Solved 2027 Specimen Paper ICSE Class 10 Chemistry

X and Y are diatomic elements. X is less reactive than Y.

What are elements X and Y?

1. X - chlorine, Y - iodine
2. X - fluorine, Y - nitrogen
3. X - iodine, Y - bromine
4. X - oxygen, Y - nitrogen

Answer

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.

The diagram given below represents the industrial process for the manufacture of concentrated sulphuric acid by the Contact process.

What is used in step R to obtain sulphuric acid from sulphur trioxide?

1. vanadium pentoxide
2. water only
3. water followed by concentrated sulphuric acid
4. concentrated sulphuric acid followed by water

Answer

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.

Which statement given below is correct with reference to the flow of electric current?

1. In all aqueous solutions, current is carried by electrons.
2. In all acidic solutions, current is carried by ions.
3. In molten electrolytes, current is carried by electrons.
4. In metal wires, current is carried by ions.

Answer

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.

The following statements are related to the properties of sulphuric acid:

1. Barium chloride reacts with sulphuric acid to give a white precipitate.
2. Sulphur dioxide gas is evolved when sulphuric acid reacts with carbon.
3. Sulphuric acid has a pH value of less than 7.0.
4. Copper (II) oxide reacts with sulphuric acid to form a blue solution.

Which two statements confirm the acidic nature of sulphuric acid?

1. 1 and 2
2. 1 and 3
3. 3 and 4
4. 2 and 4

Answer

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.

The diagram given below shows that P reacts with dilute hydrochloric acid to produce a gas which when passed through substance Q, reduces it.

What are substances P and Q?

1. P - copper, Q - lead (II) oxide
2. P - lead, Q - magnesium oxide
3. P - magnesium, Q - calcium oxide
4. P - zinc, Q - copper (II) oxide

Answer

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

Non-metals like chlorine and iodine on reaction with hydrogen give:

1. acid
2. base
3. salt
4. metal

Answer

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.

Statement I: Halogens are very reactive.

Statement II: Halogens have high electron affinity.

1. Both the statements are true.
2. Both the statements are false.
3. Statement I is true, and statement II is false.
4. Statement I is false, and statement II is true.

Answer

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.

Assertion (A): Ionic compounds have high melting points.

Reason (R): A strong electrostatic force of attraction exists between the oppositely charged ions.

1. (A) is true but (R) is false.
2. (A) is false but (R) is true.
3. Both (A) and (R) are true, and (R) is the correct explanation of (A).
4. Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Answer

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).

A student tested the nature of four different solutions P, Q, R, and S using pH paper. He noted the following observations:

Which of the above solutions is neutral in nature?

1. P
2. Q
3. R
4. S

Answer

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.

Identify the organic compound(s) with the molecular formula C₃H₈O.

1. propanol
2. propanal
3. propanoic acid

1. Only 1
2. Only 2
3. Both 2 and 3
4. Both 1 and 2

Answer

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.

An organic compound X containing carbon and hydrogen only has Vapour Density (V.D.) 14. The molecular formula of X is ............... (At. wt. of C=12, H=1)

1. CH
2. C₂H₄
3. C₂H₂
4. CH₂

Answer

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.

Identify the ion that contains one lone pair of electrons.

1. OH^-1
2. H₃O⁺
3. NH₄⁺
4. H⁺

Answer

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.

When compound X reacts with dilute hydrochloric acid, it releases a gas that turns moist lead acetate paper black and decolourises acidified potassium permanganate solution. Which of the following is compound X?

1. Sodium nitrate
2. Sodium sulphide
3. Sodium carbonate
4. Sodium sulphite

Answer

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 ↑

A teacher demonstrated four different reactions as shown in the diagram below:

Which reactions will produce water?

1. 1 and 2
2. 1 and 3
3. 3 and 4
4. 2 and 3

Answer

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.

The volume occupied by 0.2 moles of a gas at S.T.P. is:

1. 22.4 litres
2. 2.24 litres
3. 44.8 litres
4. 4.48 litres

Answer

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.

The diagram given below shows the electrolysis of acidulated water. With reference to the diagram, answer the following questions:

(a) Write an equation for the reaction that takes place at A.

(b) Give a test to confirm the product formed at A.

(c) Which gas is liberated at B?

(d) Why is water acidified for electrolysis?

(e) What is the ratio by volume of the gases liberated at A and B?

Answer

(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.

Write balanced chemical equations for the following:

(a) Chlorination of chloroform in the presence of diffused sunlight.

(b) Hydrated aluminium oxide is treated with caustic soda.

(c) Zinc oxide is treated with caustic alkali.

(d) Copper reacts with cold dilute nitric acid.

(e) Laboratory preparation of nitric acid using sodium nitrate.

Answer

(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

Complete the following sentences by choosing the correct option from the brackets:

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

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

(c) The gas liberated when alkalis are heated with ammonium salts is ............... (N₂/NH₃).

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

(e) A salt prepared by neutralization in which titration is involved is ............... (CaCl₂/ NH₄Cl).

Answer

(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.

Match Column A with Column B:

Answer

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

(b) Draw the structural diagram for the following compounds:

1. diethyl ether
2. isopentane
3. propanone

Answer

(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

A compound commonly used in the laboratory has formula H₂P.

H represents hydrogen, P represents a non-metal.

Draw a dot and cross structure for the formation of the compound H₂P.

Answer

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:

Give one significant observation when:

(a) Excess of chlorine gas reacts with ammonia.

(b) Zinc nitrate is strongly heated in a test tube.

Answer

(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 student was given a crystalline salt 'B' for analysis which was blue in colour. He added a few drops of dense oily hygroscopic liquid to it and observed a colourless neutral gas/vapour being released and the salt turning white.

(a) Identify the blue crystalline salt B.

(b) Which property of the hygroscopic liquid is demonstrated in this reaction?

(c) Write a balanced chemical equation for the above reaction.

Answer

(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 is a metal which forms a basic oxide.

N has atomic no. 11.

(a) What is the molecular formula of the oxide of N?

(b) Write the balanced equation for the reaction that takes place when the oxide of N is treated with water.

(c) State the formula of the product obtained other than water when sulphuric acid is added to the product obtained in (b) above.

Answer

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.

Name the following:

(a) The non-metallic element which is used as the anode during electrolytic reduction of fused alumina in Hall Heroult's process.

(b) An alloy which is made up of aluminium and magnesium only.

(c) The process of heating the concentrated ore in a limited supply of air or in the absence of air.

Answer

(a) Carbon (graphite)

(b) Magnalium

(c) Calcination

Rohit has three solutions X, Y and Z that have pH 2, 7 and 13 respectively. Which solution out of X, Y or Z will:

(a) liberate sulphur dioxide gas when heated with sodium sulphite?

(b) liberate ammonia gas when reacted with ammonium chloride?

(c) not have any effect on litmus paper?

Answer

(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 chloride of iron 'R', when treated with sodium hydroxide gives P and Q as a product. Q is a reddish-brown precipitate.

(a) Identify R.

(b) State the formula of the compounds P and Q formed after the reaction.

(c) What do you observe when excess NaOH is added to Q?

Answer

(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.

20 grams of a gas M₂P has lesser number of moles than 20 grams of the gas MP.

Justify the above statement using mathematical calculations.

[At.wt. M=14, P=16]

Answer

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

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

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

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

∴ Moles in 20 g of MP = $\dfrac{20}{30}$ = 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.

State giving reasons if:

(a) Zinc metal and aluminium metal can be distinguished by heating the metal powders separately in two different test tubes with concentrated sodium hydroxide solution.

(b) Zinc nitrate and lead nitrate can be distinguished by adding ammonium hydroxide solution to the salt solution.

Answer

(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.

Write balanced chemical equations for the following conversions (A to C):

Answer

A: Zn + H₂SO₄ ⟶ ZnSO₄ + H₂

B: Zn + S ⟶ ZnS

C: ZnSO₄ + 2NaOH ⟶ Zn(OH)₂ + Na₂SO₄

An element 'P' has 3 valence electrons in its L - shell and an element 'Q' has 7 valence electrons in its M - shell. Using this information, answer the following questions:

(a) Identify the exact position of element 'Q' in terms of group and period in the Modern Periodic Table.

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

Answer

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.

Identify the reactants P, Q and R in the following reactions:

(a) Copper (II) oxide + P ⟶ Copper + Water + Nitrogen

(b) Iron pyrite + Q ⟶ Iron oxide + Sulphur dioxide

(c) Ammonia + R ⟶ Nitrogen + Water

Answer

(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

Solid ammonium dichromate decomposes as under:

(NH₄)₂Cr₂O₇ ⟶ N₂ + Cr₂O₃ + 4H₂O

If 126 g of ammonium dichromate decomposes, calculate:

(a) the number of moles of ammonium dichromate that undergoes decomposition.

(b) the mass of chromic oxide formed at the same time.

(c) the volume of nitrogen gas evolved at STP.

[At. Wt: N=14, Cr=52, O=16, H=1]

Answer

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

(a) 252 g of (NH₄)₂Cr₂O₇ = 1 mole

∴ 126 g of (NH₄)₂Cr₂O₇ = $\dfrac{1}{252}$ 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 $\dfrac{152 \times 126}{252}$ = 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 $\dfrac{22.4 \times 126}{252}$ = 11.2 lit of N₂

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

Rohan has two test tubes, A containing sodium chloride and B containing ammonium chloride as shown in the diagram below. He adds sodium hydroxide solution to both the test tubes and warms them gently.

Answer the following questions:

(a) In which test tube (A or B) will Rohan observe the evolution of a gas?

(b) Name the gas that is evolved.

(c) What will be the colour of the flame when this gas burns in oxygen?

(d) Write an equation for the reaction that occurs in the test tube in which the gas has evolved.

Answer

(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

What is the ratio of methane to the acidic gaseous product obtained when methane undergoes complete combustion? Give a balanced equation for the above reaction.

Answer

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.

The diagram below illustrates the structures of two atoms P and Q.

What is the formula and mass of one mole of the compound formed when P and Q react together?

[At. wt of P is 23 and Q is 16]

Answer

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.

Give balanced equations for each of the following:

(a) Action of warm water on calcium carbide.

(b) Oxidation of sulphur with conc. nitric acid.

(c) Laboratory preparation of ethane by using sodium propionate and soda lime.

Answer

(a) CaC₂ + 2H₂O ⟶ Ca(OH)₂ + C₂H₂

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

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

In a round bottom flask, a mixture of ethanol, acetic acid and concentrated sulphuric acid was heated.

Answer the following questions:

(a) Name the type of reaction occurring in the above set up.

(b) What is the role of sulphuric acid in this reaction?

(c) State one observation that takes place during the reaction.

Answer

(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.

In the reactivity series of metals, Z is a metal above hydrogen and its oxide has the formula ZO. It was found that ZO reacted with water and formed a hydroxide as the product. With reference to the above context, answer the following questions:

(a) State the number of valence electrons present in element Z.

(b) Name the group which Z belongs to.

(c) What will be the formula of the hydroxide of Z?

Answer

(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.

Draw the electron dot structure of:

(a) ammonium ion (also label the co-ordinate bond)

(b) nitrogen molecule

[Atomic No.: N = 7, H = 1]

Answer

(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):

The structures of five organic compounds, C, D, E, F and G are shown.

Answer the following questions about these compounds.

(Each compound may be used once, more than once or not at all.)

State which compound, C, D, E, F or G:

(a) on halogenation will form F.

(b) is an alcohol.

(c) is an isomer of 2-methyl propane.

(d) is in the same homologous series as ethene.

Answer

(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.