ICSE Questions Set in Previous Years

Ammonia gas reduces hot copper (II) oxide to copper.

2Pb₃O₄ ⟶ 6PbO + O₂

Hydrogen

Oxygen

2Pb₃O₄ $\overset{\Delta}{\longrightarrow}$ 6PbO + O₂

(a) 2PbO₂ $\overset{\Delta}{\longrightarrow}$ 2PbO + O₂

(b) P₄ + 5O₂ ⟶ 2P₂O₅

Ammonia gas reduces hot copper (II) oxide to copper.

Carbon dioxide and water vapour

The reason nitrous oxide (N₂O) supports combustion and rekindles a glowing splint is that nitrous oxide decomposes on heating into nitrogen and oxygen, which supports combustion.

2N₂O ⟶ 2N₂ + O₂

Chemical test to distinguish Oxygen from N₂O :

We can use nitric oxide (NO) to distinguish between nitrous oxide (N₂O) and oxygen (O₂).

Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide which can be easily identified by its reddish brown fumes.
2NO↑ + O₂↑ ⟶ 2NO₂↑ [Reddish brown vapours]

No reaction occurs between nitric oxide (NO) and nitrous oxide (N₂O). Hence, the absence of reddish brown fumes indicates nitrous oxide (N₂O).

Nitrogen

Carbon monoxide

Sulphur dioxide

Chlorine reacts with potassium iodode to form iodine. This iodine reacts with starch to form blue black colour.

Cl₂ + 2KI ⟶ 2KCl + I₂

Starch + I₂ ⟶ Blue black colour

Tests for chlorine:

1. It turns moist blue litmus paper red and finally bleaches it.
2. It turns moist starch iodide paper (KI + starch) blue black
3. Pass the gas through silver nitrate solution, a white ppt. of silver chloride is formed.

Chlorine turns moist blue litmus paper red and finally bleaches it.

Chlorine turns moist blue litmus red and finally bleaches it i.e., decolourizes it.

When chlorine water is exposed to sunlight, it changes to colourless from greenish yellow. This colour change happens because chlorine reacts with water to form hypochlorous acid which further breaks to form hydrochloric acid and oxygen gas.

Cl₂ + H₂O ⟶ HClO + HCl

HClO ⟶ HCl + [O]

[O] + colouring matter ⟶ bleached product

CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂

Carbon dioxide gas — It turns lime water milky due to the formation of white ppt. of calcium carbonate. The milkiness disappears when carbon dioxide is passed in excess.

Ca(OH)₂ + CO₂ ⟶ CaCO₃↓ + H₂O

CaCO₃↓ + H₂O + CO₂ [excess] ⟶ Ca(HCO₃)₂ [soluble]

Hydrogen chloride gas — When a glass rod dipped in ammonia solution is brought near the gas, dense white fumes of ammonium chloride are released.

NH₃ + HCl ⟶ NH₄Cl

The incomplete combustion of coal releases carbon monoxide gas into the environment. Carbon monoxide is a highly poisonous gas. If inhaled, it binds with haemoglobin 200 times more strongly than oxygen reducing the blood's ability to carry oxygen by converting haemoglobin into carboxyhaemoglobin. So, even in small quantities, it is fatal and can lead to death.

Sodium bisulphate

Two important processes which generate or release carbon dioxide into the atmosphere are :

1. Burning of fossil fuels like coal, natural gas and petroleum.
2. Respiration by animals and plants.

Two processes which remove carbon dioxide from the atmosphere are :

1. Photosynthesis.
2. Absorption by the oceans.

Carbon dioxide present in the atmosphere turns lime water milky due to the formation of white ppt. of calcium carbonate.

CO₂ + Ca(OH)₂ ⟶ CaCO₃↓ + H₂O

Biological importance of carbon dioxide dissolved in water :

1. Aquatic plants make use of dissolved carbon dioxide for photosynthesis, i.e., to prepare their food.
   6CO₂ + 12H₂O $\xrightarrow[\text{sunlight}]{\text{Chlorophyll}}$ C₆H₁₂O₆ + 6O₂ + 6H₂O
2. Carbon dioxide dissolved in water reacts with limestone to form calcium bicarbonate.
   CaCO₃ + CO₂ + H₂O ⟶ Ca(HCO₃)₂
   Marine organisms such as snails, oysters, etc., extract calcium carbonate from calcium bicarbonate to build their shells.

(a) When carbon dioxide is passed through lime water for a short time, it turns lime water milky. This is due to the formation of insoluble calcium carbonate.

Ca(OH)₂ + CO₂ ⟶ CaCO₃ ↓ + H₂O

(b) When the gas is passed for a long time, the milkiness disappears. This is due to the formation of a soluble bicarbonate.

CaCO₃ + CO₂ + H₂O ⟶ Ca(HCO₃)₂ [soluble]

Chloride of a metal X is XCl₂

By interchanging subscript and writing as superscript:

$$\begin{aligned}\underset{\phantom{1}{1}}{\text{X}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{Cl}} \Rightarrow \overset{\phantom{2}{2}}{\text{X}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{Cl}} \\[0.5em]\end{aligned}$$

Therefore, valency of metal X = 2.

Formula of the sulphate:

$$\begin{aligned}\text{X}^{2+} \phantom{\nearrow} \text{SO}_{4}^{2-} \\[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{S}}\text{O}_{4} \\[0.5em]\end{aligned}$$

As valency of both X and SO₄ is 2 so dividing by 2 we get 1 but 1 is never written so we get the formula as $\bold{XSO}_{\bold{4}}$

Formula of the hydroxide:

$$\begin{aligned}\text{X}^{2+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}}\text{H} \\[0.5em]\end{aligned}$$

Dropping 1 and enclosing OH in brackets, we get the formula as $\bold{X(OH)}_{\bold{2}}$

Therefore, we get

Formula of Sulphate : $\bold{XSO}_{\bold{4}}$

Formula of Hydroxide : $\bold{X(OH)}_{\bold{2}}$

(a) magnesium + copper sulphate ⟶ magnesium sulphate + copper

(b) iron (II) sulphide + hydrochloric acid ⟶ iron (II) chloride + hydrogen sulphide

When lead nitrate is heated it decomposes to form lead oxide, nitrogen dioxide and oxygen. It is a decomposition reaction.

The reaction is as follows:

2Pb(NO₃)₂ $\overset{\Delta}{\longrightarrow}$ 2PbO + 4NO₂ + O₂

Constituent units of crystals of :

(a) Iodine — iodine

(b) Sodium chloride — sodium and chloride ions.

(a) The number of atoms in a molecule of an element is called its atomicity.

(b) Hydrogen (H₂) is a diatomic element.

Water exists in all the three physical states : as solid (ice), as liquid (water) and as gas (water vapour).

Heat is absorbed during the reaction of Nitrogen and Oxygen to produce nitric oxide, hence, the reaction is endothermic in nature.

N₂ + O₂ $\xrightarrow{\Delta}$ 2NO

Valency of X is 3+ and that of Oxygen is 2-

By interchanging the valency number and shifting it to the lower right side of the atom or radical, we get the formula : X₂O₃

Equation for the combustion of X in oxygen: 4X + 3O₂ ⟶ 2X₂O₃

(a) Increase in weight : Iron, conc. sulphuric acid

Reason — Increase in weight is due to the absorbed water in case of sulphuric acid. Gain in weight of iron is due to the increased weight of oxygen which has combined with the iron to form iron oxide or rust.

(b) Decrease in weight : Sodium carbonate crystals

Reason — Decrease in mass is because sodium carbonate loses its water of crystallization on exposure to dry air.

(c) No change in weight : Sodium chloride

Reason — Pure sodium chloride is neither deliquescent nor efflorescent i.e., it does not absorb moisture from atmospheric air nor does it lose it, hence there is no change in mass.

The change in any one of the parameters [pressure, volume, temperature] affects the other two parameters. Therefore, when stating the volume of a gas the pressure and temperature should also be given.

(a) Absolute temperature — Absolute temperature refers to the temperature measured in Kelvin or absolute scale that has its zero at -273°C (absolute zero) and whose each degree is equal to one degree on the Celsius scale.

(b) Boyle's law — Temperature remaining constant the volume of a given mass of dry gas is inversely proportional to it's pressure.

V ∝ $\frac{1}{\text{P}}$ [T = constant]

(c) Charles' law — Charles law states that pressure remaining constant, the volume of a given mass of dry gas increases or decreases by $\frac{1}{273}$ of its volume at 0°C for each 1°C increase or decrease in temperature, respectively.

Using gas laws:

$\frac{\text{P}_1{\text{V}_1}}{\text{T}_1}$ = $\frac{\text{P}_2{\text{V}_2}}{\text{T}_2}$

Substituting the values:

$\frac{70 \times 760}{300}$ = $\frac{76 \times {\text{V}_2}}{273}$

Therefore:

$\text{V}_2 =\frac{70 \times 760 \times 273}{300 \times 76} = 637 \text{cm}^3$

∴ The volume occupied by the gas is 637 cm³.

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = 760 mm Hg
V₁ = Initial volume of the gas = 100 cm³
T₁ = Initial temperature of the gas = 0°C = 273 K

Final conditions:

P₂ (Final pressure) = increased by one-fifth of P₁
= (1 + $\frac{1}{5}$) of 760
= $\frac{6}{5}$ x 760
= 912
V₂ (Final volume) = ?
T₂ (Final temperature) = increased by one-fifth of 273 K = 1 + $\frac{1}{5}$ of 273
= $\frac{6}{5}$ x 273 = $\frac{1638}{5}$

By Gas Law:

$\frac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \frac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$$\begin{aligned}\frac{760 \times 100 }{273} = \frac{912\times \text{V}_2}{\frac{1638}{5}} \\[1em] \frac{760 \times 100 }{273} = \frac{5 \times 912 \times \text{V}_2}{1638} \\[1em] \text{V}_2 = \frac{760 \times 100 \times 1638}{273 \times 5 \times 912} \\[1em] \text{V}_2 = 100\text{ cm}^3\end{aligned}$$

∴ The final volume of the gas = 100 cm³

Initial conditions [S.T.P.] :

P₁ = Initial pressure of the gas = 1 atm
V₁ = Initial volume of the gas = 22.4 litres
T₁ = Initial temperature of the gas = 273 K

Final conditions:

P₂ (Final pressure) = 2 atm
V₂ (Final volume) = ?
T₂ (Final temperature) = 546 K

By Gas Law:

$\frac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \frac{\text{P}_2\times\text{V}_2}{\text{T}_2}$

Substituting the values :

$$\begin{aligned}\frac{1 \times 22.4}{273} = \frac{2\times \text{V}_2}{546}\\[0.5em] \text{V}_2 = \frac{1 \times 22.4\times 546}{273\times 2} \\[0.5em] \text{V}_2 = 22.4 \text{ lit}\end{aligned}$$

∴ Final volume of the gas = 22.4 lit.

No, it is not possible. Change in any one of the parameters [pressure, volume, temperature] affects the other two parameters.

(a) Zinc nitrate + sodium carbonate ⟶ zinc carbonate + sodium nitrate

Zn(NO₃)₂ + Na₂CO₃ ⟶ ZnCO₃ + 2NaNO₃

(b) Iron (II) sulphate (ferrous sulphate) + sodium hydroxide ⟶ Iron (II) hydroxide + sodium sulphate

FeSO₄ + 2NaOH ⟶ Fe(OH)₂ + Na₂SO₄

Magnetization of a piece of steel is a physical change because no new substance is formed and the chemical composition of the piece of steel remains the same.

(a) a liquid and a solid — sugar and water

(b) two liquids — water and alcohol

Nitride of a metal X is XN.

Since valency of nitrogen is 3- so valency of X is 3+

Formula of the sulphate:

$$\begin{aligned}\text{X}^{3+} \phantom{\nearrow} \text{SO}_{4}^{2-} \\[0.5em] \overset{\phantom{2}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{S}}\text{O}_{4} \\[0.5em]\end{aligned}$$

So, we get the formula as $\bold{X}_{\bold{2}}({\bold{SO}}_{\bold{4}})_{\bold{3}}$

Formula of the hydroxide:

$$\begin{aligned}\text{X}^{3+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{2}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{O}}\text{H} \\[0.5em]\end{aligned}$$

Dropping 1, we get the formula as $\bold{X(OH)}_{\bold{3}}$

Therefore, we get

Formula of Sulphate : $\bold{X}_{\bold{2}}({\bold{SO}}_{\bold{4}})_{\bold{3}}$

Formula of Hydroxide : $\bold{X(OH)}_{\bold{3}}$

$$\begin{aligned}\text{(a)} \space \text{NO} = \underset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]\end{aligned}$$

But valency of O is 2. Multiplying by 2, we get:

$$\begin{aligned}\overset{{2 \times 1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]\end{aligned}$$

Therefore, valency of Nitrogen is 2.

$$\begin{aligned}\text{(b)} \space \text{N}_2\text{O} = \underset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]\end{aligned}$$

Therefore, valency of Nitrogen is 1.

$$\begin{aligned}\text{(c)} \space \text{NO}_2 = \underset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]\end{aligned}$$

But valency of O is 2. Multiplying by 2, we get:

$$\begin{aligned}\overset{{2 \times 2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{4}{4}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]\end{aligned}$$

Therefore, valency of Nitrogen is 4.

(a) Bromine

(b) Graphite

Lead (II) oxide (PbO)

Sublimation is the name of the process.

Iodine also undergoes sublimation process.

Hydrogen gas is evolved when dil. hydrochloric acid is added first to a mixture of iron and sulphur

Fe + 2HCl (dil.) ⟶ FeCl₂ + H₂ ↑

Hydrogen sulphide (H₂S) gas is evolved when dil. hydrochloric acid is added to compound formed by iron and sulphur.

FeS + 2HCl ⟶ FeCl₂ + H₂S

(a) Oxidation, loss of electrons

(b) Oxidation, gain of oxygen

(a) Fe²⁺ ⟶ Fe³⁺ + e^-
Oxidation, loss of electrons

(b) Cl^- ⟶ Cl + e^-
Oxidation, loss of electrons

(c) Cu²⁺ + 2e^- ⟶ Cu
Reduction, gain of electrons

(d) Ag ⟶ Ag⁺ + e^-
Oxidation, loss of electrons

(e) H ⟶ H⁺ + e^-
Oxidation, loss of electrons

(f) Al ⟶ Al³⁺ + 3e^-
Oxidation, loss of electrons

(a) Simple displacement

(b) Redox reaction

(c) Decomposition reaction

(d) Direct combination

(e) Double decomposition

Kelvin zero or absolute zero = -273°C.

(a) CaCO₃ ⟶ CaO + CO₂ [Gas A]

CaO + H₂O ⟶ Ca(OH)₂

Ca(OH)₂ + H₂O [excess] ⟶ CaO + H₂O

(i) marble — Calcium carbonate [CaCO₃]

(ii) gas 'A' — Carbon dioxide [CO₂]

(iii) solid 'B' — Calcium hydroxide [Ca(OH)₂]

(b) When drops of water are added to solid calcium oxide, a large amount of heat is liberated and a hissing sound is heard along with the formation of calcium hydroxide [Ca(OH)₂]. It is an exothermic reaction.

(c) (i) Solution C is normally called lime water

When gas 'A' is bubbled through 'C'., a white precipitate of calcium hydroxide is formed.

(ii) When excess of gas A [Carbon dioxide] is passed through the resulting suspension, the white ppt. of calcium carbonate dissolves due to the formation of calcium hydrogen carbonate which is soluble in water.

CaCO₃ + H₂O + CO₂ [excess] ⟶ Ca(HCO₃)₂ [soluble]

On boiling, calcium hydrogen carbonate decomposes to form calcium carbonate due to which, the white ppt. reappears.

Ca(HCO₃)₂ $\xrightarrow{\space \Delta \space}$ H₂O + CO₂ + CaCO₃↓