear charge experienced by the 4p electron in Ga is higher due to poor d-electron shielding, resulting in a slightly higher IE than expected.

| Gas | Source |

|—|---|

| Carbon dioxide (CO₂) | Combustion of fossil fuels, deforestation, respiration |

| Methane (CH₄) | Decomposition of organic matter, livestock farming, rice paddies, natural gas leaks |

CHM 004: ORGANIC CHEMISTRY

Question 7

(a) THREE isomers of C₃H₆O:

Propanal (aldehyde): CH₃CH₂CHO
Propan-2-one / Acetone (ketone): CH₃COCH₃
Allyl alcohol / Prop-2-en-1-ol (unsaturated alcohol): CH₂=CHCH₂OH

(Other valid answers: methyloxirane/propylene oxide, cyclopropanol)

(b) Free-radical chlorination of methane — stages:

Stage 1 — Initiation:

Cl₂ → 2Cl• (UV light)

Stage 2 — Propagation:

Cl• + CH₄ → CH₃• + HCl

CH₃• + Cl₂ → CH₃Cl + Cl•

Stage 3 — Termination:

Cl• + Cl• → Cl₂

CH₃• + Cl• → CH₃Cl

CH₃• + CH₃• → C₂H₆

(i) With bromine (Br₂):

CH₃CH=CH₂ + Br₂ → CH₃CHBrCH₂Br

(1,2-dibromopropane — electrophilic addition)

(ii) With hydrogen bromide (HBr):

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃

(2-bromopropane — Markovnikov’s rule)

(iii) With water (H₂O, H⁺ catalyst):

CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃

(propan-2-ol — Markovnikov addition)

(iv) With hydrogen (H₂, catalyst):

CH₃CH=CH₂ + H₂ →(Ni/Pt) CH₃CH₂CH₃

(propane — catalytic hydrogenation)

(d)

(i) X and Y from the reaction scheme:

CH₂=CHCH₂CH₂COOH →(H₂/Ni) X →(CH₃COCl/AlCl₃) Y

X: CH₃CH₂CH₂CH₂COOH → Pentanoic acid

(The C=C double bond is hydrogenated; the COOH is unaffected)

Y: CH₃CH₂CH₂CH₂CO—COCH₃ …

More precisely: Friedel-Crafts acylation doesn’t apply to non-aromatic chains in this context. Reconsidering the structure — CH₂=CHCH₂CH₂COOH has a terminal alkene. After hydrogenation → X = CH₃CH₂CH₂CH₂COOH (pentanoic acid). Then CH₃COCl/AlCl₃ acts as acylation: Y = CH₃COCH₂CH₂CH₂CH₃ (a ketone formed via acyl substitution of the acid to acid chloride intermediate, then reduction, or direct ketone formation).

(ii) Names of reactions:

X-producing reaction: Catalytic hydrogenation (addition of hydrogen)
Y-producing reaction: Friedel-Crafts acylation / Acylation reaction

Question 8

(a)

(i) Hybridization: The mixing of atomic orbitals of similar energies in the same atom to form a new set of equivalent orbitals (hybrid orbitals) with identical shape and energy, directed in space to minimize repulsion.

(ii) CH₂=CHCH₂CH₂COOH

Name: Pent-4-enoic acid

First carbon (CH₂=): Involved in a double bond → sp² hybridized
Last carbon (COOH): The carboxyl carbon has a C=O and C–O → sp² hybridized

(b)

(i) Type of isomerism — butanal vs 2-butanone:

Both have molecular formula C₄H₈O but differ in the position of the C=O group.

This is functional group isomerism (also described as a type of structural/constitutional isomerism — one is an aldehyde, the other a ketone).

(ii) Products with:

i. Na₂Cr₂O₇/H₂SO₄ (oxidising agent):

Butanal (aldehyde) → Butanoic acid (CH₃CH₂CH₂COOH) — aldehydes are oxidised to carboxylic acids
2-Butanone (ketone) → No reaction — ketones resist oxidation by mild oxidising agents

ii. NaBH₄, H₃O⁺ (reducing agent):

Butanal → Butan-1-ol (CH₃CH₂CH₂CH₂OH) — reduction of aldehyde to primary alcohol
2-Butanone → Butan-2-ol (CH₃CH(OH)CH₂CH₃) — reduction of ketone to secondary alcohol

(i) Polymerisation: The process by which many small molecules (monomers) join together repeatedly to form a large molecule (polymer).

(ii) TWO differences:

| Condensation Polymerisation | Addition Polymerisation |

|—|---|

| Involves monomers with two functional groups (bifunctional) | Involves monomers with C=C double bonds (alkenes) |

| A small molecule is eliminated (e.g., H₂O, HCl) during each linkage | No by-product is formed; all atoms of monomer are in the polymer |

(iii) Chemical equations:

Condensation polymer (e.g., nylon-6,6):

n H₂N(CH₂)₆NH₂ + n HOOC(CH₂)₄COOH → [–NH(CH₂)₆NHCO(CH₂)₄CO–]ₙ + 2n H₂O

Addition polymer (e.g., polyethene):

n CH₂=CH₂ → [–CH₂–CH₂–]ₙ

(d) Chemical test to distinguish reducing sugar from non-reducing sugar:

Using Fehling’s solution (or Benedict’s reagent):

Add Fehling’s solution (deep blue) to both sugar solutions and heat.
Reducing sugar (e.g., glucose, maltose): A brick-red precipitate (Cu₂O) forms → positive test. The sugar reduces Cu²⁺ to Cu⁺.
Non-reducing sugar (e.g., sucrose): No colour change (remains blue) → negative test.

To confirm the non-reducing sugar: hydrolyse it with dilute HCl, neutralise, then repeat Fehling’s test — it will now give a brick-red precipitate because hydrolysis releases reducing sugars (e.g., glucose + fructose from sucrose).

2023 JUPEB chemistry

CHM 001: GENERAL CHEMISTRY

CHM 002: PHYSICAL CHEMISTRY

CHM 003: INORGANIC CHEMISTRY

CHM 004: ORGANIC CHEMISTRY

Answers

CHM 001: GENERAL CHEMISTRY

Question 1

Question 2

CHM 002: PHYSICAL CHEMISTRY

Question 3

Question 4

CHM 003: INORGANIC CHEMISTRY

Question 5

Question 6

CHM 004: ORGANIC CHEMISTRY

Question 7

Question 8

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2023 JUPEB chemistry

CHM 001: GENERAL CHEMISTRY

CHM 002: PHYSICAL CHEMISTRY

CHM 003: INORGANIC CHEMISTRY

CHM 004: ORGANIC CHEMISTRY

Answers

CHM 001: GENERAL CHEMISTRY

Question 1

Question 2

CHM 002: PHYSICAL CHEMISTRY

Question 3

Question 4

CHM 003: INORGANIC CHEMISTRY

Question 5

Question 6

CHM 004: ORGANIC CHEMISTRY

Question 7

Question 8

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