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Carbon & Its Compounds - Practice Test, Class 10 Science - Class 10 MCQ


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25 Questions MCQ Test - Carbon & Its Compounds - Practice Test, Class 10 Science

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Carbon & Its Compounds - Practice Test, Class 10 Science - Question 1

Which of the following is an odd compound?

Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 1

Ethane is an alkane while rest all are alkene.

Carbon & Its Compounds - Practice Test, Class 10 Science - Question 2

Which one of the following is an unsaturated hydrocarbon?

Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 2
Unsaturated Hydrocarbon
An unsaturated hydrocarbon is a hydrocarbon molecule that contains at least one double or triple bond between carbon atoms. This means that it has the potential for further reactions and can undergo addition reactions to form new compounds.
Options:
A: Acetylene - C2H2
B: Butane - C4H10
C: Propane - C3H8
D: Decane - C10H22
Analysis:
To determine which one of the given options is an unsaturated hydrocarbon, we need to examine the molecular formulas and identify the presence of double or triple bonds.
- Acetylene (C2H2) has a triple bond between two carbon atoms, making it an unsaturated hydrocarbon.
- Butane (C4H10) does not have any double or triple bonds and consists of only single bonds between carbon atoms, making it a saturated hydrocarbon.
- Propane (C3H8) also does not have any double or triple bonds and consists of only single bonds between carbon atoms, making it a saturated hydrocarbon.
- Decane (C10H22) does not have any double or triple bonds and consists of only single bonds between carbon atoms, making it a saturated hydrocarbon.
Conclusion:
Based on the analysis, the answer to the question is option A: Acetylene. It is the only unsaturated hydrocarbon among the given options.
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Carbon & Its Compounds - Practice Test, Class 10 Science - Question 3

Two neighbours of homologous series differ by:

Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 3
Explanation:
To determine the difference between two neighbours of a homologous series, we need to understand the concept of homologous series and the role of the functional group in organic compounds.
A homologous series is a series of organic compounds that have the same functional group and a similar structure, with each successive member differing by a specific unit. The functional group is responsible for the characteristic properties and reactions of the compounds in the series.
In this case, we are given that the two neighbours differ by a specific unit. Let's analyze the options:
1. -CH: This represents the removal of one hydrogen atom from one compound to obtain the other. However, this would not result in a significant change in the structure or functional group of the compounds. Therefore, this option is not the correct answer.
2. -CH2: This represents the removal of two hydrogen atoms from one compound to obtain the other. The removal of two hydrogen atoms results in the formation of a double bond between two carbon atoms. This change in structure indicates a change in the functional group from a single bond to a double bond. This option is the correct answer.
3. -CH3: This represents the addition of a methyl group (CH3) to one compound to obtain the other. The addition of a methyl group does not result in a change in the functional group or the structure of the compounds. Therefore, this option is not the correct answer.
4. -CH4: This represents the addition of a methane group (CH4) to one compound to obtain the other. The addition of a methane group does not result in a change in the functional group or the structure of the compounds. Therefore, this option is not the correct answer.
In summary:
The correct answer is option B: -CH2. The two neighbours of the homologous series differ by the removal of two hydrogen atoms, resulting in the formation of a double bond between two carbon atoms.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 4

General formula of alkyne is

Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 4
General formula of alkyne is CnH2n-2
Explanation:
- Alkynes are a class of organic compounds that contain a carbon-carbon triple bond.
- The general formula for alkynes is CnH2n-2, where n represents the number of carbon atoms in the molecule.
- This formula can be derived by considering the bonding structure of alkynes.
- In an alkyne, there is one sigma bond formed by overlapping sp hybrid orbitals and two pi bonds formed by overlapping p orbitals.
- Each sigma bond involves two carbon atoms and each pi bond involves one carbon atom.
- Therefore, for a molecule with n carbon atoms, there will be n-1 sigma bonds and n-2 pi bonds.
- Since each carbon atom forms four single bonds in total, the remaining two bonds are formed by the triple bond, which consists of one sigma bond and two pi bonds.
- Hence, the general formula for alkynes is CnH2n-2.
- For example, ethyne (C2H2) is an alkyne with two carbon atoms, and its formula follows the general formula CnH2n-2: C2H2 = C(2)H(2x2-2) = C2H2.
Therefore, the correct answer is C: CnH2n-2.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 5

Which of the following represents alkynes?  

Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 5
Answer:
To identify which of the given options represents alkynes, we need to understand the structure and properties of alkynes. Alkynes are unsaturated hydrocarbons that contain a carbon-carbon triple bond. Let's examine each option:
A: -C-C-
This option represents a single bond between two carbon atoms, which is a characteristic of alkanes, not alkynes. Alkanes have only single bonds between carbon atoms.
B: C=C-
This option represents a double bond between two carbon atoms, which is a characteristic of alkenes, not alkynes. Alkenes have at least one double bond between carbon atoms.
C: -C≡C-
This option represents a triple bond between two carbon atoms, which is a characteristic of alkynes. Alkynes have at least one triple bond between carbon atoms.
D: none of these
This option is incorrect because option C (-C≡C-) represents alkynes.
Therefore, the correct answer is option C: -C≡C-, which represents alkynes.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 6
Which of the following represents ketones?
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 6
Ketones:
- Ketones are a type of organic compound that contain a carbonyl group (-C=O) bonded to two carbon atoms.
- They are characterized by the presence of a carbon double bonded to an oxygen atom (C=O) within their molecular structure.
- Ketones can be identified by their functional group, which consists of a carbon atom double bonded to an oxygen atom (C=O).
Options:
A: -C=O
- This option represents a carbonyl group, which is a characteristic feature of ketones. Therefore, option A represents ketones.
B: -OH
- This option represents a hydroxyl group, which is a characteristic feature of alcohols. It does not represent ketones.
C: -CHO
- This option represents an aldehyde functional group, which is a characteristic feature of aldehydes. It does not represent ketones.
D: -COOH
- This option represents a carboxylic acid functional group, which is a characteristic feature of carboxylic acids. It does not represent ketones.
Therefore, option A (-C=O) represents ketones, while options B, C, and D do not.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 7
Which of the following is not an aliphatic hydrocarbon?
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 7
Explanation:
Aliphatic hydrocarbons are organic compounds that consist of only carbon and hydrogen atoms arranged in open chains or branched chains. They do not contain any aromatic rings.
Given Options:
A: ethene
B: ethane
C: propyne
D: benzene
Analysis:
To determine which of the options is not an aliphatic hydrocarbon, we need to identify if any of them contain an aromatic ring (benzene ring).

D: benzene
- Benzene is a cyclic hydrocarbon with a six-membered carbon ring.
- It is an aromatic compound, meaning it contains a benzene ring.
- Therefore, benzene is not an aliphatic hydrocarbon.
Conclusion:
The option D: benzene is not an aliphatic hydrocarbon as it contains an aromatic ring.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 8

Complete combustion of a hydrocarbon gives

Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 8
Complete combustion of a hydrocarbon gives
To determine the products of complete combustion of a hydrocarbon, we need to understand the chemical reaction involved. During complete combustion, a hydrocarbon combines with oxygen to produce carbon dioxide and water.
The balanced chemical equation for the combustion of a hydrocarbon can be represented as:
Hydrocarbon + Oxygen → Carbon Dioxide + Water
Now, let's break down the options and see which one represents the correct products of complete combustion:
A: CO H2O
This option suggests that the products of complete combustion are carbon monoxide and water. However, carbon monoxide is not formed during complete combustion of a hydrocarbon.
B: CO2 H2O
This is the correct answer. In complete combustion, a hydrocarbon reacts with oxygen to produce carbon dioxide and water. Therefore, this option accurately represents the products of complete combustion.
C: CO H2
This option suggests that the products of complete combustion are carbon monoxide and hydrogen. However, carbon monoxide is not formed during complete combustion of a hydrocarbon.
D: CO2 H2
This option suggests that the products of complete combustion are carbon dioxide and hydrogen. While carbon dioxide is formed during complete combustion, hydrogen is not a product of this reaction.
Therefore, the correct answer is B: CO2 H2O.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 9
Which is NOT correct for isomers of a compound? 
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 9
Explanation:
Isomers: Isomers are compounds that have the same molecular formula but different structural arrangements.
A: They differ in physical properties: Physical properties of a compound are determined by its structure and arrangement of atoms. Isomers have different structural arrangements, so they can have different physical properties such as boiling point, melting point, density, solubility, etc.
B: They differ in chemical properties: Chemical properties of a compound are determined by its chemical structure and functional groups. Isomers can have different functional groups or different arrangements of atoms, leading to different chemical reactivity and properties.
C: They have the same molecular formula: Isomers have the same molecular formula, which means they have the same number and types of atoms.
D: They have the same structural formula: This statement is NOT correct. Isomers have different structural formulas, as the term "isomer" itself implies different structures. If two compounds have the same structural formula, they are not isomers but the same compound.
Therefore, the correct answer is D: They have the same structural formula.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 10
Buckminsterfullerene is an example of _______ of carbon.
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 10
Answer:
Introduction:
Buckminsterfullerene is a unique form of carbon that has a cage-like structure composed of 60 carbon atoms arranged in a hollow sphere. It was discovered in 1985 and is named after the architect and inventor Buckminster Fuller.
Allotrope of Carbon:
Buckminsterfullerene is an example of an allotrope of carbon. Allotropes are different structural forms of the same element. In the case of carbon, it can exist in several allotropes, each with different bonding arrangements and physical properties. Some well-known carbon allotropes include diamond, graphite, and amorphous carbon.
Unique Structure:
Buckminsterfullerene has a unique structure that distinguishes it from other carbon allotropes. Its carbon atoms form a series of hexagonal and pentagonal rings, resembling the pattern on a soccer ball. This structure is often referred to as a "buckyball."
Properties and Applications:
Buckminsterfullerene exhibits remarkable properties, such as high tensile strength, good electrical conductivity, and the ability to encapsulate other molecules within its structure. These properties have led to various potential applications, including:
1. Drug Delivery Systems: Buckminsterfullerene can encapsulate drug molecules and deliver them to specific targets in the body.
2. Photovoltaic Devices: The unique electronic properties of buckminsterfullerene make it a promising material for use in solar cells.
3. Catalysts: Buckminsterfullerene can act as a catalyst in certain chemical reactions due to its high surface area and unique reactivity.
4. Superconductors: Doped buckminsterfullerene derivatives have shown superconducting properties at low temperatures.
Conclusion:
In summary, buckminsterfullerene is an example of an allotrope of carbon. Its unique structure and properties have opened up new possibilities in various fields, making it an exciting area of research and development.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 11

Who prepared urea the first time by heating ammonium cynate?

Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 11
Who prepared urea the first time by heating ammonium cynate?
The correct answer is Wohler.
Explanation:
To provide a detailed solution, we can break down the information into several key points:
1. Friedrich Wohler, a German chemist, is credited with preparing urea for the first time by heating ammonium cyanate in 1828.
2. Prior to Wohler's experiment, it was believed that organic compounds could only be synthesized by living organisms, known as vitalism. Wohler's discovery challenged this notion.
3. In his experiment, Wohler combined silver cyanate and ammonium chloride, which resulted in the formation of a white crystalline substance - urea.
4. This experiment demonstrated that organic compounds could be produced from inorganic materials, leading to the downfall of vitalism and the development of the field of organic chemistry.
5. Wohler's discovery of the synthesis of urea is considered a landmark achievement in the history of chemistry as it marked the beginning of the chemical synthesis of organic compounds.
In summary, Friedrich Wohler prepared urea for the first time by heating ammonium cyanate, challenging the concept of vitalism and paving the way for the synthesis of organic compounds.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 12
Butanone is a four-carbon compound with the functional group?
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 12
Butanone: Functional Group Identification
Explanation:
Butanone is a four-carbon compound with the formula C4H8O. To determine the functional group present in butanone, we need to analyze its chemical structure and properties.
Ketones:
- Ketones are organic compounds that contain a carbonyl group (C=O) bonded to two alkyl or aryl groups.
- The general formula for ketones is R-CO-R', where R and R' can be alkyl or aryl groups.
- Ketones are characterized by the presence of the carbonyl group, which is a carbon-oxygen double bond.
- The carbonyl group in ketones is located in the middle of the carbon chain.
Identification of Butanone:
- Butanone has the chemical structure CH3-CO-CH2-CH3, with the carbonyl group (C=O) located in the middle of the carbon chain.
- The presence of the carbonyl group in butanone indicates that it is a ketone.
- Therefore, the correct answer is option C: ketone.
Summary:
- Butanone is a four-carbon compound with the formula C4H8O.
- The presence of the carbonyl group (C=O) in the middle of the carbon chain identifies butanone as a ketone.
- The functional group of butanone is a ketone (option C).
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 13
Major constituent of LPG is _________. 
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 14
The gas used in welding and cutting metals is:
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 14
Gas used in welding and cutting metals:
The gas used in welding and cutting metals is ethyne, also known as acetylene. Ethyne is a hydrocarbon gas that is highly flammable and used as a fuel in various industrial processes, including welding and cutting metals.
Properties of ethyne:
- Chemical formula: C2H2
- It is a colorless gas with a distinct odor.
- It is lighter than air and highly flammable.
- Ethyne has a high flame temperature, making it suitable for welding and cutting metals.
- It is produced by the reaction of calcium carbide with water.
Role of ethyne in welding:
- Ethyne is commonly used in oxy-fuel welding and cutting processes.
- When mixed with oxygen, it produces a high-temperature flame that can melt and join metal surfaces.
- The heat generated by the ethyne-oxygen flame allows for the formation of a molten pool, which can be used to weld or cut metals.
Advantages of using ethyne:
- Ethyne has a high heat content, providing intense heat for efficient metalworking.
- It has a relatively low cost compared to other gases used in welding and cutting.
- Ethyne can be easily compressed and stored in cylinders for easy transportation and use.
Disadvantages of using ethyne:
- Ethyne is highly flammable and requires careful handling and storage.
- It has a narrow range of flammability and can form explosive mixtures with air.
- The production of ethyne involves the use of calcium carbide, which can be hazardous if not handled properly.
In conclusion, ethyne (acetylene) is the gas commonly used in welding and cutting metals due to its high flame temperature and ability to generate intense heat. It is important to handle and store ethyne with caution to ensure safety in industrial processes.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 15
Which of the following is not a saturated hydrocarbon?
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 15
Not a Saturated Hydrocarbon
Definition: A saturated hydrocarbon is a hydrocarbon in which all carbon atoms are connected by single bonds and are saturated with hydrogen atoms.
Options:
A: cyclohexane
B: benzene
C: butane
D: isobutene
Explanation:
To determine which of the options is not a saturated hydrocarbon, we need to identify if any of them contain double or triple bonds between carbon atoms.
- Cyclohexane: Cyclohexane is a saturated hydrocarbon as it contains only single bonds between carbon atoms. It is composed of six carbon atoms in a ring structure, with each carbon atom bonded to two hydrogen atoms.
- Benzene: Benzene is not a saturated hydrocarbon. It is a cyclic compound consisting of a ring of six carbon atoms bonded together with alternating single and double bonds. The presence of double bonds makes benzene an unsaturated hydrocarbon.
- Butane: Butane is a saturated hydrocarbon. It is a straight-chain alkane with four carbon atoms and ten hydrogen atoms. Each carbon atom is bonded to four other atoms, including three carbon atoms and one hydrogen atom.
- Isobutene: Isobutene is not a saturated hydrocarbon. It is an unsaturated hydrocarbon as it contains a double bond. Isobutene is an isomer of butene and has the chemical formula C4H8.
Conclusion:
Among the given options, benzene is not a saturated hydrocarbon as it contains double bonds. Therefore, the correct answer is B: benzene.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 16
The bond between two identical non metallic atom has a pair of electron? 
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 16
The bond between two identical non-metallic atoms has a pair of electrons:
Detailed Explanation:
1. Covalent Bond:
- The bond between two identical non-metallic atoms is a covalent bond.
- In a covalent bond, atoms share electrons to achieve stability.
2. Electron Pair:
- The bond between two identical non-metallic atoms involves a pair of electrons.
- This means that both atoms contribute one electron to the bond, resulting in a shared pair of electrons.
3. Equally Shared:
- In the case of identical non-metallic atoms, the pair of electrons is equally shared between the atoms.
- This means that both atoms have equal ownership of the shared pair of electrons.
4. Electronegativity:
- Electronegativity is the ability of an atom to attract electrons towards itself in a chemical bond.
- In the case of identical non-metallic atoms, their electronegativity values are typically similar.
- When electronegativity values are similar, the electrons in the bond are shared equally.
5. Lewis Structure:
- The Lewis structure of a molecule can be used to represent the sharing of electrons in a covalent bond.
- In the Lewis structure, a pair of electrons is represented by a single line (-) between the atoms.
Conclusion:
- The bond between two identical non-metallic atoms has a pair of electrons that are equally shared between the atoms.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 17
Covalent compounds are generally ?
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 17
Covalent compounds are generally insoluble in water.
Covalent compounds are formed when two or more nonmetals share electrons to achieve a stable electron configuration. These compounds usually consist of molecules held together by covalent bonds, which are strong bonds resulting from the sharing of electrons.
Reasons why covalent compounds are generally insoluble in water:
1. Lack of charged particles: Covalent compounds do not ionize in water because they do not contain ions. As a result, they do not dissolve easily in water.
2. Polar vs. nonpolar nature: Covalent compounds can be classified as either polar or nonpolar, depending on the electronegativity difference between the atoms involved in the bond.
- Nonpolar covalent compounds have a symmetrical distribution of charge and do not readily dissolve in water, which is a polar solvent.
- Polar covalent compounds have an unequal sharing of electrons, resulting in partial positive and negative charges. While some polar covalent compounds may dissolve in water to a limited extent, most do not dissolve to a significant extent due to the strength of their covalent bonds.
3. Hydrophobic nature: Covalent compounds that are nonpolar or have hydrophobic functional groups tend to be insoluble in water. These compounds are repelled by water molecules and do not readily mix or dissolve in water.
Overall, the lack of charged particles, the polarity of the compound, and the hydrophobic nature of covalent compounds contribute to their general insolubility in water.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 18
Propane with the molecular formula C3H8 has:
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 18

To determine the number of covalent bonds in propane (C3H8), we need to consider the valence electrons of each atom and how they form bonds.
1. Determine the number of valence electrons:
- Carbon (C) has 4 valence electrons.
- Hydrogen (H) has 1 valence electron.
2. Calculate the total number of valence electrons in propane:
- Carbon (C) has 4 valence electrons x 3 atoms = 12 valence electrons.
- Hydrogen (H) has 1 valence electron x 8 atoms = 8 valence electrons.
3. Determine the octet rule for each atom:
- Carbon (C) wants to gain 4 electrons to achieve a stable octet.
- Hydrogen (H) wants to gain 1 electron to achieve a stable duet.
4. Form covalent bonds between carbon and hydrogen atoms:
- Carbon can share its 4 valence electrons with four hydrogen atoms to form four covalent bonds.
- Hydrogen can share its 1 valence electron with one carbon atom to form one covalent bond.
5. Calculate the total number of covalent bonds in propane:
- Carbon forms 4 covalent bonds.
- Hydrogen forms 8 covalent bonds.
Therefore, propane (C3H8) has a total of 4 + 8 = 12 covalent bonds.
Since none of the given answer options is 12, it seems that there might be an error in the question or answer choices.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 19
A hydrocarbon reacts with ammonical cuprous chloride solution to form a red precipitate .The hydrocarbon is :
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 19
Hydrocarbon Reacting with Ammonical Cuprous Chloride Solution
The hydrocarbon in question reacts with ammonical cuprous chloride solution to form a red precipitate. In order to determine the hydrocarbon, we need to consider the properties and reactions of each of the given options.
A: Ethane
- Ethane is a saturated hydrocarbon with the chemical formula C2H6.
- It does not have any double or triple bonds between carbon atoms.
- Ethane does not react with ammonical cuprous chloride solution to form a red precipitate.
B: Ethene
- Ethene is an unsaturated hydrocarbon with the chemical formula C2H4.
- It contains a double bond between the two carbon atoms.
- Ethene does not react with ammonical cuprous chloride solution to form a red precipitate.
C: Butane
- Butane is a saturated hydrocarbon with the chemical formula C4H10.
- It does not have any double or triple bonds between carbon atoms.
- Butane does not react with ammonical cuprous chloride solution to form a red precipitate.
D: 1-Propyne
- 1-Propyne is an unsaturated hydrocarbon with the chemical formula C3H4.
- It contains a triple bond between the carbon atoms.
- 1-Propyne reacts with ammonical cuprous chloride solution to form a red precipitate.
Therefore, based on the given information and the reactions observed, the hydrocarbon in question is most likely 1-Propyne (Option D).
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 20
Which of the following substance is added to denature Ethanol ? 
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 20
Denaturing Ethanol:
Ethanol is denatured by adding certain substances to make it unfit for human consumption. Denaturing ethanol involves adding toxic or unpleasant-tasting substances to discourage its consumption.
Substances Used to Denature Ethanol:
The following substances can be added to denature ethanol:
1. Methanol: Methanol is highly toxic and can cause blindness or even death if ingested. Adding methanol to ethanol makes it denatured and unfit for consumption.
2. Pyridine: Pyridine is a highly toxic liquid with a strong, unpleasant odor. It is often used as a denaturant to make ethanol undrinkable.
3. Copper Sulphate: Copper sulphate is a compound that is toxic and can cause various health issues if consumed. It can be added to ethanol to denature it.
Correct Answer:
The correct answer is D. All of them. All three substances - methanol, pyridine, and copper sulphate - can be added to denature ethanol.
Reasoning:
- Methanol is added to ethanol to make it unfit for consumption due to its high toxicity.
- Pyridine is used as a denaturant because of its strong odor and toxicity.
- Copper sulphate is a toxic compound that can be added to ethanol to make it unsafe for consumption.
By adding any of these substances individually or in combination, ethanol can be effectively denatured.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 21
Which of the following is not an allotropic form of carbon ?
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 21
Introduction:
In the given options, the allotropic forms of carbon are fluorine, fullerene, diamond, and graphite. We need to identify the option that is not an allotropic form of carbon.

The correct answer is option A: fluorine. Here's why:
Allotropic Forms of Carbon:
1. Fullerene: Also known as buckminsterfullerene or C60, it is a molecule composed entirely of carbon atoms arranged in a hollow sphere or tube. Fullerene is a unique form of carbon and has various applications in nanotechnology and materials science.
2. Diamond: Diamond is a transparent crystal made entirely of carbon atoms arranged in a rigid lattice structure. It is one of the hardest known substances and has high thermal conductivity. Diamonds are used in jewelry and industrial applications.
3. Graphite: Graphite is a soft, black, and slippery material composed of carbon atoms arranged in layers. It has a layered structure, making it useful as a lubricant and in pencil leads. Graphite can conduct electricity due to its delocalized electrons.
Fluorine:
Fluorine is not an allotropic form of carbon. It is a highly reactive non-metallic element belonging to the halogen family. Fluorine exists as diatomic molecules (F2) and is a pale yellow gas at room temperature. It is widely used in various industries, including the production of fluorocarbons.
Conclusion:
Among the given options, fluorine is the only one that is not an allotropic form of carbon. Fullerene, diamond, and graphite are different forms of carbon with unique properties and structures.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 22
Which of the following represents the correct decreasing order of hydrogen atoms ? 
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 22

To determine the correct decreasing order of hydrogen atoms, we need to understand the concept of alkanes, alkenes, and alkynes and their respective structures.
Alkanes:
- Alkanes are hydrocarbons that contain only single bonds between carbon atoms.
- They have the general formula CnH2n+2, where n is the number of carbon atoms in the molecule.
- Alkanes have the maximum number of hydrogen atoms possible for a given number of carbon atoms.
Alkenes:
- Alkenes are hydrocarbons that contain at least one carbon-carbon double bond.
- They have the general formula CnH2n, where n is the number of carbon atoms in the molecule.
- Alkenes have fewer hydrogen atoms compared to alkanes with the same number of carbon atoms.
Alkynes:
- Alkynes are hydrocarbons that contain at least one carbon-carbon triple bond.
- They have the general formula CnH2n-2, where n is the number of carbon atoms in the molecule.
- Alkynes have even fewer hydrogen atoms compared to alkenes and alkanes with the same number of carbon atoms.
Now, let's compare the number of hydrogen atoms in each type of hydrocarbon:
- Alkanes have the maximum number of hydrogen atoms, followed by alkenes and then alkynes.
- Therefore, the correct decreasing order of hydrogen atoms is A: alkanes, alkenes, alkynes.
Summary:
The correct decreasing order of hydrogen atoms is A: alkanes, alkenes, alkynes. Alkanes have the maximum number of hydrogen atoms, followed by alkenes and then alkynes.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 23
Detergents are sodium or potassium salts of long chain of :
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 23
Detergents are sodium or potassium salts of long chain of sulphonic acid.
Explanation:
Detergents are chemical compounds that are used for cleaning and washing purposes. They are typically made up of molecules with a hydrophilic (water-loving) head and a hydrophobic (water-repelling) tail. The hydrophilic head allows the detergent to dissolve in water, while the hydrophobic tail helps in removing dirt and grease from surfaces.
- Sulphonic acid: Detergents are usually made from the sulphonic acid, which is a type of organic acid. Sulphonic acid has a long chain of carbon atoms with a sulfonic acid group (SO3H) attached to it. This long carbon chain provides the hydrophobic tail of the detergent molecule.
- Sodium or potassium salts: In order to make the detergent water-soluble, the sulphonic acid is neutralized with either sodium or potassium hydroxide. This results in the formation of sodium or potassium salts of the sulphonic acid, which are more soluble in water.
So, detergents are sodium or potassium salts of long chain of sulphonic acid. These salts have both hydrophilic and hydrophobic properties, making them effective in removing dirt and stains from various surfaces.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 24

Which of the following represents the structure of N2 molecule ?

Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 24
Answer:
The correct structure of the N2 molecule is represented by option A: N ≡ N.
Explanation:
In the Lewis structure notation, the symbol ≡ denotes a triple bond between two atoms.
Here's a breakdown of each option:
A: N ≡ N
- This structure represents a triple bond between two nitrogen atoms (N≡N).
- This notation indicates that the two nitrogen atoms are sharing three pairs of electrons.
B: N = N
- This structure represents a double bond between two nitrogen atoms (N=N).
- However, N2 molecule does not exist with a double bond between nitrogen atoms. Nitrogen gas occurs as a triple bond (N≡N) due to its electronic configuration.
C: N - N
- This structure represents a single bond between two nitrogen atoms (N-N).
- However, N2 molecule does not exist with a single bond between nitrogen atoms. Nitrogen gas occurs as a triple bond (N≡N) due to its electronic configuration.
D: None of the above
- This option is incorrect because option A (N ≡ N) is the correct representation of the N2 molecule.
Therefore, the correct structure of the N2 molecule is represented by option A: N ≡ N.
Carbon & Its Compounds - Practice Test, Class 10 Science - Question 25
In double covalent bond there is sharing of :
Detailed Solution for Carbon & Its Compounds - Practice Test, Class 10 Science - Question 25
Double Covalent Bond:
In a double covalent bond, two pairs of electrons are shared between two atoms. This type of bond occurs when two atoms need to share four electrons to achieve a stable electron configuration.
Explanation:
To understand why there are 4 electrons being shared in a double covalent bond, let's consider the example of oxygen (O2) molecule.
1. Oxygen atom has 6 electrons in its outermost shell.
2. To achieve a stable electron configuration, oxygen needs a total of 8 electrons in its outermost shell (octet rule).
3. Therefore, two oxygen atoms come together and share electrons to form a double covalent bond.
4. Each oxygen atom contributes two electrons, resulting in a total of 4 electrons being shared between the two oxygen atoms.
5. This sharing of electrons allows each oxygen atom to have a complete octet, fulfilling the stability requirement.
Key Points:
- Double covalent bonds involve the sharing of two electron pairs.
- This type of bond occurs when two atoms need to share four electrons to achieve stability.
- Oxygen is a common example of a molecule that forms a double covalent bond.
- Each atom contributes two electrons to the bond, resulting in a total of four shared electrons.
- The sharing of electrons allows atoms to achieve a stable electron configuration.
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