Class 10 Science Chapter 4 Carbon and its Compounds

Key Concept: Formula Recognition

c) $C_{6}H_{12}$
[Solution Description] Cyclohexane is a cyclic hydrocarbon composed of six carbon atoms and twelve hydrogen atoms, forming a ring structure. Therefore, its molecular formula is $C_{6}H_{12}$.

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Key Concept: Reaction Conditions

c) Basic environment
[Solution Description] Saponification is a chemical reaction that involves the hydrolysis of an ester, typically triglycerides (fats/oils), using an aqueous solution of a strong base such as sodium hydroxide (NaOH). During the process, glycerol and soap (sodium or potassium salts of fatty acids) are produced. Here, the key condition is the presence of a basic medium.

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Key Concept: Triple Bond Basics

c) Triple bond
[Solution Description] Nitrogen has an atomic number of 7, which means its electronic configuration is $1s^2\, 2s^2\, 2p^3$. To attain the noble gas configuration, each nitrogen atom shares three electrons with another nitrogen atom, resulting in three shared pairs of electrons or a triple bond.

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Key Concept: Reaction Mechanism

b) Acetaldehyde
[Solution Description] The oxidation of ethanol to ethanoic acid involves the conversion of ethanol $(C_2H_5OH)$ first to acetaldehyde $(CH_3CHO)$, and then to ethanoic acid $(CH_3COOH)$. The key step in this reaction is the formation of acetaldehyde as an intermediate. Here are the steps: Ethanol $C_2H_5OH$ \xrightarrow{\text{oxidizing agent}} $CH_3CHO$ \xrightarrow{\text{oxidizing agent}} $CH_3COOH$. Thus, acetaldehyde is the intermediate formed.

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Key Concept: Electron Dot Structure

b) $ :\text{H}:\text{O}:\text{H}: $
[Solution Description] A water molecule consists of one oxygen atom bonded to two hydrogen atoms through single covalent bonds, and the oxygen atom has two lone pairs of electrons. The correct electron dot structure shows these shared electron pairs and lone pairs around the oxygen.

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Key Concept: Reaction Mechanism

a) Acid-base neutralization
[Solution Description] The reaction between ethanoic acid and sodium hydroxide is a neutralization reaction. In this reaction, an acid reacts with a base to form a salt and water. Here, the acid is ethanoic acid $(CH_3COOH)$, and the base is sodium hydroxide (NaOH). The products formed are sodium acetate

$(CH_3COONa)$

and water $(H_2O)$. This can be represented by the equation:

$\text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$

The process involves the exchange of ions where $H^+$ from the acid combines with $OH^-$ from the base to produce water.

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Key Concept: Functional Groups

b) —COOH
[Solution Description] Carboxylic acids contain the functional group —COOH, where a carbon atom is double-bonded to one oxygen atom and single-bonded to an -OH group.

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Key Concept: Comparative Analysis, Real-world Application

a) Methanol causes blindness as it oxidizes to formaldehyde in the body.
[Solution Description] Methanol and ethanol both have significant effects on human health but vary greatly in toxicity. Methanol oxidizes to methanal in the liver, which harms cells and can cause blindness or even death in small quantities. Ethanol similarly impairs body functions but requires larger amounts for severe effects.

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Key Concept: Micelle Structure

c) It interacts with oil, helping to trap it inside the micelle.
[Solution Description] The hydrophobic tails in soap molecules are oriented towards the oil droplet, forming a core around which the ionic ends face outwards into the water. This allows the soap to emulsify oils by trapping them within the hydrophobic center and making them soluble in water.

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Key Concept: Homologous Series

b) $CH_2$
[Solution Description] Alkenes are unsaturated hydrocarbons with one or more double bonds, with the general formula $C_nH_{2n}$. Successive members differ by a $—CH_2—$ group. For example, ethene $(C_2H_4)$ and propene $(C_3H_6)$ differ by $—CH_2—$.

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Key Concepts: Basic Properties

b) 290 K

[Solution Description] The melting point of pure ethanoic acid, referred to as glacial acetic acid, is a fundamental property that is often tested in basic chemistry knowledge. According to the syllabus, the melting point is listed directly.

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Key Concept: Hydrogen Source

a) Hydroxyl group
[Solution Description]
Ethanol $(C_2H_5OH)$ reacts with sodium to form sodium ethoxide $(C_2H_5O^-Na^+)$ and hydrogen gas $(H_2)$. In this reaction, hydrogen is released from the hydroxyl group $(-OH)$ of ethanol as it forms a bond with sodium.

The balanced reaction is:
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5O^-Na^+ + H_2$
Therefore, the hydrogen evolved comes from the hydroxyl group of ethanol.

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Key Concept: Functional Group Identification

c) Ketone
[Solution Description] Propanone, commonly known as acetone, has the chemical formula $CH_3COCH_3$. The structure of propanone includes a carbonyl group $$(C=O)$$ bonded to two methyl groups. In organic chemistry, this is classified as a ketone functional group because the carbonyl group is flanked by carbon atoms on either side, which is characteristic of ketones.

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Key Concept: Micelle Formation

c) Micelle
[Solution Description] When soap molecules are in water, they arrange themselves into a spherical structure called a micelle, with hydrophobic tails inside and hydrophilic heads outside.

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Key Concept: Structural Isomer Identification

b) 2-Methylbutane
[Solution Description] Pentane has the molecular formula $C_5H_{12}$. To find structural isomers, we must rearrange the carbon atoms to form different structures without changing the number of hydrogen atoms. One possible structural isomer of pentane is 2-methylbutane, which also has the formula $C_5H_{12}$ but with a branched structure.

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Key Concept: Bonding Implications, Advanced Reasoning

c) Smaller atom sizes allow better electron sharing, leading to stronger bonds.
[Solution Description] Carbon’s small atomic size allows the nucleus to hold shared electrons more tightly, resulting in strong covalent bonds. In contrast, larger atoms like silicon have weaker covalent bonds because their nuclei cannot hold shared electrons as closely or effectively.

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Key Concept: Comparative Analysis

b) Vegetable oils
[Solution Description] Unsaturated hydrocarbons like vegetable oils can be hydrogenated to form saturated fats using a nickel catalyst. This process adds hydrogen to double bonds present in unsaturated carbon chains.

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Key Concept: Soap Structure

b) Carbon Chain Tail
[Solution Description] In a soap molecule, the long hydrocarbon chain or tail is hydrophobic and interacts with oil, while the ionic head interacts with water.

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Key Concept: Substitution Reaction

c) Hydrogen atoms are replaced by chlorine atoms.
[Solution Description] In a substitution reaction between methane $(CH_4)$ and chlorine $(Cl_2)$, chlorine atoms replace hydrogen atoms one at a time. The result is various chloromethane products such as methyl chloride $(CH_3Cl)$, methylene chloride $(CH_2Cl_2)$, chloroform $(CHCl_3)$, and carbon tetrachloride $(CCl_4)$.

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Key Concept: Bond Energy, Methane Combustion

a) -694 kJ/mol
[Solution Description] To solve this problem, we need to calculate the total energy required to break the bonds in the reactants and the total energy released by forming the bonds in the products.

Step 1: Calculate the energy needed to break the bonds in the reactants:
– Breaking 4 C-H bonds: $ 4 \times 412 = 1648 \, \text{kJ/mol} $

– Breaking 2 O=O bonds: $ 2 \times 498 = 996 \, \text{kJ/mol} $

Total energy required = $ 1648 + 996 = 2644 \, \text{kJ/mol} $

Step 2: Calculate the energy released when new bonds form in the products:
– Forming 2 C=O bonds: $ 2 \times 743 = 1486 \, \text{kJ/mol} $
– Forming 4 O-H bonds: $ 4 \times 463 = 1852 \, \text{kJ/mol} $

Total energy released = $ 1486 + 1852 = 3338 \, \text{kJ/mol}

Step 3: Calculate the net energy change:
Net energy change = Total energy released – Total energy required
Net energy change = $ 3338 – 2644 = 694 \, \text{kJ/mol} $

Therefore, the reaction is exothermic with a net release of 694 kJ/mol.

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Key Concept: Reaction Optimization, Reaction Yield

d) Using a weaker acid catalyst
[Solution Description] The dehydration of ethanol to ethene requires specific conditions for optimal conversion. To increase yield, the reaction generally benefits from high temperature and continuous removal of water to shift equilibrium towards ethene formation. Using a weaker acid or lowering the concentration of reactant could reduce efficiency.

Steps:

– High Temperature: Increases rate of endothermic dehydration, favoring more ethene production.
– Continuous Water Removal: Drives equilibrium right, further increasing ethene yield.
– Weaker Acid: Decreases dehydration efficacy, as strong acids like sulfuric acid are required for efficient dehydration.
– Lower Concentration: Less reactant available shifts equilibrium back, reducing yields.

Clearly, using a weaker acid does not optimize the process.

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Key Concept: Basic Property

c) Common functional group
[Solution Description] All members of a homologous series have a similar structure and share common chemical properties due to having the same functional group.

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Key Concept: Comparative Analysis, Electron Sharing

a) Aliphatic bonds are weaker, hence more reactive
[Solution Description] The carbon-carbon bonds in aromatic compounds such as benzene have resonance stabilization due to sharing electrons over multiple atoms, which makes them stronger compared to the single carbon-carbon bonds in aliphatic hydrocarbons. Aromatic compounds have delocalized electrons which provide extra stability.

Aliphatic hydrocarbons can have saturated (single bonds) or unsaturated (double/triple bonds), but the presence of delocalized π bonds in aromatic rings provides significant stability. This usually results in lower chemical reactivity for aromatic compounds compared to aliphatic ones, making them less likely to undergo addition reactions:

$\text{Aromatic ring} \rightarrow \text{Stability via resonance}.$

Therefore, comparative analysis leads us to understand that while aliphatic C-C bonds are reactive because they lack such stabilization, aromatics react slower due to their stable structure.

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Key Concept: Complex Naming

b) 3-bromo-3-pentanol
[Solution Description]
To solve this problem, we need to consider both the bromine and alcohol groups when naming the compound. The chain must be numbered from the end closest to a substituent or functional group to ensure the lowest possible numbers are assigned. In this case, placing the alcohol group as the priority, its suffix ‘-ol’ should be used. Bromine is treated as a prefix ‘bromo-‘. Therefore, starting from the nearest end to either functional group, the proper numbering gives the compound 3-bromo-3-pentanol.

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Key Concept: Alcohol Role

b) Reactant
[Solution Description] During the esterification process involving ethanoic acid and ethanol, ethanol acts as a reactant. It combines with ethanoic acid in the presence of an acid catalyst to form an ester and water.

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Key Concept: Reaction Outcome

a) Propane
[Solution Description] Propene has the chemical formula $C_3H_6$ and contains a double bond between two of its carbon atoms. During hydrogenation, hydrogen molecules are added across the double bond, converting it into a single bond. This results in the formation of propane, which is a saturated hydrocarbon with the formula $C_3H_8$.

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Key Concept: Electron Sharing

b) 2
[Solution Description]
In ethylene $(C_2H_4)$, each carbon atom forms a double bond with the other carbon atom. A double bond consists of two pairs of shared electrons.

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Key Concept:- Carbon’s Geological Role, Carbon’s Industrial Use

b) Calcination; Carbon Dioxide
[Solution Description] The conversion of calcium carbonate (limestone) into lime (calcium oxide) is achieved through a process called calcination. This process involves heating calcium carbonate at high temperatures to drive off carbon dioxide, leaving behind calcium oxide:

$\text{CaCO}_3 \xrightarrow{\text{heat}} \text{CaO} + \text{CO}_2.$

During this reaction, carbon dioxide is released as the primary byproduct.

Notes:–

Carbon’s Geological Role:

1. Carbon Cycle:
   • Carbon is continuously exchanged among the atmosphere, oceans, and earth in the carbon cycle.
   • It exists as carbon dioxide (CO₂) in the atmosphere and carbonate rocks (limestone, chalk).
2. Fossil Fuels:
   • Carbon is a major component of fossil fuels like coal, petroleum, and natural gas, formed from decomposed plants and animals over millions of years.
3. Soil and Oceans:
   • Oceans store carbon as dissolved CO₂ and bicarbonates. Soil contains organic matter, essential for plant growth.

Carbon’s Industrial Use:

1. Fuel Source:
   • Coal, petroleum, and natural gas are carbon-based fuels powering industries and transport.
2. Chemical Industry:
   • Used to manufacture plastics, synthetic fibers, and dyes.
3. Metallurgy:
   • Carbon (as coke) is used in metal extraction (e.g., iron from ore in a blast furnace).
4. Other Uses:
   • Diamond and graphite, allotropes of carbon, are used in cutting tools, jewelry, and lubricants.

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Key Concept: Isomer Count

b) Two
[Solution Description] Butane $(C_4H_{10})$ has two structural isomers: n-butane and isobutane (also known as methylpropane). To determine the isomers, we begin with a straight chain of carbon atoms which gives us n-butane. The second structure, isobutane, involves branching by attaching one carbon atom to the middle carbon of a three-carbon chain. Both structures adhere to the molecular formula $C_4H_{10}$.

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Key Concept: Complex Reaction Pathway, Industrial Process Design

b) Chloroethane
[Solution Description]
The initial step involves the hydrogenation of ethene $\text{C}_2\text{H}_4$
to ethane

$\text{C}_2\text{H}_6$
using a nickel catalyst:

$\text{C}_2\text{H}_4 + \text{H}_2 \xrightarrow{\text{Ni}} \text{C}_2\text{H}_6$

Next, we perform a substitution reaction where chlorine replaces hydrogen atoms one at a time under sunlight. Starting with ethane:

$C_2H_6 + Cl_2 \rightarrow{\text{sunlight}} C_2H_5Cl + HCl$

Repeated interaction with chlorine can lead to multiple chlorination, but the predominant product after one substitution will be chloroethane

$\text{C}_2\text{H}_5\text{Cl}$
.Therefore, the major product after these sequential reactions is chloroethane.

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Key Concept: Functional Group Identification

c) Ketone
[Solution Description] Propanone, commonly known as acetone, has the molecular formula $$CH_3COCH_3$$. The functional group in propanone is a carbonyl group (C=O), which is characteristic of ketones.

Key Concept: Acidic Nature

d) It is a weak acid
[Solution Description] Ethanoic acid is characterized by its weak acidic nature compared to strong mineral acids like HCl. Thus, among the options provided, it is correct to say that ethanoic acid is a weak acid.

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Key Concept: Micelle Formation

b) Micelles
[Solution Description] Soap molecules arrange themselves into spherical clusters called micelles when mixed with water. The hydrophilic ends face outward towards the water, while the hydrophobic tails trap oily substances inside the micelle.

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Key Concept: Functional Group Identification

c) -CHO
[Solution Description] Propanal belongs to the aldehyde family. Aldehydes have the functional group $-CHO$.

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Key Concept: Basic Properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description] Both the assertion and the reason statements are true, but the reason does not explain why soaps are sodium or potassium salts of long-chain carboxylic acids. The statement about detergents is irrelevant to the definition of soaps.

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Key Concept: Health Effects of Ethanol

c) Various health problems
[Solution Description] Long-term consumption of ethanol can lead to various health problems such as liver damage, addiction, heart disease, and nervous system disorders. It is well-documented that chronic use of alcohol significantly impacts physical health in negative ways.

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Key Concept: Isomer Complexity, Structural Analysis, Advanced Reasoning

b) Cyclohexane
[Solution Description]
To determine which hydrocarbons can exist as structural isomers, we need to analyze possible arrangements and structures.

Begin by considering simple aliphatic hydrocarbons like hexane, which is a straight-chain alkane: $CH_{3}(CH_{2})_{4}CH_t{3}$.

By branching or forming rings, several other isomers are possible:

1. Methylpentane (branching at different positions)
2. Cyclohexane (forming a ring structure)
3. Cyclopentane derivatives with one branch

This analysis shows that cyclohexane and some cyclic compounds can be isomers alongside linear hexanes.

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Key Concept: Series Progression

b) 14 u
[Solution Description] In a homologous series like alkenes, successive members differ by a $CH_2$ unit. The atomic mass of carbon $(C)$ is 12 u and hydrogen $(H)$ is 1 u. Therefore, the additional mass from a $CH_2$ unit is $\$12 + 2(1) = 14$ u. Hence, the molecular mass increases by 14 u between any two consecutive alkenes.

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Key Concepts: Allotropes

d) Salt

[Solution Description] Diamond, graphite, and fullerenes are well-known allotropes of carbon due to their distinct structures formed by carbon atoms. Salt, however, is a compound made from sodium and chloride ions and is not an allotrope of carbon.

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Key Concept: Esterification Mechanism

c) Water
[Solution Description] The esterification reaction between ethanoic acid $(CH_3COOH)$ and ethanol $(C_2H_5OH)$ results in the formation of an ester (ethyl acetate) and water. The reaction can be represented as:
$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightarrow$ $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$

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Key Concepts: Catenation Property

d) Catenation

[Solution Description] Carbon’s ability to form long chains with itself is known as catenation. This unique property allows it to create diverse structures such as straight chains, branched chains, or rings by forming stable covalent bonds.

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Key Concept: Physical State Changes

b) 334 K
[Solution Description] According to Table 4.1, the boiling point of chloroform is given as 334 K. At this temperature, chloroform transitions from the liquid phase to the gaseous phase.

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Key Concept: Industrial Application

c) It produces margarine and improves product shelf life.
[Solution Description] The hydrogenation of vegetable oils is integral to the production of margarine and shortening. This process transforms liquid unsaturated oils into semi-solid or solid forms, thereby extending shelf life and improving texture. The adjustment of physical properties through hydrogenation makes these products versatile for various culinary applications.

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Key Concept: Micelle Formation

d) Micelles
[Solution Description] Soap molecules form structures called micelles, where the hydrophobic tails face inward and the ionic ends face outward, aiding in cleaning by trapping dirt and oil inside.

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Key Concept:- Homologous Series

c) $C_nH_{2n-2}$
[Solution Description]
To find the general formula for alkynes, let’s consider that in a homologous series, each member differs by a unit of $CH_2$. Alkanes are saturated hydrocarbons with single bonds and follow the formula $C_nH_{2n+2}$. Alkenes, which have one double bond, follow the formula $C_nH_{2n}$. Alkynes, containing a triple bond, will therefore have two fewer hydrogen atoms than alkenes to satisfy the bonding requirements. Thus, the formula for alkynes is $C_nH_{2n-2}$.

Notes:-

• Definition:
  • A homologous series is a group of organic compounds with the same functional group, similar chemical properties, and a difference of
    CH_2CH_2CH_2​$ between consecutive members.
• Characteristics:
  • General Formula: Compounds in the series follow a general formula
  • Gradation in Physical Properties: Boiling and melting points increase with molecular size.
  • Similar Chemical Properties: Due to the same functional group.
• Examples:
  • Alkanes: Methane ($CH4CH_4$​), Ethane ($C_2H_6C_2H_6$​), Propane ($C_3H_8C_3H_8$​).
  • Alkenes: Ethene ($C_2H_4C_2H_4$), Propene ($C_3H_6C_3H_6$​).
  • Alcohols: Methanol ($CH_3OHCH_3OH$), Ethanol ($C_2H_5OHC_2H_5OH$).
• Importance:
  • Simplifies the study of organic chemistry by categorizing compounds.
  • Predicts properties and reactivity of compounds in the series.
• Applications:
  • Used in systematic naming and understanding trends in organic compounds.

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Key Concept: Series Arrangement

b)$\text{CH}_3\text{OH}$, $\text{C}_2\text{H}_5\text{OH}$, $\text{C}_3\text{H}_7\text{OH}$
[Solution Description] To arrange these alcohols by increasing molecular mass, calculate the molecular mass for each.

$\text{CH}_3\text{OH}$
has a lower molecular mass compared to

$\text{C}_2\text{H}_5\text{OH}$ and $\text{C}_3\text{H}_7\text{OH}$

Thus, arrangement should be done accordingly.

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Key Concept: Formula Difference

b) $\text{CH}_2$

[Solution Description] Both C3H8 (propane) and C4H10 (butane) are alkanes, differing by one -CH2- unit. Thus, the difference in their molecular formula is

$\text{CH}_2$.

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Key Concept: Combustion Reactions

a) $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
[Solution Description]
The complete combustion of propane involves oxygen and produces carbon dioxide and water. The unbalanced reaction is:

$C_3H_8 + O_2 \rightarrow CO_2 + H_2O$

To balance, first balance the carbon atoms:

$C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$

Next, balance the hydrogen atoms (8 on the left):

$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$

Finally, balance the oxygen atoms with 10 on the right (6 from $CO_2$ and 4 from $H_2O$), requiring 5 $O_2$ molecules:

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Thus, this is the balanced equation.

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Key Concept: Synthesis Process, Real-world Application

c) Applying high pressure and high temperature to pure carbon
[Solution Description] Synthetic diamonds are created by subjecting pure carbon to high pressure and high temperature, mimicking the natural conditions under which diamonds form deep within the Earth. These extreme conditions force carbon atoms into a rigid, three-dimensional lattice identical to that of natural diamonds, resulting in similar hardness and optical properties.

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Key Concept: Real-world Application

b) They are less polluting
[Solution Description] Saturated hydrocarbons tend to undergo complete combustion, producing a clean blue flame without soot, which makes them ideal for domestic fuels as they burn efficiently and reduce indoor pollution.

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Key Concept: Complex Structures, Advanced Bonding

c) Delocalized electrons contribute to the spherical shape.
[Solution Description]
Fullerenes are a class of carbon allotropes where carbon atoms are arranged in hollow spheres. The most common fullerene,

$C_{60}$
, has 60 carbon atoms forming a spherical structure resembling a soccer ball. In fullerenes, each carbon atom forms three bonds: two single bonds with neighboring carbons in a hexagonal or pentagonal ring, and one bond contributing to a conjugated system across the structure, allowing delocalization of electrons.

Since all bonds in fullerenes can be considered as having partial double-bond character due to this delocalization, the structure is stabilized by resonance. This unique arrangement results in exceptional stability and makes them distinct from other carbon forms like graphite and diamond. Hence, the statement about delocalized electrons giving it a spherical shape is accurate.

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Key Concept: Real-world Application of Carbon Bonding, Comparative Bonding Analysis

c) Carbon shares electrons; sodium forms ionic bonds.
[Solution Description] Sodium often forms ionic bonds by losing one electron to become Na^+, thereby achieving a stable octet in its previous shell. In contrast, carbon primarily forms covalent bonds due to its four valence electrons, requiring sharing rather than transfer of electrons to satisfy the valency. In glucose, each carbon atom shares electrons with hydrogen, oxygen, or other carbons, creating a network of covalent bonds essential for molecular integrity.

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Key Concept: Hydrocarbon Classification

b) Alkene
[Solution Description] The hydrocarbon $C_{3}H_{6}$ follows the general formula for alkenes $C_{n}H_{2n}$, indicating it contains one carbon-carbon double bond.

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Key Concept: Advanced Isomerism, Complex Structures

c) 2-Methylpropan-$\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{OH}$

[Solution Description]
To determine the structural isomer of 1-butanol, we need to look for a compound with the same molecular formula $C_4H_{10}O$
but different connectivity among atoms. 1-Butanol has the structure $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$
. A potential structural isomer would rearrange the carbon chain or change the position of the hydroxyl group while maintaining the formula.

Let’s consider 2-methylpropan-1-ol: $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{OH}$:
– It maintains the molecular formula $C_4H_{10}O$
.
– The arrangement of carbon skeleton differs from 1-butanol as it has a branched structure.

Thus, 2-methylpropan-1-ol is indeed a structural isomer of 1-butanol.

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Key Concept: Molecular Formula Interpretation

d) Both single and triple bonds
[Solution Description] Acetylene $(C_2H_2)$ is known to have a triple bond between the two carbon atoms. Each carbon shares three pairs of electrons with the other, forming a strong triple bond, and each carbon also forms a single bond with one hydrogen atom.

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Key Concept:-  Bond Type Identification

c) Triple bond
[Solution Description] In a nitrogen molecule

$(N_2)$,

each nitrogen atom shares three electrons with another nitrogen atom to achieve an octet configuration. Therefore, the nitrogen atoms form a triple bond.

Notes:-

• Definition of Bond Type:
  • Refers to the way atoms share or transfer electrons in a compound, forming single, double, or triple bonds.
• Single Bond (C−CC-C):
  • Formed by the sharing of one pair of electrons.
  • Example: Methane $\text{CH}_4$
    , Ethane $C_2H_6$
  • Strong sigma (σ\sigma$\sigma$) bonds provide stability.
• Double Bond (C=CC=C):
  • Formed by the sharing of two pairs of electrons (one
    σ\sigma$\sigma$-bond and one π\pi$\pi$-bond).
  • Example: Ethene $\text{C}_2\text{H}_4$.
  • Shorter and more reactive than single bonds.
• Triple Bond (C≡CC≡C):
  • Formed by the sharing of three pairs of electrons (one
    σ\sigma$\sigma$-bond and two π\pi$\pi$-bonds).
  • Example: Ethyne $\text{C}_2\text{H}_2$
  • Strongest but most reactive.
• Steps to Identify Bond Type:
  • Analyze the molecular formula.
  • Determine the number of shared electron pairs.
• Significance:
  • Identifying bond types helps predict reactivity, stability, and properties of compounds.

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Key Concept: Simple Homologous Series

a) $C_4H_9OH$
[Solution Description] The given series is an alcohol series where each member differs by a $\text{-CH}_{2}\text{-}$
unit. Starting from

$\text{C}_3\text{H}_7\text{OH}$
(propanol), the next member would be

$C_4H_9OH$
(butanol).

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Key Concept:-  Complex Structure Drawing, Advanced Bond Analysis

b) There are two $\pi$ bonds and one $\sigma$ bond connecting the carbon atoms in ethyne.
[Solution Description] Ethyne has two carbon atoms connected by a triple bond. Each carbon atom forms one sigma $(\sigma)$ bond and two pi $(\pi)$ bonds with the other carbon atom. Additionally, each carbon also forms a sigma bond with a hydrogen atom. In total, there are three shared pairs of electrons between the carbon atoms and one shared pair between each carbon and its respective hydrogen.

Notes:-

Key Concept: Foam Observation

a) Distilled water
[Solution Description] Soap forms more foam in distilled water because it does not contain calcium and magnesium ions that react with soap to form an insoluble precipitate, reducing foam formation.

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Key Concept: Melting Points

d) Methane
[Solution Description] From the data given in the syllabus, we have the melting points of various carbon compounds: Acetic acid (290 K), Chloroform (209 K), Ethanol (156 K), Methane (90 K). Among these compounds, Methane has the lowest melting point at 90 K.

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Key Concept: Unsaturated Compounds

c) Ethene $(C_2H_4)$
[Solution Description]
Unsaturated compounds contain either double or triple bonds between carbon atoms. Ethene $(C_2H_4)$ contains a double bond, and ethyne $(C_2H_2)$ contains a triple bond, making them unsaturated. Methane $(CH_4)$ and ethane $(C_2H_6)$ have single bonds only, making them saturated.

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