Important Questions for Class 10 Science Chapter 4 – Carbon and Its Compounds
Below are some important questions for class 10 science chapter 4 – Carbon And Its Compounds that students must study thoroughly.Carbon, like hydrogen, oxygen, lead, and the other elements in the periodic table, is a chemical element. It is an abundant element. Due to its tetravalent nature and catenation property, carbon forms a wide variety of compounds. Carbon is a unique element that can form strongly bonded chains that are sealed off by hydrogen atoms. Compounds that contain carbon atoms also possess double and triple bonds. Carbon chains may take the form of straight chains, branched chains, or rings.
Students must pay attention to the following important questions topics for class 10 science chapter 4 – Carbon And Its Compounds:
- A covalent bond is bonded in carbon
- Carbon’s versatility
- Compounds of saturated and unsaturated carbon
- The nomenclature of carbon compounds
- Carbon compounds: Chemical properties
Below are some of the important questions for class 10 science chapter 4 – Carbon And Its Compounds from an examination point of view.
Class 10 Science Important Questions Chapter 4 – Carbon and its Compounds
Q1. Give reasons for the following:
(i) Element carbon forms compounds mainly by covalent bonding.
(ii) Diamond has high melting point.
(iii) Graphite is a good conductor of electricity.
(i) As carbon has four valence electrons and it can neither loose nor gain lour electrons thus, it attains noble gas configuration only by sharing of electrons. I bus, it forms covalent compounds.
(ii) In diamond, each carbon atom is bonded to four other carbon atoms forming a rigid three-dimensional structure. This makes diamond the hardest known substance. Thus, it has high melting point.
(iii) In graphite, each carbon atom is bonded to three other carbon atoms by covalent bonds in the same plane giving a hexagonal array. Thus, only three valence electrons are used for bond formation and hence, the fourth valence electron is free to move. As a result, graphite is a good conductor of electricity.
Q2. Assertion (A) : Following are the members of a homologous series :
CH3OH, CH3CH2OH, CH3CH2CH2OH
Reason (R) : A series of compounds with same functional group but differing by -CH2 unit is called homologous series.
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
(a): The given compounds are members of homologous series of alcohol.
Q3. Covalent compounds have low melting and boiling point. Why?
Covalent compounds have low melting and boiling points because the forces of attraction between molecules of covalent compounds are very weak. On applying a small amount of heat these molecular forces break.
Q4. Name the functional groups present in the following compounds
(a ) CH3COCH2CH2CH2CH3
(b ) CH3CH2CH2COOH
(c ) CH3CH2CH2CH2CHO
(d ) CH3CH2OH
(a ) A ketone functional group is present in the compound CH3COCH2CH2CH2CH3.
(b ) A carboxylic acid functional group is present in the compound CH3CH2CH2COOH.
(c ) An aldehyde functional group is present in the compound CH3CH2CH2CH2CHO.
(d ) An alcohol functional group is present in the compound CH3CH2OH.
Q5. Ethene is formed when ethanol at 443 K is heated with excess concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation of this reaction.
Concentrated sulphuric acid removes water from ethanol, thereby acting as a dehydrating agent.
CH3CH2OH + Conc H2SO4 → CH2 = CH2 + H2O
Q6. Unsaturated hydrocarbons contain multiple bonds between the two C-atoms and show addition reactions. Give the test to distinguish ethane from ethene.
The bromine water test can be used to distinguish between saturated and unsaturated hydrocarbons. Saturated compounds don’t give an addition reaction. Hence, there won’t be any change in the reaction mixture. In contrast, if an unsaturated hydrocarbon is added to bromine water, its colour will decolourise.
Saturated hydrocarbon + Br₂ → No Reaction (No Colour Change)
Unsaturated hydrocarbon + Br₂ → Reaction will occur (Decolourise)
Q7. A salt X is formed, and gas is evolved when ethanoic acid reacts with sodium hydrogen carbonate. Name the salt X and the gas evolved. Describe an activity and draw the diagram of the apparatus to prove that the evolved gas is the one you have named. Also, write a chemical equation of the reaction involved.
The salt X is sodium ethanoate (CH3COONa), and the evolved gas is carbon dioxide (CO2).
Take a test tube and add ethanoic acid (CH3COOH). Add sodium bicarbonate (NaHCO3) to the acid, close the test tube’s mouth with a cork, and attach a delivery tube.
Take lime water in another test tube and attach it to the delivery tube. The lime water turns milky. This indicates that the evolved gas is carbon dioxide.
Ca(OH)2 + CO2 → CaCO3 + H2O
The milkiness is due to the formation of CaCO3.
Reaction Involved: CH3COOH + NaHCO3 → CH3COONa + H2O + CO2 (g)
Q8. How would you bring about the following conversions? Name the process and write the
(a) Ethanol to Ethene.
(b) Propanol to Propanoic acid.
(a) Ethanol is heated at 443 K in the presence of an excess of conc. Sulphuric acid. This reaction is known as dehydrogenation.
CH3CH2OH + Conc. H2SO4 → CH2 = CH2 + H2O.
(b) Propanol is treated with alkaline potassium permanganate or acidified potassium dichromate to get propanoic acid.
CH3CH2CH2OH + Alkaline KMnO4 / Acidified K2Cr2O7 → CH3CH2COOH
Q9. An organic compound A on heating with concentrated H2SO4 forms a compound B which on the addition of one mole of hydrogen in presence of Ni forms a compound C. One mole of compound C on combustion forms two moles of CO2 and 3 moles of H2O. Identify the compounds A, B and C and write the chemical equations of the reactions involved.
Compound A is ethanol (CH3CH2OH). When it is heated with concentrated sulphuric acid, we get ethene (CH2 = CH2). Thus, compound B is ethene (CH2 = CH2).
CH3CH2OH + Conc.H2SO4 → CH2 = CH2 + H2O
When ethene (CH2 = CH2) is heated in the presence of nickel, we get ethane (CH3 - CH3).
Thus, compound C is ethane (CH3 - CH3).
CH2 = CH2 + Ni → CH3 - CH3
When 1 mole of ethane (CH3 - CH3) is burnt, we get 2 moles of carbon dioxide (CO2) and 3 moles of water (H2O).
2 CH3 - CH3 + 7 O2 → 4 CO2 + 6 H2O
Q10. A compound C (molecular formula, C2H4O2) reacts with Na – metal to form a compound R and evolves into a gas which burns with a pop sound. Compound C on treatment with an alcohol A in the presence of an acid forms a sweet-smelling compound S (molecular formula, C3H6O2). On addition of NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A. Identify C, R, A, and S and write down the reactions involved.
Here, compound C is ethanoic acid (CH3COOH), compound R is sodium ethanoate (CH3COONa), compound A is ethanol (C2H5OH), and compound S is ethyl ethanoate (CH3COOC2H5).
Ethanoic acid (CH3COOH) reacts with sodium metal to form sodium ethanoate (CH3COONa).
2 CH3COOH + 2 Na → 2 CH3COONa + H2
Ethanoic acid (CH3COOH) on treatment with ethanol (C2H5OH) in the presence of an acid forms a sweet-smelling ethyl ethanoate (CH3COOC2H5).
CH3COOH + C2H5OH → CH3COOC2H5 + H2O
On adding NaOH to ethanoic acid (CH3COOH), it also gives sodium ethanoate (CH3COONa) and water.
CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
Thus, compound C is Ethanoic acid.