VERY SHORT ANSWER -Class 10 – Science -Chemistry -Chapter 4 – Carbon and its Compounds

VERY SHORT ANSWER -Class 10 – Science -Chemistry -Chapter 4 – Carbon and its Compounds

Q1. Write the molecular formula of first two members of homologous series having functional group -Cl.

Ans:

The molecular formula of first two members of homologous series having -Cl functional group are CH₃ Cl and CH₃ CH₂ Cl.

Q2. Write the molecular formula of first two members of homologous series having functional group -OH.

Ans:

The molecular formula of first two members of homologous series having -OH functional group are CH₃ OH and CH₃ CH₂ OH.

Q3. Write the molecular formula of the 2nd and 3rd member of the homologous series whose first member is ethene.

Ans:

Homologous series of alkenes have general formula, CnH2n whose first member is ethene.

2nd member of homologous series of alkenes is C3H6 i.e., propene.

3rd member of homologous series of alkenes is C4H8 i.e., butene.

Q4. Write the molecular formula of the 2nd and 3rd member of the homologous series whose first member is methane.

Ans:

Methane, CH4 is an alkane. Alkanes have general formula, CnH2n+2.

2nd member of homologous series of alkanes is C2H6 i.e., ethane.

3rd member of homologous series of alkanes is C3H8 i.e., propane.

Q5. Write the next homologue of each of the following:

(i) C₂ H₄

(ii) C₄ H₆ 

Ans:

(i) C2H4 belongs to alkene series having general formula, CnH2n.

Thus, next homologue will be C3H2×3 = C3H6

(ii) C4H6 belongs to alkyne series having general formula, CnH2n-2.

Thus, next homologue will be C5H2×5-2 = C5H8

Q6. Select saturated hydrocarbons from the following : C₃ H₆ ; C₅ H₁₀ ; C₄ H₁₀ ; C₆ H₁₄ ; C₂ H₄

Ans:

Saturated hydrocarbons have general formula, CnH2n+2.

Among the given compounds only C4H10 and C6H14 satisfy the above formula. Thus, these are saturated hydrocarbons.

Q7. Which element exhibits the property of catenation to maximum extent and why?

Ans:

Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to large molecules. This property is called catenation. Carbon shows catenation due to its small size and Stronger carbon-carbon bond strength.

Q8. Write the name and molecular formula of the fourth member of alkane series.

Ans:

The general formula of the alkane series is CnH2n+2. For fourth member of alkane series, n = 4

∴ C4H2×4+2 = C4H10 i.e., butane.

Q9. What is homologous series of carbon compounds?

Ans:

A homologous series is the family of organic compounds having the same functional group, similar chemical properties but the successive (adjacent) members of the series differ by a -CH2 unit or 14 mass units.

Q10. Write the name and formula of the 2nd member of homologous series having general formula CnH2n-2.

Ans:

General formula, CnH2n-2 belongs to alkyne series. The second member of this series is propyne i.e., (C3H4) or CH3 – C ≡ CH

Q11. State two properties of carbon which lead to a very large number of carbon compounds.

Ans:

Carbon forms a large number of carbon compounds like long chains which may be straight or branched chains or ring of different sizes due to its tetravalency ahd unique property of catenation. Carbon due to its small size forms exceptionally stable compounds by forming strong bonds.

Q12. What are hydrocarbons? Distinguish alkanes from alkenes and each of them from alkynes, giving one example of each. Draw the structure of each compound cited as example to justify your answer.

Ans:

Hydrocarbons are the compounds of carbon and hydrogen atoms. Those hydrocarbons which contain only single carbon-carbon bonds are called alkanes (saturated hydrocarbons) while those having double and triple bonds are called alkenes and alkynes respectively (unsaturated hydrocarbon).

Q13. Give reason for the following : Kerosene does not decolourise bromine water while cooking oils do.

Ans:

Cooki4ng oils (unsaturated compounds) decolourise bromine water due to formation of addition products whereas kerosene (saturated compound) does not decolourise bromine water.

Q14. What happens when 5% alkaline KMnO4 solution is added drop by drop to warm ethanol taken in a test tube? State the role of alkaline KMnO4 solution in this reaction.

Ans:

When 5% alkaline KMnO4 solution is added drop by drop to warm ethanol then it gets oxidised to ethanoic acid.

Carbon and its Compounds Chapter Wise Important Questions Class 10 Science Img 32

CH₃ CH₂ OH –> CH₃ COOH

Here, alkaline KMnO4 acts as an oxidising agent i.e., the substance which is capable of adding oxygen to others. Thus, alkaline KMnO4 provides oxygen to ethanol to form ethanoic acid.

Q15. Two carbon compounds X and Y have the molecular formula C4H8 and C5H12 respectively. Which one of these is most likely to show addition reaction? Justify your answer. Also give the chemical equation to explain the process of addition reaction in this case.

Ans:

All unsaturated hydrocarbons (containing double or triple bonds) have tendency to get converted to saturated hydrocarbons (single bonds) by adding small molecules such as hydrogen (H2), halogens (X2), etc. Such reactions are called addition reactions.

Compound X i.e., C4H8 belongs to alkene series (CnH2n) while compound Y i.e., C5H12 belongs to alkane series (CnH2n+2). Thus, compound X will undergo addition reaction.