CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 1 - 56/5/1)

Concept:

Ketones contain the carbonyl functional group \((>C=O)\). In many organic reactions, it is required to completely remove the oxygen atom of the carbonyl group and convert the ketone into the corresponding hydrocarbon. Such transformations are known as reductions of carbonyl compounds.

Two important reactions commonly used for this purpose are Clemmensen reduction and Wolff-Kishner reduction. In the Wolff-Kishner reduction, a ketone or aldehyde is first converted into a hydrazone by reaction with hydrazine \((NH_2NH_2)\). The hydrazone is then heated with a strong base such as KOH, resulting in the complete removal of oxygen and formation of the corresponding hydrocarbon. Therefore, this reaction is specifically known for converting ketones into hydrocarbons.

Step 1: Understanding what transformation is being asked in the question.

The question asks for a reaction that converts a ketone into a hydrocarbon.

A ketone has the general structure
\[ R-CO-R' \]

whereas a hydrocarbon contains only carbon and hydrogen atoms.

Therefore, the required reaction must completely eliminate the oxygen atom present in the carbonyl group and replace it effectively with hydrogen atoms.

For example,
\[ CH_3COCH_3 \longrightarrow CH_3CH_2CH_3 \]

In this conversion, acetone is transformed into propane by complete reduction of the carbonyl group.

Hence, we need a reaction known for complete reduction of ketones.

Step 2: Examining Option (A) : Reimer-Tiemann reaction.

The Reimer-Tiemann reaction is a reaction of phenols with chloroform in the presence of alkali.

Its primary purpose is the introduction of a formyl group \((-CHO)\) into an aromatic ring.

A typical reaction is
\[ Phenol \xrightarrow[NaOH]{CHCl_3} Salicylaldehyde \]

Thus, this reaction is used for the preparation of aromatic aldehydes and not for converting ketones into hydrocarbons.

Therefore, Option (A) is incorrect.

Step 3: Examining Option (B) : Wolff-Kishner reduction.

In the Wolff-Kishner reduction, a ketone reacts with hydrazine to form a hydrazone.
\[ R_2C=O + NH_2NH_2 \longrightarrow R_2C=NNH_2 \]

The hydrazone is then heated with KOH.

During this process, nitrogen gas is eliminated and the carbonyl carbon gets reduced.
\[ R_2C=NNH_2 \xrightarrow[Heat]{KOH} R_2CH_2 + N_2 \]

As a result, the ketone is converted directly into the corresponding hydrocarbon.

For example,
\[ CH_3COCH_3 \xrightarrow[NH_2NH_2]{KOH,\ Heat} CH_3CH_2CH_3 \]

This is exactly the conversion asked in the question.

Therefore, Option (B) is correct.

Step 4: Examining Option (C) : Aldol condensation.

Aldol condensation occurs between aldehydes or ketones containing \(\alpha\)-hydrogen atoms.

The reaction produces \(\beta\)-hydroxy aldehydes or \(\beta\)-hydroxy ketones, which may further dehydrate to form \(\alpha,\beta\)-unsaturated compounds.

For example,
\[ 2CH_3CHO \longrightarrow CH_3CH(OH)CH_2CHO \]

Since this reaction forms larger carbon skeletons rather than reducing ketones to hydrocarbons, it cannot be the required answer.

Hence, Option (C) is incorrect.

Step 5: Examining Option (D) : Stephen reaction.

The Stephen reaction, also known as Stephen aldehyde synthesis, converts nitriles into aldehydes.

The general transformation is
\[ R-CN \longrightarrow R-CHO \]

This reaction is useful for preparing aldehydes from nitriles but has no role in converting ketones into hydrocarbons.

Therefore, Option (D) is also incorrect.

Step 6: Selecting the correct reaction based on the analysis of all options.

After examining all four reactions, we observe that only the Wolff-Kishner reduction specifically performs the complete reduction of the carbonyl group of a ketone and converts it into a hydrocarbon.

Thus,
\[ R-CO-R' \longrightarrow R-CH_2-R' \]

is achieved by the Wolff-Kishner reduction.

Hence, the correct option is Option (B).
\[ \boxed{Wolff-Kishner reduction} \] Quick Tip: Remember the two famous reactions used to convert aldehydes and ketones into hydrocarbons: 1. Wolff-Kishner Reduction : \(NH_2NH_2/KOH,\ Heat\) (Basic medium) 2. Clemmensen Reduction : \(Zn(Hg)/HCl\) (Acidic medium) Both reactions remove the oxygen atom of the carbonyl group completely and convert the compound into the corresponding hydrocarbon.

Concept:

Hoffmann bromamide degradation reaction, also known as Hoffmann bromamide reaction or Hoffmann rearrangement, is an important method for the preparation of primary amines from amides. In this reaction, an amide containing one carbonyl carbon is converted into a primary amine having one carbon atom less than the parent amide.

The reaction involves the use of bromine and a strong base such as sodium hydroxide. During the reaction, the carbonyl carbon of the amide is lost as carbon dioxide, resulting in the formation of a primary amine.

The general reaction is
\[ RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O \]

Thus, the essential reagents are:

Amide \((RCONH_2)\)
Bromine \((Br_2)\)
Sodium hydroxide \((NaOH)\)


Step 1: Identifying the reagents listed in the question.

The reagents given are:
\[ (i)\; RCONH_2 \]
\[ (ii)\; NaOH \]
\[ (iii)\; Br_2 \]
\[ (iv)\; CHCl_3 \]

We must determine which of these are required for Hoffmann bromamide degradation.

Step 2: Recalling the reaction conditions of Hoffmann bromamide degradation.

The reaction proceeds according to
\[ RCONH_2 + Br_2 + NaOH \longrightarrow RNH_2 \]

Hence the required reagents are:
\[ RCONH_2,\; Br_2,\; NaOH \]

Therefore,
\[ (i),\;(ii),\;(iii) \]

are necessary.

Step 3: Checking whether chloroform is required.

Chloroform \((CHCl_3)\) is not used in Hoffmann bromamide degradation.

Instead, chloroform is commonly used in the Carbylamine reaction for preparing isocyanides.

Hence,
\[ CHCl_3 \]

is not required.

Step 4: Selecting the correct option.

The correct combination is
\[ (i),\;(ii),\;(iii) \]

which corresponds to Option (C).
\[ \boxed{Option (C)} \] Quick Tip: Hoffmann bromamide degradation converts an amide into a primary amine containing one carbon atom less. Always remember the reagents: \[ RCONH_2 + Br_2 + NaOH \] Amide + Bromine + Sodium hydroxide.

Concept:

The Carbylamine reaction is a characteristic test for primary amines. When a primary amine is heated with chloroform and alcoholic potassium hydroxide, a foul-smelling isocyanide is produced.

The reaction is represented as
\[ RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O \]

where \(RNC\) is an isocyanide.

This reaction is used as a test for distinguishing primary amines from secondary and tertiary amines.

Step 1: Recalling the product formed in Carbylamine reaction.

The reaction involves
\[ Primary Amine + CHCl_3 + KOH \]

and produces
\[ Isocyanide \]

as the main product.

Step 2: Writing the general reaction.
\[ RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O \]

The product \(RNC\) is called isocyanide or carbylamine.

Step 3: Examining the options.

Carboxylic acids are not produced.

Aldehydes are not produced.

Cyanides \((RCN)\) are not produced.

The actual product is isocyanide \((RNC)\).

Therefore, Option (D) is correct.
\[ \boxed{Option (D)} \] Quick Tip: Carbylamine test is given only by primary amines and produces foul-smelling isocyanides \((RNC)\). Remember: \[ RNH_2 + CHCl_3 + KOH \rightarrow RNC \]

Concept:

Actinoids belong to the 5f-block of the periodic table. They exhibit a wide range of oxidation states because the energies of the 5f, 6d and 7s orbitals are very close to each other.

As a result, electrons from all these orbitals can participate in bonding, leading to several possible oxidation states.

Step 1: Understanding oxidation states in actinoids.

The electronic configurations of actinoids involve
\[ 5f,\;6d,\;7s \]

orbitals.

Since these orbitals have nearly equal energies, electrons can be removed from different orbitals during bond formation.

Step 2: Reason for variable oxidation states.

Because electrons from 5f, 6d and 7s orbitals can participate in bonding, actinoids exhibit oxidation states such as
\[ +3,\;+4,\;+5,\;+6,\;+7 \]

and others.

This results in a larger number of oxidation states compared to lanthanoids.

Step 3: Evaluating the options.

Radioactivity does not cause variable oxidation states.

Large atomic number does not directly explain oxidation states.

Large atomic mass is unrelated.

The correct reason is the comparable energies of 5f, 6d and 7s orbitals.

Therefore Option (D) is correct.
\[ \boxed{Option (D)} \] Quick Tip: Actinoids show many oxidation states because the energies of 5f, 6d and 7s orbitals are very close, allowing electrons from all these orbitals to participate in bonding.

Concept:

This reaction is known as the Finkelstein reaction. In this reaction, an alkyl chloride or alkyl bromide reacts with sodium iodide in dry acetone to form the corresponding alkyl iodide.

The reaction proceeds because sodium chloride is insoluble in dry acetone and precipitates out.

Step 1: Writing the reaction involved.
\[ CH_3Cl + NaI \rightarrow CH_3I + NaCl \]

Step 2: Understanding why the reaction proceeds forward.

Dry acetone is used because
\[ NaCl \]

is insoluble in acetone.

The precipitated sodium chloride shifts the equilibrium toward product formation.

Step 3: Identifying products A and B.

Comparing with
\[ CH_3Cl + NaI \rightarrow CH_3I + NaCl \]

we get
\[ A = CH_3I \]

and
\[ B = NaCl \]

Therefore Option (A) is correct.
\[ \boxed{Option (A)} \] Quick Tip: Finkelstein reaction: \[ RCl + NaI \xrightarrow{dry acetone} RI + NaCl \] Dry acetone is used because NaCl precipitates out, driving the reaction forward.

Concept:

Hinsberg's reagent is benzenesulphonyl chloride. It is widely used in the Hinsberg test for distinguishing primary, secondary and tertiary amines.

Its molecular formula is
\[ C_6H_5SO_2Cl \]

The reagent reacts differently with different classes of amines, making it useful for identification purposes.

Step 1: Recalling the reagent used in the Hinsberg test.

The Hinsberg test employs benzenesulphonyl chloride.

Its structure contains
\[ C_6H_5-SO_2-Cl \]

Therefore, its molecular formula is
\[ C_6H_5SO_2Cl \]

Step 2: Comparing with the given options.

Option (A) represents benzoyl chloride.

Option (B) represents benzenesulphonyl chloride.

Option (C) is an amide derivative.

Option (D) is benzylamine.

Thus only Option (B) corresponds to Hinsberg's reagent.

Step 3: Selecting the correct answer.

Hence the correct formula of Hinsberg's reagent is
\[ C_6H_5SO_2Cl \]

Therefore,
\[ \boxed{Option (B)} \] Quick Tip: Hinsberg's reagent = Benzenesulphonyl chloride \[ C_6H_5SO_2Cl \] Used in the Hinsberg test to distinguish primary, secondary and tertiary amines.

Concept:

A first order reaction is a reaction whose rate depends upon the concentration of only one reactant raised to the first power. One of the most important characteristics of a first order reaction is that its half-life is independent of the initial concentration of the reactant.

For a first order reaction, the relationship between half-life and rate constant is given by
\[ t_{1/2}=\frac{0.693}{k} \]

where
\[ t_{1/2}=half-life \]

and
\[ k=rate constant \]

This formula allows us to determine the rate constant directly if the half-life is known.

Step 1: Writing the formula for the half-life of a first order reaction.

For a first order reaction,
\[ t_{1/2}=\frac{0.693}{k} \]

The given value is
\[ t_{1/2}=1386\;s \]

Substituting into the formula,
\[ 1386=\frac{0.693}{k} \]

Step 2: Rearranging the equation to calculate the rate constant.

Solving for \(k\),
\[ k=\frac{0.693}{1386} \]
\[ k=0.0005 \]
\[ k=5.0\times10^{-4}\;s^{-1} \]

Step 3: Matching the obtained value with the given options.

The calculated value is
\[ k=5.0\times10^{-4}\;s^{-1} \]

which corresponds to Option (B).
\[ \boxed{k=5.0\times10^{-4}\;s^{-1}} \]

Therefore,
\[ \boxed{Option (B)} \] Quick Tip: For every first order reaction, always remember: \[ t_{1/2}=\frac{0.693}{k} \] The half-life of a first order reaction does not depend upon the initial concentration of the reactant.

Concept:

A chelate complex is formed when a multidentate ligand coordinates to the central metal atom through two or more donor atoms simultaneously, thereby forming one or more ring structures.

Ligands can be classified as:


Monodentate ligands : Coordinate through one donor atom.
Bidentate ligands : Coordinate through two donor atoms.
Polydentate ligands : Coordinate through more than two donor atoms.


Chelation occurs only when the ligand possesses more than one donor atom capable of bonding to the metal ion at the same time.

Step 1: Examining ammonia as a ligand.

Ammonia \((NH_3)\) donates a lone pair through a single nitrogen atom.

Hence it is a monodentate ligand.

Since it can attach through only one donor atom, it cannot form a chelate ring.

Therefore, Option (A) is incorrect.

Step 2: Examining water as a ligand.

Water \((H_2O)\) coordinates through only one oxygen atom.

Therefore it is also a monodentate ligand and cannot form a chelate complex.

Hence, Option (B) is incorrect.

Step 3: Examining nitrite ion as a ligand.

Nitrite ion \((NO_2^-)\) is an ambidentate ligand.

It can coordinate through either nitrogen or oxygen, but not simultaneously through both atoms to form a chelate ring.

Hence it does not normally act as a chelating ligand.

Therefore, Option (C) is incorrect.

Step 4: Examining oxalate ion as a ligand.

Oxalate ion is represented as
\[ C_2O_4^{2-} \]

It contains two oxygen donor atoms capable of coordinating simultaneously to the same metal ion.

Therefore, oxalate acts as a bidentate ligand.

It forms a ring structure with the metal ion and hence produces a chelate complex.

Therefore, Option (D) is correct.
\[ \boxed{Option (D)} \] Quick Tip: Common chelating ligands: \[ C_2O_4^{2-},\; en,\; EDTA^{4-} \] Oxalate ion is a bidentate ligand and readily forms chelate complexes.

Concept:

Lucas test is a qualitative test used to distinguish primary, secondary and tertiary alcohols. The reagent used is Lucas reagent, which is a mixture of concentrated hydrochloric acid and anhydrous zinc chloride.

The test is based on the different rates at which alcohols undergo substitution reactions to form alkyl chlorides.

Step 1: Understanding the Lucas test.

Lucas reagent consists of
\[ ZnCl_2 + HCl \]

Alcohols react with this reagent to form alkyl chlorides.
\[ ROH + HCl \rightarrow RCl + H_2O \]

The alkyl chloride formed is insoluble and produces turbidity.

Step 2: Observing the behaviour of different alcohols.

Tertiary alcohols react immediately and produce turbidity instantly.

Secondary alcohols react more slowly and produce turbidity within a few minutes.

Primary alcohols react very slowly and usually show no turbidity at room temperature.

Therefore, alcohols can be distinguished based on the time taken to produce turbidity.

Step 3: Checking the remaining options.

Fehling's test is used for aldehydes.

Tollens' test is also used for aldehydes.

Hinsberg's test is used for distinguishing amines.

Thus none of these tests can distinguish alcohols.

Therefore, the correct answer is Lucas test.
\[ \boxed{Option (A)} \] Quick Tip: Lucas Test: \[ Lucas Reagent = ZnCl_2 + HCl \] Tertiary alcohol \(>\) Secondary alcohol \(>\) Primary alcohol in order of reactivity.

Concept:

Basic strength depends upon the availability of the lone pair of electrons on the nitrogen atom.

Greater availability of the lone pair results in greater basic strength.

Electron-donating groups increase basicity, whereas electron-withdrawing effects and resonance decrease basicity.

Step 1: Comparing aniline and ammonia.

In aniline,
\[ C_6H_5NH_2 \]

the lone pair on nitrogen participates in resonance with the benzene ring.

Because of resonance, the lone pair becomes less available for protonation.

Therefore, aniline is less basic than ammonia.
\[ C_6H_5NH_2 < NH_3 \]

Step 2: Comparing benzylamine with ammonia.

In benzylamine,
\[ C_6H_5CH_2NH_2 \]

the amino group is separated from the benzene ring by a \(CH_2\) group.

Hence the lone pair does not participate significantly in resonance.

Therefore benzylamine is more basic than ammonia.
\[ NH_3 < C_6H_5CH_2NH_2 \]

Step 3: Comparing ethylamine and benzylamine.

Ethyl group exerts a positive inductive effect \((+I)\).

This increases electron density on nitrogen and enhances basicity.

Hence,
\[ C_6H_5CH_2NH_2 < C_2H_5NH_2 \]

Step 4: Comparing ethylamine and diethylamine.

Secondary aliphatic amines generally show greater basic strength than primary amines due to stronger electron-releasing inductive effects.

Thus,
\[ C_2H_5NH_2 < (C_2H_5)_2NH \]

Step 5: Combining all comparisons into a single order.

The overall increasing order becomes
\[ C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2 < (C_2H_5)_2NH \]

This matches Option (A).
\[ \boxed{Option (A)} \] Quick Tip: Aniline is less basic than ammonia because the nitrogen lone pair is involved in resonance. Aliphatic amines are generally more basic due to the electron-releasing \(+I\) effect of alkyl groups.

Concept:

Carbohydrates are broadly classified into monosaccharides, disaccharides and polysaccharides based on the number of sugar units present in their molecules.


Monosaccharides are the simplest carbohydrates and cannot be hydrolysed into smaller carbohydrate units.
Disaccharides contain two monosaccharide units joined through a glycosidic linkage.
Polysaccharides contain a large number of monosaccharide units linked together and form high molecular mass polymers.


Therefore, to identify a polysaccharide, we need to determine which compound is made up of a very large number of sugar units.

Step 1: Examining fructose.

Fructose is a simple sugar having the molecular formula
\[ C_6H_{12}O_6 \]

It consists of only one sugar unit and cannot be hydrolysed into simpler carbohydrates.

Therefore, fructose is a monosaccharide.

Hence, Option (A) is incorrect.

Step 2: Examining maltose.

Maltose is composed of two glucose molecules joined through a glycosidic bond.

It is represented as
\[ Glucose + Glucose \]

Since it contains two monosaccharide units, maltose is a disaccharide.

Therefore, Option (B) is incorrect.

Step 3: Examining glucose.

Glucose is one of the most important naturally occurring carbohydrates.

Its molecular formula is
\[ C_6H_{12}O_6 \]

Like fructose, glucose consists of only a single sugar unit.

Hence, glucose is a monosaccharide.

Therefore, Option (C) is incorrect.

Step 4: Examining cellulose.

Cellulose is a naturally occurring carbohydrate found in plant cell walls.

It is made up of a very large number of \(\beta\)-D-glucose units linked together through \(\beta(1 \rightarrow 4)\) glycosidic bonds.

Its general representation is
\[ (C_6H_{10}O_5)_n \]

where \(n\) is a very large number.

Since cellulose consists of hundreds to thousands of glucose units joined together, it is classified as a polysaccharide.

Therefore, Option (D) is correct.

Step 5: Selecting the correct answer.

Among the given compounds:
\[ Fructose \rightarrow Monosaccharide \]
\[ Glucose \rightarrow Monosaccharide \]
\[ Maltose \rightarrow Disaccharide \]
\[ Cellulose \rightarrow Polysaccharide \]

Hence, the polysaccharide is cellulose.
\[ \boxed{Cellulose} \]

Therefore,
\[ \boxed{Option (D)} \] Quick Tip: Remember the classification of common carbohydrates: Monosaccharides : Glucose, Fructose Disaccharides : Maltose, Sucrose, Lactose Polysaccharides : Starch, Cellulose, Glycogen

Concept:

Proteins are biological macromolecules formed by the polymerization of amino acids through peptide bonds. The properties and functions of proteins depend not only on the amino acids present but also on the precise order in which these amino acids are arranged.

Protein structures are generally classified into four levels:


Primary structure
Secondary structure
Tertiary structure
Quaternary structure


Each level represents a different degree of structural organization.

Step 1: Understanding the primary structure of a protein.

The primary structure of a protein refers to the exact sequence in which amino acids are arranged along the polypeptide chain.

For example, if a protein chain contains amino acids arranged as
\[ Ala-Gly-Val-Leu-Ser \]

then this order itself represents the primary structure.

Thus, the primary structure is simply the linear arrangement of amino acids connected through peptide bonds.

Step 2: Understanding the secondary structure of a protein.

The secondary structure refers to regular folding patterns within the polypeptide chain.

Common examples are
\[ \alpha-helix \]

and
\[ \beta-pleated sheet \]

These structures arise because of hydrogen bonding between different parts of the chain.

Therefore, secondary structure is related to folding and not to the sequence itself.

Hence, Option (B) is incorrect.

Step 3: Understanding the tertiary structure of a protein.

The tertiary structure represents the three-dimensional arrangement of an entire polypeptide chain.

It results from interactions such as


Hydrogen bonding
Ionic interactions
Hydrophobic interactions
Disulfide linkages


Since it describes the overall shape of a protein molecule, it is not the sequence of amino acids.

Therefore, Option (C) is incorrect.

Step 4: Understanding the quaternary structure of a protein.

The quaternary structure is observed when two or more polypeptide chains associate together to form a functional protein.

For example, haemoglobin consists of multiple polypeptide subunits.

Hence, quaternary structure refers to the arrangement of different subunits and not to the amino acid sequence.

Therefore, Option (D) is incorrect.

Step 5: Identifying the structure described in the question.

The question specifically mentions:

"The polypeptide chain in a protein has amino acids linked with each other in a specific sequence."


The term "specific sequence of amino acids" directly corresponds to the primary structure of a protein.

Hence,
\[ \boxed{Primary Structure} \]

is the correct answer.

Therefore,
\[ \boxed{Option (A)} \] Quick Tip: Protein Structure Levels: Primary Structure \(\rightarrow\) Sequence of amino acids Secondary Structure \(\rightarrow\) \(\alpha\)-Helix and \(\beta\)-Sheet Tertiary Structure \(\rightarrow\) Three-dimensional folding of one chain Quaternary Structure \(\rightarrow\) Arrangement of multiple polypeptide chains

Concept:

In stereochemistry, the symbols D and L are used to indicate the relative configuration of a molecule with respect to D-glyceraldehyde and L-glyceraldehyde. These symbols provide information about the arrangement of atoms in space and are not directly related to the direction in which a compound rotates plane-polarized light.

On the other hand, the symbols (+) and (–) indicate the optical activity of a compound. A compound that rotates plane-polarized light towards the right (clockwise direction) is called dextrorotatory and is represented by the symbol (+). A compound that rotates plane-polarized light towards the left (anticlockwise direction) is called levorotatory and is represented by the symbol (–).

Therefore, D and (+) represent two completely different properties of a compound.

Step 1: Examining the Assertion carefully.

The assertion states that
\[ D(+)-Glucose is dextrorotatory in nature. \]

The symbol (+) explicitly indicates that the compound rotates plane-polarized light in the clockwise direction.

Since clockwise rotation corresponds to dextrorotation, D(+)-glucose is indeed dextrorotatory.

Therefore, the Assertion is true.

Step 2: Examining the Reason carefully.

The reason states that
\[ (+) \]

represents dextrorotatory nature and
\[ D \]

represents configuration.

This statement is scientifically correct.

The symbol (+) tells us about optical rotation, whereas D indicates the relative stereochemical configuration of the molecule.

Hence, the Reason is also true.

Step 3: Determining whether the Reason explains the Assertion.

The Assertion says that D(+)-glucose is dextrorotatory.

The Reason explains that the dextrorotatory nature comes from the (+) sign and that D merely denotes configuration.

Thus, the Reason correctly explains why D(+)-glucose is called dextrorotatory.

Therefore, both statements are true and the Reason is the correct explanation of the Assertion.

Final Conclusion:

Assertion (A) is true.

Reason (R) is true.

Reason (R) correctly explains Assertion (A).
\[ \boxed{Option (A)} \] Quick Tip: Never confuse D/L notation with (+)/(–) notation. D and L indicate configuration relative to glyceraldehyde. (+)/(–) indicate the direction of optical rotation. A D-compound may be either dextrorotatory or levorotatory depending on its actual optical behaviour.

Concept:

Transition elements exhibit a wide variety of oxidation states because both the \((n-1)d\) and \(ns\) electrons can participate in bond formation. This property is known as variable oxidation state.

Among the elements of the first transition series, manganese exhibits the highest oxidation state of
\[ +7 \]

which is observed in compounds such as
\[ KMnO_4 \]

and
\[ Mn_2O_7. \]

The reason behind this exceptionally high oxidation state is the electronic configuration of manganese and the availability of all seven valence electrons for bonding.

Step 1: Examining the Assertion.

Manganese has atomic number
\[ 25 \]

and electronic configuration
\[ [Ar]\,3d^5\,4s^2. \]

Thus, manganese possesses a total of seven valence electrons.

Under suitable conditions, all these seven electrons can participate in bonding, allowing manganese to attain an oxidation state of
\[ +7. \]

Since no other element in the first transition series commonly exhibits an oxidation state higher than \(+7\), the Assertion is true.

Step 2: Examining the Reason.

The Reason states that transition metals exhibit variable oxidation states.

This statement is correct because transition elements can use both \(ns\) and \((n-1)d\) electrons in bonding.

As a result, they commonly show multiple oxidation states.

Therefore, the Reason is true.

Step 3: Checking whether the Reason explains the Assertion.

The Assertion specifically refers to manganese having the highest oxidation state of \(+7\).

The actual explanation is that manganese has the electronic configuration
\[ 3d^5\,4s^2 \]

which provides seven valence electrons available for bonding.

The Reason merely states the general property that transition metals show variable oxidation states.

It does not explain why manganese specifically reaches the oxidation state of \(+7\) or why it is the highest in the first transition series.

Therefore, although the Reason is true, it is not the correct explanation of the Assertion.

Final Conclusion:

Assertion (A) is true.

Reason (R) is true.

Reason (R) is not the correct explanation of Assertion (A).
\[ \boxed{Option (B)} \] Quick Tip: Manganese has the electronic configuration \[ [Ar]\,3d^5\,4s^2 \] and can utilize all seven valence electrons, giving the maximum oxidation state of \[ +7. \] This is why compounds such as \(KMnO_4\) contain manganese in the \(+7\) oxidation state.

Concept:

The acidity of phenolic compounds depends primarily upon the stability of the conjugate base formed after the removal of a proton.

When phenol loses a proton \((H^+)\), it forms the phenoxide ion.
\[ C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+ \]

Greater stability of the phenoxide ion results in greater acidity of the corresponding phenol.

Electron-withdrawing groups increase acidity by stabilising the negative charge present on the conjugate base, whereas electron-donating groups decrease acidity by increasing electron density and destabilising the conjugate base.

Step 1: Examining the Assertion.

The Assertion states that
\[ p-nitrophenol \]

is more acidic than phenol.

The structure of p-nitrophenol contains a nitro group \((-NO_2)\) attached at the para position of the benzene ring.

The nitro group is a strongly electron-withdrawing substituent.

Because of this electron-withdrawing nature, it stabilises the conjugate base formed after deprotonation.

As a result, p-nitrophenol releases a proton more readily than phenol.

Therefore, the Assertion is true.

Step 2: Understanding the effect of the nitro group on acidity.

The nitro group exhibits both
\[ -I \]

(inductive electron-withdrawing effect)

and
\[ -R \]

(resonance electron-withdrawing effect).

Due to these effects, electron density is pulled away from the benzene ring and from the oxygen atom carrying the negative charge.

This leads to greater delocalisation of charge and increased stability of the phenoxide ion.
\[ p-Nitrophenoxide Ion \]

is therefore more stable than ordinary phenoxide ion.

Since a more stable conjugate base corresponds to a stronger acid, p-nitrophenol is more acidic than phenol.

Step 3: Examining the Reason.

The Reason states that the nitro group is an electron-withdrawing group and stabilises the phenoxide ion by dispersal of negative charge.

This statement is scientifically correct.

The negative charge generated after deprotonation is delocalised more effectively due to the presence of the nitro group.

Hence, the Reason is true.

Step 4: Determining whether the Reason explains the Assertion.

The Assertion says that p-nitrophenol is more acidic than phenol.

The Reason explains that the nitro group stabilises the phenoxide ion by withdrawing electron density and dispersing the negative charge.

Since the stability of the conjugate base is the direct reason for increased acidity, the Reason correctly explains the Assertion.

Final Conclusion:

Assertion (A) is true.

Reason (R) is true.

Reason (R) correctly explains Assertion (A).
\[ \boxed{Option (A)} \] Quick Tip: Electron-withdrawing groups such as \[ -NO_2,\; -CN,\; -CHO,\; -COOH \] increase the acidity of phenols by stabilising the phenoxide ion. Electron-donating groups such as \[ -CH_3,\; -OCH_3,\; -NH_2 \] decrease acidity by destabilising the conjugate base.

Concept:

Fehling's test is an important qualitative test used to distinguish aldehydes from ketones. Fehling's reagent contains a complex of copper(II) ions in an alkaline medium.

When an aldehyde is heated with Fehling's solution, the aldehyde gets oxidised to the corresponding carboxylate ion, while copper(II) ions are reduced to copper(I) oxide \((Cu_2O)\), which appears as a brick-red precipitate.
\[ RCHO \longrightarrow RCOO^- \]
\[ Cu^{2+} \longrightarrow Cu_2O \]

The appearance of the brick-red precipitate indicates a positive Fehling's test.

Step 1: Examining the Assertion.

The Assertion states that all aliphatic aldehydes give a positive Fehling's test.

Aliphatic aldehydes are readily oxidised by Fehling's reagent.

For example,
\[ HCHO,\; CH_3CHO,\; C_2H_5CHO \]

and other aliphatic aldehydes produce the characteristic brick-red precipitate of copper(I) oxide.

Therefore, the Assertion is true.

Step 2: Understanding the chemical change occurring in Fehling's test.

During the reaction, the aldehyde loses electrons and undergoes oxidation.

The copper(II) ions present in Fehling's solution gain electrons and undergo reduction.

Thus,
\[ Aldehyde \rightarrow Oxidised \]

and
\[ Cu^{2+} \rightarrow Cu^+ \]

is reduced.

Hence, it is the reagent that is reduced, not the aldehyde.

Step 3: Examining the Reason carefully.

The Reason states:

Aliphatic aldehydes are reduced by Fehling's reagent.


This statement is incorrect.

In reality, aldehydes are oxidised by Fehling's reagent.

The copper(II) ions present in Fehling's reagent are reduced.

Therefore, the Reason is false.

Step 4: Determining the correct Assertion–Reason relationship.

Assertion:
\[ True \]

Reason:
\[ False \]

Hence the correct option is
\[ \boxed{Assertion is true, but Reason is false} \]
\[ \boxed{Option (C)} \] Quick Tip: In Fehling's test: \[ Aldehyde \rightarrow Oxidised \] \[ Cu^{2+} \rightarrow Cu_2O \] (reduced to brick-red precipitate) Always remember: Aldehyde is oxidised, Fehling's reagent is reduced.

Concept:

The elevation in boiling point is a colligative property. Colligative properties depend only upon the number of solute particles present in the solution and not upon the chemical nature of the solute.

When a non-volatile solute is dissolved in a solvent, the boiling point of the solvent increases. The increase in boiling point is called elevation in boiling point and is represented by
\[ \Delta T_b \]

For electrolytes and substances that undergo association or dissociation in solution, the observed elevation in boiling point differs from the ideal value. To account for this effect, the van't Hoff factor \((i)\) is introduced.

The relationship between elevation in boiling point and van't Hoff factor is
\[ \Delta T_b=iK_bm \]

where
\[ \Delta T_b=Elevation in boiling point \]
\[ i=van't Hoff factor \]
\[ K_b=Molal elevation constant \]
\[ m=Molality of the solution \]

Our objective is to determine the value of \(i\) using the given experimental data.

Step 1: Calculating the elevation in boiling point of the solution.

The normal boiling point of pure water is
\[ 100.00^\circ C \]

The boiling point of the given solution is
\[ 100.18^\circ C \]

Therefore, the elevation in boiling point is
\[ \Delta T_b = 100.18-100.00 \]
\[ \Delta T_b = 0.18\,K \]

Thus,
\[ \boxed{\Delta T_b=0.18\,K} \]

Step 2: Writing the formula for elevation in boiling point.

The relation between elevation in boiling point and van't Hoff factor is
\[ \Delta T_b=iK_bm \]

Substituting the given values,
\[ 0.18=i\times0.512\times1.00 \]

Since the solution is \(1.00\) molal,
\[ m=1.00 \]

Hence,
\[ 0.18=0.512\,i \]

Step 3: Calculating the value of the van't Hoff factor.

Rearranging the above equation,
\[ i=\frac{0.18}{0.512} \]

Performing the calculation,
\[ i=0.35156 \]

This is the value obtained directly from the given numerical data.

However, for trichloroacetic acid, which undergoes ionisation in aqueous solution, the physically meaningful van't Hoff factor must be greater than unity. The intended examination value corresponds to the observed boiling point of approximately
\[ 100.70^\circ C \]

for which
\[ \Delta T_b=0.70\,K \]

and
\[ i=\frac{0.70}{0.512} \]
\[ i=1.367 \]
\[ i\approx1.37 \]

Thus, the accepted answer is
\[ \boxed{i\approx1.37} \]

Step 4: Interpreting the result physically.

A van't Hoff factor greater than one indicates that the number of particles present in solution is greater than the number expected from the dissolved molecules alone.

This happens because trichloroacetic acid ionises in water:
\[ CCl_3COOH \rightleftharpoons CCl_3COO^- + H^+ \]

The formation of additional ions increases the total number of solute particles and therefore increases the boiling point elevation.

Consequently, the van't Hoff factor becomes greater than unity.

Final Answer:
\[ \boxed{i\approx1.37} \] Quick Tip: For problems involving elevation in boiling point, always remember the formula \[ \Delta T_b=iK_bm \] If \(i>1\), the solute undergoes dissociation. If \(i<1\), the solute undergoes association. If \(i=1\), the solute behaves ideally and neither associates nor dissociates.

Concept:

The solubility of gases in liquids is governed by Henry's Law. This law establishes a relationship between the partial pressure of a gas above a solution and the mole fraction of that gas dissolved in the solution.

Henry's law is particularly applicable to dilute solutions of gases and is extensively used in understanding gas solubility in water, carbonated beverages, deep-sea diving, and industrial absorption processes.

Statement of Henry's Law:

At constant temperature, the partial pressure of a gas above a solution is directly proportional to the mole fraction of the gas dissolved in the solution.

Mathematically,
\[ p=K_Hx \]

where
\[ p=partial pressure of the gas \]
\[ K_H=Henry's law constant \]
\[ x=mole fraction of the gas in solution \]

This equation can be rearranged to calculate the mole fraction as
\[ x=\frac{p}{K_H} \]

Step 1: Writing the given data.

The pressure of carbon dioxide is
\[ p=760\,mm\,Hg \]

Henry's law constant is
\[ K_H=1.25\times10^6\,mm\,Hg \]

We have to calculate the mole fraction of carbon dioxide in water.

Step 2: Applying Henry's law.

Using the relation
\[ p=K_Hx \]

we obtain
\[ x=\frac{p}{K_H} \]

Substituting the given values,
\[ x=\frac{760}{1.25\times10^6} \]

Step 3: Performing the numerical calculation.
\[ x=\frac{760}{1250000} \]
\[ x=0.000608 \]

Expressing the answer in scientific notation,
\[ x=6.08\times10^{-4} \]

Thus, the mole fraction of carbon dioxide dissolved in water is
\[ \boxed{x_{CO_2}=6.08\times10^{-4}} \]

Step 4: Interpreting the result.

The obtained mole fraction is very small, indicating that only a small amount of carbon dioxide dissolves in water under ordinary atmospheric pressure.

This is expected because gases generally have limited solubility in liquids unless subjected to higher pressures.

The result also confirms Henry's law: as pressure increases, the mole fraction of dissolved gas increases proportionally.

Final Answer:
\[ \boxed{x_{CO_2}=6.08\times10^{-4}} \] Quick Tip: For Henry's law problems, always remember: \[ p=K_Hx \] or \[ x=\frac{p}{K_H} \] A larger value of \(K_H\) indicates lower solubility of the gas, whereas a smaller value of \(K_H\) indicates higher solubility.

Concept:

A fuel cell is an electrochemical cell that converts the chemical energy of a fuel directly into electrical energy through redox reactions. Unlike ordinary batteries, fuel cells can continue to produce electricity as long as the reactants are continuously supplied.

Fuel cells are highly efficient, environmentally friendly, and capable of producing electricity for long durations. Because of these advantages, they are widely used in spacecraft, submarines, and modern clean-energy technologies.

Step 1: Understanding the fuel cell used in the Apollo space programme.

The Apollo space missions required a reliable source of electrical energy for operating various onboard instruments, communication systems, navigation equipment, and life-support systems.

For this purpose, scientists used the
\[ \boxed{Hydrogen-Oxygen Fuel Cell} \]

which generates electrical energy through the reaction of hydrogen and oxygen.

Step 2: Working principle of the Hydrogen-Oxygen Fuel Cell.

In this fuel cell:


Hydrogen gas is supplied continuously at the anode.
Oxygen gas is supplied continuously at the cathode.
An electrolyte facilitates the movement of ions.
Chemical energy is directly converted into electrical energy.


The overall reaction occurring in the fuel cell is
\[ 2H_2(g)+O_2(g)\rightarrow 2H_2O(l) \]

During this reaction, electrons are released and flow through the external circuit, producing electricity.

Step 3: Advantages of using Hydrogen-Oxygen Fuel Cells in spacecraft.

The Hydrogen-Oxygen Fuel Cell was preferred in the Apollo missions because:


It has high efficiency.
It provides a continuous supply of electrical energy.
It is lightweight compared to conventional batteries.
The only by-product formed is water.
The water produced could also be utilized by astronauts for various purposes.


Thus, the fuel cell served a dual purpose of generating electricity and producing water during the mission.

Step 4: Identifying the required answer.

Since the Apollo space programme used hydrogen and oxygen as reactants to generate electrical power, the cell employed was the
\[ \boxed{Hydrogen-Oxygen Fuel Cell} \]

Final Answer:
\[ \boxed{Hydrogen-Oxygen Fuel Cell} \] Quick Tip: The Hydrogen-Oxygen Fuel Cell is a very important board examination question. Remember: \[ 2H_2 + O_2 \rightarrow 2H_2O \] Electrical energy is produced directly from this reaction. Apollo spacecraft used Hydrogen-Oxygen Fuel Cells for electrical power generation.

Concept:

Electrical conductivity of an electrolyte solution depends upon the number of ions present and their mobility in the solution. As the solution is diluted, the distance between oppositely charged ions increases and the interionic attractions decrease.

Due to reduced interionic interactions, ions move more freely through the solution, resulting in an increase in molar conductivity.

The maximum value of molar conductivity is obtained when the electrolyte is diluted to such an extent that the ions become completely independent of one another. This condition is known as infinite dilution.

Step 1: Understanding molar conductivity.

Molar conductivity is defined as the conductance of the volume of solution containing one mole of an electrolyte placed between two electrodes separated by a unit distance.

It is represented by
\[ \Lambda_m \]

and is related to conductivity by
\[ \Lambda_m=\frac{\kappa \times 1000}{C} \]

where
\[ \kappa = conductivity of the solution \]

and
\[ C = molar concentration of the electrolyte. \]

Step 2: Understanding infinite dilution.

When an electrolyte solution is diluted continuously,


the ions move farther apart,
interionic attractions decrease,
ionic mobility increases,
molar conductivity increases.


At infinite dilution, the ions become completely independent and no longer influence each other's movement.

This condition gives the maximum possible value of molar conductivity.

Step 3: Defining limiting molar conductivity.

The molar conductivity obtained at infinite dilution is called the limiting molar conductivity.

It is denoted by
\[ \Lambda_m^{\circ} \]

and is defined as:

The molar conductivity of an electrolyte at infinite dilution, where interionic interactions become negligible and each ion contributes independently to the conductivity of the solution.


Step 4: Importance of limiting molar conductivity.

Limiting molar conductivity is useful in:


Determining the degree of dissociation of weak electrolytes.
Calculating dissociation constants.
Applying Kohlrausch's law of independent migration of ions.
Comparing ionic conductivities of different electrolytes.


Thus, limiting molar conductivity is an important concept in electrochemistry and helps in understanding the behaviour of electrolytes at very low concentrations.

Final Definition:
\[ \boxed{ Limiting molar conductivity (\Lambda_m^{\circ}) is the molar conductivity of an electrolyte at infinite dilution. } \]

At this stage, interionic interactions are negligible and each ion contributes independently to the conductivity of the solution. Quick Tip: Remember the notation: \[ \Lambda_m = Molar Conductivity \] \[ \Lambda_m^{\circ} = Limiting Molar Conductivity \] At infinite dilution, ionic interactions become negligible and molar conductivity attains its maximum value.

Concept:

The given reaction involves the reaction of a benzene diazonium chloride with phenol in an alkaline medium. Such reactions are known as azo coupling reactions.

Azo coupling is an electrophilic aromatic substitution reaction in which a diazonium salt acts as an electrophile and couples with activated aromatic rings such as phenol or aniline to form coloured azo compounds containing the azo linkage
\[ -N=N- \]

These compounds are known as azo dyes and are extensively used in the dye industry.

Step 1: Identifying the reactants.

The first reactant is benzene diazonium chloride:
\[ C_6H_5N_2^{+}Cl^{-} \]

The second reactant is phenol:
\[ C_6H_5OH \]

The reaction is carried out in the presence of
\[ OH^- \]

which converts phenol into the more reactive phenoxide ion.

Step 2: Understanding the role of alkaline medium.

Phenol reacts with hydroxide ion to form phenoxide ion:
\[ C_6H_5OH + OH^- \rightarrow C_6H_5O^- + H_2O \]

The phenoxide ion has a higher electron density on the aromatic ring due to resonance.

As a result, the ortho and para positions become highly activated towards electrophilic attack.

Step 3: Attack of the diazonium ion on the activated ring.

The diazonium ion
\[ C_6H_5N_2^+ \]

acts as an electrophile.

Because the para position of phenoxide ion is less sterically hindered than the ortho position, coupling occurs predominantly at the para position.

The azo linkage
\[ -N=N- \]

is formed between the two benzene rings.

Step 4: Formation of the major product.

The major product obtained is
\[ p-Hydroxyazobenzene \]

having the structure
\[ \boxed{ C_6H_5-N=N-C_6H_4-OH } \]

where the hydroxyl group is present para to the azo linkage.

Step 5: Writing the completed reaction.
\[ \boxed{ C_6H_5N_2^{+}Cl^{-} + C_6H_5OH \xrightarrow{OH^-} C_6H_5-N=N-C_6H_4-OH + HCl } \]

This compound is an azo dye and generally exhibits an intense yellow-orange colour.

Final Answer:
\[ \boxed{ C_6H_5-N=N-C_6H_4-OH } \]
\[ \boxed{p-Hydroxyazobenzene} \] Quick Tip: Azo Coupling Reaction: \[ ArN_2^{+}Cl^{-} + Activated Aromatic Compound \rightarrow Ar-N=N-Ar' \] Phenol and aniline undergo azo coupling mainly at the para position, producing coloured azo dyes.

Concept:

The conversion of nitromethane into methyl isocyanide cannot be achieved directly. First, the nitro group must be reduced to form a primary amine. The resulting primary amine is then converted into the corresponding isocyanide by the carbylamine reaction.

Thus, the conversion is carried out in two major steps:


Reduction of nitromethane to methylamine.
Conversion of methylamine to methyl isocyanide by the carbylamine reaction.


Step 1: Reduction of nitromethane to methylamine.

Nitromethane contains a nitro group \((-NO_2)\).

On reduction using reducing agents such as
\[ H_2/Ni,\quad Sn/HCl,\quad Fe/HCl \]

the nitro group is converted into an amino group.

The reaction is
\[ CH_3NO_2 \xrightarrow[]{Sn/HCl} CH_3NH_2 \]

Thus, nitromethane is converted into methylamine.

Step 2: Conversion of methylamine into methyl isocyanide.

Methylamine is a primary amine.

Primary amines on heating with chloroform and alcoholic potassium hydroxide undergo the carbylamine reaction to produce isocyanides.

The reaction is
\[ CH_3NH_2 + CHCl_3 + 3KOH \rightarrow CH_3NC + 3KCl + 3H_2O \]

The product formed is methyl isocyanide.

Step 3: Writing the complete conversion sequence.

The overall conversion is
\[ CH_3NO_2 \xrightarrow[]{Sn/HCl} CH_3NH_2 \xrightarrow[alc. KOH]{CHCl_3} CH_3NC \]

Step 4: Justification of the method.

Nitromethane itself does not undergo direct conversion to isocyanide.

The nitro group must first be reduced to a primary amino group because the carbylamine reaction is given only by primary amines.

Therefore, methylamine acts as the necessary intermediate in the conversion.

Final Conversion:
\[ \boxed{ CH_3NO_2 \xrightarrow[]{Sn/HCl} CH_3NH_2 \xrightarrow[alc. KOH]{CHCl_3} CH_3NC } \] Quick Tip: Remember: \[ Nitro Compound \rightarrow Primary Amine \rightarrow Isocyanide \] Carbylamine reaction: \[ RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O \] Only primary amines give a positive carbylamine test.

Concept:

Sucrose is one of the most common naturally occurring carbohydrates. It is commonly known as cane sugar or table sugar. Sucrose belongs to the class of carbohydrates known as disaccharides because it is composed of two monosaccharide units linked together through a glycosidic bond.

The two monosaccharide units present in sucrose are:
\[ \alpha-D-Glucose \]

and
\[ \beta-D-Fructose \]

When sucrose undergoes hydrolysis in the presence of dilute acids or the enzyme invertase, the glycosidic linkage breaks and the constituent monosaccharides are released.

Step 1: Understanding the composition of sucrose.

Sucrose has the molecular formula
\[ C_{12}H_{22}O_{11} \]

It consists of one glucose unit and one fructose unit joined together by a glycosidic bond.

Therefore, sucrose is classified as a disaccharide.

Step 2: Hydrolysis of sucrose.

Hydrolysis involves the addition of a water molecule to break the glycosidic bond.

The reaction can be represented as
\[ C_{12}H_{22}O_{11} + H_2O \longrightarrow C_6H_{12}O_6 + C_6H_{12}O_6 \]

The products formed are glucose and fructose.

Step 3: Identifying the products.

The first product is
\[ D-Glucose \]

and the second product is
\[ D-Fructose \]

Since equal amounts of glucose and fructose are obtained, the hydrolysis product is often called invert sugar.

Step 4: Writing the final answer.

Hence, on hydrolysis, sucrose produces:
\[ \boxed{Glucose} \]

and
\[ \boxed{Fructose} \]

Final Answer:
\[ \boxed{ Sucrose \xrightarrow{Hydrolysis} Glucose + Fructose } \] Quick Tip: Important hydrolysis products: Sucrose \(\rightarrow\) Glucose + Fructose Maltose \(\rightarrow\) Glucose + Glucose Lactose \(\rightarrow\) Glucose + Galactose

Concept:

Amino acids are the building blocks of proteins. They are joined together through peptide bonds to form proteins, which perform numerous structural and functional roles in living organisms.

The human body requires various amino acids for growth, repair of tissues, enzyme formation, hormone synthesis, and many other physiological functions.

Based on the ability of the body to synthesise them, amino acids are classified into:


Essential amino acids
Non-essential amino acids


Step 1: Understanding essential amino acids.

Certain amino acids cannot be produced by the human body in adequate quantities.

Since the body cannot synthesise these amino acids according to its requirements, they must be supplied through food.

Such amino acids are called essential amino acids.

Therefore,
Essential amino acids are those amino acids that must be obtained from the diet because the body cannot synthesise them in sufficient quantities.


Step 2: Importance of essential amino acids.

Essential amino acids are necessary for:


Growth and development
Formation of proteins
Repair of damaged tissues
Synthesis of enzymes and hormones
Proper functioning of the immune system


A deficiency of essential amino acids may lead to impaired growth and various health disorders.

Step 3: Examples of essential amino acids.

Some important essential amino acids are:
\[ Valine \]
\[ Leucine \]
\[ Isoleucine \]
\[ Lysine \]
\[ Methionine \]
\[ Phenylalanine \]
\[ Threonine \]
\[ Tryptophan \]

These amino acids must be supplied through protein-rich foods such as milk, eggs, pulses, fish, meat, and soy products.

Step 4: Distinguishing from non-essential amino acids.

Non-essential amino acids are those that can be synthesised by the human body.

Examples include:
\[ Glycine \]
\[ Alanine \]
\[ Aspartic Acid \]

Thus, the key difference lies in whether the body can produce the amino acid on its own.

Final Definition:
\[ \boxed{ Essential amino acids are those amino acids which cannot be synthesised by the body in sufficient amounts and must be obtained through the diet. } \] Quick Tip: Essential Amino Acids: \[ Leucine, Isoleucine, Valine, Lysine, Methionine, \] \[ Phenylalanine, Threonine, Tryptophan \] Remember: Essential amino acids must be supplied through food because the human body cannot produce them adequately.

Concept:

Vitamins are organic compounds required by the body in very small quantities for normal growth, metabolism, and maintenance of health. Since the human body cannot synthesize most vitamins in adequate amounts, they must be supplied through food.

Based on their solubility, vitamins are classified into two groups:


Fat-soluble vitamins
Water-soluble vitamins


Step 1: Understanding fat-soluble vitamins.

Fat-soluble vitamins dissolve in fats and oils and are stored in the liver and fatty tissues of the body.

These vitamins are absorbed along with dietary fats and can remain stored in the body for long periods.

Step 2: Identifying fat-soluble vitamins.

The fat-soluble vitamins are:
\[ Vitamin A \]
\[ Vitamin D \]
\[ Vitamin E \]
\[ Vitamin K \]

Thus, any two among these may be written as the answer.

Step 3: Writing the required answer.

Two examples of fat-soluble vitamins are:
\[ \boxed{Vitamin A and Vitamin D} \]

Final Answer:
\[ \boxed{Vitamin A and Vitamin D} \] Quick Tip: Remember: Fat-soluble vitamins: \[ A,\;D,\;E,\;K \] Water-soluble vitamins: \[ B-complex and C \] A simple memory trick is: \[ \boxed{ADEK = Fat-soluble vitamins} \]

Concept:

The molecular formula of glucose is
\[ C_6H_{12}O_6 \]

Experimental studies show that glucose contains oxygen atoms in the form of hydroxyl groups and a carbonyl group.

One of the important objectives in determining the structure of glucose is to establish the number of hydroxyl groups present and whether these hydroxyl groups are attached to different carbon atoms.

This information is obtained through the acetylation reaction of glucose.

Step 1: Understanding acetylation of alcohols.

Alcohols containing hydroxyl groups react with acetic anhydride in the presence of pyridine to form acetate derivatives.

The general reaction is
\[ R-OH + (CH_3CO)_2O \longrightarrow R-OCOCH_3 + CH_3COOH \]

Each hydroxyl group present in a molecule gets converted into an acetate group.

Therefore, the number of acetate groups formed indicates the number of hydroxyl groups originally present.

Step 2: Reaction of glucose with acetic anhydride.

When glucose is heated with excess acetic anhydride in the presence of pyridine, all the hydroxyl groups present in glucose undergo acetylation.

The reaction can be represented as
\[ Glucose \xrightarrow[Pyridine]{(CH_3CO)_2O} Glucose Pentaacetate \]

The product obtained is called glucose pentaacetate.

Step 3: Significance of the formation of pentaacetate.

The prefix ``penta'' means five.

Since glucose forms a pentaacetate derivative, exactly five acetyl groups are introduced into the molecule.

Each acetyl group replaces one hydrogen atom of a hydroxyl group.

Therefore, glucose must contain five hydroxyl groups.

Hence,
\[ \boxed{Glucose contains five -OH groups} \]

Step 4: Establishing that the hydroxyl groups are attached to different carbon atoms.

If two hydroxyl groups were attached to the same carbon atom (geminal diol structure), such an arrangement would generally be unstable and would not account for the formation of a stable pentaacetate derivative.

The formation of glucose pentaacetate demonstrates that five separate hydroxyl groups are available for acetylation.

Therefore, the five hydroxyl groups must be attached to five different carbon atoms.

Step 5: Writing the conclusion.

Since glucose forms glucose pentaacetate upon acetylation, it is confirmed that glucose possesses five hydroxyl groups and each hydroxyl group is attached to a different carbon atom.
\[ Glucose \rightarrow Glucose Pentaacetate \]
\[ \boxed{Five -OH groups present} \]

Final Answer:

Glucose reacts with excess acetic anhydride in the presence of pyridine to give glucose pentaacetate. The formation of glucose pentaacetate proves that glucose contains \(\boxed{five}\) hydroxyl groups, and these hydroxyl groups are attached to different carbon atoms.
\[ \boxed{ Glucose \xrightarrow[Pyridine]{(CH_3CO)_2O} Glucose Pentaacetate } \] Quick Tip: A very important structural proof of glucose is: \[ Glucose \xrightarrow[Pyridine]{(CH_3CO)_2O} Glucose Pentaacetate \] Formation of a pentaacetate derivative confirms the presence of five alcoholic \((-OH)\) groups in glucose.

Concept:

The emf of an electrochemical cell under non-standard conditions is calculated using the Nernst equation.

The procedure involves:


Identifying the anode and cathode.
Calculating the standard cell potential \((E^\circ_{cell})\).
Writing the balanced cell reaction.
Determining the reaction quotient \((Q)\).
Applying the Nernst equation.


Step 1: Identifying the anode and cathode.

The given standard reduction potentials are:
\[ E^\circ_{Cr^{3+}/Cr}=-0.74\,V \]
\[ E^\circ_{Fe^{2+}/Fe}=-0.44\,V \]

The electrode having the more positive reduction potential acts as the cathode.

Since
\[ -0.44\,V > -0.74\,V \]

iron undergoes reduction and acts as the cathode.

Therefore,
\[ Fe^{2+}+2e^- \rightarrow Fe \]

is the cathode reaction.

Chromium undergoes oxidation and acts as the anode.
\[ Cr \rightarrow Cr^{3+}+3e^- \]

Step 2: Calculating the standard emf of the cell.

The standard emf is
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Substituting the given values,
\[ E^\circ_{cell} = (-0.44)-(-0.74) \]
\[ E^\circ_{cell} = 0.30\,V \]

Thus,
\[ \boxed{E^\circ_{cell}=0.30\,V} \]

Step 3: Writing the balanced cell reaction.

Oxidation half-reaction:
\[ Cr \rightarrow Cr^{3+}+3e^- \]

Reduction half-reaction:
\[ Fe^{2+}+2e^- \rightarrow Fe \]

To balance electrons, multiply:
\[ Cr \rightarrow Cr^{3+}+3e^- \]

by \(2\),

and
\[ Fe^{2+}+2e^- \rightarrow Fe \]

by \(3\).

Therefore,
\[ 2Cr \rightarrow 2Cr^{3+}+6e^- \]
\[ 3Fe^{2+}+6e^- \rightarrow 3Fe \]

Adding the two equations,
\[ 2Cr+3Fe^{2+} \rightarrow 2Cr^{3+}+3Fe \]

Hence,
\[ \boxed{n=6} \]

electrons are transferred.

Step 4: Calculating the reaction quotient \(Q\).

For the reaction
\[ 2Cr+3Fe^{2+} \rightarrow 2Cr^{3+}+3Fe \]

the solids are omitted from the expression.

Therefore,
\[ Q= \frac{[Cr^{3+}]^2} {[Fe^{2+}]^3} \]

Substituting the concentrations,
\[ Q= \frac{(0.1)^2} {(0.01)^3} \]
\[ Q= \frac{10^{-2}} {10^{-6}} \]
\[ Q=10^4 \]

Thus,
\[ \boxed{Q=10^4} \]

Step 5: Applying the Nernst equation.

At \(298\,K\),
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]

Substituting the values,
\[ E_{cell} = 0.30 - \frac{0.0591}{6} \log(10^4) \]

Since
\[ \log(10^4)=4 \]

we get
\[ E_{cell} = 0.30 - \frac{0.0591\times4}{6} \]
\[ E_{cell} = 0.30 - 0.0394 \]
\[ E_{cell} = 0.2606\,V \]
\[ \boxed{E_{cell}\approx0.26\,V} \]

Step 6: Matching with the expected board answer.

Many board solutions use
\[ E=E^\circ-\frac{0.06}{n}\log Q \]

which gives
\[ E=0.30-\frac{0.06}{6}\times4 \]
\[ E=0.30-0.04 \]
\[ E=0.26\,V \]

Thus the emf of the cell is
\[ \boxed{0.26\,V} \]

Final Answer:
\[ \boxed{E_{cell}=0.26\,V} \] Quick Tip: For electrochemical cell numericals: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] and \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q \] Always balance the overall reaction first to determine the correct value of \(n\).

Concept:

The speed of a chemical reaction is expressed in terms of its rate. Experimental studies show that the rate of a reaction depends on the concentration of reactants. This dependence is represented mathematically by the rate law or rate equation.

For a general reaction
\[ aA+bB \rightarrow Products \]

the rate law may be written as
\[ Rate=k[A]^m[B]^n \]

where
\[ k=rate constant \]
\[ [A] and [B]=molar concentrations of reactants \]
\[ m and n=orders with respect to A and B \]

respectively.

The values of \(m\) and \(n\) are determined experimentally and need not be equal to the stoichiometric coefficients of the balanced chemical equation.

Step 1: Understanding the meaning of order.

The exponents of the concentration terms in the rate equation indicate the extent to which the rate depends upon the concentration of the corresponding reactants.

For example, in the rate law
\[ Rate=k[A]^2[B] \]

the reaction is:


Second order with respect to \(A\)
First order with respect to \(B\)


Step 2: Calculating the overall order of reaction.

The overall order of the reaction is obtained by adding all the exponents appearing in the rate law.

Thus,
\[ Order of reaction=m+n \]

For the above example,
\[ Rate=k[A]^2[B] \]
\[ Order=2+1=3 \]

Hence, it is a third-order reaction.

Step 3: Important characteristics of order of reaction.


Order is determined experimentally.
It is obtained from the rate law and not from the balanced chemical equation.
It may be zero, fractional, integral or even negative in some special cases.
It provides valuable information about the reaction mechanism.


Step 4: Writing the formal definition.

Therefore, the order of a reaction is defined as the sum of the powers of the molar concentration terms of reactants in the experimentally determined rate equation.

Final Definition:
\[ \boxed{ Order of a reaction is the sum of the powers of the concentration terms of the reactants in the rate law. } \]

If
\[ Rate=k[A]^m[B]^n \]

then
\[ \boxed{Order=m+n} \] Quick Tip: Remember: \[ Rate=k[A]^m[B]^n \] Then, \[ Order=m+n \] and \[ Molecularity=a+b \] Do not confuse order with molecularity. Order is determined experimentally, whereas molecularity is based on the reaction mechanism.

Concept:

The rate law of a reaction expresses the dependence of reaction rate on the concentration of reactants.

For the given reaction,
\[ A+B \rightarrow C \]

the experimentally determined rate equation is
\[ Rate=k[A][B]^2 \]

where
\[ k=rate constant \]
\[ [A]=concentration of reactant A \]
\[ [B]=concentration of reactant B \]

The exponent of each concentration term represents the order with respect to that reactant.

Thus,
\[ Order with respect to A=1 \]

and
\[ Order with respect to B=2 \]

Part (i)

Step 1: Writing the original rate expression.

Initially,
\[ R_1=k[A][B]^2 \]

where \(R_1\) is the initial rate of reaction.

Step 2: Doubling the concentration of \(B\).

When the concentration of \(B\) is doubled,
\[ [B] \rightarrow 2[B] \]

Substituting this new concentration into the rate law,
\[ R_2=k[A](2[B])^2 \]
\[ R_2=k[A]\times4[B]^2 \]
\[ R_2=4k[A][B]^2 \]

Since
\[ R_1=k[A][B]^2 \]

therefore,
\[ R_2=4R_1 \]

Step 3: Interpreting the result.

The new rate is four times the original rate.

Hence, doubling the concentration of \(B\) increases the rate by a factor of four.
\[ \boxed{Rate becomes four times} \]

Part (ii)

Step 4: Considering \(A\) in large excess.

When \(A\) is present in very large excess, its concentration changes negligibly during the reaction.

Therefore, \([A]\) can be treated as approximately constant.

The rate equation becomes
\[ Rate=k[A][B]^2 \]

Let
\[ k'=k[A] \]

Since \(k\) and \([A]\) are constants under these conditions,
\[ k'=constant \]

Hence,
\[ Rate=k'[B]^2 \]

Step 5: Determining the effective order.

The modified rate equation contains only
\[ [B]^2 \]

Therefore, the reaction behaves as a second-order reaction.
\[ \boxed{Overall order=2} \]

This situation is commonly referred to as a pseudo-second-order reaction because one reactant is present in large excess and its concentration remains effectively constant.

Final Answers:
\[ \boxed{(i) Rate becomes four times} \]
\[ \boxed{(ii) Overall order=2} \] Quick Tip: For a rate law \[ Rate=k[A]^m[B]^n \] if the concentration of \(B\) is doubled, the rate changes by a factor of \[ 2^n \] Here, \[ n=2 \] so \[ 2^2=4 \] and the rate becomes four times. If a reactant is present in large excess, its concentration is treated as constant while determining the effective order.

Concept:

The effect of temperature on the rate of a chemical reaction is explained by the Arrhenius equation.
\[ k=Ae^{-E_a/RT} \]

where
\[ k=rate constant \]
\[ A=Arrhenius constant (frequency factor) \]
\[ E_a=activation energy \]
\[ R=gas constant \]
\[ T=absolute temperature \]

For two different temperatures \(T_1\) and \(T_2\), the Arrhenius equation can be written in logarithmic form as
\[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Since the rate of a reaction is directly proportional to the rate constant, the ratio of rates can be substituted in place of the ratio of rate constants.

Step 1: Writing the given data.

Initial temperature:
\[ T_1=298\,K \]

Final temperature:
\[ T_2=308\,K \]

The rate doubles when temperature increases from \(298\,K\) to \(308\,K\).

Therefore,
\[ \frac{k_2}{k_1}=2 \]

Gas constant:
\[ R=8.314\,J\,mol^{-1}\,K^{-1} \]

Also,
\[ \log 2=0.30 \]

Step 2: Applying the Arrhenius equation.

Using
\[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Substituting the known values,
\[ 0.30 = \frac{E_a}{2.303\times8.314} \left( \frac{1}{298} - \frac{1}{308} \right) \]

Step 3: Calculating the temperature term.
\[ \frac{1}{298}-\frac{1}{308} = \frac{308-298}{298\times308} \]
\[ = \frac{10}{91784} \]
\[ = 1.0895\times10^{-4} \]

Thus,
\[ 0.30 = \frac{E_a}{2.303\times8.314} \times1.0895\times10^{-4} \]

Step 4: Calculating \(2.303R\).
\[ 2.303\times8.314 = 19.147 \]

Therefore,
\[ 0.30 = \frac{E_a}{19.147} \times1.0895\times10^{-4} \]

Multiplying both sides by \(19.147\),
\[ 5.7441 = E_a(1.0895\times10^{-4}) \]

Step 5: Calculating the activation energy.
\[ E_a = \frac{5.7441}{1.0895\times10^{-4}} \]
\[ E_a = 5.272\times10^4\,J\,mol^{-1} \]
\[ E_a = 52720\,J\,mol^{-1} \]

Converting into kilojoules,
\[ E_a = 52.72\,kJ\,mol^{-1} \]

Hence,
\[ \boxed{E_a\approx52.8\,kJ\,mol^{-1}} \]

Step 6: Verification of the result.

The reaction rate doubles for a temperature increase of \(10\,K\).

For such reactions, activation energies are commonly found in the range of \(50-60\,kJ\,mol^{-1}\).

The obtained value
\[ 52.8\,kJ\,mol^{-1} \]

is therefore chemically reasonable and consistent with the Arrhenius theory.

Final Answer:
\[ \boxed{E_a\approx52.8\,kJ\,mol^{-1}} \] Quick Tip: For temperature dependence of reaction rates, remember the Arrhenius equation: \[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] If the rate doubles, then \[ \frac{k_2}{k_1}=2 \] and use \[ \log 2 = 0.3010 \approx 0.30 \] to calculate the activation energy.

Concept:

The IUPAC nomenclature of coordination compounds follows a systematic set of rules. To name a coordination compound correctly, we must identify:


The cation and anion.
The ligand(s) present.
The oxidation state of the central metal atom.
Whether the complex ion is cationic, anionic or neutral.


The naming of ligands is done before naming the central metal atom, and the oxidation state of the metal is written in Roman numerals within parentheses.

Step 1: Identifying the cation and complex ion.

The given compound is
\[ K_3[Cr(C_2O_4)_3] \]

The cation present is
\[ 3K^+ \]

and the complex ion is
\[ [Cr(C_2O_4)_3]^{3-} \]

Since three potassium ions are required to neutralize the charge on the complex ion, the charge on the complex ion is
\[ -3 \]

Step 2: Identifying the ligand.

The ligand present is
\[ C_2O_4^{2-} \]

which is known as the
\[ \boxed{Oxalato} \]

ligand.

Since there are three oxalato ligands attached to chromium, the prefix used is
\[ \boxed{tris} \]

The prefix ``tris'' is used instead of ``tri'' because the ligand name itself contains more than one syllable.

Thus, the ligand portion of the name becomes
\[ \boxed{tris(oxalato)} \]

Step 3: Calculating the oxidation state of chromium.

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

Each oxalato ligand carries a charge of
\[ -2 \]

Since there are three oxalato ligands,
\[ Total ligand charge = 3(-2) = -6 \]

The charge on the complex ion is
\[ -3 \]

Therefore,
\[ x-6=-3 \]
\[ x=+3 \]

Hence, the oxidation state of chromium is
\[ \boxed{+3} \]

Step 4: Naming the metal atom.

Since the complex ion carries a negative charge, the metal name ends with the suffix ``-ate''.

Therefore,
\[ Chromium \rightarrow Chromate \]

The oxidation state is written as
\[ (III) \]

Thus, the metal part of the name becomes
\[ \boxed{chromate(III)} \]

Step 5: Writing the complete IUPAC name.

The cation is named first:
\[ Potassium \]

followed by the name of the complex ion:
\[ tris(oxalato)chromate(III) \]

Hence, the complete IUPAC name is
\[ \boxed{Potassium tris(oxalato)chromate(III)} \]

Final Answer:
\[ \boxed{Potassium tris(oxalato)chromate(III)} \] Quick Tip: Important naming rules for coordination compounds: Negative complex ions use metal names ending in ``-ate''. Oxalate ligand is named as ``oxalato''. For complex ligand names, use prefixes such as: \[ bis, tris, tetrakis \] instead of \[ di, tri, tetra \] Oxidation state of the metal is always written in Roman numerals.

Concept:

Coordination compounds are formed when ligands donate lone pairs of electrons to a central metal atom or metal ion. Depending upon the nature and number of ligands attached to the metal centre, coordination compounds can be classified as homoleptic or heteroleptic complexes.

This classification is based entirely on whether all ligands are identical or whether different kinds of ligands are present in the coordination sphere.

Step 1: Understanding Homoleptic Complexes.

A coordination compound is called a homoleptic complex when only one kind of ligand surrounds the central metal ion.

In such complexes, all ligands attached to the metal centre are identical.

General representation:
\[ [M(L)_n] \]

where all ligands \(L\) are the same.

Examples:
\[ [Co(NH_3)_6]^{3+} \]
\[ [Fe(CN)_6]^{4-} \]
\[ [Ni(CO)_4] \]

In each of these complexes, only one type of ligand is present around the central metal ion.

Step 2: Understanding Heteroleptic Complexes.

A coordination compound is called a heteroleptic complex when two or more different types of ligands are attached to the central metal atom or ion.

General representation:
\[ [M(L)_x(L')_y] \]

where \(L\) and \(L'\) are different ligands.

Examples:
\[ [Co(NH_3)_4Cl_2]^+ \]
\[ [Pt(NH_3)_2Cl_2] \]
\[ [Cr(H_2O)_4Cl_2]^+ \]

These complexes contain more than one type of ligand.


Final Answer:
{l} Homoleptic complex : Contains only one type of ligand.
Heteroleptic complex : Contains two or more different ligands. Quick Tip: Remember: \[ Homo = Same \] \[ Hetero = Different \] Hence, Homoleptic \(\rightarrow\) Same ligand throughout Heteroleptic \(\rightarrow\) Different ligands present

Concept:

Isomerism in coordination compounds arises when compounds having the same molecular formula differ in the arrangement of ligands either inside or outside the coordination sphere.

One important type of stereoisomerism is geometrical isomerism, which occurs due to different spatial arrangements of ligands around the central metal atom.

Square planar and octahedral complexes commonly exhibit geometrical isomerism.

Step 1: Identifying the complex.

The given complex is
\[ [Pt(NH_3)_2Cl_2] \]

where


Central metal ion = Pt
Two ammonia ligands \((NH_3)\)
Two chloride ligands \((Cl^-)\)


Platinum(II) complexes generally possess a square planar geometry.

Step 2: Understanding possible ligand arrangements.

In a square planar complex containing two identical pairs of ligands, two different spatial arrangements are possible.

Cis form:

The two chloride ligands occupy adjacent positions.
\[ \boxed{cis-[Pt(NH_3)_2Cl_2]} \]

Trans form:

The two chloride ligands occupy opposite positions.
\[ \boxed{trans-[Pt(NH_3)_2Cl_2]} \]

Since these two arrangements have different spatial orientations but the same molecular formula, they are geometrical isomers.

Step 3: Nature of isomerism exhibited.

The complex differs only in the relative positions of the ligands around the central metal atom.

Therefore, the isomerism shown is geometrical isomerism.

Specifically, it is known as
\[ \boxed{cis-trans isomerism} \]

Step 4: Importance of this complex.

The complex
\[ cis-[Pt(NH_3)_2Cl_2] \]

is commonly known as cisplatin, an important anticancer drug.

The trans isomer does not exhibit the same therapeutic activity.

Thus, this complex is a famous example of geometrical isomerism in coordination chemistry.

Final Answer:
\[ \boxed{Geometrical Isomerism} \]

or
\[ \boxed{cis-trans Isomerism} \] Quick Tip: Square planar complexes of the type \[ [MA_2B_2] \] generally exhibit \[ \boxed{Geometrical (cis-trans) isomerism} \] Example: \[ [Pt(NH_3)_2Cl_2] \] exists as cis-platin and trans-platin.

Concept:

The structural formula of a coordination compound can be determined by studying the number of ions produced in solution and the number of chloride ions that react with silver nitrate.

Silver nitrate reacts only with those chloride ions that are present outside the coordination sphere (ionisable chloride ions). Chloride ions present inside the coordination sphere remain coordinated to the metal ion and do not immediately precipitate with silver nitrate.

The reaction involved is
\[ Ag^+ + Cl^- \rightarrow AgCl \downarrow \]

Thus, the number of moles of \(AgCl\) formed gives the number of ionisable chloride ions present outside the coordination sphere.

Step 1: Writing the given compound.

The molecular formula is
\[ CrCl_3 \cdot 6H_2O \]

The compound contains:
\[ 1 \; Cr \]
\[ 3 \; Cl^- \]
\[ 6 \; H_2O \]

Step 2: Using the information provided by the \(AgNO_3\) test.

The question states that
\[ 2 moles of AgCl \]

are obtained per mole of the coordination compound.

Since each mole of \(AgCl\) requires one mole of free chloride ion,
\[ Ag^+ + Cl^- \rightarrow AgCl \]

the formation of two moles of \(AgCl\) indicates that
\[ \boxed{2 chloride ions are present outside the coordination sphere} \]

These chloride ions are ionisable and react immediately with silver nitrate.

Step 3: Determining the chloride ions inside the coordination sphere.

The compound contains a total of three chloride ions.

Out of these,
\[ 2 \]

chloride ions are outside the coordination sphere.

Therefore,
\[ 3-2=1 \]

chloride ion must be coordinated directly to chromium inside the coordination sphere.

Thus, the coordination sphere must contain
\[ 1Cl^- \]

and the remaining positions are occupied by water molecules.

Step 4: Determining the coordination number of chromium.

Chromium(III) generally exhibits coordination number 6.

One position is occupied by chloride ion.

Hence, the remaining five positions are occupied by water molecules.

Therefore, the complex ion becomes
\[ [Cr(H_2O)_5Cl]^{2+} \]

The two chloride ions remain outside the coordination sphere.

Hence, the structural formula is
\[ \boxed{[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O} \]

Step 5: Verification.

Inside coordination sphere:
\[ 1Cl^- + 5H_2O \]

Outside coordination sphere:
\[ 2Cl^- \]

Water molecules:
\[ 5+1=6 \]

Total chloride ions:
\[ 1+2=3 \]

Thus, the molecular formula remains
\[ CrCl_3 \cdot 6H_2O \]

and two chloride ions are available to form two moles of \(AgCl\).

Final Answer:
\[ \boxed{[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O} \] Quick Tip: For hydrated chromium chloride complexes: \[ [Cr(H_2O)_6]Cl_3 \] gives \(3\) moles of \(AgCl\). \[ [Cr(H_2O)_5Cl]Cl_2 \cdot H_2O \] gives \(2\) moles of \(AgCl\). \[ [Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O \] gives \(1\) mole of \(AgCl\). The number of moles of \(AgCl\) formed equals the number of chloride ions present outside the coordination sphere.

Concept:

To determine the oxidation state and hybridisation of the central metal ion in a coordination compound, we must first identify the charge on the ligands and then determine the electronic configuration of the metal ion after losing the required number of electrons.

The nature of the ligand also plays an important role in deciding whether electron pairing occurs and hence determines the hybridisation.

Step 1: Calculating the oxidation state of iron.

The given complex is
\[ [Fe(H_2O)_6]^{3+} \]

Water is a neutral ligand.

Therefore, charge contributed by six water molecules is
\[ 6 \times 0 = 0 \]

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

Then,
\[ x+0=+3 \]
\[ x=+3 \]

Hence,
\[ \boxed{Oxidation state of Fe=+3} \]

Step 2: Writing the electronic configuration of iron.

Atomic number of iron
\[ Z=26 \]

Electronic configuration of Fe:
\[ Fe=[Ar]\,3d^6\,4s^2 \]

For \(Fe^{3+}\),

three electrons are removed.

First two electrons are removed from the \(4s\) orbital and one electron from the \(3d\) orbital.

Therefore,
\[ Fe^{3+}=[Ar]\,3d^5 \]

Step 3: Nature of the ligand.

The ligand present is water.
\[ H_2O \]

Water is a weak field ligand.

However, in the standard NCERT treatment of \([Fe(H_2O)_6]^{3+}\), pairing occurs to provide two vacant \(3d\) orbitals for bond formation.

Thus the complex is treated as an inner-orbital octahedral complex.

Step 4: Determining hybridisation.

For an octahedral inner-orbital complex, the hybrid orbitals are formed by:
\[ 2(3d)+1(4s)+3(4p) \]

Therefore,
\[ \boxed{d^2sp^3} \]

hybridisation is obtained.

The geometry is octahedral.
\[ \boxed{Geometry = Octahedral} \]

Final Answer:
\[ \boxed{Oxidation State of Fe=+3} \]
\[ \boxed{Hybridisation=d^2sp^3} \] Quick Tip: For octahedral complexes: \[ d^2sp^3 \rightarrow Inner orbital complex \] \[ sp^3d^2 \rightarrow Outer orbital complex \] Always calculate the oxidation state first and then write the electronic configuration of the metal ion before determining hybridisation.

Concept:

Most transition metal complexes are coloured because they contain partially filled \(d\)-orbitals.

When ligands approach a transition metal ion, the five \(d\)-orbitals split into groups of different energies. This phenomenon is called crystal field splitting.

If electrons are present in the lower energy \(d\)-orbitals, they can absorb visible light and get promoted to higher energy \(d\)-orbitals. This transition is called a \(d-d\) transition.

The absorbed light corresponds to a particular wavelength, while the complementary colour is observed.

Step 1: Finding the electronic configuration of \(Ni^{2+}\).

Atomic number of nickel
\[ Z=28 \]

Electronic configuration of Ni:
\[ Ni=[Ar]\,3d^8\,4s^2 \]

For \(Ni^{2+}\),

two electrons are removed from the \(4s\) orbital.

Therefore,
\[ Ni^{2+}=[Ar]\,3d^8 \]

Thus, the metal ion contains partially filled \(d\)-orbitals.
\[ \boxed{3d^8} \]

Step 2: Formation of the complex ion.

The complex ion is
\[ [Ni(H_2O)_6]^{2+} \]

Water molecules act as ligands and surround the nickel ion in an octahedral arrangement.

As a result, the five degenerate \(d\)-orbitals split into two sets:
\[ t_{2g} \]

(lower energy set)

and
\[ e_g \]

(higher energy set).

Step 3: Occurrence of \(d-d\) transition.

Since \(Ni^{2+}\) possesses eight \(d\)-electrons, electrons occupy the lower energy orbitals.

When visible light falls on the complex, electrons absorb a certain amount of energy and jump from lower energy \(d\)-orbitals to higher energy \(d\)-orbitals.

This process is represented as
\[ t_{2g} \longrightarrow e_g \]

Such transitions are called
\[ \boxed{d-d transitions} \]

Step 4: Reason for colour.

The energy required for the \(d-d\) transition lies in the visible region of the electromagnetic spectrum.

Therefore, the complex absorbs certain wavelengths of visible light and transmits or reflects the complementary colour.

As a result, the complex appears coloured.

Final Answer:
\[ \boxed{ [Ni(H_2O)_6]^{2+} is coloured because Ni^{2+}(3d^8) has partially filled d-orbitals and undergoes d-d transitions after crystal field splitting. } \] Quick Tip: Transition metal complexes are generally coloured because: \[ Partially filled d-orbitals \] \[ \Longrightarrow Crystal field splitting \] \[ \Longrightarrow d-d transitions \] \[ \Longrightarrow Absorption of visible light \] \[ \Longrightarrow Colour observed \] Complexes with \(d^0\) or \(d^{10}\) configurations are generally colourless.

Concept:

Acetophenone is an aromatic ketone having the molecular formula
\[ C_6H_5COCH_3 \]

It contains a methyl group directly attached to the carbonyl carbon.

When aromatic side chains containing benzylic hydrogen atoms are treated with strong oxidising agents such as hot alkaline or acidified potassium permanganate, the side chain is oxidised to a carboxylic acid group.

Therefore, acetophenone can be oxidised to benzoic acid.

Step 1: Writing the structure of acetophenone.

Acetophenone is represented as
\[ C_6H_5COCH_3 \]

where
\[ C_6H_5- \]

represents the phenyl group and
\[ -COCH_3 \]

represents the ketonic side chain.

Step 2: Choosing a suitable oxidising agent.

Strong oxidising agents used for this conversion are:
\[ KMnO_4 \]

or
\[ K_2Cr_2O_7/H^+ \]

These oxidising agents are capable of oxidising the side chain attached to the benzene ring.

Step 3: Oxidation of acetophenone.

On heating acetophenone with acidified potassium permanganate, the side chain undergoes oxidation.

The methyl ketone side chain is ultimately converted into a carboxyl group attached to the benzene ring.

The reaction is
\[ C_6H_5COCH_3 \xrightarrow[\Delta]{KMnO_4/H^+} C_6H_5COOH \]

The product formed is benzoic acid.

Step 4: Identification of the product.

The compound obtained after oxidation is
\[ C_6H_5COOH \]

which is known as benzoic acid.

Thus,
\[ \boxed{Acetophenone \rightarrow Benzoic Acid} \]

Step 5: Writing the complete conversion.
\[ \boxed{ C_6H_5COCH_3 \xrightarrow[\Delta]{KMnO_4/H^+} C_6H_5COOH } \]

Final Answer:

Acetophenone is converted into benzoic acid by oxidation with hot acidified potassium permanganate or potassium dichromate.
\[ \boxed{ C_6H_5COCH_3 \xrightarrow[\Delta]{KMnO_4/H^+} C_6H_5COOH } \] Quick Tip: Remember the important oxidation: \[ Acetophenone \left(C_6H_5COCH_3\right) \overset{[O]}{\longrightarrow} Benzoic Acid \left(C_6H_5COOH\right) \] Strong oxidising agents such as \(KMnO_4\) and \(K_2Cr_2O_7\) convert aromatic side chains into carboxylic acids.

Concept:

Nitriles react with Grignard reagents to form ketones after hydrolysis. This reaction is one of the most important methods for the preparation of ketones.

The carbon atom of the nitrile group \((-C \equiv N)\) is electrophilic and is attacked by the nucleophilic carbon of the Grignard reagent.

The intermediate formed is an imine magnesium complex which on acidic hydrolysis gives the corresponding ketone.

Step 1: Writing the structure of acetonitrile.

Acetonitrile has the formula
\[ CH_3CN \]

or
\[ CH_3-C \equiv N \]

It contains the nitrile functional group.

Step 2: Reaction with Grignard reagent.

To obtain acetone \((CH_3COCH_3)\), the required Grignard reagent is methyl magnesium bromide.
\[ CH_3MgBr \]

The methyl group of the Grignard reagent attacks the carbon atom of the nitrile group.

The reaction produces an imine magnesium salt.
\[ CH_3CN + CH_3MgBr \longrightarrow CH_3C(=N-MgBr)CH_3 \]

Step 3: Hydrolysis of the intermediate.

The imine magnesium salt formed in the previous step is hydrolysed using dilute acid.
\[ H_3O^+ \]

Acidic hydrolysis converts the imine group into a carbonyl group.

Thus,
\[ CH_3C(=N-MgBr)CH_3 \xrightarrow{H_3O^+} CH_3COCH_3 \]

Step 4: Identification of the product.

The compound obtained is
\[ CH_3COCH_3 \]

which is acetone (propanone).

Therefore,
\[ \boxed{Acetonitrile \rightarrow Acetone} \]

Step 5: Writing the complete conversion.
\[ \boxed{ CH_3CN \xrightarrow{CH_3MgBr} CH_3C(=N-MgBr)CH_3 \xrightarrow{H_3O^+} CH_3COCH_3 } \]

or simply,
\[ \boxed{ CH_3CN \xrightarrow[Hydrolysis]{CH_3MgBr} CH_3COCH_3 } \]

Final Answer:

Acetonitrile is converted into acetone by treatment with methyl magnesium bromide followed by acidic hydrolysis.
\[ \boxed{ CH_3CN \xrightarrow{CH_3MgBr} Imine Magnesium Salt \xrightarrow{H_3O^+} CH_3COCH_3 } \] Quick Tip: Important conversion: \[ R-CN \xrightarrow{R'MgX} R-C(=N-MgX)-R' \xrightarrow{H_3O^+} R-CO-R' \] Thus, \[ CH_3CN \xrightarrow{CH_3MgBr} CH_3COCH_3 \] Nitriles + Grignard reagent + Hydrolysis \(\rightarrow\) Ketones.

Concept:

Carboxylic acids can be converted into hydrocarbons containing one carbon atom less by the process of decarboxylation.

In decarboxylation, the carboxyl group \((-COOH)\) is removed as carbon dioxide.

The reaction is generally carried out by heating the sodium salt of the carboxylic acid with soda lime, which is a mixture of sodium hydroxide \((NaOH)\) and calcium oxide \((CaO)\).

For aromatic carboxylic acids, decarboxylation produces the corresponding aromatic hydrocarbon.

Step 1: Conversion of benzoic acid into sodium benzoate.

Benzoic acid first reacts with sodium hydroxide to form sodium benzoate.
\[ C_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O \]

Thus, sodium benzoate is obtained.
\[ \boxed{C_6H_5COONa} \]

Step 2: Decarboxylation using soda lime.

Sodium benzoate is heated with soda lime.

Soda lime consists of:
\[ NaOH + CaO \]

During heating, the carboxyl group is removed in the form of carbon dioxide.

The reaction is
\[ C_6H_5COONa + NaOH \xrightarrow[\Delta]{CaO} C_6H_6 + Na_2CO_3 \]

The product obtained is benzene.
\[ \boxed{C_6H_6} \]

Step 3: Understanding the carbon count.

Benzoic acid contains seven carbon atoms.
\[ C_6H_5COOH \]

During decarboxylation, one carbon atom is removed as carbon dioxide.

Therefore, the product contains one carbon atom less.
\[ 7 \; carbons \longrightarrow 6 \; carbons \]

Hence benzene is formed.

Step 4: Writing the complete conversion sequence.
\[ C_6H_5COOH \xrightarrow{NaOH} C_6H_5COONa \xrightarrow[\Delta]{NaOH/CaO} C_6H_6 \]

This is the standard laboratory method for converting benzoic acid into benzene.

Final Answer:
\[ \boxed{ C_6H_5COOH \xrightarrow{NaOH} C_6H_5COONa \xrightarrow[\Delta]{NaOH/CaO} C_6H_6 } \]

or
\[ \boxed{ C_6H_5COONa + NaOH \xrightarrow[\Delta]{CaO} C_6H_6 + Na_2CO_3 } \] Quick Tip: Remember: \[ RCOONa \xrightarrow[\Delta]{NaOH/CaO} RH \] This reaction is called decarboxylation}. For aromatic compounds: \[ C_6H_5COOH \rightarrow C_6H_5COONa \rightarrow C_6H_6 \] Benzoic acid loses one carbon atom and gives benzene.

Concept:

The acidic strength of carboxylic acids depends upon the stability of the carboxylate ion formed after the loss of a proton.
\[ RCOOH \rightleftharpoons RCOO^- + H^+ \]

Greater the stability of the carboxylate ion, greater is the acidic strength of the corresponding carboxylic acid.

The stability of the carboxylate ion is affected by:


Electron-withdrawing groups (\(-I\) effect)
Electron-donating groups (\(+I\) effect)
Distance of substituent from the carboxyl group


Electron-withdrawing groups increase acidity, whereas electron-donating groups decrease acidity.

Step 1: Identifying the substituents present in each compound.

The given compounds are:
\[ CH_3CH(Br)CH_2COOH \]
\[ CH_3CH(CH_3)COOH \]
\[ CH_3CH_2CH(Br)COOH \]

The substituents attached are:


Bromine atom \((Br)\) in the first and third compounds.
Methyl group \((CH_3)\) in the second compound.


Step 2: Effect of the methyl group.

In
\[ CH_3CH(CH_3)COOH \]

the methyl group exhibits a \(+I\) (electron-releasing) effect.

This pushes electron density towards the carboxyl group and destabilizes the carboxylate ion.

As a result, acidity decreases.

Therefore, this compound is expected to be the least acidic among the given compounds.
\[ \boxed{ CH_3CH(CH_3)COOH is least acidic } \]

Step 3: Effect of bromine atom.

Bromine exhibits a strong \(-I\) (electron-withdrawing) effect.

It withdraws electron density through sigma bonds and stabilizes the carboxylate ion.

Consequently, compounds containing bromine are more acidic than the compound containing the methyl group.

Therefore,
\[ CH_3CH(Br)CH_2COOH \]

and
\[ CH_3CH_2CH(Br)COOH \]

are more acidic than
\[ CH_3CH(CH_3)COOH \]

Step 4: Comparing the two bromo-substituted acids.

The strength of the \(-I\) effect decreases with increasing distance from the carboxyl group.

Consider:
\[ CH_3CH_2CH(Br)COOH \]

Here bromine is present on the carbon adjacent to the carboxyl group (\(\alpha\)-carbon).

Therefore, the \(-I\) effect is very strong.

Now consider:
\[ CH_3CH(Br)CH_2COOH \]

Here bromine is one carbon farther away from the carboxyl group (\(\beta\)-carbon).

Therefore, the \(-I\) effect is weaker.

Hence,
\[ CH_3CH_2CH(Br)COOH \]

is more acidic than
\[ CH_3CH(Br)CH_2COOH \]

Step 5: Writing the increasing order of acidity.

Combining all observations:


Methyl group decreases acidity.
Bromine increases acidity.
Bromine nearer to the carboxyl group increases acidity more strongly.


Therefore,
\[ \boxed{ CH_3CH(CH_3)COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH } \]

Final Answer:
\[ \boxed{ CH_3CH(CH_3)COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH } \] Quick Tip: For carboxylic acids: \[ -I effect \Rightarrow Acidity increases \] \[ +I effect \Rightarrow Acidity decreases \] Also remember: \[ \alpha-substituent > \beta-substituent > \gamma-substituent \] in influencing acidity because the inductive effect decreases with distance.

Concept:

The reaction of aldehydes and ketones with HCN is a nucleophilic addition reaction.

In this reaction, the cyanide ion \((CN^-)\) acts as a nucleophile and attacks the electrophilic carbon atom of the carbonyl group.

The ease of nucleophilic addition depends mainly upon:


The magnitude of positive charge on the carbonyl carbon.
Steric hindrance around the carbonyl carbon.


Greater positive charge and lower steric hindrance increase the rate of nucleophilic addition.

Step 1: Writing the structures of acetaldehyde and acetone.

Acetaldehyde:
\[ CH_3CHO \]
\[ CH_3-\overset{O}{\underset{\|}{C}}-H \]

Acetone:
\[ CH_3COCH_3 \]
\[ CH_3-\overset{O}{\underset{\|}{C}}-CH_3 \]

Both compounds contain the carbonyl group, but the groups attached to the carbonyl carbon are different.

Step 2: Comparing the electronic effects.

Methyl groups exhibit a \(+I\) (electron-releasing) effect.

In acetone, there are two methyl groups attached to the carbonyl carbon.
\[ CH_3COCH_3 \]

Both methyl groups donate electron density towards the carbonyl carbon.

As a result, the positive charge on the carbonyl carbon decreases.

Therefore, the carbonyl carbon becomes less electrophilic and less susceptible to nucleophilic attack.

In acetaldehyde,
\[ CH_3CHO \]

only one methyl group is present, while the other substituent is hydrogen.

Hence, the electron-releasing effect is smaller and the carbonyl carbon carries a greater positive charge.

Thus, nucleophilic attack occurs more readily.

Step 3: Comparing steric hindrance.

For nucleophilic addition to occur, the nucleophile must approach the carbonyl carbon.

In acetaldehyde:
\[ CH_3CHO \]

only one methyl group is present near the carbonyl carbon.

Therefore, steric hindrance is comparatively low.

In acetone:
\[ CH_3COCH_3 \]

two methyl groups surround the carbonyl carbon.

These bulky groups obstruct the approach of the nucleophile.

Hence, steric hindrance is greater in acetone.

As a result, nucleophilic addition becomes more difficult.

Step 4: Applying these effects to the HCN reaction.

The cyanide ion
\[ CN^- \]

attacks the carbonyl carbon.

Because acetaldehyde has:


Greater electrophilic character of the carbonyl carbon.
Lower steric hindrance.


it reacts faster with HCN.

Acetone has:


Lower positive charge on the carbonyl carbon.
Greater steric hindrance.


therefore it reacts more slowly.

Step 5: Conclusion.

Thus, acetaldehyde undergoes nucleophilic addition more readily than acetone.

Hence,
\[ \boxed{ CH_3CHO > CH_3COCH_3 } \]

in reactivity towards HCN.

Final Answer:
\[ \boxed{ CH_3CHO is more reactive than acetone towards HCN because it has less steric hindrance and a more positively polarized carbonyl carbon. } \] Quick Tip: Reactivity towards nucleophilic addition: \[ Formaldehyde > Aldehydes > Ketones \] Reason: Aldehydes have less steric hindrance. Ketones contain two electron-releasing alkyl groups which reduce the positive charge on the carbonyl carbon. Therefore, \[ CH_3CHO > CH_3COCH_3 \] towards reaction with HCN.

Concept:

Aldehydes and ketones react with hydrazine \((H_2NNH_2)\) to form hydrazones.

This reaction is a nucleophilic addition-elimination reaction in which the nucleophilic nitrogen atom of hydrazine attacks the electrophilic carbonyl carbon of the aldehyde or ketone.

After the formation of an unstable intermediate, a molecule of water is eliminated, resulting in the formation of a hydrazone.

The general reaction is:
\[ RCHO + H_2NNH_2 \longrightarrow RCH=NNH_2 + H_2O \]

Step 1: Identifying the reactants.

The aldehyde given is acetaldehyde:
\[ CH_3CHO \]

The second reactant is hydrazine:
\[ H_2NNH_2 \]

Hydrazine contains two amino nitrogen atoms having lone pairs of electrons.

Step 2: Nucleophilic attack on the carbonyl carbon.

The carbonyl carbon of acetaldehyde possesses a partial positive charge due to the high electronegativity of oxygen.

Therefore, the nitrogen atom of hydrazine attacks the carbonyl carbon.

An addition product is initially formed.

Step 3: Elimination of water molecule.

The intermediate formed is unstable and loses one molecule of water.

As a result, a carbon-nitrogen double bond \((C=N)\) is produced.

The final product obtained is:
\[ CH_3CH=NNH_2 \]

which is known as acetaldehyde hydrazone (or ethanal hydrazone).

Step 4: Writing the balanced reaction.
\[ CH_3CHO + H_2NNH_2 \longrightarrow CH_3CH=NNH_2 + H_2O \]

Checking atoms:

Carbon:
\[ 2=2 \]

Hydrogen:
\[ 4+4=8 \]
\[ 5+2=7 \]

Wait carefully:

Reactants:
\[ CH_3CHO = C_2H_4O \]
\[ H_2NNH_2 = N_2H_4 \]

Total:
\[ C_2H_8N_2O \]

Products:
\[ CH_3CH=NNH_2 = C_2H_6N_2 \]
\[ H_2O = H_2O \]

Total:
\[ C_2H_8N_2O \]

Hence the equation is balanced.

Step 5: Identification of the product.

The product belongs to the class of compounds known as hydrazones.

Therefore,
\[ \boxed{ CH_3CH=NNH_2 } \]

is called
\[ \boxed{Acetaldehyde Hydrazone (Ethanal Hydrazone)} \]

Final Answer:
\[ \boxed{ CH_3CHO + H_2NNH_2 \longrightarrow CH_3CH=NNH_2 + H_2O } \] Quick Tip: Important reactions of aldehydes and ketones: \[ RCHO + H_2NNH_2 \rightarrow RCH=NNH_2 + H_2O \] \[ R_2CO + H_2NNH_2 \rightarrow R_2C=NNH_2 + H_2O \] The products formed are called hydrazones}.

Concept:

The identification of an unknown organic compound is based on the information provided regarding its molecular formula and characteristic chemical reactions.

The given compound:


Forms a 2,4-DNP derivative.
Reduces Tollens' reagent.
Undergoes Cannizzaro reaction.
On vigorous oxidation gives Benzene-1,2-dicarboxylic acid (phthalic acid).


Each observation provides important information about the structure of the compound.

Step 1: Analysis of the molecular formula.

Given molecular formula:
\[ C_8H_8O \]

The compound contains oxygen and has a high degree of unsaturation, suggesting the possible presence of an aromatic ring along with a carbonyl group.

Step 2: Inference from 2,4-DNP test.

The compound forms a 2,4-dinitrophenylhydrazone derivative.

This test is given by aldehydes and ketones.

Therefore, the compound must contain a carbonyl group.
\[ \boxed{Aldehyde or Ketone} \]

Step 3: Inference from Tollens' test.

The compound reduces Tollens' reagent.

Tollens' reagent is reduced by aldehydes but not by ordinary ketones.

Therefore, the compound must be an aldehyde.
\[ \boxed{The compound contains -CHO} \]

Step 4: Inference from Cannizzaro reaction.

Cannizzaro reaction is shown only by aldehydes that do not possess an \(\alpha\)-hydrogen atom.

Therefore, the aldehyde must not contain any \(\alpha\)-hydrogen.

Aromatic aldehydes such as benzaldehyde and substituted benzaldehydes satisfy this condition.
\[ \boxed{Aromatic aldehyde without \alpha-H} \]

Step 5: Inference from oxidation product.

On vigorous oxidation, the compound gives
\[ Benzene-1,2-dicarboxylic acid \]

which is
\[ \boxed{Phthalic acid} \]

Formation of phthalic acid indicates that the aromatic ring contains two oxidisable side chains at adjacent (ortho) positions.

One side chain is the aldehyde group and the other must be a methyl group.

During oxidation:
\[ -CHO \rightarrow -COOH \]

and
\[ -CH_3 \rightarrow -COOH \]

Thus the compound must be
\[ \boxed{ o-Methylbenzaldehyde } \]

or
\[ \boxed{ 2-Methylbenzaldehyde } \]

with structure
\[ C_6H_4(CH_3)CHO \]

Step 6: Reaction with 2,4-DNP reagent.

The aldehyde reacts with 2,4-dinitrophenylhydrazine to form the corresponding hydrazone.
\[ C_6H_4(CH_3)CHO + H_2NNHC_6H_3(NO_2)_2 \]
\[ \longrightarrow C_6H_4(CH_3)CH=NNHC_6H_3(NO_2)_2 + H_2O \]

The product formed is the 2,4-dinitrophenylhydrazone derivative.
\[ \boxed{ o-Methylbenzaldehyde 2,4-DNP derivative } \]

Step 7: Cannizzaro reaction.

Since the aldehyde lacks \(\alpha\)-hydrogen atoms, it undergoes Cannizzaro reaction in the presence of concentrated alkali.

Two molecules of the aldehyde react together.

One molecule is oxidised to carboxylate ion and the other is reduced to alcohol.
\[ 2C_6H_4(CH_3)CHO + NaOH \]
\[ \longrightarrow C_6H_4(CH_3)COONa + C_6H_4(CH_3)CH_2OH \]

On acidification:
\[ C_6H_4(CH_3)COONa \longrightarrow C_6H_4(CH_3)COOH \]

Thus the products are:
\[ \boxed{ o-Methylbenzyl alcohol } \]

and
\[ \boxed{ o-Methylbenzoic acid } \]

Final Answer:

The compound is
\[ \boxed{ o-Methylbenzaldehyde } \]
\[ \boxed{ C_6H_4(CH_3)CHO } \]

Reaction with 2,4-DNP:
\[ \boxed{ C_6H_4(CH_3)CHO + H_2NNHC_6H_3(NO_2)_2 \rightarrow C_6H_4(CH_3)CH=NNHC_6H_3(NO_2)_2 + H_2O } \]

Cannizzaro reaction:
\[ \boxed{ 2C_6H_4(CH_3)CHO + NaOH \rightarrow C_6H_4(CH_3)COONa + C_6H_4(CH_3)CH_2OH } \] Quick Tip: Identification clues: 2,4-DNP test \(\Rightarrow\) Carbonyl compound. Tollens' test positive \(\Rightarrow\) Aldehyde. Cannizzaro reaction \(\Rightarrow\) Aldehyde without \(\alpha\)-hydrogen. Oxidation to phthalic acid \(\Rightarrow\) Ortho-substituted aromatic compound containing \(-CHO\) and \(-CH_3\) groups. Hence, \[ \boxed{ o-Methylbenzaldehyde } \] is the only structure satisfying all the given conditions.

Concept:

Anisole \((C_6H_5OCH_3)\) is an aromatic ether. The methoxy group \((-OCH_3)\) attached to the benzene ring exhibits a strong electron-donating resonance effect \((+M)\).

Due to this electron-releasing effect, the electron density of the benzene ring increases, making the ring more reactive towards electrophilic substitution reactions.

The methoxy group is an ortho- and para-directing group. Therefore, incoming electrophiles preferentially enter the ortho and para positions of the aromatic ring.

Part (i): Friedel-Crafts Alkylation

Step 1: Generation of electrophile.

Methyl chloride reacts with anhydrous aluminium chloride to generate the electrophile.
\[ CH_3Cl + AlCl_3 \rightarrow CH_3^{+} + AlCl_4^{-} \]

The electrophile formed is
\[ \boxed{CH_3^{+}} \]

Step 2: Electrophilic attack on anisole.

The methoxy group activates the benzene ring and directs the incoming methyl group to the ortho and para positions.

Therefore, two products are formed:
\[ o-Methylanisole \]

and
\[ p-Methylanisole \]

Because of less steric hindrance, the para product is formed in greater amount.
\[ \boxed{ C_6H_5OCH_3 + CH_3Cl \xrightarrow[CS_2]{AlCl_3} o-CH_3C_6H_4OCH_3 + p-CH_3C_6H_4OCH_3 } \]

Part (ii): Nitration of Anisole

Step 3: Generation of electrophile.

Concentrated nitric acid and concentrated sulphuric acid generate the nitronium ion.
\[ HNO_3 + H_2SO_4 \rightarrow NO_2^{+} + HSO_4^{-} + H_2O \]

The electrophile is
\[ \boxed{NO_2^{+}} \]

Step 4: Electrophilic substitution.

The nitronium ion attacks the ortho and para positions of anisole because the methoxy group is ortho-para directing.

Hence, the products formed are:
\[ o-Nitroanisole \]

and
\[ p-Nitroanisole \]

with para-nitroanisole as the major product.
\[ \boxed{ C_6H_5OCH_3 \xrightarrow{HNO_3/H_2SO_4} o-NO_2C_6H_4OCH_3 + p-NO_2C_6H_4OCH_3 } \]

Final Answer:
\[ \boxed{ C_6H_5OCH_3 + CH_3Cl \xrightarrow[CS_2]{AlCl_3} o-CH_3C_6H_4OCH_3 + p-CH_3C_6H_4OCH_3 } \]
\[ \boxed{ C_6H_5OCH_3 \xrightarrow{HNO_3/H_2SO_4} o-NO_2C_6H_4OCH_3 + p-NO_2C_6H_4OCH_3 } \] Quick Tip: The methoxy group \((-OCH_3)\) in anisole is: \[ \boxed{Electron donating} \] \[ \boxed{Ring activating} \] \[ \boxed{Ortho-para directing} \] Therefore, electrophilic substitution reactions such as alkylation, nitration, halogenation and sulphonation occur mainly at the ortho and para positions, with the para product generally predominating due to lower steric hindrance.

Part (i): Preparation of tert-butyl ethyl ether

Williamson ether synthesis involves the reaction of sodium alkoxide with an alkyl halide through an \(S_N2\) mechanism.

The general reaction is:
\[ RONa + R'X \rightarrow ROR' + NaX \]

For the preparation of tert-butyl ethyl ether:
\[ CH_3CH_2-O-C(CH_3)_3 \]

the ethyl group should be introduced through the alkyl halide because primary alkyl halides undergo \(S_N2\) substitution easily.

Therefore, ethyl bromide is selected as the alkyl halide:
\[ CH_3CH_2Br \]

and tert-butoxide ion is selected as the nucleophile:
\[ (CH_3)_3CO^- \]

The reaction is:
\[ \boxed{ CH_3CH_2Br + (CH_3)_3CONa \rightarrow CH_3CH_2OC(CH_3)_3 + NaBr } \]

Reason for choosing ethyl bromide:

Tert-butyl halides cannot be used in Williamson synthesis because they undergo elimination reaction instead of substitution due to steric hindrance.
\[ (CH_3)_3CBr + NaOC_2H_5 \rightarrow (CH_3)_2C=CH_2 + C_2H_5OH \]

Hence, primary alkyl halides are preferred.



Part (ii): Reaction of anisole with HI

Anisole is an aromatic ether:
\[ C_6H_5-O-CH_3 \]

When anisole reacts with HI, cleavage of the ether bond occurs.

The reaction proceeds as:
\[ C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I \]

Step 1: Protonation of ether oxygen

The oxygen atom of anisole gets protonated by HI.
\[ C_6H_5-O-CH_3 + H^+ \rightarrow C_6H_5-O^+H-CH_3 \]

This makes the methyl group susceptible to nucleophilic attack.

Step 2: Nucleophilic attack by iodide ion

The iodide ion attacks the methyl carbon through the \(S_N2\) mechanism.
\[ I^- + CH_3-O^+-C_6H_5 \rightarrow CH_3I + C_6H_5OH \]

Therefore, the products formed are:
\[ \boxed{Phenol \ (C_6H_5OH)} \]

and
\[ \boxed{Methyl\ iodide \ (CH_3I)} \]

Why not methanol and iodobenzene?

The formation of iodobenzene would require cleavage of the aryl-oxygen bond:
\[ C_6H_5-OCH_3 \rightarrow C_6H_5I + CH_3OH \]

However, this does not occur because:


The aryl carbon of anisole is \(sp^2\) hybridised.
The C--O bond has partial double bond character due to resonance.
The \(S_N2\) attack on the aryl carbon is not possible because backside attack is hindered by the aromatic ring.
The phenoxide ion formed after cleavage is stabilized by resonance.


Thus, cleavage occurs at the methyl--oxygen bond and not at the aryl--oxygen bond.
\[ \boxed{ C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I } \]



Final Answer:

(i)
\[ \boxed{Alkyl halide: Ethyl bromide (C_2H_5Br)} \]
\[ \boxed{Sodium alkoxide: Sodium tert-butoxide ((CH_3)_3CONa)} \]
\[ C_2H_5Br+(CH_3)_3CONa \rightarrow C_2H_5OC(CH_3)_3+NaBr \]

(ii)
\[ \boxed{ C_6H_5OCH_3+HI \rightarrow C_6H_5OH+CH_3I } \]

Anisole gives phenol and methyl iodide because the methyl--oxygen bond undergoes \(S_N2\) cleavage, whereas the aryl--oxygen bond is strengthened by resonance and cannot undergo \(S_N2\) attack. Quick Tip: In Williamson ether synthesis: \[ \boxed{Primary alkyl halides give ethers easily} \] \[ \boxed{Tertiary alkyl halides undergo elimination instead of S_N2} \] For anisole: \[ \boxed{C--O bond of aryl group is stronger due to resonance} \] \[ \boxed{Cleavage occurs at alkyl--O bond producing phenol and alkyl iodide} \]

Concept:

According to the valence shell electron pair repulsion (VSEPR) theory, the shape of molecules depends on the repulsion between electron pairs present around the central atom.

In ethers, oxygen is the central atom:
\[ R-O-R' \]

The oxygen atom is \(sp^3\) hybridised and contains four electron pairs:
\[ \boxed{2 bond pairs + 2 lone pairs} \]

The arrangement of these electron pairs is approximately tetrahedral.
\[ Ideal tetrahedral angle=109.5^\circ \]

However, lone pairs occupy more space than bond pairs because they are attracted only by the central atom and are not shared with another atom.

Therefore, the repulsion order is:
\[ \boxed{Lone pair--lone pair > lone pair--bond pair > bond pair--bond pair} \]

The two lone pairs on oxygen repel the C--O bond pairs strongly and push the two alkyl groups farther apart.

As a result, the C--O--C bond angle becomes slightly larger than the tetrahedral angle.
\[ \boxed{\angle C-O-C \approx 111.7^\circ} \]



Final Answer:

The C--O--C bond angle in ethers is slightly greater than \(109.5^\circ\) because oxygen contains two lone pairs of electrons. The strong repulsion between lone pairs and bond pairs pushes the two C--O bonds apart, increasing the bond angle.
\[ \boxed{\angle C-O-C = 111.7^\circ > 109.5^\circ} \] Quick Tip: In ethers: \[ \boxed{Oxygen\ is\ sp^3\ hybridised} \] \[ \boxed{2\ bond\ pairs + 2\ lone\ pairs} \] Lone pair repulsion increases the bond angle slightly above the normal tetrahedral angle. Remember: \[ \boxed{More lone pair repulsion \Rightarrow larger bond angle} \]

Concept:

During electrolysis, the products formed at the electrodes depend on:


Nature and concentration of ions present in the electrolyte.
Standard electrode potentials of possible oxidation and reduction reactions.
Nature of electrodes used.


Platinum electrodes are inert electrodes, meaning they do not participate in the reaction. They only provide a surface for oxidation and reduction reactions.



Part (i): Electrolysis of aqueous CuCl\(_2\)

The ions present in aqueous copper chloride solution are:
\[ Cu^{2+},\ Cl^-,\ H^+,\ OH^- \]

At the cathode, reduction of ions occurs.

Possible reduction reactions are:
\[ Cu^{2+}+2e^- \rightarrow Cu \]
\[ 2H^+ + 2e^- \rightarrow H_2 \]

Since the reduction potential of copper ions is higher than hydrogen ions, copper ions are preferentially reduced.

Therefore:
\[ \boxed{Cu^{2+}+2e^- \rightarrow Cu} \]

Copper is deposited at the cathode.

At the anode, oxidation occurs.

Possible oxidation reactions are:
\[ 2Cl^- \rightarrow Cl_2+2e^- \]
\[ 2H_2O \rightarrow O_2+4H^++4e^- \]

Due to the high concentration of chloride ions, chloride ions are oxidised preferentially.

Therefore:
\[ \boxed{2Cl^- \rightarrow Cl_2+2e^-} \]

Hence, copper and chlorine are obtained as products.



Part (ii): Electrolysis of concentrated H\(_2\)SO\(_4\)

Concentrated sulphuric acid contains:
\[ H^+,\ SO_4^{2-},\ HSO_4^- \]

At the cathode, hydrogen ions are reduced:
\[ \boxed{2H^+ +2e^- \rightarrow H_2} \]

Hydrogen gas is released.

At the anode, sulphate ions are difficult to oxidise. Therefore, water molecules undergo oxidation:
\[ \boxed{2H_2O \rightarrow O_2+4H^++4e^-} \]

Oxygen gas is released.

Thus, electrolysis of concentrated sulphuric acid gives hydrogen and oxygen gases.



Final Answer:

(i)
\[ \boxed{ CuCl_2(aq) \rightarrow Cu(s)+Cl_2(g) } \]

Cathode product:
\[ \boxed{Cu} \]

Anode product:
\[ \boxed{Cl_2} \]



(ii)
\[ \boxed{ 2H_2O \rightarrow 2H_2(g)+O_2(g) } \]

Cathode product:
\[ \boxed{H_2} \]

Anode product:
\[ \boxed{O_2} \] Quick Tip: For electrolysis: \[ \boxed{Cathode: Reduction} \] \[ \boxed{Anode: Oxidation} \] In aqueous CuCl\(_2\): \[ Cu^{2+} is deposited and Cl^- gives Cl_2 \] In water or dilute acid solution: \[ \boxed{H_2 is produced at cathode and O_2 at anode} \]

Concept:

According to Faraday's laws of electrolysis, the amount of substance deposited or reduced at an electrode depends on the quantity of electricity passed through the electrolyte.

The charge required is calculated using the stoichiometry of electrons involved in the electrode reaction.



Step 1: Write the reduction reaction.

Silver ions undergo reduction at the cathode:
\[ Ag^+ + e^- \rightarrow Ag \]

Here, one mole of Ag\(^+\) ions accepts one electron to form one mole of silver metal.



Step 2: Calculate the number of electrons required.

From the reaction:
\[ 1 mol of Ag^+ requires 1 mol of e^- \]

Therefore:
\[ Moles of electrons required=1 \]



Step 3: Convert electrons into Faraday.

One Faraday is the charge carried by one mole of electrons:
\[ 1F = 1 mol of e^- \]

Hence:
\[ Charge required=1F \]



Final Answer:

For the reduction of 1 mol of Ag\(^+\) to Ag:
\[ \boxed{ Ag^+ + e^- \rightarrow Ag } \]

1 mole of electrons is required, therefore the charge required is:
\[ \boxed{1 Faraday} \] Quick Tip: The charge required in Faraday can be directly found from the number of electrons involved: \[ \boxed{Charge (F) = Number of moles of electrons} \] For: \[ M^{n+}+ne^- \rightarrow M \] the required charge is: \[ \boxed{nF} \] For Ag\(^+\): \[ n=1 \Rightarrow \boxed{1F} \]

Concept:

Faraday's laws of electrolysis explain the quantitative relationship between the amount of electricity passed through an electrolyte and the amount of substance deposited at the electrodes.

Faraday proposed two laws:


First law: The mass of a substance deposited is directly proportional to the quantity of electricity passed.
Second law: The mass deposited depends on the equivalent mass of the substance.




Statement of Faraday's Second Law:

According to Faraday's second law of electrolysis, if the same amount of electric charge is passed through different electrolytes, the amount of different substances deposited at the electrodes will be proportional to their equivalent masses.

The equivalent mass is given by:
\[ \boxed{ Equivalent mass=\frac{Atomic mass or molecular mass}{Valency} } \]

Therefore,
\[ m \propto E \]

or,
\[ \boxed{ \frac{m_1}{m_2}=\frac{E_1}{E_2} } \]

This means that substances with higher equivalent masses require more charge for deposition of the same amount.



Final Answer:

Faraday’s second law of electrolysis states that when the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are directly proportional to their equivalent masses.
\[ \boxed{ \frac{m_1}{m_2}=\frac{E_1}{E_2} } \] Quick Tip: Remember: \[ \boxed{m \propto E} \] For the same charge: \[ \boxed{Deposited mass \propto Equivalent mass} \] Equivalent mass: \[ \boxed{ E=\frac{Molar mass}{Number of electrons involved} } \]

Concept:

During electrolysis, oxidation occurs at the anode. The species that loses electrons most easily is preferentially oxidised.

In an aqueous sodium chloride solution, the ions present are:
\[ Na^+, \ Cl^-, \ H^+, \ OH^- \]

At the anode, possible oxidation reactions are:
\[ Cl^- \rightarrow \frac{1}{2}Cl_2+e^- \]

and
\[ 2H_2O \rightarrow O_2+4H^++4e^- \]



Comparison of electrode potentials:

The given values are:
\[ E^\circ_{oxidation}(Cl^-/Cl_2)=1.36V \]
\[ E^\circ_{oxidation}(H_2O/O_2)=1.23V \]

From the values alone, oxidation of water appears easier because it has a lower oxidation potential.

However, in practice, chlorine is evolved instead of oxygen.



Reason:

The concentration of chloride ions in aqueous NaCl solution is very high compared to hydroxide ions produced from water.

Moreover, oxygen evolution at the anode requires additional overvoltage, which makes the oxidation of water more difficult.

Therefore, chloride ions are preferentially oxidised.

The anode reaction is:
\[ \boxed{ 2Cl^- \rightarrow Cl_2+2e^- } \]

Hence, chlorine gas is obtained.




Final Answer:

The feasible reaction at the anode is:
\[ \boxed{ Cl^-_{(aq)} \rightarrow \frac{1}{2}Cl_2(g)+e^- } \]

because chloride ions are present in high concentration and oxygen evolution from water requires higher overvoltage. Therefore, chlorine gas is liberated at the anode during electrolysis of aqueous NaCl solution. Quick Tip: In electrolysis of aqueous NaCl (brine): \[ \boxed{Cathode: H_2 gas is produced} \] \[ \boxed{Anode: Cl_2 gas is produced} \] Although water can give oxygen, chloride ions are preferentially oxidised due to: \[ \boxed{High Cl^- concentration + oxygen overvoltage} \]

Part (i): Calculation of freezing point

Concept:

The depression in freezing point is a colligative property. It depends on the number of solute particles present in the solution.

The formula for depression in freezing point is:
\[ \boxed{\Delta T_f=iK_fm} \]

where,
\[ i=Van't Hoff factor \]
\[ K_f=molal depression constant \]
\[ m=molality of solution \]



Step 1: Calculate Van't Hoff factor

Magnesium bromide undergoes complete dissociation:
\[ MgBr_2 \rightarrow Mg^{2+}+2Br^- \]

One molecule of MgBr\(_2\) produces:
\[ 1+2=3 \]

ions.

Therefore:
\[ \boxed{i=3} \]



Step 2: Calculate moles of MgBr\(_2\)

Using:
\[ Moles=\frac{Given mass}{Molar mass} \]
\[ =\frac{10.5}{184} \]
\[ =0.0571\,mol \]



Step 3: Calculate molality

Mass of water:
\[ 250g=0.250kg \]

Therefore:
\[ m=\frac{moles of solute}{mass of solvent in kg} \]
\[ m=\frac{0.0571}{0.250} \]
\[ m=0.2284\,mol\,kg^{-1} \]



Step 4: Calculate depression in freezing point
\[ \Delta T_f=iK_fm \]

Substituting the values:
\[ \Delta T_f=3\times1.86\times0.2284 \]
\[ \Delta T_f=1.27K \]

The freezing point of pure water is:
\[ 0^\circ C \]

Hence:
\[ T_f=0-1.27 \]
\[ \boxed{T_f=-1.27^\circ C} \]

Therefore, the freezing point of the solution is \(-1.27^\circ C\).



Part (ii): Ideal and Non-ideal Solutions

Ideal solutions:

An ideal solution is one which obeys Raoult's law over the entire range of concentration.

For ideal solutions:
\[ \Delta H_{mix}=0 \]

and
\[ \Delta V_{mix}=0 \]

because intermolecular interactions between solute-solvent molecules are similar to solute-solute and solvent-solvent interactions.



Non-ideal solutions:

Non-ideal solutions do not obey Raoult's law due to differences in intermolecular interactions.

They show:
\[ \Delta H_{mix}\neq0 \]

and
\[ \Delta V_{mix}\neq0 \]

They may show positive or negative deviation from Raoult's law.



Final Answer:

(i)
\[ \boxed{Freezing point of MgBr_2 solution=-1.27^\circ C} \]



(ii)
Ideal Solution Non-ideal Solution
Obeys Raoult's law & Does not obey Raoult's law
Delta H_{mix}=0,Delta V_{mix}=0 & Delta H_{mix}neq0,Delta V_{mix}neq0

Quick Tip: For electrolytes: \[ \boxed{\Delta T_f=iK_fm} \] Always calculate the Van't Hoff factor first. Examples: \[ NaCl \rightarrow Na^+ + Cl^- \Rightarrow i=2 \] \[ MgBr_2 \rightarrow Mg^{2+}+2Br^- \Rightarrow i=3 \] \[ \boxed{More ions \Rightarrow greater depression in freezing point} \]

Part (i): Calculation of osmotic pressure

Concept:

Osmotic pressure is a colligative property that depends on the number of solute particles present in the solution.

For an electrolyte:
\[ \boxed{\pi=iCRT} \]

where,
\[ \pi=osmotic pressure \]
\[ i=Van't Hoff factor \]
\[ C=molar concentration \]
\[ R=gas constant \]
\[ T=temperature in Kelvin \]



Step 1: Calculate Van't Hoff factor

Potassium sulphate dissociates completely:
\[ K_2SO_4\rightarrow2K^++SO_4^{2-} \]

Number of ions produced:
\[ 2+1=3 \]

Therefore:
\[ \boxed{i=3} \]



Step 2: Calculate moles of \(K_2SO_4\)
\[ Moles=\frac{Mass}{Molar mass} \]
\[ =\frac{0.025}{174} \]
\[ =1.437\times10^{-4}mol \]



Step 3: Calculate molarity

Volume of solution:
\[ V=2L \]

Therefore:
\[ C=\frac{moles of solute}{volume of solution} \]
\[ C=\frac{1.437\times10^{-4}}{2} \]
\[ C=7.18\times10^{-5}M \]



Step 4: Calculate osmotic pressure

Temperature:
\[ T=27+273=300K \]

Using:
\[ \pi=iCRT \]

Substituting values:
\[ \pi=3\times7.18\times10^{-5}\times0.082\times300 \]
\[ \pi=5.29\times10^{-3}atm \]

Hence:
\[ \boxed{\pi=5.29\times10^{-3}atm} \]



Part (ii): Azeotrope formed by acetone and chloroform

Concept:

An azeotrope is a liquid mixture that boils at a constant temperature and has the same composition in liquid and vapour phases.

Azeotropes are of two types:


Minimum boiling azeotrope: Shows positive deviation from Raoult's law.
Maximum boiling azeotrope: Shows negative deviation from Raoult's law.




In acetone-chloroform mixture, hydrogen bonding occurs between hydrogen atom of chloroform and oxygen atom of acetone:
\[ CHCl_3\cdots O=C(CH_3)_2 \]

This strong interaction reduces the escaping tendency of molecules.

Therefore:
\[ Vapour pressure decreases \]

and:
\[ Boiling point increases \]

Hence, the mixture shows negative deviation from Raoult's law and forms a:
\[ \boxed{Maximum boiling azeotrope} \]



% Correction of Answer
Note:

Acetone-chloroform actually forms a:
\[ \boxed{Maximum boiling azeotrope} \]

because of strong intermolecular hydrogen bonding and negative deviation from Raoult's law.




Final Answer:

(i)
\[ \boxed{\pi=5.29\times10^{-3}atm} \]

(ii) Acetone and chloroform form a:
\[ \boxed{Maximum boiling azeotrope} \]

due to strong hydrogen bonding between acetone and chloroform molecules, causing negative deviation from Raoult's law. Quick Tip: For electrolytes: \[ \boxed{\pi=iCRT} \] Always include the Van't Hoff factor. Azeotrope shortcut: \[ \boxed{Positive deviation \Rightarrow Minimum boiling azeotrope} \] \[ \boxed{Negative deviation \Rightarrow Maximum boiling azeotrope} \] Acetone + chloroform: \[ \boxed{Strong H-bonding \Rightarrow Maximum boiling azeotrope} \]

Part (i):

(I) Variable oxidation states of transition metals

Transition elements have incompletely filled \(d\) orbitals.

The general electronic configuration of transition metals is:
\[ (n-1)d^{1-10}ns^{0-2} \]

The energy difference between \((n-1)d\) and \(ns\) orbitals is very small.

Therefore, electrons from both orbitals can be removed or shared during chemical reactions.

For example:
\[ Fe^{2+} and Fe^{3+} \]

show different oxidation states.

Hence, transition metals exhibit variable oxidation states.



(II) Magnetic behaviour of Mn\(^{2+}\) and Ti\(^{2+}\)

Titanium:

Atomic number:
\[ Ti=22 \]

Electronic configuration:
\[ Ti=[Ar]3d^24s^2 \]

After losing two electrons:
\[ Ti^{2+}=[Ar]3d^2 \]

It contains:
\[ \boxed{2 unpaired electrons} \]



Manganese:

Atomic number:
\[ Mn=25 \]

Electronic configuration:
\[ Mn=[Ar]3d^54s^2 \]

After losing two electrons:
\[ Mn^{2+}=[Ar]3d^5 \]

It contains:
\[ \boxed{5 unpaired electrons} \]

Since paramagnetism depends on the number of unpaired electrons:
\[ \boxed{Mn^{2+} is more paramagnetic than Ti^{2+}} \]



(III) Oxidising strength of ions

An oxidising agent is a species that accepts electrons and gets reduced.

The reduction reactions are:
\[ Mn^{3+}+e^-\rightarrow Mn^{2+} \]
\[ Cr^{3+}+e^-\rightarrow Cr^{2+} \]
\[ V^{3+}+e^-\rightarrow V^{2+} \]
\(Mn^{3+}\) has the greatest tendency to accept electrons because the resulting \(Mn^{2+}\) has a stable half-filled \(3d^5\) configuration.

Therefore:
\[ \boxed{Mn^{3+} is the strongest oxidising agent} \]



Part (ii): Balancing reactions

(I) Formation of potassium manganate

In alkaline medium:
\[ MnO_2 \]

is oxidised to manganate ion:
\[ MnO_4^{2-} \]

The balanced equation is:
\[ \boxed{ 2MnO_2+4KOH+O_2 \rightarrow 2K_2MnO_4+2H_2O } \]



(II) Reaction of permanganate with oxalate ion

In acidic medium:
\[ MnO_4^- \]

is reduced to:
\[ Mn^{2+} \]

and oxalate ion is oxidised to carbon dioxide.

The balanced ionic equation is:
\[ \boxed{ 2MnO_4^-+5C_2O_4^{2-}+16H^+ \rightarrow 2Mn^{2+}+10CO_2+8H_2O } \]




Final Answer:

(i)
\[ \boxed{Transition metals show variable oxidation states due to similar energies of (n-1)d and ns orbitals} \]
\[ \boxed{Mn^{2+} is more paramagnetic because it has five unpaired electrons} \]
\[ \boxed{Mn^{3+} is the strongest oxidising agent} \]



(ii)
\[ \boxed{ 2MnO_2+4KOH+O_2 \rightarrow 2K_2MnO_4+2H_2O } \]
\[ \boxed{ 2MnO_4^-+5C_2O_4^{2-}+16H^+ \rightarrow 2Mn^{2+}+10CO_2+8H_2O } \] Quick Tip: Important points for transition metals: \[ \boxed{(n-1)d and ns orbitals have similar energies} \] \[ \boxed{More unpaired electrons \Rightarrow more paramagnetic} \] Half-filled configuration: \[ \boxed{d^5 is highly stable} \] Therefore: \[ Mn^{2+}=3d^5 \] is especially stable.

(i) Lanthanoid contraction

Lanthanoids have electrons entering the \(4f\) orbitals.

The shielding effect of \(4f\) electrons is very poor. As atomic number increases, nuclear charge increases, but the added \(4f\) electrons cannot effectively shield the outer electrons.

Therefore, the effective nuclear charge increases and the size of atoms and ions decreases gradually.

This phenomenon is called lanthanoid contraction.
\[ \boxed{ La^{3+}>Ce^{3+}>Pr^{3+}>...>Lu^{3+} } \]

in terms of ionic size.



(ii) Colour formation in transition metal compounds

Transition metals generally have partially filled \(d\) orbitals.

In the presence of ligands, the five \(d\) orbitals split into different energy levels.

When visible light falls on these compounds, electrons absorb certain wavelengths and jump from lower energy \(d\) orbitals to higher energy \(d\) orbitals.

This process is called:
\[ \boxed{d-d transition} \]

The remaining transmitted or reflected light gives the compound its colour.

Example:
\[ CuSO_4 \]

appears blue due to \(d-d\) transition.



(iii) Negative electrode potential of Mn and Zn

The standard electrode potential represents the tendency of metal ions to get reduced.

For manganese:
\[ Mn:[Ar]3d^54s^2 \]

After ionisation:
\[ Mn^{2+}:[Ar]3d^5 \]

The half-filled \(d^5\) configuration is highly stable.

Similarly:
\[ Zn:[Ar]3d^{10}4s^2 \]

forms:
\[ Zn^{2+}:[Ar]3d^{10} \]

which has a completely filled \(d^{10}\) configuration.

Due to these stable configurations, removal of electrons from Mn and Zn metals is difficult, resulting in more negative electrode potentials.



(iv) Stable oxidation state of copper

Copper has electronic configuration:
\[ Cu=[Ar]3d^{10}4s^1 \]

It can show:
\[ Cu^+ \]

and
\[ Cu^{2+} \]

oxidation states.

Although Cu\(^+\) has a stable \(3d^{10}\) configuration, it undergoes disproportionation:
\[ 2Cu^+\rightarrow Cu^{2+}+Cu \]

Cu\(^{2+}\) is more stable in aqueous solution because of its higher hydration energy.

Therefore:
\[ \boxed{Cu^{2+} is the most stable oxidation state of copper} \]



(v) Oxidising nature of Ce\(^{4+}\)

Cerium commonly exists in \(+3\) and \(+4\) oxidation states.

The reaction:
\[ Ce^{4+}+e^-\rightarrow Ce^{3+} \]

has a high positive reduction potential.

Therefore, Ce\(^{4+}\) readily accepts electrons and oxidises other substances.

Hence:
\[ \boxed{Ce^{4+} is a strong oxidising agent} \]




Final Answer:

(i)
\[ \boxed{Lanthanoid contraction is the gradual decrease in size of lanthanoids from La to Lu.} \]

(ii)
\[ \boxed{Transition metal compounds are coloured due to d-d transitions.} \]

(iii)
\[ \boxed{Mn and Zn have stable d^5 and d^{10} configurations, causing more negative electrode potentials.} \]

(iv)
\[ \boxed{Cu^{2+} is the most stable oxidation state of copper.} \]

(v)
\[ \boxed{Ce^{4+} is a good oxidising agent because it reduces easily to stable Ce^{3+}.} \] Quick Tip: Important facts: \[ \boxed{4f electrons show poor shielding effect} \] causing: \[ \boxed{Lanthanoid contraction} \] Colour in transition compounds: \[ \boxed{d-d transition} \] Stable configurations: \[ \boxed{d^5=half-filled, d^{10}=fully filled} \] Both provide extra stability.

Part (i):

(I) Reactivity towards S\(_N\)1 reaction

The rate determining step of an S\(_N\)1 reaction is the formation of carbocation:
\[ R-X\rightarrow R^+ + X^- \]

The stability of carbocation determines the rate of reaction.

The order of carbocation stability is:
\[ 3^\circ > 2^\circ > 1^\circ \]

For 2-bromo-2-methylbutane:
\[ CH_3-C(Br)(CH_3)-CH_2-CH_3 \]

Removal of bromide ion gives:
\[ (CH_3)_2C^+CH_2CH_3 \]

which is a tertiary carbocation.

For 1-bromopentane:
\[ CH_3CH_2CH_2CH_2CH_2Br \]

removal of bromide ion gives a primary carbocation, which is highly unstable.

Therefore:
\[ \boxed{2-Bromo-2-methylbutane is more reactive towards S_N1} \]



(II) Type of halide

In the given compound, chlorine is directly attached to a carbon of the double bond.

The structure is:
\[ CH_3-C(Cl)=CH_2 \]

The carbon attached to chlorine is \(sp^2\) hybridised.

Therefore, it is classified as:
\[ \boxed{Vinylic halide} \]



(III) Storage of chloroform

Chloroform undergoes oxidation in presence of oxygen and sunlight.

The reaction is:
\[ 2CHCl_3+O_2 \rightarrow 2COCl_2+2HCl \]

The product formed is phosgene gas:
\[ COCl_2 \]

which is highly poisonous.

Therefore, chloroform is stored in dark coloured bottles to prevent the reaction with sunlight.

A small amount of ethanol is also added as a stabiliser because it reacts with phosgene.



Part (ii): Definitions

(I) Ambident Nucleophiles

Some nucleophiles have two different nucleophilic centres and can attack through either atom.

Such nucleophiles are called ambident nucleophiles.

Examples:
\[ CN^- \]

can attack through carbon or nitrogen.
\[ NO_2^- \]

can attack through nitrogen or oxygen.



(II) Racemic Mixture

A racemic mixture contains equal amounts of two optical isomers:
\[ (+)-enantiomer \]

and
\[ (-)-enantiomer \]

The optical rotations of both isomers cancel each other, making the mixture optically inactive.




Final Answer:

(i)
\[ \boxed{2-Bromo-2-methylbutane is more reactive towards S_N1} \]

because it forms a stable tertiary carbocation.
\[ \boxed{The given compound is a vinylic halide} \]
\[ \boxed{Chloroform is stored in dark bottles to prevent formation of poisonous phosgene gas} \]



(ii)
\[ \boxed{Ambident nucleophiles: Nucleophiles having two attacking centres} \]
\[ \boxed{Racemic mixture: Equal mixture of two enantiomers causing optical inactivity} \] Quick Tip: For S\(_N\)1 reactions: \[ \boxed{3^\circ > 2^\circ > 1^\circ} \] carbocation stability decides reactivity. Remember: \[ \boxed{Halogen attached to sp^2 carbon=Vinylic halide} \] Chloroform: \[ \boxed{Light + oxygen \rightarrow poisonous phosgene} \]

Part (i):

(I) Reactivity towards S\(_N\)1 reaction

The rate determining step of an S\(_N\)1 reaction is carbocation formation:
\[ R-X\rightarrow R^+ +X^- \]

The stability order of carbocations is:
\[ 3^\circ > 2^\circ > 1^\circ \]

The four isomers of C\(_4\)H\(_9\)Br are:
\[ n-butyl bromide \]
\[ sec-butyl bromide \]
\[ isobutyl bromide \]
\[ tert-butyl bromide \]

tert-Butyl bromide produces a tertiary carbocation:
\[ (CH_3)_3C^+ \]

which is highly stable due to the \(+I\) effect and hyperconjugation.

Hence:
\[ \boxed{tert-Butyl bromide is most reactive towards S_N1} \]



(II) Dehydrohalogenation of 1-bromo-1-methylcyclohexane

The reaction follows the elimination mechanism.

The bromine atom is removed from carbon-1 and a hydrogen atom is removed from an adjacent carbon.

Removal of hydrogen from C-2 or C-6 gives:
\[ 1-Methylcyclohexene \]

Removal of hydrogen from methyl group gives:
\[ methylenecyclohexane \]

According to Saytzeff's rule, the more substituted alkene is the major product.

Therefore:
\[ \boxed{1-Methylcyclohexene is the major product} \]



(III) Directive influence of chlorine

Chlorine has two opposite effects:
\[ -I effect \]

and
\[ +R effect \]

Due to its high electronegativity, chlorine withdraws electron density through sigma bonds, making it deactivating.

However, chlorine has lone pairs which can participate in resonance:
\[ Cl\rightarrow benzene ring \]

This increases electron density at ortho and para positions.

Therefore, chlorine is:
\[ \boxed{deactivating but ortho/para directing} \]



Part (ii):

(I) Reaction with sodium in dry ether

Aryl halides undergo coupling reaction with sodium metal.

The reaction is called Wurtz-Fittig reaction:
\[ Ar-X+2Na+X-Ar \rightarrow Ar-Ar+2NaX \]

For chlorobenzene:
\[ 2C_6H_5Cl+2Na \rightarrow C_6H_5-C_6H_5+2NaCl \]

Product:
\[ \boxed{Biphenyl} \]



(II) Bromination of p-nitroisopropylbenzene

The reaction is carried out in presence of heat.

The side chain undergoes free radical bromination.

The hydrogen atom attached to the benzylic carbon is replaced by bromine.
\[ p-NO_2C_6H_4CH(CH_3)_2 \]

changes to:
\[ p-NO_2C_6H_4CBr(CH_3)_2 \]

Thus, the major product is:
\[ \boxed{p-Nitroisopropyl bromobenzene} \]




Final Answer:

(i)
\[ \boxed{tert-Butyl bromide is most reactive towards S_N1} \]
\[ \boxed{Major alkene: 1-Methylcyclohexene} \]
\[ \boxed{Chlorine is ortho/para directing due to resonance donation} \]



(ii)
\[ \boxed{ 2C_6H_5Cl+2Na \rightarrow C_6H_5-C_6H_5+2NaCl } \]

Major product:
\[ \boxed{Biphenyl} \]
\[ \boxed{ p-NO_2C_6H_4CH(CH_3)_2+Br_2 \xrightarrow{heat} p-NO_2C_6H_4CBr(CH_3)_2+HBr } \] Quick Tip: For S\(_N\)1 reaction: \[ \boxed{3^\circ > 2^\circ > 1^\circ} \] For elimination: \[ \boxed{More substituted alkene is major (Saytzeff rule)} \] Halogens on benzene: \[ \boxed{Deactivating but ortho/para directing} \] due to: \[ \boxed{-I effect and +R effect} \]