NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

These NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Carbon and Its Compounds NCERT Solutions for Class 10 Science Chapter 4

Class 10 Science Chapter 4 Carbon and Its Compounds InText Questions and Answers

In-text Questions (Page 61)

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO2 ?
Answer:
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 1

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur ? (Hint: The eight atoms of sulphur are Joined together in the form of a ring)
Answer:
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 2

In-text Questions (Page 68-69)

Question 1.
How many structural isomers can you draw for pentane ?
Answer:
There chain isomers or structural isomers of pentane are possible namely n-pentane, 2-methyl butane and 2, 2-Dimethyl propane.
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 3

Question 2.
What are two properties of carbon which lead to the huge number of carbon compounds we see around us ?
Answer:
The two main properties of carbon are :
(i) Tetravalency : Carbon has atomic number six. Its electronic configuration, 2, 4.
Carbon has four electrons in its valence shell and, therefore, it can attain a noble gas electronic configuration either by losing or gaining or sharing four electrons. But the loss or gain of four electrons by the carbon atom to form highly charged C+4 or C-4 ions would require a very large amount of energy which is not generally available during a chemical reaction. So it is unable to form ionic bonds and as such it can participate only in the formation of covalent bond. Carbon is capable to form covalent bonds with Oxygen, Hydrogen, Nitrogen, Sulphur, Chlorine and many other elements giving rise to a large number of compounds.

(ii) Catenation : Carbon has the unique property to form bonds with atoms of carbon. This property of self linking of carbon atoms is called catenation. Due to catenation carbon atoms can form various types of straight chain, branched chain and cyclic chains, thus giving rise to a large no, of compounds.

Question 3.
What will be the formula and electron do not structure of cyclopentane ?
Answer:
Formula of cyclopentane : C5H10
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 4

Question 4.
Draw the structure for the following compounds.
(i) Ethanoic acid
(ii) Bromo pentane
(iii) Butanone
(iv) Hexanol
Are structural isomers possible for bromopentane.
Answer:
(i) Ethanoic acid : Molecular Formula, C2H5OH
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 5

(ii) Bromo pentane : Molecular Formula, C5H11-Br
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 6

(iii) Butanone, Molecular Formula, C4H8O
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 7

(iv) Hexanal: Molecular Formula: C6H12O
Structure :
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 8
Yes, structural isomers of Bromo pentane are possible. Some of them are as follows :
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 9

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Question 5.
How would you name the following compounds ?
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 10
Answer:
(i) CH3 – CH2 – Br
1 – Bromo ethane
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 11

In-text Questions (Page 71)

Question 1.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction ?
Answer:
A reaction in which a substance gain oxygen or loose hydrogen is called oxidation reaction. During the conversion of ethanol to ethanoic acid, oxidising agent like alkaline potassium permanganate or acidic potassium permanganate release oxygen atoms which are gained by ethonol and converts to ethanoic acid. So the conversion of ethanol to ethanoic acid is an oxidation reaction.
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 12

Question 2.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used ?
Answer:
A high temperature is required for welding. Ethyne burn in oxygen and produce a large amount of heat and H2O and CO2 as by product. But if we used a mixture of ethyne and air it produce oxides of nitrogen with H2O and CO2. The oxides of nitrogen are the major air pollutants so they pollute the environment, that is why a mixture of air and ethylene is not used for welding.

In-text Questions (Page 74)

Question 1.
How would you distinguish experimentally between an alcohol and a carboxylic add ?
Answer:
Carboxylic acid turns blue litmus paper or blue litmus solution to red, but alcohols do not. Carboxylic acids reacts with sodium carbonate and liberates carbondioxide, when the evolved gas passes through the lime water, lime water turns milky. Alcohol does not give this test.

Question 2.
What are oxidising agent.
Answer;
A substance which provides oxygen for oxidation and reduce itself in the reaction is called an oxidising agent e.g. Alkaline KMnO4.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

In-text Questions (Page 76)

Question 1.
Would you be able to check if water is hard by using a detergent?
Answer:
No, because detergents do not form any precipitate with hard water and form equal amount of foam with hard and soft water.

Question 2.
People use a variety of methods to wash clothes, usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer:
Soaps or detergents forms micelles with the greasy or oily material associated with the clothes. To remove oily or greasy material it is necessary to mix soaps with clothes in such a way so that it combine with greasy/oily material to form a micelles. On agitation the oily dirt tends to lift the dirty surface.

So agitation is necessary for the throughout mixing of a soap with the clothes and it give rises to the formation of micelles and we can get clean clothes.

Class 10 Science Chapter 4 Carbon and Its Compounds Textbook Questions and Answers

Page No. 77

Question 1.
Ethane with the molecular formula – C2H6 – has
(a) 6 covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds
Answer:
(b) 7 covalent bonds

Question 2.
Butanone is a four-carbon compound with the functional group
(a) Carboxylic acid
(b) aldehyde
(c) ketone
(d) alcohol
Answer:
(c) Ketone

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means.
(a) the food is not cooked completely
(b) the fuel is not burning completely.
(c) the fuel is wet
(d) the fuel is burning completely.
Answer:
(b) the fuel is not burning completely.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH3Cl.
Answer:
Carbon has four electrons in its valence shell. So it require four more electrons to complete its octet. Hydrogen has one electron in its valence shell so it form one covalent bond with carbon. CH3Cl has three hydrogen atoms and each atom will form a covalent bond with carbon. The nature of ‘C-H’ bond is purely covalent. So CH3Cl has three pure covalent bonds. Chlorine has seven electrons in its valence shell, so it require only one electron to complete its octect. Chlorine form one covalent bond with carbon and complete its octect. But the nature of ‘C-Cl’ bond is not purely covalent, it has some polar character due to the large difference in the electronegativity of carbon and chlorine. The four covalent bonds of carbon are diverted towards the four corners of a tetrahedron because the four valency of carbon is diverted towards the four corners of tetrahedron. So the geometry of CH3Cl will be tetrahedral.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Question 5.
Draw electron dot structures for:
(a) Ethanoic acid
(b) H2S
(c) Propanone
(d) F2
Answer:
(a) Ethanoic acid M.F = CH3COOH
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 13

(b) H2S
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 14

(c) Propane (CH3 – C – CH3)
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 15

(d) F2
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 16

Question 6.
Which is a homologous series ? Explain with an example.
Answer:
It is series of similarly constituted compounds in which the members have the same functional group and have similar characteristics properties and the two consecutive members differ in their molecular formula by – CH2.

The different members of the homologous series are called homologoues.
For example homologous series of alcohols:
CH3-OH
CH3-CH2-OH
CH3-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-CH2-OH

In the above series all the members have same functional group, alcoholic group (-OH) and the two consecutive members differ in their molecular formula by – CH2. So the given series is a homologous series.

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties ?
Answer:

Ethanol Physical Properties Ethanoic acid Physical Properties
1. Ethanol is commonly known as alcohol and is the active ingredient of all alcoholic drinks. 1. Ethanoic acid is commonly known as acetic acid and belongs to a group of acids called carboxylic acid.
2. It has a intense smell of alcohols. 2. It has a intense small of vinegar.
3. It is mainly used for drinking. 3. It is mainly used as a preservative in pickels.
4. It does not freezes during winter. 4. It usually freezes during winter in cold climates.
5. Melting point: 156 K 5. Melting point: 290 K
6. Boiling point: 351 K 6. Boiling point: 391 K

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 31

Question 8.
Why does micelle formation place when soap is added water ? Will a micelle be formed in other solvents such as ethanol also ?
Answer:
Soaps molecules have two ends, one is Hydrophillic, i.e., water soluble while the other end is Hydrophobic i.e., water repelling. This unique nature of soaps leads to the formation of a micelle. When soap is dissolved in water, the hydrophobic ‘tail’ of soap will not be soluble in water and the hydrocarbon ‘tail’ protruding out of water. Inside water, these molecules have a unique orientation that keeps the hydrocarbon portion out of the water. Thus a cluster is formed which is known as micelle. Ethanol also form a micelle.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Question 9.
Why are carbon and its compounds used as fuels for most applications ?
Answer:
Carbon, in all its allotropic forms, burns in oxygen to give carbonoxide along with the release of heat and light on combustion. Carbon and its compounds have a very high calorific value.
C + O2 → CO2 + heat + light
CH4 + 2O2 → CO2 + 2H2O + heat + light
CH2CH2OH + 3O2 → 2CO2 + 3H2O + heat + light
There are the oxidation reactions.
Due to high calorific value, carbon and its compounds are used as a fuel.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
Hardness of water is due to the dissolved impurities of the salts like bicarbonates, chlorides and sulphates of calcium and magnesium. Hard water does not produce lather with soap solution readily because the cations (Ca2+ and Mg2+) present in hard water react with soap to form scum (precipitates) of calcium and magnesium salts of fatty acids.
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 17
Thus scum is formed when soap is treated with hard water.

Question 11.
What change will you observe if you test soap with litmus paper (red and blue) ?
Answer:
Soap is the Sodium of Potassium salt of long chain fatty acids e.g, Sodium stearate (C17H35COONa). When soaps are dissolved in water, they are dissociated in C17H35COO and Na+ ions. Water contains H+ and OH ions. Water contains H+ and OH ions. So C17H35COO combine with H+ to form C17H35COOH and Na+ ions combine with OH ions to form NaOH. So soap changes blue litmus to red due to acidic character and red litmus to blue due to basic character.

Question 12.
What is hydrogenation ? What is its industrial application ?
Answer:
Unsaturated hydrocarbons (alkane and alkynes) combine readily with hydrogen in the presence of finely-divided nickel, platinum or palladium as catalysts. This process is known as Hydrogenation. The ultimate products of Hydrogenation are alkanes.
e.g.,
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 18
Hydrogenation is used in the manufacture of vanaspati ghee from vegetable oils.
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 19

Question 13.
Which of the following hydrocarbons undergo addition reaction; C2H6, C3H8, C3H6, C2H2, and CH4.
Answer:
Unsaturated hydrocarbons undergo addition reactions.
So C3H6 i.e. CH3 – CH = CH2 and
C2H2 i.e. CH = CH will give addition
reations but C2H6, C3H8 and CH4 will not give addition reactions because they are saturated.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Question 14.
Give a test that can be used to differentiate chemically between butter and cooking oil.
Answer:
When the butter is heated above 250°C they decompose with production of acrolein which has intense smell but oil does not give this test.

Question 15.
Explain the mechanism of the cleaning actions of soaps.
Answer:
Soaps molecules have two different ends, one is Hydrophillic (water soluble) while the other end is hydrophobic (water repelling), When soap is at the surface, the Hydorpholbic tail of soap will not be soluble in water and the soap will align along the surface of water with the ionic end in water and the hydrocarbon ‘tail’ protruding out of water.

Inside water, those molecules have unique orientation that keeps the hydrophobic end out of the water. This is achieved by forming clusters of molecules in which the hydrophobic tails are in the interior of the duster and the ionic ends are on the surface of the duster. This formation is called a micelle. Soap in the form of a micelle is able to dean, since the oily dirt will be collected in the centre of the micelle. The micelle stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion. So the dirt suspended in the micelles is also easily rinsed away.

Class 10 Science 4 Carbon and Its Compounds Textbook Activities

Activity 4.1 (Page 58)

  • Make a list of ten things you have used or consumed since the morning.
  • Compile this list with the lists made by your classmates and then sort the items into the following table.
  • If there are items which are made up of more than one material put them into bot the relevant columns.
Things made of metal Things made of glass/clay Others
Glass Glass Milk
Bucket Cup Bread
Plate Plate Butter Tea

Activity 4.2 (Page 67)

  • Calculate the difference in the formulae and molecular masses for (a) CH3OH and C2H5OH and C3H7OH and (c) C3H7OH and C4H9OH.
  • Is there any similarity in these three?
  • Arrange these alcohols in the order of increasing carbon atoms to get a family. Can we call this family a homologous series?
  • Generate the homologous series for compounds containing upto four carbons for the other functional groups given in table 4.3

Answer:
(a) CH3OH and C2H3OH
Difference in the molecular formula = CH2
The molecular mass of CH3OH
= C + 3 × H + 1 × O × 1 × H
= 12 + 3 × 1 + 16 + 1 × 1
= 12 + 3 + 16 + 1 = 32

The molecular mass of C2H5OH
= 2 × C + 5 × H + 1 × O + 1 × H
= 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1
= 24 + 5 + 16 + 1 = 46
The difference in molecular mass = 46 – 32 = 14

(b) C2H5OH and C3H7OH
Difference in the molecular formula = CH2
The molecular mass of C2H5OH
= 2 × C + 5 + H + 1 × O + 1 × H
= 2 × 12 + 5 × 1 + 1 × 16 – 1 × 1
= 24 + 5 + 16 + 1
= 46
The molecular mass of C3H7OH
= 3 × C + 7 × H + 1 × O + 1 × H
= 3 × 12 + 7 × 1 + 1 × 16 + 1 × 1
= 36 + 7 + 16 + 1
= 60
Difference in the molecular mass = 60 – 46 = 14

(c) C3H7OH and C4H9OH
Difference in the molecular formula = CH2
The molecular mass of C3H7OH
= 3 × C + 7 × H + 1 × O + 1 × H
= 3 × 12 + 7 × 1 + 1 × 16 + 1 × 1
= 36 + 7 + 16 + 1 = 60
The molecular mass of C4H9OH
= 4 × C + 9 × H + 1 × O + 1 × H
= 4 × 12 + 9 × 1 + 1 × 16 + 1 × 1
= 48 + 9 + 16 + 1
= 74
The difference in molecular mass = 74 – 60 = 14

  • Yes there is a similarity in the three organic compouds. They all have same functional group.
  • CH3OH
  • C2H5OH
  • C3H7OH
  • Yes we can call this family a homologous I series because the difference between the molecular formula of the two consecutive members differ by -“CH2” and they have same functional group. (OH)
  • Homologous series of halo-alkane

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 20
Homologous series of alcohol
CH3-OH
CH3-CH2-OH
CH3-CH2-CH2-OH2-OH
CH3-CH2-CH2-CH2-OH
Homologous series of aldehyde
H-CHO
CH3-CHO
CH3CH2-CHO
CH3CH2CH2-CHO

Homologous series of Ketons:
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 21
Homologous series of carboxylic acid :
H-COOH
CH3-COOH
CH3-CH2-COOH
CH3-CH2-CH2-COOH

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Activity 4.3 (Page 68)

Caution : This activity needs the teacher’s assitance.

  • Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatual and burn them.
  • Observe the nature of the flame and note whether smoke is produced.
  • Place a metal plate above the flame. If there a deposition on the plate in case of any of the compounds ?

Observations : Saturated hydrocarbon generally give a clean flame while unsaturated carbon compounds give a yellow flame with lots of black smoke. It results in a sooty deposit on the metal plate. Smoke is also produced in all cases and a sooty deposit on metal place.

Activity 4.4 (Page 68)

  • Light a bunsen burner and adjust the air hole at the base to get different types of flames/presence of smoke.
  • When do you get a yellow, sootry flame ?
  • When do you yet a blue flame.

Observation: During incomplete combustion a yellow sooty flame is observed and when a sufficient oxygen is supplied it gives a clean blue flame.

Activity 4.5 (Page 70)

  • Take about 3 ml of ethanol in a test tube and warm it gently in a water bath.
  • Add a 5% solution of alkaline potassium permanganate drop by drop to this solution.
  • Does the colour of potassium permanganate persist when it is added initially ?
  • Why does the colour of potassium permanganate not disappear when excess is added ?
  • Alkaline potassium permanganate is an oxidising agent so it oxidises alcohols to acids. An excess potassium permanganate, contains a lot of permanganate ions which a pink colour to the solution because all the permanganate ions are not reduced into MnO2.

Observation : Yes the colour of alkaline potassium permanganate persist initially but is disappears immediately.

Activity 4.6 (Page 72)

Teacher Demonstration:

  • Drop a small piece of sodium, about the size of a couple of grains of rice, into ethanol (absolute alcohol).
  • What do you observe ?
  • How will you test the gas evolved ?

Observation : Alcohols react with sodium metal and produce hydrogen sodium ethoxide. Put a candle near the mouth of the test tube taken, the evolved gas burn with a pop sound, it indicated the presence of hydrogen.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Activity 4.7 (Page 73)

  • Compare the pH of dilute acetic acid and dilute hydrochloric acid using both litmus paper and universal indicator.
  • Are both acids indicated by the litmus test ?
  • Does the universal indicator show them as equally strong acids?

Observation:

  • pH of oil HCl : 1.0
  • pH of dil. acetic acid : 3.5
  • Yes both the acids indicated by the litmus test. They turn blue litmus to red.
  • No, Universal indicator shows that dilute hydrochloric acid is stronger than acetic acid.

Activity 4.8 (Page 73)

  • Take I mL ethanol (absolute alcohol) and 1 mL glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
  • Warm in a water-bath for at kast five minutes as shown in Fig.
  • Pour into a beaker containing 20-50 mL of water and smell the resulting mixture.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 22
Observation : Ethanol reacts with glacial acetic acid in presence of concentrated sulphuric acid to form esters. Esters have fruity smell or sweet smell.

Activity 4.9 (Page 74)

  • Set up the apparatus as shown in chapter 2. activity 2.5
  • Take a spatula-full of sodium, carbonate in a test tube and add 2 mL of dilute ethanoic acid.
  • What do you observe?
  • Pass the gas produced through freshly prepared lime-water. What do you observe?
  • Can the gas produced by the reaction between ethanoic acid and sodium carbonate be identified by this test?
  • Repeat this activity with sodium hydrogen carbonate instead of sodium carbonate.

Observation:
Ethanoic acid reacts with sodium carbonate to give rise to a salt, carbondioxide and water.
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 23
It turns lime-water milky.
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 24
Yes the gas evolved can be identified by this test.

Ethanoic acid reacts with sodium hydrogen carbonate to give rise to salt. CO2, and water.
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 25
Same result is observed with lime-water.

Activity 4.10 (Page 74)

  • Take about 10 mL of water each in two test tubes.
  • Add a drop of oil (cooking oil) to both the test tubes and label them as A and B.
  • To test tube B, add a few drops of soap solution.
  • Now shake both the test tubes vigourously for the same period of time.
  • Can you see the oil and water layers separately in both the test tubes immediately after you stop shaking them ?
  • Leave the test tubes undisturbed for some time and observe. Does the oil layer separate out ? In which test tube does this happen first ?

Observation:

  • No, oil and water layer are not separated immediately.
  • After some time water and oil layers are separated. In test tube ‘A’ oil and water layers are separated first.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Activity 4.11 (Page 76)

  • Take about 10 mL of distilled water (or rain water) and 10 mL of hard water (from a tubewell or hand-pump in separate test tubes.
  • Add a couple of drops of soap solution to both.
  • Shake the test tubes vigourously for an equal period of time and observe the amount of foam formed.
  • In which test tube do you get more foam?
  • In which test tube do you observe a white curdy precipitate.

Note for teacher : If hardwater is not available in your locality, prepare some hard water by dissolving hydrogen carbonates/ sulphates/ chlorides of calcium or magnesium in water.

Observation : The test tube which contains distilled water has more foam than the test tube which contains hard water.

Activity 4.12 (Page 76)

  • Take two test tubes with about 10 mL of hard water in each.
  • Add five drops of soap solution to one and five drops of detergent solution to the other.
  • Shake both test tubes for the same period.
  • Do both test tubes is a curdy solid formed ?

Observation:

  • Both the test tube have different amount of foam.
  • A while curdy solid is formed in soap solution test tube.

Class 10 Science Chapter 4 Carbon and Its Compounds Additional Important Questions and Answers

Very Short Answers Type Questions

Question 1.
What is the co-valency of Carbon ?
Answer:
Four.

Question 2.
What is Vinegar ?
Answer:
5-8% solution of acetic acid in water in called vinegar.

Question 3.
What is the formula of chloroform ?
Answer:
CHCl3.

Question 4.
How many covalent bonds are present in N2?
Answer:
Three

Question 5.
Give the electron dot structure of Hydrogen ?
Answer:
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 26

Short Answer Type Questions

Question 1.
What are the characteristics of a homologous series?
Answer:
The general characteristics of a homologous series are:

  • All the members of a series can be represented by the same general formula.
    e.g. Alcohol sereis : CnH 2n+1 OH.
  • Any two consecutive members have a difference of CH2.
  • All the members of a series have same functional group.
  • All the members of a series have almost similar chemical properties.
  • All the members show a regular gradation in the physical properties.
  • All the member of a series can be synthesized by the identical methods.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Question 2.
What is longest carbon chain rule ? Explain with an example.
Answer:
During naming of organic compound certain rules are followed. Longest carbon chain rule is as follows select the longest continous chain of carbon atoms as the parent chain. If some carbon-carbon double or triple bond is present, the parent chain must contain the carbon atoms involved in it.
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 27
In this compound the three continuous chain of carbon atoms are present. One of them contain four carbon atoms, another one contain six and third chain contain seven carbon atoms. According to this rule you select the longest carbon chain which contain seven carbon atoms.

Question 3.
What is lowest number or lowest sum rule ? Explain.
Answer:
The selected longest chain is numbered using Arabic numerals and the positions of substituent groups are indicated by the number of carbon atom to which the alkyl group is attached.

“The numbering is done in the parent chain in such a way that the substituted carbon atoms have the lowest possible numbers. When series of locants containing the same number of terms are compared term by term, that series is lowest which contains the lowest number on the occasion of first difference.”
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 28
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 29

Long Answer Type Question

Question 1.
Discuss the micelles formation and cleaning action of soap with figures.
Answer:
Soaps are molecules in which the two ends have differing properties, one is Hydrophilic, that is, it dissolved in water, while the other end is Hydrophobic, that t dissolves in Hydrocarbons. When soaps is at the surface of water, the Hydrophobic ‘tall’ of soap
NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds 30
will not be soluble in water and the soap will align along the surface of water with the ionic end in water and the Hydrocarbon ‘tail’ protruding out of water. Inside water, these molecules have a unique orientation that keeps and Hydrocarbons portion out of the water. This is achieved by forming clusters of molecules in which the Hydrophobic tails are in the interior of the cluster and the ionic ends are on the surface of the cluster. This formation is called a micelle.

Soap in the form of a micelle is able to clean, since the oily dirt will be collected in the centre of the micelle. The micelles stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion. Thus, the dirt suspended in the micelles is also easily rinsed away. The soap micelles are large enough to scatter light. Hence a soap solution appears cloudy.

NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

Multiple Choice Questions

Question 1.
The IUPAC name of Acetone
(a) Propanone
(b) Butanone
(c) Alkanone
(d) Propanal
Answer:
(a) Propanone

Question 2.
Aldehydes and Ketones commonly have
(a) Carboxyl group
(b) smell
(c) Melting point
(d) Boiling point
Answer:
(a) Carboxyl group

Question 3.
Sodium/Potassium salts of fatty acid are called
(a) Soaps
(b) Detergents
(c) Micelles
(d) Hydrocarbons
Answer:
(a) Soaps

Question 4.
In porpane the no. of Carbon-Carbon single bonds is/are
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 5.
The molecular formula of benzene is
(a) C6H12
(b) C6H6
(c) C6H10
(d) C6H8
Answer:
(b) C6H6

error: Content is protected !!