CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 3 - 56/5/3)

Concept:

• The Carbylamine reaction is a diagnostic test used exclusively for primary amines (both aliphatic and aromatic).
• It involves the reaction of a primary amine with chloroform and ethanolic potassium hydroxide.
• The reaction leads to the synthesis of isocyanides, which are characterized by an extremely offensive and pungent odor.

Step 1: Generation of the reactive intermediate, Dichlorocarbene
In the presence of a strong base like \( \text{KOH} \), chloroform \( (\text{CHCl}_3) \) undergoes \( \alpha \)-elimination.
This process removes a proton and a chloride ion to generate the highly reactive neutral intermediate known as dichlorocarbene \( (:\text{CCl}_2) \).
\[ \text{CHCl}_3 + \text{OH}^- \rightarrow :\text{CCl}_2 + \text{H}_2\text{O} + \text{Cl}^- \]

Step 2: Nucleophilic attack by the primary amine and subsequent eliminations
The nitrogen lone pair of the primary amine \( (\text{R}-\text{NH}_2) \) attacks the electron-deficient dichlorocarbene.
This is followed by the elimination of two molecules of \( \text{HCl} \) (neutralized by \( \text{KOH} \)).
The overall sequence leads to the formation of the alkyl/aryl isocyanide.
\[ \text{R}-\text{NH}_2 + :\text{CCl}_2 \xrightarrow{-2\text{HCl}} \text{R}-\text{N} \equiv \text{C} \]

Step 3: Final reaction summary
The balanced chemical equation for the process is:
\[ \text{R}-\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \xrightarrow{\Delta} \text{R}-\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \]
The product \( \text{R}-\text{NC} \) is called an isocyanide or carbylamine.

Concept:

• This reaction is specifically known as the Finkelstein reaction.
• It is a nucleophilic substitution \( (\text{S}_{\text{N}}2) \) reaction used for halogen exchange.
• It is the standard method for preparing alkyl iodides from alkyl chlorides or bromides.

Step 1: Nucleophilic attack mechanism
In this reaction, the iodide ion \( (\text{I}^-) \) serves as a strong nucleophile.
Since the substrate is a primary methyl halide \( (\text{CH}_3\text{Cl}) \), the reaction proceeds via an \( \text{S}_{\text{N}}2 \) mechanism.
The \( \text{I}^- \) attacks the carbon center from the back side while the \( \text{Cl}^- \) leaves simultaneously.

Step 2: Role of dry acetone and Le Chatelier’s principle
The solvent, dry acetone, plays a critical role in driving the reaction to completion.
Sodium iodide \( (\text{NaI}) \) is soluble in acetone, but the byproduct sodium chloride \( (\text{NaCl}) \) is almost insoluble.
Consequently, \( \text{NaCl} \) precipitates out of the solution as it forms.
\[ \text{CH}_3\text{Cl} + \text{NaI} \xrightarrow{\text{acetone}} \text{CH}_3\text{I} + \text{NaCl} \downarrow \]

Step 3: Identifying products A and B
Based on the exchange, product A is Methyl Iodide \( (\text{CH}_3\text{I}) \).
Product B is the precipitated Sodium Chloride \( (\text{NaCl}) \).

Concept:

• The complete reduction of a carbonyl group \( (> \text{C=O}) \) to a methylene group \( (> \text{CH}_2) \) is termed deoxygenation.
• This transformation effectively converts aldehydes and ketones into their parent hydrocarbons.
• Two primary named reactions achieve this: Wolff-Kishner reduction (basic conditions) and Clemmensen reduction (acidic conditions).

Step 1: Formation of the hydrazone intermediate
In the Wolff-Kishner reduction, the ketone \( (\text{R}_2\text{C=O}) \) is reacted with hydrazine \( (\text{NH}_2\text{NH}_2) \).
The initial product is a hydrazone \( (\text{R}_2\text{C=N-NH}_2) \).
\[ \text{R}_2\text{C=O} + \text{NH}_2\text{NH}_2 \rightarrow \text{R}_2\text{C=N-NH}_2 + \text{H}_2\text{O} \]

Step 2: Base-catalyzed decomposition and nitrogen evolution
The hydrazone is then heated with a strong base like \( \text{KOH} \) or \( \text{NaOCH}_2\text{CH}_3 \) in a high-boiling solvent like ethylene glycol.
The base deprotonates the nitrogen, leading to the loss of Nitrogen gas \( (\text{N}_2) \) and the formation of an alkane.
\[ \text{R}_2\text{C=N-NH}_2 \xrightarrow{\text{KOH, glycol, } \Delta} \text{R}_2\text{CH}_2 + \text{N}_2 \]

Step 3: Final conclusion
This specific pathway using basic conditions and hydrazine is the Wolff-Kishner reduction.

Concept:

• Hoffmann bromamide degradation is a unique method for synthesizing primary amines.
• It involves the "degradation" of an amide, resulting in an amine with one fewer carbon atom than the starting material.
• The reaction mechanism involves an intramolecular rearrangement.

Step 1: Identifying required reagents
The reaction requires a primary amide \( (\text{R}-\text{CONH}_2) \), which matches reagent (i).
The reaction uses bromine \( (\text{Br}_2) \) as the halogen, which is reagent (iii).
An aqueous or alcoholic alkali solution is needed, typically Sodium Hydroxide \( (\text{NaOH}) \), matching reagent (ii).

Step 2: Analyzing the chemical transformation
The amide reacts with bromine and base to form an N-bromoamide.
This is followed by the migration of the alkyl group and the formation of an isocyanate.
Hydrolysis of the isocyanate yields the final primary amine and carbonate.
\[ \text{R-CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{R-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O} \]

Step 3: Reagent verification
Reagent (iv) \( \text{CHCl}_3 \) is used in the Carbylamine reaction for testing amines, not for this synthesis.
Therefore, reagents (i), (ii), and (iii) are essential.

Concept:

• Hinsberg’s reagent is the common name for Benzene Sulphonyl Chloride.
• It is used as a chemical test to classify amines into primary, secondary, or tertiary.
• The reagent reacts with the amine to form a sulphonamide derivative.

Step 1: Structural identification
Benzene sulphonyl chloride consists of a benzene ring \( (\text{C}_6\text{H}_5) \).
Attached to this ring is a sulphonyl group \( (-\text{SO}_2-) \).
The sulphonyl group is further bonded to a chlorine atom \( (-\text{Cl}) \).
Combining these parts gives the molecular formula: \( \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \).

Step 2: Differentiating from other options
Option (A) is Benzoyl Chloride \( (\text{C}_6\text{H}_5\text{COCl}) \).
Option (C) is N-Methylbenzamide \( (\text{C}_6\text{H}_5\text{CONHCH}_3) \).
Option (D) is Benzylamine \( (\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2) \).

Step 3: Application in the Hinsberg test
When a primary amine reacts with \( \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \), it forms a sulphonamide with an acidic hydrogen.
This product is soluble in alkali \( (\text{NaOH}) \).
\[ \text{C}_6\text{H}_5\text{SO}_2\text{Cl} + \text{R-NH}_2 \rightarrow \text{C}_6\text{H}_5\text{SO}_2\text{NHR} \]

Concept:

• Oxidation states in elements are determined by the number of electrons available in the valence shells for bonding.
• In f-block elements, electrons from the (n-2)f, (n-1)d, and ns subshells can potentially participate in bond formation.
• The diversity of oxidation states depends on the energy gap between these participating subshells.

Step 1: Analyzing the electronic configuration and energy levels
Actinoids are elements in which the 5f subshell is being filled.
The general electronic configuration involves the 5f, 6d, and 7s orbitals.
Unlike Lanthanoids, where the energy gap between 4f and 5d is relatively large, the energy gap between 5f, 6d, and 7s orbitals in actinoids is very small.

Step 2: Relating orbital energy to oxidation states
Because the 5f, 6d, and 7s orbitals have comparable (very similar) energies, electrons can be removed from all these levels with relatively similar amounts of energy.
This allows actinoids to utilize a larger number of electrons for bonding compared to lanthanoids.

Step 3: Conclusion
This results in a wide range of oxidation states, typically from +3 up to +7 for some elements like Neptunium and Plutonium.
The fundamental reason is the comparable energies of the 5f, 6d, and 7s orbitals.

Concept:

• Distinction tests rely on the different rates of reactivity of primary (\( 1^\circ \)), secondary (\( 2^\circ \)), and tertiary (\( 3^\circ \)) functional groups with a specific reagent.
• For alcohols, the stability of the intermediate carbocation determines the speed of the reaction with halogen acids.

Step 1: Understanding the Lucas Reagent
Lucas reagent is a mixture of concentrated Hydrochloric acid \( (\text{HCl}) \) and anhydrous Zinc Chloride \( (\text{ZnCl}_2) \).
The reaction involves the substitution of the hydroxyl \( (-\text{OH}) \) group with a chlorine \( (-\text{Cl}) \) atom to form an alkyl chloride.

Step 2: Observing the reaction rates
Alkyl chlorides are insoluble in water and produce cloudiness or turbidity in the solution.
1. Tertiary alcohols react immediately, showing turbidity instantly at room temperature.
2. Secondary alcohols react within 5 to 10 minutes to show turbidity.
3. Primary alcohols do not produce turbidity at room temperature; it only appears upon heating.

Step 3: Evaluating other options
Fehling’s and Tollens’ tests are used to identify aldehydes.
Hinsberg’s test is used to distinguish between primary, secondary, and tertiary amines.

Concept:

• Basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
• It is influenced by the +I (Inductive) effect, resonance effect, and steric hindrance.
• Alkyl groups increase basicity via +I effect, while aryl groups (phenyl) decrease it via resonance.

Step 1: Comparing Ammonia and Aniline
In Aniline \( (\text{C}_6\text{H}_5\text{NH}_2) \), the lone pair on nitrogen is delocalized into the benzene ring due to resonance.
This makes the lone pair less available for donation compared to Ammonia \( (\text{NH}_3) \).
Order: \( \text{C}_6\text{H}_5\text{NH}_2 < \text{NH}_3 \).

Step 2: Evaluating Benzylamine and Aliphatic Amines
In Benzylamine \( (\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2) \), the nitrogen is not directly attached to the ring, so no resonance occurs.
The benzyl group has a weak electron-withdrawing effect compared to ethyl groups.
In Ethylamines \( (\text{C}_2\text{H}_5\text{NH}_2 \text{ and } (\text{C}_2\text{H}_5)_2\text{NH}) \), the +I effect of ethyl groups significantly increases electron density on nitrogen.

Step 3: Ordering the ethylamines in aqueous solution
For the ethyl group, the basicity order in aqueous solution is \( 2^\circ > 1^\circ > 3^\circ > \text{NH}_3 \).
Thus, \( (\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 \).

Step 4: Final assembly of the order
Combining all factors: Aniline (weakest) < Ammonia < Benzylamine < Ethylamine < Diethylamine (strongest).
This matches the sequence: \( \text{C}_6\text{H}_5\text{NH}_2 < \text{NH}_3 < \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH} \).

Concept:

• Carbohydrates are classified based on the number of sugar units they produce upon hydrolysis.
• Monosaccharides: Single sugar units (cannot be hydrolyzed further).
• Disaccharides: Two sugar units.
• Polysaccharides: Hundreds to thousands of sugar units linked together.

Step 1: Categorizing the given options
1. Glucose: A monosaccharide (hexose).
2. Fructose: A monosaccharide (ketohexose).
3. Maltose: A disaccharide composed of two glucose units.

Step 2: Analyzing Cellulose
Cellulose is a high-molecular-weight polymer consisting of thousands of \( \text{D-glucose} \) units linked by \( \beta \)-1,4-glycosidic bonds.
It is the primary structural component of the cell walls of green plants.
Because it consists of many monosaccharide units, it is classified as a polysaccharide.

Concept:

• Protein structure is organized into four hierarchical levels: primary, secondary, tertiary, and quaternary.
• Each level describes a different aspect of the spatial arrangement of the atoms in the protein.

Step 1: Defining the levels of structure
1. Primary structure: Refers to the unique, linear sequence of amino acids in a polypeptide chain. Any change in this sequence results in a different protein.
2. Secondary structure: Refers to local folding patterns like \( \alpha \)-helices and \( \beta \)-pleated sheets held by hydrogen bonds.
3. Tertiary structure: Refers to the overall three-dimensional shape of a single polypeptide chain.
4. Quaternary structure: Refers to the spatial arrangement and interaction of multiple polypeptide subunits.

Step 2: Conclusion
Since the question specifically asks for the "specific sequence of amino acids," it corresponds to the primary structure.

Concept:

• For a first-order reaction, the rate constant \( (k) \) is independent of the initial concentration of the reactant.
• The relationship between the half-life \( (t_{1/2}) \) and the rate constant \( (k) \) is given by a specific mathematical expression derived from the integrated rate law.

Step 1: Recall the formula for the rate constant of a first-order reaction
The relationship is:
\[ k = \frac{0.693}{t_{1/2}} \]

Step 2: Substitute the given values into the formula
Given \( t_{1/2} = 1386 \text{ s} \).
\[ k = \frac{0.693}{1386} \]

Step 3: Perform the calculation
\[ \begin{aligned} k &= \frac{693 \times 10^{-3}}{1386}
k &= \frac{1}{2} \times 10^{-3}
k &= 0.5 \times 10^{-3}
k &= 5.0 \times 10^{-4} \text{ s}^{-1} \end{aligned} \]
The final answer is (B).

Concept:

• A chelate complex is formed when a polydentate (didentate or higher) ligand binds to a single metal ion through two or more donor atoms.
• This process results in the formation of a ring structure, which provides extra stability to the complex (the chelate effect).

Step 1: Analyze the denticity of the given ligands
1. Ammonia \( (\text{NH}_3) \): Monodentate ligand (binds through one N atom).
2. Water \( (\text{H}_2\text{O}) \): Monodentate ligand (binds through one O atom).
3. Nitrite \( (\text{NO}_2^-) \): Monodentate ligand (though ambidentate, it binds through one atom at a time).

Step 2: Identify the polydentate ligand
The Oxalate ion \( (\text{C}_2\text{O}_4^{2-}) \) is a didentate ligand.
It has two oxygen donor atoms that can coordinate simultaneously to the same metal ion.

Step 3: Conclusion on chelate formation
Because the oxalate ion can form a five-membered ring with the metal center, it forms a chelate complex.
The final answer is (D).

Concept:

• Acidity of phenols depends on the stability of the phenoxide ion formed after losing a proton.
• Electron-withdrawing groups (EWG) increase acidity by stabilizing the negative charge on the oxygen through inductive and resonance effects.

Step 1: Evaluate the Assertion
p-Nitrophenol has a nitro group \( (-\text{NO}_2) \) at the para position.
The nitro group is a strong electron-withdrawing group.
It facilitates the release of the proton and stabilizes the resulting phenoxide ion better than phenol itself.
Thus, p-nitrophenol is more acidic. Assertion is True.

Step 2: Evaluate the Reason
The nitro group exerts both \( -\text{I} \) and \( -\text{M} \) effects.
At the para position, the \( -\text{M} \) effect effectively delocalizes the negative charge of the phenoxide ion.
This dispersal of charge leads to higher stability. Reason is True.

Step 3: Check for logical connection
The reason specifically explains why the nitro group makes the molecule more acidic (via stabilization of the ion).
Therefore, the reason is the correct explanation.
The final answer is (A).

Concept:

• Fehling’s test is used to distinguish aldehydes from ketones.
• Aldehydes act as reducing agents in this reaction.

Step 1: Evaluate the Assertion
Aliphatic aldehydes are easily oxidized to carboxylic acids by weak oxidizing agents like Fehling’s solution.
This results in the formation of a red precipitate of \( \text{Cu}_2\text{O} \).
Assertion is True.

Step 2: Evaluate the Reason
In Fehling’s test, the aldehyde is oxidized to a carboxylate ion.
The \( \text{Cu}^{2+} \) ions in the Fehling’s reagent are reduced to \( \text{Cu}^+ \).
The reason states aldehydes are *reduced*, which is chemically incorrect.
Reason is False.

Step 3: Conclusion
Since the assertion is true and the reason is false, the correct code is C.
The final answer is (C).

Concept:

• Carbohydrate nomenclature uses specific prefixes to denote optical activity and relative configuration.
• D/L refers to the spatial arrangement (configuration) relative to glyceraldehyde.
• (+)/(-) refers to the direction in which the molecule rotates plane-polarized light (dextro or levo).

Step 1: Evaluate the Assertion
Naturally occurring Glucose rotates the plane of polarized light to the right.
It is indeed dextrorotatory, denoted as (+)-glucose.
Assertion is True.

Step 2: Evaluate the Reason
In nomenclature, the ’+’ symbol indicates dextrorotation.
The ’D’ prefix signifies that the hydroxyl group on the lowest chiral carbon is on the right side in the Fischer projection.
Reason is True.

Step 3: Check for logical connection
The configuration (D) does not dictate or explain why the molecule is dextrorotatory (+).
There is no direct causal link between relative configuration and the direction of optical rotation.
Therefore, the reason is a true statement about nomenclature but not a physical explanation for the assertion.
The final answer is (B).

Concept:

• Transition elements show variable oxidation states because of the small energy difference between \( (n-1)d \) and \( ns \) orbitals.
• The maximum oxidation state for a transition metal is often determined by the total number of \( ns \) and \( (n-1)d \) electrons.

Step 1: Analyze the Assertion
Manganese \( (\text{Mn}) \) has the electronic configuration \( [ \text{Ar} ] 3d^5 4s^2 \).
It can lose or share all 7 valence electrons (5 from \( 3d \) and 2 from \( 4s \)).
This results in a maximum oxidation state of +7, which is indeed the highest in the \( 3d \) transition series.
Assertion is True.

Step 2: Analyze the Reason
Transition metals exhibit variable oxidation states because electrons from both \( ns \) and \( (n-1)d \) subshells can participate in bonding.
Reason is True.

Step 3: Check for logical connection
While the reason is a general characteristic of transition metals, it does not specifically explain *why* Mn reaches +7.
The specific explanation for Mn being +7 is that it has the maximum number of unpaired electrons in the \( 3d \) subshell along with \( 4s \) electrons available for bonding.
The final answer is (B).

Concept:

• Sucrose is a disaccharide (\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)) commonly known as table sugar.
• Hydrolysis involves the breaking of a chemical bond by the addition of a water molecule, usually assisted by an acid catalyst or an enzyme.

Step 1: Understand the composition of the Sucrose molecule
Sucrose is composed of two monosaccharide units: one \( \alpha \text{-D-glucose} \) unit and one \( \beta \text{-D-fructose} \) unit.
These two units are joined together by a glycosidic linkage between the \( \text{C1} \) of the glucose ring and the \( \text{C2} \) of the fructose ring.

Step 2: Describe the hydrolysis process
When sucrose is heated with dilute acids (like \( \text{HCl} \)) or treated with the enzyme invertase, the glycosidic bond is hydrolyzed.
One molecule of water is consumed to restore the hydroxyl groups on the individual sugar rings.
\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+ / \text{Invertase}} \text{C}_6\text{H}_{12}\text{O}_6 \text{ (D-Glucose)} + \text{C}_6\text{H}_{12}\text{O}_6 \text{ (D-Fructose)} \]

Step 3: Explain the "Inversion" property
Sucrose is dextrorotatory \( (+66.5^\circ) \). Upon hydrolysis, it produces dextrorotatory glucose \( (+52.5^\circ) \) and strongly levorotatory fructose \( (-92.4^\circ) \).
Since the levorotation of fructose is greater than the dextrorotation of glucose, the final mixture is levorotatory. This change in sign of rotation is called the inversion of sugar.

Final Answer: The hydrolysis of sucrose yields an equimolar mixture of D-glucose and D-fructose.

Concept:

• Amino acids are the building blocks of proteins.
• They are classified into essential and non-essential based on the biological necessity of consuming them via food.

Step 1: Define Essential Amino Acids
Essential amino acids are those that the human body cannot produce on its own or cannot produce in sufficient quantities for growth and repair.
Because they are vital for biological functions but are not synthesized by internal metabolic pathways, they must be obtained through dietary sources.

Step 2: Compare with Non-essential Amino Acids
In contrast, non-essential amino acids are those that the body can synthesize itself from other nitrogen sources and carbon skeletons.
Therefore, it is not mandatory to get them specifically from the food we eat.

Step 3: Provide examples and significance
There are 10 essential amino acids for humans: Valine, Leucine, Isoleucine, Phenylalanine, Methionine, Threonine, Tryptophan, Lysine, Histidine, and Arginine.
Deficiency in any of these can lead to protein-energy malnutrition and other health disorders.

Final Answer: Essential amino acids are amino acids that the human body cannot synthesize and thus must be taken as part of the diet.

Concept:

• Low-current electronic devices require small, compact batteries that provide a very stable voltage over their entire lifespan.
• Primary cells are used where recharging is not practical or required.

Step 1: Identify the cell type and components
The cell used in small devices like hearing aids and watches is the Mercury cell.
It consists of a Zinc-Mercury amalgam as the anode and a paste of \( \text{HgO} \) and carbon as the cathode.
The electrolyte used is a paste of \( \text{KOH} \) and \( \text{ZnO} \).

Step 2: Explain the chemistry of the cell
The electrode reactions are:
Anode: \( \text{Zn(Hg)} + 2\text{OH}^- \rightarrow \text{ZnO(s)} + \text{H}_2\text{O} + 2e^- \)
Cathode: \( \text{HgO(s)} + \text{H}_2\text{O} + 2e^- \rightarrow \text{Hg(l)} + 2\text{OH}^- \)
The overall reaction is: \( \text{Zn(Hg)} + \text{HgO(s)} \rightarrow \text{ZnO(s)} + \text{Hg(l)} \).

Step 3: Discuss the advantage of using this cell
The cell potential remains constant at approximately \( 1.35 \text{ V} \) throughout its life.
The reason for this stability is that the overall reaction does not involve any ions in solution whose concentration can change over time.

Final Answer: The Mercury cell is used in low-current devices like hearing aids and watches.

Concept:

• Conductivity \( (\kappa) \) is the measure of the ease with which electric current flows through a material.
• Measuring the resistance of electrolytes involves dealing with mobile ions rather than electrons in a metal.

Step 1: Define Conductivity
Conductivity (or specific conductance) is defined as the reciprocal of resistivity \( (\rho) \).
It is numerically equal to the conductance of a solution of 1 cm length and 1 \( \text{cm}^2 \) cross-sectional area.
Formula: \( \kappa = \frac{1}{\rho} = \frac{1}{R} \times \left( \frac{l}{A} \right) \), where \( \frac{l}{A} \) is the cell constant.

Step 2: Explain the problems with using Direct Current (DC)
If a DC source is used in a Wheatstone bridge to measure the resistance of an ionic solution:
1. It causes electrolysis of the solution, which leads to a continuous change in the concentration of the electrolyte.
2. It causes "polarization" of the electrodes, where the products of electrolysis accumulate at the electrode surface, creating a back-EMF and a high resistance layer.

Step 3: Identify the standard laboratory solution
To overcome these issues, Alternating Current (AC) with a frequency of 500 to 5000 cycles per second is used.
AC prevents net chemical changes and polarization by constantly reversing the direction of ion movement.

Final Answer: Conductivity is the reciprocal of resistivity. DC is not used because it causes electrolysis and polarization, changing the solution’s composition during measurement.

Concept:

• Diazonium salts undergo coupling reactions with strongly activated aromatic rings like those of amines and phenols.
• This is a type of electrophilic aromatic substitution reaction.

Step 1: Identify the nature of the reaction
Benzenediazonium chloride acts as an electrophile due to the positive charge on the diazonium group.
Aniline \( (\text{C}_6\text{H}_5\text{NH}_2) \) is a highly activated nucleophilic aromatic compound.

Step 2: Determine the substitution position
The amino group \( (-\text{NH}_2) \) in aniline is highly activating and directs the incoming electrophile to the para position.
The reaction occurs in a weakly acidic medium \( (\text{pH } 4-5) \).

Step 3: Predict the product
The diazonium ion couples with the aniline molecule at the para position, eliminating a molecule of \( \text{HCl} \).
The product is \( p \text{-aminoazobenzene} \), which is a characteristic yellow dye.
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{H}-\text{C}_6\text{H}_4\text{NH}_2 \rightarrow \text{C}_6\text{H}_5-\text{N=N}-\text{C}_6\text{H}_4\text{-NH}_2 + \text{HCl} \]

Final Answer: The product is p-aminoazobenzene, a yellow azo dye.

Concept:

• Direct substitution of a halogen with an amino group is difficult in aryl halides.
• The transformation requires introducing a nitrogen group via electrophilic substitution and then reducing it.

Step 1: Nitration of Chlorobenzene
Chlorobenzene is treated with a mixture of concentrated \( \text{HNO}_3 \) and concentrated \( \text{H}_2\text{SO}_4 \).
Since Chlorine is ortho/para directing, nitration yields a mixture of \( o \text{-chloronitrobenzene} \) and \( p \text{-chloronitrobenzene} \).
The para isomer is isolated as the major product due to less steric hindrance.
\[ \text{C}_6\text{H}_5\text{Cl} \xrightarrow{\text{conc. HNO}_3/\text{H}_2\text{SO}_4} p\text{-Cl-C}_6\text{H}_4\text{-NO}_2 \]

Step 2: Reduction of the nitro group
The isolated \( p \text{-chloronitrobenzene} \) is then reduced to the corresponding amine.
Common laboratory reagents for this are \( \text{Fe / HCl} \), \( \text{Sn / HCl} \), or \( \text{H}_2 / \text{Pd} \).
\[ p\text{-Cl-C}_6\text{H}_4\text{-NO}_2 \xrightarrow{\text{Fe/HCl}} p\text{-Cl-C}_6\text{H}_4\text{-NH}_2 \]

Step 3: Conclusion
This two-step sequence successfully converts the hydrogen atom at the para position of the chlorobenzene ring into an amino group.

Final Answer: Chlorobenzene is first nitrated to get p-chloronitrobenzene, which is then reduced with Fe/HCl to obtain p-chloroaniline.

Concept:

• Elevation in boiling point \( (\Delta T_b) \) is a colligative property.
• For a solute that dissociates or associates, the relationship is \( \Delta T_b = i \cdot K_b \cdot m \), where \( i \) is the van’t Hoff factor.

Step 1: Calculate the observed elevation in boiling point
Normal boiling point of pure water \( T_b^\circ = 100^\circ\text{C} \).
Boiling point of the solution \( T_b = 100.18^\circ\text{C} \).
\[ \Delta T_b = T_b - T_b^\circ = 100.18 - 100 = 0.18^\circ\text{C} = 0.18 \text{ K} \]

Step 2: Set up the mathematical equation
Given:
Molality \( (m) = 1.00 \text{ m} \)
Ebullioscopic constant \( (K_b) = 0.512 \text{ K kg mol}^{-1} \)
Using the formula: \( \Delta T_b = i \cdot K_b \cdot m \)
\[ 0.18 = i \cdot (0.512) \cdot (1.00) \]

Step 3: Calculate the van’t Hoff factor \( (i) \)
\[ i = \frac{0.18}{0.512 \times 1.00} \]
\[ i = \frac{0.18}{0.512} \approx 0.3515 \]

Final Answer: The van’t Hoff factor for trichloroacetic acid under these conditions is approximately 0.35.

Concept:

• Henry’s Law relates the partial pressure of a gas to its solubility (mole fraction) in a liquid.
• Statement: The partial pressure of the gas in vapor phase \( (p) \) is proportional to the mole fraction of the gas \( (x) \) in the solution.

Step 1: State the law and formula
Henry’s Law is expressed as: \( p = K_H \cdot x \).
Where \( p \) is the partial pressure of the gas above the liquid, \( x \) is its mole fraction in the solution, and \( K_H \) is the Henry’s Law constant.

Step 2: Identify and check units of given data
Pressure \( (p) = 760 \text{ mm Hg} \).
Henry’s Law constant \( (K_H) = 1.25 \times 10^6 \text{ mm Hg} \).
Since both are in ’mm Hg’, we can substitute them directly into the formula.

Step 3: Perform the calculation
\[ x = \frac{p}{K_H} \]
\[ x = \frac{760}{1.25 \times 10^6} \]
\[ x = \frac{760}{1250000} \]
\[ x = 0.000608 = 6.08 \times 10^{-4} \]

Final Answer: The mole fraction of \( \text{CO}_2 \) in water is \( 6.08 \times 10^{-4} \).

Concept:

• Vitamins are essential organic compounds required in small quantities for the normal growth and maintenance of the body.
• Vitamin A, also known as Retinol, is a fat-soluble vitamin that plays a critical role in maintaining healthy vision, skin, and immune function.

Step 1: Identify the biological role of Vitamin A
Vitamin A is a precursor to rhodopsin, the photopigment found in the rods of the retina.
Rhodopsin is necessary for seeing in low-light conditions (scotopic vision).

Step 2: Determine the effect of deficiency
When there is a lack of Vitamin A, the production of rhodopsin is impaired.
This leads to a condition called Night Blindness (Nyctalopia), where the individual cannot see clearly in the dark or in dim light.

Step 3: Identify clinical conditions
Prolonged deficiency leads to Xerophthalmia, characterized by extreme dryness of the conjunctiva and cornea.
If untreated, it can lead to corneal ulceration and permanent blindness.

Final Answer: The disease caused by the deficiency of Vitamin A is Night Blindness or Xerophthalmia.

Concept:

• Glucose contains a carbonyl group. This group could potentially be an aldehyde or a ketone.
• Aldehydes are much easier to oxidize than ketones. Mild oxidizing agents can be used to distinguish between them.

Step 1: Choose the specific reagent for aldehyde confirmation
To confirm that the carbonyl group is specifically an aldehyde (and not a ketone), we use a mild oxidizing agent like bromine water \( (\text{Br}_2 / \text{H}_2\text{O}) \).
Bromine water is not strong enough to oxidize ketones or secondary alcohols under normal conditions.

Step 2: Analyze the chemical reaction
When glucose is treated with bromine water, the aldehyde group at the C1 position is oxidized to a carboxylic acid group.
This results in the formation of gluconic acid.
\[ \text{CHO-(CHOH)}_4\text{-CH}_2\text{OH} + [\text{O}] \xrightarrow{\text{Br}_2 / \text{H}_2\text{O}} \text{COOH-(CHOH)}_4\text{-CH}_2\text{OH} \]

Step 3: Conclusion
The successful oxidation of glucose to a six-carbon monocarboxylic acid (gluconic acid) by a mild reagent confirms that the carbonyl group is an aldehyde.

Final Answer: The presence of an aldehydic group in glucose is confirmed by its reaction with bromine water to yield gluconic acid.

Concept:

• The emf of a cell under non-standard conditions is calculated using the Nernst Equation.
• The overall cell reaction must be determined to identify the number of electrons \( (n) \) transferred.

Step 1: Identify the half-reactions and determine n
Anode (Oxidation): \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2e^- \)
Cathode (Reduction): \( 2\text{Ag}^+\text{(aq)} + 2e^- \rightarrow 2\text{Ag(s)} \)
Overall reaction: \( \text{Zn(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{Ag(s)} \)
Number of electrons transferred, \( n = 2 \).

Step 2: Calculate the standard cell potential \( (E^\circ_{\text{cell}}) \)
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = 0.80 \text{ V} - (-0.76 \text{ V}) = 1.56 \text{ V} \]

Step 3: Apply the Nernst Equation
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} \right) \]
Substituting the given values:
\[ E_{\text{cell}} = 1.56 - \frac{0.059}{2} \log \left( \frac{0.1}{(0.01)^2} \right) \]
\[ E_{\text{cell}} = 1.56 - 0.0295 \log \left( \frac{10^{-1}}{10^{-4}} \right) \]
\[ E_{\text{cell}} = 1.56 - 0.0295 \log(10^3) = 1.56 - (0.0295 \times 3) \]
\[ E_{\text{cell}} = 1.56 - 0.0885 = 1.4715 \text{ V} \]

Final Answer: The emf of the cell is 1.4715 V.

Concept:

• The rate of a chemical reaction is expressed mathematically as a function of the concentrations of the reactants.
• This expression is known as the Rate Law.

Step 1: Understanding the Rate Law expression

For a general reaction: \( a\text{A} + b\text{B} \rightarrow \text{Products} \)

The rate law is determined experimentally as: \( \text{Rate} = k[\text{A}]^x[\text{B}]^y \)

Here, \( k \) is the rate constant.

Step 2: Defining the order

The powers \( x \) and \( y \) represent the order of the reaction with respect to reactants A and B, respectively.

The overall order of the reaction is defined as the sum of these individual exponents: \( n = x + y \).

Step 3: Characteristics of reaction order

The order of a reaction can be zero, whole numbers, or even fractions.

It is an experimental quantity and cannot be predicted simply by looking at the balanced chemical equation.

Final Answer: The order of a reaction is the sum of the powers of the concentration terms of the reactants in the rate law expression.

Concept:

• The dependence of rate on a specific reactant is defined by the power to which its concentration is raised in the rate law.
• In this case, the reaction is second order with respect to B.

Step 1: Set up the initial rate expression
Let the initial rate be \( R_1 \):
\[ R_1 = k[\text{A}][\text{B}]^2 \]

Step 2: Apply the change to concentration
If the concentration of B is doubled, the new concentration becomes \( [\text{B}]' = 2[\text{B}] \).
The new rate \( R_2 \) will be:
\[ R_2 = k[\text{A}](2[\text{B}])^2 \]

Step 3: Calculate the ratio of rates
\[ R_2 = k[\text{A}](4[\text{B}]^2) = 4(k[\text{A}][\text{B}]^2) \]
\[ R_2 = 4R_1 \]

Final Answer: Doubling the concentration of B makes the reaction rate increase by a factor of 4.

Concept:

• When a reactant is present in large excess, its concentration effectively remains constant throughout the reaction.
• This leads to a simplified rate law where the order with respect to the excess reactant becomes effectively zero.

Step 1: Analyze the rate law with excess conditions
The given rate law is: \( \text{Rate} = k[\text{A}][\text{B}]^2 \).
If ’A’ is in large excess, then \( [\text{A}] \approx \text{constant} \).

Step 2: Simplify the rate law
We can combine the constant concentration of A with the rate constant \( k \) to form a new effective rate constant \( k' \).
\[ \text{Rate} = (k \cdot [\text{A}]) \cdot [\text{B}]^2 = k'[\text{B}]^2 \]

Step 3: Identify the new overall order
The rate now depends only on the square of the concentration of B.
Therefore, the overall order of the reaction is 2.

Final Answer: If A is in large excess, the overall order of the reaction is 2.

Concept:

• The temperature dependence of the rate constant is given by the Arrhenius Equation.
• For two different temperatures, the integrated form of the Arrhenius equation is used.

Step 1: Identify the variables and the formula
Given:
\( T_1 = 300 \text{ K}, T_2 = 320 \text{ K} \).
Since the rate quadruples, \( k_2/k_1 = 4 \).
\( R = 8.314 \text{ JK}^{-1} \text{ mol}^{-1} \).
The formula is:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]

Step 2: Substitute values into the equation
\[ \log(4) = \frac{E_a}{2.303 \times 8.314} \left( \frac{320 - 300}{320 \times 300} \right) \]
\[ 0.60 = \frac{E_a}{19.147} \left( \frac{20}{96000} \right) \]
\[ 0.60 = \frac{E_a}{19.147} \left( \frac{1}{4800} \right) \]

Step 3: Solve for \( E_a \)
\[ E_a = 0.60 \times 19.147 \times 4800 \]
\[ E_a = 11.4882 \times 4800 = 55143.36 \text{ J/mol} \]
To convert to kJ/mol: \( E_a = 55.14 \text{ kJ/mol} \).

Final Answer: The energy of activation for the reaction is 55.14 kJ/mol.

Concept:

• IUPAC naming of coordination compounds follows specific rules for cations, anions, ligands, and central metals.
• If the complex part is an anion, the metal name ends in the suffix ’-ate’.

Step 1: Identify the cation and the anionic complex

The cation is Potassium \( (\text{K}^+) \).

The complex part is the anion: \( [\text{Zn(OH)}_4]^{2-} \).

Step 2: Determine the oxidation state of the central metal Zinc

Let the oxidation state of Zn be \( x \).

Hydroxyl ligand \( (\text{OH}^-) \) has a charge of -1.

In the complex anion: \( x + 4(-1) = -2 \)
\( x - 4 = -2 \Rightarrow x = +2 \).

Step 3: Assemble the name

Name the cation first: Potassium.

In the anion, name the ligands in alphabetical order with prefix: tetrahydroxo.

Name the metal with the suffix ’-ate’ (since it is in an anion) followed by the oxidation state in Roman numerals: zincate(II).

Full name: Potassium tetrahydroxozincate(II).

Final Answer: The IUPAC name is Potassium tetrahydroxozincate(II).

Concept:

• Both double salts and complexes are formed by the combination of two or more stable salts.
• The fundamental difference lies in their behavior when dissolved in a solvent like water.

Step 1: Define and describe Double Salts
Double salts exist only in the solid state.
When dissolved in water, they completely dissociate into their constituent simple ions.
They give the test for all the ions they are made of. Example: Mohr’s Salt \( [\text{FeSO}_4 \cdot (\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O}] \).

Step 2: Define and describe Coordination Complexes
Complexes (coordination compounds) retain their chemical identity even in solution.
The central metal and the ligands are bonded by coordinate bonds and remain together as a complex ion when dissolved.
They do not give the individual tests for the metal or the ligands. Example: \( \text{K}_4[\text{Fe(CN)}_6] \).

Step 3: Summary table of differences
1. Bonding: Double salts have ionic bonds; complexes have coordinate covalent bonds.
2. Dissociation: Double salts dissociate into simple ions; complexes dissociate into a counter-ion and a complex ion.

Final Answer: Double salts dissociate into simple ions in aqueous solution, while complexes do not dissociate into their constituent parts and maintain their identity.

Concept:

• Isomerism in coordination compounds involves different spatial arrangements of ligands around the central metal.
• For square planar complexes of the type \( [\text{MA}_2\text{BC}] \), different arrangements of similar ligands relative to each other are possible.

Step 1: Determine the geometry of the complex
Platinum(II) complexes with coordination number 4 are typically square planar.
The formula \( [\text{Pt(NH}_3\text{)(H}_2\text{O)Cl}_2] \) is of the type \( [\text{MA}_2\text{BC}] \), where A = Cl.

Step 2: Identify the possible isomers
Because there are two identical ligands (Cl), they can be positioned differently:
1. Cis isomer: The two chlorine ligands are adjacent to each other.
2. Trans isomer: The two chlorine ligands are opposite to each other.

Step 3: Classification
This type of isomerism, based on the spatial orientation of ligands, is called Geometrical Isomerism.

Final Answer: The complex exhibits Geometrical Isomerism (cis and trans isomers).

Concept:

• According to Werner’s theory, ionizable groups (primary valence) are those located outside the coordination sphere.
• Only the chloride ions present outside the coordination sphere react with silver nitrate to form a precipitate of silver chloride (\( \text{AgCl} \)).

Step 1: Determine the number of ionizable chloride ions
The problem states that 3 moles of \( \text{AgCl} \) are precipitated per mole of the complex.
This directly implies that there are 3 chloride (\( \text{Cl}^- \)) ions present outside the coordination bracket.

Step 2: Assemble the coordination sphere
The total formula given is \( \text{CrCl}_3 \cdot 6\text{H}_2\text{O} \).
If all 3 chlorine atoms are outside, the 6 water molecules must be ligands inside the coordination sphere to satisfy the typical coordination number of 6 for Chromium(III).

Step 3: Write the final structural formula
The metal \( \text{Cr} \) is the central atom, the 6 \( \text{H}_2\text{O} \) molecules are ligands, and 3 \( \text{Cl} \) atoms are counter-ions.
Formula: \( [\text{Cr(H}_2\text{O)}_6]\text{Cl}_3 \).

Final Answer: The structural formula of the compound is \( [\text{Cr(H}_2\text{O)}_6]\text{Cl}_3 \).

Concept:

• Oxidation state is calculated by the sum of charges of metal and ligands equal to the total complex charge.
• Valence Bond Theory (VBT) helps determine the hybridisation based on the electronic configuration of the metal ion.

Step 1: Calculate the oxidation state
Let the oxidation state of Cr be \( x \).
\( \text{NH}_3 \) is a neutral ligand (charge = 0).
\( x + 6(0) = +3 \Rightarrow x = +3 \).

Step 2: Determine electronic configuration
Neutral \( \text{Cr (Z=24)} \): \( [\text{Ar}] 3d^5 4s^1 \).
\( \text{Cr}^{3+} \) ion: \( [\text{Ar}] 3d^3 4s^0 4p^0 \).

Step 3: Identify available orbitals for hybridisation
In \( \text{Cr}^{3+} \), the three electrons occupy three \( 3d \) orbitals (\( t_{2g} \)).
Two \( 3d \) orbitals are empty, along with one \( 4s \) and three \( 4p \) orbitals.
These six orbitals (\( 2 \times 3d, 1 \times 4s, 3 \times 4p \)) hybridise to form six \( d^2sp^3 \) hybrid orbitals.

Final Answer: The oxidation state of Cr is +3 and its hybridisation is \( d^2sp^3 \) (Inner orbital octahedral complex).

Concept:

• Colour in transition metal complexes arises due to \( d-d \) transitions.
• A \( d-d \) transition occurs when an electron from a lower energy \( d \)-orbital is excited to a higher energy \( d \)-orbital after absorbing visible light.

Step 1: Determine the oxidation state of Scandium

In \( [\text{Sc(H}_2\text{O)}_6]^{3+} \), water is neutral, so \( \text{Sc} \) is in the +3 oxidation state.

Step 2: Examine the electronic configuration

Neutral \( \text{Sc (Z=21)} \): \( [\text{Ar}] 3d^1 4s^2 \).
\( \text{Sc}^{3+} \): \( [\text{Ar}] 3d^0 4s^0 \).

Step 3: Conclude based on electron availability

The \( \text{Sc}^{3+} \) ion has an empty \( 3d \) subshell (\( 3d^0 \)).
Since there are no electrons in the \( d \)-orbitals, \( d-d \) transitions are impossible. Therefore, the complex does not absorb visible light and appears colourless.

Final Answer: The complex is colourless because \( \text{Sc}^{3+} \) has a \( d^0 \) configuration, which prevents any \( d-d \) electronic transitions.

Concept:

• Direct nitration of benzene gives nitrobenzene, which is difficult to acetylate.
• Acetylation of benzene gives acetophenone. The acetyl group is a meta-directing deactivating group.

Step 1: Friedel-Crafts Acetylation
React Benzene with Acetyl chloride \( (\text{CH}_3\text{COCl}) \) in the presence of anhydrous \( \text{AlCl}_3 \). This produces Acetophenone.
\( \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{anh. AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \)

Step 2: Nitration
Treat Acetophenone with a nitrating mixture (concentrated \( \text{HNO}_3 \) and concentrated \( \text{H}_2\text{SO}_4 \)).
Since the \( -\text{COCH}_3 \) group is meta-directing, the nitro group attaches to the meta position.
\( \text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{conc. HNO}_3 / \text{H}_2\text{SO}_4} m\text{-NO}_2\text{-C}_6\text{H}_4\text{COCH}_3 \)

Final Answer: Benzene \( \xrightarrow{\text{CH}_3\text{COCl / AlCl}_3} \) Acetophenone \( \xrightarrow{\text{HNO}_3 / \text{H}_2\text{SO}_4} \) m-Nitroacetophenone.

Concept:

• Grignard reagents (\( \text{RMgX} \)) react with aldehydes to form secondary alcohols.
• 1-Phenylethanol is a secondary alcohol with a phenyl and a methyl group on the alpha carbon.

Step 1: Formation of Phenylmagnesium bromide
React Bromobenzene with Magnesium metal in dry ether.
\( \text{C}_6\text{H}_5\text{Br} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{MgBr} \)

Step 2: Reaction with Acetaldehyde
The Grignard reagent attacks the carbonyl carbon of acetaldehyde (\( \text{CH}_3\text{CHO} \)).
\( \text{C}_6\text{H}_5\text{MgBr} + \text{CH}_3\text{CHO} \rightarrow \text{CH}_3\text{CH(OMgBr)C}_6\text{H}_5 \)

Step 3: Hydrolysis
Acidic hydrolysis of the adduct yields the final alcohol.
\( \text{CH}_3\text{CH(OMgBr)C}_6\text{H}_5 \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH(OH)C}_6\text{H}_5 \)

Final Answer: Bromobenzene \( \xrightarrow{\text{Mg/ether}} \text{PhMgBr} \xrightarrow{\text{1. CH}_3\text{CHO, 2. H}_3\text{O}^+} \) 1-Phenylethanol.

Concept:

• The transformation requires the conversion of a methyl ketone group \( (-\text{COCH}_3) \) into a carboxylic acid group \( (-\text{COOH}) \).
• A major constraint is that the starting material contains a carbon-carbon double bond \( (\text{C=C}) \), which must remain intact.
• Standard strong oxidizing agents like \( \text{KMnO}_4 \) or \( \text{K}_2\text{Cr}_2\text{O}_7 \) are unsuitable because they would oxidatively cleave the alkene.
• The Haloform reaction is a specific oxidation for methyl ketones that does not affect isolated carbon-carbon double bonds.

Step 1: Identification of the reactant and required change

But-3-en-2-one has the structure: \( \text{CH}_2\text{=CH}-\text{CO}-\text{CH}_3 \).

The target molecule, Prop-2-en-1-oic acid (Acrylic acid), has the structure: \( \text{CH}_2\text{=CH}-\text{COOH} \).

We need to remove the methyl group attached to the carbonyl and replace the resulting system with a hydroxyl group.

Step 2: Application of the Haloform Reaction

The methyl ketone is treated with iodine \( (\text{I}_2) \) in the presence of sodium hydroxide \( (\text{NaOH}) \). This generates sodium hypoiodite \( (\text{NaOI}) \) in situ.

The hypoiodite ion specifically oxidizes the methyl group of the ketone to form a triiodomethyl intermediate, which is then cleaved by the base to form a carboxylate salt and iodoform \( (\text{CHI}_3) \).

\[ \text{CH}_2\text{=CH}-\text{CO}-\text{CH}_3 + 3\text{NaOI} \rightarrow \text{CH}_2\text{=CH}-\text{COONa} + \text{CHI}_3 \downarrow + 2\text{NaOH} \]

Step 3: Acidification to yield the final acid

The resulting sodium salt (sodium acrylate) is then treated with a dilute mineral acid (like \( \text{HCl} \)) to liberate the free carboxylic acid.

\[ \text{CH}_2\text{=CH}-\text{COONa} + \text{HCl} \rightarrow \text{CH}_2\text{=CH}-\text{COOH} + \text{NaCl} \]

Final Answer: But-3-en-2-one is converted to Prop-2-en-1-oic acid by reacting it with \( \text{I}_2/\text{NaOH} \) (Haloform reaction) to produce the sodium salt, followed by acidification with dilute \( \text{HCl} \).

Concept:

• Acidic strength of carboxylic acids depends on the stability of the resulting carboxylate ion.
• Electron-withdrawing groups (EWG) like halogens (\( -I \) effect) increase acidity by stabilizing the negative charge.
• Electron-donating groups (EDG) like alkyl groups (\( +I \) effect) decrease acidity by destabilizing the negative charge.
• The Inductive effect is distance-dependent; it weakens as the substituent moves further from the \( -\text{COOH} \) group.

Step 1: Evaluate the effect of alkyl vs halogen substituents
In \( \text{CH}_3\text{CH(CH}_3\text{)COOH} \) (2-methylpropanoic acid), the methyl group exerts a \( +I \) effect.
This increases electron density on the carboxylate carbon, destabilizing the anion. Thus, it is the weakest acid among the three.

Step 2: Compare the distance of the Bromine atom
In \( \text{CH}_3\text{CH(Br)CH}_2\text{COOH} \), the Bromine atom is at the \( \beta \)-carbon (3rd position).
In \( \text{CH}_3\text{CH}_2\text{CH(Br)COOH} \), the Bromine atom is at the \( \alpha \)-carbon (2nd position).

Step 3: Apply the distance rule of the Inductive effect
The \( -I \) effect of Bromine is stronger when it is closer to the carboxylic group.
Therefore, the \( \alpha \)-bromo acid stabilizes the carboxylate ion more effectively than the \( \beta \)-bromo acid.
Order: \( \text{CH}_3\text{CH(CH}_3\text{)COOH} < \text{CH}_3\text{CH(Br)CH}_2\text{COOH} < \text{CH}_3\text{CH}_2\text{CH(Br)COOH} \).

Final Answer: The increasing order of acidic strength is 2-methylpropanoic acid followed by 3-bromobutanoic acid, and 2-bromobutanoic acid is the strongest.

Concept:

• The reaction of \( \text{HCN} \) with carbonyl compounds is a Nucleophilic Addition reaction.
• The rate of reaction depends on how easily the nucleophile \( (\text{CN}^-) \) can attack the carbonyl carbon.
• This is governed by steric (space) and electronic (charge) factors.

Step 1: Analyze the Steric Factor
Acetaldehyde \( (\text{CH}_3\text{CHO}) \) has one small hydrogen atom and one methyl group attached to the carbonyl carbon.
Acetone \( (\text{CH}_3\text{COCH}_3) \) has two relatively bulky methyl groups.
The presence of two bulky groups in acetone hinders the approach of the \( \text{CN}^- \) nucleophile more than in acetaldehyde.

Step 2: Analyze the Electronic Factor
Alkyl groups (methyl) exert a \( +I \) (inductive) effect, which pushes electron density towards the carbonyl carbon.
Acetaldehyde has only one methyl group reducing the partial positive charge on the carbon.
Acetone has two methyl groups, which significantly reduce the electrophilicity (positive character) of the carbonyl carbon, making it less attractive to the nucleophile.

Step 3: Conclusion
Both the lesser steric hindrance and the greater electrophilicity of the carbonyl carbon make acetaldehyde more reactive than acetone.

Final Answer: Acetaldehyde is more reactive because it is less sterically hindered and has a more electrophilic carbonyl carbon compared to acetone.

Concept:

• This is a Nucleophilic Addition-Elimination reaction.
• Carbonyl compounds react with ammonia derivatives \( (\text{NH}_2-\text{Z}) \) to form compounds containing \( \text{C=N} \) bonds.
• In this case, the reagent is Hydrazine \( (\text{NH}_2\text{NH}_2) \).

Step 1: Identify the nucleophilic attack
The lone pair on the nitrogen of hydrazine attacks the carbonyl carbon of acetaldehyde.
This forms an unstable intermediate addition product.

Step 2: Dehydration (Elimination of water)
Under acidic catalysis, a molecule of water is eliminated from the intermediate.
The oxygen from the carbonyl group and two hydrogen atoms from the \( -\text{NH}_2 \) group are lost as \( \text{H}_2\text{O} \).

Step 3: Write the final product
The resulting product contains a double bond between the carbon and nitrogen.
\[ \text{CH}_3\text{CHO} + \text{NH}_2\text{NH}_2 \xrightarrow{\text{H}^+} \text{CH}_3\text{CH=N-NH}_2 + \text{H}_2\text{O} \]
The product is called acetaldehyde hydrazone.

Final Answer: The product of the reaction is acetaldehyde hydrazone.

Concept:

• Positive 2,4-DNP test \( \rightarrow \) Carbonyl group present.
• Reduces Tollens’ reagent \( \rightarrow \) Aldehyde group \( (-\text{CHO}) \) present.
• Cannizzaro reaction \( \rightarrow \) Aldehyde with no \( \alpha \)-hydrogens.
• Oxidation to Benzene-1,2-dicarboxylic acid \( \rightarrow \) Benzene ring with two side chains at ortho positions.

Step 1: Determine the structure from molecular formula and oxidation product
The formula is \( \text{C}_8\text{H}_8\text{O} \). Vigorous oxidation gives Phthalic acid (Benzene-1,2-dicarboxylic acid).
This means the benzene ring (\( \text{C}_6 \)) has two carbon-containing groups at the 1 and 2 positions.
One group is \( -\text{CHO} \) (from Tollens’ test). To total \( \text{C}_8 \), the other must be \( -\text{CH}_3 \).
The compound is 2-Methylbenzaldehyde. It has no \( \alpha \)-hydrogen (the carbon adjacent to \( \text{CHO} \) is part of the ring and is fully bonded).

Step 2: Reaction with 2,4-DNP
The aldehyde reacts with 2,4-dinitrophenylhydrazine to form a 2,4-DNP hydrazone.
\[ \text{CH}_3\text{C}_6\text{H}_4\text{CHO} + \text{H}_2\text{NNH-C}_6\text{H}_3(\text{NO}_2)_2 \rightarrow \text{CH}_3\text{C}_6\text{H}_4\text{CH=N-NH-C}_6\text{H}_3(\text{NO}_2)_2 + \text{H}_2\text{O} \]

Step 3: Cannizzaro Reaction
Reaction with conc. \( \text{NaOH} \) results in disproportionation.
\[ 2\text{CH}_3\text{C}_6\text{H}_4\text{CHO} \xrightarrow{\text{conc. NaOH}} \text{CH}_3\text{C}_6\text{H}_4\text{CH}_2\text{OH} + \text{CH}_3\text{C}_6\text{H}_4\text{COONa} \]
Products are 2-Methylbenzyl alcohol and Sodium 2-methylbenzoate.

Final Answer: The compound is 2-Methylbenzaldehyde. Its reactions yield a 2,4-DNP hydrazone and undergo disproportionation in the Cannizzaro reaction.

Concept:

• This is an example of the Friedel-Crafts Alkylation reaction.
• The methoxy group \( (-\text{OCH}_3) \) in anisole is an activating group that directs the incoming electrophile to the ortho and para positions.
• Activation occurs because the lone pair of electrons on the oxygen atom is in conjugation with the benzene ring, increasing electron density at these specific positions.

Step 1: Generation of the methyl electrophile
The reaction starts with the interaction between methyl chloride \( (\text{CH}_3\text{Cl}) \) and the Lewis acid catalyst, anhydrous aluminum chloride \( (\text{AlCl}_3) \).
The catalyst accepts a pair of electrons from the chlorine atom, generating a reactive methyl carbocation \( (\text{CH}_3^+) \).
\[ \text{CH}_3\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3^+ + \text{AlCl}_4^- \]

Step 2: Electrophilic attack and intermediate formation
The methyl carbocation attacks the electron-rich ortho and para carbons of the anisole ring.
The resulting carbocation intermediate (sigma complex) is stabilized by the electron-donating resonance effect of the methoxy group.

Step 3: Formation of the final products and steric considerations
Loss of a proton from the intermediate restores the aromaticity of the ring.
Substitution at the para position gives 4-Methoxytoluene, and at the ortho position gives 2-Methoxytoluene.
The para-isomer is the major product because it is less sterically hindered than the ortho-isomer, where the methoxy and methyl groups are in close proximity.

Concept:

• Nitration is an electrophilic aromatic substitution reaction.
• The "nitrating mixture" \( (\text{concentrated HNO}_3 \text{ and H}_2\text{SO}_4) \) is used to generate the active electrophile, the nitronium ion \( (\text{NO}_2^+) \).
• The methoxy group is an ortho/para directing group.

Step 1: Formation of the nitronium ion electrophile
Sulfuric acid, being a stronger acid than nitric acid, protonates the \( \text{HNO}_3 \) molecule.
Subsequent loss of water results in the formation of the highly reactive nitronium ion.
\[ \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- \]

Step 2: Mechanism of substitution
The \( \text{NO}_2^+ \) ion attacks the anisole ring at the ortho and para positions.
The presence of the methoxy group stabilizes the transition state specifically at these positions through resonance, making the reaction faster than it would be with benzene.

Step 3: Identifying the major and minor isomers
The reaction yields a mixture of 2-nitroanisole and 4-nitroanisole.
Consistent with the general trend for disubstituted benzenes, the para-isomer (4-nitroanisole) is the major product because there is less steric repulsion between the \( -\text{OCH}_3 \) and \( -\text{NO}_2 \) groups.

Concept:

• This specific ether synthesis is performed using the Williamson ether synthesis method.
• The reaction is a nucleophilic substitution \( (\text{S}_{\text{N}}2) \) between an alkoxide ion \( (\text{RO}^-) \) and an alkyl halide \( (\text{R'X}) \).
• For a successful \( \text{S}_{\text{N}}2 \) reaction, the alkyl halide must be primary to minimize steric hindrance and avoid elimination.

Step 1: Analyzing possible reactant pairs
To prepare tert-butyl ethyl ether \( (\text{CH}_3)_3\text{C-O-CH}_2\text{CH}_3 \), two combinations are possible:
Path 1: Sodium ethoxide \( (\text{CH}_3\text{CH}_2\text{ONa}) \) + tert-butyl bromide \( [(\text{CH}_3)_3\text{CBr}] \).
Path 2: Sodium tert-butoxide \( [(\text{CH}_3)_3\text{CONa}] \) + Ethyl bromide \( (\text{CH}_3\text{CH}_2\text{Br}) \).

Step 2: Evaluating Path 1 (Elimination competition)
In Path 1, the alkyl halide is tertiary. Alkoxide ions are strong bases.
With a tertiary halide, elimination \( (\text{E}2) \) occurs almost exclusively, yielding isobutylene (2-methylpropene) rather than the desired ether.

Step 3: Selecting the correct reactants (Path 2)
In Path 2, the alkyl halide is primary (Ethyl bromide).
Primary halides undergo \( \text{S}_{\text{N}}2 \) substitution with alkoxides with very high yields.
\[ (\text{CH}_3)_3\text{CO}^- + \text{CH}_3\text{CH}_2\text{Br} \rightarrow (\text{CH}_3)_3\text{C-O-CH}_2\text{CH}_3 + \text{Br}^- \]

Concept:

• Cleavage of ethers by hydrogen halides involves protonation of the ether followed by a nucleophilic attack.
• In alkyl aryl ethers, the oxygen atom is connected to one \( sp^3 \) carbon and one \( sp^2 \) carbon of the benzene ring.

Step 1: Analyze the resonance effect in Anisole
In anisole \( (\text{C}_6\text{H}_5\text{OCH}_3) \), the lone pair of electrons on the oxygen atom is delocalized into the benzene ring through resonance.
This gives the carbon (phenyl)-oxygen bond partial double bond character.

Step 2: Compare bond dissociation energies
Due to resonance and the higher \( s \)-character of the \( sp^2 \) hybridized carbon, the phenyl-oxygen bond is much shorter and stronger than the \( \text{O-CH}_3 \) bond.
The methyl-oxygen bond is a pure single bond and is easier to cleave.

Step 3: Predict the mechanism of attack
Upon treatment with \( \text{HI} \), the ether is protonated to form an oxonium ion.
The iodide ion \( (\text{I}^-) \) acts as a nucleophile and attacks the less hindered methyl group via an \( \text{S}_{\text{N}}2 \) mechanism.
It cannot attack the benzene ring carbon because of high electron density repulsion and the high energy required to break the resonance-stabilized bond.
\[ \text{C}_6\text{H}_5\text{OCH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{I} \]

Concept:

• The oxygen atom in ethers is \( sp^3 \) hybridized.
• Standard tetrahedral geometry has an angle of \( 109.5^\circ \).
• Lone pairs and the bulkiness of attached groups affect the actual bond angles in a molecule.

Step 1: Analyze the electronic environment of the oxygen atom
Like water, the oxygen atom in ethers has two lone pairs of electrons and two bond pairs.
According to VSEPR theory, lone pair-lone pair repulsions would normally tend to compress the bond angle to less than \( 109.5^\circ \).

Step 2: Evaluate the effect of bulky alkyl groups
Unlike water (which has small hydrogen atoms), ethers contain alkyl groups \( (\text{R-O-R}) \).
These alkyl groups are significantly bulkier than hydrogen atoms and occupy more space.

Step 3: Determine the dominant repulsive force
The two alkyl groups experience mutual steric repulsion (Van der Waals repulsion) because they are forced into close proximity.
This steric repulsion between the bulky alkyl groups is stronger than the compressive force of the lone pairs.
Consequently, the alkyl groups push each other further apart, resulting in a bond angle slightly larger than \( 109.5^\circ \) (e.g., \( \approx 111.7^\circ \) in dimethyl ether).

Concept:

• Electrolysis involves the migration of ions to electrodes where they undergo redox reactions.
• At the cathode, the species with the higher standard reduction potential is reduced.
• At the anode, the species with the lower standard reduction potential is oxidized, though kinetic factors like overvoltage can alter the outcome.

Step 1: Identify ions present in the aqueous solution
In aqueous \( \text{CuCl}_2 \), the ions present are \( \text{Cu}^{2+} \), \( \text{Cl}^- \), \( \text{H}^+ \), and \( \text{OH}^- \).
The electrodes are platinum (Pt), which are inert and do not participate in the reaction.

Step 2: Analyze the reaction at the Cathode (Reduction)
Competing reactions:
1. \( \text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu}(\text{s}) \quad E^\circ = +0.34 \text{ V} \)
2. \( 2\text{H}^+(\text{aq}) + 2e^- \rightarrow \text{H}_2(\text{g}) \quad E^\circ = 0.00 \text{ V} \)
Since \( E^\circ \) for copper is higher, \( \text{Cu}^{2+} \) is reduced to Cu metal.

Step 3: Analyze the reaction at the Anode (Oxidation)
Competing reactions:
1. \( 2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2e^- \quad E^\circ = -1.36 \text{ V} \)
2. \( 2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+ + 4e^- \quad E^\circ = -1.23 \text{ V} \)
Thermodynamically, water oxidation is preferred. However, due to the overvoltage of oxygen evolution, chloride oxidation is kinetically favored. Thus, \( \text{Cl}_2 \) gas is evolved.

Final Answer: Copper is deposited at the cathode and Chlorine gas is evolved at the anode.

Concept:

• In electrolysis, the concentration of the electrolyte can change the path of the reaction.
• Concentrated acids provide a high density of anionic species that may undergo oxidation instead of water.

Step 1: Analyze the reaction at the Cathode
The reduction of protons is the only major reduction process:
\[ 2\text{H}^+(\text{aq}) + 2e^- \rightarrow \text{H}_2(\text{g}) \]
Hydrogen gas is liberated at the cathode.

Step 2: Analyze the reaction at the Anode for concentrated solution
In dilute \( \text{H}_2\text{SO}_4 \), water is oxidized to \( \text{O}_2 \).
In concentrated \( \text{H}_2\text{SO}_4 \), the oxidation of sulphate/bisulphate ions is preferred at high current densities.
\[ 2\text{SO}_4^{2-}(\text{aq}) \rightarrow \text{S}_2\text{O}_8^{2-}(\text{aq}) + 2e^- \]

Step 3: Identify the chemical product
The peroxodisulphate ion reacts with \( \text{H}^+ \) to form peroxodisulphuric acid (\( \text{H}_2\text{S}_2\text{O}_8 \)), also known as Marshall’s acid.

Final Answer: The products are \( \text{H}_2 \) gas at the cathode and \( \text{H}_2\text{S}_2\text{O}_8 \) at the anode.

Concept:

• 1 Faraday (F) is the charge of 1 mole of electrons.
• The amount of charge required for a redox reaction is equal to the number of moles of electrons transferred multiplied by Faraday’s constant.

Step 1: Write the balanced half-reaction
The reduction of silver ions to silver metal is:
\[ \text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag}(\text{s}) \]

Step 2: Determine the number of moles of electrons
For the reduction of 1 mole of \( \text{Ag}^+ \), the equation shows that exactly 1 mole of electrons is required.

Step 3: Convert to Faraday unit
Since the charge of 1 mole of electrons is defined as 1 Faraday, the total charge required is 1 F.

Final Answer: 1 Faraday.

Concept:

• Faraday’s second law relates the amounts of different substances deposited by the same amount of electricity.

Step 1: Formulate the definition
Faraday’s second law states that when the same quantity of electricity is passed through different electrolytes connected in series, the masses of the substances liberated at the electrodes are directly proportional to their chemical equivalent weights.

Step 2: State the mathematical relationship
\[ \frac{W_1}{W_2} = \frac{E_1}{E_2} \]
Where \( W \) is the mass of substance and \( E \) is the equivalent weight.

Step 3: Define Equivalent Weight
Equivalent weight is the ratio of Atomic Mass to the number of electrons transferred (valency).
\[ E = \frac{\text{Atomic Mass}}{\text{Valency}} \]

Final Answer: The law states that mass deposited is proportional to the chemical equivalent weight for a constant charge.

Concept:

• During the electrolysis of aqueous sodium chloride, both chloride ions and water can undergo oxidation at the anode.
• The actual reaction depends on both thermodynamic and kinetic factors.

Step 1: Identify the feasible reaction

Although the standard oxidation potential of water is lower than that of chloride ions, the oxidation of water requires a high overvoltage.

Therefore, the oxidation of chloride ions occurs more readily.

\[ \boxed{\mathrm{Cl^- (aq)\rightarrow \frac{1}{2}Cl_2(g)+e^-}} \]

Hence, Reaction (I) is feasible at the anode.

Step 2: Why is Reaction (I) preferred over Reaction (II) at the anode?

Thermodynamically, oxidation of water appears more favourable because it has a lower standard oxidation potential (\(1.23\,\mathrm{V}\)) than chloride ion oxidation (\(1.36\,\mathrm{V}\)).

Step 3: Explain the actual observation

The oxidation of water to oxygen is kinetically slow and requires a high overvoltage.

As a result, chloride ions are oxidized more easily, producing chlorine gas.

Hence, Reaction (I) occurs preferentially at the anode despite its higher standard potential.

Concept:

• Oxidation states in elements are determined by the participation of valence electrons in bond formation.
• In transition elements, the valence shell consists of both the outermost \( ns \) orbital and the penultimate \( (n-1)d \) orbital.

Step 1: Analyze the electronic configuration and energy levels
Transition metals have a general electronic configuration of \( (n-1)d^{1-10} ns^{1-2} \).
The energy levels of the \( (n-1)d \) and \( ns \) orbitals are very close to each other. Because of this comparable energy, electrons from both subshells can be utilized for chemical bonding.

Step 2: Relate orbital availability to oxidation states
In the beginning of the series, only a few electrons are used, typically showing lower oxidation states. As we move across the series, more \( d \)-electrons become available.
Depending on the nature of the reacting atoms or ligands, the metal can lose or share different numbers of electrons from both the \( s \) and \( d \) subshells, leading to a variety of oxidation states (e.g., Manganese shows states from +2 to +7).

Step 3: Final Conclusion
The variable oxidation states are a direct consequence of the small energy gap that allows the sequential involvement of \( (n-1)d \) electrons along with \( ns \) electrons.

Final Answer: Transition metals show variable oxidation states because the energy difference between \( (n-1)d \) and \( ns \) orbitals is very small, allowing electrons from both to participate in bonding.

Concept:

• Paramagnetism arises from the presence of unpaired electrons in a species.
• The magnitude of paramagnetism (magnetic moment) is directly proportional to the number of unpaired electrons.
• Magnetic moment \( (\mu) \) is calculated as \( \sqrt{n(n+2)} \) Bohr Magnetons, where \( n \) is the number of unpaired electrons.

Step 1: Determine electronic configuration of \( \text{Ti}^{2+} \)
Atomic number of Ti is 22. Configuration: \( [\text{Ar}] 3d^2 4s^2 \).
For \( \text{Ti}^{2+} \), two electrons are removed from the 4s orbital.
Configuration: \( [\text{Ar}] 3d^2 \). Number of unpaired electrons \( (n) = 2 \).

Step 2: Determine electronic configuration of \( \text{Mn}^{2+} \)
Atomic number of Mn is 25. Configuration: \( [\text{Ar}] 3d^5 4s^2 \).
For \( \text{Mn}^{2+} \), two electrons are removed from the 4s orbital.
Configuration: \( [\text{Ar}] 3d^5 \). Number of unpaired electrons \( (n) = 5 \).

Step 3: Compare paramagnetism
Since \( \text{Mn}^{2+} \) has 5 unpaired electrons and \( \text{Ti}^{2+} \) has only 2, \( \text{Mn}^{2+} \) will exhibit a much stronger magnetic field interaction.
Magnetic moment for \( \text{Mn}^{2+} \approx 5.92 \text{ BM} \); for \( \text{Ti}^{2+} \approx 2.84 \text{ BM} \).

Final Answer: \( \text{Mn}^{2+} \) is more paramagnetic because it has a greater number of unpaired electrons (\( n=5 \)) than \( \text{Ti}^{2+} \) (\( n=2 \)).

Concept:

• An oxidising agent is a species that undergoes reduction (gains electrons).
• The tendency of an ion to be reduced depends on the stability of the resulting electronic configuration.
• Half-filled (\( d^5 \)) and fully-filled (\( d^{10} \)) subshells provide extra stability.

Step 1: Examine electronic configurations of the given ions

1. \( \text{V}^{3+} \): \( [\text{Ar}] 3d^2 \).
2. \( \text{Cr}^{3+} \): \( [\text{Ar}] 3d^3 \).
3. \( \text{Mn}^{3+} \): \( [\text{Ar}] 3d^4 \).

Step 2: Analyze the stability of the reduced products

When \( \text{Mn}^{3+} \) gains one electron, it becomes \( \text{Mn}^{2+} \) with a \( 3d^5 \) configuration.

This \( 3d^5 \) configuration is exceptionally stable because it is exactly half-filled.

In contrast, \( \text{Cr}^{3+} \) is already quite stable due to its half-filled \( t_{2g}^3 \) level (in octahedral fields), and its reduction to \( \text{Cr}^{2+} (d^4) \) is not energetically favorable.

Step 3: Conclusion

Due to the high stability of the resulting \( \text{Mn}^{2+} \) ion, \( \text{Mn}^{3+} \) has a very high tendency to accept an electron, making it a powerful oxidising agent.

Final Answer: \( \text{Mn}^{3+} \) is the strongest oxidising agent because it readily reduces to \( \text{Mn}^{2+} \), which has a stable half-filled \( 3d^5 \) configuration.

Concept:

• This reaction is the first step in the industrial preparation of Potassium Permanganate \( (\text{KMnO}_4) \).
• Pyrolusite ore \( (\text{MnO}_2) \) is fused with an alkali metal hydroxide in the presence of an oxidising agent like air (\( \text{O}_2 \)).

Step 1: Identify the oxidation change
Manganese in \( \text{MnO}_2 \) is in the +4 oxidation state.
In the presence of alkali and oxygen, it is oxidized to the +6 state.

Step 2: Identify the product
The +6 oxidation state of Manganese in an alkaline medium forms the manganate ion \( (\text{MnO}_4^{2-}) \).
The specific salt formed with \( \text{KOH} \) is Potassium Manganate \( (\text{K}_2\text{MnO}_4) \), which is dark green in color.

Step 3: Balanced equation
\[ 2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \rightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O} \]

Final Answer: \( 2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \rightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O} \)

Concept:

• This is a classic redox titration reaction between permanganate and oxalate ions in an acidic medium.
• Permanganate \( (\text{MnO}_4^-) \) acts as a strong oxidising agent.
• Oxalate \( (\text{C}_2\text{O}_4^{2-}) \) acts as a reducing agent.

Step 1: Identify half-reactions
Reduction: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \). (Mn goes from +7 to +2).
Oxidation: \( \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^- \). (C goes from +3 to +4).

Step 2: Equalize electrons and combine
Multiply reduction half-reaction by 2 and oxidation half-reaction by 5 to equalize electrons (10 \( e^- \)).
\[ 2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \]
\[ 5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10e^- \]

Step 3: Final balanced equation
Combine the two:
\[ 5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O} \]

Final Answer: \( 5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O} \)

Concept:

• Lanthanoids are a series of 14 elements from Cerium to Lutetium where the 4f subshell is being filled.
• Atomic size is determined by the balance between nuclear charge and electron shielding.

Step 1: Identify the cause of contraction
As we move across the lanthanoid series, the nuclear charge increases by one unit at each step. The new electron is added to the 4f subshell.
The shape of f-orbitals is highly diffused, resulting in very poor shielding of the outer electrons from the nuclear charge.

Step 2: Analyze the effect of poor shielding
Due to poor shielding, the effective nuclear charge experienced by the outer electrons increases steadily. This pulls the electron cloud closer to the nucleus.

Step 3: Observe the consequence
This results in a gradual and cumulative decrease in the size (atomic and ionic radii) of the elements as the atomic number increases. This unique phenomenon is termed the "Lanthanoid Contraction."

Final Answer: Lanthanoid contraction is the progressive decrease in the size of lanthanoid atoms and ions as atomic number increases, caused by the poor shielding effect of 4f electrons.

Concept:

• Colour in transition metals is generally attributed to the presence of partially filled d-orbitals.
• In an isolated atom, all five d-orbitals have the same energy (degenerate).

Step 1: Understand Crystal Field Splitting
When ligands approach a transition metal ion, the degeneracy of the d-orbitals is removed.
In an octahedral field, they split into two sets: lower energy \( t_{2g} \) and higher energy \( e_g \).

Step 2: Describe the electronic transition
When white light falls on the complex, an electron from a lower energy d-orbital absorbs energy and is excited to a higher energy d-orbital. This is called a "d-d transition."
The energy absorbed corresponds to a specific wavelength in the visible spectrum.

Step 3: Explain the observed colour
The light that is transmitted or reflected is the complementary colour of the light absorbed. For example, if yellow light is absorbed, the compound appears violet.

Final Answer: Transition metals form coloured compounds because the d-orbitals split into different energy levels, allowing electrons to undergo d-d transitions by absorbing specific wavelengths of visible light.

Concept:

• The electrode potential \( (E^\circ) \) of a metal depends on its sublimation energy, ionisation energy, and hydration enthalpy.
• A more negative reduction potential indicates that the metal is more easily oxidized and the resulting ion is very stable.

Step 1: Examine the stability of \( \text{Mn}^{2+} \)
Neutral Mn is \( [\text{Ar}] 3d^5 4s^2 \). After losing two electrons, \( \text{Mn}^{2+} \) has a \( 3d^5 \) configuration.
This half-filled subshell is exceptionally stable due to symmetry and exchange energy. This stability makes the loss of 4s electrons very favorable.

Step 2: Examine the stability of \( \text{Zn}^{2+} \)
Neutral Zn is \( [\text{Ar}] 3d^{10} 4s^2 \). After losing two electrons, \( \text{Zn}^{2+} \) has a \( 3d^{10} \) configuration.
The fully-filled d-subshell is highly stable. The removal of two electrons from the 4s orbital results in this stable configuration.

Step 3: Relate stability to potential
The high stability of these ions leads to lower (more negative) standard reduction potentials compared to the general trend in the series.

Final Answer: The \( E^\circ \) values for Mn and Zn are more negative than expected because of the extra stability associated with the half-filled \( 3d^5 \) subshell of \( \text{Mn}^{2+} \) and the fully-filled \( 3d^{10} \) subshell of \( \text{Zn}^{2+} \).

Concept:

• The stability of an oxidation state in solution depends on the net energy change involving ionisation enthalpy and hydration enthalpy.
• Although \( \text{Cu}^+ \) has a stable \( d^{10} \) configuration, its stability in aqueous medium is lower than \( \text{Cu}^{2+} \).

Step 1: Compare electronic configurations
\( \text{Cu}^+ \): \( [\text{Ar}] 3d^{10} \).
\( \text{Cu}^{2+} \): \( [\text{Ar}] 3d^9 \).
Based on configuration alone, one might expect \( \text{Cu}^+ \) to be more stable.

Step 2: Evaluate Energetics in aqueous solution
To form \( \text{Cu}^{2+} \) from \( \text{Cu}^+ \), a high second ionisation enthalpy is required.
However, the \( \text{Cu}^{2+} \) ion has a much smaller size and higher charge density than \( \text{Cu}^+ \). Consequently, it releases a much larger amount of energy when hydrated (more negative hydration enthalpy).

Step 3: Determine Net Stability
The large negative hydration enthalpy of \( \text{Cu}^{2+} \) effectively compensates for the energy required for the second ionisation.
In fact, \( \text{Cu}^+ \) ions are unstable in aqueous solution and undergo disproportionation: \( 2\text{Cu}^+(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + \text{Cu}(\text{s}) \).

Final Answer: The +2 oxidation state of Copper is the most stable in aqueous solution because the high hydration enthalpy of \( \text{Cu}^{2+} \) outweighs the second ionisation enthalpy needed to form it.

Concept:

• Cerium \( (\text{Z}=58) \) is the first element of the lanthanoid series.
• For all lanthanoids, the most stable and common oxidation state is +3.

Step 1: Identify the electronic configurations

Neutral Ce: \( [\text{Xe}] 4f^1 5d^1 6s^2 \).

\( \text{Ce}^{4+} \): \( [\text{Xe}] 4f^0 5d^0 6s^0 \) (Noble gas configuration).

\( \text{Ce}^{3+} \): \( [\text{Xe}] 4f^1 \).

Step 2: Evaluate stability vs. configuration

While \( \text{Ce}^{4+} \) achieves a stable noble gas configuration, the thermodynamics of lanthanoids strongly favor the +3 state in solution.

Step 3: Predict behavior

Because \( \text{Ce}^{3+} \) is much more stable than \( \text{Ce}^{4+} \) in aqueous conditions, \( \text{Ce}^{4+} \) has a strong tendency to gain one electron to achieve the +3 state.

Acting as an electron acceptor, it behaves as a powerful oxidising agent.

Final Answer: \( \text{Ce}^{4+} \) is a good oxidising agent because it readily reduces to the more stable +3 oxidation state, which is the preferred state for all lanthanoids in solution.

Concept:

• The \( \text{S}_{\text{N}}1 \) (Substitution Nucleophilic Unimolecular) reaction mechanism proceeds via the formation of a carbocation intermediate.
• The rate of an \( \text{S}_{\text{N}}1 \) reaction is directly proportional to the stability of the carbocation formed in the rate-determining step.
• Carbocation stability follows the order: Tertiary \( (3^\circ) \) > Secondary \( (2^\circ) \) > Primary \( (1^\circ) \) > Methyl.

Step 1: Identify the nature of the alkyl halides
2-Bromo-2-methylbutane is a tertiary \( (3^\circ) \) alkyl halide because the carbon attached to the bromine is bonded to three other carbon atoms.
1-Bromopentane is a primary \( (1^\circ) \) alkyl halide because the carbon attached to the bromine is bonded to only one other carbon atom.

Step 2: Analyze carbocation stability
In the first step of \( \text{S}_{\text{N}}1 \), 2-bromo-2-methylbutane loses a bromide ion to form a stable \( 3^\circ \) carbocation.
1-bromopentane would form a highly unstable \( 1^\circ \) carbocation.
Due to inductive effects and hyperconjugation, the tertiary carbocation is significantly more stable.

Final Answer: 2-Bromo-2-methylbutane is more reactive towards \( \text{S}_{\text{N}}1 \) due to the formation of a more stable tertiary carbocation.

Concept:

• Halides are classified based on the hybridization of the carbon atom to which the halogen is attached.
• Vinylic halides: The halogen atom is bonded to an \( sp^2 \)-hybridized carbon atom of a carbon-carbon double bond \( (\text{C=C}) \).
• Allylic halides: The halogen atom is bonded to an \( sp^3 \)-hybridized carbon atom adjacent to a carbon-carbon double bond.

Step 1: Identify the carbon atom bonded to the halogen
In the given structure \( \text{CH}_3 - \text{CH}(\text{CH}_3) - \text{C}(\text{Cl}) = \text{CH}_2 \), the chlorine atom \( (\text{Cl}) \) is directly attached to the second carbon of the main chain.

Step 2: Determine the hybridization and environment of that carbon
The carbon atom bonded to Cl is part of a double bond \( (\dots \text{C(Cl)=CH}_2) \).
Since the halogen is directly attached to one of the doubly bonded carbons, it falls into the vinylic category.

Final Answer: The compound is a vinylic halide.

Concept:

• Chloroform \( (\text{CHCl}_3) \) is chemically sensitive to oxygen in the presence of light.
• Photochemical oxidation leads to the formation of carbonyl chloride, commonly known as phosgene.

Step 1: Analyze the chemical reaction
In the presence of light and atmospheric oxygen, chloroform undergoes a slow oxidation process.
\[ 2\text{CHCl}_3 + \text{O}_2 \xrightarrow{\text{light}} 2\text{COCl}_2 + 2\text{HCl} \]

Step 2: Identify the hazard of the product
Phosgene \( (\text{COCl}_2) \) is an extremely poisonous and toxic gas.
Storing chloroform in clear bottles would allow light to catalyze this reaction, making the stored chloroform dangerous to use.

Step 3: Explain the storage method
Dark coloured (amber) bottles absorb visible light and prevent it from reaching the liquid, thereby inhibiting the photochemical reaction. Often, a small amount of ethanol is also added to convert any formed phosgene into harmless diethyl carbonate.

Final Answer: Chloroform is stored in dark bottles to block light and prevent its conversion into the toxic gas phosgene.

Concept:

• A nucleophile is a species that donates a pair of electrons to form a chemical bond.
• While most nucleophiles bond through a single specific atom, some have multiple sites of high electron density.

Step 1: Explain the structural characteristic
Ambident nucleophiles contain more than one nucleophilic center (donor atom). However, during a specific reaction, they only bond through one of these sites at a time.

Step 2: Provide examples
1. Cyanide ion \( (\text{CN}^-) \): It can bond through carbon to form a cyanide \( (\text{R-CN}) \) or through nitrogen to form an isocyanide \( (\text{R-NC}) \).
2. Nitrite ion \( (\text{NO}_2^-) \): It can bond through nitrogen to form a nitroalkane \( (\text{R-NO}_2) \) or through oxygen to form an alkyl nitrite \( (\text{R-ONO}) \).

Final Answer: Ambident nucleophiles are groups that have two nucleophilic centers but coordinate through only one site during a reaction.

Concept:

• Enantiomers are non-superimposable mirror images that rotate plane-polarized light in opposite directions by equal amounts.
• Optical activity is the ability of a substance to rotate plane-polarized light.

Step 1: Explain the composition
A racemic mixture (or racemate) contains equal amounts of both the dextrorotatory \( (+) \) and levorotatory \( (-) \) isomers of a chiral compound.

Step 2: Discuss optical properties
Because the mixture contains equal concentrations of both enantiomers, the rotation caused by one isomer is exactly cancelled by the rotation caused by the other. This is known as "external compensation."

Step 3: Identify the result
As a result of this cancellation, the overall racemic mixture does not rotate plane-polarized light and is described as optically inactive. It is often denoted by the prefix \( (\pm) \).

Final Answer: A racemic mixture is a 1:1 mixture of two enantiomers that shows no net optical rotation.

Concept:

• The \( \text{C}_4\text{H}_9\text{Br} \) formula has four structural isomers: n-butyl bromide \( (1^\circ) \), isobutyl bromide \( (1^\circ) \), sec-butyl bromide \( (2^\circ) \), and tert-butyl bromide \( (3^\circ) \).
• \( \text{S}_{\text{N}}1 \) reactivity is determined by the stability of the carbocation intermediate.

Step 1: Compare carbocation stability of the isomers
tert-Butyl bromide on ionization forms the tert-butyl carbocation, which is a tertiary \( (3^\circ) \) carbocation.
Secondary and primary isomers form \( 2^\circ \) and \( 1^\circ \) carbocations respectively.

Step 2: Relate stability to reactivity
Since the tertiary carbocation is the most stable due to maximum \( +I \) effect and hyperconjugation from three methyl groups, tert-butyl bromide reacts the fastest in an \( \text{S}_{\text{N}}1 \) pathway.

Final Answer: tert-Butyl bromide is the most reactive isomer for \( \text{S}_{\text{N}}1 \) reactions.

Concept:

• Dehydrohalogenation is an elimination reaction \( (E1 \text{ or } E2) \) where a hydrogen halide is removed.
• Saytzeff’s Rule states that in an elimination reaction, the preferred product is the alkene that has the greater number of alkyl groups attached to the doubly bonded carbon atoms (more substituted alkene).

Step 1: Analyze the structure and possible hydrogens
In 1-bromo-1-methylcyclohexane, the bromine is at the C1 position. There are two types of beta-hydrogens:
1. Hydrogens on the ring carbons (C2 and C6).
2. Hydrogens on the methyl group attached to C1.

Step 2: Compare the possible alkene products
Elimination using a ring hydrogen gives 1-methylcyclohexene (trisubstituted alkene).
Elimination using a methyl hydrogen gives methylenecyclohexane (disubstituted alkene).

Step 3: Apply Saytzeff’s Rule
Since 1-methylcyclohexene is more highly substituted (more stable), it is the major product of the reaction.

Final Answer: The predicted major product is 1-methylcyclohexene.

Concept:

• This reaction is known as the Fittig reaction.
• It involves the coupling of two aryl halide molecules in the presence of sodium metal and dry ether.
• It is the aromatic counterpart of the Wurtz reaction.

Step 1: Analyze the reaction conditions and reagents
Chlorobenzene \( (\text{C}_6\text{H}_5\text{Cl}) \) is treated with metallic sodium in an anhydrous medium (dry ether).
The use of dry ether is essential to prevent the sodium metal from reacting with moisture, which would lead to the formation of sodium hydroxide and hydrogen gas.

Step 2: Describe the coupling mechanism
Sodium metal reacts with the carbon-chlorine bond of two separate chlorobenzene molecules.
The chlorine atoms are removed as sodium chloride \( (\text{NaCl}) \), and the two resulting phenyl radicals (or equivalent organometallic intermediates) join together to form a new carbon-carbon bond.
\[ 2\text{C}_6\text{H}_5\text{Cl} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5-\text{C}_6\text{H}_5 + 2\text{NaCl} \]

Final Answer: The major product of the reaction is Biphenyl.

Concept:

• This reaction is a free radical substitution reaction of an alkyl side chain on a benzene ring.
• In the presence of heat \( (\Delta) \), bromine \( (\text{Br}_2) \) undergoes homolytic cleavage to form bromine radicals \( (\text{Br}^\cdot) \).
• The reaction site is determined by the stability of the intermediate alkyl radical formed.

Step 1: Identify the reactive sites in the isopropyl group
The isopropyl group has six primary hydrogens on the two methyl groups and one tertiary hydrogen at the benzylic position.
The benzylic position is the carbon atom directly attached to the aromatic ring.

Step 2: Determine the most stable radical intermediate
Loss of a hydrogen atom from the benzylic position yields a tertiary benzylic radical.
This radical is highly stabilized due to resonance with the benzene ring and the inductive effect (and hyperconjugation) of the two methyl groups.
Radical stability order: Tertiary benzylic > Secondary benzylic > Primary.

Step 3: Predict the final major product
Substitution occurs at the most stable radical center. The bromine atom replaces the tertiary benzylic hydrogen to form the major product.
\[ p\text{-O}_2\text{N-C}_6\text{H}_4\text{-CH(CH}_3)_2 + \text{Br}_2 \xrightarrow{\text{heat}} p\text{-O}_2\text{N-C}_6\text{H}_4\text{-C(Br)(CH}_3)_2 + \text{HBr} \]

Final Answer: The major product is 2-bromo-2-(4-nitrophenyl)propane.

Concept:

• The depression in freezing point \( (\Delta T_f) \) is a colligative property given by the formula \( \Delta T_f = i \cdot K_f \cdot m \).
• For electrolytes, the van’t Hoff factor \( (i) \) accounts for the dissociation of the solute.

Step 1: Calculate the molality (\( m \)) of the solution
Moles of \( \text{MgBr}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{10.5 \text{ g}}{184 \text{ g mol}^{-1}} \approx 0.0571 \text{ mol} \).
Molality \( (m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.0571 \text{ mol}}{0.250 \text{ kg}} = 0.2284 \text{ mol kg}^{-1} \).

Step 2: Determine the van’t Hoff factor (\( i \))
\( \text{MgBr}_2 \) dissociates as: \( \text{MgBr}_2 \rightarrow \text{Mg}^{2+} + 2\text{Br}^- \).
Since complete dissociation is assumed, the total number of ions formed from one formula unit is 3.
Therefore, \( i = 3 \).

Step 3: Calculate the depression in freezing point (\( \Delta T_f \))
\[ \Delta T_f = i \cdot K_f \cdot m \]
\[ \Delta T_f = 3 \times 1.86 \text{ K kg mol}^{-1} \times 0.2284 \text{ mol kg}^{-1} \approx 1.274 \text{ K} \]

Step 4: Calculate the freezing point of the solution (\( T_f \))
The freezing point of pure water (\( T_f^\circ \)) is 273.15 K (or 0\(^\circ\)C).
\[ T_f = T_f^\circ - \Delta T_f = 273.15 \text{ K} - 1.274 \text{ K} = 271.876 \text{ K} \]

Final Answer: The freezing point of the solution is approximately 271.88 K or -1.27\(^\circ\)C.

Concept:

• Solutions are classified based on their compliance with Raoult’s Law and the energetic/volumetric changes upon mixing.

Step 1: Compliance with Raoult’s Law
Ideal Solutions: These solutions obey Raoult’s Law (\( P = P^\circ \chi \)) over the entire range of concentration and temperature.
Non-ideal Solutions: These solutions do not obey Raoult’s Law and show either positive or negative deviations from the expected vapor pressure.

Step 2: Enthalpy and Volume change on mixing
Ideal Solutions: The enthalpy of mixing (\( \Delta H_{\text{mix}} \)) is zero, and the volume change on mixing (\( \Delta V_{\text{mix}} \)) is zero. This means no heat is absorbed or evolved, and there is no expansion or contraction during mixing.
Non-ideal Solutions: For these solutions, \( \Delta H_{\text{mix}} \neq 0 \) and \( \Delta V_{\text{mix}} \neq 0 \).

Final Answer: Two key differences are that ideal solutions obey Raoult’s Law and have zero enthalpy of mixing, whereas non-ideal solutions deviate from Raoult’s Law and have non-zero enthalpy of mixing.

Concept:

• Osmotic pressure (\( \pi \)) is a colligative property defined by the equation \( \pi = i \cdot M \cdot R \cdot T \).
• The van’t Hoff factor (\( i \)) must be used for dissociating salts like potassium sulphate.

Step 1: Calculate the molarity (\( M \)) of the solution
Moles of \( \text{K}_2\text{SO}_4 = \frac{0.025 \text{ g}}{174 \text{ g mol}^{-1}} \approx 1.437 \times 10^{-4} \text{ mol} \).
Molarity \( (M) = \frac{\text{Moles}}{\text{Volume in L}} = \frac{1.437 \times 10^{-4} \text{ mol}}{2 \text{ L}} = 7.184 \times 10^{-5} \text{ mol L}^{-1} \).

Step 2: Determine van’t Hoff factor and Temperature
\( \text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \). For complete dissociation, \( i = 3 \).
Temperature \( (T) = 27 + 273 = 300 \text{ K} \).

Step 3: Calculate osmotic pressure (\( \pi \))
\[ \pi = i \cdot M \cdot R \cdot T \]
\[ \pi = 3 \times (7.184 \times 10^{-5} \text{ mol L}^{-1}) \times 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 300 \text{ K} \]
\[ \pi \approx 0.00530 \text{ atm} \]

Final Answer: The osmotic pressure of the solution is approximately 5.3 \(\times\) 10\(^{-3}\) atm.

Concept:

• Azeotropes are constant boiling mixtures that have the same composition in liquid and vapor phase.
• Solutions showing large negative deviations from Raoult’s law form maximum boiling azeotropes.

Step 1: Identify the type of deviation
Acetone and chloroform molecules interact through strong intermolecular hydrogen bonding.
The A-B interactions (Acetone-Chloroform) are stronger than the A-A (Acetone-Acetone) and B-B (Chloroform-Chloroform) interactions.

Step 2: Explain the effect on vapor pressure
Because the new interactions are stronger, the escaping tendency of molecules decreases.
This results in a negative deviation from Raoult’s Law, meaning the total vapor pressure of the solution is lower than predicted.

Step 3: Relate to boiling point
A decrease in vapor pressure leads to an increase in boiling point.
The point of minimum vapor pressure corresponds to the maximum boiling temperature in the mixture. Thus, it forms a maximum boiling azeotrope.

Final Answer: A solution of acetone and chloroform forms a maximum boiling azeotrope because it shows negative deviation from Raoult’s law due to strong hydrogen bonding between the components.