--- title: "Carbon and its Compounds: CBSE Class 10 Science previous year questions" url: https://www.swavid.com/cbse/class-10/science/pyq/carbon-and-its-compounds --- # Carbon and its Compounds: CBSE Class 10 Science previous year questions 17 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Science Question Paper 2026 (Set 31/1/1) with Solutions ### Question 20 *1 mark · MCQ* The formula of functional group for aldehyde is : - $-\text{COOH}$ - $-\text{CHO}$ - $-\text{C}=\text{O}$ - $-\text{OH}$ **Solution** 1. The functional group for aldehydes is represented by $-\text{CHO}$. 2. $-\text{COOH}$ is for carboxylic acids, $-\text{C}=\text{O}$ is for ketones, and $-\text{OH}$ is for alcohols. **Answer:** (b) $-\text{CHO}$ > Common mistake: Confusing the aldehyde group $-\text{CHO}$ with the ketone carbonyl group $-\text{C}=\text{O}$. ### Question 24 *1 mark · Assertion and reason* Assertion (A) : Carbon shares its valence electrons with other atoms of carbon or with atoms of other elements. Reason (R) : The shared electrons belong to the outermost shells of both the atoms and lead to both atoms attaining the noble gas configuration. - Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). - Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). - Assertion (A) is true, but Reason (R) is false. - Assertion (A) is false, but Reason (R) is true. **Solution** 1. Carbon has 4 valence electrons and achieves noble gas configuration by sharing electrons. 2. Both Assertion and Reason are true, and Reason correctly explains the Assertion. **Answer:** Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). > Common mistake: Thinking carbon forms ionic bonds by losing or gaining four electrons. ### Question 29 *5 marks · Long answer* (a) (i) Give reasons for the following : (I) Covalent compounds are poor conductor of electricity. (II) Soap does not form lather in hard water. (III) Carbon shows catenation but silicon does not. (ii) Write chemical equations for the following : (I) Oxidation of ethanol by acidified $\text{K}_2\text{Cr}_2\text{O}_7$. (II) Hydrogenation of ethene. **Part (a)(I)** 1. Covalent compounds do not form ions when dissolved in water. 2. They lack charged particles (ions or free electrons) responsible for conducting electricity. Answer (a)(I): They do not form ions in solution and lack free electrons. **Part (a)(II)** 1. Hard water contains calcium and magnesium salts. 2. Soap reacts with these calcium and magnesium ions to form an insoluble precipitate called scum. 3. This consumes the soap and prevents lather formation. Answer (a)(II): Soap reacts with calcium and magnesium ions in hard water to form insoluble scum. **Part (a)(III)** 1. Carbon has a very small size, which enables its nucleus to hold onto the shared pair of electrons strongly. 2. Silicon has a larger atomic size, making its $\text{Si}-\text{Si}$ bonds much weaker and less stable than $\text{C}-\text{C}$ bonds. Answer (a)(III): Carbon has a smaller size and strong inter-atomic $\text{C}-\text{C}$ bonds compared to silicon. **Part (b)(I)** 1. Ethanol is oxidized to ethanoic acid when treated with acidified potassium dichromate. 2. $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Acidified } \text{K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH}$ Answer (b)(I): $\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \xrightarrow{\text{Acidified } \text{K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH} + \text{H}_2\text{O}$ **Part (b)(II)** 1. Ethene undergoes addition reaction with hydrogen in the presence of nickel or palladium catalyst to form ethane. 2. $\text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni catalyst}} \text{CH}_3-\text{CH}_3$ Answer (b)(II): $\text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_3-\text{CH}_3$ **Answer:** Carbon compounds show characteristic bonding, soap action, and chemical reactions as explained in the parts. > Common mistake: Confusing catenation property between carbon and silicon due to atomic size differences. ### Question 29 (OR) *5 marks · Long answer* (b) Mohan heated ethanol with a compound 'X' in the presence of a few drops of conc. $\text{H}_2\text{SO}_4$ and observed a sweet smelling compound 'Y' is formed. When 'Y' is treated with sodium hydroxide it gives back ethanol and a compound 'Z'. (i) Identify 'X', 'Y' and 'Z'. (ii) Write the role of conc. $\text{H}_2\text{SO}_4$ in the reaction. (iii) Write the chemical equations involved and name the reactions. **Part (i)** 1. Compound 'X' is ethanoic acid ($\text{CH}_3\text{COOH}$), which reacts with ethanol to form a sweet-smelling ester. 2. Compound 'Y' is ethyl ethanoate ($\text{CH}_3\text{COOCH}_2\text{CH}_3$), an ester. 3. Compound 'Z' is sodium ethanoate ($\text{CH}_3\text{COONa}$), formed during alkaline hydrolysis of the ester. Answer (i): X: Ethanoic acid, Y: Ethyl ethanoate, Z: Sodium ethanoate. **Part (ii)** 1. Conc. $\text{H}_2\text{SO}_4$ acts as a dehydrating agent in the esterification reaction. 2. It removes the water molecule formed during the reaction, shifting the equilibrium forward to increase the yield of the ester. Answer (ii): It acts as a dehydrating agent to remove water and drive the reaction forward. **Part (iii)** 1. Esterification reaction: $\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Conc. } \text{H}_2\text{SO}_4} \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}$ 2. Saponification reaction: $\text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{CH}_3\text{CH}_2\text{OH}$ Answer (iii): Esterification and Saponification equations with proper names. **Answer:** X is ethanoic acid, Y is ethyl ethanoate, and Z is sodium ethanoate along with ethanol. > Common mistake: Forgetting to name the reactions (esterification and saponification) alongside writing the chemical equations. ## CBSE Class 10 Science Question Paper 2025 (Set 31/1/1) with Solutions ### Question 34 *5 marks · Long answer* (a) (i) Draw two isomeric structures of Butene ($C_4H_8$). (ii) Name the following compounds : (I) $CH_3-CH_2-CH_2-CH_2-Cl$ (II) $CH_3-CH_2-CH_2-CHO$ (iii) Write the chemical equations for the following reactions. Mention one essential condition each for these reactions to take place. (I) Ethanol undergoes complete oxidation (II) Propene undergoes hydrogenation (III) Ethanoic acid reacts with ethanol **Part (a) (i)** 1. Draw But-1-ene: $CH_3-CH_2-CH=CH_2$ 2. Draw But-2-ene: $CH_3-CH=CH-CH_3$ Answer (a) (i): But-1-ene and But-2-ene **Part (a) (ii)** 1. (I) $CH_3-CH_2-CH_2-CH_2-Cl$ is 1-chlorobutane. 2. (II) $CH_3-CH_2-CH_2-CHO$ is butanal. Answer (a) (ii): (I) 1-Chlorobutane, (II) Butanal **Part (a) (iii)** 1. (I) $CH_3CH_2OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$ (Condition: Heat/Combustion). 2. (II) $CH_3-CH=CH_2 + H_2 \xrightarrow{Ni/Pd catalyst} CH_3-CH_2-CH_3$ (Condition: Presence of nickel or palladium catalyst and heat). 3. (III) $CH_3COOH + CH_3CH_2OH \xrightarrow{Conc. H_2SO_4} CH_3COOCH_2CH_3 + H_2O$ (Condition: Presence of acid catalyst). Answer (a) (iii): Equations and conditions for oxidation, hydrogenation, and esterification written. **Answer:** Isomeric structures of butene, IUPAC names of given compounds, and chemical equations for oxidation, hydrogenation, and esterification. > Common mistake: Forgetting to write the reaction conditions such as catalysts or temperatures in chemical equations. ### Question 34 (OR) *5 marks · Long answer* (b) (i) A carbon compound $X$ is a good solvent. On reaction with sodium, $X$ forms two products $Y$ and $Z$. $Z$ is used to convert vegetable oil into vegetable ghee. Identify and name $X$, $Y$ and $Z$. Also write the equation of reaction of $X$ with sodium to justify your answer. (ii) Write chemical equation to show what happens when ethanol : (I) burns in oxygen/air. (II) is heated at 443 K in excess conc. $H_2SO_4$. (III) reacts with acidified potassium dichromate. **Part (b) (i)** 1. Compound X is ethanol ($CH_3CH_2OH$), which is a good solvent. 2. Ethanol reacts with sodium to form sodium ethoxide (Y) and hydrogen gas (Z). 3. Hydrogen ($Z$) is used in the hydrogenation of vegetable oils to vegetable ghee. 4. Equation: $2CH_3CH_2OH + 2Na \rightarrow 2CH_3CH_2ONa + H_2$ Answer (b) (i): X: Ethanol, Y: Sodium ethoxide, Z: Hydrogen gas **Part (b) (ii)** 1. (I) $CH_3CH_2OH + 3O_2 \rightarrow 2CO_2 + 3H_2O + \text{Heat and light}$ 2. (II) $CH_3CH_2OH \xrightarrow[443\text{ K}]{\text{Conc. } H_2SO_4} CH_2=CH_2 + H_2O$ 3. (III) $CH_3CH_2OH + \text{Acidified } K_2Cr_2O_7 + \text{Heat} \rightarrow CH_3COOH$ Answer (b) (ii): Chemical equations for burning, dehydration, and oxidation of ethanol. **Answer:** Identification of X as ethanol, Y as sodium ethoxide, Z as hydrogen gas, along with chemical equations. > Common mistake: Writing incorrect temperature or catalyst for the dehydration of ethanol. ## CBSE Class 10 Science Question Paper 2024 (Set 31/1/1) with Solutions ### Question 4 *1 mark · MCQ* Carbon compounds : (i) are good conductors of electricity. (ii) are bad conductors of electricity. (iii) have strong forces of attraction between their molecules. (iv) have weak forces of attraction between their molecules. The correct statements are : - (i) and (ii) - (ii) and (iii) - (ii) and (iv) - (i) and (iii) **Solution** 1. Carbon compounds form covalent bonds and do not contain ions, making them bad conductors of electricity. 2. Due to weak intermolecular forces between their molecules, they have low melting and boiling points. **Answer:** (c) (ii) and (iv) > Common mistake: Assuming carbon compounds conduct electricity because graphite is a good conductor. ### Question 34 *5 marks · Long answer* (i) Define a homologous series of carbon compounds. (ii) Why is the melting and boiling points of $\text{C}_4\text{H}_8$ higher than that of $\text{C}_3\text{H}_6$ or $\text{C}_2\text{H}_4$ ? (iii) Why do we NOT see any gradation in chemical properties of a homologous series compounds ? (iv) Write the name and structures of (i) aldehyde and (ii) ketone with molecular form $\text{C}_3\text{H}_6\text{O}$. **Part (i)** 1. A homologous series is a series of carbon compounds in which the same functional group substitutes for hydrogen in a carbon chain. 2. Successive members differ by a $-\text{CH}_2-$ unit and a molecular mass difference of $14\text{ u}$. Answer (i): A series of compounds with the same functional group and similar chemical properties, differing by a $-\text{CH}_2-$ group. **Part (ii)** 1. Melting and boiling points increase with increasing molecular mass in a homologous series. 2. As the molecular size and surface area of $\text{C}_4\text{H}_8$ is greater than $\text{C}_3\text{H}_6$ or $\text{C}_2\text{H}_4$, intermolecular forces of attraction are stronger. Answer (ii): Higher molecular mass leads to stronger intermolecular forces, requiring more energy to boil. **Part (iii)** 1. Chemical properties of carbon compounds in a homologous series are determined by the functional group. 2. Since all members of a series possess the same functional group, their chemical properties remain identical or show no gradation. Answer (iii): Chemical properties depend on the functional group, which is the same for all members of a homologous series. **Part (iv)** 1. The aldehyde with molecular formula $\text{C}_3\text{H}_6\text{O}$ is Propanal ($\text{CH}_3-\text{CH}_2-\text{CHO}$). 2. The ketone with molecular formula $\text{C}_3\text{H}_6\text{O}$ is Propanone ($\text{CH}_3-\text{CO}-\text{CH}_3$). Answer (iv): Propanal: $\text{CH}_3\text{CH}_2\text{CHO}$; Propanone: $\text{CH}_3\text{COCH}_3$. **Answer:** Homologous series defined, boiling point explanation, chemical properties uniformity, and structures for propanal and propanone given. > Common mistake: Confusing the structures of aldehyde and ketone having the same molecular formula. ### Question 34 (OR) *5 marks · Long answer* (B) (i) Write the name and structure of an organic compound 'X' having two carbon atoms in its molecule and its name is suffixed with '-ol'. (ii) What happens when 'X' is heated with excess concentrated sulphuric acid at $443\text{ K}$? Write chemical equation for the reaction stating the conditions for the reaction. Also state the role played by concentrated sulphuric acid in the reaction. (iii) Name and draw the electron dot structure of hydrocarbon produced in the above reaction. **Part (i)** 1. The organic compound having two carbon atoms with '-ol' suffix is ethanol. 2. Its molecular formula is $\text{C}_2\text{H}_5\text{OH}$ and structure is $\text{CH}_3-\text{CH}_2-\text{OH}$. Answer (i): Ethanol, $\text{CH}_3-\text{CH}_2-\text{OH}$. **Part (ii)** 1. When ethanol is heated with excess concentrated sulphuric acid at $443\text{ K}$, it undergoes dehydration to form ethene gas. 2. Chemical equation: $\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4, \, 443\text{ K}} \text{C}_2\text{H}_4 + \text{H}_2\text{O}$. 3. Concentrated sulphuric acid acts as a dehydrating agent which removes water molecules from ethanol. Answer (ii): Ethene is formed and concentrated $\text{H}_2\text{SO}_4$ acts as a dehydrating agent. **Part (iii)** 1. The hydrocarbon produced is ethene ($\text{C}_2\text{H}_4$). 2. Its electron dot structure shows a double covalent bond between the two carbon atoms and single bonds with four hydrogen atoms. Answer (iii): Ethene, with a double bond between carbon atoms. **Answer:** Compound X is ethanol, dehydration reaction produces ethene and acts as a dehydrating agent. > Common mistake: Forgetting to mention the temperature $443\text{ K}$ or the role of concentrated sulphuric acid. ## CBSE Class 10 Science Question Paper 2023 (Set 31/1/1) with Solutions ### Question 7 *1 mark · MCQ* Consider the structures of the three cyclic carbon compounds $\text{A}, \text{B}$ and $\text{C}$ given below and select the correct option from the following : [Structures of A, B, and C shown] - $\text{A}$ and $\text{C}$ are isomers of hexane and $\text{B}$ is benzene. - $\text{A}$ is an isomer of hexene, $\text{B}$ is benzene and $\text{C}$ is an isomer of hexene. - $\text{A}$ is a saturated cyclic hydrocarbon and $\text{B}$ and $\text{C}$ are unsaturated cyclic hydrocarbons. - $\text{A}$ is cyclohexane and $\text{B}$ and $\text{C}$ are the isomers of benzene. **Solution** 1. Compound A is cyclohexane ($\text{C}_6\text{H}_{12}$), which is a saturated cyclic hydrocarbon. 2. Compounds B (benzene, $\text{C}_6\text{H}_6$) and C are unsaturated cyclic hydrocarbons with alternating double bonds. **Answer:** (c) $\text{A}$ is a saturated cyclic hydrocarbon and $\text{B}$ and $\text{C}$ are unsaturated cyclic hydrocarbons. > Common mistake: Confusing the degree of saturation in ring structures. ### Question 34 *5 marks · Long answer* (a) A saturated organic compound 'A' belongs to the homologous series of alcohols. On heating 'A' with concentrated sulphuric acid at $443\text{ K}$, it forms an unsaturated compound 'B' with molecular mass $28\text{ u}$. The compound 'B' on addition of one mole of hydrogen in the presence of Nickel, changes to a saturated hydrocarbon 'C'. (i) Identify A, B and C. (ii) Write the chemical equations showing the conversion of A into B. (iii) What happens when compound C undergoes combustion? (iv) State one industrial application of hydrogenation reaction. (v) Name the products formed when compound A reacts with sodium. **Part (i)** 1. Compound B is an alkene with molecular mass $28\text{ u}$, which corresponds to ethene ($\text{C}_2\text{H}_4$). 2. Compound A is an alcohol that forms ethene on heating with concentrated $\text{H}_2\text{SO}_4$, so A is ethanol ($\text{C}_2\text{H}_5\text{OH}$). 3. Compound C is formed by the hydrogenation of B, so C is ethane ($\text{C}_2\text{H}_6$). Answer (i): A: Ethanol ($\text{C}_2\text{H}_5\text{OH}$), B: Ethene ($\text{C}_2\text{H}_4$), C: Ethane ($\text{C}_2\text{H}_6$). **Part (ii)** 1. Ethanol is heated with excess concentrated sulphuric acid at $443\text{ K}$. 2. $\text{CH}_3-\text{CH}_2-\text{OH} \xrightarrow{\text{Conc. }\text{H}_2\text{SO}_4, 443\text{ K}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$ Answer (ii): $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Conc. }\text{H}_2\text{SO}_4, 443\text{ K}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$ **Part (iii)** 1. Compound C is ethane, which is a saturated hydrocarbon. 2. It undergoes combustion in the presence of air to give carbon dioxide, water, heat, and light. 3. $2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} + \text{Heat and light}$ Answer (iii): It burns with a clean flame to produce carbon dioxide, water, heat, and light. **Part (iv)** 1. Hydrogenation is used in the food industry for the hardening of vegetable oils. 2. Vegetable oils are converted into vegetable ghee using nickel catalyst. Answer (iv): Hydrogenation of vegetable oils to form vegetable ghee (vanaspati fat). **Part (v)** 1. Ethanol reacts with sodium to form sodium ethoxide and hydrogen gas. 2. $2\text{CH}_3\text{CH}_2\text{OH} + 2\text{Na} \rightarrow 2\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2\uparrow$ Answer (v): Sodium ethoxide and hydrogen gas. **Answer:** Identified compounds A, B, C along with their chemical reactions, combustion behavior, industrial application, and reaction with sodium. > Common mistake: Confusing the molecular mass of ethene with ethane or writing incorrect balancing for combustion reactions. ### Question 34 (OR) *5 marks · Long answer* (b) (i) With the help of diagram, show the formation of micelles, when soap is applied on oily dirt. (ii) Take two test tubes X and Y with $10\text{ mL}$ of hard water in each. In test tube 'X', add few drops of soap solution and in test tube 'Y' add a few drops of detergent solution. Shake both the test tubes for the same period. (1) In which test tube the formation of foam will be more? Why? (2) In which test tube is a curdy solid formed? Why? **Part (i)** 1. Diagram: Draw a spherical aggregate of soap molecules (micelle). Show the hydrophobic hydrocarbon tail pointing towards the oily dirt at the centre and the hydrophilic ionic head ($-\text{COO}^-\text{Na}^+$) facing outward towards water. 2. When soap is dissolved in water, its hydrophobic tails attach to the oily dirt and the hydrophilic heads dissolve in water, forming a structure called a micelle. Answer (i): Micelle formation diagram showing hydrophobic tails towards oil and hydrophilic heads towards water. **Part (ii)** 1. 1. Foam formation will be more in test tube 'Y' containing detergent solution because detergents do not form insoluble precipitates (scum) with calcium and magnesium ions present in hard water. 2. 2. A curdy solid (scum) is formed in test tube 'X' containing soap solution because soap reacts with calcium and magnesium salts in hard water to form insoluble salts. Answer (ii): 1. More foam in test tube Y (detergent). 2. Curdy solid in test tube X (soap due to scum formation). **Answer:** Explained micelle formation with diagram and compared the action of soap and detergent in hard water. > Common mistake: Failing to label the hydrophilic head and hydrophobic tail correctly in the micelle diagram. ## CBSE Class 10 Science Question Paper 2022 (Set 31/1/1) with Solutions ### Question 1 *2 marks · Short answer* (a) Write the molecular formula of the following carbon compounds : (i) Methane (ii) Propane (b) Carbon compounds have low melting and boiling points. Why ? **Part (a)(i)** 1. Methane has one carbon atom, so its molecular formula is $\text{CH}_4$. Answer (a)(i): $\text{CH}_4$ **Part (a)(ii)** 1. Propane has three carbon atoms, so its molecular formula is $\text{C}_3\text{H}_8$. Answer (a)(ii): $\text{C}_3\text{H}_8$ **Part (b)** 1. Carbon compounds are generally covalent compounds formed by the sharing of electrons. 2. They have weak forces of attraction between their molecules, requiring very little energy to break. Answer (b): Weak intermolecular forces result in low melting and boiling points. **Answer:** Methane is $\text{CH}_4$, propane is $\text{C}_3\text{H}_8$, and carbon compounds have low melting and boiling points due to weak intermolecular forces. > Common mistake: Confusing molecular formulas of alkanes or attributing ionic bonding to carbon compounds. ### Question 2 *2 marks · Short answer* The electrons in the atoms of two elements X and Y are distributed in three shells having 1 and 7 electrons respectively in their outermost shells. (a) Write the group numbers of these elements in the Modern Periodic Table. (b) Write the molecular formula of the compound formed when X and Y combine with each other. (c) Which of the two is electropositive ? **Part (a)** 1. Element X has 1 valence electron and 3 shells, placing it in Period 3 and Group 1. 2. Element Y has 7 valence electrons and 3 shells, placing it in Period 3 and Group 17. Answer (a): Group 1 for X and Group 17 for Y **Part (b)** 1. Element X has a valency of 1 and element Y has a valency of 1. 2. Combining them gives the molecular formula XY. Answer (b): XY **Part (c)** 1. Element X is a metal with 1 valence electron which it tends to lose easily to form a positive ion. Answer (c): Element X **Answer:** Element X is in Group 1, element Y is in Group 17, the formula is XY, and X is electropositive. > Common mistake: Writing group numbers without considering the total number of valence electrons. ### Question 8 *3 marks · Short answer* (a) List two advantages of adopting the atomic number of an element as the basis of classification of elements in the Modern Periodic Table. (b) Write the electronic configurations of the elements X (atomic number 13) and Y (atomic number 20). **Part (a)** 1. Atomic number resolves the anomaly regarding the position of isotopes as they have the same atomic number. 2. It clearly explains the periodic recurrence of properties based on the number of valence electrons. Answer (a): Isotopes are placed at the same position, and properties repeat due to identical valence shell electronic configurations. **Part (b)** 1. For element X with atomic number 13, the electronic configuration is $2, 8, 3$. 2. For element Y with atomic number 20, the electronic configuration is $2, 8, 8, 2$. Answer (b): Element X: 2, 8, 3 and Element Y: 2, 8, 8, 2 **Answer:** (a) Atomic number removes anomalies like position of isotopes; properties repeat periodically. (b) X: 2, 8, 3; Y: 2, 8, 8, 2. > Common mistake: Writing incorrect electronic configuration by exceeding shell capacities. ### Question 9 *3 marks · Short answer* (a) Draw two different possible structures of a saturated hydrocarbon having four carbon atoms in its molecule. What are these two structures of the hydrocarbon having same molecular formula called ? Write the molecular formula and the common name of this compound. Also write the molecular formula of its alkyne. **Part (a)** 1. The two possible structures for $\text{C}_4\text{H}_{10}$ are straight-chain butane and branched-chain isobutane (2-methylpropane). 2. Compounds with the same molecular formula but different structures are called isomers. 3. The molecular formula is $\text{C}_4\text{H}_{10}$ and the common name is butane. 4. The molecular formula of the corresponding alkyne (butyne) is $\text{C}_4\text{H}_6$. Answer (a): Isomers: butane and isobutane; Molecular formula: $\text{C}_4\text{H}_{10}$; Common name: butane; Alkyne formula: $\text{C}_4\text{H}_6$ **Answer:** Structures: straight chain and branched chain (isomers). Molecular formula: $\text{C}_4\text{H}_{10}$, common name: butane. Alkyne formula: $\text{C}_4\text{H}_6$. > Common mistake: Confusing structural isomers with allotropes. ### Question 9 (OR) *3 marks · Short answer* (b) (i) Write the molecular formula of benzene and draw its structure. (ii) Write the number of single and double covalent bonds present in a molecule of benzene. (iii) Which compounds are called alkynes ? **Part (i)** 1. The molecular formula of benzene is $\text{C}_6\text{H}_6$. 2. Its structure is a six-carbon ring with alternating single and double bonds. Answer (i): Molecular formula: $\text{C}_6\text{H}_6$; ring structure with alternating double bonds. **Part (ii)** 1. A molecule of benzene contains 6 carbon-carbon bonds and 6 carbon-hydrogen bonds. 2. In total, there are 9 single covalent bonds and 3 double covalent bonds. Answer (ii): 9 single covalent bonds and 3 double covalent bonds. **Part (iii)** 1. Unsaturated hydrocarbons which contain at least one carbon-carbon triple bond are called alkynes. Answer (iii): Unsaturated hydrocarbons with at least one triple bond. **Answer:** (i) $\text{C}_6\text{H}_6$, ring structure with alternating double bonds. (ii) 9 single bonds and 3 double bonds. (iii) Unsaturated hydrocarbons with a triple bond. > Common mistake: Counting incorrect number of bonds in benzene ring. ## Related pages - [Carbon and its Compounds: NCERT solutions](https://www.swavid.com/science/class/10/chapter/carbon-and-its-compounds/ncert-solutions) - [All CBSE Class 10 Science papers](https://www.swavid.com/cbse/class-10/science/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student