NCERT Exemplar Solutions Class 10 Science Chemical Reactions and Equations — Multiple Choice Questions Q1–18

NCERT Exemplar Problems, Class 10 Science — Chemical Reactions and Equations, Multiple Choice Questions. All 18 questions solved.

This is the exemplar’s opening run of eighteen multiple-choice questions on the first chapter of the Class 10 Science course, and almost every one of them turns on a single skill: deciding what type a reaction is, and whether energy goes in or comes out. Three definitions carry the whole exercise.

• The five reaction types. Combination ($\mathrm{A} + \mathrm{B} \rightarrow \mathrm{AB}$, one product), decomposition ($\mathrm{AB} \rightarrow \mathrm{A} + \mathrm{B}$, one reactant), displacement ($\mathrm{A} + \mathrm{BC} \rightarrow \mathrm{AC} + \mathrm{B}$, one element swapped in), double displacement ($\mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB}$, two partners exchanged) and redox, which cuts across all of them.
• Oxidation is gain of oxygen or loss of hydrogen; reduction is the reverse. The substance that causes oxidation is the oxidising agent and is itself reduced — which is the swap the exercise tests over and over.
• Exothermic releases heat, endothermic absorbs it. Reactions are not the only candidates: dissolution, dilution, evaporation and sublimation all count as processes here.

Key insight. Nearly every wrong option in this exercise is a correct statement about the wrong substance — water reduced instead of water reducing, the oxidising agent named as the reducing agent, the exothermic beaker named as the endothermic one. So read each statement twice: once for whether it is true, and once for whether it is true of the thing the question asked about.

Choose the correct answer from the given four options in the following questions:

Question 1

Which of the following is not a physical change?

Solution. A physical change alters the state or the appearance of a substance but leaves its chemical identity alone — the same molecules are still there afterwards, and the change can usually be reversed by heating or cooling.

Boiling, melting and dissolving all pass that test. Water vapour is still $\mathrm{H_2O}$, melted ice is still $\mathrm{H_2O}$, and salt dissolved in water can be recovered unchanged by evaporating the water. Burning LPG is different: the butane and propane molecules are destroyed and new ones — carbon dioxide and water — appear in their place, with a large release of heat. You cannot get the LPG back by cooling the exhaust.

Question 2

The following reaction is an example of a

$$4\mathrm{NH_3}(g) + 5\mathrm{O_2}(g) \rightarrow 4\mathrm{NO}(g) + 6\mathrm{H_2O}(g)$$

(i) displacement reaction (ii) combination reaction (iii) redox reaction (iv) neutralisation reaction

Solution. Statement (iii) is certain, and it is the anchor. Ammonia loses hydrogen and gains oxygen as it becomes $\mathrm{NO}$, so it is oxidised; oxygen gains hydrogen as it becomes water, so it is reduced. Both happen in the same reaction, which is exactly what makes it a redox reaction. That immediately kills option (a), which does not contain (iii).

Statement (iv) is false — neutralisation is an acid reacting with a base to give a salt and water, and there is no acid and no base here. So (d) goes too.

Statement (ii) is false. A combination reaction produces a single product; this one produces two. So (b) goes, and only (c) survives.

That leaves statement (i), displacement, which is the debatable one. The textbook’s own pattern for displacement is $\mathrm{A} + \mathrm{BC} \rightarrow \mathrm{AC} + \mathrm{B}$, and this reaction does not fit it cleanly — two products carry oxygen. The reading the exemplar intends is that oxygen has displaced hydrogen from ammonia, and that is how the key marks it. It is worth being honest that the argument from elimination above is the reliable one: three of the four statements can be settled beyond doubt, and they leave exactly one option standing.

Question 3

Which of the following statements about the given reaction are correct?

$$3\mathrm{Fe}(s) + 4\mathrm{H_2O}(g) \rightarrow \mathrm{Fe_3O_4}(s) + 4\mathrm{H_2}(g)$$

(i) Iron metal is getting oxidised (ii) Water is getting reduced (iii) Water is acting as reducing agent (iv) Water is acting as oxidising agent

Solution. Track the oxygen and the hydrogen separately.

Iron starts as the free metal and ends up in $\mathrm{Fe_3O_4}$ — it has gained oxygen, so iron is oxidised. Statement (i) is true.

Water starts as $\mathrm{H_2O}$ and ends up as $\mathrm{H_2}$ — it has lost oxygen, so water is reduced. Statement (ii) is true.

Now the agents, which is where the question is really aimed. The substance that is reduced is the one that took oxygen away from nothing and gave it to something else; it is the oxidising agent. Water gave its oxygen to the iron, so water oxidised the iron: statement (iv) is true and statement (iii) is false.

A useful way to keep this straight is that the agent and its own fate are always opposite. The oxidising agent gets reduced; the reducing agent gets oxidised. Here iron is the reducing agent and water is the oxidising agent.

Question 4

Which of the following are exothermic processes?

(i) Reaction of water with quick lime (ii) Dilution of an acid (iii) Evaporation of water (iv) Sublimation of camphor (crystals)

Solution. Two of these release heat and two absorb it, and the split falls along an easy line: things that pull particles together release energy, things that pull them apart need energy supplied.

Quick lime slaking, $\mathrm{CaO} + \mathrm{H_2O} \rightarrow \mathrm{Ca(OH)_2}$, is the standard classroom example of a violently exothermic reaction — the mixture hisses and steams. Diluting a concentrated acid is exothermic too, because the acid molecules become hydrated and that hydration releases energy; it is why the safety rule is to add acid to water and never the reverse.

Evaporation and sublimation both take a substance from a condensed state to a gas, which means breaking the forces holding the particles together. Both absorb heat, and both are endothermic. Evaporating sweat cooling your skin is the everyday proof.

Question 5

Three beakers labelled as A, B and C each containing 25 mL of water were taken. A small amount of NaOH, anhydrous $\mathrm{CuSO_4}$ and NaCl were added to the beakers A, B and C respectively. It was observed that there was an increase in the temperature of the solutions contained in beakers A and B, whereas in case of beaker C, the temperature of the solution falls. Which one of the following statement(s) is(are) correct?

(i) In beakers A and B, exothermic process has occurred. (ii) In beakers A and B, endothermic process has occurred. (iii) In beaker C exothermic process has occurred. (iv) In beaker C endothermic process has occurred.

Solution. The experiment hands you the answer directly, provided you have the direction of the definition the right way round. A process that releases heat into its surroundings makes those surroundings warmer; a process that absorbs heat makes them cooler.

The solution in a beaker is the surroundings for the dissolving solid. Beakers A and B warmed up, so heat was released into them: sodium hydroxide dissolving and anhydrous copper sulphate hydrating are both exothermic. Statement (i) is true and (ii) is false.

Beaker C cooled down, so heat was drawn out of it: dissolving sodium chloride is slightly endothermic. Statement (iv) is true and (iii) is false.

Only option (c) pairs (i) with (iv).

Question 6

A dilute ferrous sulphate solution was gradually added to the beaker containing acidified permanganate solution. The light purple colour of the solution fades and finally disappears. Which of the following is the correct explanation for the observation?

Solution. The purple colour belongs to the permanganate ion, $\mathrm{MnO_4^-}$. It disappearing means the permanganate is being consumed — converted into manganese in a lower, colourless oxidation state.

Acidified potassium permanganate is the standard strong oxidising agent of the school laboratory. It oxidises iron(II) to iron(III), and in doing so is itself reduced, which is precisely what decolourises it. That is option (a).

Option (b) has the roles exactly reversed, which is the trap this question is built on. Option (c) fails an easy observational test: dilution would fade the colour gradually and evenly, not remove it, and the question says it finally disappears. Option (d) is wrong about the chemistry — permanganate is perfectly stable in acid on its own, and it is the reaction with the iron(II), not any spontaneous decomposition, that removes it.

The fact that the colour vanishes only when the ferrous solution is added, and stops fading when you stop adding, is the decisive evidence that a reaction is doing it.

Question 7

Which among the following is(are) double displacement reaction(s)?

(i) $\mathrm{Pb} + \mathrm{CuCl_2} \rightarrow \mathrm{PbCl_2} + \mathrm{Cu}$ (ii) $\mathrm{Na_2SO_4} + \mathrm{BaCl_2} \rightarrow \mathrm{BaSO_4} + 2\mathrm{NaCl}$ (iii) $\mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2}$ (iv) $\mathrm{CH_4} + 2\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O}$

Solution. Double displacement means two compounds swap partners: $\mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB}$. The test is easy to apply — both sides must have two compounds, and the ions must have changed partners.

In (i) the left-hand side has a free element, lead, so it cannot be a double displacement; lead displaces copper and this is a single displacement. In (iii) two reactants give one product, a combination. In (iv) methane burns, which is a combustion and a redox reaction, not a partner swap.

In (ii) sodium sulphate and barium chloride both arrive as compounds and leave as different compounds: barium has taken the sulphate and sodium has taken the chloride. It is also a precipitation reaction, since barium sulphate is the insoluble white solid that drops out — and precipitation reactions are always a special case of double displacement.

Question 8

Which among the following statement(s) is(are) true? Exposure of silver chloride to sunlight for a long duration turns grey due to

(i) the formation of silver by decomposition of silver chloride (ii) sublimation of silver chloride (iii) decomposition of chlorine gas from silver chloride (iv) oxidation of silver chloride

Solution. The reaction is the photochemical decomposition

$$2\mathrm{AgCl}(s) \xrightarrow{\text{Sunlight}} 2\mathrm{Ag}(s) + \mathrm{Cl_2}(g)$$

and the grey colour is the finely divided silver metal left behind. Statement (i) says exactly that, and it is true.

Statement (iii) is the one that catches people, because chlorine really is released. But read what it claims: the decomposition of chlorine gas. Chlorine gas is a product here, not a reactant, and nothing decomposes it. The compound that decomposes is silver chloride. A statement can name a real part of the reaction and still be false about what is happening to it.

Statement (ii) is wrong because sublimation would take the silver chloride away as a vapour and leave nothing grey behind. Statement (iv) is wrong in direction: silver is going from $\mathrm{Ag^+}$ to $\mathrm{Ag}$, a gain of electrons, which is reduction.

Question 9

Solid calcium oxide reacts vigorously with water to form calcium hydroxide accompanied by liberation of heat. This process is called slaking of lime. Calcium hydroxide dissolves in water to form its solution called lime water. Which among the following is(are) true about slaking of lime and the solution formed?

(i) It is an endothermic reaction (ii) It is an exothermic reaction (iii) The pH of the resulting solution will be more than seven (iv) The pH of the resulting solution will be less than seven

Solution. The question has already told you the answer to the first half — it says the reaction is accompanied by liberation of heat, which is the definition of exothermic. Statement (ii) is true, (i) false.

$$\mathrm{CaO}(s) + \mathrm{H_2O}(l) \rightarrow \mathrm{Ca(OH)_2}(aq) + \text{heat}$$

For the second half, calcium hydroxide is a base — a metal hydroxide that releases $\mathrm{OH^-}$ ions in water. A basic solution has a pH above seven, so statement (iii) is true and (iv) false. Lime water is in fact strongly basic, around pH 12.

Note that (a) and (d) are self-contradictory pairs: a reaction cannot be both endothermic and exothermic, and a solution cannot have pH both above and below seven. Spotting that eliminates half the options before any chemistry is done.

Question 10

Barium chloride on reacting with ammonium sulphate forms barium sulphate and ammonium chloride. Which of the following correctly represents the type of the reaction involved?

(i) Displacement reaction (ii) Precipitation reaction (iii) Combination reaction (iv) Double displacement reaction

Solution. Write the equation out before classifying it:

$$\mathrm{BaCl_2}(aq) + (\mathrm{NH_4})_2\mathrm{SO_4}(aq) \rightarrow \mathrm{BaSO_4}(s)\downarrow + 2\mathrm{NH_4Cl}(aq)$$

Barium has swapped its chloride for a sulphate and ammonium has swapped its sulphate for a chloride, so this is a double displacement — statement (iv) is true. And one of the products, barium sulphate, is insoluble and comes out of solution as a white solid, so it is also a precipitation reaction — statement (ii) is true.

This is the point of the question. Precipitation and double displacement are not alternatives to choose between; a precipitation reaction is a double displacement in which one product happens to be insoluble. Both labels apply at once, which is why the answer is a pair and not a single statement. Options (b) and (c) are each half right, and being half right is what makes them tempting.

Statement (i) is false — there is no free element anywhere — and (iii) is false, since two products are formed.

Question 11

Electrolysis of water is a decomposition reaction. The mole ratio of hydrogen and oxygen gases liberated during electrolysis of water is

Solution. Read the ratio straight off the balanced equation:

$$2\mathrm{H_2O}(l) \xrightarrow{\text{Electricity}} 2\mathrm{H_2}(g) + \mathrm{O_2}(g)$$

Two moles of hydrogen for every one mole of oxygen, so the ratio is $2:1$. This is why, in the classroom electrolysis of water, the test tube over the cathode collects twice the volume of gas the one over the anode does — equal volumes of gases at the same temperature and pressure contain equal numbers of moles.

Option (d) is the same ratio written backwards, and it is the mistake this question is set to catch. Attach the number to the gas as you read: hydrogen is the one there is more of, because each water molecule carries two hydrogen atoms and only one oxygen atom.

Question 12

Which of the following is(are) an endothermic process(es)?

(i) Dilution of sulphuric acid (ii) Sublimation of dry ice (iii) Condensation of water vapours (iv) Evaporation of water

Solution. Apply the same rule as Question 4, and note that this question and that one are near-mirror images: here you are asked for the endothermic ones.

Diluting sulphuric acid releases a great deal of heat — enough to make the mixture spit if water is poured onto the acid — so (i) is exothermic. Condensation is the reverse of evaporation, so it must release the heat that evaporation absorbed; (iii) is exothermic.

Sublimation of dry ice takes solid carbon dioxide directly to gas, breaking all the forces holding the solid together, and it absorbs heat — which is why dry ice keeps things cold. Evaporation of water does the same. Both are endothermic.

Question 13

In the double displacement reaction between aqueous potassium iodide and aqueous lead nitrate, a yellow precipitate of lead iodide is formed. While performing the activity if lead nitrate is not available, which of the following can be used in place of lead nitrate?

Solution. The substitute has to do the same job lead nitrate does, and that job is to supply $\mathrm{Pb^{2+}}$ ions in solution:

$$\mathrm{Pb(NO_3)_2}(aq) + 2\mathrm{KI}(aq) \rightarrow \mathrm{PbI_2}(s)\downarrow + 2\mathrm{KNO_3}(aq)$$

Options (c) and (d) contain no lead at all, so neither can give a lead iodide precipitate; they are out immediately.

That leaves two lead compounds, and the word in brackets settles it. Lead sulphate is insoluble — the question tells you so — which means it will not dissociate into free $\mathrm{Pb^{2+}}$ ions in water, and an ion that stays locked in an undissolved solid cannot react. Lead acetate is soluble, releases $\mathrm{Pb^{2+}}$ freely, and gives the same yellow precipitate:

$$\mathrm{(CH_3COO)_2Pb}(aq) + 2\mathrm{KI}(aq) \rightarrow \mathrm{PbI_2}(s)\downarrow + 2\mathrm{CH_3COOK}(aq)$$

Question 14

Which of the following gases can be used for storage of fresh sample of an oil for a long time?

Solution. Oils and fats go rancid because they are slowly oxidised by the oxygen in the air, which produces the stale smell and taste. To keep a sample fresh you flush the container with a gas that will not oxidise the oil — an unreactive one.

Oxygen is the culprit itself, so any option containing it is self-defeating: (a) and (b) are out at once.

Between the two remaining, carbon dioxide is the problem. It is reasonably unreactive, but it dissolves in fats and in any water present, forming carbonic acid, and it is not the gas used commercially. Nitrogen is, and helium is the textbook inert-gas example. Packets of crisps are flushed with nitrogen for exactly this reason.

Question 15

The following reaction is used for the preparation of oxygen gas in the laboratory

$$2\mathrm{KClO_3}(s) \xrightarrow[\text{Catalyst}]{\text{Heat}} 2\mathrm{KCl}(s) + 3\mathrm{O_2}(g)$$

Which of the following statement(s) is(are) correct about the reaction?

Solution. Two features have to be got right, and the options are built so that each wrong one gets exactly one of them wrong.

The type is clear from the shape of the equation: one reactant, two products. That is decomposition, so (b) is out.

The energy direction is given away by the word Heat over the arrow. Heat written above the arrow means heat must be supplied continuously for the reaction to proceed — the reaction absorbs it. So this is endothermic, which rules out (c) and (d), both of which claim heat is released.

Option (d) has a second error worth naming. A photochemical decomposition is one driven by light, as the silver chloride of Question 8 is. This one is driven by heat, so it is a thermal decomposition. The three kinds — thermal, electrolytic and photochemical — are distinguished by what supplies the energy, and the arrow tells you which.

Question 16

Which one of the following processes involve chemical reactions?

Solution. Three of these are physical changes dressed up to look like processes.

Compressing oxygen into a cylinder only pushes the same molecules closer together. Liquefying air cools it until the nitrogen and oxygen condense, but they are still nitrogen and oxygen. Petrol left in an open dish evaporates — it leaves, it does not change.

Heating copper in air is the odd one out, because a new substance appears:

$$2\mathrm{Cu}(s) + \mathrm{O_2}(g) \xrightarrow{\text{Heat}} 2\mathrm{CuO}(s)$$

and you can see it happen: the shiny red-brown wire acquires a black coating of copper(II) oxide. A visible, permanent change of colour with no way back by cooling is a reliable sign that a chemical reaction has occurred.

Question 17

In which of the following chemical equations, the abbreviations represent the correct states of the reactants and products involved at reaction temperature?

Solution. The phrase that decides this is at reaction temperature, and it is easy to read past.

Hydrogen and oxygen are both gases under ordinary conditions, so any equation writing either as a liquid is wrong about the reactants. That removes (a), which liquefies both, and (b), which liquefies the oxygen.

That leaves (c) and (d), differing only in the state of the water. Burning hydrogen in oxygen is strongly exothermic and the flame is extremely hot — far above $100^\circ\mathrm{C}$. At that temperature water cannot exist as a liquid; it comes off as steam. So the product is $\mathrm{H_2O}(g)$, and (d) is correct.

Option (c) would be right if the question asked about the state after everything had cooled to room temperature. It does not.

Question 18

Which of the following are combination reactions?

(i) $2\mathrm{KClO_3} \xrightarrow{\text{Heat}} 2\mathrm{KCl} + 3\mathrm{O_2}$ (ii) $\mathrm{MgO} + \mathrm{H_2O} \rightarrow \mathrm{Mg(OH)_2}$ (iii) $4\mathrm{Al} + 3\mathrm{O_2} \rightarrow 2\mathrm{Al_2O_3}$ (iv) $\mathrm{Zn} + \mathrm{FeSO_4} \rightarrow \mathrm{ZnSO_4} + \mathrm{Fe}$

Solution. A combination reaction has two or more reactants and exactly one product. Counting products is enough to sort this question out, and it takes a few seconds.

(i) has one reactant and two products — that is decomposition, the reverse of what is wanted. (iv) has two reactants and two products, with zinc displacing iron from its sulphate — a displacement reaction.

(ii) and (iii) each produce a single substance: magnesium hydroxide from magnesium oxide and water, aluminium oxide from aluminium and oxygen. Both are combinations.

It is worth noticing that (iii) is also a redox reaction — aluminium gains oxygen — and that this does not stop it being a combination. The five types are not mutually exclusive, and redox in particular overlaps freely with the other four.

Common mistakes

• Swapping the agent and its fate in Question 3 and Question 6. The oxidising agent is the substance that gets reduced, and the reducing agent gets oxidised. Both questions place the reversed statement among the options deliberately, and in Question 6 it is option (b) — the single most-chosen wrong answer in the whole exercise.
• Calling Question 10’s reaction a precipitation reaction instead of a double displacement. It is both, and the answer is a pair. A precipitation reaction is a double displacement whose product happens to be insoluble; the two labels describe different things about the same event.
• Writing the ratio backwards in Question 11. Hydrogen to oxygen is $2:1$, not $1:2$. Attach the number to the gas as you read the balanced equation rather than remembering “two to one” as a bare phrase.
• Ignoring “at reaction temperature” in Question 17. Without that phrase option (c) would be right. Hydrogen burns at well over $1000^\circ\mathrm{C}$, and water is steam at that temperature, not liquid.
• Treating the arrow label as decoration in Question 15. Heat written over an arrow means energy is being supplied, so the reaction is endothermic — and it also tells you the decomposition is thermal rather than photochemical or electrolytic. Two of the four options fail on exactly this reading.
• Missing that lead sulphate’s insolubility is the whole of Question 13. A reagent that will not dissolve cannot supply ions, and a reaction between ions needs them free in solution. The bracketed word “(insoluble)” is the question, not a description.
• Accepting statement (iii) in Question 8 because chlorine really is involved. It says chlorine gas decomposes; chlorine gas is a product. A statement that names a real participant can still be false about what happens to it.
• Assuming a reaction has one type. Question 18’s aluminium reaction is a combination and a redox reaction; Question 7’s barium reaction is a double displacement and a precipitation. Only the pairing combination/decomposition is genuinely exclusive.

Practise next

• Exemplar Short and Long Answer Questions — the same classification skill, but now you have to write and balance the equation yourself instead of recognising one, which is where the marks actually are.
• NCERT textbook Chapter 1 — Chemical Reactions and Equations for the end-of-chapter exercise, including the balancing practice Questions 15 and 18 above assume.
• Textbook in-text questions for the short checks that follow each section, which cover corrosion and rancidity — the background to Question 14.
• Textbook activities for the experiments behind Questions 5, 6 and 13, where the observations these questions describe are actually made.