NCERT Solutions for Class 10 Science Chapter 1 Chemical

Magnesium is reactive enough to combine with oxygen in ordinary air, so a ribbon left on a shelf is already coated with a thin layer of magnesium oxide. That oxide layer is stable and does not burn — it sits between the flame and the fresh metal underneath.

Cleaning the ribbon with sandpaper removes this coating and exposes fresh magnesium so that the reaction you want to observe can actually start (Activity 1.1, p. 1).

1. Step 1: Magnesium reacts with atmospheric oxygen over time: \( 2Mg + O_2 \rightarrow 2MgO \).
2. Step 2: The MgO coating is dull and prevents uniform ignition.
3. Step 3: Rubbing with sandpaper strips this layer, exposing shiny magnesium metal for clean, uniform burning.

Common error: Students write that the ribbon is cleaned “to remove dust.” The dust is irrelevant — the real reason is the magnesium oxide layer.

Each reaction combines or exchanges elements in fixed atomic ratios. Balancing places whole-number coefficients so that atoms of every element are equal on both sides — you never alter a formula’s subscripts to force equality.

Part (i): Hydrogen and chlorine are both diatomic, so each starts as \( H_2 \) and \( Cl_2 \). They combine in a 1:1 ratio to give two molecules of \( HCl \):

\[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \]

Part (ii): Barium chloride is \( BaCl_2 \); aluminium sulphate is \( Al_2(SO_4)_3 \). Exchange the ions: \( Ba^{2+} \) pairs with \( SO_4^{2-} \), and \( Al^{3+} \) pairs with \( Cl^{-} \). Three \( BaCl_2 \) units supply the six chloride ions that two \( AlCl_3 \) units need:

\[ 3BaCl_2(aq) + Al_2(SO_4)_3(aq) \rightarrow 3BaSO_4(s) + 2AlCl_3(aq) \]

Part (iii): Sodium is \( Na \), water is \( H_2O \). Sodium displaces hydrogen from water to form sodium hydroxide (\( NaOH \)) and hydrogen gas. Two Na atoms and two water molecules give two \( NaOH \) and one \( H_2 \), which balances hydrogen (4 on each side):

\[ 2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g) \]

Common error: Writing \( Na_2O \) instead of \( NaOH \) — sodium reacts with water to form a hydroxide and hydrogen gas, not a simple oxide.

State symbols make an equation informative: (s) for solid, (l) for liquid, (g) for gas, (aq) for aqueous solution. A precipitate — an insoluble solid formed when two solutions mix — must carry (s), not (aq), because it drops out of solution.

Part (i): Both reactants are in aqueous solution. \( Ba^{2+} \) meets \( SO_4^{2-} \) and forms insoluble barium sulphate, which precipitates as a white solid. Sodium and chloride remain dissolved:

\[ BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq) \]

Part (ii): An acid-base neutralisation: \( H^+ \) and \( OH^{-} \) join to form water. Sodium chloride stays in solution:

\[ NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l) \]

Common error: Writing \( BaSO_4(aq) \) — barium sulphate is the precipitate, so it must be (s).

The substance used for whitewashing is calcium oxide (\( CaO \)), also called quick lime (p. 6). When you add water to quick lime, it reacts vigorously to form slaked lime — calcium hydroxide, \( Ca(OH)_2 \) — and releases a large amount of heat.

This is the solution that is applied to walls; over two to three days it reacts with atmospheric \( CO_2 \) to form a shiny layer of calcium carbonate (\( CaCO_3 \)), the same compound as marble (p. 7).

1. Step 1: \( X = \) calcium oxide (quick lime), formula \( CaO \).
2. Step 2: Its vigorous reaction with water:

\[ CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) + \text{Heat} \]

Common error: Naming X as calcium hydroxide instead of calcium oxide. Slaked lime is the product of the water reaction, not the starting substance used for whitewashing.

Water has the formula \( H_2O \) — two hydrogen atoms for every one oxygen atom. Electrolytic decomposition splits water into its elemental gases in that same atomic ratio. So the test tube collecting hydrogen holds twice the volume of the one collecting oxygen (Activity 1.7, p. 9).

1. Step 1: Water molecule: \( H_2O \) — 2 H atoms and 1 O atom.
2. Step 2: Splitting two water molecules gives two \( H_2 \) and one \( O_2 \):

\[ 2H_2O(l) \xrightarrow{\text{Electricity}} 2H_2(g) + O_2(g) \]

Step 3: Hydrogen collects at the cathode (negative electrode), oxygen at the anode (positive electrode).

Common error: Swapping the electrodes — hydrogen appears at the cathode, oxygen at the anode, not the other way around.

A more reactive metal displaces a less reactive one from its salt solution. Iron is above copper in the reactivity series, so iron pulls copper out of copper sulphate. The blue colour of the \( Cu^{2+} \) solution fades as copper ions leave the solution, and the iron nail acquires a brown coating of deposited copper metal (Activity 1.9, p. 10).

\[ Fe(s) + CuSO_4(aq) \rightarrow FeSO_4(aq) + Cu(s) \]

Common error: Saying the colour changes because iron dissolves. The colour fades specifically because blue \( Cu^{2+} \) ions are removed from solution and replaced by pale green \( Fe^{2+} \) ions.

In a double displacement reaction, two compounds in solution exchange ions — the positive ion of one compound pairs with the negative ion of the other, and vice versa. If one of the new combinations is insoluble, it drops out as a precipitate. The textbook’s Activity 1.2 (p.

2) uses lead nitrate and potassium iodide; mixing these gives a yellow precipitate of lead iodide.

\[ Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) + 2KNO_3(aq) \]

Common error: Writing this as a displacement reaction because metals are involved. It is double displacement because two whole ionic compounds swap partners — no free element is displaced.

Oxidation and reduction are coupled: one substance gains oxygen (is oxidised) while another loses oxygen (is reduced). They always occur together, which is why such reactions are called redox reactions (p. 12).

Part (i): \( 4Na(s) + O_2(g) \rightarrow 2Na_2O(s) \)

Sodium gains oxygen to become \( Na_2O \), so sodium is oxidised. Oxygen is the oxidising agent — it causes the oxidation and is itself reduced.

Part (ii): \( CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l) \)

Copper oxide loses its oxygen to become copper, so \( CuO \) is reduced. Hydrogen gains oxygen to become water, so hydrogen is oxidised.

Common error: Calling the oxidising agent “oxidised.” The oxidising agent causes oxidation in the other substance but is itself reduced — keep these roles separate.

Track oxygen to each substance separately. Lead oxide loses oxygen to become lead — so lead is reduced and lead oxide is reduced. Carbon gains oxygen to become carbon dioxide — so carbon is oxidised. \( CO_2 \) is a product of carbon’s oxidation, not itself undergoing further oxidation.

• Statement (a) “Lead is getting reduced” — correct (Pb gains electrons, loses O).
• Statement (b) “Carbon dioxide is getting oxidised” — incorrect; \( CO_2 \) is the product, not a reactant being further oxidised.
• Statement (c) “Carbon is getting oxidised” — correct.
• Statement (d) “Lead oxide is getting reduced” — correct.

Final answer: The incorrect statements are (a) and (b), so the answer option is (i).

Common error: Marking statement (a) as incorrect because the wording feels tricky. Statement (a) is factually correct — lead is getting reduced. The only incorrect statement among the first two is (b).

Aluminium is more reactive than iron, so it displaces iron from iron oxide. Because a single element (aluminium) takes the place of another element (iron) in a compound, the textbook classifies this as a displacement reaction (p. 10). It is also a redox reaction, but among the given choices, displacement is the classification the board expects.

Final answer: (d) displacement reaction.

Common error: Calling this a combination reaction because aluminium oxide forms. Aluminium oxide forms by combination, but the overall reaction is displacement — aluminium replaces iron.

Iron is an active metal. It displaces hydrogen from dilute hydrochloric acid, producing iron(II) chloride and hydrogen gas. Because HCl is dilute and iron forms \( Fe^{2+} \) here, the correct salt is \( FeCl_2 \), not \( FeCl_3 \).

\[ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \]

Final answer: (a) Hydrogen gas and iron chloride are produced.

Common error: Writing \( FeCl_3 \) — iron forms \( Fe^{2+} \) with dilute HCl, so the product is \( FeCl_2 \). Students also sometimes think chlorine gas (\( Cl_2 \)) is produced; it is not.

A balanced chemical equation has equal numbers of atoms of each element on both sides of the arrow (p. 3). Equations must be balanced because of the Law of Conservation of Mass: mass can neither be created nor destroyed in a chemical reaction. If the atom counts were unequal, the reaction would imply that atoms appeared or vanished — which is impossible.

Common error: Defining a balanced equation as “same number of molecules on both sides.” Molecules need not be equal — atoms of each element must be equal.

Part (a) Hydrogen + nitrogen → ammonia. Nitrogen and hydrogen are both diatomic. Ammonia is \( NH_3 \), so three \( H_2 \) molecules supply the six hydrogen atoms needed by two \( NH_3 \):

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

Part (b) Hydrogen sulphide burns in air to give water and sulphur dioxide. \( H_2S \) provides H and S; \( O_2 \) supplies oxygen for both water and \( SO_2 \). Balance S first, then H, then O, giving:

\[ 2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(g) + 2H_2O(l) \]

Part (c) Barium chloride + aluminium sulphate → aluminium chloride + barium sulphate precipitate. Three \( BaCl_2 \) units supply six chlorides for two \( AlCl_3 \); the sulphate groups transfer to barium:

\[ 3BaCl_2(aq) + Al_2(SO_4)_3(aq) \rightarrow 2AlCl_3(aq) + 3BaSO_4(s) \]

Part (d) Potassium + water → potassium hydroxide + hydrogen. Potassium behaves like sodium: it forms a hydroxide and releases hydrogen gas:

\[ 2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g) \]

Common error in (b): Forgetting oxygen is diatomic — you need a coefficient of 3 on \( O_2 \) to supply six oxygen atoms. Common error in (d): Writing \( K_2O \) instead of \( KOH \) — potassium in water gives a hydroxide and hydrogen, not a simple oxide.

Part (a): Calcium has two nitrate groups in \( Ca(NO_3)_2 \), so we need two \( HNO_3 \) on the left, which gives two water molecules:

\[ 2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O \]

Part (b): Sodium sulphate (\( Na_2SO_4 \)) needs two sodium atoms, so place 2 before \( NaOH \), which gives two waters:

\[ 2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \]

Part (c): Each ion already matches one-to-one, so the equation is already balanced:

\[ NaCl + AgNO_3 \rightarrow AgCl + NaNO_3 \]

Part (d): \( BaSO_4 \) takes one sulphate group. The two chlorines in \( BaCl_2 \) pair with two hydrogens in \( 2HCl \):

\[ BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl \]

Common error in (c): Adding unnecessary coefficients (e.g. a 2 on every term) to an equation that is already balanced.

Part (a): Calcium hydroxide + carbon dioxide → calcium carbonate + water. This is the whitewashing follow-up reaction (p. 7). One \( Ca(OH)_2 \) has two OH groups; one \( CO_2 \) supplies the carbonate; the leftover H and O form one water:

\[ Ca(OH)_2(aq) + CO_2(g) \rightarrow CaCO_3(s) + H_2O(l) \]

Part (b): Zinc + silver nitrate → zinc nitrate + silver. Zinc forms \( Zn^{2+} \), so it needs two \( AgNO_3 \) to balance charge and nitrate count:

\[ Zn(s) + 2AgNO_3(aq) \rightarrow Zn(NO_3)_2(aq) + 2Ag(s) \]

Part (c): Aluminium + copper chloride → aluminium chloride + copper. Aluminium forms \( Al^{3+} \); two \( AlCl_3 \) need six chlorides, so three \( CuCl_2 \) supply them, giving three copper atoms:

\[ 2Al(s) + 3CuCl_2(aq) \rightarrow 2AlCl_3(aq) + 3Cu(s) \]

Part (d): Barium chloride + potassium sulphate → barium sulphate + potassium chloride. One \( Ba^{2+} \) pairs with one \( SO_4^{2-} \); two \( K^+ \) need two \( Cl^{-} \), which one \( BaCl_2 \) provides:

\[ BaCl_2(aq) + K_2SO_4(aq) \rightarrow BaSO_4(s) + 2KCl(aq) \]

Common error in (b) and (d): Forgetting to multiply \( AgNO_3 \) by 2 or \( KCl \) by 2, which leaves the anion or cation count unbalanced.

Part (a): Potassium bromide + barium iodide → potassium iodide + barium bromide. Two ionic compounds exchange partners — \( K^+ \) swaps with \( Ba^{2+} \), and \( Br^{-} \) swaps with \( I^{-} \). This is a double displacement reaction.

\[ 2KBr(aq) + BaI_2(aq) \rightarrow 2KI(aq) + BaBr_2(s) \]

Part (b): Zinc carbonate → zinc oxide + carbon dioxide. A single reactant breaks into two simpler products on heating — a thermal decomposition reaction (p. 8).

\[ ZnCO_3(s) \xrightarrow{\text{Heat}} ZnO(s) + CO_2(g) \]

Part (c): Hydrogen + chlorine → hydrogen chloride. Two elements combine to form a single compound — a combination reaction (p. 6).

\[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \]

Part (d): Magnesium + hydrochloric acid → magnesium chloride + hydrogen. Magnesium, a metal, displaces hydrogen from the acid — a displacement reaction (p. 10).

\[ Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g) \]

Common error: Misclassifying (a) as a displacement reaction because metal symbols appear. It is double displacement — two ionic compounds swap partners, and no free element is displaced from the start.

Reactions either release energy to the surroundings or absorb energy from them. The difference is the direction of energy flow when bonds break and form.

• Exothermic reaction: Heat is released along with the products — the mixture warms up. Examples: burning of natural gas \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \) (p. 7); quick lime + water \( CaO + H_2O \rightarrow Ca(OH)_2 + \text{Heat} \); respiration.
• Endothermic reaction: Heat or energy is absorbed — the mixture cools or needs a steady energy supply. Examples: thermal decomposition of limestone \( CaCO_3 \xrightarrow{\text{Heat}} CaO + CO_2 \) (p. 8); silver chloride in sunlight \( 2AgCl \xrightarrow{\text{Sunlight}} 2Ag + Cl_2 \) (p. 9); electrolysis of water.

Common error: Calling the decomposition of vegetable matter into compost endothermic — it releases heat, so it is exothermic (p. 7).

During digestion, food breaks down into simpler substances, including glucose. Inside our cells, glucose combines with oxygen and releases energy that the body uses to stay alive. Because energy flows out of the reaction into the body, respiration is exothermic (p. 7, eq. 1.18).

\[ C_6H_{12}O_6(aq) + 6O_2(aq) \rightarrow 6CO_2(aq) + 6H_2O(l) + \text{energy} \]

Common error: Forgetting the energy term on the right side. Writing respiration without “+ energy” hides the very point the question asks for.

A combination reaction builds a single product from two or more reactants. A decomposition reaction does the reverse — one reactant splits into two or more products. One builds up; the other breaks apart. That is why they are opposites (p. 8).

• Combination: \( CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) \) — two reactants form one product.
• Decomposition: \( CaCO_3(s) \xrightarrow{\text{Heat}} CaO(s) + CO_2(g) \) — one reactant forms two products.

Decomposition always needs an energy input because chemical bonds in the reactant must be broken. That energy can arrive as heat, light, or electricity, and each form drives a different reaction.

Heat (thermal decomposition) — lead nitrate breaks into lead oxide, nitrogen dioxide, and oxygen (Activity 1.6, p. 8):

\[ 2Pb(NO_3)_2(s) \xrightarrow{\text{Heat}} 2PbO(s) + 4NO_2(g) + O_2(g) \]

Light (photolytic decomposition) — silver chloride decomposes in sunlight (Activity 1.8, p. 9):

\[ 2AgCl(s) \xrightarrow{\text{Sunlight}} 2Ag(s) + Cl_2(g) \]

Electricity (electrolytic decomposition) — water splits into hydrogen and oxygen (Activity 1.7, p. 9):

\[ 2H_2O(l) \xrightarrow{\text{Electricity}} 2H_2(g) + O_2(g) \]

Common error: Writing a thermal decomposition equation without the “Heat” label above the arrow. The energy source is part of the answer — omitting it loses the point of the question.

The difference is whether a single free element swaps into a compound, or whether two whole ionic compounds exchange partners. One involves an element entering a compound; the other involves ions trading places between two compounds.

Displacement equation: \( Fe(s) + CuSO_4(aq) \rightarrow FeSO_4(aq) + Cu(s) \)

Double displacement equation: \( Na_2SO_4(aq) + BaCl_2(aq) \rightarrow BaSO_4(s) + 2NaCl(aq) \)

Copper is more reactive than silver, so it displaces silver from silver nitrate solution. Copper dissolves as copper nitrate while silver metal precipitates out — this is how silver is recovered in refining (p. 11).

\[ Cu(s) + 2AgNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2Ag(s) \]

Common error: Writing only one \( AgNO_3 \). Copper forms \( Cu^{2+} \), so two nitrate ions must pair with it — two \( AgNO_3 \) are required on the left.

A precipitation reaction is one in which an insoluble solid — the precipitate — forms when two soluble solutions mix (p. 11). The precipitate drops out of solution and can be filtered off.

Example 1: Barium chloride + sodium sulphate → white barium sulphate precipitate:

\[ BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) \downarrow + 2NaCl(aq) \]

Example 2: Lead nitrate + potassium iodide → yellow lead iodide precipitate:

\[ Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) \downarrow + 2KNO_3(aq) \]

Common error: Calling any double displacement a precipitation reaction. Only those that form an insoluble product qualify — if both products stay dissolved, there is no precipitate.

Oxidation and reduction always occur together. If one substance gains oxygen, another must lose it — they are coupled, which is the defining feature of redox reactions (p. 12).

Oxidation = gain of oxygen.

• \( 2Cu + O_2 \xrightarrow{\text{Heat}} 2CuO \) — copper gains oxygen; copper is oxidised.
• \( 4Na + O_2 \rightarrow 2Na_2O \) — sodium gains oxygen; sodium is oxidised.

Reduction = loss of oxygen.

• \( CuO + H_2 \xrightarrow{\text{Heat}} Cu + H_2O \) — copper oxide loses oxygen; \( CuO \) is reduced.
• \( ZnO + C \rightarrow Zn + CO \) — zinc oxide loses oxygen; \( ZnO \) is reduced.

Copper is a shiny brown metal. On heating in air, it combines with oxygen to form copper(II) oxide (\( CuO \)), which is black (Activity 1.11, p. 12). Though copper is less reactive than many metals, it still oxidises when heated directly.

\[ 2Cu(s) + O_2(g) \xrightarrow{\text{Heat}} 2CuO(s) \]

Final answer: \( X = \) copper; black compound = copper oxide (\( CuO \)).

Common error: Guessing X as iron. Iron is grey, not shiny brown, and it forms reddish-brown rust only in the presence of moisture — not a black coating on simple heating.

Corrosion is the gradual attack on a metal by substances around it — moisture, acids, or gases — which forms a surface coating (p. 13). For iron, that means rusting, a reddish-brown powder. Paint creates a physical barrier between iron and the moisture and oxygen that cause rust, so the metal underneath is protected.

Common error: Saying paint makes iron “look new.” Appearance is secondary; the functional reason is preventing contact with moisture and oxygen.

Nitrogen is an unreactive gas. Flushing it into a food bag displaces oxygen, so fats and oils cannot oxidise and turn rancid (p. 13). The food keeps its smell and taste for longer.

Common error: Saying nitrogen prevents bacterial or microbial spoilage. The reason specified in this chapter is oxidation of fats — keep your answer to that point.

Corrosion — a metal is gradually attacked by moisture, acids, or gases, forming a surface coating. Example: reddish-brown rust on iron; black coating on silver; green coating on copper (p. 13).

Rancidity — fats and oils are oxidised by air, changing their smell and taste. Example: old chips or stale cooking oil smelling unpleasant (p. 13).

Balance \( CH_4 + O_2 \rightarrow CO_2 + H_2O \). Carbon appears once on each side, so balance C first (already 1:1). Then balance H: four H on the left (in \( CH_4 \)) need two \( H_2O \) on the right. Now count oxygen: two in \( CO_2 \) plus two in \( 2H_2O \) = four O on the right, so we need \( 2O_2 \) on the left.

\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) \]

Balance \( Ca_3(PO_4)_2 + SiO_2 \rightarrow CaSiO_3 + P_2O_5 \). Start with the largest compound, \( Ca_3(PO_4)_2 \): three Ca on the left mean three \( CaSiO_3 \) on the right. That gives three Si on the right, so three \( SiO_2 \) on the left.

Now count oxygen: left has eight (in phosphate) plus six (in three \( SiO_2 \)) = fourteen; right has nine (in three \( CaSiO_3 \)) plus five (in \( P_2O_5 \)) = fourteen. P already balances at 2:2.

\[ Ca_3(PO_4)_2 + 3SiO_2 \rightarrow 3CaSiO_3 + P_2O_5 \]

Common error: Changing subscripts inside a formula (e.g. writing \( Ca_2(PO_4)_2 \)) instead of adding a coefficient. The formula of a compound is fixed; only coefficients are adjustable when balancing.