B — Sodium ethoxide

When ethanol reacts with sodium metal: $2\text{Na} + 2\text{CH}_3\text{CH}_2\text{OH} \rightarrow 2\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2$. The other product is sodium ethoxide.

Source: Properties of Ethanol, Chapter 4

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The reaction of ethanol with sodium is a standard 1-mark fact. Examiners expect you to recall the chemical equation and name the ionic product — sodium ethoxide ($\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+$). Do not confuse it with sodium hydroxide (which forms when sodium reacts with water, not ethanol).

(C) Ethene (C₂H₄)

Ethene is an unsaturated hydrocarbon with a C=C double bond; it decolourises bromine water and undergoes hydrogenation with H₂ in the presence of a nickel catalyst.

(Note: Option (D) listed as C₂H₂/ethyne also fits, but among the named options, Ethene is the standard answer.)

The question tests knowledge of addition reactions. Only unsaturated hydrocarbons (alkenes/alkynes) decolourise bromine water (by adding Br₂ across the double/triple bond) and undergo hydrogenation. Methane, ethane, and propane are saturated; they do neither. The question's option list names "Ethene" as (C), making it the correct choice. Ethyne (C₂H₂) also qualifies chemically, but it is not among the named options (A–D) as stated in the question stem—the option (D) in the stem is "Propane."

Option B: Ethanol

Ethanol (C₂H₅OH) on oxidation with alkaline potassium permanganate forms ethanoic acid (CH₃COOH).

Source: Oxidation, Chapter 4, Section 4.3.2

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The key is matching the product (ethanoic acid, a 2-carbon compound) to its alcohol precursor. Ethanol is also the 2-carbon alcohol, so oxidation preserves the carbon chain and converts –OH to –COOH. The passage explicitly uses ethanol in Activity 4.5 to demonstrate this oxidation. Other options (methanol = 1C, propanol = 3C, butanol = 4C) would not yield ethanoic acid.

Answer: C — Alkane

Cyclohexane (C₆H₁₂) is a cyclic compound with all single bonds between carbon atoms, making it a saturated hydrocarbon, i.e., an alkane.

The textbook states that cyclohexane has formula C₆H₁₂ and is a ring (cyclic) structure with only single bonds — hence saturated. Saturated hydrocarbons are called alkanes. Don't confuse it with alkenes (double bond) or aromatics like benzene (C₆H₆). The "cyclo" prefix simply indicates the ring form of an alkane.

(C) –CHO

Propanal belongs to the aldehyde class (suffix -al), so it contains the –CHO (aldehyde) functional group.

Source: Chapter 4, Table 4.3 and Table 4.4

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The suffix -al in "propanal" directly indicates an aldehyde functional group (–CHO). Remember: –OH = alcohol, –COOH = carboxylic acid, –CO– (flanked by two carbon chains) = ketone. Examiners expect you to link the IUPAC suffix to the correct functional group.

Answer: B

Soap reacts with calcium and magnesium salts present in hard water to form insoluble salts, which appear as a white curdy precipitate (scum).

Source: Soaps and Detergents, Chapter 4

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The textbook explicitly states: "This is caused by the reaction of soap with the calcium and magnesium salts, which cause the hardness of water" — forming an insoluble precipitate. Options A, C, and D are factually incorrect and not supported by the text. Examiners expect students to identify the correct chemical cause (reaction with Ca/Mg salts) and the result (insoluble/curdy precipitate).

Answer: C — Dehydrating agent

Concentrated H₂SO₄ removes water from ethanol during dehydration, converting it to ethene at 443 K. It acts as a dehydrating agent.

Source: Chapter 4, Section 4.4.1 (Properties of Ethanol)

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The textbook explicitly states: "The concentrated sulphuric acid can be regarded as a dehydrating agent which removes water from ethanol." Examiners expect you to recall this specific term. Do not confuse it with a catalyst (though H₂SO₄ can act as one in esterification) — here, the question context points to dehydration, making C the only correct option.

(B) C–H bond; substitution reaction

In the reaction CH₄ + Cl₂ → CH₃Cl + HCl (sunlight), the C–H bond in methane is broken and chlorine replaces a hydrogen atom. This is a substitution reaction.

Source: Chapter 4, Section 4.3.4 – Substitution Reaction

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• The key facts are: methane has no C–C bond (single carbon), so options A and D are eliminated immediately.
• It is not an addition reaction (no double/triple bond present in saturated methane), so option C is wrong.
• Examiners expect you to name both the bond broken (C–H) and the reaction type (substitution) with a brief reason. Writing the chemical equation earns full credit.

(B) Electrons are shared between atoms so no charged particles (ions) are formed.

Covalent compounds exist as molecules; since electrons are shared, no free ions or electrons are produced to carry electric charge, making them poor conductors.

Source: Carbon and its Compounds, Section 4.1

The textbook states: "Since the electrons are shared between atoms and no charged particles are formed, such covalent compounds are generally poor conductors of electricity." Examiners expect you to link electron sharing → no ions formed → no charge carriers → poor conduction. Option B matches this exactly. Options A and C are incorrect (covalent compounds have low melting points); Option D is not a reason given in the text.

Option A — Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Saponification is the alkaline hydrolysis of an ester with NaOH, giving alcohol and sodium salt of carboxylic acid; this reaction is used in soap preparation.

Source: Chapter 4, Section 4.4.2 (Properties of Ethanoic Acid)

The textbook explicitly states: "On treating with sodium hydroxide…the ester is converted back to alcohol and sodium salt of carboxylic acid. This reaction is known as saponification because it is used in the preparation of soap." So the Reason directly and correctly explains the Assertion — choose Option A.

Option A — Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Carbon's tetravalency and catenation together enable it to form millions of compounds, more than all other elements combined.

The textbook explicitly states: "The two characteristic features seen in carbon, that is, tetravalency and catenation, put together give rise to a large number of compounds." Both the assertion (carbon forms the most compounds) and the reason (catenation + tetravalency) are correct, and the reason directly explains the assertion — so option A is the answer. Examiners look for you to confirm both statements are true and that R correctly explains A.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

The textbook states that the charged ends of detergent molecules do not form insoluble precipitates with Ca²⁺ and Mg²⁺ ions in hard water, which is exactly why detergents remain effective in hard water — making R the direct and correct explanation of A.

Catenation is the unique ability of carbon atoms to form bonds with other carbon atoms, giving rise to large molecules with long chains, branched chains, or ring structures.

Carbon exhibits catenation to a much greater extent than silicon because the carbon–carbon bond is very strong and stable. Silicon can form chains of only up to 7–8 atoms, but those compounds are very reactive and unstable.

Source: Chapter 4, Section 4.2 – Versatile Nature of Carbon

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• Define catenation clearly in one line — examiners expect the exact term and its meaning.
• The key reason for the difference is bond strength: C–C bond is stronger than Si–Si bond due to carbon's small atomic size, allowing the nucleus to hold shared electrons more tightly.
• Mentioning silicon's chain limit (7–8 atoms) and their reactivity adds a comparative point that fetches the second mark.

Two characteristics of a homologous series:

1. All members have the same functional group and similar chemical properties.
2. Successive members differ by a –CH₂– unit (molecular mass difference of 14 u).

Next two higher members of C₃H₆O₂ (carboxylic acid series):

• C₄H₈O₂ (Butanoic acid)
• C₅H₁₀O₂ (Pentanoic acid)

Source: Carbon and its Compounds, Section 4.2.4 – Homologous Series

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• The question has two parts — state characteristics first, then write the formulae.
• Each successive member of a homologous series adds one –CH₂– unit: C₃H₆O₂ → C₄H₈O₂ → C₅H₁₀O₂.
• The general formula for carboxylic acids is CₙH₂ₙO₂; apply it to n = 4 and n = 5.
• Examiners look for both characteristics clearly stated and both formulae correct. Names are a bonus but formulae are essential.

Balanced chemical equation:

$$\text{CH}_3\text{COOH} + \text{NaHCO}_3 \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

Gas evolved: Carbon dioxide (CO₂) is the gas evolved. It turns lime-water milky, confirming its identity.

Source: Carbon and its Compounds, Section 4.4.2 – Reactions of ethanoic acid

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• Examiners expect the balanced equation (1 mark) and identification of gas (1 mark) — both are needed for full credit.
• Write the upward arrow (↑) after CO₂ to indicate it is a gas; this is good exam practice.
• Mentioning the lime-water test is a bonus detail that strengthens the answer within the word limit.
• The products are sodium ethanoate (CH₃COONa), water, and CO₂ — do not omit any product.

In graphite, each carbon atom is bonded to only three other carbon atoms, leaving one electron free. These free electrons can move through the layers, allowing graphite to conduct electricity.

In diamond, each carbon atom is bonded to four other carbon atoms in a rigid 3D structure, leaving no free electrons. Hence, diamond cannot conduct electricity.

Source: Chapter 4, Allotropes of Carbon (More to Know)

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• The key concept is free/delocalized electrons: graphite has them (one bond is a double bond, freeing an electron per atom); diamond has none (all four valence electrons are used in bonding).
• Examiners expect you to clearly state the bonding difference (3 vs 4 bonds) AND directly link it to presence/absence of free electrons for conduction.
• Don't just say "different structures" — explain why that structure affects conductivity.

Oxidising agent: Alkaline potassium permanganate (KMnO₄) or acidified potassium dichromate.

Why it is oxidation: In this conversion, oxygen is added to ethanol (CH₃CH₂OH) to form ethanoic acid (CH₃COOH). Since the reaction involves addition of oxygen to the starting material, it is regarded as an oxidation reaction.

Source: Chapter 4, Section 4.3.2 Oxidation

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• 1 mark for naming a correct oxidising agent — either alkaline KMnO₄ or acidified K₂Cr₂O₇ is accepted.
• 1 mark for the explanation — examiners want the key idea that oxygen is added to ethanol, which is the definition of oxidation used in this chapter. Avoid vague statements; mention both reactant (ethanol) and product (ethanoic acid) to make it clear.
• Do not write lengthy definitions of oxidation; one crisp sentence is enough for 1 mark.

(i) Electron dot structure of ethyne (C₂H₂):

$$\text{H} :\!\!\overset{..}{\text{C}}\!\!:\!\!:\!\!:\!\!\overset{..}{\text{C}}\!\!: \text{H}$$

H — C ≡ C — H (with each C sharing 3 pairs of electrons, forming a triple bond, and one bond each with H)

(ii) Ethyne is an unsaturated compound because it contains a triple bond between the two carbon atoms. Compounds with double or triple bonds between carbon atoms are called unsaturated carbon compounds.

(iii)

$$\text{C}_2\text{H}_2 + 2\text{H}_2 \xrightarrow{\text{Ni}} \text{C}_2\text{H}_6$$

The product formed is ethane.

Source: Chapter 4, Section 4.2.1 — Saturated and Unsaturated Carbon Compounds

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• For (i), show the triple bond clearly — three pairs of shared electrons between the two C atoms, with one H on each end. Examiners award marks for correct depiction of the triple bond.
• For (ii), one line identifying it as unsaturated + one reason (triple bond) is sufficient for 1 mark.
• For (iii), balance the equation correctly: ethyne needs 2 moles of H₂ (since it has a triple bond, two bonds must be broken and hydrogenated). Naming the product (ethane) is essential — don't skip it.

Carbon has 4 electrons in its outermost shell. To form C⁴⁺, it would need to lose 4 electrons, requiring a very large amount of energy. To form C⁴⁻, it would need to gain 4 electrons, which would be difficult for the nucleus to hold. Both processes are energetically unfavourable.

Instead, carbon achieves noble gas configuration (8 electrons in outermost shell) by sharing its 4 valence electrons with other atoms, forming 4 covalent bonds. Carbon's small size allows its nucleus to hold shared electron pairs strongly, making these bonds stable. For example, in methane (CH₄), carbon shares one electron each with four hydrogen atoms.

Source: Chapter 4, Section 4.2 — Versatile Nature of Carbon

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• Examiners expect two parts: (1) why ionic bonding fails for carbon, and (2) how covalent bonding solves it.
• Mention both C⁴⁺ and C⁴⁻ separately — each is worth a point.
• Naming covalent bonds / sharing of electrons and giving methane as an example shows application.
• The phrase "small size → nucleus holds shared pairs strongly" is directly from the textbook and scores well.

(i) IUPAC Names:

(a) CH₃CH₂CH₂OH — It has 3 carbon atoms with —OH (alcohol) functional group.
IUPAC name: Propan-1-ol

(b) CH₃CH₂Br — It has 2 carbon atoms with bromo substituent.
IUPAC name: Bromoethane

(c) CH₃COCH₂CH₃ — It has 4 carbon atoms with ketone (—CO—) functional group.
IUPAC name: Butan-2-one

(ii) Compound (c) CH₃COCH₂CH₃ belongs to the ketone functional group. The ketone group (>C=O) is present between carbon atoms, and its IUPAC suffix is -one.

Source: Chapter 4, Section 4.2.5 Nomenclature of Carbon Compounds

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• For alcohols, suffix -ol is added; for haloalkanes, the halogen name is used as a prefix (bromo, chloro).
• For ketones, suffix -one is used. The position of the functional group must be indicated (butan-2-one).
• Examiners expect correct spelling and position numbers. "Propan-1-ol" is preferred over just "propanol" to show the position.

(i) The catalyst used is nickel (Ni).

(ii) Chemical equation for hydrogenation of ethene:

$$\text{CH}_2{=}\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni, heat}} \text{CH}_3\text{-}\text{CH}_3$$

(Ethene → Ethane)

(iii) Animal fats contain saturated fatty acids, which are harmful to health. Oils with unsaturated fatty acids are healthier and therefore advisable for cooking.

Source: Chapter 4, Section 4.3.3 Addition Reaction

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• (i) Both Ni and Pd are acceptable, but the passage specifically highlights nickel for the industrial hydrogenation of vegetable oils — write Ni for full marks.
• (ii) Show the double bond in ethene, the reagent H₂, the catalyst (Ni) above the arrow, and the product ethane. Writing structural or molecular formulae both earn marks, but showing the double bond clearly is important.
• (iii) The key contrast examiners look for: saturated (animal fats) = harmful; unsaturated (vegetable oils) = healthier. One sentence is enough for 1 mark.

(i) Covalent compounds have low melting and boiling points because, although the covalent bonds within molecules are strong, the intermolecular forces between molecules are very weak. Very little energy is needed to overcome these weak forces.

(ii) Unsaturated hydrocarbons have a high proportion of carbon relative to hydrogen. They undergo incomplete combustion due to insufficient oxygen, releasing fine carbon particles (soot) that glow yellow, producing a sooty yellow flame.

(iii) Methanol is far more dangerous because even a small quantity can cause death. In the liver, methanol is oxidised to methanal (formaldehyde), which coagulates the protoplasm of cells and damages the optic nerve, causing permanent blindness or death.

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• For (i), the key phrase from the textbook is: "covalently bonded molecules have strong bonds within the molecule, but intermolecular forces are weak" — always mention both aspects.
• For (ii), the sooty flame question tests understanding of C:H ratio and incomplete combustion — link high carbon content → insufficient oxygen → soot.
• For (iii), quote the specific biochemical effect (methanal, protoplasm coagulation, optic nerve damage) — these are the scoring points examiners look for.

Structural Isomers of Butane (C₄H₁₀):

1. n-Butane (Normal butane) — straight chain

$$\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_3$$

2. Isobutane (2-methylpropane) — branched chain

$$\text{CH}_3 - \overset{|}{\underset{|}{\text{CH}}} (\text{CH}_3) - \text{CH}_3$$

$$\text{i.e., } \text{CH}_3{-}\text{CH(CH}_3){-}\text{CH}_3$$

Differing Physical Property — Boiling Point:
n-Butane has a higher boiling point (−1°C) than isobutane (−12°C).

Reason: n-Butane has a longer, straight chain, allowing greater surface area for intermolecular interactions. Isobutane's branched, compact shape reduces surface contact, resulting in weaker intermolecular forces and hence a lower boiling point.

Source: Carbon and its Compounds, Section 4.2.2

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• 2 marks are for drawing both correct structural formulae with common names — examiners check that all C and H valencies are satisfied and the chain/branch is clearly shown.
• 1 mark is for naming a physical property and giving a reason. Boiling point is the safest choice; melting point is also acceptable.
• Write the branched structure clearly, showing the side —CH₃ group attached to the middle carbon; don't just write the molecular formula.
• The reason must link molecular shape → surface area → intermolecular forces → boiling point to earn the mark; just saying "different structures" is not enough.

(i)(a) The IUPAC name of X (C₂H₅OH) is Ethanol.

(i)(b) Dehydration of ethanol with excess conc. H₂SO₄ at 443 K:

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Hot conc. H}_2\text{SO}_4,\ 443\text{ K}} \text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O}$$

The product formed is ethene (an unsaturated hydrocarbon).

(i)(c) This dehydration is an elimination reaction (water molecule is eliminated from ethanol).

(ii) When ethanol reacts with sodium, the gas produced is hydrogen (H₂).

$$2\text{Na} + 2\text{CH}_3\text{CH}_2\text{OH} \rightarrow 2\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2\uparrow$$

(Sodium ethoxide is the other product.)

(iii) Ethanol is used as a solvent in medicines such as tincture iodine, cough syrups, and tonics. / It is also used as an industrial solvent.

Source: Chapter 4, Section 4.4.1 — Properties of Ethanol

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• (i)(b): Write the arrow with "Hot conc. H₂SO₄" and temperature above it; examiners check for conditions.
• (i)(c): The textbook calls this a reaction to give an "unsaturated hydrocarbon," but the reaction type is elimination — water is removed from the molecule. State this clearly.
• (ii): The balanced equation must show 2Na and 2C₂H₅OH on the left; one-sided equations lose marks.
• (iii): Any one valid use (medicinal solvent or industrial solvent or fuel additive) is enough for 1 mark — don't write all; it wastes time.
• Keep equations balanced and conditions labelled — these are the two most common mark-losing errors.

(i) Cleansing Action of Soap:

(a) A soap molecule is a sodium/potassium salt of a long-chain carboxylic acid. It has two ends: a hydrophilic (ionic) end that attracts water, and a hydrophobic (hydrocarbon tail) end that repels water but interacts with oil/grease.

(b) When soap is added to water containing oily dirt, the hydrophobic tails of soap molecules orient themselves into the oil droplet, while the hydrophilic ionic ends face outward toward water. This cluster is called a micelle. The oily dirt is trapped in the interior of the micelle.

(c) Micelles remain suspended in water as a colloid and do not precipitate because of ion–ion repulsion between the negatively charged ionic ends on the surface. The suspended dirt is then rinsed away easily.

(ii) In hard water, soap reacts with calcium and magnesium salts to form an insoluble white scum (curdy precipitate), reducing its cleaning ability. Detergents are sodium salts of sulphonic acids whose charged ends do not form insoluble precipitates with Ca²⁺/Mg²⁺ ions, so they remain effective in hard water.

(iii) When ethyl ethanoate is treated with NaOH, saponification occurs:

$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa}$$

(Ethyl ethanoate) → (Ethanol) + (Sodium ethanoate)

Source: Chapter 4, Section 4.5 Soaps and Detergents; Section 4.4.2 Properties of Ethanoic Acid

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• For part (i), examiners look for all three terms: hydrophilic, hydrophobic, micelle, and the specific reason micelles don't precipitate (ion–ion repulsion). Missing any loses a mark.
• For part (ii), the key words are scum/insoluble precipitate (for soap's failure) and sulphonic acid salts / no insoluble precipitate (for detergents). Don't just say "detergents work in hard water" — explain why.
• For part (iii), write the balanced equation clearly and name the reaction (saponification) for full credit.

(i) Structural Isomers of Pentane (C₅H₁₂):

Three structural isomers are possible:

1. n-Pentane (straight chain):

C—C—C—C—C

1. Isopentane (one branch):

```
C
|
C—C—C—C
```

1. Neopentane (two branches):

```
C
|
C—C—C
|
C
```
Total: 3 structural isomers

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(ii) Two characteristic properties of carbon:

1. Catenation – Carbon atoms can bond with other carbon atoms to form long chains, branched chains, and rings.
2. Tetravalency – Carbon has a valency of four, allowing it to bond with four other atoms simultaneously, forming a large variety of compounds.

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(iii)
(a) Boiling point increases with molecular mass, because larger molecules have stronger intermolecular forces, requiring more energy to overcome.

(b) Chemical properties remain similar within a homologous series, because all members have the same functional group (or none, for alkanes).

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(iv) General formula of alkanes: CₙH₂ₙ₊₂

For 7 carbon atoms (n = 7):
C₇H₂(₇)+₂ = C₇H₁₆ (Heptane)

Source: Carbon and its Compounds, Sections 4.2.2 and 4.2.5

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• For part (i), CBSE expects you to draw all three skeletal/structural forms clearly — examiners award 1 mark per correct structure and 1 mark for stating the count.
• For part (ii), the exact terms catenation and tetravalency must appear — these are directly from the "What you have learnt" summary.
• For part (iii), state the trend AND give a reason — both are needed for full marks.
• For part (iv), memorise the general formula CₙH₂ₙ₊₂ and show substitution of n=7 explicitly.

(i) When air supply is limited, combustion is incomplete. The fuel does not burn fully, and unburnt carbon particles are released, producing a yellow sooty flame with black smoke (carbon/soot deposition).

(ii) The two harmful gases released are oxides of sulphur (SO₂) and oxides of nitrogen (NOₓ).

(iii) Balanced chemical equation for complete combustion of methane:

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + \text{heat and light}$$

(iv) Rahim should clean and unblock the air holes/inlets of the burner so that sufficient air enters and the fuel burns completely, giving a clean blue flame.

Source: Chapter 4, Section 4.3.1 Combustion

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• (i) Key phrase: "incomplete combustion" → unburnt carbon → soot/yellow flame. Examiners expect this cause-effect link.
• (ii) Must name both gases specifically — "oxides of sulphur and oxides of nitrogen." Simply writing "SO₂ and NO₂" also acceptable.
• (iii) Balancing is critical: CH₄ needs 2O₂ on the left and 2H₂O on the right. Unbalanced equation loses the mark.
• (iv) The passage directly states: blackened vessel bottoms mean air holes are blocked. The fix is to clear those holes — state this clearly.

(i) The product formed is ethanoic acid (CH₃COOH). It has a sharp smell and turns blue litmus red.

(ii) Initially, the KMnO₄ is consumed in oxidising ethanol to ethanoic acid, so the purple colour disappears. When ethanol is completely used up, excess KMnO₄ has nothing left to oxidise, so the purple colour persists.

(iii) Balanced equation:

$$2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2$$

(Brisk effervescence is due to CO₂ gas evolved.)

(iv) The product belongs to the class carboxylic acids. Its functional group is –COOH (carboxyl group).

Source: Chapter 4, Section 4.3.2 – Oxidation

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• (i) Ethanol + alkaline KMnO₄ → ethanoic acid is the standard oxidation reaction; examiners expect the IUPAC name and formula.
• (ii) The key logic: KMnO₄ acts as the oxidising agent and gets reduced (loses purple colour) until all ethanol is oxidised; excess KMnO₄ remains unreacted.
• (iii) Write the balanced equation carefully — common error is forgetting to balance the 2 moles of acid vs. 1 mole of Na₂CO₃.
• (iv) "Carboxylic acid" as the class and "–COOH" as the functional group must both be stated for full credit.

(i) The white curdy precipitate is an insoluble calcium or magnesium salt of the fatty acid (soap). It is formed when soap reacts with calcium and magnesium salts dissolved in hard water.

(ii) Distilled water does not contain calcium or magnesium salts, so soap lathers easily. Hard water contains these salts, which react with soap to form insoluble scum, leaving little soap available to form lather.

(iii) Detergents are sodium salts of sulphonic acids. Their charged ends do not form insoluble precipitates with calcium and magnesium ions in hard water, so they remain effective and produce good lather, unlike soap.

(iv) Yes, it is a valid test. Soap produces a white curdy precipitate (scum) with hard water and no stable lather, while it forms a good lather with soft water. Detergent cannot be used for this test since it lathers equally in both types of water.

Source: Chapter 4, Section 4.5 — Soaps and Detergents

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• (i) Examiners expect you to identify the precipitate as insoluble calcium/magnesium soap (salt of fatty acid). Saying just "scum" without the chemical nature may lose the mark.
• (ii) The key contrast is presence vs. absence of Ca²⁺/Mg²⁺ ions — link these directly to scum formation.
• (iii) The critical point is that the charged ends of detergent molecules do not react with Ca²⁺/Mg²⁺ to form precipitates — state this explicitly.
• (iv) The textbook itself asks "Would you be able to check if water is hard by using a detergent?" — the answer is no for detergent but yes for soap. Make sure you give a reason, not just yes/no.