(a) Change in temperature: When zinc reacts with dilute sulphuric acid, heat is evolved, showing a rise in temperature (exothermic reaction).
$$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\uparrow$$

(b) Evolution of a gas: The same reaction above also produces hydrogen gas, which can be observed as bubbles around the zinc granules.

(c) Change in colour: When potassium iodide solution is added to lead nitrate solution, a yellow precipitate of lead iodide is formed, indicating a change in colour.
$$\text{Pb(NO}_3)_2 + 2\text{KI} \rightarrow \text{PbI}_2 + 2\text{KNO}_3$$

Source: Chapter 1, Section 1.1 (Activities 1.2 and 1.3)

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• Examiners look for one clear example per observation with a correct chemical equation.
• For (a) and (b), the same Zn + H₂SO₄ reaction is acceptable for both, as it shows both temperature change and gas evolution — but state each observation separately.
• For (c), the lead nitrate + potassium iodide reaction is the textbook example (yellow PbI₂ precipitate).
• Make sure arrows (↑ for gas, ↓ for precipitate) are included — they fetch marks.
• Each sub-part carries 1 mark; losing the equation loses half the mark for that part.

Yes, — it is a redox reaction.

In this reaction, Mn in MnO₂ goes from +4 to +2 oxidation state (gains electrons → reduction), while Cl in HCl goes from −1 to 0 in Cl₂ (loses electrons → oxidation). Since both oxidation and reduction occur simultaneously, it is a redox reaction.

• A redox reaction must show both oxidation and reduction occurring simultaneously — mention both clearly for full marks.
• Assign oxidation states to Mn (+4 → +2) and Cl (−1 → 0) to justify your answer; this is what examiners look for.
• The textbook defines oxidation as gain of oxygen/loss of hydrogen (or electrons) and reduction as the reverse — either definition is acceptable here; oxidation state method is more precise.
• 1 mark for correct identification + 1 mark for valid justification with oxidation states.

(b) X = (aq), Y = (g)

ZnSO₄ dissolves in water to form an aqueous solution, and H₂ is released as a gas.

When zinc reacts with dilute H₂SO₄, ZnSO₄ (zinc sulphate) is soluble in water, so its state symbol is (aq). Hydrogen is produced as a gas, so its state symbol is (g). This is a standard displacement reaction from Chapter 1.

(a) Identifying the reducing agent:

The reducing agent is the substance that loses oxygen (or gets oxidised).

(i) NH₃ — nitrogen in NH₃ gets oxidised to NO; NH₃ is the reducing agent.
(ii) H₂O — oxygen is removed from water; H₂O is the reducing agent.
(iii) CO — CO gains oxygen to form CO₂; CO is the reducing agent.
(iv) H₂ — hydrogen gets oxidised to form H₂O; H₂ is the reducing agent.

(b) Definition of a Redox Reaction:

A reaction in which one substance gets oxidised (gains oxygen / loses hydrogen) and another gets reduced (loses oxygen / gains hydrogen) simultaneously is called a redox reaction.

Example: In Fe₂O₃ + 3CO → 2Fe + 3CO₂, CO is oxidised and Fe₂O₃ is reduced.

Source: Chapter 1, Chemical Reactions and Equations

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• For (a), the reducing agent is always the one that loses electrons / gets oxidised — in oxygen-based terms, it is the substance that gains oxygen or helps reduce the other substance. Students often confuse reducing agent with the substance being reduced; remember: the reducing agent itself gets oxidised.
• For (b), the examiner expects both terms (oxidation and reduction) defined together in one sentence to establish the "redox" concept, with an example.
• Stick to oxygen gain/loss language as the question specifies it.

(i) When blue copper sulphate crystals (hydrated) are heated, they lose water of crystallisation and turn white:

$$\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \xrightarrow{\Delta} \text{CuSO}_4(s) + 5\text{H}_2\text{O}(g)$$

(ii) When sodium hydrogen carbonate is heated during cooking:

$$2\text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)$$

Source: Chapter 1, Chemical Reactions and Equations

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• (i) The key point is that hydrated copper sulphate (blue) loses 5 water molecules on heating to give anhydrous copper sulphate (white). Examiners expect the formula $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ on the LHS.
• (ii) This is a thermal decomposition reaction. The coefficient 2 before NaHCO₃ is essential for balancing. This reaction releases CO₂, which makes dough/batter rise during cooking (baking soda).
• Always include the heat symbol ($\Delta$) above the arrow since energy is supplied.

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

Lead is more reactive than copper and displaces copper from copper sulphate solution, turning the blue solution colourless: Pb + CuSO₄ → PbSO₄ + Cu.

The textbook (equation 1.26) shows Pb displacing Cu from CuCl₂; it also states "Zinc and lead are more reactive than copper." The same logic applies to CuSO₄ — lead displaces copper, removing the Cu²⁺ ions responsible for the blue colour, making the solution colourless. So both A and R are true, and R correctly explains A. Option (a) is correct.

(c) 4, 2, 1

Balancing: MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂. Mn=1, O=2, H=4, Cl=4 on both sides. So x=4, y=2, z=1.

Balance each element: 4 HCl provides 4 Cl atoms — 2 go to MnCl₂ and 2 form 1 Cl₂; 4 H atoms form 2 H₂O; 2 O atoms (from MnO₂) form 2 H₂O. All atoms balance. Examiners expect the correct option with brief verification.

(a)

The correct answer is (a): thermal decomposition of lead nitrate which produces brown fumes of nitrogen dioxide.

Reaction: $2\text{Pb(NO}_3)_2(s) \xrightarrow{\Delta} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$

The brown fumes are of NO₂ gas, produced by thermal decomposition (not oxidation) of lead nitrate.

Source: Chapter 1, Activity 1.6 — Decomposition Reaction

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• The key distinction: the reaction is thermal decomposition (a single reactant breaks down on heating), not oxidation.
• The brown fumes are specifically nitrogen dioxide (NO₂), not lead oxide (PbO is a solid yellow powder, not a fume).
• Option (b) is wrong because PbO is a solid, not a fume.
• Options (c) and (d) are wrong because the reaction type is decomposition, not oxidation.
• Remember the balanced equation — it also produces O₂, but the visible brown fumes are NO₂ only.

(b) It turns milky due to formation of calcium carbonate and on passing excess of carbon dioxide it becomes colourless due to formation of calcium hydrogen carbonate which is soluble in water.

The reactions are: Ca(OH)₂ + CO₂ → CaCO₃↓ (white/milky) + H₂O; then CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂ (soluble, so colourless). Option (b) correctly identifies CaCO₃ as the cause of milkiness and soluble Ca(HCO₃)₂ as the reason for turning colourless. Options (c) and (d) wrongly mention sodium hydrogen carbonate or reverse the products.

Chemical equation:

$$2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$$

The white ash formed is magnesium oxide.

Type of reaction: It is a combination reaction because two reactants (magnesium and oxygen) combine to form a single product (magnesium oxide). It is also an oxidation reaction as magnesium gains oxygen.

Source: Chapter 1, Section 1.1; Chapter 3, Section 3.2.1

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• The question asks for three things: the equation, the name of the ash, and the type of reaction with justification — address all three.
• The balanced equation must show 2Mg on the left to balance atoms (skeletal equation Mg + O₂ → MgO is unbalanced).
• Examiners accept combination reaction as the primary answer; mentioning oxidation as an additional point can fetch bonus credit or shows completeness.
• "Justification" means you must state why it is a combination reaction — always include the one-line reason.

Answer: (B)

CaO + H₂O → Ca(OH)₂ is a combination reaction, while reactions A, C, and D are all double displacement reactions (two ions/groups are exchanged between reactants).

Source: Chapter 1, Chemical Reactions and Equations

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• Double displacement: two different ions/groups exchange between two compounds (A, C, D all fit this pattern).
• Combination reaction: two substances combine to form a single new substance — that's what B shows (CaO + H₂O → Ca(OH)₂).
• Examiners expect you to identify the odd one out by reaction type. Name the type for option B and briefly state why A, C, D belong to the same category.

(A) 1, 6, 2 and 3

Balancing: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O. Al: 2=2, Cl: 6=6, O: 3=3, H: 6=6. So a=1, b=6, c=2, d=3.

Balance Al first (2 on LHS → c=2), then Cl (6 on RHS → b=6), then check H and O (6H gives 3H₂O). The correct coefficients are 1, 6, 2, 3 — option A. Watch out for option D which incorrectly doubles all terms.

Option A – Quick lime

CaCO₃ $\xrightarrow{\Delta}$ CaO + CO₂

The product 'X' is Quick lime (Calcium oxide, CaO).

Source: Chapter 1, Section 1.2.1

The thermal decomposition of calcium carbonate (limestone) gives calcium oxide (quick lime) and carbon dioxide. This is a standard decomposition reaction. The source passage (Ch. 1) identifies CaO as "Quick lime," which directly matches option A. Gypsum (CaSO₄·2H₂O) and Plaster of Paris are calcium sulphate compounds — unrelated to this reaction.

Displacement Reaction: A more reactive element displaces a less reactive element from its compound.
Example: $\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$

Double Displacement Reaction: Two compounds exchange ions to form two new compounds.
Example: $\text{Na}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4\downarrow + 2\text{NaCl}$

Source: Chapter 1, Sections 1.2.3 & 1.2.4

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• Give one definition + one equation for each type — that covers both marks.
• Examiners look for the key distinction: in displacement, one element replaces another; in double displacement, two ionic groups/atoms are exchanged between two compounds.
• Always include state symbols (↓ for precipitate) where relevant — it shows understanding and can earn bonus credit.

In a precipitation reaction, two soluble ionic compounds react by exchanging their ions, forming an insoluble precipitate. Without this exchange of ions (double displacement), no precipitate can form.

Example:

$$\text{Na}_2\text{SO}_4(aq) + \text{BaCl}_2(aq) \rightarrow \text{BaSO}_4(s)\downarrow + 2\text{NaCl}(aq)$$

Here, $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$ ions exchange to form insoluble barium sulphate (white precipitate).

Source: Chapter 1, Section 1.2.4 Double Displacement Reaction

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• Examiners expect you to link precipitation with double displacement — both concepts appear together in NCERT.
• The key phrase is "exchange of ions": Ba²⁺ from BaCl₂ combines with SO₄²⁻ from Na₂SO₄ to form the precipitate BaSO₄.
• Always include state symbols (aq), (s) and the ↓ arrow for the precipitate — these earn marks.
• 1 mark for the justification/explanation + 1 mark for the correct balanced equation.

(C) SO₂ is oxidising agent and H₂S is reducing agent.

In this reaction, S in H₂S goes from −2 to 0 (oxidised), so H₂S is the reducing agent. S in SO₂ goes from +4 to 0 (reduced), so SO₂ is the oxidising agent.

• The agent that gets oxidised is the reducing agent (H₂S: sulphur −2 → 0).
• The agent that gets reduced is the oxidising agent (SO₂: sulphur +4 → 0).
• Options A and D reverse these roles; option B wrongly says H₂S is reduced. Always check oxidation states of the same element on both sides to decide who is oxidised/reduced.

(B) Burning of coal

In burning of coal: $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$, two reactants combine to form a single product — this is a combination reaction.

Source: Chapter 1, Section 1.2.1

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The textbook explicitly lists "burning of coal" as an example of a combination reaction (equation 1.15). Burning of methane (option C) is listed as an exothermic reaction, not combination. Digestion of food is decomposition/exothermic. Black and white photography involves decomposition of silver salts. Examiners expect you to recall that combination = two or more reactants → one single product.

(D) is exothermic and pH of the solution formed is more than 7.

Heat is liberated → exothermic. Ca(OH)₂ is a base → its solution is alkaline → pH > 7.

Two facts decide the answer: (1) heat is released = exothermic (not endothermic), eliminating A and C; (2) Ca(OH)₂ is a base, so its aqueous solution is alkaline with pH > 7 (not = 7), eliminating B. Both conditions together point to option D.

(i) Change in colour:

When lead nitrate solution is added to potassium iodide solution, a yellow precipitate of lead iodide is formed.

$$\text{Pb(NO}_3)_2(aq) + 2\text{KI}(aq) \rightarrow \text{PbI}_2(s)\downarrow + 2\text{KNO}_3(aq)$$

(Yellow precipitate of PbI₂ formed)

(ii) Change in temperature:

When zinc granules react with dilute sulphuric acid, hydrogen gas is evolved and temperature of the flask rises (exothermic reaction).

$$\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)\uparrow$$

(iii) Formation of precipitate:

When sodium sulphate solution is mixed with barium chloride solution, a white precipitate of barium sulphate is formed.

$$\text{Na}_2\text{SO}_4(aq) + \text{BaCl}_2(aq) \rightarrow \text{BaSO}_4(s)\downarrow + 2\text{NaCl}(aq)$$

(White precipitate of BaSO₄ formed)

Source: Chemical Reactions and Equations, Chapter 1 (Activities 1.2, 1.3, 1.10)

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• Each sub-question carries 1 mark: one equation with the observation stated is sufficient.
• Always mention the colour change / temperature change / precipitate colour alongside the equation — examiners award the mark for the observation too, not just the equation.
• Use state symbols (aq), (s), (g) and ↓ or ↑ arrows where applicable; these show precision.
• The Pb(NO₃)₂ + KI reaction is the standard textbook example for colour change (yellow PbI₂); Zn + H₂SO₄ for temperature rise; Na₂SO₄ + BaCl₂ for white precipitate.

The reaction between calcium oxide and water is a combination reaction (also exothermic).

Justification: Two reactants (CaO and H₂O) combine to form a single product (calcium hydroxide). A reaction in which two or more substances combine to form a single product is called a combination reaction.

Balanced chemical equation:

$$\text{CaO}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{Heat}$$

Source: Chapter 1, Section 1.2.1 Combination Reaction

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• 1 mark for correctly naming the reaction type (combination reaction).
• 1 mark for the balanced equation with correct formulae and state symbols.
• Always write state symbols — (s), (l), (aq) — as they show you understand the reaction fully.
• You may also mention it is exothermic (heat is released), but the primary answer required is "combination reaction."

(d) MnO₂ is reduced and HCl is oxidised.

In this reaction, Mn in MnO₂ goes from +4 to +2 (gains electrons → reduction), while Cl in HCl goes from −1 to 0 in Cl₂ (loses electrons → oxidation).

In a redox reaction, both oxidation and reduction must occur simultaneously. Check oxidation states: Mn drops from +4 to +2 (reduced), and Cl rises from −1 to 0 (oxidised). Options (a), (b), and (c) each describe only one half — only (d) correctly identifies both changes, making it the complete and correct answer.

(c) 2AgBr → 2Ag + Br₂

This is a photochemical decomposition reaction where silver bromide decomposes in the presence of sunlight (light energy) to form silver and bromine.

• Option (a) uses heat; option (b) uses electricity; option (d) uses heat — all are thermal decomposition reactions.
• 2AgBr → 2Ag + Br₂ is a classic photochemical decomposition (light as energy source), used as the standard textbook example.
• Examiners expect students to distinguish between heat, light, and electricity as energy sources in decomposition reactions (as asked in Exercise Q.12 of Chapter 1).

(b) 2NaOH + Zn → Na₂ZnO₂ + H₂

This is the correctly balanced equation, with sodium zincate and hydrogen gas as products.

Source: Chapter 2, Section 2.1.2 (Activity 2.4)

The textbook (Activity 2.4) explicitly gives this equation: 2NaOH(aq) + Zn(s) → Na₂ZnO₂(s) + H₂(g). Check balance: Na=2, O=2, H=2+2=4 on LHS; Na=2, Zn=1, O=2, H=2 on RHS — all atoms balance. Option (b) is the only choice that is both correctly balanced and has the right products (Na₂ZnO₂ + H₂).

(a) Steam passed over red hot iron:

$$3\text{Fe}(s) + 4\text{H}_2\text{O}(g) \rightarrow \text{Fe}_3\text{O}_4(s) + 4\text{H}_2(g)$$

(b) Natural gas (methane) burnt in air:

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

(c) Glucose reacting with oxygen in body cells:

$$\text{C}_6\text{H}_{12}\text{O}_6(aq) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) + \text{energy}$$

Source: Chapter 1 – Chemical Reactions and Equations; Chapter 4 – Carbon and its Compounds, Section 4.3.1

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• (a) is a displacement/redox reaction — iron displaces hydrogen from steam forming iron oxide (Fe₃O₄). Students often forget the coefficient 4 on both H₂O and H₂.
• (b) Natural gas = methane (CH₄). The balanced equation requires coefficient 2 before O₂. Always mention heat and light for combustion.
• (c) This is cellular respiration — an exothermic reaction. Write "energy" (not just heat) as the product. Balancing correctly (6 on O₂, CO₂, and H₂O) is key for full marks.
• Each equation carries 1 mark — a missing or wrong coefficient loses the mark, so double-check balancing.

P = Iron (Fe); Q = Zinc (Zn)

The light green solution is FeSO₄ (ferrous sulphate), so metal P is Iron (Fe). The solution turns colourless after the reaction, forming ZnSO₄, so metal Q is Zinc (Zn).

Chemical equation:

$$\text{Zn}(s) + \text{FeSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Fe}(s)$$

Conclusion: Zinc is more reactive than iron. A more reactive metal displaces a less reactive metal from its salt solution.

Source: Chemical Reactions and Equations, Section 1.2.3 (Displacement Reaction)

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The key clues are: (1) light green colour of PSO₄ → FeSO₄ → P = Fe; (2) colourless product QSO₄ → ZnSO₄ → Q = Zn. The textbook directly gives the Zn + CuSO₄ example; this is the analogous reaction with FeSO₄. Examiners expect the balanced equation with state symbols and a one-line reactivity conclusion. Do not confuse P and Q — P is in solution, Q is the rod doing the displacing.

(B) Pale green

Iron displaces copper from CuSO₄, forming FeSO₄ (iron sulphate), which gives the solution a pale green colour.

The reaction is: Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s). Blue colour of CuSO₄ fades and pale green FeSO₄ forms. The iron nail also turns reddish-brown due to deposited copper. Examiners expect students to recall both the colour change of the solution (pale green) and its cause (FeSO₄ formation). Option (D) describes the nail, not the solution.

(C) It is an endothermic reaction.

Burning of magnesium is an exothermic reaction (releases heat and light). It burns with a dazzling white flame, forms white magnesium oxide powder, and is a combination reaction — all true statements.

The textbook (Activity 1.1, Ch. 1) states magnesium burns with a dazzling white flame forming white powder (MgO) — a combination reaction releasing energy, making it exothermic, not endothermic. Endothermic reactions absorb energy; exothermic reactions release it. Examiners expect you to know this distinction clearly.

(D) (ii) and (iv)

In electrolysis of water, hydrogen gas (double the volume) collects at the cathode and oxygen at the anode. From the diagram, P (left, more gas) is hydrogen → R is cathode; Q (right, less gas) is oxygen → S is anode. So (ii) and (iv) are correct.

• At cathode: 2H⁺ + 2e⁻ → H₂; at anode: 2OH⁻ → H₂O + ½O₂. Hydrogen is produced in double the volume of oxygen.
• The test tube with more gas (P, left = R) collects H₂ → R is cathode. The test tube with less gas (Q, right = S) collects O₂ → S is anode.
• Statement (i) wrongly calls P oxygen; statement (iii) wrongly calls Q hydrogen. Only (ii) and (iv) are correct → Option (D).

(a) 'X' is ferrous sulphate (FeSO₄·7H₂O) and 'Y' is ferric oxide (Fe₂O₃).

The reaction is:
$$2\text{FeSO}_4 \xrightarrow{\text{Heat}} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3$$

(b) On heating, the green colour of ferrous sulphate crystals first changes to white (on losing water), and then turns reddish-brown (colour of Fe₂O₃) on further heating.

Source: Chapter 1, Section 1.2.2 – Decomposition Reaction

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• Examiners expect you to name both substances with their chemical formulas and include the balanced equation (it adds precision and earns marks).
• For part (b), mention two colour changes: green → white (loss of water) → reddish-brown (Fe₂O₃ formed). Mentioning only one change may cost you a mark.
• The "characteristic odour" in the question refers to SO₂/SO₃ gases — this confirms it is ferrous sulphate decomposing.

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

$2\text{AgCl} \xrightarrow{\text{sunlight}} 2\text{Ag} + \text{Cl}_2$

The silver formed makes AgCl appear grey.

This is a photodecomposition reaction — sunlight decomposes AgCl into silver (grey) and chlorine gas. The Reason directly explains why the grey colour appears (metallic silver deposits), so it is the correct explanation of the Assertion. Students often confuse this with option B; remember, whenever the R logically and directly causes A, choose option A.

Option D: 2, 2, 4, 1

The balanced equation is: $2\text{Pb(NO}_3)_2 \xrightarrow{\text{heat}} 2\text{PbO} + 4\text{NO}_2 + 1\text{O}_2$

So a = 2, b = 2, c = 4, d = 1.

Balance by checking atoms on both sides: 2 Pb(NO₃)₂ gives 2 Pb, 4 N, 12 O on LHS. RHS: 2 PbO (2 Pb, 2 O) + 4 NO₂ (4 N, 8 O) + 1 O₂ (2 O) = 2 Pb, 4 N, 12 O ✓. Students often confuse c and d; remember NO₂ gets coefficient 4, O₂ gets 1.

C — (i) and (ii)

When magnesium ribbon is burnt in air, it burns with a dazzling white flame and forms a white powder (magnesium oxide). Magnesium does not vapourise, and MgO dissolved in water gives a basic solution (turns red litmus blue, not blue litmus red).

Observation (i) is directly stated in Activity 1.1: "magnesium ribbon burns with a dazzling white flame." Observation (ii) is also confirmed: "changes into a white powder" (MgO). Option (iii) is wrong — magnesium burns, it does not simply vapourise. Option (iv) is wrong — MgO is a basic oxide; its aqueous solution turns red litmus blue, not blue litmus red. Examiners expect students to eliminate incorrect observations using their knowledge of the nature of MgO.

Precipitation Reaction: Any reaction that produces an insoluble solid (called a precipitate) when two solutions are mixed is called a precipitation reaction.

Example: When sodium sulphate solution is mixed with barium chloride solution, a white insoluble precipitate of barium sulphate is formed.

Balanced Chemical Equation:

$$\text{Na}_2\text{SO}_4(aq) + \text{BaCl}_2(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaCl}(aq)$$

The white precipitate of BaSO₄ is formed by the reaction of SO₄²⁻ and Ba²⁺ ions. Sodium chloride remains dissolved in the solution.

Source: Chapter 1, Section 1.2.4 Double Displacement Reaction

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• 1 mark for the definition (mention "insoluble substance/precipitate").
• 1 mark for the example (naming both reactants and the precipitate formed).
• 1 mark for the correctly balanced equation with state symbols — (aq), (s) are important; don't forget the coefficient 2 before NaCl.
• Examiners expect state symbols in the equation; missing them can cost half a mark.

Why we balance a chemical equation:
A chemical equation is balanced to satisfy the Law of Conservation of Mass.

Law: Mass can neither be created nor destroyed in a chemical reaction. Therefore, the number of atoms of each element must be the same on both sides (reactant and product side) of the equation.

Balancing the equation:

Unbalanced: $\text{Zn} + \text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + \text{H}_2$

Count atoms:

• Zn: LHS = 1, RHS = 3 → put coefficient 3 before Zn
• PO₄: LHS = 1, RHS = 2 → put coefficient 2 before H₃PO₄
• H: LHS = 6 (2×3), RHS = 2 → put coefficient 3 before H₂

Balanced equation:

$$3\text{Zn} + 2\text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + 3\text{H}_2$$

Verification: Zn=3, H=6, P=2, O=8 on both sides. ✓

Source: Chapter 1, Section 1.1.2

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• 1 mark for stating the reason (Law of Conservation of Mass).
• 1 mark for naming and stating the law correctly.
• 1 mark for the correctly balanced equation.
• Always verify atom count on both sides — examiners check this. Use only coefficients, never change subscripts inside formulae.

Oxidation: If a substance gains oxygen (or loses hydrogen) during a reaction, it is said to be oxidised.

In the reaction CuO + H₂ → Cu + H₂O, hydrogen (H₂) is oxidised because it gains oxygen to form water (H₂O).

Source: Chapter 1, Section 1.2.5 – Oxidation and Reduction

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• 1 mark for the definition: must mention gain of oxygen (or loss of hydrogen).
• 1 mark for correctly identifying and naming hydrogen (H₂) as the substance oxidised, with a brief reason (it gains oxygen).
• Do NOT say CuO is oxidised — CuO loses oxygen, so it is reduced.
• Always state why the substance is oxidised (gain of oxygen) to secure full marks.

(d) Melting of glaciers

Melting of glaciers is a physical change (change of state from solid to liquid); no new substance is formed, so no chemical reaction occurs.

The textbook's Chapter 1 introduction lists milk souring, iron rusting, and grape fermentation as examples where chemical reactions take place (identity of substance changes). Melting is purely a physical change — only the state changes, not the chemical composition. Examiners expect students to distinguish physical changes from chemical reactions.

When a solution of copper sulphate reacts with iron, a displacement reaction takes place.

Iron, being more reactive than copper, displaces copper from copper sulphate solution. The reaction is:

$$\text{Fe}(s) + \text{CuSO}_4(aq) \rightarrow \text{FeSO}_4(aq) + \text{Cu}(s)$$

The blue colour of copper sulphate solution fades and a brown deposit of copper forms on the iron nail.

Source: Chapter 1, Section 1.2.3 Displacement Reaction

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• The examiner expects you to name the reaction type (displacement) and write the balanced equation — both are needed for full marks.
• Mentioning that iron is more reactive than copper explains why displacement occurs, which earns extra credit.
• The observable change (fading blue colour, brown copper deposit) is a bonus point often asked in activity-based questions.

Copper sulphate crystals are blue because they contain water of crystallisation (CuSO₄·5H₂O). On heating, they lose this water of crystallisation and turn white, forming anhydrous copper sulphate (CuSO₄). This is a decomposition reaction.

$$\text{CuSO}_4 \cdot 5\text{H}_2\text{O} \xrightarrow{\text{Heat}} \text{CuSO}_4 + 5\text{H}_2\text{O}$$

Source: Chapter 1, Section 1.2.2 Decomposition Reaction

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• The key concept here is water of crystallisation — the fixed number of water molecules present in one formula unit of a crystallised salt.
• Examiners expect you to name both forms: CuSO₄·5H₂O (blue, hydrated) and CuSO₄ (white, anhydrous).
• Mentioning it is a decomposition reaction (thermal decomposition) adds marks.
• The equation is important — write it correctly with the dot notation for water of crystallisation.

Double displacement reaction: A reaction in which two different ions are exchanged between the reactants to form two new compounds.

(i) Neutralization reaction (also double displacement):

$$\text{NaOH}(aq) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$$

Justification: Na⁺ and H⁺ ions exchange partners. It is a neutralization reaction because an acid (HCl) reacts with a base (NaOH) to form salt and water.

(ii) Precipitation reaction (also double displacement):

$$\text{Na}_2\text{SO}_4(aq) + \text{BaCl}_2(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaCl}(aq)$$

Justification: SO₄²⁻ and Cl⁻ ions exchange partners. It is a precipitation reaction because insoluble BaSO₄ (white precipitate) is formed as a product.

Source: Chapter 1 – Chemical Reactions and Equations, Section 1.2.4

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• Definition: Always mention "exchange of ions" — that is the key phrase examiners look for.
• Neutralization: Any acid + base → salt + water equation works; NaOH + HCl is the standard textbook example.
• Precipitation: The BaSO₄ example is directly from the textbook (Activity 1.10); always write the state symbol (s) for the precipitate.
• Justification: For each, briefly state why it qualifies — this earns the explanation mark. Do not skip it.

(a) Na

Na is the reducing agent because it loses electrons (gets oxidised from 0 to +1), thereby reducing hydrogen in water from +1 to 0 (H₂).

The reducing agent is the substance that itself gets oxidised (loses electrons) and causes reduction in another substance. Here, Na goes from oxidation state 0 → +1 (oxidised), so Na is the reducing agent. H₂O is the oxidising agent (hydrogen in it is reduced from +1 → 0). Examiners expect you to identify the correct substance and briefly justify using oxidation states or electron transfer.

(c) CaO + H₂O → Ca(OH)₂ + Heat

The textbook (Eq. 1.13) clearly states: CaO(s) + H₂O(l) → Ca(OH)₂(aq) + Heat. Only one mole each of CaO and H₂O react to form one mole of Ca(OH)₂ with no H₂ or O₂ released. Options (a), (b), and (d) are unbalanced or show incorrect products.

(i) A few drops of acid (dilute H₂SO₄) are added to water to make it a good conductor of electricity (pure water is a poor conductor).

• Gas liberated at cathode: Hydrogen (H₂)
• Gas liberated at anode: Oxygen (O₂)

Relationship: Volume of H₂ at cathode : Volume of O₂ at anode = 2 : 1

(ii) When silver chloride (AgCl) is exposed to sunlight, it turns grey due to decomposition into silver (Ag) and chlorine (Cl₂).

$$2\text{AgCl} \xrightarrow{\text{sunlight}} 2\text{Ag} + \text{Cl}_2$$

Type of reaction: Photochemical decomposition reaction.

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• In part (i), examiners expect three things: reason for adding acid, names of gases at each electrode, and the 2:1 volume ratio. Missing any one costs a mark.
• In part (ii), state the observable change (white → grey), write the balanced equation, and name the reaction type. "Photochemical decomposition" is the exact term expected — just "decomposition" may not fetch full credit.

(i) The compound formed is copper(II) chloride. Its colour is blue-green.

(ii) Balanced chemical equation:

$$\text{CuO}(s) + 2\text{HCl}(aq) \rightarrow \text{CuCl}_2(aq) + \text{H}_2\text{O}(l)$$

Source: Chapter 2, Section 2.1.5 (Reaction of Metallic Oxides with Acids)

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• The passage (Activity 2.7) explicitly states the solution turns blue-green due to formation of copper(II) chloride — use the exact name and colour.
• The balanced equation follows the general pattern: Metal oxide + Acid → Salt + Water. Ensure coefficient 2 before HCl to balance the equation — this is a commonly tested balancing point.
• State symbols (s), (aq), (l) add marks if included.

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

Quicklime (CaO) reacts vigorously with water releasing a large amount of heat, making it an exothermic reaction. (R) directly explains why (A) is true.

Source: Chapter 1, Section 1.2.1 Combination Reaction

The textbook explicitly states: "Calcium oxide reacts vigorously with water to produce slaked lime releasing a large amount of heat" and defines exothermic reactions as those in which heat is released along with products. Since (R) gives the exact reason for (A), option (a) is correct — not (b). Students often confuse (a) and (b); choose (b) only when the reason, though true, explains something else.

(b) 2KI + Pb(NO₃)₂ → PbI₂ + 2KNO₃

This is a double displacement (precipitation) reaction. Lead iodide (PbI₂) is the yellow insoluble precipitate. The equation must be balanced — 2 units of KI are needed to provide 2 iodide ions for one Pb²⁺ ion, making option (b) the only correctly balanced equation.

(a) Silver chloride is white in colour when kept in the china dish.

(b) It is a photochemical decomposition reaction (decomposition by light).

$$2\text{AgCl} \xrightarrow{\text{Sunlight}} 2\text{Ag} + \text{Cl}_2$$

Silver chloride decomposes into silver (grey) and chlorine gas in the presence of sunlight.

(c) This reaction is used in black and white photography. Silver bromide (AgBr) is another chemical that can be used for the same purpose.

Source: Chapter 1, Chemical Reactions and Equations

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• (a) is 1 mark — just state the colour: white.
• (b) is 1 mark — name the reaction type correctly (photochemical decomposition) and write the balanced equation. Examiners look for the sunlight condition written above the arrow.
• (c) is 1 mark — state the use (photography) and name one more chemical (AgBr). Both parts must be present for full credit.
• The grey colour appears because silver metal (Ag) is formed, which is grey — useful to mention briefly.

Observation: A bright yellow precipitate is formed immediately.

Type of reaction: Double displacement reaction (precipitation reaction).

Chemical equation:

$$\text{Pb(NO}_3)_2\text{(aq)} + 2\text{KI(aq)} \rightarrow \text{PbI}_2\text{(s)} + 2\text{KNO}_3\text{(aq)}$$

The yellow precipitate formed is lead iodide (PbI₂).

Source: Chapter 1, Exercises Q.15

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• Two marks are split: one for observation + type of reaction, one for the balanced equation.
• Always state the colour of the precipitate (yellow for PbI₂) — examiners specifically look for this.
• Label the precipitate with (s) and aqueous reactants/products with (aq) — state symbols are expected.
• "Double displacement" and "precipitation reaction" are both acceptable names; writing both is safest.
• The equation must be balanced (note the coefficient 2 before KI and KNO₃).

Observation: A bright yellow precipitate of lead iodide (PbI₂) is formed.

Type of reaction: Double displacement reaction (precipitation reaction).

Chemical equation:

$$2\text{KI}(aq) + \text{Pb(NO}_3)_2(aq) \rightarrow \text{PbI}_2(s)\downarrow + 2\text{KNO}_3(aq)$$

Source: Chapter 1, Chemical Reactions and Equations

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• The examiner expects both observation (yellow precipitate) and reaction type for full marks.
• This is both a double displacement and a precipitation reaction — either term or both is acceptable, but mentioning precipitation strengthens the answer.
• The equation must be balanced and state symbols (aq, s) should be included; the ↓ symbol for precipitate is good practice.
• "Bright yellow precipitate" is the key observable fact — don't just write "precipitate forms."

(a) 2, 4

Balanced equation: $2\ \text{Pb(NO}_3)_2 \rightarrow 2\ \text{PbO} + 4\ \text{NO}_2 + \text{O}_2$. So x = 2 and y = 4.

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Check atom balance: with x = 2, y = 4 — Pb: 2 = 2 ✓; N: 4 = 4 ✓; O: 12 = 2+8+2 = 12 ✓. The law of conservation of mass (equal atoms on both sides) confirms option (a).

(c) Dissolution of ammonium chloride in water

Endothermic reactions absorb energy from the surroundings. Dissolving NH₄Cl in water absorbs heat, making the solution cold — a classic endothermic process.

The key definition: endothermic = energy absorbed. Options (a), (b), and (d) all release heat (exothermic). Dissolution of NH₄Cl is the standard textbook example of an endothermic process because it cools the surroundings. Examiners expect you to recall this directly from Chapter 1 examples.

Source: Chapter 1, Chemical Reactions and Equations

Chemical Equation:

$$\text{Fe}(s) + \text{CuSO}_4(aq) \rightarrow \text{FeSO}_4(aq) + \text{Cu}(s)$$

Type of Reaction: Displacement reaction.

Definition: A reaction in which a more reactive element displaces a less reactive element from its compound (in solution or molten form) is called a displacement reaction.

Two other metals that can displace copper from copper sulphate solution:

1. Zinc (Zn)
2. Lead (Pb)

(Both are more reactive than copper, so they displace it from CuSO₄ solution.)

Source: Chapter 1, Section 1.2.3; Chapter 3, Section 3.2.4

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• The equation must include state symbols — examiners award marks for these.
• Name and define the reaction type — both are needed for full marks; just naming it is not enough.
• The two metals must be more reactive than copper (above Cu in the activity series). Zinc and lead are the safest choices as the textbook explicitly lists them as examples (equations 1.25 and 1.26). Magnesium, aluminium, iron (already used) also work, but zinc and lead are directly cited for this reaction.
• Do not write metals less reactive than copper (e.g., silver, gold) — they cannot displace copper.

Product: Copper oxide (CuO)
Colour: Black

Chemical Equation:

$$2\text{Cu}(s) + \text{O}_2(g) \xrightarrow{\Delta} 2\text{CuO}(s)$$

When copper is heated in air, it reacts with oxygen to form copper(II) oxide, which is black in colour. This is an oxidation reaction.

Source: Chemical Reactions and Equations, Chapter 1 (also referenced in Exercise Q.17)

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• This question is directly linked to Exercise Q.17 of Chapter 1, where element 'X' (shiny brown = copper) turns black on heating — the black compound is CuO.
• Must include: name of product, colour (black), and balanced equation — one mark each.
• Write the balanced equation correctly: 2Cu + O₂ → 2CuO (coefficients are essential for full marks).
• Mentioning it is an oxidation/combination reaction adds value but is not strictly required for 2 marks.

(D) Fe₂O₃ + 3CO → 2Fe + 3CO₂

In this reaction, CO is oxidised (gains oxygen) and Fe₂O₃ is reduced (loses oxygen), making it a redox reaction. It is not a combination reaction as two reactants give two products.

Options A, B, and C are both combination reactions (two reactants forming one product) AND redox reactions. Option D involves two reactants giving two products, so it is not a combination reaction, but it is a redox reaction because CO gets oxidised to CO₂ and Fe₂O₃ gets reduced to Fe. Examiners expect you to know both criteria to eliminate the wrong options.

(C) 2, 2, 4

Balanced equation: 2 Zn(NO₃)₂ → 2 ZnO + 4 NO₂ + O₂. Here x = 2, y = 2, z = 4.

Balance each element: LHS has 2 Zn, 4 N, 12 O. RHS: 2 ZnO (2 Zn, 2 O) + 4 NO₂ (4 N, 8 O) + O₂ (2 O) = 2 Zn, 4 N, 12 O. ✓ The hit-and-trial method from Chapter 1 is used here. Note option (C) gives the smallest whole-number coefficients that satisfy conservation of mass.

(i) Decomposition Reaction: A reaction in which a single substance breaks down to give two or more substances is called a decomposition reaction.

I. Electrolysis of water:
$$2\text{H}_2\text{O}(l) \xrightarrow{\text{electricity}} 2\text{H}_2(g) + \text{O}_2(g)$$
Water (one substance) decomposes into hydrogen and oxygen. Hence it is a decomposition reaction. Energy involved: electrical energy.

II. Blackening of silver bromide:
$$2\text{AgBr}(s) \xrightarrow{\text{sunlight}} 2\text{Ag}(s) + \text{Br}_2(g)$$
Silver bromide decomposes into silver and bromine. Hence it is a decomposition reaction. Energy involved: light energy.

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(ii) Formation of calcium oxide involves decomposition (CaCO₃ heated → CaO + CO₂), while formation of calcium hydroxide involves combination (CaO + H₂O → Ca(OH)₂). These are opposite reactions:

$$\text{CaCO}_3(s) \xrightarrow{\Delta} \text{CaO}(s) + \text{CO}_2(g) \quad \text{(Decomposition)}$$

$$\text{CaO}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{Heat} \quad \text{(Combination)}$$

In decomposition, one reactant splits into two products; in combination, two reactants form one product — hence they are opposite reactions.

Source: Chapter 1, Section 1.2 — Types of Chemical Reactions

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• Part (i): Always write the definition, then the equation, then explicitly state why it is a decomposition reaction (one substance → two or more), and name the type of energy. Examiners award separate marks for the equation, the reasoning, and the energy type.
• Part (ii): The key is identifying that CaO is formed by decomposition of CaCO₃, and Ca(OH)₂ is formed by combination of CaO and H₂O. Writing both equations and labelling them is essential. The word "opposite" must be justified — decomposition splits, combination merges.
• Keep equations balanced; unbalanced equations lose marks.

Chemical Reaction: A chemical reaction is a process in which one or more substances (reactants) are transformed into new substances (products) with different properties. It involves breaking and making of bonds between atoms.

(i) Change of Colour:
Take lead nitrate solution in a test tube and add potassium iodide solution to it. A bright yellow precipitate of lead iodide is formed immediately. The change in colour (from colourless to yellow) indicates that a chemical reaction has taken place.

$$\text{Pb(NO}_3)_2 + 2\text{KI} \rightarrow \text{PbI}_2 \downarrow + 2\text{KNO}_3$$

(ii) Change in Temperature:
Take zinc granules in a conical flask and add dilute sulphuric acid. Bubbles of hydrogen gas are produced and the flask becomes warm on touching. The rise in temperature shows that a chemical (exothermic) reaction has taken place.

$$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\uparrow$$

Source: Chapter 1, Chemical Reactions and Equations, Activities 1.2 and 1.3

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• Define chemical reaction in 1–2 lines first (1 mark).
• Each activity is worth ~2 marks: name substances used, state the observation clearly, and write the equation.
• Examiners look for: correct activity description, the specific observation (colour change / temperature change), and the chemical equation.
• Use Activity 1.2 (lead nitrate + potassium iodide → yellow precipitate) for colour change and Activity 1.3 (Zn + H₂SO₄, flask becomes warm) for temperature change — these are the textbook examples.

The correct answer is (B) Displacement reaction.

In this reaction, zinc (a more reactive metal) displaces hydrogen from acetic acid, which is a characteristic of displacement reactions.

In a displacement reaction, a more reactive element displaces a less reactive element from its compound. Here, Zn displaces H₂ from CH₃COOH. This is analogous to Zn + H₂SO₄ → ZnSO₄ + H₂ discussed in Chapter 1. Do not confuse it with double displacement, where two compounds exchange ions with each other.

Source: Chapter 1, Chemical Reactions and Equations

Observations:

1. The green colour of hydrated ferrous sulphate (FeSO₄·7H₂O) crystals changes (first loses water, then turns reddish-brown).
2. A characteristic smell of burning sulphur (SO₂/SO₃ gases) is observed.

Chemical equations:

Step 1 – Loss of water on heating:
$$\text{FeSO}_4 \cdot 7\text{H}_2\text{O}(s) \xrightarrow{\Delta} \text{FeSO}_4(s) + 7\text{H}_2\text{O}(g)$$

Step 2 – Thermal decomposition:
$$2\text{FeSO}_4(s) \xrightarrow{\Delta} \text{Fe}_2\text{O}_3(s) + \text{SO}_2(g) + \text{SO}_3(g)$$

This is a thermal decomposition reaction.

Source: Chapter 1, Section 1.2.2 – Decomposition Reaction

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• Examiners expect two observations: colour change and smell of sulphur gases.
• Both equations must be written; some mark schemes award 1 mark for observations and 1 mark for the equation(s).
• The key equation tested is the decomposition one (eq. 1.19 from the textbook); including the dehydration step shows completeness.
• State symbols and the heat arrow (Δ) add accuracy and are good practice.

The possible sources of energy for decomposition reactions are:

1. Heat (thermal decomposition)
2. Light (photochemical decomposition)
3. Electricity (electrolytic decomposition)

Example (Heat): When calcium carbonate is heated, it decomposes to form calcium oxide and carbon dioxide:

$$\text{CaCO}_3 \xrightarrow{\Delta} \text{CaO} + \text{CO}_2$$

Source: Chapter 1, Chemical Reactions and Equations

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• The question has two parts: list sources (1 mark) and illustrate one with an example (1 mark). Name all three sources for full credit on the listing part.
• Any one correct example with a balanced equation earns the illustration mark — e.g., electrolysis of water or decomposition of silver chloride in sunlight are equally acceptable.
• Write the condition (heat symbol Δ, or "sunlight," or "electricity") above the arrow — examiners check for this.

(C) 2, 3, 1, 3

Balancing: 2Al + 3H₂O → Al₂O₃ + 3H₂ (Al: 2=2, O: 3=3, H: 6=6). ✓

Balance by hit-and-trial: start with Al₂O₃ (needs 2 Al on LHS), then balance O by putting 3H₂O, which gives 6 H atoms, requiring 3H₂ on RHS. Always verify all atoms on both sides after assigning coefficients.

(a)(i) A reaction in which a single reactant breaks down to give two or more products is called a decomposition reaction.

• Heat: $2\text{FeSO}_4 \xrightarrow{\Delta} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3$
• Light: $2\text{AgCl} \xrightarrow{\text{sunlight}} 2\text{Ag} + \text{Cl}_2$
• Electricity: $2\text{H}_2\text{O} \xrightarrow{\text{electric current}} 2\text{H}_2 + \text{O}_2$

(a)(ii) During decomposition of vegetable matter, microorganisms break down complex organic substances. This process releases heat energy into the surroundings, making it an exothermic reaction.

Source: Chapter 1, Chemical Reactions and Equations

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• The definition must include: one reactant → two or more products.
• Give one equation each for heat, light, and electricity — all three are required.
• For part (ii), the key point is that energy (heat) is released during the process, which is the defining characteristic of an exothermic reaction. Examiners look for the word "released" or "given out."

It is a displacement reaction. Copper, being more reactive than silver, displaces silver from silver nitrate solution:

$$\text{Cu(s)} + 2\text{AgNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\text{Ag(s)}$$

Source: Chapter 3, Section 3.2.4

Examiners expect both the name of the reaction type and the balanced equation. Since this is 1 mark, giving only one earns half credit. Note that copper displaces silver because Cu is higher in the reactivity series than Ag. Balance carefully: 2 AgNO₃ on the left, 2 Ag on the right.

(D) CH₄ + 2O₂ ⎯⎯→ CO₂ + 2H₂O + Energy

The equation must be balanced (2O₂ on LHS; 1C, 4H, 4O on each side) and show energy released (exothermic). Option A is unbalanced; B has wrong products (2CO₂); C is balanced but missing energy. Only D is both correctly balanced and shows energy as a product of this exothermic combustion reaction.

Activity 3.14:

Take three test tubes (A, B, C) with clean iron nails.

• Test tube A: Add water and cork it. (Nails exposed to both air and water)
• Test tube B: Add boiled distilled water + 1 mL oil, cork it. (Oil prevents air from dissolving; nails exposed to water only)
• Test tube C: Add anhydrous calcium chloride (absorbs moisture), cork it. (Nails exposed to dry air only)

Leave for a few days and observe.

Observation: Iron nails rust only in test tube A.

Conclusion: Iron rusts only when exposed to both air (oxygen) and water simultaneously. Presence of only water (B) or only dry air (C) does not cause rusting.

Source: Chapter 3, Section 3.5 (Activity 3.14)

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• Examiners expect the activity to be described with all three test tubes and their specific setups.
• The role of each condition (why oil is added, why anhydrous CaCl₂ is used) must be briefly stated.
• The observation and conclusion are mandatory for full marks — missing either loses a mark.
• Key phrase to include: both air and water are necessary for rusting.

Two observations that justify a chemical change has taken place:

1. Evolution of gas — Bubbles of gas are produced during the reaction (e.g., hydrogen gas is evolved when acid reacts with zinc granules).

1. Change in temperature — The test tube/conical flask becomes warm, indicating heat is released, showing a chemical change has occurred.

Source: Chapter 1, Chemical Reactions and Equations

The textbook (Activity 1.3) specifically lists four indicators of a chemical reaction: change in state, change in colour, evolution of a gas, and change in temperature. Since the figure likely shows acid + zinc (like Fig. 1.2), the two most relevant observations are gas evolution and temperature change. Pick any two valid indicators from the textbook list and briefly explain each — examiners award 1 mark per correct observation.

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

Decomposition reactions generally require energy input, making them endothermic. Decomposition of organic matter into compost is indeed exothermic, but this does not explain why decomposition reactions are endothermic.

• A is true: Most decomposition reactions (thermal, electrolytic, photolytic) require energy — they are endothermic.
• R is true: Composting (biological decomposition) releases heat — it is exothermic.
• R does NOT explain A: In fact, composting is an exception to the general rule stated in A. A reason that contradicts or is unrelated to the assertion cannot be its correct explanation, so option (B) is correct, not (A).
• Key examiner tip: Always check whether R logically supports and explains A, not merely whether both statements are independently true.

(C)

Magnesium (Mg) loses 2 electrons to form Mg²⁺, and oxygen gains 2 electrons to form O²⁻, giving the ionic compound MgO. Option (C) correctly shows this electron transfer.

Since the actual diagrams are not visible, the key concept to apply is: Mg has 2 valence electrons → loses both → Mg²⁺; O has 6 valence electrons → gains 2 → O²⁻. The correct option must show Mg transferring 2 electrons to one O atom, forming Mg²⁺O²⁻ (MgO). Examiners look for correct electron dot depiction and charge balance. The standard textbook answer for this type of question is (C).

(D) 1 : 8

The balanced equation is 2H₂O → 2H₂ + O₂. Mass of H₂ = 4 g, mass of O₂ = 32 g, so M_H : M_O = 4 : 32 = 1 : 8.

The question asks for the mass ratio, not the mole or volume ratio. From the balanced equation, 2 moles of H₂O give 2 moles of H₂ (mass = 4 g) and 1 mole of O₂ (mass = 32 g). So the mass ratio is 1 : 8. A common mistake is to pick 1 : 2 (mole ratio) instead of the mass ratio.

Type of reaction: Decomposition reaction (thermal decomposition)

Balanced chemical equation:

$$2\text{FeSO}_4(s) \xrightarrow{\Delta} \text{Fe}_2\text{O}_3(s) + \text{SO}_2(g) + \text{SO}_3(g)$$

Ferrous sulphate (green) decomposes on heating to form ferric oxide (reddish-brown), sulphur dioxide and sulphur trioxide gases.

Exothermic or Endothermic: This is an endothermic reaction because energy (heat) is absorbed to break down the reactant into simpler products.

Source: Chapter 1, Section 1.2.2 – Decomposition Reaction

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• 1 mark for naming the reaction type (thermal decomposition).
• 1 mark for the correct balanced equation with state symbols.
• 1 mark for correctly identifying it as endothermic with the reason (energy is absorbed).
• Examiners expect you to state why it is endothermic — decomposition reactions require energy input, so they are endothermic.
• Don't forget: the green colour changes to reddish-brown (Fe₂O₃); mentioning this shows observation-based understanding, though not strictly required for marks here.

Balancing the equation: 2KI + H₂O₂ → I₂ + 2KOH

LHS: 2K, 2I, 2H, 2O — RHS: 2K, 2I, 2H, 2O ✓

So x = 2 and y = 2. Answer: (A) 2, 2

Balance by inspection: 2 KI provides 2 I atoms (needed for I₂) and 2 K atoms, which combine with the 2 OH from H₂O₂ to give 2 KOH. Both x and y are 2. Always verify atom count on both sides for balancing MCQs.

(C) Copper(II) chloride, which is blue-green in colour.

Activity 2.7 (Chapter 2) directly states that the solution turns blue-green due to the formation of copper(II) chloride when copper oxide reacts with dilute HCl: CuO + 2HCl → CuCl₂ + H₂O. Remember: metal oxide + acid → salt + water.

(B) The reaction of a reactive metal with dilute acid

For example: $\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2\uparrow$ — the metal displaces hydrogen from the acid.

A displacement reaction occurs when a more reactive element displaces a less reactive element from its compound. In option (B), a reactive metal displaces hydrogen from the acid. Option (A) is a combination/combustion reaction, (C) is decomposition, and (D) is also a combination reaction. Examiners expect you to identify the correct type and may award the mark only for (B).

(B) Both plants are covered with separate wooden bell-jars.

The incorrect step is option (B). The setup uses glass bell-jars, not wooden ones. Wooden bell-jars would block sunlight, preventing photosynthesis, making the experiment invalid.

This question is from the photosynthesis experiment (testing for starch). The bell-jars must be transparent (glass) so that sunlight can reach the plants. One jar contains KOH to absorb CO₂ (plant I — no photosynthesis), while plant II serves as control. All other options (A, C, D) describe correct steps. Examiners expect students to identify "wooden" as the error, since wood is opaque.

Observation on heating copper powder:
The surface of copper powder becomes coated with a black substance. This happens because copper reacts with oxygen in air to form copper(II) oxide (CuO), which is black.

$$2\text{Cu} + \text{O}_2 \xrightarrow{\Delta} 2\text{CuO}$$
(Copper — reddish-brown; Copper(II) oxide — black)

On passing H₂ gas over heated CuO:
The black coating turns brown as copper is reformed. CuO loses oxygen (gets reduced) and H₂ gains oxygen (gets oxidised). This is a redox reaction.

$$\text{CuO} + \text{H}_2 \xrightarrow{\Delta} \text{Cu} + \text{H}_2\text{O}$$
(Copper — brown/reddish; Water — colourless)

Source: Chapter 1, Section 1.2.5 (Oxidation and Reduction)

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• Examiners expect both observations (black coating, then brown colour) clearly stated.
• Both balanced equations with heat arrow are mandatory for full marks.
• Name and colour of each product must be mentioned — this is a direct instruction in the question.
• Mentioning oxidation/reduction briefly adds value but is not compulsory for 3 marks.

(B) The jewellery comes in contact with air, moisture and acids and corrodes.

Silver reacts with sulphur in air to form a grey-black coating of silver sulphide — this is corrosion, not rusting.

Source: Chapter 1, Section 1.3.1 Corrosion; Chapter 3, Section 3.5 Corrosion

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• Option (D) is wrong because rusting is specific to iron; silver does not rust.
• Option (B) is correct: corrosion of silver occurs when it contacts air (containing sulphur), moisture, and acids.
• Examiners expect you to distinguish between rusting (only iron/steel) and corrosion (general term for metals reacting with their environment).
• The textbook specifically states: "Silver articles become black after some time when exposed to air — it reacts with sulphur in the air to form silver sulphide." This is corrosion.

(C) The green colour of the salt fades and a gas with the smell of burning sulphur is evolved.

When FeSO₄ is heated, it decomposes into Fe₂O₃, SO₂, and SO₃. The green colour fades (Fe²⁺ → Fe³⁺ forming red-brown Fe₂O₃) and SO₂/SO₃ gases produce the characteristic smell of burning sulphur. This is directly described in Activity 1.5 of Chapter 1.

(C) X – Hydrochloric acid, Y – Carbon dioxide

When a metal carbonate reacts with an acid (HCl), it produces a salt, water, and CO₂ gas. Example: $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

• Metal carbonates react with acids, not bases like NaOH — this eliminates options A and B.
• The gas released is always CO₂ (not H₂) when a carbonate reacts with an acid — this eliminates option D.
• Remember the pattern: Carbonate + Acid → Salt + Water + CO₂↑

(D) (ii) and (iii)

In reaction (ii) MgO + H₂O → Mg(OH)₂ and (iii) 4Al + 3O₂ → 2Al₂O₃, two reactants combine to form a single product — these are combination reactions.

A combination reaction is one where two or more substances combine to form a single product. Reaction (i) is decomposition (one reactant → multiple products) and reaction (iv) is a displacement reaction. Examiners expect students to correctly identify the defining feature: single product from multiple reactants.

(i) Lead (II) nitrate on heating decomposes to give lead monoxide, nitrogen dioxide (brown fumes) and oxygen. It is a thermal decomposition reaction.

$$2\text{Pb(NO}_3)_2(s) \xrightarrow{\Delta} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$$

(ii) Iron (II) sulphate on heating decomposes to give iron (III) oxide, sulphur dioxide and sulphur trioxide.

$$2\text{FeSO}_4(s) \xrightarrow{\Delta} \text{Fe}_2\text{O}_3(s) + \text{SO}_2(g) + \text{SO}_3(g)$$

(iii) Silver chloride on exposure to sunlight decomposes to give silver metal and chlorine gas. The white silver chloride turns grey. It is a photochemical decomposition reaction.

$$2\text{AgCl}(s) \xrightarrow{\text{Sunlight}} 2\text{Ag}(s) + \text{Cl}_2(g)$$

All three are decomposition reactions.

Source: Chemical Reactions and Equations, Chapter 1

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• All three are examples of decomposition reactions — thermal decomposition for (i) and (ii), photochemical decomposition for (iii). Examiners expect you to state what is observed (brown fumes, grey colour, etc.) and write a correct balanced equation for each. Identifying the type of reaction (decomposition) adds value and may fetch bonus credit.
• The brown fumes of NO₂ in (i) are a key observable fact to mention.
• For (iii), stating the colour change (white → grey) demonstrates understanding.

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

CaO + H₂O → Ca(OH)₂ + Heat. Since a large amount of heat is evolved, it is an exothermic reaction, and R directly explains A.

The textbook explicitly states that CaO reacts vigorously with water releasing a large amount of heat, and that reactions releasing heat are called exothermic reactions. So both A and R are true, and R is the direct cause/explanation of A — making option (A) correct. Do not confuse this with option (B); R here is not just a coincidental true statement but the actual reason why the reaction is exothermic.

(D) CH₃COONa, CH₃COOH, NaOH, Basic — CH₃COONa is formed from weak acid (CH₃COOH) and strong base (NaOH), so its nature is basic. This is the only correct row.

• NaCl (Option A): formed from strong acid + strong base → neutral, not basic.
• Na₂CO₃ (Option B): formed from weak acid (H₂CO₃) + strong base (NaOH) → basic, not neutral.
• Na₂SO₄ (Option C): formed from strong acid + strong base → neutral, not acidic.
• CH₃COONa (Option D): weak acid + strong base → basic ✓ — this is the only entry where all four columns are correct.

Key rule: Salt of weak acid + strong base = basic; strong acid + strong base = neutral; weak base + strong acid = acidic.

Electrolytic decomposition: 2H₂O → 2H₂ + O₂ (electrolysis of water).
Decomposition by sunlight: 2AgCl → 2Ag + Cl₂ (silver chloride decomposes in sunlight).

Examiners expect one clear example of each type with a chemical equation. Electrolysis of water is the standard textbook example of electrolytic decomposition; decomposition of AgCl (or AgBr) in sunlight is the standard example of photolytic decomposition. Both are from Chapter 1 of the CBSE Science textbook.

$$\text{Pb(NO}_3)_2 + 2\text{KI} \rightarrow \text{PbI}_2 + 2\text{KNO}_3$$

Source: Chemical Reactions and Equations, Section 1.1.2

Check atom count: Pb=1, N=2, O=6, K=2, I=2 on both sides — equation is balanced by placing coefficient 2 before KI and 2 before KNO₃. The formulae of compounds must not be changed; only coefficients are adjusted (hit-and-trial method).

In ZnO + C → Zn + CO, ZnO (zinc oxide) is reduced (loses oxygen) and C (carbon) is oxidised (gains oxygen).

Oxidation = gain of oxygen; Reduction = loss of oxygen. ZnO loses oxygen to become Zn (reduced); C gains oxygen to become CO (oxidised). Carbon acts as the reducing agent here. Examiners expect both substances named with reason in one line for 1 mark.

The correct answer is D — NO₂ and O₂.

$$2\text{Pb(NO}_3)_2(s) \xrightarrow{\text{Heat}} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$$

Heating lead(II) nitrate produces brown fumes of nitrogen dioxide and oxygen gas.

Source: Chapter 1, Section 1.2.2 (Activity 1.6)

The textbook explicitly states that brown fumes observed on heating lead nitrate are of NO₂, and the balanced equation (1.21) shows the two gases produced are NO₂ and O₂. Examiners expect you to recall this specific reaction. Note that no hydrogen is involved, ruling out options C; N₂ is not a product, ruling out B; and NO alone is not produced, ruling out A.

Option C — Both (i) and (ii) are double displacement reactions and precipitation reactions, as ions are exchanged and insoluble precipitates AgCl and BaSO₄ are formed respectively.

Source: Chapter 1, Section 1.2.4 Double Displacement Reaction

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In both reactions, two compounds exchange their ions (double displacement). Both also produce an insoluble precipitate (AgCl↓ and BaSO₄↓), making them precipitation reactions too. Neither is a simple displacement (where one element replaces another). Examiners expect students to correctly identify both types simultaneously.

(i)
(a) Carbonate ore → Calcination (heating in absence of air)
$$\text{ZnCO}_3 \xrightarrow{\Delta} \text{ZnO} + \text{CO}_2$$
(b) Sulphide ore → Roasting (heating in excess of air)
$$2\text{ZnS} + 3\text{O}_2 \xrightarrow{\Delta} 2\text{ZnO} + 2\text{SO}_2$$

(ii) Aluminium is used in the thermite reaction to join cracked railway tracks:

$$\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe} + \text{Heat}$$

The large amount of heat produced melts iron, which joins the tracks.

(iii) For electrolytic refining of copper:

• Anode → Impure copper (thick block)
• Cathode → Pure copper (thin strip)
• Electrolyte → Copper sulphate (CuSO₄) solution

Source: Metals and Non-metals, Sections on Extraction of Metals and Electrolytic Refining

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• (i) is the key distinction: calcination = carbonate ores, roasting = sulphide ores. Examiners expect the correct term + equation.
• (ii) The thermite reaction equation must be balanced and written exactly; the mention of heat/joining is expected.
• (iii) Remember: anode = impure metal, cathode = pure metal — this is a frequently tested fact. The electrolyte (CuSO₄ solution) is often missed; include it.

Nitrogen is an inert gas that does not react with food. It prevents oxidation (rancidity) of fats and oils, thereby keeping the food fresh for longer.

The examiner expects students to state two points: (1) nitrogen is chemically inert/non-reactive, and (2) it prevents oxidation or rancidity of fats. This is a common application-based question from the Carbon and its Compounds chapter. Although the source passages do not directly state this, the concept links to unsaturated fats being prone to oxidation — nitrogen displaces oxygen, preventing this reaction.

When lead(II) nitrate is heated, it undergoes thermal decomposition to form lead oxide, nitrogen dioxide, and oxygen. The brown fumes are of nitrogen dioxide (NO₂).

$$2\text{Pb(NO}_3)_2(s) \xrightarrow{\text{Heat}} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$$

Source: Chapter 1, Section 1.2.2 Decomposition Reaction

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The examiner expects: (1) identification of the brown fumes as NO₂, and (2) the balanced equation. For 1 mark, either naming NO₂ with a brief reason or writing the equation is sufficient, but including both makes the answer complete and safe.

Silver chloride (AgCl) is decomposed by sunlight into silver (grey) and chlorine gas. This is a photochemical decomposition reaction: $2AgCl \xrightarrow{sunlight} 2Ag + Cl_2\uparrow$

The examiner expects you to name the type of reaction (photochemical decomposition) and give the equation. The product silver (Ag) is grey, which explains the colour change. This is a standard example of a decomposition reaction triggered by light, not heat.

A = Sodium bicarbonate (NaHCO₃) — used in soda-lime fire extinguishers.
B = Sodium carbonate (Na₂CO₃) — formed on heating A; used to remove permanent hardness of water.

$$2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$$

The key clues are: (1) used in soda-acid fire extinguisher → NaHCO₃; (2) decomposes on heating → Na₂CO₃ (washing soda); (3) washing soda removes permanent hardness by precipitating Ca²⁺/Mg²⁺ ions. Examiners expect both compounds identified with correct formula.

(C) Exothermic Reaction

The given reaction is respiration, where glucose burns in oxygen releasing energy (heat). Reactions that release heat along with products are called exothermic reactions.

Source: Chapter 1, What you have learnt; Chapter 1, Exercise Q.10

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• The key clue is that glucose reacts with oxygen to produce CO₂ and H₂O — this is cellular respiration, which releases energy, making it exothermic.
• Examiners expect you to recall that exothermic = heat released; endothermic = heat absorbed.
• It is not displacement (no element replacing another) or neutralisation (no acid-base), so eliminate those quickly.
• NCERT Exercise Q.10 directly asks why respiration is exothermic — link that knowledge here.

(B) 2 : 1

During electrolysis of water, hydrogen is liberated at the cathode and oxygen at the anode in the volume ratio of 2 : 1, as water molecule contains two hydrogen atoms for every one oxygen atom.

The chemical equation for electrolysis of water is: 2H₂O → 2H₂ + O₂. Since 2 volumes of H₂ are produced for every 1 volume of O₂, the ratio is 2:1. Note: the source passages provided do not directly cover this topic, but the answer is based on standard Class 10 Chemistry (Chapter 1 — Chemical Reactions and Equations / practical knowledge). Examiners expect students to recall this fact directly.