ICSE Class 10 Chemistry — All Questions

X - iodine, Y - bromine

Reason — Halogens exist as diatomic molecules and their chemical reactivity decreases on moving down the group. As iodine lies below bromine in Group 17, iodine is less reactive than bromine. Hence, X is iodine and Y is bromine.

concentrated sulphuric acid followed by water

Reason — Sulphur trioxide is not dissolved directly in water as the reaction is highly exothermic and forms a dense mist of sulphuric acid. So, SO₃ is first absorbed in concentrated sulphuric acid to form oleum (H₂S₂O₇), which is then diluted with water to obtain sulphuric acid.

In acidic solutions, current is carried by ions.

Reason — In metallic conductors, current is carried by free electrons, whereas in electrolytes (acidic solutions, molten electrolytes and aqueous ionic solutions) current is carried by free ions. Hence, in acidic solutions current is carried by ions.

3 and 4

Reason — An acid has a pH value less than 7 (statement 3) and reacts with a basic oxide like copper(II) oxide to form a salt and water (statement 4). Statement 1 is only a test for the sulphate ion and statement 2 shows the oxidising property of sulphuric acid.

P - zinc, Q - copper (II) oxide

Reason — Zinc reacts with dilute hydrochloric acid to liberate hydrogen gas. Hydrogen is a reducing agent, so on passing over heated copper(II) oxide it reduces the black oxide to reddish-brown copper.

Zn + 2HCl ⟶ ZnCl₂ + H₂

CuO + H₂ ⟶ Cu + H₂O

acid

Reason — Chlorine and iodine react with hydrogen to form the hydrogen halides hydrogen chloride and hydrogen iodide. These dissolve in water to form hydrochloric acid and hydroiodic acid respectively.

Both the statements are true.

Reason — Halogens have seven electrons in their valence shell and need only one electron to complete their octet. This gives them a very high electron affinity, which makes them very reactive. Hence, both statements are true.

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

Reason — In ionic compounds, the oppositely charged ions are held together by strong electrostatic forces of attraction. A large amount of heat energy is required to break these forces, so ionic compounds have high melting points. Thus, (R) correctly explains (A).

Q

Reason — On the pH scale, a neutral solution (pH 7) turns pH paper green. Red indicates an acidic solution, while blue and violet indicate basic (alkaline) solutions. Hence, solution Q (green) is neutral.

Only 1

Reason — Propanol (C₃H₇OH) has the molecular formula C₃H₈O. Propanal (C₂H₅CHO) is C₃H₆O and propanoic acid (C₂H₅COOH) is C₃H₆O₂. Hence, only propanol has the formula C₃H₈O.

C₂H₄

Reason — Molecular mass = 2 × Vapour Density = 2 × 14 = 28.

The molecular mass of C₂H₄ = 2(12) + 4(1) = 24 + 4 = 28, which matches. Hence, the molecular formula of X is C₂H₄.

H₃O⁺

Reason — A pair of electrons that is not shared in bond formation is called a lone pair of electrons. H₃O⁺ has one lone pair of electrons. OH^1- has three lone pairs, whereas NH₄⁺ and H⁺ have no lone pairs of electrons.

Sodium sulphide

Reason — Sodium sulphide reacts with dilute hydrochloric acid to release hydrogen sulphide gas. H₂S turns moist lead acetate paper black (due to the formation of black lead sulphide) and being a reducing gas, it also decolourises acidified potassium permanganate solution.

Na₂S + 2HCl ⟶ 2NaCl + H₂S ↑

3 and 4

Reason — Reaction 3 is the combustion of ethanol, which produces carbon dioxide and water. Reaction 4 is dilute sulphuric acid reacting with magnesium carbonate, which produces a salt, water and carbon dioxide. Hence, reactions 3 and 4 produce water.

4.48 litres

Reason — One mole of any gas occupies 22.4 litres at S.T.P.

∴ Volume of 0.2 mole of gas = 0.2 × 22.4 = 4.48 litres.

(a) A is the anode (positive electrode) where oxygen gas is liberated.

4OH^- - 4e^- ⟶ 2H₂O + O₂

(b) The gas at A is oxygen. It rekindles a glowing splinter brought near it.

(c) Hydrogen gas is liberated at B.

(d) Pure water is a very poor conductor of electricity as it has very few ions. Water is acidified by adding a little dilute sulphuric acid to provide ions, which makes it a good conductor so that electrolysis can take place.

(e) The ratio by volume of gases at A (oxygen) and B (hydrogen) is 1 : 2.

(a) CHCl₃ + Cl₂ $\xrightarrow{\text{diffused sunlight}}$ CCl₄ + HCl

(b) Al₂O₃.2H₂O + 2NaOH ⟶ 2NaAlO₂ + 3H₂O

(c) ZnO + 2NaOH ⟶ Na₂ZnO₂ + H₂O

(d) 3Cu + 8HNO₃ (dilute) ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(e) NaNO₃ + H₂SO₄ $\xrightarrow{\text{below 200°C}}$ NaHSO₄ + HNO₃

(a) The general formula of an alkyl group is C_nH_2n+1.

(b) The aqueous solution of sodium sulphate gives white precipitate with barium nitrate.

(c) The gas liberated when alkalis are heated with ammonium salts is NH₃.

(d) The alloy that contains a non-metal as one of its constituents is steel.

(e) A salt prepared by neutralization in which titration is involved is NH₄Cl.

Note: In part (b), the chemically correct answer is sodium sulphate, since barium nitrate reacts with the sulphate ion of sodium sulphate to give a white precipitate of barium sulphate (Na₂SO₄ + Ba(NO₃)₂ ⟶ BaSO₄ ↓ + 2NaNO₃). Zinc chloride would not give any precipitate with barium nitrate as both products would be soluble.

(a) IUPAC names of the given organic compounds are:

1. butanoic acid
2. 2-methylpentane

(b) Structural diagrams:

1. diethyl ether

2. isopentane

3. propanone

P is a divalent non-metal having six electrons in its valence shell. Each of the two hydrogen atoms shares one electron with P to form two covalent bonds, so that P completes its octet while each H completes its duplet. The dot and cross structure for the formation of H₂P is shown below:

(a) Colourless ammonia gas reacts with excess greenish-yellow chlorine to form a yellow, explosive liquid (nitrogen trichloride).
NH₃ + 3Cl₂ [excess] ⟶ 3HCl + NCl₃

(b) Reddish-brown nitrogen dioxide gas is evolved on heating zinc nitrate crystals.
$2\text{Zn(NO}_3)_2 \xrightarrow{\Delta} 2\text{ZnO} + \text{O}_2 + 4\text{NO}_2$

(a) The blue crystalline salt B is copper(II) sulphate pentahydrate, CuSO₄·5H₂O.

(b) The dehydrating property of the hygroscopic liquid (concentrated sulphuric acid) is demonstrated, as it removes the water of crystallisation from the blue salt, leaving behind white anhydrous copper sulphate.

(c) CuSO₄·5H₂O $\xrightarrow{\text{conc. H}_2\text{SO}_4}$ CuSO₄ + 5H₂O

N has atomic number 11, so its electronic configuration is 2, 8, 1 and its valency is 1.

(a) The molecular formula of the oxide is N₂O (i.e., Na₂O).

(b) N₂O + H₂O ⟶ 2NOH

(c) The formula of the product other than water is N₂SO₄.

(a) Carbon (graphite)

(b) Magnalium

(c) Calcination

(a) X

Reason — Sodium sulphite reacts with an acid to liberate sulphur dioxide gas. Hence, solution X with pH 2 is acidic and will liberate sulphur dioxide.

SO₃^2- + 2H⁺ ⟶ SO₂ + H₂O

(b) Z

Reason — Ammonium salts liberate ammonia gas on reaction with an alkali. Hence, solution Z with pH 13 is alkaline and will liberate ammonia from ammonium chloride.

NH₄⁺ + OH^- ⟶ NH₃ + H₂O

(c) Y

Reason — The solution Y with pH 7 will not have any effect on litmus paper as it is neither acidic nor basic.

(a) R is ferric chloride (iron(III) chloride), FeCl₃.

(b) P is NaCl and Q is Fe(OH)₃.

FeCl₃ + 3NaOH ⟶ Fe(OH)₃ ↓ + 3NaCl

(c) On adding excess sodium hydroxide, the reddish-brown precipitate of Fe(OH)₃ remains insoluble, i.e., there is no change.

Molecular mass of M₂P = 2(14) + 16 = 44 g/mol

Molecular mass of MP = 14 + 16 = 30 g/mol

Number of moles = $\frac{\text{Mass}}{\text{Molecular mass}}$

∴ Moles in 20 g of M₂P = $\frac{20}{44}$ = 0.45 mol

∴ Moles in 20 g of MP = $\frac{20}{30}$ = 0.67 mol

Since 0.45 < 0.67, 20 grams of M₂P has a lesser number of moles than 20 grams of MP. Hence, the statement is justified.

(a) No, zinc and aluminium cannot be distinguished by this method. Both metals react with concentrated sodium hydroxide solution to form soluble sodium salts and liberate hydrogen gas, which burns with a pop sound.

Zn + 2NaOH ⟶ Na₂ZnO₂ + H₂

2Al + 2NaOH + 2H₂O ⟶ 2NaAlO₂ + 3H₂

(b) Yes, they can be distinguished. With zinc nitrate, ammonium hydroxide gives a white precipitate of Zn(OH)₂ which is soluble in excess of ammonium hydroxide. With lead nitrate, a white precipitate of Pb(OH)₂ is formed which is insoluble in excess of ammonium hydroxide.

A: Zn + H₂SO₄ ⟶ ZnSO₄ + H₂

B: Zn + S ⟶ ZnS

C: ZnSO₄ + 2NaOH ⟶ Zn(OH)₂ + Na₂SO₄

Element P has the L shell as its outermost shell with 3 valence electrons, so its configuration is 2, 3 (period 2). Element Q has the M shell as its outermost shell with 7 valence electrons, so its configuration is 2, 8, 7 (period 3).

(a) Element Q has 7 valence electrons, so it belongs to Group 17, and its outermost shell is the M (third) shell, so it belongs to Period 3.

Position of Q: Group 17, Period 3.

(b) The atomic radius of P is less than that of Q.

Element P has only two shells whereas Q has three shells. As the number of shells increases, the atomic radius increases. Hence, P (fewer shells) has a smaller atomic radius than Q.

(a) P is Ammonia (NH₃).

3CuO + 2NH₃ ⟶ 3Cu + 3H₂O + N₂

(b) Q is Oxygen (O₂).

4FeS₂ + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂

(c) R is Oxygen (O₂).

4NH₃ + 3O₂ ⟶ 2N₂ + 6H₂O

$$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{N}_2 & + & 4\text{H}_2\text{O} & + & \text{Cr}_2\text{O}_3 \\ 2[14 + 4(1)] & & & & & & 2(52) \\ + 2(52) + 7(16) & & & & & & + 3(16) \\ = 36 + 104 & & & & & & = 104 + 48 \\ + 112 = 252 \text{ g} & & & & & & = 152 \text{ g} \\ \end{matrix}$$

(a) 252 g of (NH₄)₂Cr₂O₇ = 1 mole

∴ 126 g of (NH₄)₂Cr₂O₇ = $\frac{1}{252}$ x 126 = 0.5 moles

Hence, no. of moles = 0.5 moles

(b) 252 g of (NH₄)₂Cr₂O₇ gives 152 g of Cr₂O₃

126 g of (NH₄)₂Cr₂O₇ will give $\frac{152 \times 126}{252}$ = 76 g of Cr₂O₃

Hence, mass in gms of Cr₂O₃ formed = 76 g.

(c) 252 g of (NH₄)₂Cr₂O₇ produces 22.4 lit of N₂

126 g of (NH₄)₂Cr₂O₇ will produce $\frac{22.4 \times 126}{252}$ = 11.2 lit of N₂

Hence, volume of N₂ evolved at s.t.p = 11.2 lit.

(a) The gas is evolved in test tube B (containing ammonium chloride).

(b) The gas evolved is ammonia (NH₃).

(c) Ammonia burns in oxygen with a greenish-yellow flame.

(d) NH₄Cl + NaOH ⟶ NaCl + NH₃ + H₂O

On complete combustion, the acidic gaseous product is carbon dioxide.

CH₄ + 2O₂ ⟶ CO₂ + 2H₂O

The ratio of methane (CH₄) to carbon dioxide (CO₂) is 1 : 1.

From the diagram, P has 11 protons (2, 8, 1) and is a monovalent metal, while Q has 8 protons (2, 6) and is a divalent non-metal. When they react, P loses one electron and Q gains two electrons, so two atoms of P combine with one atom of Q.

Formula of the compound = P₂Q

Mass of one mole = 2 × (At. wt of P) + (At. wt of Q)

= 2 × 23 + 16

= 46 + 16

= 62 g

∴ The formula is P₂Q and one mole has a mass of 62 g.

(a) CaC₂ + 2H₂O ⟶ Ca(OH)₂ + C₂H₂

(b) S + 6HNO₃ (conc.) ⟶ H₂SO₄ + 6NO₂ + 2H₂O

(c) C₂H₅COONa + NaOH $\xrightarrow{\text{CaO}}$ C₂H₆ + Na₂CO₃

(a) Esterification reaction

$\underset{\text{ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} + \text{CH}_3\text{COOH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \underset{\text{ethyl acetate}}{{\text{CH}_3\text{COOC}_2\text{H}_5}} + \text{H}_2\text{O}$

(b) Dehydrating agent

(c) Fruity smell is observed due to the formation of ester by the reaction of ethanol and acetic acid.

(a) The oxide has the formula ZO, so Z is divalent. Hence, Z has 2 valence electrons.

(b) Z belongs to Group 2 (alkaline earth metals).

(c) The formula of the hydroxide of Z is Z(OH)₂.

(a) Electron dot structure of the ammonium ion (with the co-ordinate bond labelled):

(b) Electron dot structure of the nitrogen molecule (which contains a triple bond between the two nitrogen atoms):

(a) G — ethene (G) on halogenation, i.e., addition of bromine, forms 1,2-dibromoethane (F).

(b) D — it contains the –OH (hydroxyl) functional group, so it is an alcohol.

(c) E — 2-methyl propane has the formula C₄H₁₀. Compound E (n-butane) is a saturated hydrocarbon with the same molecular formula C₄H₁₀, so it is an isomer of 2-methyl propane.

(d) C — ethene belongs to the alkene series (contains a C=C double bond). Compound C also contains a carbon-carbon double bond, so it belongs to the same homologous series as ethene.

Carbon

Reason — Carbon (a non-metal) reacts with hot concentrated sulphuric acid to form two gases — carbon dioxide (CO₂) and sulphur dioxide (SO₂). Both these gases turn lime water milky.

C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2H₂O + 2SO₂

R

Reason — An ionic bond is formed between a metal (low ionisation potential) and a non-metal (high electron affinity). Group 2 elements are metals which lose 2 electrons, while Group 17 elements are non-metals which gain 1 electron. Hence, the elements of Group 2 and Group 17 form ionic bond (e.g., MgCl₂, CaCl₂).

Pair P consists of two metals (no bond formation), Q consists of two non-metals (covalent bond), and S consists of a non-metal and a noble gas (no bond formed).

2

Reason — An element with electronic configuration 2, 8, 2 is magnesium (Mg) with valency 2. Its hydroxide is Mg(OH)₂, which contains 2 hydroxyl (OH^-) ions per molecule.

Mg(OH)₂ ⟶ Mg²⁺ + 2OH^-

Copper metal is deposited at the negative electrode.

Reason — During electrolysis of copper(II) sulphate solution using copper electrodes:

• At the cathode (negative electrode): Cu²⁺ + 2e^- ⟶ Cu, so copper is deposited here and the cathode increases in mass.
• At the anode (positive electrode): Cu - 2e^- ⟶ Cu²⁺, so copper of the anode dissolves and the anode decreases in mass.

No oxygen gas is produced because the copper is preferentially oxidised over the OH^- ions.

C₂H₆ + Cl₂ ⟶ C₂H₅Cl + HCl

Reason — In the presence of ultraviolet (UV) light, ethane (a saturated hydrocarbon) undergoes a substitution reaction with chlorine. One hydrogen atom of ethane is replaced by a chlorine atom forming chloroethane (C₂H₅Cl) and hydrogen chloride (HCl) is released.

C₂H₆ + Cl₂ $\xrightarrow{\text{UV light}}$ C₂H₅Cl + HCl

Y only

Reason — The reaction shown is cracking (thermal decomposition) of butane (X = C₄H₁₀), a saturated hydrocarbon, into propene (Y = C₃H₆) and methane (Z = CH₄).

• X (butane, C₄H₁₀) — saturated (only single bonds).
• Y (propene, C₃H₆) — unsaturated (contains one C=C double bond).
• Z (methane, CH₄) — saturated (only single bonds).

Hence, only Y is unsaturated.

Both 2 and 4

Reason — The reactivity of these metals with dilute HCl depends on their position in the activity series:

• Copper lies below hydrogen in the activity series, so it does not react with dilute HCl. Hence, no gas is given off — observation (1) is incorrect.
• Iron is above hydrogen in the activity series, hence it reacts with dilute HCl liberating hydrogen gas — observation (2) is correct.
  Fe + 2HCl ⟶ FeCl₂ + H₂ ↑
• Magnesium is highly reactive and reacts vigorously with dilute HCl liberating hydrogen gas — observation (3) is incorrect.
  Mg + 2HCl ⟶ MgCl₂ + H₂ ↑
• Zinc is above hydrogen in the activity series and reacts with dilute HCl liberating hydrogen gas — observation (4) is correct.
  Zn + 2HCl ⟶ ZnCl₂ + H₂ ↑

Hence, observations 2 and 4 are correct.

12

Reason — An element with atomic number 12 has the electronic configuration 2, 8, 2. It loses its 2 valence electrons easily to attain a stable octet configuration (2, 8) and forms a divalent positive ion X²⁺. This element is magnesium (Mg).

Mg ⟶ Mg²⁺ + 2e^-

Whereas, atomic number 16 (Sulphur) gains 2 electrons to form S^2-, atomic number 10 (Neon) is a noble gas which doesn't form ions, and atomic number 14 (Silicon) shares electrons.

3

Reason — Copper lies below hydrogen in the activity series, so it does not displace hydrogen from dilute acids. Hence, copper salts cannot be prepared by the action of dilute acids on the metal copper.

However, copper salts can be prepared from copper bases like Cu(OH)₂, copper carbonate (CuCO₃), and copper sulphite by the action of dilute acid.

Sodium chloride

Reason — Sodium chloride (NaCl) is an ionic compound. Ionic compounds have very strong electrostatic forces of attraction between oppositely charged ions, requiring a large amount of energy to break, so they have very high melting points.

Whereas, methane, ammonia and ethanol are covalent compounds with weak intermolecular forces, so they have low melting points.

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

Reason — Dilute sulphuric acid is a stronger electrolyte because the presence of water causes greater ionisation, producing more mobile ions. Concentrated sulphuric acid contains comparatively fewer free ions, hence Assertion (A) is true.
The strength of an electrolyte depends on the concentration of mobile ions. In dilute sulphuric acid greater ionisation and ion mobility make it a better conductor of electricity, hence Reason (R) is true and is the correct explanation of (A).

H₂SO₄ ⟶ 2H⁺ + SO₄^2-

15 litres

Reason — According to Gay-Lussac's law of gaseous volumes, when gases react they do so in volumes which bear a simple whole number ratio to one another, given temperature and pressure remain constant.

From the equation: 1 volume of C₃H₈ produces 3 volumes of CO₂.

∴ 5 litres of C₃H₈ will produce 5 × 3 = 15 litres of CO₂.

propene

Reason — Propene is an alkene with molecular formula C₃H₆, which contains 3 carbon atoms and 6 hydrogen atoms with a C=C double bond, making it unsaturated. It follows the general formula of alkenes (C_nH_2n).

CH₂=CH-CH₃ (propene)

Whereas, propyne (C₃H₄) has 4 hydrogens, propane (C₃H₈) is saturated, and propanol (C₃H₈O) is an alcohol containing oxygen.

(A) is true but (R) is false.

Reason — During the electrolysis of acidified water:

• At cathode: 4H⁺ + 4e^- ⟶ 2H₂
• At anode: 4OH^- - 4e^- ⟶ 2H₂O + O₂

The ratio of H₂ to O₂ liberated is 2:1, so the volume of hydrogen liberated is twice the volume of oxygen formed. Hence, assertion (A) is true.

However, on electrolysis water gives hydrogen and oxygen in the ratio of 2:1 by volume, and not 1:2 as stated. Hence, reason (R) is false.

1

Reason — The negative electrode is the cathode where reduction takes place. At the cathode, the positively charged Al³⁺ ions get attracted, gain 3 electrons, and get deposited as aluminium metal.

Al³⁺ + 3e^- ⟶ Al

Reaction (2) takes place at the anode (oxidation), reaction (3) is the reaction of carbon anode with oxygen, and reaction (4) is the dissociation of alumina.

(a) Haber's process.

(b) Finely divided iron.

(c) The mixture of ammonia and unreacted gases (N₂ and H₂) coming out of the catalytic chamber is passed through the cooling pipes (condenser). Ammonia liquefies first, whereas nitrogen and hydrogen do not liquefy easily. The ammonia produced is stored as liquid under pressure.

(d) The two properties of ammonia gas demonstrated by the Fountain Experiment are:

1. Ammonia gas is highly soluble in water.
2. Ammonia is a basic gas (its aqueous solution turns red litmus blue / phenolphthalein turns pink).

(a) Calcium carbide (CaC₂)

(b) Ethane (C₂H₆)

(c) Ferric hydroxide [Fe(OH)₃]

(d) Lone pair of electrons

(e) Gram molecular mass

(a) The oxide that dissolves in caustic potash (KOH) is ZnO.
Reason — ZnO is amphoteric in nature, so it dissolves in both acids and alkalis like KOH.
ZnO + 2KOH ⟶ K₂ZnO₂ + H₂O

(b) The reaction that takes place at the anode during the electrolysis of molten lead bromide is 2Br^- - 2e^- ⟶ Br₂.

(c) The volume occupied by 8 grams of oxygen gas at STP is 5.6 litres.
Reason —
Molar mass of O₂ = 32 g
32 g of O₂ at STP occupies 22.4 L
∴ 8 g of O₂ occupies $\frac{22.4}{32}$ × 8 = 5.6 L

(d) Cu does not give hydrogen gas when it reacts with cold and very dilute nitric acid.
Reason — Copper lies below hydrogen in the activity series, so it does not liberate hydrogen with nitric acid.

(e) HCl is a polar covalent compound.
Reason — HCl is formed by sharing of electrons between H and Cl, but due to large electronegativity difference, the bonding electrons are unequally shared, giving the molecule a polar character. CCl₄ is a non-polar covalent compound due to symmetric structure.

(a) IUPAC names of the given compounds:

1. Butan-2-ol
2. 3-methylpentane

(b) Structural diagrams:

1. butanal

2. Pent-2-yne (CH₃—C≡C—CH₂—CH₃)

3. The isomer of n-butane is 2-methylpropane (iso-butane).

(a) Anion — Chloride (Cl^-)

(b) Cation — Copper(II) (Cu²⁺)

Reason —

(a) The white precipitate formed with silver nitrate solution which is soluble in ammonium hydroxide is silver chloride (AgCl). This indicates the presence of chloride ion (Cl^-) as the anion.

AgNO₃ + Cl^- ⟶ AgCl ↓ [white ppt.] + NO₃^-

AgCl + 2NH₄OH ⟶ [Ag(NH₃)₂]Cl + 2H₂O

(b) The pale blue precipitate formed with sodium hydroxide solution is copper(II) hydroxide [Cu(OH)₂]. This indicates the presence of copper(II) ion (Cu²⁺) as the cation.

Cu²⁺ + 2NaOH ⟶ Cu(OH)₂ ↓ [pale blue ppt.] + 2Na⁺

Hence, the salt Y is copper(II) chloride (CuCl₂).

(a) The temperature is maintained below 200°C because nitric acid decomposes at higher temperature to give nitrogen dioxide, water and oxygen.

4HNO₃ ⟶ 4NO₂ + 2H₂O + O₂

(b) Concentrated sulphuric acid is used because it is a non-volatile acid with a high boiling point. It displaces the more volatile nitric acid from sodium nitrate and does not distil over with nitric acid vapours.

(a) Element A is Nitrogen (N) with electronic configuration 2, 5.

(b) Nitrogen has a valency of 3 and combines with hydrogen (valency 1) to form ammonia (NH₃).

(c) Dot (•) and cross (×) structure of NH₃:

(a) C₂H₄Br₂ + 2KOH (alcoholic) ⟶ C₂H₂ + 2KBr + 2H₂O

(b) 8NH₃ (excess) + 3Cl₂ ⟶ N₂ + 6NH₄Cl

(c) C₁₂H₂₂O₁₁ $\xrightarrow{\text{(conc.) H}_2\text{SO}_4}$ 12C + 11H₂O

(a)

Acetic acid	Sulphuric acid
Acetic acid (CH3COOH) is a monobasic acid, having only one replaceable hydrogen ion per molecule.	Sulphuric acid (H2SO4) is a dibasic acid, having two replaceable hydrogen ions per molecule.

(b)

Electrolyte	Metallic conductor
In an electrolyte, electricity is conducted by the movement of ions (cations and anions) in the molten state or in aqueous solution.	In a metallic conductor, electricity is conducted by the movement of free electrons through the metal lattice.

According to Avogadro's Law, equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.

Hence, the number of molecules of chlorine gas = number of molecules of ammonia gas.

Molar mass of NH₃ = 14 + 3(1) = 17 g

Number of moles of NH₃ = $\frac{\text{Mass}}{\text{Molar mass}}$ = $\frac{34}{17}$ = 2 moles

Number of molecules in 1 mole = 6.022 × 10²³

∴ Number of molecules in 2 moles of NH₃ = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules

Hence, the number of molecules in chlorine gas = 1.2044 × 10²⁴ molecules.

P loses one valence electron, so P is a Group 1 metal (most electropositive). R has high electronegativity, so R is a non-metal on the right side of the periodic table. Q shares electrons (covalent), so Q is in Group 14 (between P and R).

(a) P would be most difficult to reduce.
Reason — Reduction means gain of electrons. Element P readily loses its valence electron and is the most electropositive (strongest reducing agent itself). Hence, its tendency to gain electrons (be reduced) is the lowest, so it is most difficult to reduce.

(b) R has the smallest atomic radius.
Reason — Across a period from left to right, the atomic radius decreases due to increasing nuclear charge. R is on the rightmost side, so it has the smallest atomic radius.

(c) Decreasing ionisation potential: R > Q > P
Reason — Across a period, ionisation potential increases from left to right because atomic size decreases and the effective nuclear charge increases.

(a) Pb(NO₃)₂ + Na₂CO₃ ⟶ PbCO₃ ↓ + 2NaNO₃

(b) CuCO₃ + 2HCl ⟶ CuCl₂ + H₂O + CO₂ ↑

(c) Fe + 2HCl ⟶ FeCl₂ + H₂ ↑

(a) The type of reaction that occurred is addition.

(b) Nickel (Ni)

(c) C₂H₂ + 2H₂ $\xrightarrow{\text{Ni}}$ C₂H₆

(a) Sulphur trioxide

(b) Sulphuric acid (H₂SO₄)

(c) Sulphur trioxide (SO₃) does not dissolve in water satisfactorily and it gives a lot of heat and forms misty droplets of sulphuric acid which are difficult to condense and escape into the atmosphere. Hence, it is not directly dissolved in water.

(a) Ag⁺ + e^- ⟶ Ag

(b) Silver nitrate solution is not preferred because it ionises very readily releasing a high concentration of Ag⁺ ions. This causes silver to be deposited too rapidly on the spoon, producing a non-uniform, loose, and rough coating with poor adhesion. Sodium argentocyanide, on the other hand, releases Ag⁺ ions slowly, producing a smooth, uniform, and shiny coating.

(c) Alternating current (AC) is not used because in AC, the polarity of electrodes keeps changing. As a result, silver deposited on the spoon at one moment would dissolve back into the electrolyte at the next moment. Hence, direct current (DC) is used to ensure uni-directional flow of ions and continuous deposition of silver on the spoon.

(d) The silver anode gradually dissolves into the electrolyte as silver ions:

Ag - e^- ⟶ Ag⁺

The mass of the anode decreases during the process.

(a) Hydrochloric acid (HCl) is a monobasic acid, i.e., it has only one replaceable hydrogen ion per molecule. Acid salts are formed by the partial replacement of replaceable hydrogen ions of an acid by a metal, which requires the acid to be at least dibasic (with two or more replaceable H⁺). Since HCl has only one replaceable hydrogen, it can only form normal salts and not acid salts.

(b) On moving across a period from left to right, the nuclear charge increases while the number of shells (energy levels) remains the same. As a result, the atomic radius decreases and the effective pull of the nucleus on the bonding electrons increases. Hence, the tendency of an atom to attract the shared pair of electrons (electronegativity) increases across a period.

(a) False
Reason — Isomers are organic compounds having the same molecular formula but different structural formulae.

(b) True
Reason — A salt is indeed formed when the hydrogen ion of an acid is partially or completely replaced by a metal or electropositive radical (such as the ammonium ion).

(a) The error made by Anil was that he stored the nitric acid in a transparent container exposed to sunlight. Nitric acid should be stored in a dark-coloured (brown) reagent bottle kept in a cool and dark place to prevent decomposition.

(b) Nitric acid is unstable and decomposes on exposure to light and heat to release reddish-brown nitrogen dioxide (NO₂) gas, water and oxygen:

4HNO₃ $\xrightarrow{\text{light/heat}}$ 4NO₂ ↑ + 2H₂O + O₂ ↑

The brown fumes seen in the container are due to NO₂ gas.

(c) The yellowish tinge in nitric acid (caused by dissolved NO₂) can be removed by bubbling air or carbon dioxide gas through the acid. This drives out the dissolved nitrogen dioxide and the acid becomes colourless again.

$$\begin{matrix} \text{Zn} & + & 4\text{HNO}_3 & \longrightarrow \\ 1 \text{ mole} && 4 \text{ moles} & \\ 65 && 4[1 + 14 + 3(16)] & \\ = 65 \text{ g} & & = 252 \text{ g} & \\ & & & \\ \text{Zn(NO}_3)_2 & + & 2\text{H}_2\text{O} & + & 2\text{NO}_2 \\ 1\text{ mole}&& 2 \text{ moles} && 2\text{ moles} \\ 65 + 2[14 + 3(16)] & & 2[2(1) + 16] && 2[14 + 2(16)] \\ = 189 \text{ g} & & = 36 \text{ g} & & = 92 \text{ g} \\ \end{matrix}$$

(a) Number of moles of Zn = $\frac{\text{Mass}}{\text{Molar mass}}$ = $\frac{32.5}{65}$ = 0.5 mole

(b) From the equation, 65 g of Zn reacts with 252 g of HNO₃

∴ 32.5 g of Zn reacts with $\frac{252}{65}$ × 32.5 = 126 g of HNO₃

(c) From the equation, 65 g of Zn produces 2 × 22.4 = 44.8 L of NO₂ at STP

∴ 32.5 g of Zn produces $\frac{44.8}{65}$ × 32.5 = 22.4 L of NO₂ at STP

Hence, 0.5 mole of zinc is required, 126 g of nitric acid is needed, and 22.4 L of NO₂ is liberated at STP.

(a) Concentrated Nitric Acid (HNO₃)

(b) C + 4HNO₃ [conc.] ⟶ CO₂ + 2H₂O + 4NO₂

(a) Bauxite contains alumina along with impurities such as ferric oxide and sand. Alumina is amphoteric and dissolves in hot concentrated caustic soda to form soluble sodium meta-aluminate, while the insoluble impurities remain as red mud and are removed by filtration.

(b) Al₂O₃·2H₂O + 2NaOH ⟶ 2NaAlO₂ + 3H₂O

(a) A white precipitate of calcium hydroxide is formed, which is insoluble in excess sodium hydroxide.
$Ca(NO_{3})_{2} + 2NaOH \longrightarrow \underset{[\text{white ppt.}]}{Ca(OH)_{2} \downarrow} + 2NaNO_{3}$

(b) A colourless gas with the smell of rotten eggs, hydrogen sulphide (H₂S), is evolved with effervescence.
FeS + 2HCl ⟶ FeCl₂ + H₂S ↑

(c) The amphoteric metal (e.g., zinc or aluminium) dissolves in the hot concentrated alkali showing vigorous effervescence (bubbling) due to the evolution of a colourless, odourless gas. This gas burns with a "pop" sound when a burning splinter is brought near the mouth of the test tube, confirming it is hydrogen.

$Zn + 2NaOH \longrightarrow \underset{\text{(sodium zincate)}}{Na_{2}ZnO_{2}} + H_{2} \uparrow$

A → Concentrated Sulphuric Acid (conc. H₂SO₄)
B → Acetic acid (CH₃COOH)
C → Oxygen (O₂)

Explanation

• Ethanol ⟶ Ethene (with reactant A): This is dehydration of ethanol. A is concentrated sulphuric acid (conc. H₂SO₄) at 170°C.
  C₂H₅OH $\xrightarrow[170^{\circ}\text{C}]{\text{conc. H}_2\text{SO}_4}$ C₂H₄ + H₂O

• Ethanol ⟶ Ethyl acetate (with reactant B): This is esterification. B is acetic acid (CH₃COOH) in the presence of conc. H₂SO₄ (acts as a catalyst).
  C₂H₅OH + CH₃COOH $\xrightarrow{\text{conc. H}_2\text{SO}_4}$ CH₃COOC₂H₅ + H₂O

• Ethanol ⟶ Carbon dioxide + Water (with reactant C): This is combustion. C is oxygen (O₂).
  C₂H₅OH + 3O₂ ⟶ 2CO₂ + 3H₂O

Hence, A = Concentrated H₂SO₄, B = Acetic acid (CH₃COOH), and C = Oxygen (O₂).

At STP, 22.4 L of any gas = 1 mole

∴ Number of moles of Z = $\frac{5.6}{22.4}$ = 0.25 mole

Molecular weight of Z = $\frac{\text{Mass}}{\text{Number of moles}}$ = $\frac{32}{0.25}$ = 128 g/mol

Vapour density = $\frac{\text{Molecular weight}}{2}$ = $\frac{128}{2}$ = 64

Hence, the molecular weight of Z is 128 and its vapour density is 64.

(a) Fluorine (F)

(b) Sodium (Na)

(a) Fluorine (F)

(b) Calcium (Ca)

(c) Potassium (K)

Explanation

(a) Fluorine (F) has the least atomic radius. It is on the rightmost side of period 2, where atomic radius is minimum.

(b) Calcium (Ca) has two valence electrons (electronic configuration: 2, 8, 8, 2). It belongs to Group 2 of the periodic table.

(c) Potassium (K) is the most electropositive. Within Group 1, electropositive character increases down the group; potassium is below lithium and hence is more electropositive.

(a) L — Sodium chloride is a strong electrolyte and ionises completely in water, dissociating fully into Na⁺ and Cl^- ions.

NaCl ⟶ Na⁺ + Cl^-

(b) M — Distilled water consists almost entirely of H₂O molecules and contains no ions (its self-ionisation is negligible).

(c) K — Aqueous acetic acid is acidic in nature (it is a weak acid which partially ionises to give H⁺ ions), and hence has a pH less than 7.

CH₃COOH ⇌ CH₃COO^- + H⁺

salt + water

Reason — A metal oxide is basic in nature. When it reacts with an acid, neutralisation takes place to give a salt and water only.

CuO + H₂SO₄ ⟶ CuSO₄ + H₂O

water

Reason — In a water molecule (H₂O), oxygen forms two polar O—H covalent bonds and still carries two unshared (lone) pairs of electrons. The bent shape makes the molecule polar. Ammonia has only one lone pair, while chlorine (Cl₂) and oxygen (O₂) are non-polar molecules.

A pale blue precipitate is formed that dissolves in excess of ammonium hydroxide to form a deep blue solution.

Reason — On adding ammonium hydroxide dropwise, a pale blue precipitate of copper hydroxide is formed. In excess ammonium hydroxide, the precipitate dissolves to form a deep blue solution of tetraammine copper(II) sulphate.

CuSO₄ + 2NH₄OH ⟶ Cu(OH)₂ ↓ [pale blue] + (NH₄)₂SO₄

Both 3 and 4

Reason — Group 1 elements (alkali metals) are good conductors of heat and electricity, so statements 1 and 2 are true. They readily lose their single valence electron, so they are strong reducing agents (not oxidising agents); hence statement 3 is false. They form ionic (electrovalent) compounds, not covalent compounds, with non-metals; hence statement 4 is also false.

C₂H₆

Reason — C₂H₆ (ethane) is a saturated hydrocarbon (alkane) containing only single bonds, and saturated hydrocarbons undergo substitution reactions. C₂H₄, C₂H₂ and C₁₀H₁₈ are unsaturated and undergo addition reactions.

A greenish yellow gas is liberated that turns moist starch iodide paper blue-black.

Reason — Manganese dioxide oxidises concentrated hydrochloric acid on heating to liberate chlorine, a greenish-yellow gas. Chlorine liberates iodine from starch iodide paper, turning it blue-black.

MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

3 moles of hydrogen

Reason — At STP, 1 mole of any gas occupies 22.4 L.

• 3 moles of hydrogen = 3 × 22.4 = 67.2 L
• 1 mole of helium = 22.4 L
• 64 g of oxygen = $\frac{64}{32}$ = 2 moles = 44.8 L
• 44 g of carbon dioxide = $\frac{44}{44}$ = 1 mole = 22.4 L

Hence, 3 moles of hydrogen occupy the maximum volume (67.2 L).

X and Y

Reason — An electrovalent (ionic) compound is formed between a metal and a non-metal. X (atomic number 13, configuration 2, 8, 3) is a metal that loses 3 electrons to form X³⁺, and Y (atomic number 7, configuration 2, 5) is a non-metal that gains 3 electrons to form Y^3-. They combine to form an electrovalent compound (XY). Z (atomic number 10) is a noble gas and does not form compounds.

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

Reason — Noble gases have completely filled valence shells (a stable octet, or a duplet in the case of helium). Because of this stable configuration they have almost no tendency to gain or lose electrons, and so they are unreactive under standard conditions. Thus the reason correctly explains the assertion.

Froth flotation method

Reason — In the froth flotation method, the powdered ore is stirred with water and pine oil. The ore particles are preferentially wetted by the oil and rise up with the froth, while the gangue particles are wetted by water and settle down. This separation based on preferential wetting is the froth flotation method.

3 shells and 2 valence electrons

Reason — The period number gives the number of shells and, for Groups 1 and 2, the group number gives the number of valence electrons. Hence an element in Period 3 and Group 2 has 3 shells and 2 valence electrons (electronic configuration 2, 8, 2).

23

Reason — Vapour density = $\frac{\text{Molecular mass}}{2}$. Molecular mass of NO₂ = 14 + 2(16) = 46. So vapour density = $\frac{46}{2}$ = 23.

Hydrogen

Reason — Reduction (gain of electrons) takes place at the cathode. Hydrogen ions (H⁺) migrate to the cathode and gain electrons to be reduced to hydrogen gas. Bromine, chlorine and oxygen are formed by oxidation (loss of electrons) at the anode.

2H⁺ + 2e^- ⟶ H₂

Sodium atom loses an electron to form Na⁺ and chlorine atom gains an electron to form Cl^-

Reason — Sodium has one valence electron which it loses to attain a stable octet, forming Na⁺. Chlorine has seven valence electrons and gains one electron to complete its octet, forming Cl^-. These oppositely charged ions are held together by an ionic bond.

polar covalent compound

Reason — HCl is a polar covalent compound formed by the unequal sharing of electrons between hydrogen and the more electronegative chlorine. Being polar, it dissolves readily in water and ionises to form H₃O⁺ and Cl^- ions.

(a)

1. R — Concentrated sulphuric acid (H₂SO₄)
2. S — Sodium nitrate (NaNO₃) [potassium nitrate, KNO₃, is also acceptable]

(b) Only glass apparatus is used because nitric acid vapours attack rubber and cork.

(c) 4HNO₃ $\xrightarrow{\text{heat / light}}$ 4NO₂ + 2H₂O + O₂

(d) Dense reddish-brown fumes of nitrogen dioxide are evolved and a blue solution of copper(II) nitrate is formed.

Cu + 4HNO₃ [conc.] ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

(a) Lead(II) chloride (PbCl₂)

(b) Ammonium hydroxide (NH₄OH)

(c) Ferric ion / Iron(III) ion (Fe³⁺)

(d) Avogadro's Law

(e) Electrodes

(a) The IUPAC name for formaldehyde is methanal.

(b) A black solid, copper oxide reacts with dilute sulphuric acid to form a blue-coloured solution.

(c) The drying agent not used for drying ammonia is P₂O₅.

(d) The salt formed when iron reacts with dilute HCl is FeCl₂.

(e) Ammonia reacts with excess chlorine to form NCl₃.

(a) Structural diagrams:

1. ethanal (CH₃CHO)

2. acetic acid (CH₃COOH)

3. 2-bromopropane (CH₃CHBrCH₃)

(b) IUPAC names:

1. 2-methylbutane
2. Prop-1-ene (propene)

(a) Element D (configuration 2, 8, 1) is an alkali metal (sodium). It reacts with water to form an alkali (sodium hydroxide).

(b) D has a valency of 1 (forms D⁺) and C has a valency of 3 (forms C^3-). Therefore the formula of the compound is D₃C.

(a) X — It reacts with both an acid (HCl) and a base (NaOH), which is the property of an amphoteric oxide.

(b) Z — It reacts with a base (NaOH) but not with an acid, which is the property of an acidic oxide.

$$\begin{matrix} 2\text{KNO}_3 & \longrightarrow & 2\text{KNO}_2 & + & \text{O}_2 \\ 2 \text{ moles} && 2 \text{ moles} && 1 \text{ mole} \\ 2[39 + 14 + 3(16)] && 2[39 + 14 + 2(16)] && 2(16) \\ = 202 \text{ g} && = 170 \text{ g} && = 32 \text{ g} \\ \end{matrix}$$

(a) From the equation, 202 g of KNO₃ gives 22.4 L of O₂ at STP.

∴ 20.2 g of KNO₃ gives $\frac{22.4}{202}$ × 20.2 = 2.24 L of O₂ at STP.

(b) From the equation, 202 g of KNO₃ gives 170 g of KNO₂.

∴ 20.2 g of KNO₃ gives $\frac{170}{202}$ × 20.2 = 17 g of KNO₂.

Hence, 2.24 L of oxygen is obtained at STP and 17 g of KNO₂ is formed.

(a) Set-up Q — It uses sodium argentocyanide as the electrolyte, which releases silver ions slowly and gives a uniform, smooth and adherent coating of silver on the spoon (cathode).

(b) Set-up P — It uses silver nitrate solution, which ionises readily and releases a high concentration of silver ions, so silver is deposited too quickly, giving an uneven and rough coating.

(c) At the anode, the silver plate dissolves into the electrolyte:

Ag - e^- ⟶ Ag⁺

(a) O₂ — Oxygen is a colourless gas evolved during the thermal decomposition of metallic nitrates.

(b) NO — Nitric oxide (colourless) on oxidation gives the reddish-brown gas nitrogen dioxide.

2NO + O₂ ⟶ 2NO₂

(a)

Hydronium ion (H3O+)	Ammonium ion (NH4+)
It has one lone pair of electrons (on the oxygen atom).	It has no lone pair of electrons; all four electron pairs are bonded.

(b)

Calcium chloride	Calcium nitrate
On adding silver nitrate solution, a white precipitate of silver chloride is formed.	On adding silver nitrate solution, no precipitate is formed (no visible change).

CaCl₂ + 2AgNO₃ ⟶ 2AgCl ↓ [white ppt.] + Ca(NO₃)₂

(a) C₅H₈ and C₈H₁₄ — Both follow the general formula of alkynes, C_nH_2n-2, which contain a carbon–carbon triple bond.

(b) C₆H₆ — It represents benzene, a cyclic (aromatic) hydrocarbon.

(a) Aluminium oxide — Bauxite is hydrated aluminium oxide, Al₂O₃·2H₂O.

(b) Iron(II) chloride — Fe²⁺ ions form a dirty green precipitate of iron(II) hydroxide with NaOH.

(c) Sodium bisulphite — Sodium hydrogen sulphite (NaHSO₃) still has a replaceable hydrogen ion, so it is an acid salt.

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	26.7	12	$\frac{26.7}{12}$ = 2.225	$\frac{2.225}{2.2}$ = 1
Hydrogen	2.2	1	$\frac{2.2}{1}$ = 2.2	$\frac{2.2}{2.2}$ = 1
Oxygen	71.1	16	$\frac{71.1}{16}$ = 4.44	$\frac{4.44}{2.2}$ = 2

Simplest ratio C : H : O = 1 : 1 : 2

Hence, empirical formula is CHO₂.

Empirical formula weight = 12 + 1 + 2(16) = 45

n = $\frac{\text{Relative molecular mass}}{\text{Empirical formula weight}}$ = $\frac{90}{45}$ = 2

Molecular formula = 2 × (CHO₂) = C₂H₂O₄

(a) The solution is neutral, so its pH = 7.

(b) The solution is an acid (an acid reacts with ferrous sulphide to give hydrogen sulphide), so its pH is less than 7.

(c) The solution is an alkali (an alkali on heating with an ammonium salt gives pungent-smelling ammonia gas), so its pH is greater than 7.

(a) Electronic configuration of P (atomic number 19) = 2, 8, 8, 1

(b) Q has 6 valence electrons and gains 2 electrons to complete its octet, so its valency is 2.

(c) Element R has a tendency to gain one electron, so it is a non-metal.

(d) P has 4 electron shells (2, 8, 8, 1), so it belongs to Period 4.

(a) False — Brass is an alloy of copper and zinc, both of which are metals; it does not contain a non-metal.

(b) True — Duralumin is an alloy of aluminium, copper, magnesium and manganese, so it contains copper.

(c) False — Vapour density is half the molecular mass (Molecular mass = 2 × Vapour density), not twice it.

(a) Na₂CO₃ + 2HNO₃ ⟶ 2NaNO₃ + H₂O + CO₂

(b) Carbon dioxide (CO₂)

(c) C + 4HNO₃ [conc.] ⟶ CO₂ + 2H₂O + 4NO₂

(a) Fluorspar (calcium fluoride, CaF₂) — it contains the divalent metal calcium.

(b)

1. Y (anode): 2O^2- - 4e^- ⟶ O₂
2. Z (cathode): Al³⁺ + 3e^- ⟶ Al

(c) The powdered coke sprinkled over the electrolyte prevents the carbon anodes from burning in air.

(a) Non-volatile nature — being a non-volatile (less volatile) acid, it displaces the more volatile hydrogen chloride from a metal chloride.

(b) Dehydrating property — it removes the water of crystallisation from blue copper sulphate crystals, leaving behind white anhydrous copper sulphate.

(a) Carbonate ion (CO₃^2-) — the gas evolved is carbon dioxide, which turns lime water milky but has no effect on potassium dichromate.

(b) Chloride ion (Cl^-) — the gas evolved is hydrogen chloride, which gives a white precipitate of silver chloride (insoluble in dilute nitric acid) with silver nitrate.

(a) Zinc sulphate — G: Simple displacement (zinc, being above hydrogen, displaces hydrogen from dilute sulphuric acid).

(b) Silver chloride — F: Precipitation (silver chloride is an insoluble salt).

(c) Ammonium nitrate — E: Neutralization by titration (a soluble salt of the alkali ammonium hydroxide with nitric acid).

(a) Metal X is lead.

(b) The bromide is in the molten state.

(c) The sulphate of metal X will have the formula XSO₄.

(a) P, Si, Al, Mg, Na

(b) I, Br, Cl, F

(c) He, Ne, Ar, Ca

(a) L belongs to Group 2 (valency 2, forms L²⁺) and P belongs to Group 17 (valency 1, forms P^-). One atom of L transfers one electron each to two atoms of P, forming the electrovalent compound LP₂.

(b) Q — It is a noble gas placed at the top of Group 18 (rightmost and highest in the table), where the atomic size is smallest and the outermost electrons are most tightly held, giving it the highest ionization potential.

(c) Both M and N belong to Group 1. Going down the group, the atomic size increases and the outermost electron is held less tightly by the nucleus, so it is lost more easily. Since N lies below M in the group, N loses its valence electron more readily and is therefore more reactive than M.

(a) P is ethyne (acetylene), C₂H₂.

(b) The general formula of the homologous series (alkynes) is C_nH_2n-2.

(c) CaC₂ + 2H₂O ⟶ Ca(OH)₂ + C₂H₂

(d) It undergoes an addition reaction with hydrogen.

NaCl

Reason — NaCl dissociates in water into positive ion Na⁺ other than hydronium ion, and a negative ion Cl^- other than hydroxyl ion. Whereas, KOH gives hydroxyl ion and HCl and CH₃COOH gives hydronium ion in water.

Ammonium sulphate

Reason — Ammonium sulphate reacts with sodium hydroxide to produce sodium sulphate, ammonia gas, and water. The red litmus turns blue due to the basic ammonia gas.

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

Reason — KCl is an ionic compound. In water, it dissociates completely into K⁺ and Cl^- ions. These free moving ions carry electric current. Hence, the assertion (A) is true. The principle behind electrolytic conduction in solutions is the presence of free ions. Aqueous solution of KCl conduct electricity due to the presence of free ions. Hence, the reason (R) is true and it correctly explains assertion (A).

H₃O⁺

Reason — The pair of electrons not shared by any other atom are called lone pair of electrons. H₃O⁺ has only one lone pair of electrons. Whereas, OH^1- contains three lone pair of electrons and NH₄⁺ and H⁺ contains no lone pair of electrons.

3 and 4

Reason — Reaction 3 shows burning of ethanol, which is an example of combustion and combustion produces water and carbon dioxide. Reaction 4 has dilute sulphuric acid and magnesium carbonate, a reaction of acid and carbonate produces salt, water and carbon dioxide. Hence, reaction 3 and reaction 4 produces water.

Potassium sulphite

Reason — Potassium sulphite reacts with dilute sulphuric acid, to release sulphur dioxide. SO₂ is a reducing gas, that turns acidified potassium dichromate solution from orange to green.

K₂SO₃ + H₂SO₄ ⟶ SO₂ ↑ + H₂O + K₂SO₄

44.8L

Reason — One mole of gas occupies 22.4 L of volume at STP

∴ Volume occupied by 2 moles of gas = 2 x 22.4 L = 44.8 L

Aluminium oxide

Reason — Amphoteric oxides like aluminium oxide react with acids as well as alkalis giving salt and water.

2

Reason — Among the four hydrocarbon structure show, structure 3 and 4 are isomers of butene. Structure 3 is But-1-ene and structure 4 is 2-methyl propene.

Na

Reason — Aluminium (Al), Sulphur (S), Magnesium (Mg) and Sodium (Na) belong to third period. Atomic size decreases on moving across the period as new electrons are added to the same shell. Sodium is the first member of period 3, hence sodium has the largest atomic radius of the given elements.

4.

6

2

Reason — According to the pH scale, lesser the pH more is the acidity. When pH decreases from 6 to 2, the amount of H⁺ increases 10,000 times.

pH= −log₁₀ [H⁺ ] ⟶ [H⁺ ] = 10^-pH ​

Initial [H⁺] = 10^-6 = 0.000001 mol/L
Final [H⁺] = 10^-2 = 0.01 mol/L

Change: $\frac{0.01}{0.000001}$ = 10,000 times more acidic

Both positive and negative ions

Reason — Sulphuric acid will dissociate into H⁺ and SO₄^2-. Also, in aqueous solution, the compound XSO₄ would dissociate into X⁺ and SO₄^2-. In both cases, both positive and negative ions are responsible for conducting electricity.

They have the same functional group.

Reason — Methanol and ethanol belong to the same homologous series. They share the general formula and same functional group. Ethanol differs methanol by CH₂ group. Relative molecular mass and number of carbon atoms in successive members of homologous series increases.

1:10

Reason — Molar mass of CO₂ = 12 + 2 x 16 = 44 g/mole

∴ Moles of 22g CO₂ = $\frac{22}{44}$ = 0.5

Molar mass of H₂ = 2 x 1 = 2 g/mole

∴ Moles of 10g H₂ = $\frac{10}{2}$ = 5

Under the same conditions of temperature and pressure, the volume of a gas is directly proportional to the number of moles. Therefore, the volume ratio will be the same as the mole ratio.

Volume ratio = $\frac{\text{Vol. of CO}_2}{\text{Vol. of H}_2}$ = $\frac{0.5}{5}$ = 1 : 10

Copper ions from the anode move to the cathode and get deposited as pure Copper.

Reason — At the anode, impure copper dissolves into the solution as Cu²⁺ ions. These Cu²⁺ ions migrate through the solution to the cathode, where they gain electrons and are deposited as pure copper.

(a) Coordinate bond

(b) Normal salt

(c) Substitution/ Halogenation

(d) Ionisation potential

(e) Alloy

(a) When dilute sulphuric acid reacts with zinc granules, the gas evolved is hydrogen, which can be tested using a burning splint.

(b) A solution of copper(II) sulphate in sodium hydroxide solution forms a pale blue precipitate.

(c) In methane, each hydrogen atom share one electron with the central carbon atom to complete its valence shell.

(d) The electron affinity of element X is greater than that of element Y.The oxidising power of X is likely to be more than that of element Y.

(e) The naturally occurring compound of a metal from which the metal can be extracted is called its ore.

(a) IUPAC name of the following organic compounds are :

1. propanol
2. pentene

(b) Structural formula

1. But-2-yne

2. 1, 1, 1, trichloro methane

3. Pentan-2-ol

(a) Colourless ammonia gas reacts with greenish yellow excess chlorine giving a yellow explosive liquid (Nitrogen trichloride).
NH₃ + 3Cl₂ [excess] ⟶ 3HCl + NCl₃

(b) Reddish brown nitrogen dioxide gas is evolved on heating zinc nitrate crystals.
$2\text{Zn(NO}_3)_2 \xrightarrow{\Delta} 2\text{ZnO} + \text{O}_2 + 4\text{NO}_2$

(a) When ammonia gas is passed over black copper oxide in a combustion tube a reddish-brown substance is left behind because ammonia gas reduces black copper [II] oxide to brown Copper.
2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ ↑

(b) Quick lime is not used to dry hydrogen chloride gas because quick lime reacts with hydrogen chloride gas

(a) Oxidising power of X > Y. Elements with high electron affinity accept electrons more easily hence have greater oxidising power.

(b) Electronegativity of X > Y. Element with high electron affinity has high electronegativity.

(c) X is to the right side of Y as electron affinity increases from left to right in a period.

(a) 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

(b) 2HNO₃ + ZnCO₃ ⟶ Zn(NO₃)₂ + H₂O + CO₂

(c) H₂SO₄ + CuSO₄. 5H₂O ⟶ CuSO₄ + 5H₂O + H₂SO₄

(a) Aluminium

(b) Copper

(a) NaCl + conc. H₂SO₄ ⟶ NaHSO₄ + HCl

(b) C₂H₄ + Br₂ ⟶ C₂H₄Br₂

(a) Nitrogen dioxide
Reason — The reddish-brown colour indicates that the gas is Nitrogen dioxide.

(b) Lead ion (Pb²⁺)
Reason — The chalky white insoluble precipitate with excess of ammonium hydroxide indicates that the cation present is Pb²⁺.

(c) Lead nitrate[Pb(NO₃)₂]
Reason — The given salt is a nitrate salt as nitrogen dioxide (NO₂), is a common product obtained from heating nitrates. As the cation is lead, hence the salt is Lead nitrate.

(a) Esterification reaction

$\underset{\text{ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} + \text{CH}_3\text{COOH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \underset{\text{ethyl acetate}}{{\text{CH}_3\text{COOC}_2\text{H}_5}} + \text{H}_2\text{O}$

(b) Dehydrating agent

(c) Fruity smell is observed due to the formation of ester by the reaction of ethanol and acetic acid.

Q is Calcium carbonate or Calcium bicarbonate

CaCO₃ + 2HNO₃ ⟶ Ca(NO₃)₂ + H₂O + CO₂

or

Ca(HCO₃)₂ + 2HNO₃ ⟶ Ca(NO₃)₂ + 2H₂O + 2CO₂

Molecular mass of O₂ = 2 × 16 = 32 g/mol

Given mass = 3.2 g of Oxygen

Moles of O₂ = $\frac{\text{Mass}}{\text{Molar mass}}$ = $\frac{3.2}{32}$ = 0.1 mol

∴ 0.1 mol of O₂ contain = 0.1 x 6.023 × 10²³ of molecules.

We need the same number of molecules of CO₂

∴ Moles of CO₂ will also be 0.1 mol

Molar mass of CO₂ = 12 + 2(16) = 12 + 32 = 44 g/mol

Mass = moles x molar mass

Mass of CO₂ = 0.1 x 44 = 4.4 g.

∴ Mass of CO₂ = 4.4 g

(a) Oxidising property

(b) Acidic property

(c) Dehydrating property

(a) CaC₂ + 2H₂O ⟶ Ca(OH)₂ + C₂H₂

(b) CH₃COOH + C₂H₅OH ⟶ CH₃COOC₂H₅ + H₂O

(C) NaNO₃ + conc. H₂SO₄ ⟶ NaHSO₄ + 2HNO₃

(a) White precipitate of barium sulphate is formed.

Reason — When a solution of zinc sulphate is added to a solution of barium chloride, a double displacement reaction occurs. This reaction involves the exchange of ions between the two compounds and a white precipitate of barium sulphate is formed.
ZnSO₄ (aq) + BaCl₂ (aq) ⟶ BaSO₄ (s) + ZnCl₂ (aq).

(b) Blue precipitate dissolves to form an inky blue/deep blue solution.

Reason — The light blue precipitate of copper(II) hydroxide (Cu(OH)₂) dissolves in excess of ammonium hydroxide due to the formation of a deep blue complex, tetraamminecopper(II) ([Cu(NH₃)₄]²⁺), making it soluble.

(a) CuCO₃ + H₂SO₄ ⟶ CuSO₄ + H₂O + CO₂

(b) Zn + 2NaOH ⟶ Na₂ZnO₂ + H₂

$$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{N}_2 & + & 4\text{H}_2\text{O} & + & \text{Cr}_2\text{O}_3 \\ 2[14 + 4(1)] & & & & & & 2(52) \\ + 2(52) + 7(16) & & & & & & + 3(16) \\ = 36 + 104 & & & & & & = 104 + 48 \\ + 112 = 252 \text{ g} & & & & & & = 152 \text{ g} \\ \end{matrix}$$

(a) 252 g of (NH₄)₂Cr₂O₇ = 1 mole

∴ 126 g of (NH₄)₂Cr₂O₇ = $\frac{1}{252}$ x 126 = 0.5 moles

Hence, no. of moles = 0.5 moles

(b) 252 g of (NH₄)₂Cr₂O₇ gives 152 g of Cr₂O₇

126 g of (NH₄)₂Cr₂O₇ will give $\frac{152 \times 126}{252}$ = 76 g of Cr₂O₇

Hence, mass in gms of Cr₂O₇ formed = 76 g.

(c) 252 g of (NH₄)₂Cr₂O₇ produces 22.4 lit of N₂

126 g of (NH₄)₂Cr₂O₇ will produce $\frac{22.4 \times 126}{252}$ = 11.2 lit of N₂

Hence, volume of N₂ evolved at s.t.p = 11.2 lit.

(a) Hydrogen

(b) Oxygen

(c) Silver Nitrate

(a) Nitric Acid is a very strong oxidising agent and hence oxidises hydrogen to water. So, it is not used for obtaining hydrogen from metals.

(b) Silver plated cutlery is not considered as pure silver because it is not made fully of silver only its top coating is of silver.

(a) Cryolite

(b) 2Al³⁺ + 6e^- ⟶ 2Al

(a) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃

(b) C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

(c) C₂H₅Cl + aq. NaOH ⟶ C₂H₅OH + NaCl

(a) X

Reason — Sodium sulphite can react with acids to liberate sulphur dioxide gas. So, the solution X with pH 2 (acidic) will react with sodium sulphite to liberate sulphur dioxide gas. Na₂SO₃ + 2H^- ⟶ SO ₂+ H₂O + 2Na⁺

(b) Z

Reason — Ammonium chloride is a salt that, in the presence of a base, will release ammonia gas. Therefore, a basic solution (with a high pH) will react with ammonium chloride to liberate ammonia gas. Thus, the solution Z with pH 13 (basic) will liberate ammonia gas when reacted with ammonium chloride. NH₄Cl + OH^- ⟶ NH₃ + H₃O + Cl^-

(c) Y

Reason — The solution Y with pH 7 will not have any effect on litmus paper as it is neither acidic nor basic.

(a) Zinc and aluminium cannot be distinguished by heating the metal powder with concentrated sodium hydroxide solution as they both react with conc. alkalis to form soluble sodium salts and hydrogen gas.

Zinc reacts to form sodium zincate

Zn + 2NaOH ⟶ H₂ + Na₂ZnO₂ [sodium zincate]

Aluminium reacts to form sodium aluminate

2Al + 2NaOH + 2H₂O ⟶ 3H₂ + 2NaAlO₂ [sodium aluminate]

(b) Yes, calcium nitrate and lead nitrate can be distinguished using ammonium hydroxide solution. Ammonium hydroxide on reaction with lead nitrate gives chalky white precipitate of Pb(OH)2. No precipitation occurs on adding Ammonium hydroxide to calcium nitrate even when it is added in excess.

Pb(NO₃)₂ + 2NH₄OH ⟶ Pb(OH)₂ + 2NH₄NO₃

Electron dot diagram of ammonium ion is shown below:

A = Zn + H₂SO₄ ⟶ ZnSO₄ + H₂

B = Zn + S ⟶ ZnS

C = ZnSO₄ + 2NaOH ⟶ Zn(OH)₂ + Na₂SO₄

(a) M
Reason — Atomic number of element M is 7 so its electronic configuration is 2,5.Hence, it is a non-metal with 5 valence electrons. It combines with hydrogen to form a covalent hydride, MH₃, which is a basic gas. MH₃ reacts with concentrated HCl to give dense white fumes of ammonium salt.

(b) N
Reason — Atomic number of element N is 10 so its electronic configuration is 2, 8. It has electron affinity zero as it's octet is complete and it is a stable element.

(c) L
Reason — Atomic number of element L is 7 so its electronic configuration is 2, 8, 3. Hence, it is a metal with 3 valence electrons. Metals tend to lose electrons, and non-metals like oxygen tend to gain electrons, so L can transfer its 3 valence electrons to oxygen. This forms L³⁺ ions and O^2- ions, giving an ionic compound with oxygen, generally of the form L₂O₃. So, L is the element that can form an ionic compound with oxygen.

Sulphur dioxide

Reason — Sulphur dioxide gas turns acidified potassium permanganate from pink to clear colourless.

2KMnO₄ + 2H₂O + 5SO₄ ⟶ K₂SO₄ + MnSO₄ + H₂SO₄

C₅H₁₂

Reason — C₅H₁₂ is an alkane that represents a saturated hydrocarbon. It is a simplest open chain hydrocarbon represented by the formula C_nH_2n+2.

Copper

Reason — Copper lies below hydrogen in the reactivity (activity) series, so its oxide is not very stable. Therefore, common chemical reducing agents such as carbon, carbon monoxide, or hydrogen can readily reduce copper(II) oxide to metallic copper, e.g.

CuO + H₂ → Cu + H₂O

In contrast, the oxides of sodium, aluminium and potassium are extremely stable because those metals are highly reactive and have a very strong affinity for oxygen. Their oxides can be reduced only by electrolysis and not by ordinary chemical reducing agents.

C₃H₈

Reason

Given,

Vapour density of organic compound = 22

Molecular weight = 2 x V.D. = 2 x 22 = 44

Compound	Molecular mass
1. CH4	16
2. C2H4	28
3. C2H6	30
4. C3H8	44

Hence, C₃H₈ is a organic compound that has the molecular mass of 44.

Oxidising agent

Reason — Oxidising agents are substances which add oxygen atoms to others compounds or remove hydrogen atoms from others compounds. Sulphuric acid accepts electrons and donates oxygen atoms to sulphur, leading to the formation of SO₂ and water (H₂O). Thus, sulphuric acid acts as an oxidising agent.

(A) is true but (R) is false.

Reason — As we move down any group in the periodic table, atomic size increases and the outermost electron is held less tightly by the nucleus. Therefore the loss of this electron becomes easier, so the tendency to lose electrons increases; Assertion (A) is true. In group 1 the reactivity of metals increases down the group for the same reason. Consequently, the most reactive metal is found at the bottom, not at the top; Reason (R) is false.

Haematite

Reason — Magnetic separation works when either the ore or the gangue shows magnetic behaviour. Haematite, an iron ore, is weakly magnetic, whereas its common impurity (silica) is non-magnetic. When the crushed mixture is passed over a high-intensity magnetic separator, the haematite particles are attracted and collected separately from the non-magnetic gangue, concentrating the ore.

4

2, 6

2, 8, 7

Reason

• Atom A has six valence electrons (configuration 2, 6) and therefore needs two more electrons to complete its octet.
• Each atom B has seven valence electrons (configuration 2, 8, 7) and needs one additional electron to attain a stable, noble-gas configuration.
• In the AB₂ molecule, atom A forms a single covalent bond with each of two B atoms. Through this sharing, atom A gains two electrons (one from each B), while each B gains one electron from A, so all three atoms achieve an octet.

CH₄, C₂H₆, C₃H₈

Reason — These three molecules are consecutive members of the alkane series, which has the general formula C_nH_2n+2. Each successive member differs from the previous one by a CH₂ group, satisfying the defining characteristic of a homologous series.

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

Reason — When sulphur trioxide is brought into direct contact with water,

SO₃(g) + H₂O(l) ⟶ H₂SO₄(l)

the reaction is highly exothermic. The heat released vaporises some of the acid and produces a dense fog of minute H₂SO₄ droplets, which are difficult to condense and collect. Therefore, in the Contact Process the SO₃ is first absorbed in concentrated sulphuric acid to form oleum, which is later diluted with water to give the required concentration of H₂SO₄.

Cl^- & K⁺

Reason

• Chlorine has atomic number 17. Its neutral atom contains 17 electrons with the configuration 2, 8, 7. When it gains one electron to form Cl⁻, it has 18 electrons, giving the configuration 2, 8, 8 (the same as argon).
• Potassium has atomic number 19. A neutral K atom has the configuration 2, 8, 8, 1. When it loses one electron to form K⁺, it is left with 18 electrons, also giving the configuration 2, 8, 8.

Because both Cl⁻ and K⁺ possess 18 electrons and share the electron configuration 2, 8, 8, they form the required pair with identical electronic configurations.

Sodium sulphate + Lead nitrate

Reason — Both sodium sulphate (Na₂SO₄) and lead(II) nitrate [Pb(NO₃)₂] are readily soluble in water. When their aqueous solutions are mixed, a double-decomposition reaction occurs, forming insoluble lead(II) sulphate (PbSO₄) as a white precipitate:

Pb(NO₃)₂(aq) + Na₂SO₄(aq) ⟶ PbSO₄(s) ↓ + 2NaNO₃(aq)

Because the desired product precipitates out directly and the by-product (sodium nitrate) remains in solution, this pair of reactants is the most convenient for preparing pure lead(II) sulphate.

The number of copper (II) ions in the electrolyte decreases.

Reason — Cu²⁺ ions deposited at the cathode are replaced by Cu²⁺ ions produced at the anode, so the overall Cu²⁺ concentration stays constant.

Z

Reason — The greater the tendency to lose electrons, the greater the metallic character and the stronger the reducing power of a metal. As we move down a group, atomic size increases and ionisation energy decreases, so the tendency to lose electrons and thus the reducing power also increases. Therefore, of the three atoms, Z (the lowest member of the group) is the best reducing agent.

40 cm³

Reason — From the equation :

CH₄ + 2O₂ ⟶ CO₂ + 2H₂O

1 volume of CH₄ reacts with 2 volumes of O₂

For 40 cm³ of CH₄ the required O₂ is 40 cm³ x 2 = 80 cm³

Since only 60 cm³ of O₂ is available.

The amount of CH₄ that reacts with 60 cm³ of O₂ is = $\frac{60}{2}$ = 30 cm³

∴ The excess CH₄ remaining is 40 cm³ - 30 cm³ = 10 cm³

From the balanced equation, 2 volumes of O₂ produce 1 volume of CO₂.

∴ The volume of CO₂ produced from 60 cm of O₂ is $\frac{60}{2}$ = 30 cm³

The remaining gases are excess CH₄ and produced CO₂.

∴ Total volume = 10 cm³ (excess CH₄) + 30 cm³ (CO₂) = 40 cm³

(a) X — Water

(b) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃↑

(c) Y — Quicklime (CaO)

(d) Substance Y — Quicklime (CaO) is used as a drying agent

(e) Ammonia gas is highly soluble in water so it cannot be collected over water hence it is collected by the downward displacement of air.

(a) Spurious alcohol

(b) Basic salt

(c) Isomers

(d) Electronegativity

(e) Polar covalent bond.

(a) Copper nitrate solution forms a coloured precipitate with ammonium hydroxide which is soluble in excess of ammonium hydroxide.

(b) Zinc blende is converted to zinc oxide by roasting.

(c) Molten sodium chloride conducts electricity by the movement of ions.

(d) The reaction that takes place at the anode during the electrolysis of aqueous Sodium argentocyanide with silver electrodes is Ag ⟶ Ag⁺ + e^-

(e) The salt formed when ZnO reacts with hot concentrated NaOH is sodium zincate.

(a)

1. 2-methyl propene

2. butanal

(b) IUPAC names

1. 1,1,2,2-Tetrachloroethane
2. Butanoic acid
3. Pent-2-ene

(a) The atomic number of X is 14
Electronic configuration ⟶ 2, 8, 4

Number of electron shells = 3, hence X belongs to period 3

(b) The atomic number of Y is 8
Electronic configuration ⟶ 2, 6

X has a valency of 4 and Y has valency of 2

$$\begin{aligned}\overset{\phantom{4}{4}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{Y}} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{4}{4}}{\text{Y}} \\[1em] \text{X}_2\text{Y}_4\end{aligned}$$

∴ Simplified formula of the compound is XY₂

(a) Oxidation is the loss of electrons. In an electrolytic cell the anode is the positive electrode that attracts anions. These anions give up electrons to the anode, so oxidation takes place there. Because the anode accepts electrons and thereby causes the ions to be oxidised, it is called the oxidising electrode.

(b) As graphite is unaffected by the reactive bromine vapours released at the anode hence, a graphite anode is preferred to other inert electrodes like platinum during the electrolysis of molten lead bromide.

$$\begin{matrix} \text{Mg} & + & 2\text{H}_2 \text{SO}_4 & \longrightarrow & \text{MgSO}_4 & + & 2\text{H}_2\text{O} & + & \text{SO}_2 \\ 1 \text{ mole} && 2 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \\ 24 && 2[2(1) + 32 + 4(16)] & & 24 + 32 + 4(16) & & 2[2(1) + 16] && 32 + 2(16) \\ = 24 \text{ g} & & = 196 \text{ g} & & = 120 \text{ g} & & 36 \text{ g} & & = 64\text{ g} \\ \end{matrix}$$

(a) 24 g of Mg reacts with 196 g of H₂SO₄

∴ 60 g of Mg reacts with $\frac{196}{24}$ x 60 = 490 g of H₂SO₄

Hence, 490 g of H₂SO₄ is required.

(b) 24 g of Mg gives 22.4 L volume of SO₂

∴ 60 g of Mg gives $\frac{22.4}{24}$ x 60 = 56 L

Hence, the volume of sulphur dioxide gas liberated at S.T.P. is 56 L.

Thus, 490 g of sulphuric acid are needed and 56 L of SO₂ will be liberated at S.T.P.

(a) When barium chloride solution is added to a solution of Zinc sulphate, white ppt. of barium sulphate is obtained which is insoluble in dil. HCl or nitric acid.

ZnSO₄ + BaCl₂ ⟶ BaSO₄ ↓ [white ppt.] + ZnCl₂

(b) When lead nitrate is heated in a test tube, brown fumes of nitrogen dioxide (NO₂) are evolved and a yellow residue of lead(II) oxide remains.

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

(c) The moist starch-iodide paper turns blue-black because chlorine liberates iodine from iodide.

Cl₂ + 2KI ⟶ 2KCl + I₂

(a ) V.D. = $\frac{\text{mass of certain vol. of gas under similar conditions}}{\text{mass of equal vol. of H}_2\text{ under similar conditions}}$ = $\frac{450}{150}$ = 3

(b) Molecular weight = 2 x V.D. = 2 x 3 = 6 a.m.u.

(a) CH₃COONa + NaOH $\xrightarrow{\text{CaO, heat}}$ CH₄ + Na₂CO₃

(b) 2CH₃COOH + Mg ⟶ (CH₃COO)₂Mg + H₂ ↑

(a) Nitrogen dioxide (NO₂)

(b) Nitrogen (N₂)

(c) Hydrogen sulphide (H₂S)