(i) Carbon is confirmed by the formation of CO₂ gas. When this gas is passed through lime water [Ca(OH)₂], it turns milky due to formation of calcium carbonate (CaCO₃). This confirms the presence of carbon.

(ii) Hydrogen is confirmed by the formation of water (H₂O). Water droplets appear on a cold surface/cobalt chloride paper turns from blue to pink, confirming hydrogen.

General equation for complete combustion of a hydrocarbon:

$$C_xH_y + O_2 \rightarrow CO_2 + H_2O + \text{heat and light}$$

Source: Chapter 4, Section 4.3.1 Combustion

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• Examiners expect two named products with one test each — lime water test for CO₂ and cobalt chloride paper / condensation for water.
• The general equation need not be balanced but writing it symbolically scores the mark.
• Keep the lime water test wording precise: "turns milky" is the key phrase; writing "white precipitate of CaCO₃" adds value.
• This answer covers all three marks: product for C (1), product for H (1), general equation (1).

Carbon has 4 valence electrons and needs 4 more to achieve noble gas configuration. Forming ions is not feasible because:

(i) Gaining 4 electrons to form C⁴⁻ would be difficult, as the nucleus with only 6 protons cannot hold 10 electrons stably.

(ii) Losing 4 electrons to form C⁴⁺ would require a very large amount of energy, leaving 6 protons holding just 2 electrons.

Therefore, carbon overcomes this problem by sharing its 4 valence electrons with other atoms, forming covalent bonds. This allows both atoms to achieve noble gas configuration without gaining or losing electrons.

Source: Chapter 4, Section 4.1 — Bonding in Carbon: The Covalent Bond

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Examiners look for three specific points matching the 3 marks:

1. Why gaining 4 electrons is not possible (nuclear attraction too weak for C⁴⁻).
2. Why losing 4 electrons is not possible (too much energy needed for C⁴⁺).
3. The conclusion — carbon shares electrons, forming covalent bonds instead.

Avoid vague answers like "carbon is non-metal." Use the exact reasoning from the textbook. The terms C⁴⁻, C⁴⁺, and sharing are key scoring terms.

Each hydrogen atom has 1 electron in its outermost shell and needs 2 electrons (like helium, a noble gas) to have a completely filled shell.

When two hydrogen atoms come together, each contributes 1 electron to form a shared pair — a covalent bond. This shared pair of electrons is counted by both atoms simultaneously.

Thus, each hydrogen atom effectively has 2 electrons in its outermost shell, achieving the electronic configuration of helium (a noble gas). The molecule formed is H₂.

H• + •H → H:H (shared pair)

Source: Chapter 4, "What you have learnt" — Covalent bonds are formed by the sharing of electrons between two atoms so that both can achieve a completely filled outermost shell.

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• Examiners award marks for three points: (1) what each H atom needs, (2) how sharing works, and (3) the noble gas configuration achieved.
• Always name the noble gas (helium) whose configuration H achieves — this earns the specific mark.
• The electron dot diagram (H:H) is a good addition in a 3-mark answer; it shows understanding without taking extra time.
• Do not confuse "shared pair counted by both" — this is the key idea of covalent bonding.

(B) The inter-molecular forces between covalent molecules are weak, even though intra-molecular bonds are strong.

Source: Chapter 4, Section 4.1 (Bonding in Carbon – The Covalent Bond)

The textbook explicitly states: "Covalently bonded molecules are seen to have strong bonds within the molecule, but inter-molecular forces are weak. This gives rise to the low melting and boiling points of these compounds." So it is the weak inter-molecular forces (not weak covalent bonds) that explain low melting/boiling points. Options A, C, and D are factually incorrect as per the passage.

Ethanol (liquid): Ethanol is a covalent compound. Its molecules are held together by weak intermolecular forces. When electrodes are dipped in liquid ethanol, no charged particles (ions) are present — electrons are shared between atoms, not transferred. Since there are no free ions to carry charge, no current flows.

Molten NaCl: Sodium chloride is an ionic compound made of Na⁺ and Cl⁻ ions held in a crystal lattice. On melting, these ions become free to move. These freely moving charged ions carry electric current, so current flows readily through molten NaCl.

Source: Chapter 4, Section 4.1 (Bonding in Carbon – The Covalent Bond)

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• Key contrast examiners look for: covalent (no ions → no current) vs. ionic (free ions in molten state → conducts).
• Mention shared electrons / no charged particles for ethanol; mention free ions for NaCl — these are the scoring phrases.
• Don't write about aqueous solutions; the question specifies liquid ethanol and molten NaCl.
• 3 marks typically split as: 1 mark (ethanol covalent/no ions) + 1 mark (NaCl ionic/ions) + 1 mark (clear explanation of current flow linked to free ions).

Hydrogen (H₂ — single bond):
Hydrogen has only 1 valence electron. It needs just 1 more electron to complete its duplet (stable configuration of 2 electrons). So each hydrogen atom shares one electron with the other, forming a single bond (one shared pair), and both atoms achieve a duplet.

Oxygen (O₂ — double bond):
Oxygen has 6 valence electrons. It needs 2 more electrons to complete its octet (stable configuration of 8 electrons). Sharing just one pair would leave each oxygen atom with only 7 electrons — octet unfulfilled. So each oxygen atom shares two pairs of electrons, forming a double bond, and both atoms achieve a complete octet.

This is why H₂ has a single bond while O₂ has a double bond.

Source: Chapter 4, Covalent Bonds / Versatile Nature of Carbon

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• The examiner expects you to clearly link number of valence electrons → electrons needed → bonds formed.
• Key terms to use: duplet (for hydrogen), octet (for oxygen), single bond, double bond, shared pair.
• Do NOT just state the result — briefly show the reasoning (how many electrons needed and why one/two pairs are shared). That's where the 3 marks lie: ~1 mark per element's reasoning + 1 mark for the conclusion/contrast.

Nitrogen has 3 shared pairs of electrons (triple bond). Each nitrogen atom has 5 valence electrons; sharing 3 pairs gives each atom a complete octet (2+6=8 electrons), which one shared pair cannot achieve.

The key points examiners look for: (1) state the number — 3 shared pairs, and (2) briefly justify using valence electrons / octet rule. Nitrogen has 5 valence electrons, so it needs 3 more to complete its octet; sharing one pair would give only 6 electrons per atom, which is insufficient. A triple bond (3 shared pairs) satisfies both atoms' octets.

No, the student is not correct.

Oxygen needs two more electrons to complete its octet, but this does not mean it must form double bonds. In a water molecule (H₂O), oxygen shares one electron with each of the two hydrogen atoms, forming two single covalent bonds. Each hydrogen needs only one electron to complete its duplet. This satisfies oxygen's octet (2 + 2 shared + 4 non-bonding = 8 electrons) without any double bond.

Double bonds form only when a single bond leaves valencies unsatisfied, as in ethene (C₂H₄).

Source: Chapter 4, Section 4.2 / 4.2.1

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• Examiners expect you to clearly state the student is wrong, then explain why using electron sharing logic.
• Key point: needing 2 more electrons means forming 2 bonds, not necessarily 1 double bond. H₂O forms two separate single bonds with two H atoms.
• Mentioning octet/duplet satisfaction shows conceptual clarity — that earns full marks.
• Contrast with a genuine double-bond situation (like ethene) to strengthen your answer — useful for the third mark.

A double bond requires both atoms involved to share two pairs of electrons. Hydrogen has only one electron in its outermost shell (electronic configuration: 1), so it can share only one electron — forming one single bond. It cannot form a double bond. Therefore, carbon cannot form two double bonds with hydrogen atoms in CH₄.

Source: Chapter 4, Section 4.2 & 4.2.1

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• The key point examiners expect: hydrogen's valency is 1 because it has only 1 electron and needs just 1 more to complete its shell (duplet). It can only ever form one covalent bond.
• A double bond needs each atom to contribute 2 electrons to the shared pairs — hydrogen simply doesn't have enough electrons to do this.
• Always link the electronic configuration (2,1 → only 1 electron available for sharing) to the conclusion. Writing "hydrogen has valency 1" alone without explaining why may lose a mark.

In diamond, each carbon atom forms covalent bonds with four other carbon atoms in a rigid three-dimensional network. This extensive cross-linking leaves no free electrons and creates an extremely strong, compact structure with no weak points, making diamond the hardest known substance.

In graphite, each carbon atom bonds to only three other carbon atoms in flat hexagonal layers. The fourth valence electron forms a weak force between the layers. These layers can slide easily over one another, making graphite smooth and slippery.

Thus, the difference in bonding pattern — 4 bonds (3D) vs. 3 bonds (layered) — explains their contrasting physical properties.

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

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• Examiners look for: (1) diamond — 4 bonds, 3D rigid network → hardness; (2) graphite — 3 bonds, layered planes → slipperiness. Award 1 mark each, with 1 mark for linking bond type to property.
• Do not say "ionic bonds" — these are covalent structures.
• Mentioning the "fourth valence electron / double bond" in graphite as the inter-layer weak force is a bonus point but keep it brief.
• Avoid over-writing; the passage itself is your authority here.

Covalent compounds are generally poor conductors because electron sharing produces no free charged particles (ions). However, graphite is an exception due to its unique structure.

In graphite, each carbon atom is bonded to three other carbon atoms in the same plane in a hexagonal arrangement, with one double bond. This leaves each carbon atom with one electron not used in bonding. These electrons are free to move across the layers, allowing graphite to conduct electricity.

Thus, graphite conducts electricity not because it forms ions, but because of its special layered structure that allows the movement of free electrons.

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

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• The examiner wants you to first acknowledge the general rule (covalent compounds don't conduct), then explain graphite as an exception using its structure.
• Key facts to mention: 3 bonds per carbon atom → one free/delocalized electron per atom → free electrons carry charge.
• The passage states graphite is "a very good conductor of electricity unlike other non-metals" — always quote or reference the textbook reason.
• Do not use the word "delocalized" if you haven't been taught it formally; instead say "free electrons" which is sufficient for CBSE Class 10.

Carbon's small atomic size allows its nucleus to hold shared electron pairs very strongly, making the C–C bond very strong and stable. This property, called catenation, allows carbon to form long, stable chains.

Silicon, though also tetravalent, has a larger atom, so Si–Si bonds are much weaker, making silicon hydride chains (up to 7–8 atoms) very reactive and unstable.

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

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• Examiners expect two key points: (1) small size → strong bond, and (2) contrast with silicon's weak/reactive chains.
• The term catenation must appear.
• Avoid over-explaining; two tight sentences covering both points earn full marks.
• "Strong bond = stable chain" is the core logic the examiner is looking for.

Ethanol is a covalent compound. In covalent compounds, the molecules are held together by weak intermolecular forces, which require little energy to overcome. Hence ethanol has a low melting point (156 K).

Sodium chloride is an ionic compound made up of oppositely charged Na⁺ and Cl⁻ ions held together by strong electrostatic forces of attraction. A large amount of energy is needed to break these forces, resulting in a very high melting point (above 1000 K).

Therefore, the large difference in melting points shows that intermolecular forces in ethanol are much weaker than the ionic forces in sodium chloride.

Source: Chapter 4, Section 4.1 – Bonding in Carbon – The Covalent Bond

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• Examiners want two clear contrasts: weak intermolecular forces (covalent/ethanol) vs. strong electrostatic/ionic forces (NaCl).
• Mention both compounds by name and link the force to the melting point.
• The key phrase from the textbook is: "covalently bonded molecules have strong bonds within the molecule, but inter-molecular forces are weak" — use this logic explicitly.
• Do not just say "covalent is weak, ionic is strong" without explaining why that leads to different melting points.

If carbon loses all four valence electrons to form C⁴⁺, the K shell (first shell) would become its outermost shell, containing 2 electrons.

This arrangement is energetically unfavourable because removing four electrons from carbon requires a very large amount of energy. The resulting C⁴⁺ cation would have six protons holding only two electrons, making it highly unstable. Therefore, carbon prefers sharing its four valence electrons to form stable covalent bonds instead.

Source: Chapter 4, Section 4.1 — Bonding in Carbon: The Covalent Bond

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• Examiners expect you to correctly identify the K shell with 2 electrons (carbon's configuration is 2, 4).
• The key reasoning from the textbook: forming C⁴⁺ needs a large amount of energy — state this explicitly.
• Mention the instability (6 protons holding 2 electrons) to earn full marks.
• Don't write more than this — 2 marks = roughly 2 key points.

No, the student's claim is incorrect.

More bonds do not mean more reactivity — in fact, the opposite is true here.

• H₂ has a single bond — one shared electron pair between the two hydrogen atoms. This bond is relatively weaker and easier to break, making H₂ fairly reactive.
• N₂ has a triple bond — three shared electron pairs between the two nitrogen atoms. This makes the N≡N bond extremely strong and stable, so it is very difficult to break.

Bond strength determines reactivity: a stronger bond requires more energy to break, making the molecule less reactive, not more. This is similar to how unsaturated carbon compounds (with double/triple bonds between carbons) are more reactive than saturated ones, but in N₂, the triple bond is so strong that the molecule is largely inert under normal conditions.

Therefore, N₂ is actually much less reactive than H₂, despite having more bonds.

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Examiners look for: (1) correctly rejecting the claim, (2) defining single vs triple bond in terms of shared electron pairs, (3) linking bond strength to reactivity (more bonds = stronger, harder to break = less reactive for N₂), and (4) contrast with H₂'s weaker bond making it more reactive. The source passages establish that stronger bonds (as in carbon's stable C–C bonds) mean greater stability/less reactivity — apply the same logic to N₂ vs H₂. Avoid writing that "more bonds = more reactive" — that is the misconception being tested.

Two reasons why carbon's catenation produces far more stable compounds than silicon:

1. Strength of C–C bond: The carbon–carbon bond is very strong and stable due to carbon's small atomic size, which allows the nucleus to hold shared electron pairs firmly. Silicon has larger atoms, so Si–Si bonds are much weaker, making silicon hydride chains (up to 7–8 atoms) highly reactive and unstable.

1. Variety of bonding in chains: Carbon can form single, double, or triple bonds between atoms, and can arrange into straight chains, branched chains, or rings. Silicon lacks this ability, severely limiting the variety of compounds it can form.

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

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• The examiner expects two distinct, well-explained points — each worth ~1.5 marks.
• Key terms to use: catenation, C–C bond strength, small atomic size, single/double/triple bonds, chains/rings.
• Don't just say "silicon is reactive" — explain why (larger atom → weaker bonds). That's the scoring detail.
• Avoid padding; two tight points with a reason each is all that's needed for 3 marks.

Ethyne (C₂H₂) is an unsaturated hydrocarbon containing a carbon–carbon triple bond (C≡C). The extra bonds (pi bonds) are weaker and electron-rich, making them available for addition reactions. Bromine from bromine water adds across the triple bond, decolourising it.

Ethane (C₂H₆) is a saturated hydrocarbon with only single bonds. All valencies are fully satisfied, leaving no electron-rich bonds available for addition. Hence, it does not react with bromine water and cannot decolourise it.

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

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• The key distinction examiners want: saturated vs. unsaturated, and linking that to addition reaction capability.
• Mention the triple bond in ethyne specifically — one mark is typically for bond type, one for the reaction type (addition), one for why ethane does not react.
• Avoid over-explaining; the passage confirms unsaturated compounds "are more reactive than the saturated carbon compounds" — use this directly.

(A) n-Butane (C₄H₁₀) and 2-methylpropane (C₄H₁₀) are structural isomers as both have the same molecular formula C₄H₁₀ but different carbon skeletons (straight chain and branched chain).

Structural isomers must have the same molecular formula but different structures. Only option (A) satisfies this — both compounds are C₄H₁₀. Options B, C, D pair compounds with different molecular formulae, so they cannot be isomers. The textbook explicitly uses butane and 2-methylpropane (isobutane) as the example of structural isomers.

(i) Two pieces of evidence that CH₃OH, C₂H₅OH and C₃H₇OH form a homologous series:

1. Same functional group: All three compounds contain the –OH (hydroxyl) functional group, giving them similar chemical properties.

1. Successive members differ by –CH₂– unit: CH₃OH → C₂H₅OH → C₃H₇OH; each successive compound differs by one –CH₂– unit (molecular mass difference = 14 u), giving the general formula CₙH₂ₙ₊₁OH.

(ii) Physical property that differs — Boiling point:

Propanol (C₃H₇OH) has a higher boiling point than methanol (CH₃OH). As molecular mass increases in a homologous series, intermolecular forces increase, so more heat energy is needed to separate the molecules. Since propanol has a greater molecular mass (60 u) than methanol (32 u), its boiling point is higher.

Source: Chapter 4, Section 4.2.4 — Homologous Series

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• Part (i) is worth ~2 marks: examiners expect same functional group AND –CH₂– difference — give both clearly.
• Part (ii) is worth ~3 marks: name the property, state the comparative difference, and give the reason (increasing molecular mass → stronger intermolecular forces → higher boiling point). All three steps are needed for full credit.
• The textbook explicitly states: "As molecular mass increases in any homologous series, a gradation in physical properties is seen… melting and boiling points increase with increasing molecular mass." Quote or paraphrase this reasoning.
• Do not just say "boiling point differs" — always state which is higher and why.

Cyclohexane (C₆H₁₂) — Saturated hydrocarbon:
Its molecular formula matches the general formula CₙH₂ₙ, with all single bonds between carbon atoms. All valencies are satisfied by single bonds, so no scope for further addition of atoms.

Cyclohexene (C₆H₁₀) — Unsaturated hydrocarbon:
Its formula (CₙH₂ₙ₋₂) indicates fewer hydrogen atoms, which means it contains a carbon–carbon double bond.

Bromine water test:
Cyclohexane does not decolourize bromine water (saturated; unreactive). Cyclohexene decolourizes bromine water immediately, as the double bond allows addition of bromine across it. This confirms cyclohexene is unsaturated.

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

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• Examiners expect: correct class identified for each compound + molecular formula justification + bromine water result linked to reactivity.
• Key rule: saturated hydrocarbons do not decolourize bromine water; unsaturated ones do (addition reaction at the double/triple bond).
• Cyclohexane's formula C₆H₁₂ = CₙH₂ₙ (same as alkenes, but ring structure with all single bonds — still saturated). Cyclohexene has one double bond, making it unsaturated.
• Don't just state the result — link the bromine water observation to the bond type for full marks.

Ethyne (C₂H₂) contains a triple bond between its two carbon atoms (C≡C). This means its carbon atoms are not bonded to the maximum possible number of hydrogen atoms, making it an unsaturated compound. Ethane (C₂H₆), on the other hand, has only single bonds with all valencies satisfied by hydrogen atoms, so it is saturated.

The triple bond in ethyne makes it more reactive because it can undergo addition reactions, where other atoms or groups add across the multiple bond. Saturated ethane, having no such multiple bonds, is fairly unreactive and undergoes only slow substitution reactions.

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

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• Key structural point (1 mark): Ethyne has a C≡C triple bond; ethane has only C–C single bonds.
• Definition link (1 mark): Unsaturated = double or triple bond present; saturated = only single bonds.
• Reactivity reason (1 mark): Multiple bonds allow addition reactions; single bonds allow only slow substitution — examiners expect this contrast explicitly stated.
• Don't just say "more bonds = more reactive"; name the type of reaction (addition vs. substitution).

(i) The general formula for alkanes is $C_nH_{2n+2}$ and for alkenes is $C_nH_{2n}$.

• $C_3H_8$: For $n=3$, $2(3)+2 = 8$ → matches alkane formula. It is saturated (propane).
• $C_3H_6$: For $n=3$, $2(3) = 6$ → matches alkene formula. It is unsaturated (propene).

(ii) Test: Pass each compound through bromine water (or bromine solution).

• $C_3H_6$ (unsaturated): Bromine water gets decolourised due to addition reaction across the double bond.
• $C_3H_8$ (saturated): Bromine water shows no colour change as saturated hydrocarbons do not undergo addition reactions.

Source: Carbon and its Compounds, Section 4.2.2 and Exercise Q.13–14

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• Examiners expect you to explicitly apply the general formulas $C_nH_{2n+2}$ (alkanes) and $C_nH_{2n}$ (alkenes) and substitute $n=3$ — don't just state the answer without showing the check.
• For the chemical test, bromine water (decolourisation) is the standard NCERT-expected answer. The textbook Exercise Q.14 directly asks for a test to differentiate saturated and unsaturated hydrocarbons — this is it.
• Avoid writing about the combustion/flame test here; it is less reliable as a distinguishing test compared to bromine water.

(i) Structural formulae:

n-Butane (straight chain):
$$\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_3$$

Isobutane (branched chain):
$$\text{CH}_3 - \overset{|}{\underset{|}{\text{CH}}} (\text{CH}_3) - \text{CH}_3$$
i.e., a central carbon bonded to three –CH₃ groups and one H.

(ii) Such compounds with identical molecular formula but different structures are called structural isomers.

(iii) Two differing properties:

• Boiling point: n-Butane has a higher boiling point than isobutane because its straight chain allows closer packing between molecules.
• Reactivity: The branched structure of isobutane makes it slightly more reactive than n-butane in certain reactions due to steric and structural differences.

Source: Carbon and its Compounds, Section 4.2.2 — Chains, Branches and Rings

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• Part (i): Draw both clearly — a 4-carbon straight chain and a branched structure with a central carbon carrying three methyl groups.
• Part (ii): The exact NCERT term is "structural isomers" — use it precisely.
• Part (iii): The passage does not explicitly list physical properties of these two isomers, so use general knowledge of isomers: branched chains have lower boiling points due to less surface contact. Examiners accept any two valid, relevant points. Avoid vague statements — be specific.

Hexane: C₆H₁₄; Cyclohexane: C₆H₁₂. Hexane is an open chain (straight chain), so it has 2 extra H atoms compared to cyclohexane, which has carbon atoms arranged in a ring, reducing the hydrogen count by 2.

Source: Carbon and its Compounds, Section 4.2.2

The key point is the structural difference: open chain vs. ring. When a chain closes into a ring, two terminal hydrogen atoms are lost to form the extra C–C bond, hence the formula drops from C₆H₁₄ to C₆H₁₂. The examiner wants both formulae stated and a one-line structural reason.

(i) The –Cl group is called a functional group (specifically, the halo/chloro functional group). It is also referred to as a heteroatom when considered as the replacing atom.

(ii) The chemist's statement is correct because chemical properties of a carbon compound are determined solely by its functional group, regardless of the length of the carbon chain. Whether the chloro (–Cl) group is attached to a 2-carbon or 10-carbon chain, it confers the same characteristic chemical properties to the compound. This is the basis of a homologous series — members differ in chain length but show similar chemical behaviour.

Source: Chapter 4, Sections 4.2.3 and 4.2.4

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• For part (i), examiners accept "functional group" or "heteroatom"; both terms appear in the passage. Writing both is safest.
• For part (ii), the key phrase to use is: "chemical properties are determined solely by the functional group, regardless of the length of the carbon chain" — this is almost directly from the textbook and will earn full marks.
• Mentioning "homologous series" shows higher-order understanding and impresses examiners.
• Do not write about physical properties changing — that would contradict the chemist's statement and cost marks.

Both CH₃CH₂OH (ethanol) and CH₃CH₂CH₂CH₂OH (butan-1-ol) belong to the homologous series of alcohols — they share the same functional group (–OH). Since chemical properties are determined solely by the functional group, both compounds show very similar chemical reactions (e.g., reaction with sodium, oxidation).

However, their boiling points differ because, in a homologous series, boiling point increases with increasing molecular mass. Butan-1-ol has a higher molecular mass (74 u) than ethanol (46 u), so it has a higher boiling point.

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

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• Part 1 (similar chemical properties): Examiners want the key phrase: "functional group determines chemical properties, regardless of chain length." Name the functional group (–OH) and state it's a homologous series.
• Part 2 (different boiling points): The textbook explicitly states "melting and boiling points increase with increasing molecular mass" in a homologous series. Mentioning the actual molecular masses (46 u vs 74 u) adds precision and earns full marks.
• Don't write more than this — padding loses time without gaining marks.

Two successive members of a homologous series differ in molecular mass by 14 u.

This is because they differ by one –CH₂– unit, and the mass of C = 12 u and 2 × H = 2 u, giving 12 + 2 = 14 u.

This value is constant throughout the series because each successive member differs by exactly one –CH₂– unit, no more and no less.

Source: Chapter 4, Section 4.2.4 – Homologous Series

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• The examiner expects you to state the value (14 u) and justify it using atomic masses (C = 12, H = 1 each, so CH₂ = 14).
• Always show the arithmetic: 12 + 2(1) = 14 u — this is where the mark lies.
• The second mark is for explaining why it is constant: every successive pair differs by exactly one –CH₂– unit, making the difference uniform across the series.
• Do not just say "14 u" without justification — that will cost you the reasoning mark.

A homologous series is a group of compounds with the same functional group, differing by a –CH₂– unit each time.

Different boiling points: As molecular mass increases in a homologous series, physical properties (such as boiling point) show a gradation. Pentanol (C₅H₁₁OH, mol. mass = 88 u) has a much higher molecular mass than methanol (CH₃OH, mol. mass = 32 u), so its boiling point is significantly higher.

Same chemical behaviour with sodium: Chemical properties in a homologous series are determined solely by the functional group. Both pentanol and methanol contain the –OH (hydroxyl) functional group, so they react identically with sodium metal.

Source: Chapter 4, Section 4.2.4 – Homologous Series

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• Examiners expect two separate, clearly labelled observations explained using the two key properties of a homologous series.
• Physical properties (boiling point, melting point, solubility) show gradation with increasing molecular mass — this is the principle for the first part.
• Chemical properties depend only on the functional group, not on chain length — this explains the second part.
• Mention actual molecular masses or the –CH₂– difference to show you understand the series; this earns the third mark.

The student's argument is partially correct but partially wrong.

What is correct: Both butanol and ethanol belong to the same homologous series (alcohols) and share the same functional group (–OH). Therefore, their chemical properties are similar — both react with sodium to release hydrogen, and undergo similar reactions.

What is incorrect: Physical properties such as boiling point and solubility change across a homologous series. As molecular mass increases (butanol > ethanol), boiling point increases. Solubility in water also shows a gradation — ethanol is completely miscible with water, while butanol has lower solubility. These physical properties differ because each successive member differs by a –CH₂– unit, increasing molecular mass.

Source: Chapter 4, Section 4.2.4 — Homologous Series

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Examiners look for three things here:

1. Identifying what is correct — same functional group → similar chemical properties.
2. Identifying what is wrong — boiling point and solubility are physical properties that change with molecular mass (gradation).
3. Reason — each member differs by –CH₂–, so molecular mass increases → boiling point increases, solubility decreases.

The key textbook line is: "As the molecular mass increases in any homologous series, a gradation in physical properties is seen… But the chemical properties… remain similar." Quote or paraphrase this for full marks.

Reasoning Steps:

1. The compound has 3 carbon atoms → base name is propane.
2. The functional group is a ketone → suffix is "-one" (Table 4.4).
3. Since "-one" begins with a vowel, drop the final 'e' from propane:

Propane − 'e' + one = Propanone

IUPAC Name: Propanone

Source: Chapter 4, Section 4.2.5 — Nomenclature of Carbon Compounds

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Examiners award 1 mark for correct reasoning (identifying base chain + applying suffix rule) and 1 mark for the correct IUPAC name. The key rule to remember: when a suffix starts with a vowel (like "-ol", "-al", "-one"), drop the final 'e' of the parent chain name before adding it. Writing the steps clearly secures full marks even if the final name has a minor slip.

(i) Structural formulae:

Butane (C₄H₁₀): CH₃—CH₂—CH₂—CH₃
Bond type: Only single bonds (C–C and C–H). It is a saturated compound.

But-1-yne (C₄H₆): CH≡C—CH₂—CH₃
Bond type: Contains one triple bond (C≡C) and single bonds. It is unsaturated.

(ii) Chemical reactivity:

• Butane undergoes substitution reactions (e.g., with chlorine in sunlight).
• But-1-yne undergoes addition reactions (e.g., with H₂, Cl₂, HCl) due to the triple bond. Unsaturated compounds are more reactive than saturated ones.

(iii) Homologous series:

• Butane belongs to alkanes; but-1-yne belongs to alkynes.
• Characteristic property: Successive members of a homologous series differ by a –CH₂– unit (molecular mass difference of 14 u), and members show a gradation in physical properties but similar chemical properties.

Source: Chapter 4, Section 4.2.4 – Homologous Series; Section 4.2.2 – Chains, Branches and Rings

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• Examiners expect the structural formula drawn as a condensed/chain structure — full electron dot structure is not required here unless asked.
• The key contrast for reactivity is substitution (saturated) vs. addition (unsaturated) — these exact terms score marks.
• For homologous series property, state both the –CH₂– difference AND similar chemical properties; either alone may lose half a mark.
• "Alkanes" and "alkynes" are the required series names — do not write just "hydrocarbons."

(i) Ethyne is an unsaturated hydrocarbon with a high carbon-to-hydrogen ratio. When burnt in air, the oxygen supply is insufficient for complete combustion, so it burns with a yellow, sooty flame and produces less heat. Pure oxygen ensures complete combustion, giving a much hotter, cleaner flame.

(ii) Ethyne (C₂H₂) has a higher proportion of carbon relative to hydrogen compared to ethane (C₂H₆), so it requires more oxygen per carbon atom for complete combustion.

Balanced equations:

Ethane: $2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$

Ethyne: $2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}$

Ethyne needs 2.5 mol O₂ per carbon, while ethane needs 1.75 mol O₂ per carbon, confirming unsaturated hydrocarbons demand more oxygen.

Source: Chapter 4, Section 4.3.1 Combustion; Section 4.2.1 Saturated and Unsaturated Carbon Compounds

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• Part (i) is directly from the textbook: unsaturated hydrocarbons burn with a sooty, yellow flame due to insufficient oxygen; this is exactly why pure oxygen (not air) is used in oxy-acetylene welding.
• Part (ii): examiners expect both balanced equations AND a comparison of O₂ required per carbon atom. Show the ratio explicitly — this earns the justification mark.
• Common mistake: forgetting to balance equations properly or not relating the result back to the question's claim.

Observation: A sooty (black) deposit will appear on the metal plate held above the unsaturated hydrocarbon's flame. The plate held above the saturated hydrocarbon's flame will remain comparatively clean (little or no soot).

Reason: Unsaturated hydrocarbons contain multiple bonds (C=C or C≡C) and have a higher carbon-to-hydrogen ratio. They undergo incomplete combustion, producing a yellow, sooty flame with unburnt carbon particles that deposit as soot on the plate. Saturated hydrocarbons combust more cleanly, giving a blue flame with no sooty deposit.

Source: Chapter 4, Section 4.3.1 – Combustion

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• Examiners award marks for: (1) correct observation for each hydrocarbon, (2) linking unsaturated compounds to sooty/yellow flame and incomplete combustion, and (3) explaining why (higher C:H ratio / multiple bonds → incomplete combustion → carbon deposit).
• Use the textbook's exact language: "saturated hydrocarbons give a clean flame; unsaturated carbon compounds give a yellow flame with lots of black smoke / sooty deposit."
• Do not write a lengthy essay — two short paragraphs covering observation + reason is ideal for 3 marks.

Cause: The air holes of the gas stove are blocked, causing insufficient oxygen supply. This leads to incomplete combustion of the fuel (LPG).

Blue → Yellow flame: When the air supply is adequate, complete combustion occurs giving a clean blue flame. With blocked air holes, incomplete combustion produces a yellow, sooty flame.

Black deposit: The black substance deposited on the bottom of the vessel is carbon (soot). During incomplete combustion, the carbon in the fuel is not fully oxidised to CO₂, and unburnt carbon particles are deposited as soot.

Source: Chapter 4, Section 4.3.1 Combustion

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• Examiners expect three clear points: (1) cause — blocked air holes/insufficient air, (2) incomplete combustion producing yellow flame, (3) black deposit = unburnt carbon/soot.
• The textbook explicitly states: "If you observe the bottoms of cooking vessels getting blackened, it means that the air holes are blocked and fuel is getting wasted."
• Use the terms incomplete combustion and soot/unburnt carbon — these are the key scoring words.

In combustion of ethanol, oxygen is added to it, converting it into carbon dioxide and water. Addition of oxygen to a substance is defined as oxidation.

The examiner expects you to link the definition of oxidation (addition of oxygen) directly to what happens during combustion. One crisp sentence stating that ethanol gains oxygen to form CO₂ and H₂O is sufficient for 1 mark. Avoid writing the full balanced equation unless asked.

(i) Pure oxygen is preferred over air because air contains about 78% nitrogen, which dilutes the oxygen and lowers the flame temperature. Pure oxygen ensures a more complete and intense combustion of ethyne, producing a much hotter flame — sufficient for welding metals.

(ii) When sufficient oxygen is available, complete combustion occurs, producing CO₂ and H₂O with a clean blue flame. When oxygen is insufficient, incomplete combustion occurs, producing carbon (soot) and CO, with a yellow, sooty flame. The observable difference is a clean blue flame vs. a smoky, yellow, sooty flame.

Source: Chapter 4, Section 4.3.1 Combustion

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• The key contrast examiners expect: complete vs. incomplete combustion tied to oxygen supply.
• Mention flame colour (blue vs. yellow/sooty) — this comes directly from Activity 4.4 and the textbook paragraph on saturated/unsaturated combustion.
• For part (i), the point about nitrogen diluting oxygen and reducing flame temperature is the core reason stated in the textbook question ("why a mixture of ethyne and air is not used?").
• Don't forget to name the products of both types of combustion for full marks.

(i) Structure, Hardness, and Conductivity:

In diamond, each carbon atom is bonded to four others in a rigid tetrahedral 3D network. All four valence electrons are used in bonding, so no free electrons exist → diamond is a non-conductor and extremely hard.

In graphite, each carbon atom bonds to three others in flat hexagonal layers. The fourth electron is free to move between layers → graphite is a good conductor. The layers are held by weak forces and can slide over each other → graphite is soft.

(ii) Combustion behaviour:

Despite structural differences, both diamond and graphite are allotropes of carbon. On combustion, both burn in oxygen to produce only carbon dioxide with release of heat and light:
$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2 + \text{heat and light}$$

Their chemical behaviour during combustion is identical because the product depends on the element (carbon), not its structural form.

(iii) Disagreement with the student:

No, the student is incorrect. Hardness is a physical property related to structure, not a chemical property. Since diamond and graphite are both pure carbon, they produce the same product — CO₂ — on combustion. The textbook states: "Carbon, in all its allotropic forms, burns in oxygen to give carbon dioxide."

Source: Chapter 4, Section 4.3.1 (Combustion)

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• Examiners expect you to clearly contrast diamond (3D network, non-conductor, hard) and graphite (layered, conductor, soft) using the concept of free electrons and bonding.
• The key textbook line for parts (ii) and (iii) is: "Carbon, in all its allotropic forms, burns in oxygen to give CO₂." Quote or paraphrase it — it directly answers the student's argument.
• Part (iii) tests whether you can distinguish physical properties (hardness) from chemical properties (combustion products). Always clarify this distinction.
• Keep the equation balanced and include "heat and light" — examiners often check for this.

(i) When ethanol is treated with alkaline potassium permanganate, it is oxidised to form ethanoic acid (CH₃COOH).

(ii) When ethanol is heated with excess concentrated sulphuric acid at 443 K, dehydration occurs and ethene (CH₂=CH₂) is formed along with water.

Different categories:

• Reaction (i) is an oxidation reaction — oxygen is added to ethanol by the oxidising agent (alkaline KMnO₄).
• Reaction (ii) is a dehydration (elimination) reaction — water is removed from ethanol by the dehydrating agent (conc. H₂SO₄), producing an unsaturated hydrocarbon.

Thus, one reaction involves addition of oxygen, while the other involves removal of water — making them fundamentally different types.

Source: Chapter 4, Sections 4.3.2 and 4.4.1

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• Examiners expect both products named correctly — ethanoic acid and ethene — for full credit.
• You must clearly contrast the two reaction types: oxidation vs dehydration/elimination. Just naming the products is not enough.
• Use key terms: oxidising agent, dehydrating agent, oxygen added, water removed.
• The 3-mark split is typically: 1 mark for each product + 1 mark for explaining the two different categories.

(i) Vegetable oils have long unsaturated carbon chains (contain C=C double bonds), whereas vanaspati ghee has saturated carbon chains (all single bonds), similar to animal fats.

(ii) During hydrogenation, hydrogen is added across the double bonds of unsaturated vegetable oil, converting it into a saturated fat (vanaspati ghee). Saturated fatty acids are considered harmful to health, making the hydrogenated product less healthy than the original oil.

(iii) The catalyst used is nickel (Ni). It speeds up the addition of hydrogen to the unsaturated carbon chains without itself being consumed in the reaction.

Source: Chapter 4, Section 4.3.3 Addition Reaction

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• The examiner expects three distinct points matching the three sub-questions — don't merge them.
• Key terms to use: unsaturated, saturated, double bonds, hydrogenation, nickel catalyst.
• Avoid saying oils are "healthy" in general — the point is that saturated fats (product of hydrogenation) are harmful, not unsaturated ones.
• The role of catalyst is always: speeds up the reaction / causes it to proceed without being used up — state both parts for full credit.

Although saturated hydrocarbons are generally inert to most reagents, they can still undergo substitution reactions under specific conditions. In this reaction, a chlorine atom replaces one hydrogen atom of methane to form chloromethane (CH₃Cl) and HCl.

Sunlight is essential because it provides the energy needed to activate the chlorine molecules (Cl₂), enabling the reaction to proceed. Without sunlight, the activation energy is not supplied, so the reaction does not occur. This is why the same mixture of methane and chlorine shows no reaction in the dark.

$$\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{sunlight}} \text{CH}_3\text{Cl} + \text{HCl}$$

Source: Chapter 4, Section 4.3.4 – Substitution Reaction

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• Examiners expect two clear points: (1) saturated hydrocarbons are inert under normal conditions but not absolutely unreactive, and (2) sunlight provides the activation energy needed to initiate the reaction.
• Name the reaction type (substitution reaction) — this is a key term the examiner will look for.
• Writing the balanced equation with the condition (sunlight) shown earns a mark on its own.
• Don't write about free radicals or chain mechanisms — that is beyond CBSE Class 10 scope and wastes word budget.

What determines the product:
The reaction conditions determine the product. At 443 K with excess concentrated H₂SO₄, dehydration of ethanol occurs giving ethene. When ethanol is warmed with glacial acetic acid and concentrated H₂SO₄, esterification occurs giving ethyl ethanoate (ester).

Role of concentrated H₂SO₄:

• In dehydration: It acts as a dehydrating agent, removing water from ethanol to form ethene.

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

• In esterification: It acts as an acid catalyst, speeding up the reaction between ethanoic acid and ethanol to form the ester.

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Acid}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

Source: Chapter 4, Sections 4.4.1 and 4.4.2

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• Examiners want two clear points: (1) what determines the product — temperature/conditions/reactant present, and (2) the dual role of conc. H₂SO₄ (dehydrating agent vs. acid catalyst).
• Writing both equations earns full credit and shows understanding.
• Do not confuse "dehydrating agent" (removes water from a single molecule) with "acid catalyst" (speeds up a reaction between two molecules). This distinction is what the question tests.

When ethanol reacts with sodium metal, hydrogen gas is evolved and sodium ethoxide is formed:

$$2\text{Na} + 2\text{C}_2\text{H}_5\text{OH} \rightarrow 2\text{C}_2\text{H}_5\text{O}^-\text{Na}^+ + \text{H}_2\uparrow$$

This shows that only the O–H bond in ethanol is broken during this reaction, releasing hydrogen. The C–H bonds remain intact. This indicates that the O–H bond in ethanol is weaker and more reactive than the C–H bonds. The oxygen atom makes the O–H bond polar, allowing sodium to displace hydrogen easily, whereas C–H bonds are non-polar and relatively strong, making them unreactive with sodium.

Source: Chapter 4, Section 4.4.1 – Properties of Ethanol

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• Examiners expect the chemical equation for full marks — always include it.
• The key comparison is O–H (polar/reactive) vs C–H (non-polar/unreactive) — this is the core of the 3-mark answer.
• Mention that C–H bonds are not broken in this reaction; only the O–H bond reacts.
• The word "polar" is a bonus if you remember it, but saying "O–H is weaker/more reactive" is sufficient for CBSE.

(i) Concentrated sulphuric acid acts as a dehydrating agent — it removes water from ethanol.

(ii) The product formed is ethene (CH₂=CH₂). It is an unsaturated hydrocarbon (contains a C=C double bond).

Chemical equation:

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

Source: Carbon and its Compounds, Section 4.4.1 (Properties of Ethanol)

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• The key term examiners look for is "dehydrating agent" — not catalyst or oxidising agent.
• Ethene must be correctly identified as unsaturated because of its C=C double bond; saturated compounds have only single bonds.
• Always write the balanced chemical equation with the condition (hot conc. H₂SO₄, 443 K) shown above the arrow.
• This reaction is called dehydration of ethanol.

Both methanol and ethanol belong to the same homologous series, but they differ in toxicity due to their metabolic products.

• Methanol is oxidised in the liver to methanal (formaldehyde), which reacts rapidly with cell components. It coagulates the protoplasm (similar to how heat coagulates egg white) and also damages the optic nerve, causing blindness and even death in very small quantities.
• Ethanol, when consumed in excess, slows metabolic processes and depresses the central nervous system, but it does not produce such immediately lethal metabolites in dilute amounts.

Thus, the extreme toxicity of methanol is due to its oxidation product — methanal — which is highly destructive to cells.

Source: Chapter 4, Section 4.4.1 (Do You Know? — How do alcohols affect living beings?)

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Examiners look for three key points here (1 mark each):

1. Methanol is oxidised to methanal in the liver.
2. Methanal coagulates protoplasm / reacts with cell components.
3. It also damages the optic nerve, causing blindness/death.

Do not write a general comparison of physical properties — the question is specifically about toxicity mechanism. Always anchor your answer to the metabolic/chemical difference, not just "methanol is poisonous."

Option B is correct. Methanol and dyes are added to industrial ethanol to make it poisonous and identifiable; this product is called denatured alcohol. Methanol causes blindness and can be lethal even in small quantities.

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

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Although this question has three sub-parts, it is a 1-mark MCQ, so you only need to identify the correct option and briefly justify it. Option B matches the passage directly: the process is denaturation, the product is denatured alcohol, it is done to prevent misuse, and methanol (the added substance) affects the optic nerve causing blindness. The other options describe unrelated reactions (dehydration, oxidation, dilution) and are not supported by the passage.

Ethanol (C₂H₅OH) has two parts in its structure: a polar –OH (hydroxyl) group and a non-polar hydrocarbon (C₂H₅–) part.

(i) Miscibility with water: The –OH group is similar to water (H–OH) and can form hydrogen bonds with water molecules. This "like dissolves like" interaction allows ethanol to mix with water in all proportions.

(ii) Dissolving organic substances: The non-polar hydrocarbon part of ethanol can interact with non-polar organic substances (which water cannot dissolve, as water is polar). Thus ethanol acts as a bridge solvent, dissolving both polar and non-polar substances.

Source: Chapter 4, Section 4.4.1 — Properties of Ethanol

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The examiner expects students to link structure → property. The key concept is the dual nature of ethanol's molecule: the –OH end makes it water-compatible (polar), and the –C₂H₅ end makes it organic-solvent-compatible (non-polar). The textbook directly states ethanol is soluble in water in all proportions and is a good solvent used in medicines. For full marks, both parts (i) and (ii) must be addressed separately. The phrase "like dissolves like" is a useful principle to mention, even if not explicitly in the textbook passage.

(i) The reaction is esterification. The product formed is ethyl ethanoate (an ester) and water.
$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

(ii) Here, conc. H₂SO₄ acts as an acid catalyst. When ethanol alone is heated with excess conc. H₂SO₄ at 443 K, H₂SO₄ acts as a dehydrating agent, removing water to form ethene.

(iii) The ester (ethyl ethanoate) is identified by its sweet smell. When treated with NaOH solution, it undergoes saponification, breaking back into ethanol and sodium ethanoate (sodium salt of ethanoic acid):
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa}$$

Source: Chapter 4, Section 4.4.1 & 4.4.2

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• Examiners expect the equation for esterification and saponification — even in short answers, these earn marks.
• The key contrast for part (ii) is: catalyst vs. dehydrating agent — this distinction is frequently tested.
• For identification, "sweet smell" is the textbook answer; always link it to the ester's property.
• "Saponification" as the name of the NaOH reaction is a scoring keyword — include it.

Ethanoic acid is a weak acid because it is partially (incompletely) ionised in water — only a few H⁺ ions are released. HCl, being a mineral (strong) acid, is completely ionised, releasing far more H⁺ ions at the same concentration. Since the universal indicator responds to H⁺ ion concentration (pH), it shows ethanoic acid at a higher pH than HCl, indicating weaker acidity. Litmus only detects whether a substance is acidic or basic, so both turn it red.

Source: Chapter 4, Section 4.4.2 — Properties of Ethanoic Acid

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• The key concept is degree of ionisation: strong acids ionise completely; weak acids ionise partially.
• Litmus is a qualitative test (acid/base only); universal indicator is quantitative — it reflects actual H⁺ concentration, so it distinguishes strong from weak acids.
• Examiners award 1 mark for "partial/incomplete ionisation of ethanoic acid" and 1 mark for the comparison with complete ionisation of HCl. Don't just say "weak acid" — explain why (partial ionisation = fewer H⁺ ions).

(i)

• P = Ethyl ethanoate (an ester); formed by esterification reaction.
• Q = Sodium ethanoate (sodium acetate); formed by saponification reaction.

(ii)

Formation of P (Esterification):
$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

Formation of Q (Saponification):
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH}$$

(iii) Saponification is used industrially in soap making — fats/oils are hydrolysed with sodium hydroxide to produce soap (sodium salts of fatty acids) and glycerol.

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• Examiners expect you to correctly name both compounds and both reaction types — these fetch separate marks.
• The balanced equations must show the catalyst (conc. H₂SO₄) for esterification; forgetting it costs marks.
• For the industrial application, "soap making / saponification of fats and oils" is the standard accepted answer. Do not confuse this with the saponification of the ester in part (ii) — the principle is the same but applied to triglycerides in industry.

(i)

• With NaOH: Salt formed is Sodium ethanoate (sodium acetate), CH₃COONa

$$\text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$

• With Na₂CO₃: Salt formed is Sodium ethanoate (sodium acetate), CH₃COONa

$$2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2$$

(ii)

The student's claim is incorrect. The fact that ethanoic acid reacts with Na₂CO₃ only shows it is acidic enough to do so, not that it is equally strong as HCl. Unlike HCl, which is completely ionised, ethanoic acid is a weak acid and only partially ionises.

Experimental observation: On testing equal concentrations of both acids with universal indicator, HCl shows a much lower pH (strong acid colour) than ethanoic acid, proving ethanoic acid is weaker.

Source: Chapter 4, Section 4.4.2

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• Part (i) is worth ~2 marks: name the salt, give the formula, write both balanced equations — examiners award marks for each correct equation.
• Part (ii) is ~1 mark: clearly state the claim is wrong, give the reason (partial vs complete ionisation), and name a specific, observable experiment (universal indicator / pH comparison). Litmus alone won't distinguish them — universal indicator/pH is the key point from Activity 4.7.

Soap molecules are sodium or potassium salts of long-chain carboxylic acids. When added to hard water (containing dissolved Ca²⁺ and Mg²⁺ salts), the soap reacts with these ions to form insoluble calcium and magnesium salts of the fatty acid. This insoluble substance appears as a white curdy precipitate (scum), and lather is not formed until all the Ca²⁺/Mg²⁺ ions are used up.

Detergents are sodium salts of sulphonic acids or ammonium salts with chloride/bromide ions. The charged ends of detergent molecules do not form insoluble precipitates with calcium and magnesium ions in hard water. Therefore, detergents produce lather readily in hard water and remain effective as cleansing agents.

Source: Chapter 4, Section 4.5 – Soaps and Detergents

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• The key contrast examiners look for: soap → insoluble precipitate with Ca²⁺/Mg²⁺ vs. detergent → no such precipitate.
• Mention the chemical nature of both (soap = salt of carboxylic acid; detergent = salt of sulphonic acid) for full marks.
• Avoid vague phrases like "detergent is better" without the chemical reason — that's where students lose marks.

(i) Tyndall effect — scattering of light by colloidal particles.
(ii) Soap solution is a colloid (not a true solution); soap micelles are large enough to scatter light.

Source: Chapter 4, Section 4.5 Soaps and Detergents

The key phrase in the passage is: "The soap micelles are large enough to scatter light. Hence a soap solution appears cloudy." Examiners expect both parts answered: name the phenomenon (Tyndall effect) and state what it reveals (colloidal nature). Since this is 1 mark, keep it to one or two short lines — don't over-explain.

When soap is dissolved in a hydrocarbon solvent, the micelle orientation is reversed compared to water. The ionic (hydrophilic) ends of the soap molecules point inward (toward the centre of the micelle), while the hydrophobic hydrocarbon tails point outward (toward the hydrocarbon solvent).

Such a micelle, with its ionic interior, could be used to clean substances that are water-soluble or ionic in nature (e.g., water-based stains or salts), since the ionic core can trap and dissolve them, even in a non-polar/hydrocarbon medium.

Source: Carbon and its Compounds, Section 4.5 – Soaps and Detergents

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• The key concept is the reversal of micelle structure in a non-polar solvent: tails go out, ionic heads go in — the opposite of the aqueous micelle.
• Examiners expect students to clearly state which end faces inward/outward and why (hydrophobic tails are compatible with the hydrocarbon solvent).
• For the cleaning part, students must logically deduce that since the ionic/hydrophilic core is now exposed to trapped material, such a micelle would clean water-soluble or ionic dirt — this is the direct application of the reversed structure.
• This question is based on the textbook Activity 4.10 discussion and the "More to Know" micelle box. Do not confuse the orientation with the standard aqueous micelle.

The outer surface of a micelle is covered with ionic (hydrophilic) ends of soap molecules, which cause ion-ion repulsion between micelles, preventing them from coming together and precipitating.

Source: Chapter 4, Section 4.5 – Soaps and Detergents

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The key phrase from the passage is: "The micelles stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion." Examiners expect you to identify the ionic/hydrophilic outer surface and link it directly to ion-ion repulsion as the cause of colloidal stability. Mentioning only "hydrophilic" without the repulsion mechanism may lose the mark.

(i) Distilled water is soft water. When soap solution is added and shaken, it produces abundant lather, because soap molecules form micelles freely in soft water.

Tubewell water is hard water — it contains dissolved calcium and magnesium salts. When soap is added to it, the soap reacts with these salts to form an insoluble white curdy precipitate (scum) with very little lather:

> Ca²⁺/Mg²⁺ ions + soap → insoluble calcium/magnesium salt (scum)

This scum is the white precipitate observed. Soap is wasted in forming scum, so little lather is produced.

(ii) When detergent solution is used with tubewell (hard) water, it produces good lather with no white precipitate. This is because detergents are sodium salts of sulphonic acids whose charged ends do not form insoluble precipitates with calcium and magnesium ions. Hence detergents remain effective in hard water, unlike soaps.

Source: Chapter 4, Section 4.5 Soaps and Detergents (Activities 4.11 and 4.12)

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• The key distinction examiners want: distilled water = soft, tubewell = hard; soap forms scum with hard water, detergent does not.
• Name the precipitate as insoluble calcium/magnesium salt (scum) — just saying "white precipitate" alone may lose a mark.
• For part (ii), explicitly state two things: lather is formed AND no precipitate forms, then give the reason (charged ends of detergent don't react with Ca²⁺/Mg²⁺).
• Avoid over-explaining micelle mechanism here — it's not asked; focus on hard/soft water behaviour.

(i) Hydrophilic end (ionic/carboxylate end) and Hydrophobic end (long hydrocarbon chain tail).
(ii) The hydrophilic (ionic) end interacts with water; the hydrophobic (hydrocarbon) end interacts with oily substances.

Source: Chapter 4, Section 4.5 Soaps and Detergents

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Though this is a 1-mark question, it has two parts (i) and (ii), so keep each part to one short phrase. Examiners look for the two key terms — hydrophilic and hydrophobic — and the correct pairing: ionic end ↔ water, hydrocarbon tail ↔ oil. Don't write full paragraphs; two compact lines are enough.

Soap molecules form micelles around oily dirt — the hydrophobic tails surround the oil droplet while the hydrophilic ionic ends face water. However, micelles form only when there is sufficient contact between soap and dirt. Mechanical agitation (beating/scrubbing) breaks oil into smaller droplets, increases surface area, and helps soap molecules quickly surround each droplet, forming stable micelles that remain suspended as a colloid and are easily rinsed away.

Source: Chapter 4, Section 4.5 – Soaps and Detergents

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• Examiners expect you to mention: (1) micelle structure (hydrophobic tail inward, ionic end outward), and (2) how agitation aids micelle formation by increasing contact/surface area.
• The textbook Activity 4.10 and Question 2 at the end of the section directly address this — agitation breaks oil into droplets so soap can act effectively.
• Avoid writing vague statements like "soap cleans better" — link it explicitly to micelle formation for full marks.

Carbon forms far more stable compounds than silicon due to two main reasons:

(i) Catenation: Carbon has a unique ability to form strong bonds with other carbon atoms, producing long chains, branched chains, and rings. The carbon–carbon bond is very strong and stable, allowing millions of compounds. Silicon also shows catenation, but silicon–silicon bonds are much weaker, so silicon chains are limited to only seven or eight atoms and are highly reactive.

(ii) Small atomic size: Carbon's nucleus is small, so it holds the shared pairs of electrons very strongly, forming exceptionally stable covalent bonds. Silicon has a larger atom, so its nucleus cannot hold shared electrons as firmly, making its bonds weaker and its compounds less stable.

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

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• Examiners expect two distinct, labelled points — catenation and atomic size — directly from the textbook.
• For catenation, you must contrast carbon (stable, unlimited chains) with silicon (reactive, max ~7–8 atoms).
• For atomic size, the key phrase is: small nucleus → stronger hold on shared electrons → stronger, more stable bonds.
• Avoid vague statements like "carbon is special"; use precise textbook language.

(i) Reagents and Conditions:

• Ethanol → Ethene (dehydration): Reagent — excess concentrated H₂SO₄; Condition — heated at 443 K.
• Ethanol → Ethanoic acid (oxidation): Reagent — alkaline KMnO₄ or acidified K₂Cr₂O₇ (oxidising agent); Condition — heated.

(ii) Meaning of each term:

• Dehydration means removal of water from a molecule. Here, conc. H₂SO₄ removes H and OH from ethanol to form ethene (CH₂=CH₂) and H₂O.
• Oxidation means addition of oxygen (or removal of hydrogen). Here, an oxidising agent adds oxygen to ethanol (–OH group) to form ethanoic acid (–COOH), increasing the oxygen content.

(iii) Different homologous series:

Ethanol belongs to the alcohol series (functional group: –OH), while ethanoic acid belongs to the carboxylic acid series (functional group: –COOH). A homologous series is defined by its functional group, and since the two compounds have different functional groups with different chemical properties, they belong to different homologous series.

Source: Chapter 4, Section 4.4.1 (Properties of Ethanol); Section 4.2.4 (Homologous Series)

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• For (i), the textbook explicitly states conc. H₂SO₄ at 443 K for dehydration. For oxidation, the textbook question asks why it is oxidation — oxidising agents like KMnO₄/K₂Cr₂O₇ are the standard expected reagents; mention either.
• For (ii), examiners want the general definition applied to these reactions specifically — don't just define in the abstract.
• For (iii), the key concept is from Section 4.2.4: a homologous series is identified by its functional group. Two compounds with different functional groups (–OH vs –COOH) cannot belong to the same series, regardless of carbon chain length. State both functional groups clearly for full marks.

(i)

• Gas 1 (clean blue flame): Saturated hydrocarbon. Saturated hydrocarbons burn with a clean blue flame due to complete combustion.
• Gas 2 (yellow, sooty flame): Unsaturated hydrocarbon. Unsaturated hydrocarbons burn with a yellow, sooty flame due to their high carbon content and incomplete combustion.

(ii)
Unsaturated hydrocarbons contain double or triple bonds (C=C or C≡C). These bonds break easily and react with bromine water in an addition reaction, decolourising it. Saturated hydrocarbons have only single bonds and are fairly unreactive; they cannot undergo addition reactions, so bromine water remains orange-brown.

Source: Chapter 4, Sections 4.2.2 and 4.3.1

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• For (i), examiners want the correct label (saturated/unsaturated) and the reason linked to the type of flame — both parts needed for full credit.
• For (ii), the key terms examiners look for are double/triple bonds, addition reaction, and the contrast with single bonds in saturated hydrocarbons. Avoid just saying "they react" — name the reaction type.
• The bromine water test is the standard differentiation test between saturated and unsaturated hydrocarbons (Q14 in exercises).

Low melting points of covalent compounds: In covalent compounds, electrons are shared between atoms, forming strong bonds within molecules. However, the inter-molecular forces between molecules are weak. Therefore, little energy is needed to separate the molecules, resulting in low melting and boiling points.

Why graphite is an exception: In graphite, each carbon atom is bonded to three other carbon atoms in the same plane, forming hexagonal layers. One bond is a double bond. These layers are stacked one above the other. The remaining electrons are free to move between the layers, allowing graphite to conduct electricity well, unlike typical covalent compounds.

Source: Chapter 4 – Bonding in Carbon (Covalent Bond); Allotropes of Carbon (More to Know)

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• Key distinction: Examiners want you to separate intra-molecular bonds (strong, within the molecule) from inter-molecular forces (weak, between molecules) — this is the textbook reason for low melting points.
• Graphite exception: The textbook only states graphite conducts electricity due to its layered hexagonal structure with a double bond. At Class 10 level, you are not required to explain "delocalised electrons" explicitly — but noting free/mobile electrons between layers earns the mark.
• Stick to source passages: Do not add extra detail (e.g., specific band theory) that goes beyond the textbook.
• Aim: ~2 marks for covalent low melting point explanation + ~1 mark for graphite exception.

(B) Soap reacts with calcium and magnesium ions in hard water to form insoluble salts (scum), while the charged ends of detergent molecules do not form such precipitates, so detergents remain effective in hard water.

The textbook explicitly states: "The charged ends of these (detergent) compounds do not form insoluble precipitates with the calcium and magnesium ions in hard water." Soap forms a curdy scum with Ca²⁺/Mg²⁺ ions, wasting it and reducing cleaning. Options A, C, and D are not supported by the textbook and describe incorrect mechanisms.

Source: Chapter 4, Section 4.5 Soaps and Detergents

(i) Observations with NaHCO₃:

• Ethanoic acid reacts with sodium hydrogencarbonate to produce brisk effervescence (CO₂ gas), which turns lime water milky.

CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂

• Ethanol shows no reaction with sodium hydrogencarbonate; no effervescence is observed.

(ii) Explanation in terms of functional groups:
Ethanoic acid contains the carboxyl group (–COOH), which is acidic enough to react with weak bases like NaHCO₃, releasing CO₂. Ethanol contains the hydroxyl group (–OH), which is only weakly acidic — too weak to react with NaHCO₃. However, both –COOH and –OH can donate H⁺ to sodium metal (a strong reducing agent), so both produce hydrogen gas with Na.

Source: Chapter 4, Section 4.4.2 – Properties of Ethanoic Acid

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• Examiners award marks for: (a) correct observation for each compound (1 mark), (b) functional group explanation linking –COOH to acidity and NaHCO₃ reaction, and –OH to insufficient acidity (2 marks).
• Always write the chemical equation for ethanoic acid + NaHCO₃ — it shows understanding and secures full credit.
• The key contrast: carboxylic acids are acidic enough to displace CO₂ from hydrogencarbonates; alcohols are not. Both react with Na because Na is a far stronger base/reducing agent.

(i) The reaction is called saponification. The reagent required is sodium hydroxide (NaOH) — a strong alkali (or potassium hydroxide, KOH).

(ii) The other organic product formed alongside soap is glycerol (propane-1,2,3-triol).

(iii) Reagent: concentrated sulphuric acid (H₂SO₄) as catalyst; Condition: heating. The reaction is reversible because the ester and water formed can react with each other to regenerate the original acid and alcohol, so neither reaction goes to completion — both forward and backward reactions occur simultaneously.

Source: Chapter 4, Section 4.5 Soaps and Detergents

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• Saponification is the key term examiners expect for (i); simply writing "hydrolysis" may lose the mark.
• For (ii), glycerol is the specific organic by-product — do not write water (it is inorganic).
• For (iii), the two marks cover: naming the reagent/condition and explaining reversibility. The reversibility point must mention that the backward reaction (hydrolysis of ester) also occurs, making it an equilibrium — this is what earns the mark.

(i) As we move from methane to butane (or methanol to butanol), each successive compound has one extra –CH₂– unit, increasing molecular mass. Greater molecular mass means stronger intermolecular forces, so more energy (higher boiling point) is needed to separate molecules. Chemical properties remain similar because they are determined solely by the functional group (e.g., –OH in alcohols), which stays the same throughout the series.

(ii) The student's error is that ethane (C₂H₆) and ethanol (C₂H₅OH) do not differ merely by one atom — they belong to different homologous series (alkanes and alcohols respectively) with different functional groups. For two compounds to belong to the same homologous series, they must: (a) have the same functional group, (b) differ by a –CH₂– unit, and (c) follow the same general formula.

Source: Carbon and its Compounds, Section 4.2.4 Homologous Series

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• Examiners look for: molecular mass → intermolecular forces → boiling point (for part i), and functional group → same chemical properties.
• For part (ii), clearly name the error (different functional groups / different series) and then state the three conditions of a homologous series. Don't just say "they differ by one atom is wrong" — explain why and state the correct conditions.
• Key phrase to use: "chemical properties are determined solely by the functional group."