Model Paper 3

Anode gets thicker

Reason — During electro-refining, the impure copper anode dissolves, so it gradually becomes thinner, while pure copper is deposited on the cathode, making the cathode thicker.

A colourless gas which gives dense white fumes with conc. HCI

Reason — When caustic soda (sodium hydroxide) is heated with an ammonium salt, ammonia gas (NH₃) is evolved. Ammonia is colourless, has a pungent smell, turns damp red litmus blue, and produces dense white fumes of ammonium chloride (NH₄Cl) when it comes in contact with concentrated hydrochloric acid.

X₂Y₄

Reason — Vapour density = Empirical formula weight,

∴ Molecular weight = 2 × Vapour density = 2 × (Empirical formula weight)

Therefore,

Molecular formula = 2 × (Empirical formula)

So, the correct molecular formula should be X₂Y₄

Alumina is not reduced by reducing agents.

Reason — Electrolytic reduction is chosen as the method for reducing alumina. Since, aluminium oxide due to its great affinity for oxygen is a very stable compound. It is not reduced easily by common reducing agents like carbon, carbon monoxide or hydrogen.

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

Reason — Alkenes undergo addition reactions due to the presence of a reactive double bond (C=C). Hence, the assertion (A) is true. Alkenes are compounds with at least one carbon-carbon double bond. Addition reaction in alkenes is due to the presence of double bond. Hence, the reason (R) is true and it is the correct explanation of (A).

Polar covalent bond

Reason — Hydrogen chloride is formed by the sharing of one pair of electrons between hydrogen and chlorine, so the bond is covalent. Because chlorine is much more electronegative than hydrogen, the shared electron pair is pulled closer to chlorine, producing a partial negative charge (δ–) on chlorine and a partial positive charge (δ+) on hydrogen. This unequal sharing of electrons leads to the formation of a polar covalent bond.

Methanal

Reason — Formaldehyde contains one carbon atom and an aldehyde functional group (-CHO). Hence, the name methanal.

Finely divided iron

Reason — The speed of the reaction is improved by using a catalyst, which is finely divided iron, obtained by the reduction of iron oxide. It provides a surface for the reaction to occur and lowers the activation energy, allowing the reaction to proceed faster at given temperature and pressure

Sulphur dioxide

Reason — Concentrated sulphuric acid acts as a strong oxidising agent when it reacts with zinc; zinc is oxidised to zinc sulphate (ZnSO₄) and the acid is reduced, liberating sulphur dioxide (SO₂) gas.

Zn + 2H₂SO₄ ⟶ ZnSO₄ + 2H₂O + SO₂ ↑

Reason — When sodium hydrogen carbonate (NaHCO₃), also known as baking soda, is heated, it decomposes to produce carbon dioxide (CO₂) gas, along with sodium carbonate (Na₂CO₃) and water (H₂O).

2NaHCO₃ ⟶ Na₂CO₃ + H₂O + CO₂ ↑

A is true but R is false.

Reason — During electrolysis, cations gain electrons at the cathode and are reduced, while anions lose electrons at the anode and are oxidised. Because oxidation and reduction take place simultaneously, the overall process is a redox reaction, so the assertion is true. However, reduction actually occurs at the cathode and oxidation at the anode, making the stated reason false.

Ethylene

Reason — Ethylene is a common name of compound ethene, which consist of two carbon atoms attached by double covalent bond and four C-H single covalent bond.

Lead hydroxide

Reason — Lead hydroxide, a white precipitate dissolves with excess of sodium hydroxide and gives colourless Sodium plumbite (Na₂[Pb(OH)₄]).

${\text{Pb(OH)}_2} + \underset{\text{excess}}{2\text{NaOH}} \longrightarrow \underset{\text{sodium plumbite - colourless}}{\text{Na}_2\text{PbO}_2↓} + 2\text{H}_2\text{O}$

Whereas, lead hydroxide is insoluble in excess of ammonium hydroxide (NH₄OH).

$\underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2} ↓ + 2\text{NH}_4\text{NO}_3$

Haematite

Reason — Haematite (Fe₂O₃) is the most common ore of iron. It contains a high percentage of iron and is widely used in the extraction of iron in the blast furnace.

Pentanal

Reason — The –CHO group is the characteristic functional group of aldehydes. It contains a carbonyl group (C=O) bonded to a hydrogen atom, and aldehyde names end with the suffix “-al”. Pentanal is an aldehyde that consists of five carbon atoms and the –CHO functional group, so it fits the description.

(a)

(b) Given,

112 mL of gaseous fluoride has mass = 0.63 g

∴ 22400 cm³ of gaseous fluoride will have mass = $\dfrac{0.63}{112}$ x 22400

= 126 g

∴ Relative molecular mass of fluoride = 126 g

(c) Let the formula be PF_n

Atomic mass of P + n × atomic mass of F = 126

⇒ 31 + 19n = 126

⇒ 19n = 126 - 31

⇒ 19n = 95

⇒ n = $\dfrac{95}{19}$

⇒ n = 5

∴ The molecular formula is PF₅

(a) Atom Y
Reason — Atom Y is most likely to form a cation, as it has only two electrons in its outermost shell. It will readily donate these two electrons to form a cation.

(b) Valency of X = 1
Valency of Z = 4

$\text{Z}^{4+} \space \text{X}^{1-} \\[1em] \overset{{4}}{\text{Z}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{X}} \Rightarrow \underset{1}{\text{Z}} \space {\swarrow}\mathllap{\searrow} \underset{4}{\text{X}} \\[1em] \text{Z}\text{X}_4$

∴ The formula of the compound formed between X with valency 1 and Z with valency 4 will be ZX₄.

(c)

(d) Two properties of the compound formed between X and Z i.e, ZX₄ are:

1. It is a non-electrolyte as it contains no free ions or electrons to conduct electricity.
2. It is a covalent compound having low melting and boiling point.

(a) Water

(b) Sodium nitrate

(c) Ammonia

(d) Ammonium chloride

(e) Acetylene

(a) Acid salts are the salts formed by partial replacement of the replaceable hydrogen ion of an acid molecule by a basic radical [metallic or ammonium ion]. Examples — Sodium Hydrogen Sulphate [NaHSO₄], Disodium Hydrogen Phosphate [Na₂HPO₄]

(b) The tendency of an atom in a molecule to attract the shared pair of electrons towards itself is called its electronegativity.

(c) Ammonia is a basic gas and a strong reducing agent. It reduces the black Copper [II] oxide to reddish brown Copper with the evolution of nitrogen gas.

(d) 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

(e) The ion formed is Hydroxyl ion. Its electron dot diagram is shown below:

Completed table is given below:

(a) Element A has electronic configuration of 2, 8, 8, 1, its atomic number is 19. Thus, element A will be potassium (K), and potassium is a metal, whereas oxygen is a non-metal, so the bonding in its oxide will be ionic bond.

(b) A has valency of = 1

Oxygen (O) has valency = 2

So, two A atoms combine with one O atom to form A₂O.

(a) A mixture of dry air (free from carbon dioxide and dust particles) and dry ammonia gas in the ratio of 10 : 1 by volume.

(b) The reaction takes place in a catalytic chamber containing platinum gauze as a catalyst at about 800°C.

(a) oxidation

(b) reduced

(c) hydrogen
Reason — Since M is above lead in the reactivity series, it is reactive enough to displace hydrogen from dilute acids. Therefore, it will liberate hydrogen gas (H₂) when reacting with dilute sulphuric acid.

(a) Ca(HCO₃)₂ + H₂SO₄ ⟶ CaSO₄ ↓ + 2H₂O + 2CO₂ ↑

(b) $\underset{\text{Cane Sugar}}{\text{C}<em>{12}\text{H}</em>{22}\text{O}_{11}} \text{ (s)} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{Sugar Charcoal}}{12\text{C (s)}} + 11\text{H}_2\text{O}$

(c) S + 2H₂SO₄ ⟶ 3SO₂ + 2H₂O

Reaction at anode :

6O^2- - 12e- ⟶ 6[O]

3[O] + 3[O] ⟶ 3O₂

Anode is oxidised to carbon monoxide which further forms carbon dioxide

2C + O₂ ⟶ 2CO

2CO + O₂ ⟶ 2CO₂

Reaction at cathode : 4Al³⁺ + 12e^- ⟶ 4Al

Simplest ratio of whole numbers Pb : Cl = 1 : 4

Hence, empirical formula is PbCl₄

(a) C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

(b) Pb(NO₃)₂ + 2HCl (dil.) ⟶ PbCl₂+ 2HNO₃

(c) $\underset{\text{ethanol}}{\text{C}_2\text{H}_5\text{OH}} + \underset{\text{acetic acid}}{\text{CH}_3\text{COOH}} \xrightarrow[\Delta]{\text{Conc. H}_2\text{SO}_4} \underset{\text{ethyl acetate}}{\text{CH}_3-\text{COO}-\text{C}_2\text{H}_5} + \text{H}_2\text{O}$

(a) Cathode

(b) Sodium argentocyanide Na[Ag(CN)₂] or Potassium argentocyanide K[Ag(CN)₂]

(c) If silver nitrate solution is used directly instead of double cyanide of sodium and silver, the deposition of silver will be very fast and hence not very smooth and uniform.

Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1

Hence, empirical formula is CHO₂

(b) Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45

V.D. = 45

Molecular weight = 2 x V.D. = 2 x 45 = 90

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{90}{45} = 2$

So, molecular formula = 2[E.F.] = 2(CHO₂) = C₂H₂O₄

(a)

1. Ammonium chloride (NH₄Cl) and calcium hydroxide (Ca(OH)₂) are mixed together. (Excess of calcium hydroxide is mixed well with ammonium chloride).

2. Reactants are finely grounded and taken in a round-bottom flask fitted in a slanting position, mouth downwards.

3. On heating the mixture, ammonia gas is evolved.
   2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

4. Ammonia gas is collected in inverted gas jars by the downward displacement of air because it is lighter than air and highly soluble in water and therefore, it cannot be collected over water.

5. Quicklime (CaO) is used as a drying agent.

(b)

1. Nitric acid is obtained by distilling conc. sulphuric acid with potassium nitrate KNO₃ (nitre) or sodium nitrate NaNO₃ (Chile saltpetre).
2. A mixture of equal parts, by weight, of potassium/sodium nitrate and concentrated sulphuric acid is heated gently to 180°C-200°C in a glass retort.
3. Sulphuric acid is a non-volatile acid and produces volatile nitric acid on reacting with potassium nitrate or sodium nitrate.

$\underset{[conc.]}{\text{KNO}_3} + \underset{[conc.]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 \degree\text{C}} \underset{[acid \space salt]}{\text{KHSO}_4} + \text{HNO}_3$

$\underset{[conc.]}{\text{NaNO}_3} + \underset{[conc.]}{\text{H}_2\text{SO}_4} \xrightarrow{\gt 200 \degree\text{C}} \underset{[acid \space salt]}{\text{NaHSO}_4} + \text{HNO}_3$

4. The vapours of nitric acid are condensed to a light yellow liquid by chilling the receiver with running cold water.

(a) Ammonia solution turns moist red litmus blue. Whereas, hydrogen chloride gas turns moist blue litmus red.

(b) On adding bromine water, unsaturated compound decolourises bromine water due to addition reaction across the double or triple bond. However, saturated compound shows no reaction. So, there is no change in colour.

(c) Add a few drops of aqueous NaOH:

• Ferrous chloride gives a dirty-green precipitate of ferrous hydroxide, Fe(OH)₂, which soon turns brown on standing.
• Ferric chloride gives a reddish-brown precipitate of ferric hydroxide, Fe(OH)₃.

$\text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{copper (II) sulphate}}{\text{Tetraammine}}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$

(b)

$\underset{\text{ methane} }{\text{CH}_4} + \text{Cl}_2 \underset{\Delta}{\xrightarrow{\text{diffused sunlight}}} \underset{\text{monochloromethane}}{\text{CH}_3\text{Cl}} + \text{HCl}$

$\underset{\text{monochloromethane}}{\text{CH}_3\text{Cl}} + \underset{\text{[excess]}}{\text{Cl}_2} \longrightarrow \underset{\text{dichloromethane}}{\text{CH}_2\text{Cl}_2} + \text{HCl}$

$\underset{\text{dichloromethane}}{\text{CH}_2\text{Cl}_2} + \text{Cl}_2 \longrightarrow \underset{\text{trichloromethane or chloroform}}{\text{CHCl}_3}+ \text{HCl}$

(a) A freshly prepared ferrous sulphate solution is used, because on exposure to the atmosphere, it is oxidised to ferric sulphate, which will not give the brown ring test.

(b) The brown ring of nitroso ferrous sulphate is formed at the junction of the two liquids because the conc. sulphuric acid being heavier settles down and the ferrous sulphate layer remains above it resulting in the formation of a brown ring at the junction.

(a) Acetic acid

(b) Acetaldehyde

(c)

(a) When ethylene gas is passed through bromine water, the reddish brown colour of bromine water disappears due to the formation of the colourless dibromoethane.

$\underset{\text{ ethene}}{\text{C}_2\text{H}_4} + \text{Br}_2 \xrightarrow[\text{[inert solvent]}]{\text{CCl}_4} \underset{\text{1,2, dibromooethane [ethylene bromide]}}{\text{C}_2\text{H}_4\text{Br}_2}$

(b) When silver nitrate reacts with Hydrochloric acid it gives a white precipitate of AgCl which is insoluble in nitric acid but soluble in ammonium hydroxide.

AgNO₃ + HCl ⟶ AgCl ↓ + HNO₃

(c) Barium chloride reacts with potassium sulphate to form barium sulphate (a white insoluble solid) and potassium chloride.

BaCl₂ + K₂SO₄ ⟶ BaSO₄ + 2KCl

$\begin{matrix} \text{CH}_4 & + & 2\text{Cl}_2 & \longrightarrow & \text{CH}_2\text{Cl}_2 & + & 2\text{HCl} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1 \text{ vol.} & : & 2 \text{ vol.}\\ \end{matrix}$

(a) Volume of chlorine gas used = ?

For 1 Vol of methane = 2V of Cl₂ required

∴ For 40 cm³ of methane = 40 x 2 = 80 cm³ of Cl₂ is required.

(b) Volume of HCl gas formed = ?

[By Gay Lussac's law]

1 Vol of methane produces = 2 Vol. HCl

∴ 40 cm³ of methane produces = 80 cm³ HCl

Hence, volume of HCl gas formed = 80 cm³ and chlorine gas required = 80 cm³

(c) Hydrogen chloride gas is dried by passing through a washer bottle containing concentrated sulphuric acid.

(a) Across a period the nuclear (effective) charge on the atoms increases while the number of electron shells remains the same. As a result, atomic radius decreases and the attraction of the nucleus for the bonding electrons becomes stronger, so electronegativity increases from left to right.

(b) When treated with a strong alkali (like sodium hydroxide, NaOH), both aluminium and zinc powder dissolve and form soluble complex salts with the evolution of hydrogen gas. As both metals behave in the same way, the alkali test cannot distinguish between them.

(c) Water is never poured on acid to dilute it as large amount of heat is evolved which changes poured water to steam. The steam so formed causes spurting of acid which can cause burn injuries.

(a) Quicklime (CaO)

(b) Hydrochloric acid is prepared by dissolving hydrogen chloride gas in water using a special funnel arrangement in which an inverted funnel, connected to the hydrogen chloride gas supply is placed in the beaker in such a way that it just touches the water taken in the trough. This arrangement minimizes back suction.

(c) Reaction at anode:

6O^2- - 12e- ⟶ 6[O]

3[O] + 3[O] ⟶ 3O₂

Anode is oxidised to carbon monoxide which further forms carbon dioxide

2C + O₂ ⟶ 2CO

2CO + O₂ ⟶ 2CO₂

(d) In the laboratory preparation of HCl, concentrated sulphuric acid (H₂SO₄) must be added to sodium chloride (NaCl) to obtain HCl gas.

1. $\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_4 + \text{HCl [g]}$ ↑

2. $\text{2NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\gt 200 \degree\text{C}} \text{Na}_2\text{SO}_4 + \text{2HCl [g}]$ ↑

Isomerism is the phenomenon due to which two or more compounds have the same molecular formula but differ in molecular arrangement or in structural formula.

Position isomers — When two or more compounds with the same molecular formula differ in the position of substituent atom or functional group on the carbon atom, they are called position isomers.

Example : But-1-yne and But-2-yne

But-1-yne

But-2-yne

(a) Add copper sulphate solution (CuSO₄) to both the solutions. With sodium hydroxide, a pale blue precipitate is formed which is insoluble in excess of sodium hydroxide. With ammonium hydroxide it forms a pale blue precipitate which dissolves in excess of ammonium hydroxide and forms a deep/inky blue solution.

(b) When ammonium hydroxide (NH₄OH) is added dropwise to Zinc chloride (ZnCl₂), white gelatinous precipitate of zinc hydroxide forms initially. On adding excess NH₄OH, the precipitate dissolves, forming a clear solution.

With calcium chloride, no precipitate forms with NH₄OH because calcium hydroxide is not precipitated by ammonium hydroxide in cold solution.

(a) Carbon dioxide (CO₂)

(b) Magnesium (Mg)

(c) Zinc sulphide (ZnS)

(a)

(b) Bromine solution in CCl₄ has a reddish brown colour. When added dropwise to ethene, the reddish brown colour of bromine disappears due to the formation of the colourless dibromoethane.

(c) When ethene (X) reacts with steam, a water molecule is added in the presence of acids like sulphuric acid to form alcohols.

$\underset{ \text{ethene}}{\text{C}_2\text{H}_4} + \text{H}_2\text{O} \xrightarrow {\text{Conc. H}_2\text{SO}_4} \underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \\$