NCERT Exemplar Class 10 Science Solutions Chapter 4 Carbon And Its Compounds 2021

NCERT Exemplar Class 10 Science Solutions Chapter 4 Carbon And Its Compounds

Edited By Sumit Saini | Updated on Sep 09, 2022 01:53 PM IST | #CBSE Class 10th

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NCERT exemplar Class 10 Science solutions chapter 4 explores various properties of carbon whether it be in elemental form or part of a compound. The chapter on Carbon and its Compounds is thoroughly analyzed by our experts in order to prepare NCERT exemplar Class 10 Science chapter 4 solutions for the students. These Class 10 Science NCERT exemplar solutions chapter 4 develop a better understanding of concepts of carbon and its compounds as they are detailed and expressive. The CBSE syllabus for Class 10 chapter 4 is covered in the NCERT exemplar Class 10 Science chapter 4 solutions.

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NCERT Exemplar Class 10 Science Solutions Chapter 4-MCQ
NCERT Exemplar Class 10 Science Solutions Chapter 4-Short Answer
NCERT Exemplar Class 10 Science Solutions Chapter 4-Long Answer
NCERT Exemplar Solutions Class 10 Science Chapter 4 Important Topics:
NCERT Class 10 Science Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 10 Science Solutions Chapter 4:

Also read - NCERT solutions for Class 10

NCERT Exemplar Class 10 Science Solutions Chapter 4-MCQ

Question:1

Carbon exists in the atmosphere in the form of
(a) carbon monoxide only
(b) carbon monoxide in traces and carbon dioxide
(c) carbon dioxide only
(d) coal
Answer: (c)
Solution:
Carbon exists in the atmosphere in the form of carbon dioxide gas (CO₂) in air (only 0.03%).
Carbon occurs in the earth's crust in the form of minerals likes carbonates.
Carbon occurs in the form of fossil fuels, organic compounds, wood, cotton and wool
Therefore, option (c) is correct.

Question:2

Which of the following statements are usually correct for carbon compounds? These
(i) are good conductors of electricity
(ii) are poor conductors of electricity
(iii) have strong forces of attraction between their molecules
(iv) do not have strong forces of attraction between their molecules
(a) (i) and (iii) (b) (ii) and (iii)
(c) (i) and (iv) (d) (ii) and (iv)

Answer:(d)
Solution:
Carbon predominantly forms covalent compounds, due to valency 4. Generally carbon is a poor conductor of electricity as its compounds are predominantly covalent in nature. The forces of attraction in between their molecules are weak as they are not covalent in nature.
Statement (ii) and (iv) are correct.
Therefore, option (d) is correct.

Question:3

A molecule of ammonia $(NH_3)$ has
(a) only single bonds
(b) only double bonds
(c) only triple bonds
(d) two double bonds and one single bond
Answer:(a)
Solution:
A molecule of ammonia (NH3) has only single bonds and these are covalent bonds.

Therefore, option (a) is correct.

Question:4

Buckminsterfullerene is an allotropic form of
(a) phosphorus
(b) sulphur
(c) carbon
(d) tin
Answer:(c)
Solution:
Buckminsterfullerene is an allotrope of carbon containing clusters of 60 carbon atoms joined together to form spherical molecules. Its formula is C60, (C-Sixty)
It has 20 hexagons and 12 pentagons. Other allotropic forms of carbon are graphite, diamond, coal, coke…etc.

Therefore, option (c) is correct.

Question:5

Which of the following are correct structural isomers of butane?

(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer:(c)
Solution: Structural isomers: two or more organic compounds having the same molecular formulas but different structures.
Butane: $C_4H_{10}$

Structure (i) is n-butane. It is a structural isomer of butane.
Structure (iii) is iso-butane. It is a structural isomer of butane.
Other structures, i.e., (ii) and (iv) have different molecular formula, hence incorrect
Therefore, option (c) is correct.

Question:6

$CH_{3}-CH_{2}-OH\xrightarrow[Heat]{Alkaline KMnO_{4}}CH_{3}-COOH$
In the above given reaction, alkaline $KMnO_4$ acts as
(a) reducing agent
(b) oxidising agent
(c) catalyst
(d) dehydrating agent
Answer:(b)
Solution: Here ethanol is converted to ethanoic acid.
$KMnO_4$ acts as oxidizing agent as it removes hydrogen from $CH_3CH_2OH$ and adds one oxygen to it.
Therefore, option (b) is correct.

Question:7

Oils on treating with hydrogen in the presence of palladium or nickel catalyst form fats. This is an example of
(a) Addition reaction
(b) Substitution reaction
(c) Displacement reaction
(d) Oxidation reaction
Answer: (a)
Solution:
Oils are unsaturated compounds containing double bonds. Unsaturated compounds are the only one which can undergo addition reaction.
The reaction is known as hydrogenation reaction. For example:

Hydrogenation reactions are addition reactions.
Therefore, option (a) is correct.

Question:8

In which of the following compounds, $-OH$ is the functional group?
(a) Butanone
(b) Butanol
(c) Butanoic acid
(d) Butanal
Answer:(b)
Solution:
(a) Butanone: $C_4H_8O$

The functional group is $-C=O$
(b) Butanol: $CH_3-CH_2-CH_2-CH_2-OH$
The general formula of alcohols is $C_nH_{(2n+1)}- OH.$
For Butanol, n = 4
Butanol Formula = $C_4H_9OH$
The functional group is $-OH$ .
(c) Butanoic acid: $C_4H_8O_2$
The functional group is $-COOH.$
(d) Butanal: $C_4H_8O$
The functional group is $-CHO$ .
Therefore, option (b) is correct.

Question:9

The soap molecule has a
(a) hydrophilic head and a hydrophobic tail
(b) hydrophobic head and a hydrophilic tail
(c) hydrophobic head and a hydrophobic tail
(d) hydrophilic head and a hydrophilic tail
Answer:(a)
Solution:
A soap molecule is made up of two parts: A long hydrocarbon part and a short ionic part $-COO-Na^+$ group. The long hydrocarbon chain is hydrophobic (water repelling) and ionic portion is hydrophilic (water attracting).

Therefore, option (a) is correct.

Question:10

Which of the following is the correct representation of electron dot structure of nitrogen?

Answer:(d)

Electronic configuration of N (atomic number 7) is 2, 5
Therefore, it needs three more electrons to complete its octet. Each nitrogen atom shares three electrons to form a molecule of $N_{2}$ as:
$N\equiv N$
Hence the correct electron dot structure is:

Therefore, option (d) is correct.

Question:11

Structural formula of ethyne is

Answer:(a)
Ethyne: $C_2H_2$
‘eth’ shows presence of two carbon atoms and ‘yne' shows presence of a triple bond,
Thus, ethyne structure is:
$H-C\equiv C-H$
It is also known as acetylene.
Therefore, option (a) is correct.

Question:12

Identify the unsaturated compounds from the following
(i) Propane
(ii) Propene
(iii) Propyne
(iv) Chloropropane
(a) (i) and (ii) (b) (ii) and (iv)
(c) (iii) and (iv) (d) (ii) and (iii)
Answer:

A hydrocarbon in which the two carbon atoms are connected by a ‘double bond’ or ‘triple bond' are categorized as unsaturated hydrocarbon
(i) Propane: $C_3H_8$

(ii) Propene: $C_3H_6$

(iii) Propyne: $C_3H_4$
$CH_3-C\equiv CH$
(iv) Chloropropane: $C_3H_7Cl$

So, Propene and propyne has double bonds and triple bonds respectively which means they are unsaturated.
Propane and chloropropane are saturated hydrocarbons which contain only single bonds.
Therefore, option (d) is correct.

Question:13

Chlorine reacts with saturated hydrocarbons at room temperature in the
(a) absence of sunlight
(b) presence of sunlight
(c) presence of water
(d) presence of hydrochloric acid
Answer: (b)
Chlorine reacts with saturated hydrocarbon at room temperature in the presence of sunlight,
$CH_{4}+Cl_{2}\overset{sunlight}{\rightarrow}CH_{3}Cl+HCl$
$CH_3Cl+Cl_{2}\overset{sunlight}{\rightarrow}CH_{2}Cl_{2}+HCl$
$CH_2Cl_{2}+Cl_{2}\overset{sunlight}{\rightarrow}CHCl_{3}+HCl$
$CHCl_{3}+Cl_{2}\overset{sunlight}{\rightarrow}CCl_{4}+HCl$
Therefore, option (b) is correct.

Question:14

In the soap micelles
(a) the ionic end of soap is on the surface of the cluster while the carbon chain is in the interior of the cluster.
(b) ionic end of soap is in the interior of the cluster and the carbon chain is out of the cluster.
(c) both ionic end and carbon chain are in the interior of the cluster
(d) both ionic end and carbon chain are on the exterior of the cluster
Answer:

A group of soup molecules aggregated in spherical arrangement surrounding the dirt or oil in the soap solution in water is called a ‘micelle’. In a soap micelle, the soap molecules are arranged readily with hydrocarbon ends directed towards the centre and ionic ends directed outwards.

Therefore, option (a) is correct.

Question:15

Pentane has the molecular formula $C_5 H_{12}$ . It has
(a) 5 covalent bonds
(b) 12 covalent bonds
(c) 16 covalent bonds
(d) 17 covalent bonds
Answer:

Structure of pentane:

It contains 16 covalent bonds.
Therefore, option (c) is correct.

Question:16

Structural formula of benzene is

Answer:(c)
Benzene molecule contains alternate single and double bonds in hexagonal carbon chain. Each carbon is attached with a hydrogen atom which makes formula of benzene as $C_6H_6$ in structure.

Therefore, option (c) is correct.

Question:17

Ethanol reacts with sodium and forms two products. These are
(a) sodium ethanoate and hydrogen
(b) sodium ethanoate and oxygen
(c) sodium ethoxide and hydrogen
(d) sodium ethoxide and oxygen
Answer:(c)
Ethanol reacts with sodium to form sodium ethoxide and release hydrogen gas.
$C_2H_5OH + 2Na \rightarrow 2C_2H_5O^{-}Na^{+} + H_2$
Therefore, option (c) is correct.

Question:18

The correct structural formula of butanoic acid is:

Answer:(d)
Butanoic acid formula = $C_4H_8O_2$
$C_3H_7COOH$
The general formula of carboxylic acid is $R-COOH$ where R is an alkyl group.
General formula of carboxylic acid = $C_{n-1}H_{2n-1}COOH$
Butanoic acid means 4 carbons

Therefore, option (d) is correct.

Question:19

Vinegar is a solution of
(a) 50% – 60% acetic acid in alcohol
(b) 5% – 8% acetic acid in alcohol
(c) 5% – 8% acetic acid in water
(d) 50% – 60% acetic acid in water
Answer:(c)
Vinegar: $CH_3COOH$
Vinegar is 5%-8% solution of acetic acid in water.
It is also known as acetic acid.

Therefore, option (c) is correct.

Question:20

Mineral acids are stronger acids than carboxylic acids because
(i) mineral acids are completely ionised
(ii) carboxylic acids are completely ionised
(iii) mineral acids are partially ionised
(iv) carboxylic acids are partially ionised
(a) (i) and (iv) (b) (ii) and (iii)
(c) (i) and (ii) (d) (iii) and (iv)
Answer:(a)
Solution:
Mineral acids are strong acids which ionizes almost completely and carboxylic acids are weak acids as they don’t ionize easily and only a little percentage of it gets ionized.
Therefore, option (a) is correct.

Question:21

Carbon forms four covalent bonds by sharing its four valence electrons with four univalent atoms, e.g. hydrogen. After the formation of four bonds, carbon attains the electronic configuration of
(a) helium
(b) neon
(c) argon
(d) krypton
Answer:(b)
Electronic configuration of carbon (C) = 2, 4
When it forms four covalent bonds by sharing its four valence electrons with hydrogen, it forms CH₄

So it has total 10 electrons which is configuration of Neon.
Therefore, option (b) is correct.

Question:22

The correct electron dot structure of a water molecule is

Answer:(c)
Oxygen atom has 6 valence electrons, out of which 2 are involved in bonding with the 2 hydrogen atoms. Therefore water molecule has 2 bonded pair of electrons and 2 lone pairs.

Hence the answer is

Therefore, option (c) is correct.

Question:23

Which of the following is not a straight chain hydrocarbon?

Answer: (d)

We can see that all the carbon atoms are attached by covalent bonds in a continuous straight chain.
But in structure (d), 5 carbons in a chain with another carbon connected to 2^nd carbon from the right.
Therefore, option (d) is correct.

Question:24

Which among the following are unsaturated hydrocarbons?

(a) (i) and (iii) (b) (ii) and (iii)
(c) (ii) and (iv) (d) (iii) and (iv)
Answer:(c)
Unsaturated hydrocarbons have double or triple bond in their structures.
We can see that structure (ii) has a triple bond.
Structure (iv) has a double bond
Therefore, option (c) is correct.

Question:25

Which of the following does not belong to the same homologous series?
(a) $CH_4$
(b) $C_2 H_6$
(c) $C_3 H_8$
(d) $C_4 H_{8}$
Answer:(d)
A homologous series is a group of organic compounds having similar structures and similar chemical properties in which the successive compounds differ by $CH_2$ group.
So, homologous series of alkanes is
Methane $CH_4$
Ethane $C_2 H_6$
Propane $C_3 H_8$
Butane $C_4 H_{8}$
So, $C_4H_8$ does not belong to the homologous series
Therefore, option (d) is correct.

Question:26

The name of the compound $CH_3 - CH_2 - CHO$ is
(a) Propanal
(b) Propanone
(c) Ethanol
(d) Ethanal
Answer:(a)
Functional group present in $CH_3 - CH_2 - CHO$ is aldehyde.
The compound has 3 carbons so -Prop is the prefix
Aldehyde has a suffix of -al
Prefix + an + Suffix = Propanal
Therefore, option (a) is correct.

Question:27

The heteroatoms present in
$CH_3 - CH_2 - O - CH_2 - CH_2Cl$ are
(i) oxygen
(ii) carbon
(iii) hydrogen
(iv) chlorine
(a) (i) and (ii) (b) (ii) and (iii)
(c) (iii) and (iv) (d) (i) and (iv)
Answer:(d)
Atoms other than C and H, if present in organic compound, are called heteroatoms.
Oxygen and chlorine are the elements found in the given compound apart from carbon and hydrogen.
Therefore, option (d) is correct.

Question:28

Which of the following represents saponification reaction?
(a) $CH_3COONa + NaOH \overset{CaO}{\rightarrow} CH_4 + Na_2CO_3$
(b) $CH_3COOH + C_2H_5OH \overset{H_2SO_{4}}{\rightarrow}CH_3 COOC_2H_5 + H_2O$
(c) $2CH_3COOH + 2Na \rightarrow 2CH_3COONa + H_2$
(d) $CH_3COOC_2H_5 + NaOH \rightarrow CH_3 COONa + C_2H_5OH$

Answer: (d)
When an ester is heated with sodium hydroxide solution, ester gets hydrolyzed (breaks down) to form the parent alcohol and sodium salt of carboxylic acid. This process is called as saponification. Equation (d) has alcohol and sodium salt of carboxylic acid.
Therefore, option (d) is correct.

Question:29

The first member of alkyne homologous series is
(a) ethyne
(b) ethene
(c) propyne
(d) methane
Answer: (a)
A homologous series is a group of organic compounds having similar structures and similar chemical properties in which the successive compounds differ by $CH_2$ group.
The general formula of alkyne is $C_nH_{2n-2}$ where n is the number of carbon atoms. The first member of alkyne homologous series is ethyne $C_2H_2$ as you need 2 carbons minimum to have a carbon connected with double bond.
Therefore, option (a) is correct.

NCERT Exemplar Class 10 Science Solutions Chapter 4-Short Answer

Question:30

Draw the electron dot structure of ethyne and also draw its structural formula.
Answer:

Molecular formula of ethyne is $C_2H_2$ .
Electronic configuration of C= 2, 4
Electronic configuration of H = 1
Electron dot structure is:

Structural formula is
$H-C\equiv C-H$

Question:31

Write the names of the following compounds

Answer:

(a) Pentane —e + oic acid = Pentanoic acid
As the structure has 5 carbons and carboxylic acid group.
(b) Butane—ane + yne = Butyne
As the structure has 5 carbons and triple bond.
(c) Heptane —e +al = Heptanal
As the structure has 7 carbons and aldehyde group
(d) Pentane —e + ol = Pentanol
As the structure has 5 carbon atoms and one alcohol group

Question:32

Identify and name the functional groups present in the following compounds.

Answer:

(a) —OH (alcohol)
(b)
(carboxylic acid)
(c) Ketone - Wikipedia
(ketone)
(d) —C= C—
(alkene)

Question:33

A compound X is formed by the reaction of a carboxylic acid $C_2H_4O_2$ and an alcohol in presence of a few drops of $H_2SO_4$ . The alcohol on oxidation with alkaline $KMnO_4$ followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of:
(a) carboxylic acid, (b) alcohol and (c) the compound X. Also write the reaction.
Answer:

Carboxylic acid is ethanoic acid

Alcohol is ethanol

X is ethyl ethanoate

Reaction involved:
$CH_{3}COOH(l)+C_{2}H_{5}OH(l)\overset{Conc. H_{2}SO_{4}}{\rightarrow}CH_{3}COOC_{2}H_{5}(l)+H_{2}O(l)$

Question:34

Why detergents are better cleansing agents than soaps? Explain.
Answer:

Detergents are better cleansing agents than soaps because they can be used even with hard water. The charged ends of detergents do not form insoluble precipitates with calcium and magnesium ions in hard water but soap doesn’t work in hard water. Hard water contains Ca²⁺ and Mg²⁺ ions which react with soap to form Insoluble salts of calcium and magnesium.
Detergents have a stronger cleansing action than soaps and are more soluble in water than soaps.

Question:35

Name the functional groups present in the following compounds
(a) $CH_3 CO CH_2 CH_2 CH_2 CH_3$
(b) $CH_3 CH_2 CH_2 COOH$
(c) $CH_3 CH_2 CH_2 CH_2 CHO$
(d) $CH_3 CH_2 OH$
Answer:

(a) Ketone
Ketone - Wikipedia
(b) Carboxylic acid, —COOH

(c) Aldehyde, —CHO
(d) Alcohol, —OH

Question:36

How is ethene prepared from ethanol? Give the reaction involved in it.
Answer:

Ethene is prepared by the dehydration of ethyl alcohol in the presence of $conc. H_{2}SO_{4}$ at $160^{\circ}C$ .
$C_2H_5OH \overset{conc. H_{2}SO_{4}}{\rightarrow} CH_2 = CH_2 + H_2O$

Question:37

Intake of small quantity of methanol can be lethal. Comment.
Answer:

Methanol is very poisonous because it gets oxidized to methanol (formaldehyde) in the liver. Methanal reacts aggressively with the components of cells and causes the protoplasm to get coagulated (In the same way, as an egg gets coagulated on boiling). Methanol also affects the optic nerve causing blindness.
$CH_{3}OH+[O]\xrightarrow[H_{2}SO_{4}]{Na_{2}Cr_{2}O_{7}}CH_{2}O+H_{2}O$

Question:38

A gas is evolved when ethanol reacts with sodium. Name the gas evolved and also write the balanced chemical equation of the reaction involved.

Answer:

When ethanol reacts with sodium, the gas evolved is hydrogen.
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5O^-Na^+ + H_2$

Question:39

Ethene is formed when ethanol at 443 K is heated with excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation of this reaction.
Answer:

When ethanol is heated with excess of concentrated sulphuric acid at $170^{\circ}C$ , it gets dehydrated to form ethene.
The reaction is as follows:
$CH_3CH_2OH \overset{Conc. H_{2}SO_{4}}{\rightarrow} CH_2 = CH_2 + H_2O$
Concentrated sulphuric acid removes water from ethanol, thereby, acting as a dehydrating agent.

Question:40

Carbon, Group (14) element in the Periodic Table, is known to form compounds with many elements.
Write an example of a compound formed with
(a) chlorine (Group 17 of Periodic Table)
(b) oxygen (Group 16 of Periodic Table)
Answer:

(a) Carbon forms carbon tetrachloride when reacted with chlorine in the presence of sunlight.
Carbon tetrachloride: $CCl_4$
(b) Carbons forms carbon dioxide or carbon monoxide when reacted with oxygen based on the availability of oxygen.
Carbon dioxide: $CO_2$
Carbon monoxide: CO

Question:41

In electron dot structure, the valence shell electrons are represented by crosses or dots.
(a) The atomic number of chlorine is 17. Write its electronic configuration
(b) Draw the electron dot structure of chlorine molecule.
Answer:

(a) Electronic configuration of CI (atomic number 17)
2 8 7
(b) Electron dot structure of chlorine molecule is

Question:42

Catenation is the ability of an atom to form bonds with other atoms of the same element. It is exhibited by both carbon and silicon. Compare the ability of catenation of the two elements. Give reasons.
Answer:

Both carbon (C) and silicon (Si) have similar valence shell electronic configuration
C (atomic number 6) – Electronic configuration 2, 4
Si (atomic number 14) – Electronic configuration 2, 8, 4
Both have four electrons in the valence shell and hence show the phenomenon of catenation.
Carbon exhibits catenation much more than silicon or any other element due to its smaller size which makes the C—C bonds strong while the Si—Si bonds are comparatively weaker due to its large size.
Thus, due to greater strength of C—C over Si—Si bonds, carbon shows catenation to a greater extent than silicon.

Question:43

Unsaturated hydrocarbons contain multiple bonds between the two C-atoms and show addition reactions. Give the test to distinguish ethane from ethene.
Answer:

When an unsaturated hydrocarbon reacts with bromine the double or triple bonds between the carbons get broken and bromine gets attached to those carbons.
Conformation test:
Ethane when burnt in air will give clear flame whereas ethene will give lots of smoke.
$2 C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$
$C_{2}H_{4}(g)+3O_{2}(g) \rightarrow 2CO_{2}(g)+2H_{2}O(l)+Smoke$

Question:44

Match the reactions given in Column (A) with the names given in column (B).

Column A	Column B
(a) $CH_3OH + CH_3COOH \overset{H^{+}}{\rightarrow} CH_3COOCH_3 + H_2O$	(i) Addition reaction
(b) $CH_2 = CH_2 + H_2 \overset{Ni}{\rightarrow} CH_3 - CH_3$	(ii) Substitution reaction
(c) $CH_4 + Cl_2 \overset{sunlight}{\rightarrow} CH_3Cl + HCl$	(iii) Neutralisation reaction
(d) $CH_3COOH+NaOH \rightarrow CH_3COONa+H_2O$	(iv) Esterification reaction

Answer:

A – iv
B – i
C – ii
D – iii
Solution:
Addition reaction - a reaction in which one molecule combines with another to form a larger molecule with no other products.
$CH_2 = CH_2 + H_2 \overset{Ni}{\rightarrow} CH_3 - CH_3$
Substitution reaction – in which one functional group in a chemical compound is replaced by another functional group.
$CH_4 + Cl_2 \overset{sunlight}{\rightarrow} CH_3Cl + HCl$
Neutralisation reaction – in which an acid and base react to give salt and water
$CH_3COOH+NaOH \rightarrow CH_3COONa+H_2O$
Esterification reaction – in which an organic acid reacts with an alcohol to form an ester and water
$CH_3OH + CH_3COOH \overset{H^{+}}{\rightarrow} CH_3COOCH_3 + H_2O$

Question:45

Write the structural formulae of all the isomers of hexane.
Answer:

(a) Hexane

(b) 2-methyl pentane

(c) 2,2-dimethyl butane

(d) 2,3-dimethyl butane

(e) 3-methyl pentane

Question:46

What is the role of metal or reagents written on arrows in the given chemical reactions?
(a)
(b) $CH_{3}COOH +CH_{3}CH_{2}OH \overset{Conc. H_{2}SO_{4}}{\rightarrow} CH_{3}COOC_{2}H_{5}+H_{2}O$
(c) $CH_{3}CH_{2}OH\xrightarrow[Heat]{Alk. KMnO_{4}} CH_{3}COOH$
Answer:

(a) Nickel (Ni) acts as the catalyst during the reaction. It first absorbs the hydrogen molecule on its surface as hydrogen atoms and then the alkene molecule side by side. The two hydrogen atoms then add across the double bond of the alkene to form the addition product, i.e., 2, 3- dimethyl butane.
(b) Conc. $H_{2}SO_{4}$ increases the rate of the forward reaction by removing $H_{2}O$ formed during the reaction. In other words, conc. $H_{2}SO_{4}$ acts as a dehydrating agent.
(c) Alkaline $KMnO_{4}$ acts as an oxidizing agent and oxidizes ethanol to ethanoic acid

NCERT Exemplar Class 10 Science Solutions Chapter 4-Long Answer

Question:47

A salt X is formed and a gas is evolved when ethanoic acid reacts with sodium hydrogencarbonate. 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 which you have named. Also, write chemical equation of the reaction involved.
Answer:

When sodium hydrogen carbonate reacts with ethanoic acid, it forms sodium ethanoate salt, water and carbon dioxide.
$CH_3COOH + NaHCO_3\rightarrow CH_3COONa + H_2O + CO_2$
So, salt X is Sodium ethanoate = $CH_3COONa$
Activity
1. Take sodium hydrogen carbonate in a test tube and add ethanoic acid
2. Carbon dioxide is evolved with brisk effervescence.
3. Pass the gas into lime water and if it turns white (milky), it means carbon dioxide gas is present. This is the conformation test for carbon dioxide.
$Ca(OH)_2 (aq) + CO_2 (g) \rightarrow CaCO_3 (s) + H_2O (l)$

Question:48

(a) What are hydrocarbons? Give examples.
(b) Give the structural differences between saturated and unsaturated hydrocarbons with two examples each.
(c) What is a functional group? Give examples of four different functional groups.
Answer:

a) Hydrocarbons are compounds made of carbon and hydrogen.
Example- Methane: $CH_4$
Ethene: $CH_2 = CH_2$
Benzene: $C_6H_6$
b) In saturated hydrocarbons, all the bonds of carbon are single bonds but in unsaturated hydrocarbon compounds, carbon chain has either double bond or triple bond or both in some cases.
Example
Saturated hydrocarbons:
Methane

Ethane

Unsaturated hydrocarbons:
Ethene

Ethyne
$H-C\equiv C-H$
Propene

c) A functional group is a substituent or moiety in a molecule that causes the molecule's characteristic chemical reactions.
Example:
Alcohol -OH
Carboxylic Acid -COOH
Chlorine -Cl
Fluorine -F
Iodine -I
Aldehyde -CHO
Cyanide -CN

Question:49

Name the reaction which is commonly used in the conversion of vegetable oils to fats. Explain the reaction involved in detail.
Answer:

Conversion of vegetable oil to fat is called as hydrogenation reaction.
Vegetable oils (unsaturated hydrocarbons) are treated with hydrogen and passed over nickel and palladium at 200 $^{\circ}C$ . The unsaturated compounds turns into saturated.
Example:
$H_{2}C=CH_{2}+H_{2}\overset{Ni}{\rightarrow}C_{2}H_{6}$
This process is used in industries to convert oils to vanaspati ghee as it is more stable and has a lot of shelf-life.

Question:50

(a) Write the formula and draw electron dot structure of carbon tetrachloride.
(b) What is saponification? Write the reaction involved in this process.
Answer:

a) Carbon tetrachloride
CCl₄

(b) When an ester is heated with sodium hydroxide solution, ester gets hydrolyzed (breaks down) to form the parent alcohol and sodium salt of carboxylic acid.
Example:
$CH_3COOCH_3 + NaOH \rightarrow CH_3COONa + CH_3OH$

Question:51

Esters are sweet-smelling substances and are used in making perfumes. Suggest some activity and the reaction involved for the preparation of an ester with well labeled diagram.
Answer:

Reaction Set up:
Take 1 mL ethanol (absolute alcohol) and 1 mL glacial acetic acid and few drops of concentrated sulphuric acid in a test tube.
Warm the test tube in a water-bath for at least five minutes.
Pour the contents of the test tube into a beaker containing 20-50 mL of water and smell the resulting mixture.
Sweet smell confirms presence of an ester.

Reaction mixture:
Ethanoic acid + Alcohol in presence of concentrated Sulphuric Acid
Reaction:
$CH_3COOH + CH_3CH_2OH \overset{H_{2}SO_4}{\rightarrow}CH_3COOC_2H_5 + H_2O$
Conformation Test for ester formation is sweet smell.

Question:52

A compound C (molecular formula, $C_2H_4O_2$ ) reacts with Na - metal to form a compound R and evolves a gas which burns with a pop sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, $C_3H_6O_2$ ). 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, S and write down the reactions involved.
Answer:

Since Compound C has two oxygen atoms it can be an ester or carboxylic acid.
Given that the gas evolved (on the reaction with sodium) burns with a pop sound in an indication that evolved gas is hydrogen.
Compound R is going to be a salt with sodium being the positive ion and the rest of the compound C which lost hydrogen must be negatively charged. This is only possible with carboxylic acid and not ester.
So C is carboxylic acid and based on number of carbons it is Ethanoic acid.
Compound C is $CH_3COOH.$
Compound R is $CH_3COONa.$ .
Compound A is $CH_3OH$ .
Compound S is $CH_3COOCH_3$
Reactions involved:
$2CH_3 COOH + 2Na \rightarrow 2CH_{3}COONa+H_{2}$
(C) (R)
$CH_3COOH + CH_{3}OH \rightarrow CH_3COOCH_{3}+H_2O?$
(C) (A) (R)
$CH_3COOOH +NaOH\rightarrow CH_{3}COONa?+H_{2}O$
(C) (R)
$CH_3COOCH_{3}+NaOH\rightarrow CH_{3}COONa +CH_{3}OH?$
(R) (A)

Question:53

Look at Figure 4.1 and answer the following questions
(a) What change would you observe in the calcium hydroxide solution taken in tube B?
(b) Write the reaction involved in test tubes A and B respectively.
(c) If ethanol is given instead of ethanoic acid, would you expect the same change?
(d) How can a solution of lime water be prepared in the laboratory?

Answer:

(a) Calcium hydroxide or Limewater $(Ca(OH)_2)$ taken in tube B turns milky when carbon dioxide passes through it as it forms Calcium carbonate $(CaCO_3)$
$Ca(OH)_2 (aq) + CO_2 (g) \rightarrow CaCO_3 (s) + H_2O (l)$
(b) Test Tube A
Sodium Carbonate reacts with ethanoic acid to form sodium ethanoate, carbon dioxide and water
$Na_2CO_3 + CH_3COOH \rightarrow CH_3COONa + CO_2 + H_2O$
Test Tube B
Calcium hydroxide turns milky white due to formation of Calcium carbonate (CaCO3) when Carbon dioxide is passed through it.
$Ca(OH)_2 (aq) + CO_2 (g) \rightarrow CaCO_3 (s) + H_2O (l)$
(c) No change will take place when we take ethanol instead of ethanoic acid.
$CH_3CH_2OH + Na_2CO_3 \rightarrow \text{ NO REACTION}$
(d) Take a small piece of Calcium oxide in a test tube. Add water to it, part of it gets dissolved but majority of it gets suspended. So pass the complete solution into a filter paper and collect it in a test tube and this solution is lime water.
$CaO + H_2O \rightarrow Ca(OH)_2$

Question:54

How would you bring about the following conversions? Name the process and write the reaction involved.
(a) ethanol to ethene.
(b) propanol to propanoic acid.
Write the reactions.
Answer:

(a) When Ethanol is reacted with concentrated Sulphuric acid in the presence of heat it will lead to removal of water molecule and forms ethene. This type of reaction is called dehydrogenation reaction.
$CH_3CH_2OH \overset{conc. H_{2}SO_{4}}{\rightarrow} CH_2=CH_2 + H_2O$
(b) When propanol is heated in the presence of potassium permanganate, it will convert to propanoic acid. This type of reaction is called oxidation reaction.
$CH_3CH_2CH_2OH \overset{Alk. KMnO_{4}}{\rightarrow} CH_3CH_2COOH$

Question:55

Draw the possible isomers of the compound with molecular formula $C_3H_6O$ and also give their electron dot structures.
Answer:

For $C_3H_6O$
1) Propanal

2) Acetone

3) Methoxyethane
$H_{2}C=CH-O-CH_{3}$
4) trans-1-propenol

5) cis-1-propenol

6) 2-propen-2-ol

7) 2-propen-1-ol

8) oxacyclobutane

9) cyclopropanol

10 and 11) Enantiomers of methyl oxirane:

(S)-Methyloxirane (R)-Methyloxirane

Question:56

Explain the given reactions with the examples
(a) Hydrogenation reaction
(b) Oxidation reaction
(c) Substitution reaction
(d) Saponification reaction
(e) Combustion reaction
Answer:

(a) Hydrogenation reaction is addition of hydrogen to unsaturated molecule of a hydrocarbon.
Oils are unsaturated compounds containing double bonds. Unsaturated compounds are the only one which can undergo addition reaction.
The reaction is known as hydrogenation reaction. For example:

Hydrogenation reactions are addition reactions

(b) Oxidation reaction: A reaction in which an organic compound is reacted with an oxidizing agent where hydrogen is removed or oxygen is added to the compound.
For example:
Alcohol $\rightarrow$ Aldehyde/Ketone $\rightarrow$ Carboxylic acid
$CH_{3}CH_{2}OH \overset{Alk. KMnO_{4}}{\rightarrow}CH_{3}COOH$

(c) Substitution reaction is when atom or group of atoms are replaced with other atom or group of atoms.
$CH_4 + Cl_2 \overset{hv}{\rightarrow} CH_3Cl + HCl$
(d) When an ester is heated with sodium hydroxide solution, ester gets hydrolyzed (breaks down) to form the parent alcohol and sodium salt of carboxylic acid.
Example:
$CH_3COOCH_3 + NaOH \overset{Alk. KMnO_{4}}{\rightarrow} CH_3COONa + CH_3OH$
(e) Combustion Reaction is where an organic compound reacts with oxygen to form carbon dioxide and water vapor.
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + Heat + Light$

Question:57

An organic compound A on heating with concentrated $H_2SO_4$ forms a compound B which on 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 $CO_2$ and 3 moles of $H_2O$ . Identify the compounds A, B and C and write the chemical equations of the reactions involved.
Answer:

As the compound on combustion forms two moles of carbon dioxide and 3 moles of water, it means that 2 carbon atoms and 6 hydrogen atoms are present. So molecular formula is C2H6 (ethane).
$C_2H_6 + 3.5 O_2 \rightarrow 2CO_2 + 3H_2O$
Compound A is ethanol (CH3CH2OH). On dehydrogenation, it gives ethene.
$C_2H_5OH \overset{Hot conc. H_{2}SO_{4}}{\rightarrow}C_{2}H_{4}+H_{2}O$
Compound B is ethene $(CH_2=CH_2)$
Compound is C is obtained by addition of 1 mole hydrogen in presence of nickel to B which means C is ethane
$C_{2}H_{4}+H_{2}\overset{Ni}{\rightarrow}C_{2}H_{6}$
Compound C is $CH_3-CH_3$
$C_{2}H_{6}+7O_{2}\rightarrow4CO_{2}+6H_{2}O+Heat+Light$

NCERT Exemplar Solutions Class 10 Science Chapter 4 Important Topics:

NCERT exemplar Class 10 Science solutions chapter 4 explores the following topics:

Hydrocarbons and their basic nomenclature.
NCERT exemplar Class 10 Science solutions chapter 4 discusses some other organic compounds and their interesting properties.
Different allotropic forms of carbon.

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NCERT Class 10 Exemplar Solutions for Other Subjects:

NCERT Class 10 Science Exemplar Solutions for Other Chapters:

Chapter 1 Chemical Reactions and Equations

Chapter 2 Acids, Bases and Salts

Chapter 3 Metals and Non-metals

Chapter 5 Periodic Classification of Elements

Chapter 6 Life Processes

Chapter 7 Control and Coordination

Chapter 8 How do Organisms Reproduce?

Chapter 9 Heredity and Evolution

Chapter 10 Light Reflection and Refraction

Chapter 11 Human Eye and Colourful World

Chapter 12 Electricity

Chapter 13 Magnetic Effects of Electric Current

Chapter 14 Sources of Energy

Chapter 15 Our Environment

Chapter 16 Management of Natural Resources

Features of NCERT Exemplar Class 10 Science Solutions Chapter 4:

These Class 10 Science NCERT exemplar chapter 4 solutions provide an understanding of carbon and its compounds. We will learn that mostly carbon compounds form a covalent bond. This can be treated as fundamental of organic chemistry which will be taught in detail in higher classes. The learning of this chapter will be useful if students perceive engineering or medical entrance examinations. The solutions of Carbon and its compounds-based practice problems will provide valuable insights along with a comprehensive approach. The Class 10 Science NCERT exemplar solutions chapter 4 Carbon and its compounds are sufficient to attempt and answer additional publications such as S.Chand by Lakhmir Singh and Manjit Kaur, Chemistry question bank, NCERT Class 10 Science et cetera.

Students can refer to the NCERT exemplar Class 10 Science solutions chapter 4 pdf download in order to resolve the difficulties met while venturing the NCERT exemplar Class 10 Science chapter 4.

Check NCERT Solutions for questions given in the book

Chapter No.	Chapter Name
Chapter 1	Chemical Reactions and Equations
Chapter 2	Acids, Bases, and Salts
Chapter 3	Metals and Non-metals
Chapter 4	Carbon and its Compounds
Chapter 5	Periodic Classification of Elements
Chapter 6	Life Processes
Chapter 7	Control and Coordination
Chapter 8	How do Organisms Reproduce?
Chapter 9	Heredity and Evolution
Chapter 10	Light Reflection and Refraction
Chapter 11	The Human Eye and The Colorful World
Chapter 12	Electricity
Chapter 13	Magnetic Effects of Electric Current
Chapter 14	Sources of Energy
Chapter 15	Our Environment
Chapter 16	Sustainable Management of Natural Resources

Must check NCERT Solution Subject Wise

Read more NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Q1. What is an organic compound?

A1. Generally, all carbon compounds are known as organic compounds

2. Q2. In how many forms carbon is found?

A2. Carbon in nature has three forms; amorphous carbon, graphite, and diamond

3. Q3. What is the percentage of carbon in any human body?

A3. Most of the human body contains protein, sugar, etc. These all are carbon compounds, and almost 18% of the mass is because of these carbon compounds.

4. Q4. How many questions should you expect from Carbon and its compounds in the board examination?

A4. Carbon and its compounds weigh around 6-7% marks of the final paper. The NCERT exemplar Class 10 Science solutions chapter 4 on Carbon and its compounds are sufficient to prepare for the board examination.

5. Q5. Can carbon conduct electricity?

A5. Graphite is a form of carbon and it is an exception among all non-metals which can conduct electricity as well as heat.

6. Q6. What is carbon dating?

A6. With the help of the radioactive form of carbon which is also known as radiocarbon, the age of any organic material is calculated. This method is called carbon dating

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Questions related to CBSE Class 10th

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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

Starting Point:
- Determine your starting point in Aligarh or the nearby area.
Use Google Maps:
- Open Google Maps on your phone or computer.
- Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
- Follow the navigation instructions provided by Google Maps.
By Local Transport:
- Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
- Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.
Landmarks to Look For:
- As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.
Ask for Directions:
- If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.
Final Destination:
- Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

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sai raja 01 September,2024

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024

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Central Board of Secondary Education Class 10th Examination

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 10 Science Solutions Chapter 4 Carbon And Its Compounds

NCERT Exemplar Class 10 Science Solutions Chapter 4-MCQ

NCERT Exemplar Class 10 Science Solutions Chapter 4-Short Answer

NCERT Exemplar Class 10 Science Solutions Chapter 4-Long Answer

NCERT Exemplar Solutions Class 10 Science Chapter 4 Important Topics:

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