CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 3 - 56/4/3)

Concept:

• The \( \text{S}_{\text{N}}1 \) (Substitution Nucleophilic Unimolecular) reaction is a two-step nucleophilic substitution mechanism.
• It is characterized by the formation of a carbocation intermediate and follows first-order kinetics where the rate depends only on the substrate concentration.

Step 1: Analyze the kinetics of the reaction
In an \( \text{S}_{\text{N}}1 \) mechanism, the rate-determining step is the slow ionization of the alkyl halide to form a carbocation. The rate law for this reaction is: \[ \text{Rate} = k[\text{RX}] \] Since the rate depends only on the concentration of the substrate and is independent of the nucleophile’s concentration, it follows first-order kinetics. Therefore, statement (A) is correct.

Step 2: Analyze the effect of the solvent
Polar protic solvents (such as water, alcohols, and carboxylic acids) have high dielectric constants and the ability to form hydrogen bonds. These solvents stabilize the transition state and the resulting carbocation intermediate through ion-dipole interactions (solvation). This stabilization reduces the activation energy required for the formation of the carbocation. Thus, statement (B) is correct.

Step 3: Examine the reaction intermediate
The mechanism proceeds as follows: \[ \text{R-X (slow)} \rightarrow \text{R}^+ + \text{X}^- \] \[ \text{R}^+ + \text{Nu}^- \text{ (fast)} \rightarrow \text{R-Nu} \] The first and rate-limiting step involves the heterolytic cleavage of the C-X bond to produce a carbocation intermediate. Therefore, statement (C) is correct.

Step 4: Evaluate the stereochemical outcome
The intermediate carbocation is \( sp^2 \) hybridized and has a planar geometry. The nucleophile can attack the planar carbocation from either the front side or the back side with roughly equal probability. This leads to a mixture of products: one with retention of configuration and one with inversion of configuration, a process known as racemization. Statement (D) claims it results only in inversion, which is a characteristic of \( \text{S}_{\text{N}}2 \) reactions, not \( \text{S}_{\text{N}}1 \).

Concept:

• When vapours of different types of alcohols are passed over heated copper at \( 573 \text{ K} \), they undergo different types of chemical transformations.
• Primary and secondary alcohols typically undergo dehydrogenation (a form of oxidation), while tertiary alcohols undergo dehydration.

Step 1: Compare primary, secondary, and tertiary alcohols
Primary alcohols (\( 1^\circ \)) lose hydrogen (dehydrogenation) to form aldehydes: \[ \text{RCH}_2\text{OH} \xrightarrow{\text{Cu}/573\text{K}} \text{RCHO} + \text{H}_2 \] Secondary alcohols (\( 2^\circ \)) lose hydrogen (dehydrogenation) to form ketones: \[ \text{R}_2\text{CHOH} \xrightarrow{\text{Cu}/573\text{K}} \text{R}_2\text{CO} + \text{H}_2 \]

Step 2: Identify why tertiary alcohols behave differently
Tertiary alcohols (\( 3^\circ \)) have the general structure \( \text{R}_3\text{COH} \). The carbon atom bearing the \( \text{-OH} \) group (the \( \alpha \)-carbon) does not have any hydrogen atoms attached to it. Because there is no \( \alpha \)-hydrogen, the standard dehydrogenation reaction cannot occur.

Step 3: Determine the product for tertiary alcohols
Under the high-temperature conditions (\( 573 \text{ K} \)) with the copper catalyst, the tertiary alcohol undergoes the elimination of a water molecule (dehydration) instead of losing hydrogen. For example, with tert-butyl alcohol: \[ (\text{CH}_3)_3\text{COH} \xrightarrow{\text{Cu}/573\text{K}} (\text{CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O} \] The resulting product is an alkene (2-methylpropene).

Concept:

• The coordination number (CN) of a central metal atom or ion in a complex is defined as the total number of ligand donor atoms to which the metal is directly bonded.
• It is not necessarily equal to the number of ligands; one must account for the "denticity" (the number of donor sites) of each ligand.

Step 1: Identify the ligands in the complex
The complex is \( [\text{Co}(\text{en})_2\text{Cl}_2]^+ \). The ligands coordinated to the cobalt ion are:

1. Two \( \text{en} \) (ethylenediamine) molecules.
2. Two \( \text{Cl}^- \) (chloride) ions.

Step 2: Determine the denticity of each ligand

• Ethylenediamine (en): It is a bidentate ligand (\( \text{H}_2\text{N-CH}_2\text{-CH}_2\text{-NH}_2 \)). Each \( \text{en} \) molecule has two nitrogen donor atoms that can bond to the metal simultaneously.
• Chloride ion (\( \text{Cl}^- \)): It is a monodentate ligand. Each ion has only one donor atom bonded to the metal.

Step 3: Calculate the total coordination number
Total CN = (Number of \( \text{en} \) ligands \( \times \) 2) + (Number of \( \text{Cl}^- \) ligands \( \times \) 1)
\[ \text{CN} = (2 \times 2) + (2 \times 1) \] \[ \text{CN} = 4 + 2 = 6 \] Thus, the cobalt ion is bonded to 6 donor atoms in an octahedral arrangement.

Concept:

• Crystal Field Theory (CFT) explains that in an octahedral field, the five degenerate d-orbitals of a metal ion split into two sets of different energies due to repulsion between the electrons of the metal and the ligands.
• Repulsion is strongest when the orbital lobes point directly at the approaching ligands.

Step 1: Identify ligand approach in octahedral geometry
In an octahedral complex, the six ligands approach the central metal ion along the Cartesian axes (\( \pm x, \pm y, \text{ and } \pm z \)).

Step 2: Evaluate the spatial orientation of d-orbitals
The five d-orbitals are oriented differently in space:

• \( e_g \) set (\( d_{x^2-y^2} \text{ and } d_{z^2} \)): These orbitals have their lobes pointing directly along the axes.
• \( t_{2g} \) set (\( d_{xy}, d_{yz}, \text{ and } d_{xz} \)): These orbitals have their lobes pointing between the axes.

Step 3: Determine where repulsion is maximum
Since the ligands are approaching along the axes, they encounter the lobes of the \( d_{x^2-y^2} \) and \( d_{z^2} \) orbitals head-on. This direct interaction leads to maximum electrostatic repulsion, causing the energy of these two orbitals to increase more than the others. The \( t_{2g} \) orbitals experience less repulsion because the ligands pass between their lobes.

Concept:

• Transition metals show variable oxidation states because they can use both \( ns \) and \( (n-1)d \) electrons for bonding.
• The stability of a particular oxidation state often depends on achieving a half-filled (\( d^5 \)) or fully filled (\( d^{10} \)) configuration, or on hydration/lattice energies.

Step 1: Analyze the electronic configurations and common states

• Zn (\( Z=30 \)): Configuration is \( [Ar] 3d^{10} 4s^2 \). It only shows \( +2 \) by losing the \( 4s \) electrons to reach a stable \( d^{10} \) state.
• Fe (\( Z=26 \)): Configuration is \( [Ar] 3d^6 4s^2 \). Common states are \( +2 \) and \( +3 \).
• Cr (\( Z=24 \)): Configuration is \( [Ar] 3d^5 4s^1 \). It shows states from \( +2 \) to \( +6 \).
• Cu (\( Z=29 \)): Configuration is \( [Ar] 3d^{10} 4s^1 \).

Step 2: Examine Copper’s oxidation states
Copper can lose the single \( 4s \) electron to form the \( \text{Cu}^+ \) (cuprous) ion. In the \( \text{Cu}^+ \) state, the ion has a stable, fully filled \( 3d^{10} \) configuration. Copper can also lose one \( 3d \) electron along with the \( 4s \) electron to form the \( \text{Cu}^{2+} \) (cupric) ion (\( 3d^9 \)). While \( \text{Cu}^+ \) is stable in some solid compounds, \( \text{Cu}^{2+} \) is typically more stable in aqueous solution due to its higher hydration enthalpy compared to \( \text{Cu}^+ \).

Step 3: Conclusion
Among the 3d transition series, Copper is the characteristic metal that exhibits both \( +1 \) and \( +2 \) oxidation states.

Concept:

• The elevation of boiling point (\( \Delta T_b \)) is a colligative property given by the formula \( \Delta T_b = i \cdot K_b \cdot m \).
• The van’t Hoff factor (\( i \)) accounts for the dissociation of the solute.

Step 1: Determine the van’t Hoff factor (i) for NaCl
Since \( \text{NaCl} \) is assumed to be completely dissociated, it breaks into two ions: \[ \text{NaCl (aq)} \rightarrow \text{Na}^+ \text{(aq)} + \text{Cl}^- \text{(aq)} \] Total number of ions per formula unit = 2. Therefore, \( i = 2 \).

Step 2: Calculate the elevation in boiling point (\( \Delta T_b \))
Given: Molality (\( m \)) = \( 1\text{ m} \) Molal elevation constant (\( K_b \)) = \( 0.52\text{ K kg mol}^{-1} \) Using the formula: \[ \Delta T_b = i \cdot K_b \cdot m \] \[ \Delta T_b = 2 \cdot 0.52 \cdot 1 = 1.04\text{ K} \text{ (or } 1.04^\circ\text{C)} \]

Step 3: Calculate the boiling point of the solution
The standard boiling point of pure water (\( T_b^\circ \)) is \( 100^\circ\text{C} \). The boiling point of the solution (\( T_b \)) is: \[ T_b = T_b^\circ + \Delta T_b \] \[ T_b = 100^\circ\text{C} + 1.04^\circ\text{C} = 101.04^\circ\text{C} \]

Concept:

• \( pK_b \) is inversely proportional to the basic strength (\( K_b \)). A higher \( pK_b \) means a weaker base.
• Basic strength of amines depends on the availability of the lone pair on the nitrogen atom.

Step 1: Compare aliphatic and aromatic amines
Ethylamine (\( \text{C}_2\text{H}_5\text{NH}_2 \)) is an aliphatic amine. The \( \text{+I} \) effect of the ethyl group increases electron density on Nitrogen, making it a very strong base. Aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)) is an aromatic amine. The lone pair on Nitrogen is delocalized into the benzene ring due to resonance, making it significantly less available for donation. Thus, aniline is a much weaker base than aliphatic amines.

Step 2: Compare Aniline and N-methylaniline
In N-methylaniline (\( \text{C}_6\text{H}_5\text{NHCH}_3 \)), the methyl group (\( \text{-CH}_3 \)) attached to the nitrogen exerts a \( \text{+I} \) effect. This increases the electron density on Nitrogen compared to aniline. While the lone pair is still delocalized, the electron density is higher than in aniline. Therefore, basic strength order: \( \text{C}_2\text{H}_5\text{NH}_2 > \text{C}_6\text{H}_5\text{NHCH}_3 > \text{C}_6\text{H}_5\text{NH}_2 \).

Step 3: Determine the order of \( pK_b \)
Since \( pK_b = -\log K_b \), the order for \( pK_b \) will be the exact opposite of the basic strength order. Decreasing order of \( pK_b \) (from highest to lowest): \[ \text{C}_6\text{H}_5\text{NH}_2 > \text{C}_6\text{H}_5\text{NHCH}_3 > \text{C}_2\text{H}_5\text{NH}_2 \]

Concept:

• The cell potential (\( E_{\text{cell}} \)) is related to the concentrations of the ions involved by the Nernst Equation.

Step 1: Write the cell reaction and Nernst equation
The overall cell reaction is: \[ \text{Mg(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu(s)} \] The number of electrons transferred (\( n \)) is 2. The Nernst equation at \( 298\text{ K} \) is: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \]

Step 2: Analyze how to increase \( E_{\text{cell}} \)
To increase \( E_{\text{cell}} \), we must decrease the value of the term being subtracted: \( \frac{0.059}{n} \log \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \). This happens when the ratio \( \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \) decreases. Ways to decrease this ratio:

1. Decrease the concentration of product ions (\( [\text{Mg}^{2+}] \)).
2. Increase the concentration of reactant ions (\( [\text{Cu}^{2+}] \)).

Step 3: Evaluate the options
Option (A): Decreasing \( [\text{Mg}^{2+}] \) to \( 0.1\text{ M} \) decreases the numerator of the ratio, thus decreasing the ratio. This increases \( E_{\text{cell}} \). Option (B): Decreasing \( [\text{Cu}^{2+}] \) increases the ratio, which decreases \( E_{\text{cell}} \). Option (D): Increasing \( [\text{Mg}^{2+}] \) increases the ratio, which decreases \( E_{\text{cell}} \). Option (C): Increasing both proportionally keeps the ratio the same (at standard state).

Concept:

• Phase equilibria respond to changes in pressure according to Le Chatelier’s Principle.
• Evaporation is an endothermic process (\( \Delta H > 0 \)).

Step 1: Analyze the effect of pressure decrease on equilibrium
Consider the equilibrium: \( \text{Liquid} \rightleftharpoons \text{Vapour} \). Vapour occupies a much larger volume than the liquid. According to Le Chatelier’s Principle, if the pressure is decreased, the system will try to oppose this change by shifting the equilibrium toward the side with more gaseous moles. Therefore, the equilibrium shifts to the right, promoting more evaporation.

Step 2: Consider the energetics of phase change
Evaporation (conversion of liquid to gas) is an endothermic process. It requires energy to overcome intermolecular forces. When the pressure is suddenly dropped, rapid evaporation occurs. The energy required for this evaporation is taken from the internal energy of the liquid itself.

Step 3: Determine the temperature change
As the liquid loses internal energy (heat) to fuel the evaporation process, its temperature decreases. This results in a "cooling" effect.

Concept:

• Protein structure is organized into four levels: primary, secondary, tertiary, and quaternary.

Step 1: Define primary structure
The primary structure of a protein refers to the unique, linear sequence of amino acids in a polypeptide chain. Amino acids are linked by covalent peptide bonds. Any change in this specific sequence (e.g., in genetic mutations like sickle cell anemia) fundamentally alters the protein.

Step 2: Distinguish other levels of structure

• Secondary structure: Refers to the local folding of the chain into regular patterns like \( \alpha \)-helices or \( \beta \)-pleated sheets, stabilized by hydrogen bonds.
• Tertiary structure: Refers to the overall three-dimensional folding of the entire polypeptide chain.
• Quaternary structure: Refers to the spatial arrangement and interaction of multiple polypeptide chains (subunits).

Step 3: Match with the question
The question explicitly asks about the "specific sequence" of amino acids. This corresponds directly to the definition of the primary structure.

Concept:

• For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant raised to the power of one.
• The mathematical expression is: \( \text{Rate} = k[A] \), where \( k \) is the rate constant and \( [A] \) is the concentration.

Step 1: Identify the given values
Rate of reaction (\( R \)) = \( 5.6 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \)
Concentration of reactant (\( [A] \)) = \( 0.2 \text{ mol L}^{-1} \)

Step 2: Apply the first-order rate law equation
For a first-order reaction: \[ \text{Rate} = k \cdot [A] \] We need to find the rate constant \( k \), so we rearrange the formula: \[ k = \frac{\text{Rate}}{[A]} \]

Step 3: Calculate the value of k
Substitute the given values into the equation: \[ k = \frac{5.6 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}}{0.2 \text{ mol L}^{-1}} \] \[ k = \frac{5.6 \times 10^{-4}}{2 \times 10^{-1}} \text{ s}^{-1} \] \[ k = 2.8 \times 10^{-3} \text{ s}^{-1} \]

Concept:

• Aromatic diazonium salts react with electron-rich aromatic compounds like phenols and amines to form brightly coloured azo compounds.
• This is known as a "Coupling Reaction" and is an example of electrophilic aromatic substitution.

Step 1: Identify the nature of the reactants
Benzene diazonium chloride (\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \)) acts as an electrophile due to the positively charged diazonium group.
Phenol (\( \text{C}_6\text{H}_5\text{OH} \)) acts as a nucleophile because the \( \text{-OH} \) group is strongly activating and increases electron density on the ring.

Step 2: Analyze the reaction conditions
The reaction with phenol is carried out in a weakly basic medium (\( \text{pH } 9\text{--}10 \)). In basic conditions, phenol is converted to the phenoxide ion (\( \text{C}_6\text{H}_5\text{O}^- \)), which is an even stronger nucleophile, facilitating the attack of the diazonium cation.

Step 3: Determine the site of attack and product
The diazonium cation attacks the para-position of the phenol (unless it is blocked) due to steric reasons and the directing nature of the hydroxyl group. The reaction is: \[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{OH} + \text{HCl} \] The product is p-hydroxyazobenzene, which is an orange-coloured dye.

Concept:

• Cleavage of ethers with \( \text{HI} \) typically follows the \( \text{S}_{\text{N}}2 \) mechanism for primary or secondary alkyl groups.
• However, if one of the alkyl groups is tertiary (\( 3^\circ \)), the mechanism shifts to \( \text{S}_{\text{N}}1 \).

Step 1: Evaluate the Assertion (A)

The given ether is Methyl tert-butyl ether. When reacted with \( \text{HI} \), the ether oxygen is first protonated. Because one of the groups is a tertiary butyl group, the C-O bond associated with the tertiary carbon breaks to form a highly stable tertiary carbocation (\( (\text{CH}_3)_3\text{C}^+ \)).

The iodide ion (\( \text{I}^- \)) then attacks this carbocation to form \( (\text{CH}_3)_3\text{C-I} \) (tert-butyl iodide), while the other fragment becomes \( \text{CH}_3\text{OH} \) (methanol). Thus, Assertion (A) is true.

Step 2: Evaluate the Reason (R)

The formation of the products described above is only possible if the reaction proceeds through a carbocation intermediate. In \( \text{S}_{\text{N}}2 \) reactions, the nucleophile would attack the smaller methyl group.

But here, the stability of the tertiary carbocation drives the \( \text{S}_{\text{N}}1 \) mechanism. Thus, Reason (R) is true.

Step 3: Determine the relationship

The reason perfectly explains why the iodide ends up on the bulky tertiary group rather than the smaller methyl group. Therefore, (R) is the correct explanation for (A).

Concept:

• Stability of diazonium salts (\( \text{R-N}_2^+\text{Cl}^- \)) depends on the nature of the group attached to the nitrogen.

Step 1: Evaluate the Assertion (A)

Primary aliphatic diazonium salts are extremely unstable and decompose immediately, even at \( 0^\circ\text{C} \), to release nitrogen gas and form carbocations.

In contrast, benzene diazonium salts are stable for a short time in cold aqueous solution (\( 0\text{--}5^\circ\text{C} \)). Therefore, aromatic diazonium salts are more stable than aliphatic ones.

Assertion (A) is false.

Step 2: Evaluate the Reason (R)

The relative stability of aromatic diazonium ions is due to the delocalization of the positive charge on the nitrogen atom into the benzene ring through resonance. This resonance stabilization is not possible in aliphatic systems. Therefore, Reason (R) is true.

Step 3: Conclusion

Since the assertion is false and the reason is true, the correct code is (D).

Concept:

• A complex reaction proceeds through a series of elementary steps (a mechanism).
• The overall rate of the reaction is determined by the slowest step in the mechanism, known as the Rate Determining Step (RDS).

Step 1: Evaluate the Assertion (A)

In a multistep reaction, the product cannot be formed faster than the slowest step allows. The stoichiometric coefficients of the reactants in the balanced equation of this slowest elementary step define the order of the overall reaction (with respect to those reactants). Thus, Assertion (A) is true.

Step 2: Evaluate the Reason (R)

Order of reaction is an experimental quantity that applies to all reactions, whether they are simple (elementary) or complex. It defines the relationship between the rate and concentrations. On the other hand, molecularity is only defined for elementary steps and has no meaning for a complex reaction. Reason (R) is false.

Step 3: Conclusion

Since the assertion is true and the reason is false, the correct code is (C).

Concept:

• Standard electrode potential (\( \text{E}^\circ \)) values for the reduction of transition metal ions to their metallic state depend on sublimation enthalpy, ionization enthalpy, and hydration enthalpy.
• Stability of specific oxidation states is often linked to extra stability associated with empty (\( d^0 \)), half-filled (\( d^5 \)), or completely filled (\( d^{10} \)) d-subshells.

Step 1: Evaluate Assertion (A)

The standard reduction potential for Manganese (\( \text{Mn}^{2+} + 2e^- \rightarrow \text{Mn} \)) is \( -1.18 \text{ V} \). This value is indeed "highly negative" compared to other members of the 3d transition series like Fe (\( -0.44 \text{ V} \)) or Co (\( -0.28 \text{ V} \)).

A negative value indicates that \( \text{Mn}^{2+} \) ions are very stable and resistant to reduction to the metallic state. Thus, Assertion (A) is true.

Step 2: Evaluate Reason (R)

The electronic configuration of Manganese (\( Z=25 \)) is \( [Ar] 3d^5 4s^2 \). Upon losing two electrons to form \( \text{Mn}^{2+} \), the configuration becomes \( [Ar] 3d^5 \). The \( d^5 \) subshell is exactly half-filled, which provides exceptional stability due to maximum exchange energy and symmetrical distribution of electrons. Thus, Reason (R) is true.

Step 3: Determine the relationship

Because the \( \text{Mn}^{2+} \) state is so energetically favorable (stable), the energy required to convert it back to neutral \( \text{Mn} \) is significantly higher. This stability of the ion directly translates to a more negative reduction potential. Therefore, the reason correctly explains the assertion.

Concept:

• This is a double-complex compound where both the cation and the anion are complex ions.
• Name the cation first, then the anion.
• In the anionic part, the metal name ends with the suffix ’-ate’.

Step 1: Analyze the cation

The cationic part is \( [\text{Ag}(\text{NH}_3)_2]^+ \). Ligand: ammine (two of them, so ’diammine’). Metal: Silver.

Oxidation state of Silver here is \( +1 \). Name: diamminesilver(I).

Step 2: Analyze the anion

The anionic part is \( [\text{Ag}(\text{CN})_2]^- \). Ligand: cyanido (two of them, so ’dicyanido’). Metal: Silver (must use Latin name with ’-ate’ suffix: argentate). Oxidation state of Silver here is also \( +1 \). Name: dicyanidoargentate(I).

Step 3: Combine the names

The full IUPAC name is: diamminesilver(I) dicyanidoargentate(I).

Concept:

• Identify the complex cation and simple anion.
• For polydentate ligands that already have numerical prefixes (like ethylenediamine), use ’bis’, ’tris’, etc.

Step 1: Identify the ions

Cation: \( [\text{Co}(\text{en})_3]^{3+} \).

Anion: \( \text{SO}_4^{2-} \) (sulphate).

Step 2: Name the complex cation

Ligand: \( \text{en} \) is ethane-1,2-diamine (or ethylenediamine). Since there are three, use the prefix ’tris’. Metal: Cobalt.

Oxidation state: Let Co be \( x \). \( 2x + 3(0) + 3(-2) = 0 \Rightarrow 2x = 6 \Rightarrow x = +3 \). Name: tris(ethane-1,2-diamine)cobalt(III).

Step 3: Name the overall compound

Combine the cation name with the anion name. Full Name: tris(ethane-1,2-diamine)cobalt(III) sulphate.

Concept:

• Denaturation involves the disruption of the native structural organization of proteins.
• Secondary structure refers to the localized folding of the polypeptide backbone.

Step 1: Explain the effect of denaturation

Denaturation occurs when a protein is subjected to physical changes (heat) or chemical changes (pH changes, salts).

• Structural change: Hydrogen bonds are disturbed, causing globule structures to unfold and helices to uncoil.
• Impact on levels: Secondary and tertiary structures are destroyed, but the primary structure (sequence of amino acids) remains intact.
• Result: The protein loses its biological activity (e.g., coagulation of egg white, curdling of milk).

Step 2: Identify types of secondary structures

Secondary structure describes the spatial arrangement of the polypeptide chain. The two most common types are:

1. \( \alpha \)-Helix: The polypeptide chain is coiled into a right-handed screw, stabilized by intramolecular hydrogen bonds between \( \text{-NH} \) and \( \text{-CO} \) groups of the peptide bond.
2. \( \beta \)-Pleated Sheet: Polypeptide chains are stretched out side-by-side and held together by intermolecular hydrogen bonds, creating a sheet-like appearance.

Concept:

• The order of a reaction defines how the rate changes with concentration changes.

Step 1: Identify the order

A zero-order reaction is one where the rate of the reaction is independent of the concentration of the reactants.

Mathematically: \( \text{Rate} = k[A]^0 = k \).

Step 2: Provide an example

A classic example is the decomposition of gaseous ammonia on a hot platinum surface at high pressure:

\[ 2\text{NH}_3\text{(g)} \xrightarrow{\text{Pt catalyst}} \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \]

At high pressures, the metal surface becomes completely saturated with ammonia molecules. Any further increase in ammonia concentration cannot increase the rate as there are no free sites on the catalyst surface.

Concept:

• In many reactions involving two molecules (bimolecular), the kinetics can be simplified under specific experimental conditions.

Step 1: State the condition

A bimolecular reaction follows first-order kinetics when one of the reactants is present in a large excess compared to the other. Because its concentration is so large, the change in its concentration during the reaction is negligible and can be treated as constant.

Step 2: Explain the kinetics

Consider: \( A + B \rightarrow \text{Product} \). The rate law is \( \text{Rate} = k[A][B] \). If \( [B] \) is in large excess, the rate becomes: \[ \text{Rate} = k'[A] \text{ where } k' = k[B] \] These reactions are termed pseudo-first-order reactions.

Step 3: Give an example

Acid-catalyzed hydrolysis of an ester: \[ \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \] Since water is the solvent and is in huge excess, the rate depends only on the ester concentration.

Concept:

• Henry’s Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Step 1: Analyze the condition at high pressure

When scuba divers descend into the ocean, they experience increased atmospheric and hydrostatic pressure. According to Henry’s Law, this high pressure increases the solubility of atmospheric gases (especially Nitrogen) in the diver’s blood and tissues.

Step 2: Analyze the effect of decompression

When the diver ascends rapidly to the surface, the external pressure decreases quickly. This causes the dissolved Nitrogen gas to become less soluble and rapidly escape from the blood.

Step 3: Identify the medical consequence

The released Nitrogen forms bubbles in the bloodstream. These bubbles can block capillaries and affect nerve impulses, a painful and dangerous medical condition known as Bends. Divers use tanks diluted with Helium (which is less soluble) to minimize this risk.

Concept:

• Solubility of Oxygen in blood is dependent on its partial pressure.

Step 1: Analyze the condition at high altitude

At high altitudes (mountain peaks), the atmospheric pressure is significantly lower than at sea level. Consequently, the partial pressure of Oxygen (\( p\text{O}_2 \)) is also low.

Step 2: Apply Henry’s Law

Since the partial pressure of oxygen is low, Henry’s Law predicts that the solubility of oxygen in the blood and tissues of people living at high altitudes or climbers will be lower.

Step 3: Identify the consequence

Low blood oxygen levels lead to a condition called Anoxia, which makes people feel weak and impairs their ability to think clearly.

Concept:

• Colligative properties depend on the number of solute particles in a solution.

Step 1: Identify the colligative property

The phenomenon involved is the Depression in Freezing Point.

Step 2: Explain the process

When salt (like \( \text{NaCl} \) or \( \text{CaCl}_2 \)) is sprinkled on snow-covered roads, it acts as a non-volatile solute and dissolves in the thin layer of moisture on the surface of the ice. This lowers the freezing point of water below its normal value of \( 0^\circ\text{C} \).

Step 3: State the result

If the atmospheric temperature is higher than this new, lowered freezing point of the salt-water mixture, the snow and ice will melt and turn into liquid water, effectively clearing the roads.

Concept:

• When alkyl halides react with strong bases like aqueous KOH, they undergo substitution. However, with alcoholic KOH and heat, they undergo dehydrohalogenation (elimination).
• Tertiary (\( 3^\circ \)) alkyl halides are highly prone to elimination (E2 mechanism) because the resulting alkene is stabilized and the central carbon is sterically hindered for substitution.

Step 1: Identify the nature of the reactants

The substrate is tert-butyl bromide (\( 3^\circ \) alkyl halide). The reagent is KOH in ethanol (alcoholic KOH). Alcoholic KOH acts as a strong base rather than a nucleophile.

Step 2: Determine the reaction mechanism

Since the substrate is a tertiary halide and a strong base is used with heating, the E2 elimination mechanism is favored. A proton is removed from one of the \( \beta \)-carbon atoms (methyl groups), and the bromide ion (leaving group) is expelled simultaneously.

Step 3: Draw the product structure

Removing \( \text{HBr} \) from \( (\text{CH}_3)_3\text{CBr} \) results in the formation of a double bond between the central carbon and one of the terminal methyl carbons. The major product is 2-methylpropene (isobutylene). \[ (\text{CH}_3)_2\text{C=CH}_2 \]

Concept:

• Reaction with \( \text{Br}_2 \) and heat (or light) in the presence of an alkyl side chain on a benzene ring favors free radical substitution at the benzylic position.
• Benzylic radicals are highly stable due to resonance with the benzene ring.

Step 1: Analyze the substrate structure

The substrate is 1-isopropyl-4-nitrobenzene. It has an isopropyl group at the para position relative to a nitro group. The carbon atom directly attached to the ring is the benzylic carbon. In an isopropyl group, this carbon is tertiary.

Step 2: Identify the site of substitution

Substitution occurs at the benzylic position because the intermediate tertiary benzylic radical is exceptionally stable.

The nitro group (\( \text{-NO}_2 \)) is a deactivating group for electrophilic aromatic substitution, further ensuring that substitution occurs on the side chain rather than the ring under these conditions.

Step 3: Draw the product

The hydrogen atom at the benzylic carbon is replaced by a Bromine atom. The product is 1-(2-bromopropan-2-yl)-4-nitrobenzene.

Concept:

• Reduction of nitro groups to amines.
• Diazotization of primary aromatic amines.
• Hydrolysis of diazonium salts to phenols.

Step 1: Determine structure A

The starting material is p-nitrotoluene. Treatment with \( \text{Fe/HCl} \) reduces the nitro (\( \text{-NO}_2 \)) group to an amino (\( \text{-NH}_2 \)) group.

Structure A is p-toluidine (4-methylaniline).

Step 2: Determine structure B

p-toluidine reacts with \( \text{NaNO}_2 \) and \( \text{HCl} \) at \( 0\text{--}5^\circ\text{C} \) (diazotization). The amino group is converted into a diazonium group (\( \text{-N}_2^+\text{Cl}^- \)). Structure B is p-toluenediazonium chloride.

Step 3: Determine structure C

When the diazonium salt is warmed with water (\( \text{H}_2\text{O} \)), the diazonium group is replaced by a hydroxyl (\( \text{-OH} \)) group. Structure C is p-cresol (4-methylphenol).

Concept:

• Substitution of diazonium group by cyanide (Sandmeyer reaction).
• Partial hydrolysis of nitriles to amides.
• Hofmann Bromamide Degradation of amides to primary amines.

Step 1: Determine structure A

Benzenediazonium chloride reacts with \( \text{CuCN} \). The diazonium group is replaced by the nitrile (\( \text{-CN} \)) group. Structure A is benzonitrile (phenyl cyanide).

Step 2: Determine structure B

Partial hydrolysis of benzonitrile in the presence of \( \text{OH}^- \) (alkaline medium) converts the nitrile group to an amide group. Structure B is benzamide (\( \text{C}_6\text{H}_5\text{CONH}_2 \)).

Step 3: Determine structure C

Benzamide is treated with \( \text{NaOBr} \) (which is \( \text{Br}_2 + \text{NaOH} \)). This is the Hofmann Bromamide Degradation reaction, which removes the carbonyl group and shortens the chain by one carbon. Structure C is aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)).

Concept:

• The reactivity of alkyl halides in \( \text{S}_{\text{N}}1 \) reactions depends on the stability of the carbocation intermediate formed in the slow, rate-determining step.
• Carbocation stability follows the order: Tertiary (\( 3^\circ \)) \( > \) Secondary (\( 2^\circ \)) \( > \) Primary (\( 1^\circ \)).

Step 1: Identify the four possible isomers of \( \text{C}_4\text{H}_9\text{Br} \)
The four structural isomers are:

1. n-Butyl bromide: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \) (Primary, \( 1^\circ \))
2. Isobutyl bromide: \( (\text{CH}_3)_2\text{CHCH}_2\text{Br} \) (Primary, \( 1^\circ \))
3. sec-Butyl bromide: \( \text{CH}_3\text{CH}_2\text{CH(Br)CH}_3 \) (Secondary, \( 2^\circ \))
4. tert-Butyl bromide: \( (\text{CH}_3)_3\text{CBr} \) (Tertiary, \( 3^\circ \))

Step 2: Analyze the stability of the corresponding carbocations
In the \( \text{S}_{\text{N}}1 \) mechanism, the C-Br bond breaks first.

• tert-Butyl bromide forms the tert-butyl carbocation (\( (\text{CH}_3)_3\text{C}^+ \)).
• This is a tertiary carbocation stabilized by the \( \text{+I} \) effect of three methyl groups and 9 hyperconjugative structures.
• It is much more stable than the secondary or primary alternatives.

Step 3: Conclusion on reactivity
Since the formation of the tert-butyl carbocation has the lowest activation energy due to its high stability, tert-butyl bromide (\( 2 \text{-bromo-2-methylpropane} \)) reacts the fastest.

Concept:

• Dehydrohalogenation is an elimination reaction where a hydrogen halide is removed from a substrate.
• According to Zaitsev’s Rule, the major product is the more highly substituted alkene (the one with the greater number of alkyl groups attached to the double-bonded carbons).

Step 1: Analyze the structure of the reactant
1-Bromo-1-methylcyclohexane has a bromine atom and a methyl group attached to the same carbon (C1) of the cyclohexane ring. The carbons adjacent to C1 (the \( \beta \)-carbons) have hydrogens that can be eliminated. There are two types:

1. The \( \text{CH}_2 \) groups at C2 and C6 within the ring.
2. The \( \text{CH}_3 \) group (methyl) attached to C1.

Step 2: Identify the possible elimination products

• Elimination involving a ring hydrogen (from C2 or C6) leads to 1-methylcyclohexene. This double bond is "trisubstituted" (attached to three carbons).
• Elimination involving a methyl hydrogen leads to methylenecyclohexane. This double bond is "disubstituted".

Step 3: Apply Zaitsev’s Rule to determine the major product
Since 1-methylcyclohexene is more substituted and therefore more thermodynamically stable, it will be the major product of the reaction.

Concept:

• Directing influence in aryl halides is determined by the competition between the Inductive effect (\( -\text{I} \)) and the Resonance effect (\( +\text{R} \)).

Step 1: Explain the deactivating nature (\( -\text{I} \) effect)

Chlorine is highly electronegative. It withdraws electron density from the benzene ring through the sigma (\( \sigma \)) bond. This overall reduction in electron density makes the ring less reactive towards electrophiles compared to benzene, hence it is "deactivating".

Step 2: Explain the directing nature (\( +\text{R} \) effect)

Chlorine has three lone pairs of electrons. These can be donated into the ring system via resonance.

When an electrophile attacks at the ortho or para positions, one of the resonance structures of the resulting carbocation (sigma complex) allows the positive charge to be placed directly on the carbon bearing the chlorine atom.

Step 3: Analyze the stabilization of the intermediate

In this specific structure, the lone pairs of chlorine can be shared with the carbocation, providing an additional stable resonance structure (where all atoms have octets).

This stabilization is not possible for meta-attack. Thus, while the \( -\text{I} \) effect slows down the reaction overall, the \( +\text{R} \) effect specifically lowers the energy of the ortho and para transition states, making them the preferred pathways.

Concept:

• For a multi-step (complex) reaction, the overall rate of the reaction is determined by the slowest elementary step, known as the Rate Determining Step (RDS).

Step 1: Identify the Rate Determining Step

The mechanism provided is:

• Step I: \( \text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{H}_2\text{O} + \text{IO}^- \) (slow)
• Step II: \( \text{H}_2\text{O}_2 + \text{IO}^- \rightarrow \text{H}_2\text{O} + \text{I}^- + \text{O}_2 \) (fast)

Step 2: Formulate the rate expression from the RDS

The rate law is derived from the stoichiometry of the reactants in the slow elementary step.

The reactants in Step I are one molecule of \( \text{H}_2\text{O}_2 \) and one ion of \( \text{I}^- \).

Step 3: Write the final expression

The rate law is expressed as:

\[ \text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-] \] where \( k \) is the rate constant for the overall reaction.

Concept:

• The order of reaction with respect to a reactant is the power to which its concentration term is raised in the rate law.
• The overall order is the sum of these powers.

Step 1: Recall the rate law expression
From the previous analysis of the slow step, the rate law was determined to be: \[ \text{Rate} = k[\text{H}_2\text{O}_2]^1[\text{I}^-]^1 \]

Step 2: Identify individual orders

• The exponent of the \( [\text{H}_2\text{O}_2] \) term is 1. Therefore, the reaction is first order with respect to \( \text{H}_2\text{O}_2 \).
• The exponent of the \( [\text{I}^-] \) term is 1. Therefore, the reaction is first order with respect to \( \text{I}^- \).

Step 3: Calculate the overall order
Overall Order = (Order w.r.t \( \text{H}_2\text{O}_2 \)) + (Order w.r.t \( \text{I}^- \)) \[ \text{Overall Order} = 1 + 1 = 2 \] The reaction follows second order kinetics overall.

Concept:

• Molecularity is defined as the number of reacting molecules, atoms, or ions that collide simultaneously to bring about a chemical reaction in an elementary step.

Step 1: Identify the reactants in Step II
The equation for Step II is: \[ \text{H}_2\text{O}_2 + \text{IO}^- \rightarrow \text{H}_2\text{O} + \text{I}^- + \text{O}_2 \]

Step 2: Count the reacting species
In this specific elementary step, the following species are reacting:

1. One molecule of hydrogen peroxide (\( \text{H}_2\text{O}_2 \)).
2. One hypoiodite ion (\( \text{IO}^- \)).

Step 3: State the molecularity
Since two species are involved in the collision for this step, the molecularity of Step II is 2. It is a bimolecular step.

Concept:

• Relative lowering of vapour pressure is a colligative property: \( \frac{P^\circ - P_s}{P^\circ} = i \cdot \chi_{\text{solute}} \).
• Dimerization affects the van’t Hoff factor (\( i \)).

Step 1: Calculate the number of moles of solute and solvent
Moles of Benzoic acid (\( n_{\text{solute}} \)) = \( \frac{61\text{ g}}{122\text{ g mol}^{-1}} = 0.5 \text{ mol} \). Moles of Benzene (\( n_{\text{solvent}} \)) = \( \frac{500\text{ g}}{78\text{ g mol}^{-1}} \approx 6.41 \text{ mol} \).

Step 2: Determine the van’t Hoff factor (i)
Benzoic acid dimerizes in benzene: \( 2\text{C}_6\text{H}_5\text{COOH} \rightarrow (\text{C}_6\text{H}_5\text{COOH})_2 \). For association, \( i = 1 - \alpha + \frac{\alpha}{n} \). With complete dimerization, \( \alpha = 1 \) and \( n = 2 \): \[ i = 1 - 1 + \frac{1}{2} = 0.5 \]

Step 3: Apply the relative lowering of vapour pressure formula
\[ \frac{P^\circ - P_s}{P^\circ} = i \cdot \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \] Substitute values: \[ \frac{66 - P_s}{66} = 0.5 \cdot \frac{0.5}{0.5 + 6.41} \] \[ \frac{66 - P_s}{66} = 0.5 \cdot \frac{0.5}{6.91} \approx 0.5 \cdot 0.07236 = 0.03618 \] \[ 66 - P_s = 66 \times 0.03618 = 2.388 \] \[ P_s = 66 - 2.388 = 63.612 \text{ torr} \]

Concept:

• Werner’s coordination theory states that primary valency is ionizable and satisfied by negative ions outside the coordination sphere.
• Valence Bond Theory (VBT) explains the bonding in complexes using hybridization of atomic orbitals.

Step 1: Determine the coordination formula from the precipitation data

The reaction with excess \( \text{AgNO}_3 \) yields 3 moles of \( \text{AgCl} \) per mole of the complex. This implies that there are 3 chloride ions (\( \text{Cl}^- \)) present outside the coordination sphere as ionizable primary valencies. The remaining 6 \( \text{NH}_3 \) molecules must be coordinated directly to the Cobalt central metal. Structural Formula: \( [\text{Co}(\text{NH}_3)_6]\text{Cl}_3 \)

Step 2: Determine the IUPAC name

The complex cation is \( [\text{Co}(\text{NH}_3)_6]^{3+} \). Ligands: six ammine groups \( \rightarrow \) hexaammine. Metal: Cobalt with oxidation state +3 (calculated as \( x + 6(0) = +3 \)). Full IUPAC Name: hexaamminecobalt(III) chloride

Step 3: Apply Valence Bond Theory for hybridisation

In \( [\text{Co}(\text{NH}_3)_6]^{3+} \), the central metal ion is \( \text{Co}^{3+} \). Electronic configuration of \( \text{Co} \): \( [Ar] 3d^7 4s^2 \) Electronic configuration of \( \text{Co}^{3+} \): \( [Ar] 3d^6 \) \( \text{NH}_3 \) is a strong field ligand. It causes the pairing of electrons in the \( 3d \) orbitals. The six electrons in \( 3d \) pair up in three orbitals, leaving two \( 3d \) orbitals empty. Hybridization: \( d^2sp^3 \) (inner orbital complex).

Step 4: Determine the magnetic behavior

Since all the electrons in the \( 3d \) subshell are paired due to the effect of the strong field ammine ligands, there are no unpaired electrons. The complex is diamagnetic.

Concept:

• Geometrical isomerism arises in coordination compounds due to different possible spatial arrangements of ligands.

Step 1: Analyze the complex \( [\text{Cr}(\text{C}_2\text{O}_4)_3]^{3-} \)

This is a complex of the type \( [\text{M}(\text{AA})_3] \), where \( \text{AA} \) is a symmetrical bidentate ligand (oxalate).

In this octahedral geometry, all positions are equivalent relative to one another in terms of the connectivity of these bidentate rings. Number of geometrical isomers = 0 (Zero). (Note: It only shows optical isomerism, existing as ’d’ and ’l’ forms).

Step 2: Analyze the complex \( [\text{Co}(\text{NH}_3)_3\text{Cl}_3] \)

This is a complex of the type \( [\text{MA}_3\text{B}_3] \). For such octahedral complexes, two distinct geometrical arrangements are possible:

1. Facial (fac) isomer: Three identical ligands occupy adjacent positions at the corners of one octahedral face.

2. Meridional (mer) isomer: Three identical ligands occupy positions around the meridian of the octahedron. Number of geometrical isomers = 2.

Concept:

• Inner orbital complexes use \( (n-1)d \) orbitals for hybridization (\( d^2sp^3 \)).
• Outer orbital complexes use \( nd \) orbitals for hybridization (\( sp^3d^2 \)).

Step 1: Explain the case of \( [\text{Co}(\text{NH}_3)_6]^{3+} \)

\( \text{Co}^{3+} \) has the configuration \( 3d^6 \). In the presence of the strong field ligand \( \text{NH}_3 \), the six electrons in the \( 3d \) subshell pair up. This leaves two \( 3d \) orbitals vacant. These two \( 3d \) orbitals, along with one \( 4s \) and three \( 4p \) orbitals, hybridize to form \( d^2sp^3 \) hybrid orbitals. Since \( (n-1)d \) orbitals are used, it is an inner orbital complex.

Step 2: Explain the case of \( [\text{Ni}(\text{NH}_3)_6]^{2+} \)

\( \text{Ni}^{2+} \) has the configuration \( 3d^8 \). Even if the electrons were to pair up as much as possible, there would be 4 pairs of electrons in the \( 3d \) subshell. This would only leave one \( 3d \) orbital vacant. For octahedral hybridization, two vacant \( d \)-orbitals are mandatory. Therefore, the metal must use the higher energy \( 4d \) orbitals for hybridization. The resulting hybridization is \( sp^3d^2 \). Since \( nd \) orbitals are used, it is an outer orbital complex.

Concept:

• Wilkinson’s catalyst is a coordination compound of Rhodium used in organic synthesis.

Step 1: State the formula

The formula for Wilkinson’s catalyst is \( [\text{RhCl}(\text{PPh}_3)_3] \).

Its chemical name is chlorotris(triphenylphosphine)rhodium(I).

Step 2: Describe its use

It is used as a homogeneous catalyst for the hydrogenation of alkenes. It facilitates the addition of hydrogen across carbon-carbon double bonds to form alkanes at room temperature and atmospheric pressure.

Concept:

• Conductance of ionic solutions depends on the mobility and concentration of ions.

Step 1: Identify the effect of DC on ionic solutions

When a direct current is passed through an ionic solution, electrolysis occurs.

The ions move toward the electrodes and undergo chemical reactions (reduction and oxidation).

Step 2: Explain the consequences of electrolysis

1. Change in Concentration: The chemical reactions change the composition and concentration of the solution near the electrodes.

2. Polarization: The products of electrolysis accumulate at the electrode surface, creating a counter-potential (polarization effect) that increases the resistance.

Step 3: State the alternative

Because of these changes, the measured resistance would be inaccurate and would vary over time. To avoid these issues, Alternating Current (AC) is used for measuring electrolytic resistance.

Concept:

• The products of electrolysis depend on the nature of the electrodes used.

Step 1: Analyze electrolysis with Platinum (Inert) electrodes

Platinum is an inert electrode and does not participate in chemical reactions.

• At Cathode: \( \text{Ag}^+ \) ions have a higher reduction potential than water, so Silver is deposited: \( \text{Ag}^+ + e^- \rightarrow \text{Ag(s)} \).
• At Anode: Water is oxidized preferentially over nitrate ions to release oxygen gas: \( 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \).

Step 2: Analyze electrolysis with Silver (Active) electrodes

Silver is an active electrode and can participate in the redox process.

• At Cathode: \( \text{Ag}^+ \) ions are reduced and deposited as Silver: \( \text{Ag}^+ + e^- \rightarrow \text{Ag(s)} \).
• At Anode: The silver metal of the anode itself gets oxidized because its oxidation potential is higher than that of water or nitrate ions: \( \text{Ag(s)} \rightarrow \text{Ag}^+ + e^- \).

Step 3: Conclusion

The products differ at the anode. With Pt, oxygen is produced. With Ag, the anode dissolves to provide more \( \text{Ag}^+ \) ions.

Concept:

• This is an application of "Cathodic Protection" or "Sacrificial Protection" against corrosion.

Step 1: Compare the reactivity of Mg and Fe

Magnesium is a more reactive metal than Iron (it is higher in the electrochemical series and has a more negative reduction potential).

Step 2: Describe the sacrificial process

When Magnesium blocks are connected to the Iron pipeline, the Magnesium acts as the Anode and the Iron acts as the Cathode. Magnesium undergoes oxidation (corrosion) instead of Iron:

\[ \text{Mg(s)} \rightarrow \text{Mg}^{2+} + 2e^- \]

Step 3: Explain the result

As long as the Magnesium is present, the Iron pipeline is protected from rusting. The Magnesium blocks are "sacrificed" to save the more expensive iron structure and are periodically replaced.

Concept:

• Grignard reagents (\( \text{RMgX} \)) react with nitriles to form imine salts, which on acid hydrolysis yield ketones.

Step 1: Write the reaction between the Grignard reagent and Nitrile

The nucleophilic phenyl group (\( \text{Ph}^- \)) from \( \text{C}_6\text{H}_5\text{MgBr} \) attacks the electrophilic carbon of the nitrile group in \( \text{CH}_3\text{CH}_2\text{CN} \).

\[ \text{CH}_3\text{CH}_2\text{C} \equiv \text{N} + \text{C}_6\text{H}_5\text{MgBr} \rightarrow \text{CH}_3\text{CH}_2\text{C}(\text{C}_6\text{H}_5)=\text{N-MgBr} \]

Step 2: Perform the hydrolysis

The imine salt intermediate is hydrolyzed in the presence of water and acid to form a ketone.

\[ \text{CH}_3\text{CH}_2\text{C}(\text{C}_6\text{H}_5)=\text{N-MgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{CO-C}_6\text{H}_5 + \text{NH}_3 + \text{Mg(OH)Br} \]

Step 3: Identify the product

The major product is Propiophenone (1-phenylpropan-1-one).

Concept:

• This is a modified Etard’s reaction used to oxidize a benzylic methyl group to an aldehyde group.

Step 1: Identify the transformation

Chromium trioxide (\( \text{CrO}_3 \)) in acetic anhydride reacts with p-fluorotoluene to form a gem-diacetate intermediate. This prevents further oxidation to a carboxylic acid.

Step 2: Show the hydrolysis step

The gem-diacetate intermediate is subsequently hydrolyzed with aqueous acid to yield the aldehyde.

\[ \text{F-C}_6\text{H}_4\text{-CH}_3 \xrightarrow[\text{(CH}_3\text{CO)}_2\text{O}]{\text{CrO}_3} \text{F-C}_6\text{H}_4\text{-CH(OCOCH}_3)_2 \xrightarrow{\text{H}_3\text{O}^+} \text{F-C}_6\text{H}_4\text{-CHO} \]

Step 3: Identify the product

The major product is p-fluorobenzaldehyde.

Concept:

• Carboxylic acids react with ammonia to form ammonium salts, which on heating yield amides. Dicarboxylic acids like phthalic acid can undergo cyclization.

Step 1: Form the ammonium salt

Phthalic acid reacts with two moles of ammonia to form ammonium phthalate. \[ \text{C}_6\text{H}_4(\text{COOH})_2 + 2\text{NH}_3 \rightarrow \text{C}_6\text{H}_4(\text{COONH}_4)_2 \]

Step 2: Identify the product after moderate heating

Heating leads to the loss of two water molecules to form phthalamide. \[ \text{C}_6\text{H}_4(\text{COONH}_4)_2 \xrightarrow{\Delta, -2\text{H}_2\text{O}} \text{C}_6\text{H}_4(\text{CONH}_2)_2 \]

Step 3: Identify the product after strong heating

Further strong heating causes the loss of one ammonia molecule via intramolecular cyclization.

\[ \text{C}_6\text{H}_4(\text{CONH}_2)_2 \xrightarrow{\text{strong } \Delta, -\text{NH}_3} \text{C}_6\text{H}_4(\text{CO})_2\text{NH} \] The final product is Phthalimide.

Concept:

• The presence of functional groups in glucose is proven by its chemical reactions.

Step 1: Identify the reagents for the carbonyl test

Glucose reacts with hydroxylamine (\( \text{NH}_2\text{OH} \)) and hydrogen cyanide (\( \text{HCN} \)).

Step 2: Describe the reaction with Hydroxylamine

Glucose reacts with hydroxylamine to form an oxime. This reaction involves the addition-elimination of the carbonyl oxygen. \[ \text{C}_5\text{H}_{11}\text{O}_5\text{-CHO} + \text{NH}_2\text{OH} \rightarrow \text{C}_5\text{H}_{11}\text{O}_5\text{-CH=N-OH} + \text{H}_2\text{O} \]

Step 3: Describe the reaction with HCN

Glucose adds a molecule of hydrogen cyanide to form a cyanohydrin. Both these reactions are characteristic of a carbonyl group (\( \text{>C=O} \)).

Concept:

• Acetylation is used to count the number of hydroxyl groups in a molecule.

Step 1: Identify the reagent for the hydroxyl test

Glucose is reacted with acetic anhydride (\( (\text{CH}_3\text{CO)}_2\text{O} \)).

Step 2: Describe the reaction outcome

Acetylation of glucose with acetic anhydride yields glucose pentaacetate.

Each \( \text{-OH} \) group reacts with one molecule of acetic anhydride to form an ester (\( \text{-OCOCH}_3 \)) group.

Step 3: Explain the stability implication

Since five acetate groups are added and the resulting pentaacetate molecule is stable, it confirms that glucose contains five \( \text{-OH} \) groups. Furthermore, because compounds with more than one \( \text{-OH} \) group on a single carbon are generally unstable, these five groups must be attached to five different carbon atoms.

Concept:

• Reducing sugars are defined by their ability to reduce specific mild oxidizing agents.

Step 1: State the chemical definition

Carbohydrates which reduce Fehling’s solution (to red \( \text{Cu}_2\text{O} \)) and Tollens’ reagent (to a silver mirror) are called reducing sugars.

Step 2: Identify the structural requirement

In these sugars, the carbonyl group (aldehyde or ketone) is free or in equilibrium with a free form in solution. This means the anomeric carbon is not involved in a glycosidic linkage.

Step 3: Give examples

All monosaccharides (like glucose, fructose) are reducing sugars. Disaccharides like maltose and lactose are also reducing sugars. Sucrose is not a reducing sugar because its carbonyl groups are bonded in the glycosidic linkage.

Concept:

• Configuration and optical rotation are two independent properties of a molecule.

Step 1: Explain the letter ’D’

The letter ’D’ represents the relative configuration of the molecule with respect to glyceraldehyde. It indicates that the \( \text{-OH} \) group on the lowest asymmetric carbon (C5 in glucose) is on the right side in the Fischer projection.

Step 2: Explain the sign ’(+)’

The sign ’(+)’ represents the optical activity or rotation of the plane of polarized light. It indicates that glucose is dextrorotatory, meaning it rotates the plane of polarized light to the right (clockwise).

Step 3: Clarify the distinction

It is important to note that ’D’ configuration is not necessarily linked to ’(+)’ rotation; they must be determined independently (configuration by comparison, rotation by polarimeter).

Concept:

• Glucose exists primarily in a cyclic pyranose form (six-membered ring containing an oxygen atom).

Step 1: Determine the ring type

The ring is a pyranose ring (similar to pyran), consisting of 5 carbons and 1 oxygen.

Step 2: Define the \( \alpha \)-anomer
In the \( \alpha \) form of D-glucose, the \( \text{-OH} \) group at the anomeric carbon (C1) is in the downward position (trans to the \( \text{-CH}_2\text{OH} \) group at C5) in the Haworth projection.

Step 3: Draw the Haworth structure

Concept:

• Acid-catalyzed dehydration of alcohols to alkenes proceeds via a carbocation intermediate.

Step 1: Formation of protonated alcohol

Ethanol reacts with a proton from the acid catalyst (\( \text{H}_2\text{SO}_4 \) or \( \text{H}_3\text{PO}_4 \)) to form an ethyloxonium ion. This makes the \( \text{-OH} \) group a better leaving group (\( \text{H}_2\text{O} \)).

\[ \text{CH}_3\text{CH}_2\text{OH} + \text{H}^+ \rightleftharpoons \text{CH}_3\text{CH}_2\text{O}^+\text{H}_2 \]

Step 2: Formation of carbocation

This is the slow, rate-determining step. The C-O bond breaks, releasing a water molecule and forming an ethyl carbocation.

\[ \text{CH}_3\text{CH}_2\text{O}^+\text{H}_2 \xrightarrow{\text{slow}} \text{CH}_3\text{CH}_2^+ + \text{H}_2\text{O} \]

Step 3: Formation of ethene by elimination of a proton

A proton is lost from the \( \beta \)-carbon to form the carbon-carbon double bond. The catalyst \( \text{H}^+ \) is regenerated. \[ \text{CH}_3\text{CH}_2^+ \rightarrow \text{CH}_2=\text{CH}_2 + \text{H}^+ \]

Concept:

• Oxidation of alcohols to aldehydes or ketones requires the removal of two hydrogen atoms: one from the hydroxyl group and one from the \( \alpha \)-carbon.

Step 1: Analyze the structure of tertiary alcohols

In a tertiary (\( 3^\circ \)) alcohol, the carbon atom attached to the hydroxyl group (the \( \alpha \)-carbon) is bonded to three other carbon atoms.

Example: \( (\text{CH}_3)_3\text{C-OH} \).

Step 2: Identify the missing requirement for oxidation

Since the \( \alpha \)-carbon is tertiary, there is no hydrogen atom attached to the \( \alpha \)-carbon.

Without an \( \alpha \)-hydrogen, standard mild oxidizing agents (like \( \text{PCC} \) or \( \text{K}_2\text{Cr}_2\text{O}_7 \)) cannot perform the dehydrogenation required to form a carbonyl group.

Step 3: State the result under strong conditions

Under very strong acidic oxidizing conditions and high temperatures, tertiary alcohols may eventually oxidize, but only by the breaking of C-C bonds to produce a mixture of ketones and carboxylic acids with fewer carbon atoms.

Concept:

• Electrophilic aromatic substitution is directed by the groups already present on the ring.

Step 1: Analyze the directing influence of existing groups

3-methylphenol (m-cresol) has:

1. \( \text{-OH} \) group at C1: Strongly activating and ortho/para directing.

2. \( \text{-CH}_3 \) group at C3: Weakly activating and ortho/para directing.

Step 2: Identify the most reactive sites

The \( \text{-OH} \) group is a much stronger activator than the methyl group and will dominate the directing influence.

Sites activated by \( \text{-OH} \) are ortho (C2 and C6) and para (C4). C2 is sterically hindered (situated between two groups).

Step 3: Determine the dinitration product

For dinitration, two nitro groups (\( \text{-NO}_2 \)) will enter the ring. The preferred positions are C4 (para to \( \text{-OH} \)) and C6 (ortho to \( \text{-OH} \)). Structure: 3-methyl-4,6-dinitrophenol.

Concept:

• Acidity of phenols is increased by electron-withdrawing groups and decreased by electron-donating groups.

Step 1: Analyze the effect of the Nitro group

In ortho-nitrophenol, the nitro (\( \text{-NO}_2 \)) group is a strong electron-withdrawing group (both \( -\text{I} \) and \( -\text{R} \) effects). It withdraws electron density from the phenoxide ion, stabilizing the negative charge. This facilitates the release of the proton.

Step 2: Analyze the effect of the Methoxy group

In ortho-methoxyphenol, the methoxy (\( \text{-OCH}_3 \)) group is an electron-donating group through resonance (\( +\text{R} \)). It increases electron density on the ring and the phenoxide ion, making the negative charge less stable. This hinders the release of the proton.

Step 3: Conclusion

Due to the strong stabilization of the phenoxide ion by the nitro group and the destabilization by the methoxy group, ortho-nitrophenol is significantly more acidic.

Concept:

• Grignard reagents are versatile intermediates in organic synthesis.
• They are prepared from alkyl or aryl halides and react with carbonyl compounds to form alcohols.
• Reaction with acetaldehyde specifically yields a secondary alcohol.

Step 1: Preparation of the Grignard Reagent
Bromobenzene is reacted with Magnesium metal in the presence of dry ether. The Magnesium atom inserts itself between the carbon and bromine to form Phenylmagnesium bromide. \[ \text{C}_6\text{H}_5\text{Br} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{MgBr} \]

Step 2: Nucleophilic addition to Acetaldehyde
The Grignard reagent acts as a nucleophile. The phenyl group (\( \text{C}_6\text{H}_5^- \)) attacks the electrophilic carbonyl carbon of acetaldehyde (ethanal). This forms an intermediate magnesium alkoxide adduct. \[ \text{C}_6\text{H}_5\text{MgBr} + \text{CH}_3\text{CHO} \rightarrow \text{CH}_3\text{CH}(\text{OMgBr})\text{C}_6\text{H}_5 \]

Step 3: Acid Hydrolysis
The magnesium alkoxide adduct is treated with dilute acid (hydrolysis). The \( \text{-OMgBr} \) group is converted into a hydroxyl (\( \text{-OH} \)) group. \[ \text{CH}_3\text{CH}(\text{OMgBr})\text{C}_6\text{H}_5 \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{CH(OH)CH}_3 \] The final product is 1-phenylethanol.

Concept:

• To obtain a meta-disubstituted product, the first group introduced must be a deactivating, meta-directing group.
• Acetyl groups are meta-directing, while nitro groups are also meta-directing.

Step 1: Friedel-Crafts Acylation of Benzene
Benzene is reacted with acetyl chloride (\( \text{CH}_3\text{COCl} \)) in the presence of anhydrous aluminium chloride (\( \text{AlCl}_3 \)) as a catalyst. This introduces an acetyl group onto the ring to form Acetophenone. \[ \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{anh. AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \]

Step 2: Nitration of Acetophenone
Acetophenone is treated with a nitrating mixture consisting of concentrated Nitric acid and concentrated Sulphuric acid. The carbonyl group (\( \text{-COCH}_3 \)) already present on the ring is electron-withdrawing and deactivating. Therefore, it directs the incoming electrophile (\( \text{NO}_2^+ \)) to the meta position. \[ \text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{conc. HNO}_3 / \text{conc. H}_2\text{SO}_4} \text{m-NO}_2\text{-C}_6\text{H}_4\text{-COCH}_3 \]

Step 3: Conclusion
The sequence of reactions must be Acylation followed by Nitration. If Nitration were done first, the resulting nitrobenzene would be too deactivated to undergo Friedel-Crafts Acylation.

Concept:

• 2,4-DNP test confirms the presence of a carbonyl group.
• Tollens’ test confirms the presence of an aldehyde group.
• Cannizzaro reaction is characteristic of aldehydes with no \( \alpha \)-hydrogen atoms.
• Final oxidation product determines the relative positions of substituents on the benzene ring.

Step 1: Deduce functional groups and structure
The formula \( \text{C}_8\text{H}_8\text{O} \) suggests high unsaturation (likely a benzene ring). 1. Positive 2,4-DNP test indicates a carbonyl group. 2. Reduction of Tollens’ reagent specifies that the carbonyl is an aldehyde (\( \text{-CHO} \)). 3. Participation in the Cannizzaro reaction indicates the absence of \( \alpha \)-hydrogen atoms. This implies the \( \text{-CHO} \) group is attached directly to the benzene ring or a quaternary carbon. 4. Vigorous oxidation to Benzene-1,2-dicarboxylic acid (phthalic acid) proves that the compound is an ortho-disubstituted benzene derivative. One substituent is \( \text{-CHO} \). To satisfy the \( \text{C}_8\text{H}_8\text{O} \) formula, the second substituent must be a methyl group (\( \text{-CH}_3 \)) at the ortho position. Therefore, the compound is 2-methylbenzaldehyde (also known as o-tolualdehyde).

Step 2: Write the Cannizzaro reaction products
The Cannizzaro reaction involves the disproportionation of two molecules of the aldehyde in the presence of concentrated alkali (e.g., \( \text{KOH} \)). One molecule is reduced to an alcohol and the other is oxidized to a carboxylic acid salt. Reaction: \[ 2(\textit{o}\text{-CH}_3\text{C}_6\text{H}_4\text{CHO}) \xrightarrow{\text{conc. KOH}} \textit{o}\text{-CH}_3\text{C}_6\text{H}_4\text{CH}_2\text{OH} + \textit{o}\text{-CH}_3\text{C}_6\text{H}_4\text{COOK} \] The products are 2-methylbenzyl alcohol and Potassium 2-methylbenzoate.

Concept:

• Acid strength increases with the presence of electron-withdrawing groups (\( -\text{I} \) and \( -\text{M} \)) which stabilize the carboxylate anion.
• Acid strength decreases with electron-donating groups.

Step 1: Evaluate the inductive effects of substituents

• \( \text{CF}_3\text{-} \): Three fluorine atoms exert a massive \( -\text{I} \) effect, making \( \text{CF}_3\text{COOH} \) the strongest acid.
• \( \text{NO}_2\text{-} \): The nitro group is a very strong electron-withdrawing group (\( -\text{I} \)). Thus, \( \text{O}_2\text{NCH}_2\text{COOH} \) is very acidic, but less so than the trifluoro derivative.
• \( \text{H-} \): In \( \text{HCOOH} \), there is no significant inductive effect.
• \( \text{C}_6\text{H}_5\text{-} \): While the phenyl ring has a \( -\text{I} \) effect, its \( +\text{R} \) effect in the benzoic acid system (and the large size of the group) makes it slightly weaker than formic acid (\( pK_a \text{ of } \text{PhCOOH} \approx 4.2 \text{ vs } \text{HCOOH} \approx 3.75 \)).

Step 2: Determine the final order
Based on the stability of the resulting carboxylate ions, the increasing order of acid strength is: \[ \text{C}_6\text{H}_5\text{COOH} < \text{HCOOH} < \text{O}_2\text{N} - \text{CH}_2 - \text{COOH} < \text{CF}_3 - \text{COOH} \]

Concept:

• Aldehydes and ketones undergo nucleophilic addition reactions with ammonia derivatives like 2,4-Dinitrophenylhydrazine (2,4-DNP).
• These reactions involve the addition of the nucleophile followed by the elimination of a water molecule to form a carbon-nitrogen double bond (\( \text{C=N} \)).

Step 1: Analyze the structure of the reactants
Benzaldehyde is an aromatic aldehyde with the formula \( \text{C}_6\text{H}_5\text{CHO} \). 2,4-Dinitrophenylhydrazine is a substituted hydrazine. The lone pair on the terminal amino group of the hydrazine acts as the nucleophile.

Step 2: Describe the reaction mechanism
The reaction is typically catalyzed by traces of acid. 1. The nucleophilic nitrogen atom of 2,4-DNP attacks the electrophilic carbonyl carbon of benzaldehyde. 2. A proton transfer occurs within the intermediate. 3. A molecule of water is eliminated from the intermediate to establish a \( \text{C=N} \) double bond, resulting in the formation of a hydrazone.

Step 3: Identify the final product and its characteristics
The reaction is: \[ \text{C}_6\text{H}_5\text{CHO} + \text{H}_2\text{NNH-C}_6\text{H}_3(\text{NO}_2)_2 \rightarrow \text{C}_6\text{H}_5\text{CH=NNH-C}_6\text{H}_3(\text{NO}_2)_2 + \text{H}_2\text{O} \] The product is Benzaldehyde 2,4-dinitrophenylhydrazone. It usually appears as an orange or yellow crystalline precipitate, which is used to identify the presence of a carbonyl group.

Concept:

• This is a Cross-Aldol Condensation reaction.
• Specifically, when an aromatic aldehyde reacts with an aliphatic or aromatic ketone having \( \alpha \)-hydrogens, it is known as the Claisen-Schmidt reaction.

Step 1: Identify the role of the reactants
Benzaldehyde (\( \text{C}_6\text{H}_5\text{CHO} \)) lacks \( \alpha \)-hydrogen atoms and therefore cannot form an enolate ion. Acetophenone (\( \text{C}_6\text{H}_5\text{COCH}_3 \)) has three \( \alpha \)-hydrogen atoms on the methyl group.

Step 2: Describe the condensation mechanism
1. The base (\( \text{OH}^- \)) removes an \( \alpha \)-hydrogen from acetophenone to form a resonance-stabilized enolate ion: \( \text{C}_6\text{H}_5\text{COCH}_2^- \). 2. The enolate ion acts as a nucleophile and attacks the carbonyl carbon of benzaldehyde. 3. Protonation of the resulting alkoxide yields a \( \beta \)-hydroxy ketone (aldol).

Step 3: Explain the effect of heating
Upon heating, the \( \beta \)-hydroxy ketone undergoes spontaneous dehydration (loss of water) to form a stable, conjugated \( \alpha,\beta \)-unsaturated ketone. The overall reaction is: \[ \text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COC}_6\text{H}_5 \xrightarrow{\text{dil. NaOH, } \Delta} \text{C}_6\text{H}_5\text{CH=CH-COC}_6\text{H}_5 + \text{H}_2\text{O} \]

Step 4: Identify the final product
The major product is 1,3-diphenylprop-2-en-1-one, commonly known as Benzalacetophenone or Chalcone.

Concept:

• Friedel-Crafts reactions (alkylation and acylation) involve the attack of a carbocation electrophile on an electron-rich aromatic ring.
• The presence of strongly electron-withdrawing groups on the ring significantly inhibits these reactions.

Step 1: Identify the deactivating nature of the carboxyl group

In benzoic acid, the carboxyl group (\( \text{-COOH} \)) is directly attached to the benzene ring.

The carbonyl carbon within the \( \text{-COOH} \) group is electron-deficient and exerts a powerful electron-withdrawing effect through both inductive (\( -\text{I} \)) and resonance (\( -\text{M} \)) effects. This drastically reduces the electron density on the benzene ring, making it a poor nucleophile for the Friedel-Crafts electrophile.

Step 2: Analyze the interaction with the Lewis acid catalyst

The Friedel-Crafts reaction requires a Lewis acid catalyst, typically anhydrous aluminium chloride (\( \text{AlCl}_3 \)). The oxygen atoms in the carboxyl group possess lone pairs of electrons. Because \( \text{AlCl}_3 \) is a strong Lewis acid, it coordinates with the oxygen atoms of the \( \text{-COOH} \) group rather than reacting with the alkyl or acyl halide.

This "complexation" not only consumes the catalyst but also puts a positive charge near the ring, further deactivating it.

Concept:

• Boiling point depends on the strength of intermolecular forces.
• Both alcohols and carboxylic acids exhibit hydrogen bonding, but the extent and nature of this bonding differ.

Step 1: Compare the nature of hydrogen bonding

While both functional groups form hydrogen bonds, the carboxyl group (\( \text{-COOH} \)) is more polarized than the hydroxyl group (\( \text{-OH} \)) of an alcohol.

The carbonyl oxygen (\( \text{C=O} \)) in carboxylic acids is strongly electronegative and acts as an excellent hydrogen bond acceptor, while the \( \text{-OH} \) group acts as a donor.

Step 2: Describe the formation of cyclic dimers

Due to the specific arrangement of the \( \text{C=O} \) and \( \text{O-H} \) groups, carboxylic acid molecules associate in pairs to form stable cyclic dimers through two hydrogen bonds.

\[ \text{R-C} \begin{pmatrix} \text{O} \cdots \text{H-O}
\text{O-H} \cdots \text{O} \end{pmatrix} \text{C-R} \] These dimers are so stable that they often persist even in the vapour phase.

Step 3: Explain the effect on boiling point

Alcohols only form linear chains of hydrogen-bonded molecules. Because carboxylic acids form these double-bridged dimeric structures, the effective molecular mass of the associated units is doubled, and the intermolecular forces are much harder to break. Consequently, a significantly higher amount of thermal energy (temperature) is required to vaporize carboxylic acids compared to alcohols of similar size.

Concept:

• Aldehydes can be distinguished from ketones using mild oxidizing agents that only aldehydes can reduce.

Step 1: Perform the Tollens’ Test
Add Tollens’ reagent (ammoniacal silver nitrate) to both compounds in separate test tubes and warm them.

• Propanal (Aldehyde): It is oxidized to propionate ion and reduces \( \text{Ag}^+ \) to metallic silver, forming a silver mirror on the walls of the tube. \[ \text{CH}_3\text{CH}_2\text{CHO} + 2[\text{Ag}(\text{NH}_3)_2]^+ + 3\text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + 2\text{Ag(s)} + 4\text{NH}_3 + 2\text{H}_2\text{O} \]
• Propanone (Ketone): It does not react with Tollens’ reagent, and no silver mirror is formed.

Step 2: Alternative (Iodoform Test)
Propanone is a methyl ketone and will give a positive Iodoform test (yellow precipitate of \( \text{CHI}_3 \) with \( \text{I}_2/\text{NaOH} \)), whereas propanal will not.

Concept:

• The standard cell potential (\( \text{E}^\circ_{\text{cell}} \)) is related to the equilibrium constant (\( \text{K}_{\text{c}} \)) of the cell reaction.
• At equilibrium, the cell potential (\( \text{E}_{\text{cell}} \)) is zero.
• The relationship is given by the Nernst equation at \( 298\text{ K} \): \( \text{E}^\circ_{\text{cell}} = \frac{0.059}{n} \log \text{K}_{\text{c}} \).

Step 1: Identify the number of electrons transferred (n)
The reaction is: \( \text{Cu (s)} + 2\text{Ag}^+ \text{(aq)} \rightarrow \text{Cu}^{2+} \text{(aq)} + 2\text{Ag (s)} \).
Oxidation half-reaction: \( \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \)
Reduction half-reaction: \( 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag} \)
Total electrons transferred, \( n = 2 \).

Step 2: Apply the relationship formula
Using the formula: \[ \text{E}^\circ_{\text{cell}} = \frac{0.059}{n} \log \text{K}_{\text{c}} \] Substitute the values: \( n = 2 \) and \( \text{K}_{\text{c}} = 10^{15} \). \[ \text{E}^\circ_{\text{cell}} = \frac{0.059}{2} \log(10^{15}) \]

Step 3: Calculate the final value
Using the property \( \log(a^b) = b \log a \): \[ \log(10^{15}) = 15 \log 10 = 15 \times 1 = 15 \] Now substitute back: \[ \text{E}^\circ_{\text{cell}} = \frac{0.059}{2} \times 15 \] \[ \text{E}^\circ_{\text{cell}} = 0.0295 \times 15 = 0.4425 \text{ V} \]

Concept:

• A lead storage battery is a secondary cell consisting of a lead anode and a grid of lead packed with lead dioxide (\( \text{PbO}_2 \)) as the cathode.
• A \( 38\% \) solution of sulphuric acid (\( \text{H}_2\text{SO}_4 \)) is used as the electrolyte.

Step 1: Write the anode reaction
At the anode, lead metal is oxidized to lead sulphate. \[ \text{Pb(s)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{PbSO}_4\text{(s)} + 2e^- \]

Step 2: Write the cathode reaction
At the cathode, lead dioxide is reduced to lead sulphate in the presence of acid. \[ \text{PbO}_2\text{(s)} + \text{SO}_4^{2-}\text{(aq)} + 4\text{H}^+\text{(aq)} + 2e^- \rightarrow \text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)} \]

Step 3: Write the overall cell reaction
Summing the anode and cathode reactions: \[ \text{Pb(s)} + \text{PbO}_2\text{(s)} + 2\text{H}_2\text{SO}_4\text{(aq)} \rightarrow 2\text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)} \]

Concept:

• Faraday’s first law states that the amount of substance produced or consumed at an electrode is proportional to the quantity of electricity passed.
• The quantity of electricity required (\( Q \)) is given by \( Q = n \times F \), where \( n \) is the number of moles of electrons and \( F \) is Faraday’s constant (\( \approx 96500\text{ C} \)).

Step 1: Determine the oxidation state change
In \( \text{FeO} \), iron is in the \( +2 \) state (\( \text{Fe}^{2+} \)).
In \( \text{Fe}_2\text{O}_3 \), iron is in the \( +3 \) state (\( \text{Fe}^{3+} \)).
The oxidation process for 1 mole of iron is: \[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + 1e^- \]

Step 2: Identify the moles of electrons
For the oxidation of 1 mole of \( \text{FeO} \) to \( \text{Fe}_2\text{O}_3 \), 1 mole of electrons (\( n = 1 \)) is released.

Step 3: Calculate the electricity in coulombs
Electricity (\( Q \)) = \( n \times F \)
\( Q = 1 \text{ mol } e^- \times 96500 \text{ C mol}^{-1} \)
\( Q = 96500 \text{ C} \).

Concept:

• Molar conductivity (\( \Lambda_{\text{m}} \)) is calculated using the formula \( \Lambda_{\text{m}} = \frac{\kappa \times 1000}{C} \).
• Degree of dissociation (\( \alpha \)) is the ratio of molar conductivity at a given concentration to the limiting molar conductivity: \( \alpha = \frac{\Lambda_{\text{m}}}{\Lambda^\circ_{\text{m}}} \).

Step 1: Calculate the molar conductivity (\( \Lambda_{\text{m}} \))
Given:
Conductivity (\( \kappa \)) = \( 7.2 \times 10^{-5} \text{ S cm}^{-1} \)
Concentration (\( C \)) = \( 0.0024\text{ M} \) \[ \Lambda_{\text{m}} = \frac{7.2 \times 10^{-5} \text{ S cm}^{-1} \times 1000 \text{ cm}^3\text{/L}}{0.0024 \text{ mol/L}} \] \[ \Lambda_{\text{m}} = \frac{7.2 \times 10^{-2}}{0.0024} = \frac{0.072}{0.0024} = 30 \text{ S cm}^2 \text{ mol}^{-1} \]

Step 2: Calculate the degree of dissociation (\( \alpha \))
Given: \( \Lambda^\circ_{\text{m}} = 390.5 \text{ S cm}^2 \text{ mol}^{-1} \) \[ \alpha = \frac{\Lambda_{\text{m}}}{\Lambda^\circ_{\text{m}}} \] \[ \alpha = \frac{30}{390.5} \approx 0.0768 \]

Concept:

• The potential of a half-cell depends on the concentration of the ions involved, as described by the Nernst equation.
• For the reduction half-reaction \( \text{M}^{n+} + n e^- \rightarrow \text{M} \), the equation is: \( \text{E} = \text{E}^\circ - \frac{0.059}{n} \log \frac{1}{[\text{M}^{n+}]} \).

Step 1: Identify parameters
\( \text{E}^\circ = +0.80 \text{ V} \)
\( n = 1 \) (one electron transferred)
\( [\text{Ag}^+] = 0.01 \text{ M} = 10^{-2} \text{ M} \)

Step 2: Apply the Nernst equation
\[ \text{E} = \text{E}^\circ - \frac{0.059}{1} \log \frac{1}{[\text{Ag}^+]} \] \[ \text{E} = 0.80 - 0.059 \log \frac{1}{10^{-2}} \] \[ \text{E} = 0.80 - 0.059 \log(10^2) \]

Step 3: Calculate the final potential
\[ \log(10^2) = 2 \log 10 = 2 \times 1 = 2 \] \[ \text{E} = 0.80 - 0.059 \times 2 \] \[ \text{E} = 0.80 - 0.118 = 0.682 \text{ V} \]

Concept:

• Electrolytes are substances that provide ions to facilitate the flow of current within a cell or battery.

Step 1: Identify electrolyte for Dry cell

In a standard dry cell (Leclanché cell), the electrolyte is a moist paste of Ammonium chloride (\( \text{NH}_4\text{Cl} \)) and Zinc chloride (\( \text{ZnCl}_2 \)). Starch is often added to maintain the paste-like consistency.

Step 2: Identify electrolyte for Fuel cell

In the hydrogen-oxygen fuel cell used in spacecraft, the electrolyte is typically a concentrated aqueous solution of Potassium hydroxide (\( \text{KOH} \)) or Sodium hydroxide (\( \text{NaOH} \)).

Concept:

• The acidity and basicity of transition metal oxides are closely linked to the oxidation state of the metal.
• As the oxidation state of a metal increases, its electronegativity and the covalent character of its oxides increase, leading to higher acidity.

Step 1: Compare the oxidation states

In \( \text{MnO} \), the oxidation state of Manganese is \( +2 \).
In \( \text{Mn}_2\text{O}_7 \), the oxidation state of Manganese is \( +7 \).

Step 2: Analyze the bonding and chemical nature

1. Lower Oxidation State (+2): In \( \text{MnO} \), the metal has a low positive charge and low electronegativity. It easily loses electrons and forms ionic bonds. Consequently, it reacts with acids to form salts and water, behaving as a basic oxide.

2. Higher Oxidation State (+7): In \( \text{Mn}_2\text{O}_7 \), the metal has a very high positive charge. This high charge density pulls electron density away from the oxygen atoms (Fajans’ rule), increasing covalent character. The high oxidation state makes the metal atom more electronegative, causing it to react with water to form permanganic acid (\( \text{HMnO}_4 \)). Thus, it is an acidic oxide.

Concept:

• Enthalpy of atomization is the energy required to break the metallic bonds and convert one mole of a solid metal into isolated gaseous atoms.
• The strength of metallic bonding depends on the number of unpaired d-electrons available for interatomic d-d overlapping.

Step 1: Compare the electronic configurations

1. Iron (\( Z=26 \)): Configuration is \( [Ar] 3d^6 4s^2 \). In the \( 3d \) subshell, it has 4 unpaired electrons.

2. Copper (\( Z=29 \)): Configuration is \( [Ar] 3d^{10} 4s^1 \). In the \( 3d \) subshell, it has 0 unpaired electrons (fully filled).

Step 2: Relate configuration to bonding strength

In Iron, the 4 unpaired d-electrons participate strongly in interatomic metallic bonding through d-d orbital overlapping in addition to \( 4s \) metallic bonding.

In Copper, the \( 3d \) shell is completely filled, so d-d overlapping is absent or very weak. Bonding is primarily due to the single \( 4s \) electron.

Step 3: Conclusion

Since Iron has more unpaired electrons, its metallic bonds are significantly stronger than those in Copper. Therefore, it requires more energy to atomize Iron.

Concept:

• An oxidizing agent is a species that tends to gain electrons (is reduced).
• The stability of the resulting electronic configuration determines the ease of reduction.

Step 1: Analyze \( \text{Mn}^{3+} \)

Electronic configuration of \( \text{Mn}^{3+} \): \( [Ar] 3d^4 \).

When \( \text{Mn}^{3+} \) acts as an oxidizing agent, it gains one electron to become \( \text{Mn}^{2+} \).

Configuration of \( \text{Mn}^{2+} \): \( [Ar] 3d^5 \).

The \( 3d^5 \) configuration is exactly half-filled, which is exceptionally stable. Thus, \( \text{Mn}^{3+} \) has a strong tendency to gain an electron.

Step 2: Analyze \( \text{Cr}^{3+} \)

Electronic configuration of \( \text{Cr}^{3+} \): \( [Ar] 3d^3 \).

In aqueous solution or coordination complexes, the \( 3d \) orbitals split into \( t_{2g} \) and \( e_g \) levels.

For \( \text{Cr}^{3+} \), the configuration is \( t_{2g}^3 \). This is a half-filled \( t_{2g} \) level, which is a stable state.

\( \text{Cr}^{3+} \) is more stable in its current state and actually tends to be oxidized further to \( \text{Cr}^{6+} \) (making it a reducing agent in some conditions).

Step 3: Conclusion

Because the reduction of \( \text{Mn}^{3+} \) leads to a highly stable \( d^5 \) state, it is a much stronger oxidizing agent than \( \text{Cr}^{3+} \).

Concept:

• The industrial preparation of dichromates involves the conversion of chromate to dichromate in acidic medium followed by a displacement reaction.

Step 1: Conversion of Sodium chromate to Sodium dichromate

Sodium chromate solution is acidified with concentrated sulphuric acid. This converts the yellow chromate ion into the orange dichromate ion.

\[ 2\text{Na}_2\text{CrO}_4\text{(aq)} + \text{H}_2\text{SO}_4\text{(conc)} \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7\text{(aq)} + \text{Na}_2\text{SO}_4\text{(aq)} + \text{H}_2\text{O(l)} \]

Step 2: Conversion of Sodium dichromate to Potassium dichromate

Sodium dichromate is highly soluble in water, making it difficult to crystallize. It is treated with a calculated amount of Potassium chloride (\( \text{KCl} \)). Since Potassium dichromate is much less soluble than sodium dichromate, it precipitates as orange crystals upon cooling. \[ \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl} \]

Concept:

• The complexity of actinoid chemistry arises from their electronic structure and radioactive properties.

Step 1: Explain the oxidation states

In lanthanoids, the energy gap between \( 4f \) and \( 5d \) subshells is large, making \( +3 \) the overwhelmingly dominant oxidation state.

In actinoids, the \( 5f, 6d, \text{ and } 7s \) energy levels are very close to each other. Consequently, more electrons can participate in bonding, leading to a much wider variety of oxidation states (up to \( +7 \)).

Step 2: Explain the radioactive nature

All actinoids are radioactive elements. Many have very short half-lives, making it extremely difficult to isolate and study them in large quantities. The high radioactivity also leads to self-irradiation damage to the compounds being studied.

Step 3: Mention the actinoid contraction

The contraction in atomic and ionic sizes across the actinoid series is more pronounced than the lanthanoid contraction due to the poorer shielding provided by \( 5f \) electrons compared to \( 4f \) electrons.

Concept:

• The standard electrode potential (\( \text{E}^\circ \)) is not a single property; it is the net result of several thermodynamic processes.

Step 1: Identify the energy components

The conversion of a solid metal atom to an aqueous divalent ion involves:

1. Enthalpy of sublimation (\( \Delta_{\text{sub}}\text{H} \)): Converting solid metal to gaseous atoms.

2. Ionization Enthalpy (\( \Delta_{\text{i}}\text{H}_1 + \Delta_{\text{i}}\text{H}_2 \)): Removing two electrons to form the ion.

3. Enthalpy of Hydration (\( \Delta_{\text{hyd}}\text{H} \)): Stability of the ion in water.

Step 2: Explain the lack of regularity

Across the 3d series, these three energies do not change in a smooth, monotonic fashion.

For example, there is a sudden jump in sublimation enthalpy for some metals, or an unusual ionization energy due to stable half-filled shells (like \( \text{Mn} \)).

Because these terms vary independently and non-linearly, the total sum (\( \text{E}^\circ \)) shows irregular trends across the series.

Concept:

• Chromium exhibits a variety of oxoanions, primarily the chromate ion (\( \text{CrO}_4^{2-} \)) and the dichromate ion (\( \text{Cr}_2\text{O}_7^{2-} \)).
• These two ions exist in a pH-dependent equilibrium in aqueous solution.

Step 1: Understand the chromate-dichromate equilibrium
The interconversion between chromate (yellow) and dichromate (orange) is controlled by the concentration of hydrogen ions (\( \text{H}^+ \)) in the solution. The equilibrium reaction is represented as follows: \[ 2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \]

Step 2: Analyze the effect of an acidic medium
In an acidic medium, the concentration of \( \text{H}^+ \) ions is high. According to Le Chatelier’s Principle, adding more reactant (\( \text{H}^+ \)) will shift the equilibrium toward the right side. This shift converts the yellow chromate ions into orange dichromate ions. Therefore, at low pH (acidic conditions), the dichromate ion is the predominant and stable species.

Step 3: Identify the ion
The oxoanion of chromium stable in acidic medium is the Dichromate ion (\( \text{Cr}_2\text{O}_7^{2-} \)). In this ion, chromium is in its highest oxidation state of \( +6 \), which makes it a powerful oxidizing agent in acidic solutions.

Concept:

• The most common and stable oxidation state for all lanthanoids is \( +3 \).
• However, some elements exhibit additional oxidation states like \( +2 \) or \( +4 \) to achieve particularly stable electronic configurations, such as empty (\( f^0 \)), half-filled (\( f^7 \)), or completely filled (\( f^{14} \)) subshells.

Step 1: Analyze the electronic configuration of Cerium
Cerium (\( \text{Ce} \)) is the first element of the lanthanoid series with atomic number 58. The ground state electronic configuration of Cerium is: \[ \text{Ce: } [\text{Xe}] 4f^1 5d^1 6s^2 \]

Step 2: Explain the formation of the \( +4 \) state
By losing all four of its valence electrons (one from \( 4f \), one from \( 5d \), and two from \( 6s \)), Cerium forms the \( \text{Ce}^{4+} \) ion. Configuration of \( \text{Ce}^{4+}: [\text{Xe}] 4f^0 \). This state is exceptionally stable because it corresponds to the noble gas configuration of Xenon (\( 4f^0 \)).

Step 3: Describe its chemical behavior
Although \( \text{Ce}^{4+} \) is stable due to its empty \( f \)-subshell, it is a very strong oxidizing agent in aqueous solution. This is because it has a strong tendency to return to the more favored and common \( +3 \) oxidation state of the lanthanoid series. \[ \text{Ce}^{4+} \text{(aq)} + e^- \rightarrow \text{Ce}^{3+} \text{(aq)} \quad (\text{E}^\circ = +1.74 \text{ V}) \] Because of this high reduction potential, Cerium(IV) salts (like ceric ammonium sulphate) are widely used as volumetric oxidizing agents in analytical chemistry.

Step 4: Conclusion
The lanthanoid element that exhibits a prominent \( +4 \) oxidation state is Cerium (\( \text{Ce} \)).

Concept:

• Potassium permanganate is thermally unstable and decomposes upon strong heating.

Step 1: Write the reaction

When dark purple crystals of Potassium permanganate are heated to about \( 513 \text{ K} \), they decompose to form potassium manganate, manganese dioxide, and oxygen gas.

\[ 2\text{KMnO}_4 \xrightarrow{\Delta} \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2 \]

Step 2: Identify the observations

The purple crystals turn into a black/green residue (\( \text{K}_2\text{MnO}_4 + \text{MnO}_2 \)) and a gas that supports combustion (\( \text{O}_2 \)) is evolved.

Concept:

• The color of transition metal ions is generally due to d-d transitions.
• For an ion to be colored, it must have a partially filled d-subshell (unpaired d-electrons). Ions with \( d^0 \) or \( d^{10} \) configurations are colorless.

Step 1: Analyze the electronic configurations

1. \( \text{Cu}^+ \): \( [Ar] 3d^{10} \). Completely filled \( d \)-subshell. Colorless.

2. \( \text{Sc}^{3+} \): \( [Ar] 3d^0 \). Empty \( d \)-subshell. Colorless.

3. \( \text{Fe}^{2+} \): \( [Ar] 3d^6 \). Partially filled \( d \)-subshell (4 unpaired electrons). Coloured (Pale Green).

4. \( \text{Ti}^{3+} \): \( [Ar] 3d^1 \). Partially filled \( d \)-subshell (1 unpaired electron). Coloured (Purple/Violet).

5. \( \text{Zn}^{2+} \): \( [Ar] 3d^{10} \). Completely filled \( d \)-subshell. Colorless.

Step 2: Conclusion

The colored ions are \( \text{Fe}^{2+} \) and \( \text{Ti}^{3+} \).