2<\/sub> ?
\nAnswer:
\n<\/p>\nQuestion 2.
\nWhat 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)
\nAnswer:
\n<\/p>\n
In-text Questions (Page 68-69)<\/span><\/p>\nQuestion 1.
\nHow many structural isomers can you draw for pentane ?
\nAnswer:
\nThere chain isomers or structural isomers of pentane are possible namely n-pentane, 2-methyl butane and 2, 2-Dimethyl propane.
\n<\/p>\n
Question 2.
\nWhat are two properties of carbon which lead to the huge number of carbon compounds we see around us ?
\nAnswer:
\nThe two main properties of carbon are :
\n(i) Tetravalency : Carbon has atomic number six. Its electronic configuration, 2, 4.
\nCarbon 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<\/sup> or C-4<\/sup> 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.<\/p>\n(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.<\/p>\n
Question 3.
\nWhat will be the formula and electron do not structure of cyclopentane ?
\nAnswer:
\nFormula of cyclopentane : C5<\/sub>H10<\/sub>
\n<\/p>\nQuestion 4.
\nDraw the structure for the following compounds.
\n(i) Ethanoic acid
\n(ii) Bromo pentane
\n(iii) Butanone
\n(iv) Hexanol
\nAre structural isomers possible for bromopentane.
\nAnswer:
\n(i) Ethanoic acid : Molecular Formula, C2<\/sub>H5<\/sub>OH
\n<\/p>\n(ii) Bromo pentane : Molecular Formula, C5<\/sub>H11<\/sub>-Br
\n<\/p>\n(iii) Butanone, Molecular Formula, C4<\/sub>H8<\/sub>O
\n<\/p>\n(iv) Hexanal: Molecular Formula: C6<\/sub>H12<\/sub>O
\nStructure :
\n
\nYes, structural isomers of Bromo pentane are possible. Some of them are as follows :
\n<\/p>\n<\/p>\n
Question 5.
\nHow would you name the following compounds ?
\n
\nAnswer:
\n(i) CH3<\/sub> – CH2<\/sub> – Br
\n1 – Bromo ethane
\n<\/p>\nIn-text Questions (Page 71)<\/span><\/p>\nQuestion 1.
\nWhy is the conversion of ethanol to ethanoic acid an oxidation reaction ?
\nAnswer:
\nA 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.
\n<\/p>\n
Question 2.
\nA mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used ?
\nAnswer:
\nA high temperature is required for welding. Ethyne burn in oxygen and produce a large amount of heat and H2<\/sub>O and CO2<\/sub> as by product. But if we used a mixture of ethyne and air it produce oxides of nitrogen with H2<\/sub>O and CO2<\/sub>. 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.<\/p>\nIn-text Questions (Page 74)<\/span><\/p>\nQuestion 1.
\nHow would you distinguish experimentally between an alcohol and a carboxylic add ?
\nAnswer:
\nCarboxylic 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.<\/p>\n
Question 2.
\nWhat are oxidising agent.
\nAnswer;
\nA substance which provides oxygen for oxidation and reduce itself in the reaction is called an oxidising agent e.g. Alkaline KMnO4<\/sub>.<\/p>\n<\/p>\n
In-text Questions (Page 76)<\/span><\/p>\nQuestion 1.
\nWould you be able to check if water is hard by using a detergent?
\nAnswer:
\nNo, because detergents do not form any precipitate with hard water and form equal amount of foam with hard and soft water.<\/p>\n
Question 2.
\nPeople use a variety of methods to wash clothes, usually after adding the soap, they \u2018beat\u2019 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?
\nAnswer:
\nSoaps 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.<\/p>\n
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.<\/p>\n
Class 10 Science Chapter 4 Carbon and Its Compounds Textbook Questions and Answers<\/h3>\n
Page No. 77<\/span><\/p>\nQuestion 1.
\nEthane with the molecular formula – C2<\/sub>H6<\/sub> – has
\n(a) 6 covalent bonds
\n(b) 7 covalent bonds
\n(c) 8 covalent bonds
\n(d) 9 covalent bonds
\nAnswer:
\n(b) 7 covalent bonds<\/p>\nQuestion 2.
\nButanone is a four-carbon compound with the functional group
\n(a) Carboxylic acid
\n(b) aldehyde
\n(c) ketone
\n(d) alcohol
\nAnswer:
\n(c) Ketone<\/p>\n
Question 3.
\nWhile cooking, if the bottom of the vessel is getting blackened on the outside, it means.
\n(a) the food is not cooked completely
\n(b) the fuel is not burning completely.
\n(c) the fuel is wet
\n(d) the fuel is burning completely.
\nAnswer:
\n(b) the fuel is not burning completely.<\/p>\n
Question 4.
\nExplain the nature of the covalent bond using the bond formation in CH3<\/sub>Cl.
\nAnswer:
\nCarbon 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. CH3<\/sub>Cl has three hydrogen atoms and each atom will form a covalent bond with carbon. The nature of ‘C-H’ bond is purely covalent. So CH3<\/sub>Cl 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 CH3<\/sub>Cl will be tetrahedral.<\/p>\n<\/p>\n
Question 5.
\nDraw electron dot structures for:
\n(a) Ethanoic acid
\n(b) H2<\/sub>S
\n(c) Propanone
\n(d) F2<\/sub>
\nAnswer:
\n(a) Ethanoic acid M.F = CH3<\/sub>COOH
\n<\/p>\n(b) H2<\/sub>S
\n<\/p>\n(c) Propane (CH3<\/sub> – C – CH3<\/sub>)
\n<\/p>\n(d) F2<\/sub>
\n<\/p>\nQuestion 6.
\nWhich is a homologous series ? Explain with an example.
\nAnswer:
\nIt 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<\/sub>.<\/p>\nThe different members of the homologous series are called homologoues.
\nFor example homologous series of alcohols:
\nCH3<\/sub>-OH
\nCH3<\/sub>-CH2<\/sub>-OH
\nCH3<\/sub>-CH2<\/sub>-CH2<\/sub>-OH
\nCH3<\/sub>-CH2<\/sub>-CH2<\/sub>-CH2<\/sub>-OH
\nCH3<\/sub>-CH2<\/sub>-CH2<\/sub>-CH2<\/sub>-CH2<\/sub>-OH<\/p>\nIn 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<\/sub>. So the given series is a homologous series.<\/p>\nQuestion 7.
\nHow can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties ?
\nAnswer:<\/p>\n
\n\n\nEthanol Physical Properties<\/td>\n | Ethanoic acid Physical Properties<\/td>\n<\/tr>\n |
\n1. Ethanol is commonly known as alcohol and is the active ingredient of all alcoholic drinks.<\/td>\n | 1. Ethanoic acid is commonly known as acetic acid and belongs to a group of acids called carboxylic acid.<\/td>\n<\/tr>\n |
\n2. It has a intense smell of alcohols.<\/td>\n | 2. It has a intense small of vinegar.<\/td>\n<\/tr>\n |
\n3. It is mainly used for drinking.<\/td>\n | 3. It is mainly used as a preservative in pickels.<\/td>\n<\/tr>\n |
\n4. It does not freezes during winter.<\/td>\n | 4. It usually freezes during winter in cold climates.<\/td>\n<\/tr>\n |
\n5. Melting point: 156 K<\/td>\n | 5. Melting point: 290 K<\/td>\n<\/tr>\n |
\n6. Boiling point: 351 K<\/td>\n | 6. Boiling point: 391 K<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n <\/p>\n Question 8. \nWhy does micelle formation place when soap is added water ? Will a micelle be formed in other solvents such as ethanol also ? \nAnswer: \nSoaps 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.<\/p>\n <\/p>\n Question 9. \nWhy are carbon and its compounds used as fuels for most applications ? \nAnswer: \nCarbon, 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. \nC + O2<\/sub> \u2192 CO2<\/sub> + heat + light \nCH4<\/sub> + 2O2<\/sub> \u2192 CO2<\/sub> + 2H2<\/sub>O + heat + light \nCH2<\/sub>CH2<\/sub>OH + 3O2<\/sub> \u2192 2CO2<\/sub> + 3H2<\/sub>O + heat + light \nThere are the oxidation reactions. \nDue to high calorific value, carbon and its compounds are used as a fuel.<\/p>\nQuestion 10. \nExplain the formation of scum when hard water is treated with soap. \nAnswer: \nHardness 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+<\/sup> and Mg2+<\/sup>) present in hard water react with soap to form scum (precipitates) of calcium and magnesium salts of fatty acids. \n \nThus scum is formed when soap is treated with hard water.<\/p>\nQuestion 11. \nWhat change will you observe if you test soap with litmus paper (red and blue) ? \nAnswer: \nSoap is the Sodium of Potassium salt of long chain fatty acids e.g, Sodium stearate (C17<\/sub>H35<\/sub>COONa). When soaps are dissolved in water, they are dissociated in C17<\/sub>H35<\/sub>COO–<\/sup> and Na+<\/sup> ions. Water contains H+<\/sup> and OH–<\/sup> ions. Water contains H+<\/sup> and OH–<\/sup> ions. So C17<\/sub>H35<\/sub>COO combine with H+<\/sup> to form C17<\/sub>H35<\/sub>COOH and Na+<\/sup> ions combine with OH–<\/sup> ions to form NaOH. So soap changes blue litmus to red due to acidic character and red litmus to blue due to basic character.<\/p>\nQuestion 12. \nWhat is hydrogenation ? What is its industrial application ? \nAnswer: \nUnsaturated 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. \ne.g., \n \nHydrogenation is used in the manufacture of vanaspati ghee from vegetable oils. \n<\/p>\n Question 13. \nWhich of the following hydrocarbons undergo addition reaction; C2<\/sub>H6<\/sub>, C3<\/sub>H8<\/sub>, C3<\/sub>H6<\/sub>, C2<\/sub>H2<\/sub>, and CH4<\/sub>. \nAnswer: \nUnsaturated hydrocarbons undergo addition reactions. \nSo C3<\/sub>H6<\/sub> i.e. CH3<\/sub> – CH = CH2<\/sub> and \nC2<\/sub>H2<\/sub> i.e. CH = CH will give addition \nreations but C2<\/sub>H6<\/sub>, C3<\/sub>H8<\/sub> and CH4<\/sub> will not give addition reactions because they are saturated.<\/p>\n<\/p>\n Question 14. \nGive a test that can be used to differentiate chemically between butter and cooking oil. \nAnswer: \nWhen the butter is heated above 250\u00b0C they decompose with production of acrolein which has intense smell but oil does not give this test.<\/p>\n Question 15. \nExplain the mechanism of the cleaning actions of soaps. \nAnswer: \nSoaps 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.<\/p>\n 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.<\/p>\n Class 10 Science 4 Carbon and Its Compounds Textbook Activities<\/h3>\nActivity 4.1 (Page 58)<\/span><\/p>\n\n- Make a list of ten things you have used or consumed since the morning.<\/li>\n
- Compile this list with the lists made by your classmates and then sort the items into the following table.<\/li>\n
- If there are items which are made up of more than one material put them into bot the relevant columns.<\/li>\n<\/ul>\n
\n\n\nThings made of metal<\/td>\n | Things made of glass\/clay<\/td>\n | Others<\/td>\n<\/tr>\n | \nGlass<\/td>\n | Glass<\/td>\n | Milk<\/td>\n<\/tr>\n | \nBucket<\/td>\n | Cup<\/td>\n | Bread<\/td>\n<\/tr>\n | \nPlate<\/td>\n | Plate<\/td>\n | Butter Tea<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Activity 4.2 (Page 67)<\/span><\/p>\n\n- Calculate the difference in the formulae and molecular masses for (a) CH3<\/sub>OH and C2<\/sub>H5<\/sub>OH and C3<\/sub>H7<\/sub>OH and (c) C3<\/sub>H7<\/sub>OH and C4<\/sub>H9<\/sub>OH.<\/li>\n
- Is there any similarity in these three?<\/li>\n
- Arrange these alcohols in the order of increasing carbon atoms to get a family. Can we call this family a homologous series?<\/li>\n
- Generate the homologous series for compounds containing upto four carbons for the other functional groups given in table 4.3<\/li>\n<\/ul>\n
Answer: \n(a) CH3<\/sub>OH and C2<\/sub>H3<\/sub>OH \nDifference in the molecular formula = CH2<\/sub> \nThe molecular mass of CH3<\/sub>OH \n= C + 3 \u00d7 H + 1 \u00d7 O \u00d7 1 \u00d7 H \n= 12 + 3 \u00d7 1 + 16 + 1 \u00d7 1 \n= 12 + 3 + 16 + 1 = 32<\/p>\nThe molecular mass of C2<\/sub>H5<\/sub>OH \n= 2 \u00d7 C + 5 \u00d7 H + 1 \u00d7 O + 1 \u00d7 H \n= 2 \u00d7 12 + 5 \u00d7 1 + 1 \u00d7 16 + 1 \u00d7 1 \n= 24 + 5 + 16 + 1 = 46 \nThe difference in molecular mass = 46 – 32 = 14<\/p>\n(b) C2<\/sub>H5<\/sub>OH and C3<\/sub>H7<\/sub>OH \nDifference in the molecular formula = CH2<\/sub> \nThe molecular mass of C2<\/sub>H5<\/sub>OH \n= 2 \u00d7 C + 5 + H + 1 \u00d7 O + 1 \u00d7 H \n= 2 \u00d7 12 + 5 \u00d7 1 + 1 \u00d7 16 – 1 \u00d7 1 \n= 24 + 5 + 16 + 1 \n= 46 \nThe molecular mass of C3<\/sub>H7<\/sub>OH \n= 3 \u00d7 C + 7 \u00d7 H + 1 \u00d7 O + 1 \u00d7 H \n= 3 \u00d7 12 + 7 \u00d7 1 + 1 \u00d7 16 + 1 \u00d7 1 \n= 36 + 7 + 16 + 1 \n= 60 \nDifference in the molecular mass = 60 – 46 = 14<\/p>\n(c) C3<\/sub>H7<\/sub>OH and C4<\/sub>H9<\/sub>OH \nDifference in the molecular formula = CH2<\/sub> \nThe molecular mass of C3<\/sub>H7<\/sub>OH \n= 3 \u00d7 C + 7 \u00d7 H + 1 \u00d7 O + 1 \u00d7 H \n= 3 \u00d7 12 + 7 \u00d7 1 + 1 \u00d7 16 + 1 \u00d7 1 \n= 36 + 7 + 16 + 1 = 60 \nThe molecular mass of C4<\/sub>H9<\/sub>OH \n= 4 \u00d7 C + 9 \u00d7 H + 1 \u00d7 O + 1 \u00d7 H \n= 4 \u00d7 12 + 9 \u00d7 1 + 1 \u00d7 16 + 1 \u00d7 1 \n= 48 + 9 + 16 + 1 \n= 74 \nThe difference in molecular mass = 74 – 60 = 14<\/p>\n\n- Yes there is a similarity in the three organic compouds. They all have same functional group.<\/li>\n
- CH3<\/sub>OH<\/li>\n
- C2<\/sub>H5<\/sub>OH<\/li>\n
- C3<\/sub>H7<\/sub>OH<\/li>\n
- 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<\/sub>” and they have same functional group. (OH)<\/li>\n
- Homologous series of halo-alkane<\/li>\n<\/ul>\n
\nHomologous series of alcohol \nCH3<\/sub>-OH \nCH3<\/sub>-CH2<\/sub>-OH \nCH3<\/sub>-CH2<\/sub>-CH2<\/sub>-OH2<\/sub>-OH \nCH3<\/sub>-CH2<\/sub>-CH2<\/sub>-CH2<\/sub>-OH \nHomologous series of aldehyde \nH-CHO \nCH3<\/sub>-CHO \nCH3<\/sub>CH2<\/sub>-CHO \nCH3<\/sub>CH2<\/sub>CH2<\/sub>-CHO<\/p>\nHomologous series of Ketons: \n \nHomologous series of carboxylic acid : \nH-COOH \nCH3<\/sub>-COOH \nCH3<\/sub>-CH2<\/sub>-COOH \nCH3<\/sub>-CH2<\/sub>-CH2<\/sub>-COOH<\/p>\n<\/p>\n Activity 4.3 (Page 68)<\/span><\/p>\nCaution : This activity needs the teacher’s assitance.<\/p>\n \n- Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatual and burn them.<\/li>\n
- Observe the nature of the flame and note whether smoke is produced.<\/li>\n
- Place a metal plate above the flame. If there a deposition on the plate in case of any of the compounds ?<\/li>\n<\/ul>\n
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.<\/p>\n Activity 4.4 (Page 68)<\/span><\/p>\n\n- Light a bunsen burner and adjust the air hole at the base to get different types of flames\/presence of smoke.<\/li>\n
- When do you get a yellow, sootry flame ?<\/li>\n
- When do you yet a blue flame.<\/li>\n<\/ul>\n
Observation: During incomplete combustion a yellow sooty flame is observed and when a sufficient oxygen is supplied it gives a clean blue flame.<\/p>\n Activity 4.5 (Page 70)<\/span><\/p>\n\n- Take about 3 ml of ethanol in a test tube and warm it gently in a water bath.<\/li>\n
- Add a 5% solution of alkaline potassium permanganate drop by drop to this solution.<\/li>\n
- Does the colour of potassium permanganate persist when it is added initially ?<\/li>\n
- Why does the colour of potassium permanganate not disappear when excess is added ?<\/li>\n
- 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<\/sub>.<\/li>\n<\/ul>\n
Observation : Yes the colour of alkaline potassium permanganate persist initially but is disappears immediately.<\/p>\n Activity 4.6 (Page 72)<\/span><\/p>\nTeacher Demonstration:<\/p>\n \n- Drop a small piece of sodium, about the size of a couple of grains of rice, into ethanol (absolute alcohol).<\/li>\n
- What do you observe ?<\/li>\n
- How will you test the gas evolved ?<\/li>\n<\/ul>\n
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.<\/p>\n <\/p>\n Activity 4.7 (Page 73)<\/span><\/p>\n\n- Compare the pH of dilute acetic acid and dilute hydrochloric acid using both litmus paper and universal indicator.<\/li>\n
- Are both acids indicated by the litmus test ?<\/li>\n
- Does the universal indicator show them as equally strong acids?<\/li>\n<\/ul>\n
Observation:<\/p>\n
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