CBSE Class 12 Chemistry Question Paper 2024 (Set 3 - 56/4/3) with Answer Key

Step 1: The molar conductivity for \( Al_2(SO_4)_3 \) is the sum of the molar conductivities of \( Al^{3+} \) and \( SO_4^{2-} \).
Using the given data, the limiting molar conductivity can be calculated as: \[ \Lambda_m = 2 \times \Lambda_{Al^{3+}} + 3 \times \Lambda_{SO_4^{2-}} = 2 \times 189 + 3 \times 160 = 858 \, S cm^2 \, mol^{-1}. \] Quick Tip: When calculating the limiting molar conductivity, add the ionic conductivities of the individual ions in the compound, accounting for the number of ions per formula unit.

Vitamin C is chemically known as ascorbic acid (C\(_6\)H\(_8\)O\(_6\)). It is a water-soluble vitamin that plays an important role in the body as an antioxidant and in collagen synthesis. The other acids listed, such as saccharic acid, gluconic acid, and benzoic acid, do not represent Vitamin C.

Step 1: Ascorbic acid is widely known as Vitamin C.
Step 2: It is essential for the proper functioning of the immune system and collagen formation. Quick Tip: Ascorbic acid (Vitamin C) is an essential nutrient for humans, playing a role in tissue repair and antioxidant activity.

The Rosenmund reduction is used to reduce acyl chlorides (RCOCl) to aldehydes. The catalyst used in this reaction is palladium on barium sulfate (Pd – BaSO\(_4\)). The presence of barium sulfate acts as a poison to the catalyst, which prevents over-reduction to the corresponding alcohol, allowing the selective formation of the aldehyde.

Step 1: Palladium (Pd) is the catalyst that facilitates the reduction of acyl chlorides.
Step 2: Barium sulfate (BaSO\(_4\)) is used to poison the palladium and prevent over-reduction. Quick Tip: Rosenmund reduction requires Pd – BaSO\(_4\) as the catalyst to selectively reduce acyl chlorides to aldehydes.

Manganese (Mn) has the highest number of oxidation states among the given elements of the 3d series. Manganese can exhibit oxidation states from \( +2 \) to \( +7 \) in its compounds, giving it a total of 7 possible oxidation states. These oxidation states include \( +2, +3, +4, +5, +6, +7 \), and sometimes \( 0 \) (in its elemental form).

- Scandium (Sc) shows a maximum oxidation state of \( +3 \),

- Chromium (Cr) shows a maximum oxidation state of \( +6 \),

- Titanium (Ti) shows a maximum oxidation state of \( +4 \).


Therefore, Manganese has the most oxidation states among the 3d transition elements in this group. Quick Tip: Manganese exhibits the maximum number of oxidation states in the 3d transition series, ranging from \( +2 \) to \( +7 \).

The reaction described is a Formose reaction, where two molecules of formaldehyde (HCHO) react with concentrated KOH under heat to form methanol (CH\(_3\)OH) and potassium formate (HCOOK). The formose reaction typically involves the condensation of formaldehyde in the presence of an alkaline catalyst, producing methanol and potassium formate.

Step 1: Two molecules of formaldehyde undergo condensation.
Step 2: Methanol and potassium formate are the expected products. Quick Tip: The Formose reaction with formaldehyde and KOH produces methanol and potassium formate.

S\(_N\)1 reactions are favored by the formation of a stable carbocation. Among the given options, (CH\(_3\))\(_3\)C–Br (tert-butyl bromide) forms a highly stable tertiary carbocation, making it the most reactive towards S\(_N\)1 reactions. The tertiary carbocation is more stable than the secondary or primary carbocations formed from the other alkyl halides.

Step 1: Stability of carbocations is crucial in S\(_N\)1 reactions.
Step 2: Tertiary carbocations are the most stable, thus (CH\(_3\))\(_3\)C–Br undergoes the reaction faster. Quick Tip: S\(_N\)1 reactions are favored by tertiary carbocations due to their stability.

The fraction of molecules with energy equal to or greater than the activation energy is given by the Arrhenius equation: \[ f = e^{-\frac{E_a}{RT}} \] Quick Tip: The Arrhenius equation describes the fraction of molecules that have enough energy to overcome the activation energy barrier for a reaction to occur.

The primary amine contains one alkyl group attached to the nitrogen. Among the options, \( C_2H_5NH_2 \) (ethanamine) is the primary amine because it has one ethyl group attached to the nitrogen. Quick Tip: A primary amine has one alkyl group attached to the nitrogen atom, whereas secondary and tertiary amines have two and three alkyl groups, respectively.

The general electronic configuration of d-block elements is \( (n - 1) d^{1-10}ns^{1-2} \), where \( n \) represents the principal quantum number, and the electrons fill the \( (n-1) \) d-orbitals and the \( n \) s-orbitals. Quick Tip: D-block elements fill the \( (n-1) \) d-orbitals, and their electronic configuration follows the pattern \( (n - 1) d^{1-10}ns^{1-2} \).

- The oxidation of primary alcohols to aldehydes is done using PCC Tribromophenol.

- Butan-2-one is reduced to Butan-2-ol using \(NaBH_4\).

- Bromination of Phenol to 2, 4, 6-tribromophenol is done using Bromine water propene.

- Dehydration of propan-2-ol leads to propene when treated with 85% phosphoric acid at 440 K


Step 1: Match each reaction with its respective reagent. Quick Tip: For oxidation reactions, NaBH\(_4\) is commonly used, and for reduction reactions, phosphoric acid is often the reagent.

For a zero-order reaction, the integrated rate law is given by:
\[ [R] = [R]_0 - kt \]

Where:

- \([R]\) is the concentration of the reactant at time \(t\),

- \([R]_0\) is the initial concentration of the reactant,

- \(k\) is the rate constant,

- \(t\) is the time.


The graph of \([R]\) vs. \(t\) is a straight line with a slope of \(-k\) and an intercept of \([R]_0\).


Step 1: The slope of the line is the rate constant \(-k\).

Step 2: The intercept is the initial concentration of the reactant \([R]_0\). Quick Tip: For a zero-order reaction, the slope of the plot of [R] vs. time is –k, and the intercept is the initial concentration [R]\(_0\).

The reaction of a Grignard reagent (RMgX) with a ketone results in the formation of a tertiary alcohol. The mechanism involves the nucleophilic attack of the Grignard reagent on the carbonyl carbon of the ketone, followed by hydrolysis to yield a tertiary alcohol. This is because the Grignard reagent adds a second alkyl group to the carbonyl carbon, resulting in a tertiary alcohol.

Step 1: Grignard reagent attacks the carbonyl carbon of the ketone to form a tetrahedral intermediate.

Step 2: Hydrolysis of the intermediate yields a tertiary alcohol. Quick Tip: Grignard reagents add to the carbonyl carbon of ketones to form tertiary alcohols after hydrolysis.

The Assertion (A) is true because phenols are stronger acids than alcohols due to the ability of the phenoxide ion to stabilize the negative charge through resonance, whereas the alkoxide ion does not have such stabilization.
The Reason (R) is false because the phenoxide ion is actually more stable than the alkoxide ion due to resonance stabilization in phenoxide. Quick Tip: Phenols are stronger acids than alcohols because their conjugate base (phenoxide ion) is stabilized by resonance, whereas alkoxide ions lack this stabilization.

- Assertion (A) is correct because, in a Daniell cell, when the external opposing potential exceeds 1.1 V, the electron flow is reversed, causing electrons to flow from Cu to Zn.

- Reason (R) is false because a galvanic cell operates without an external opposing potential. A Daniell cell is a type of galvanic cell, but the external potential in this case is artificially introduced, which is not a typical condition for a galvanic cell.


Step 1: In a Daniell cell, electron flow is from the anode (Zn) to the cathode (Cu), but if an external voltage is applied, it can reverse the flow of electrons.

Step 2: A galvanic cell generates electrical energy spontaneously, but external voltages disrupt the normal flow.
Quick Tip: In a Daniell cell, if the external potential is greater than the cell potential, the direction of electron flow is reversed.

- Assertion (A) is correct because benzoic acid does not undergo Friedel – Crafts reactions due to the deactivating effect of the carboxyl group.

- Reason (R) is also true. The carboxyl group (-COOH) is electron-withdrawing, which deactivates the aromatic ring, making it less reactive towards electrophilic substitution reactions, including Friedel – Crafts reactions.


Step 1: The carboxyl group in benzoic acid deactivates the aromatic ring and prevents the reaction with electrophiles.

Step 2: The catalyst, AlCl\(_3\), is unable to bond with the deactivated ring for the Friedel – Crafts reaction to occur.
Quick Tip: Electron-withdrawing groups such as carboxyl groups deactivate the aromatic ring, preventing Friedel – Crafts reactions.

- Assertion (A) is true because fructose is a reducing sugar. It contains a carbonyl group in the form of a ketone, which can reduce Fehling's solution and Tollen's reagent under certain conditions.

- Reason (R) is false because fructose, being a reducing sugar, can reduce Tollen’s reagent and Fehling’s solution. The statement that fructose does not reduce these reagents is incorrect.


Step 1: Fructose, though a ketose, has reducing properties due to its ability to undergo isomerization into an aldose form.

Step 2: Fructose is a reducing sugar and can reduce Fehling’s solution and Tollen’s reagent.
Quick Tip: Fructose is a reducing sugar because it can isomerize to an aldose form, which is capable of reducing Tollen's and Fehling's solutions.

(a) Half-life period (\( t_{1/2} \)):

The half-life period is the time required for the concentration of a reactant to decrease to half of its initial value in a chemical reaction. For first-order reactions, the half-life period is constant and independent of the initial concentration. It is given by the formula: \[ t_{1/2} = \frac{\ln 2}{k} \]
where \( k \) is the rate constant of the reaction. For reactions of other orders, the half-life varies with concentration.


(b) Effective collisions:

Effective collisions are collisions between reacting particles that result in the formation of products. For a collision to be effective, two conditions must be met:

1. The molecules must collide with proper orientation, so that the reacting atoms can rearrange.

2. The molecules must collide with enough energy to overcome the activation energy barrier of the reaction. This energy is called the activation energy (Eₐ).

Only a small fraction of all collisions result in effective collisions, contributing to the rate of the reaction. Quick Tip: The concept of effective collisions is crucial in understanding the rate of a reaction. Increasing temperature or concentration often increases the number of effective collisions, thus speeding up the reaction.

The freezing point depression (\( \Delta T_f \)) is given by the formula: \[ \Delta T_f = T_f^{pure solvent} - T_f^{solution} \]
where \( T_f^{pure solvent} \) is the freezing point of the pure solvent (water) and \( T_f^{solution} \) is the freezing point of the solution.

The freezing point of pure water is 273.15 K.
Thus, \[ \Delta T_f = 273.15 \, K - 270.67 \, K = 2.48 \, K \]

The freezing point depression is also related to the molality (\( m \)) of the solution: \[ \Delta T_f = K_f \times m \]
where \( K_f \) is the cryoscopic constant (given as 1.86 K kg mol\(^{-1}\)) and \( m \) is the molality of the solution.

Rearranging for molality: \[ m = \frac{\Delta T_f}{K_f} = \frac{2.48 \, K}{1.86 \, K kg mol^{-1}} = 1.33 \, mol kg^{-1} \]

Molality is defined as the number of moles of solute per kilogram of solvent. Using the molality, we can calculate the number of moles of solute:

\(m = \frac{moles of solute}{kg of solvent} \quad \Rightarrow \quad moles\ of\ solute = m \times kg \ of \ solvent = 1.33 \, mol \ kg^-1 \times 0.250 \, kg = 0.3325 \, mol\)

Now, the molar mass of the solute is given by: \[ Molar mass = \frac{mass of solute}{moles of solute} = \frac{60 \, g}{0.3325 \, mol} = 180 \, g mol^{-1} \] Quick Tip: When solving problems involving freezing point depression, remember to use the relationship \( \Delta T_f = K_f \times m \), where molality (\( m \)) is the number of moles of solute per kilogram of solvent.

The first compound, \( CH_3C \, Br \) (isopropyl bromide), will react more rapidly in an \( S_N1 \) reaction compared to \( CH_3CH_2Br \) (ethyl bromide). The key factor that influences this is the stability of the carbocation intermediate that is formed during the reaction.

In an \( S_N1 \) reaction, the reaction proceeds in two steps:

1. Formation of the carbocation: The leaving group (Br-) departs, forming a positively charged carbocation.
2. Nucleophilic attack: The nucleophile attacks the carbocation, leading to the formation of the product.

The rate-determining step in the \( S_N1 \) reaction is the formation of the carbocation. The more stable the carbocation, the faster the reaction proceeds. This is because a more stable carbocation is less likely to undergo side reactions and can form more easily.

- Isopropyl bromide (\( CH_3C \, Br \)) forms a tertiary carbocation when the bromine leaves. Tertiary carbocations are highly stabilized due to two effects:
- Inductive effect: The electron-donating nature of the two methyl groups (attached to the central carbon) helps stabilize the positive charge.

- Hyperconjugation: The overlapping of the bonds of the adjacent C-H or C-C groups with the empty p-orbital of the carbocation also provides stabilization.

- Ethyl bromide (\( CH_3CH_2Br \)) forms a secondary carbocation when the bromine leaves. Secondary carbocations are less stable than tertiary carbocations because there is less electron donation from neighboring groups to stabilize the positive charge. The secondary carbocation does not benefit as much from inductive and hyperconjugation effects as the tertiary carbocation does.

Thus, the tertiary carbocation formed from isopropyl bromide is more stable than the secondary carbocation formed from ethyl bromide, making the \( S_N1 \) reaction faster in the case of isopropyl bromide. Quick Tip: In \( S_N1 \) reactions, the rate of reaction increases with the stability of the carbocation. Tertiary carbocations are the most stable, followed by secondary, and primary carbocations are the least stable.

Boiling points are influenced by the type of intermolecular forces present in the compound. The primary factors affecting boiling points are:

1. Molecular size: Larger molecules generally have higher boiling points due to increased Van der Waals forces.

2. Polarity: More polar molecules tend to have higher boiling points due to stronger dipole-dipole interactions.

3. Halogen substituent: The presence of more halogen atoms increases the boiling point due to the increase in molecular weight and the associated Van der Waals forces.

Let's examine the compounds:

- Chloromethane \((CH_3Cl)\) has the lowest boiling point because chlorine is smaller and less electronegative than bromine, leading to weaker Van der Waals forces and dipole-dipole interactions.

- Bromomethane \((CH_3Br)\) has a higher boiling point than chloromethane because bromine is larger, which increases the molecular size and leads to stronger Van der Waals forces.
- Dibromomethane \((CH_2Br_2)\) has an even higher boiling point due to the presence of two bromine atoms, which further increase the molecular size and intermolecular forces.
- Bromoform \((CHBr_3)\) has the highest boiling point because it contains three bromine atoms, resulting in the largest molecular size and the strongest intermolecular forces.

Thus, the increasing order of boiling points is: \[ Chloromethane < Bromomethane < Dibromomethane < Bromoform \] Quick Tip: In general, as the size of the halogen atom increases in a molecule, the boiling point increases due to stronger Van der Waals forces. Additionally, the more halogen atoms present, the higher the boiling point.

Nucleophilic addition reactions occur when a nucleophile attacks the electrophilic carbonyl carbon in aldehydes or ketones. The general mechanism is as follows:

Step 1: The nucleophile (e.g., a hydride ion, alkoxide, or Grignard reagent) attacks the carbonyl carbon of the aldehyde or ketone, breaking the double bond and forming a tetrahedral intermediate.

Step 2: The intermediate is protonated (usually by a water molecule or proton donor) to form the final addition product. For example, if the nucleophile is a hydride (H\(^-\)), the product will be an alcohol.

Step 3: The product after protonation is a carbonyl addition product, such as an alcohol.


Quick Tip: In nucleophilic addition reactions, the nucleophile adds to the electrophilic carbonyl carbon, forming a tetrahedral intermediate that is protonated to yield the product.

(i) Toluene to Benzoic acid:

Toluene (C\(_6\)H\(_5\)CH\(_3\)) can be oxidized to benzoic acid (C\(_6\)H\(_5\)COOH) using strong oxidizing agents such as potassium permanganate (KMnO\(_4\)) or chromic acid (H\(_2\)CrO\(_4\)). These reagents will oxidize the methyl group (-CH\(_3\)) into a carboxyl group (-COOH), converting toluene into benzoic acid.






(ii) Ethanol to 3-Hydroxybutanal:

Ethanol (CH\(_3\)CH\(_2\)OH) can be oxidized to acetaldehyde (CH\(_3\)CHO) by using an oxidizing agent like PCC (Pyridinium chlorochromate) or mild oxidizers. The acetaldehyde then undergoes a nucleophilic addition with a Grignard reagent (like CH\(_3\)MgBr) to form the intermediate 3-hydroxybutanal.


Quick Tip: Toluene can be oxidized using strong oxidizing agents to form benzoic acid, while ethanol can be oxidized to acetaldehyde, which then reacts to form 3-hydroxybutanal.

When glucose reacts with HI, the reaction produces an alkyl iodide. The hydroxyl group (-OH) of glucose is replaced by iodine. The reaction mechanism involves the cleavage of the -OH group and its substitution with iodine.
\[ C_6H_{12}O_6 \xrightarrow{HI} C_6H_{12}I + H_2O \]

Thus, glucose undergoes substitution with iodine, forming an alkyl iodide. Quick Tip: HI reacts with glucose by replacing the hydroxyl group (-OH) with iodine, forming an alkyl iodide.

One key structural difference between DNA and RNA lies in the sugar component of their nucleotides:

- DNA (Deoxyribonucleic acid) contains the sugar deoxyribose, which has a hydrogen atom (-H) attached to the 2' carbon atom of the sugar ring. This absence of an -OH group at the 2' position is where the name "deoxyribo-" comes from, signifying "lacking oxygen."

- RNA (Ribonucleic acid), on the other hand, contains the sugar ribose, which has a hydroxyl group (-OH) attached to the 2' carbon atom of the sugar ring.

This structural difference, where DNA contains deoxyribose and RNA contains ribose, is significant because it affects the stability and function of these molecules. The presence of the -OH group in RNA makes it more reactive and less stable than DNA, which is crucial for RNA's role in protein synthesis and other cellular processes where it is required to be more transient. Quick Tip: The presence of the -OH group in ribose (RNA) makes RNA more prone to hydrolysis, leading to its relatively short lifespan compared to DNA, which is more stable due to the lack of the -OH group at the 2' carbon.

The complex \([Pt(NH_3)_2Cl_2]^{2+}\) consists of platinum (Pt) as the central metal ion, two ammonia (\(NH_3\)) ligands, and two chloride (\(Cl^-\)) ligands. This is a square planar complex (since Pt\(^{2+}\) is a d\(^8\) ion and typically forms square planar complexes). The two chloride ligands can be arranged in different ways, leading to two possible geometrical isomers:

1. Cis-isomer: In this isomer, the two chloride ions are adjacent to each other (i.e., positioned at 90° to each other), which results in a square planar arrangement where both chloride ions are next to each other. The two ammonia ligands will be on opposite sides of the complex.

\[ Cis- [Pt(NH_3)_2Cl_2]^{2+} \]
In this form, the chloride ions are placed next to each other in a cis position.

2. Trans-isomer: In this isomer, the two chloride ions are positioned opposite each other (i.e., 180° apart). The ammonia ligands will be adjacent to the chloride ions, forming a linear arrangement.

\[ Trans- [Pt(NH_3)_2Cl_2]^{2+} \]
In this form, the chloride ions are placed on opposite sides in a trans position.

These two isomers arise because the ligands are free to rotate about the central metal ion in square planar complexes, which allows them to adopt different spatial arrangements.

Quick Tip: In coordination complexes, cis and trans isomerism occurs due to different spatial arrangements of ligands around the central metal ion.

The electronic configuration of a \(d^5\) ion is influenced by the splitting of the \(d\)-orbitals in an octahedral field. In an octahedral crystal field, the five degenerate \(d\)-orbitals split into two sets:

- The lower-energy \(t_{2g}\) orbitals (which are of lower energy due to the ligand field splitting).

- The higher-energy \(e_g\) orbitals (which are of higher energy due to the repulsion from the ligands).

The splitting energy between the two sets of orbitals is represented by \(\Delta_0\), and the pairing energy of electrons is denoted by \(P\). If \(\Delta_0 < P\) (i.e., the splitting energy is less than the pairing energy), the complex will adopt a high-spin configuration, meaning that the electrons will occupy all the orbitals before any pairing occurs.

For a \(d^5\) configuration with \(\Delta_0 < P\), the electronic configuration is:
\[ t_{2g}^3 e_g^2 \]

In this configuration:

- The three electrons will fill the lower-energy \(t_{2g}\) orbitals (3 electrons in 3 orbitals).

- The two remaining electrons will fill the higher-energy \(e_g\) orbitals (2 electrons in 2 orbitals).

This configuration corresponds to the high-spin state, which is typical when \(\Delta_0\) is smaller than the pairing energy. Quick Tip: When \(\Delta_0 > P\), electrons prefer to pair in the lower-energy \(t_{2g}\) orbitals, leading to low-spin configurations.

An ambidentate ligand is a ligand that can bind to a metal center through two different donor atoms. The ligand has more than one donor atom, and depending on the reaction conditions, the ligand can attach to the metal through one donor atom or another. The donor atoms are typically different atoms with lone pairs of electrons that can coordinate with the metal ion.

For example:
- Thiocyanate (\(SCN^-\)) is an ambidentate ligand because it can coordinate through either the sulfur (S) atom or the nitrogen (N) atom.

- When it binds through sulfur, it forms a thiol (S) complex.

- When it binds through nitrogen, it forms a nitrile (N) complex.

Similarly, the cyanate ion (\(OCN^-\)) can also act as an ambidentate ligand, coordinating through either the oxygen or the nitrogen atom.

The ability of ambidentate ligands to bind via different donor atoms can lead to different isomeric forms of the complex, and this ability plays a crucial role in the chemistry of coordination compounds. Quick Tip: Ambidentate ligands can lead to structural isomerism depending on which donor atom is used to bind to the metal. This feature is important in the study of coordination chemistry and the formation of isomers.

(a) Hydroboration – Oxidation reaction:

The hydroboration-oxidation of an alkene results in an alcohol. In this reaction, alkene (CH\(_3\)CH=CH\(_2\)) undergoes hydroboration with diborane (B\(_2\)H\(_6\)), followed by oxidation with hydrogen peroxide to form an alcohol. The reaction is as follows: \[ CH_3CH= CH_2 + B_2H_6 \xrightarrow{H_2O_2} CH_3CH_2OH \]
Here, the alkene is converted to the alcohol with anti-Markovnikov addition.


(b) Williamson Synthesis:

The Williamson synthesis involves the reaction of an alkoxide ion with an alkyl halide to form an ether. The general reaction is: \[ R-X + R'O^- \rightarrow R-R' + NaX \]
Where \( R \)-X is the alkyl halide, \( R' \)-O\(^-\) is the alkoxide ion, and \( R \)-R' is the ether product.


(c) Friedel-Crafts Alkylation of Anisole:

In Friedel-Crafts alkylation, anisole (C\(_6\)H\(_5\)OCH\(_3\)) reacts with an alkyl halide (CH\(_3\)Cl) in the presence of AlCl\(_3\) catalyst to form an alkylated product. The reaction is as follows:





Here, the methyl group is added to the aromatic ring, forming methyl anisole.


(d) Reimer-Tiemann Reaction:

The Reimer-Tiemann reaction involves the reaction of phenol (C\(_6\)H\(_5\)OH) with chloroform (CHCl\(_3\)) in the presence of aqueous NaOH, followed by acidification to form salicylaldehyde (C\(_7\)H\(_6\)O\(_2\)). The reaction is as follows:






Here, the formyl group (-CHO) is introduced to the phenol to form salicylaldehyde.
Quick Tip: These reactions are classic organic transformations. Hydroboration-oxidation adds alcohol across alkenes, Williamson synthesis forms ethers, Friedel-Crafts alkylation alkylates aromatic rings, and Reimer-Tiemann introduces a formyl group to phenol.

(i) Phenol and Benzoic acid:

- To distinguish phenol and benzoic acid, add neutral FeCl\(_3\).

- Phenol: When neutral FeCl\(_3\) is added to phenol, a violet color complex is formed due to the coordination of the Fe\(^{3+}\) ion with the oxygen of the phenol's hydroxyl group.

- Benzoic acid: Benzoic acid, which is a carboxylic acid, does not form a complex with FeCl\(_3\), hence no color change is observed.

- Alternatively, you can add bromine water to each compound:

- Phenol: Phenol reacts with bromine water to form 2,4,6-tribromophenol, a white precipitate.

- Benzoic acid: Benzoic acid does not react with bromine water.


(ii) Propanal and Propanone:

- Tollens' reagent: This reagent is a mild oxidizer used to distinguish aldehydes from ketones. Propanal, being an aldehyde, will reduce Tollens' reagent to silver metal, which forms a silver mirror.

- Propanone: Propanone, a ketone, does not react with Tollens' reagent. Hence, no silver mirror will form.

- Fehling’s solution: This is another test to distinguish between aldehydes and ketones.

- Propanal: It will reduce Fehling’s solution, which contains Cu\(^{2+}\) ions, to a red precipitate of Cu\(_2\)O.

- Propanone: Being a ketone, it will not reduce Fehling’s solution, and no precipitate will form.
Quick Tip: For distinguishing phenol and benzoic acid, FeCl\(_3\) is used to form a violet color with phenol. For distinguishing aldehydes from ketones, Tollens' and Fehling's reagents are useful.

The compound CH\(_3\)CHFCH\(_2\)COOH is the stronger acid. This is because:

- The fluorine atom in CH\(_3\)CHFCH\(_2\)COOH is more electronegative than hydrogen and pulls electron density away from the carboxyl group (-COOH) through the inductive effect. This results in the stabilization of the conjugate base (carboxylate anion, RCOO\(^-\)), making the release of a proton (H\(^+\)) easier, thus increasing the acidity.

- In contrast, CH\(_2\)FCH\(_2\)CH\(_2\)COOH has the fluorine atom further from the carboxyl group, resulting in a weaker electron-withdrawing effect. Therefore, the carboxyl group is less stabilized, making CH\(_2\)FCH\(_2\)CH\(_2\)COOH a weaker acid than CH\(_3\)CHFCH\(_2\)COOH.
Quick Tip: Fluorine is a highly electronegative element. The closer it is to the carboxyl group, the stronger the acid due to the electron-withdrawing inductive effect.

The integrated rate law for a first-order reaction is: \[ \ln \left( \frac{[R_0]}{[R]} \right) = kt \]
Where:

- \( [R_0] \) is the initial concentration,

- \( [R] \) is the concentration at time \( t \),

- \( k \) is the rate constant,

- \( t \) is the time.


Step 1: For 99.9% completion, \( [R] = 0.1 \times [R_0] \), so: \[ \ln \left( \frac{[R_0]}{0.1 [R_0]} \right) = kt_{99.9%} \] \[ \ln 10 = kt_{99.9%} \] \[ t_{99.9%} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} \]
Step 2: For half-life \( t_{1/2} \), \( [R] = 0.5 [R_0] \), so: \[ \ln \left( \frac{[R_0]}{0.5 [R_0]} \right) = kt_{1/2} \] \[ \ln 2 = kt_{1/2} \] \[ t_{1/2} = \frac{2.303}{k} \log 2 = \frac{2.303}{k} \times 0.3010 = \frac{2.303}{k} \times 0.3010 \]

Step 3: Now, dividing the two expressions: \[ \frac{t_{99.9%}}{t_{1/2}} = \frac{\frac{2.303}{k}}{\frac{2.303 \times 0.3010}{k}} = \frac{1}{0.3010} = 10 \]

Final Answer: The time required for 99.9% completion is 10 times the half-life of the reaction.
Quick Tip: For first-order reactions, the time for 99.9% completion is approximately 10 times the half-life (t\(_{1/2}\)).

(a) Glycosidic linkage:
A glycosidic linkage is a covalent bond formed between two monosaccharide units. This bond forms through a dehydration reaction, in which a water molecule is removed when a hydroxyl group from one monosaccharide reacts with the anomeric carbon (the carbon that was part of the carbonyl group in the monosaccharide) of another monosaccharide. The bond formed between the two sugars is typically between the 1st carbon of one sugar and the hydroxyl group attached to another carbon in the second sugar. For example, in disaccharides like sucrose, glucose and fructose are joined by a glycosidic bond.


(b) Primary structure of protein:
The primary structure of a protein refers to the unique sequence of amino acids that are linked by peptide bonds to form a polypeptide chain. This sequence is encoded by the genetic information in DNA and determines the specific characteristics of the protein. The primary structure is essential because it dictates how the protein will fold into its higher-order structures (secondary, tertiary, and quaternary structures), ultimately determining its function. A single change in the amino acid sequence can result in a dysfunctional protein, as seen in various genetic disorders like sickle cell anemia.


(c) Disaccharides:
Disaccharides are carbohydrates composed of two monosaccharide units that are bonded together by a glycosidic linkage. Common disaccharides include sucrose (table sugar), which is composed of glucose and fructose, and lactose (milk sugar), which consists of glucose and galactose. Disaccharides are important in energy storage and transport. Upon hydrolysis, disaccharides break down into their component monosaccharides, which can then be used by the body for energy. Quick Tip: The primary structure of a protein is the sequence of amino acids in the polypeptide chain. This sequence directly influences how the protein will fold and function.

The given compound is a simple organic molecule with a chlorine atom attached to a carbon chain. The molecule contains a double bond between the first and second carbon atoms, and a chlorine atom attached to the third carbon.

- The molecule contains four carbon atoms, so it is based on the butane chain \((C_4)\).

- The double bond between the first and second carbon atoms means the name should include "ene" to indicate the presence of a double bond.

- The chlorine atom is attached to the third carbon, so the prefix "4-chloro" indicates the position of the chlorine.

Thus, the IUPAC name of the compound is 4-chlorobut-1-ene.

Thionyl chloride (SOCl₂) is commonly used for converting alcohols to alkyl chlorides because it offers several advantages:

1. High efficiency: Thionyl chloride reacts with alcohols to form alkyl chlorides in high yield, with minimal side reactions.

2. Volatile by-products: The by-products of the reaction, sulfur dioxide (SO₂) and hydrogen chloride (HCl), are gaseous, making them easy to remove from the reaction mixture. This reduces the need for further purification steps.

3. Mild reaction conditions: The reaction can be performed under relatively mild conditions, often requiring only a solvent (such as pyridine) and moderate temperatures. This makes it more convenient than some other chlorinating reagents that may require harsher conditions.

Thionyl chloride is particularly preferred for sensitive alcohols or when a clean conversion to alkyl chloride is desired.

When methyl bromide \((CH_3Br)\) reacts with potassium cyanide (KCN), a nucleophilic substitution reaction occurs. The cyanide ion \((CN^-)\) from KCN acts as a nucleophile and attacks the electrophilic carbon atom in methyl bromide, which is attached to the bromine atom. This leads to the displacement of the bromine atom, resulting in the formation of methyl cyanide \((CH_3CN)\), also known as acetonitrile.

The reaction proceeds via the \(S_N2\) mechanism, where the cyanide ion directly displaces the bromine atom in a single step, producing the product without any intermediate formation of carbocations.

The overall reaction is: \[ CH_3Br + KCN \rightarrow CH_3CN + KBr \] Quick Tip: When methyl bromide reacts with KCN, it undergoes an \(S_N2\) substitution to form methyl cyanide \((CH_3CN)\), which is a useful organic compound in synthesis.

We are given the following data:

- The cell consists of two half-reactions:

1. \( Sn^{2+} + 2e^- \rightarrow Sn \) with \( E^\circ = -0.14 \, V \)

2. \( 2H^+ + 2e^- \rightarrow H_2 \) with \( E^\circ = 0.00 \, V \)

- Concentration of Sn\(^{2+}\) = 0.001 M

- Concentration of H\(^+\) = 0.01 M

- Standard pressure for H\(_2\) = 1 bar


Step 1: Write the Nernst equation for the cell:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q \]
Where:

- \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \)

- \( n \) is the number of electrons transferred (in this case, \( n = 2 \)).

- \( Q \) is the reaction quotient, which is calculated as:
\[ Q = \frac{[Sn^{2+}]}{[H^+]^2} \]
Substitute the given values:
\[ Q = \frac{0.001}{(0.01)^2} = \frac{0.001}{0.0001} = 10 \]

Step 2: Calculate the standard cell potential \( E^\circ_{cell} \):
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00 \, V - (-0.14 \, V) = 0.14 \, V \]

Step 3: Apply the values to the Nernst equation:
\[ E_{cell} = 0.14 \, V - \frac{0.0592}{2} \log 10 \]
Since \( \log 10 = 1 \), we get:
\[ E_{cell} = 0.14 \, V - \frac{0.0592}{2} \times 1 \] \[ E_{cell} = 0.14 \, V - 0.0296 \, V = 0.1104 \, V \]

Final Answer: The emf of the cell at 25°C is \( E_{cell} = 0.1104 \, V \).
Quick Tip: Use the Nernst equation to calculate the emf of electrochemical cells, considering concentrations and standard electrode potentials.

Function of a salt bridge in a galvanic cell:

The salt bridge allows the flow of ions between the two half-cells, which maintains electrical neutrality. Without the salt bridge, the flow of electrons in the external circuit would stop as the charges would accumulate in the solutions, thus preventing the redox reactions from continuing. It completes the circuit and ensures the flow of charge, allowing the electrochemical reactions to continue.
Quick Tip: A salt bridge maintains electrical neutrality in the galvanic cell by allowing ions to flow between the two half-cells.

A galvanic cell behaves like an electrolytic cell when an external potential greater than the emf (\( E_{cell} \)) is applied to the system. This causes the reverse of the natural spontaneous reaction, i.e., a non-spontaneous reaction occurs. In this situation, the cell starts to consume electrical energy rather than produce it. The external voltage must exceed the cell's electromotive force (\( E_{ext} > E_{cell} \)) for the reaction to reverse.
Quick Tip: When \( E_{ext} > E_{cell} \), a galvanic cell behaves like an electrolytic cell, and the electrochemical reaction is driven in reverse.

To determine if copper sulphate solution can be stored in a pot made of zinc, we need to calculate the \( E^\circ_{cell} \) for the cell. The \( E^\circ_{cell} \) is given by: \[ E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Zn^{2+}/Zn} \]
Substituting the given values: \[ E^\circ_{cell} = 0.34 \, V - (-0.76 \, V) = 1.10 \, V \]
Since \( E^\circ_{cell} \) is positive (\(+1.10 \, V\)), the reaction will proceed spontaneously, with zinc being oxidized and copper being reduced. This means that copper sulphate solution cannot be stored in a pot made of zinc because zinc will react with the copper ions in the solution, causing zinc to dissolve and copper to deposit on the surface.
Quick Tip: A positive \( E^\circ_{cell} \) indicates that the reaction is spontaneous, so zinc will be oxidized in the presence of copper ions, making copper sulphate unsuitable for storage in a zinc pot.

To calculate the charge required, we use Faraday's first law of electrolysis, which states that one mole of electrons corresponds to 1 Faraday (F) of charge, and the amount of charge required for a reaction depends on the number of moles of electrons involved in the reaction.

(i) 1 mol of \( MnO_4^- \) to \( Mn^{2+} \):

The reaction for the reduction of \( MnO_4^- \) to \( Mn^{2+} \) involves 5 electrons, so the charge required is: \[ Charge = 5 \times F = 5F \]

(ii) 1 mol of \( H_2O \) to \( O_2 \):

The reaction for the oxidation of \( H_2O \) to \( O_2 \) involves 2 electrons, so the charge required is: \[ Charge = 2 \times F = 2F \] Quick Tip: Faraday's law relates the charge passed through a solution to the amount of substance decomposed or deposited. The number of electrons involved in a reaction determines the total charge required.

(a) Paramagnetic or Diamagnetic Nature of \([CoF_6]^{3-}\):

To predict whether \([CoF_6]^{3-}\) is paramagnetic or diamagnetic, let's look at the electronic configuration of \(Co^{3+}\). The electronic configuration of \(Co\) is \([Ar]\, 3d^7\, 4s^2\). For \(Co^{3+}\), the electrons are removed from the 4s and 3d orbitals, leaving the configuration \(3d^6\). In the case of \([CoF_6]^{3-}\), fluorine is a weak field ligand and does not cause pairing of the electrons. Therefore, the 6 electrons in the \(3d\) orbitals remain unpaired, making \([CoF_6]^{3-}\) paramagnetic.


Step 1: Determine the electron configuration of \(Co^{3+}\)
\[ Co^{3+} : 3d^6 \]
Step 2: The presence of unpaired electrons in the \(3d\) orbitals means that the complex is paramagnetic.
Quick Tip: A complex is paramagnetic if it contains unpaired electrons in its metal's d-orbitals. If all electrons are paired, it is diamagnetic.

The coordination number of a metal in a coordination complex is determined by the number of ligand atoms directly bonded to the metal ion. In \([Co(en)_2Cl_2]^+\), ethylenediamine (en) is a bidentate ligand, meaning each en ligand forms two bonds with the metal. Chloride (Cl) is a monodentate ligand, forming one bond with the metal. Therefore, the total coordination number is: \[ Coordination number = 2 \times 2 \, (from two en ligands) + 2 \, (from two Cl ligands) = 6 \] Quick Tip: The coordination number is determined by counting the number of bonds formed between the metal and the ligands. Bidentate ligands contribute two bonds.

In this complex, platinum (Pt) is the central metal ion, and it is surrounded by two ammonia (NH\(_3\)) ligands and two chloride (Cl\(^-\)) ligands. According to IUPAC nomenclature:

- Ammonia is a neutral ligand and is named as "ammine."

- Chloride is an anionic ligand and is named as "chloro."

- The metal ion (platinum) is named first, followed by the ligands in alphabetical order.


The IUPAC name of the complex is: \[ Diamminedichloroplatinum(IV) \, ion \]

Step 1: Identify the ligands and their names (ammine and chloro).

Step 2: Name the metal and its oxidation state (Pt(IV)).

Step 3: Put the ligands and metal together following IUPAC naming rules.
Quick Tip: In IUPAC naming, the ligands are named alphabetically before the metal, and the oxidation state of the metal is indicated in parentheses.

The complex \([Co(NH_3)_6]^{3+}\) involves the coordination of six ammine (NH\(_3\)) ligands to a \(Co^{3+}\) ion. For \(Co^{3+}\) (with electron configuration \(3d^6\)), the hybridization occurs using the inner \(3d\), \(4s\), and \(4p\) orbitals, as \(NH_3\) is a weak field ligand and does not cause significant pairing of electrons. Therefore, the complex uses the inner \(d\)-orbitals for hybridization and is considered an inner orbital complex.

Step 1: Consider the electron configuration of \(Co^{3+}\) and the ligands involved.

Step 2: Since \(NH_3\) is a weak field ligand, \(Co^{3+}\) does not undergo pairing, and the metal uses its inner orbitals for hybridization.
Quick Tip: Inner orbital complexes form when the metal ion uses lower energy \(d\)-orbitals for bonding. Outer orbital complexes use higher energy \(d\)-orbitals.

For \([Ni(NH_3)_6]^{2+}\), the coordination number is 6, and the ligand is ammine (NH\(_3\)), which is a weak field ligand. Thus, the metal ion, \(Ni^{2+}\), will undergo octahedral hybridization using \(3d\), \(4s\), and \(4p\) orbitals. Since the coordination number is 6, the complex must have an octahedral shape.



Step 1: The \(Ni^{2+}\) ion has the electron configuration \(3d^8 4s^2\), and upon losing two electrons, it forms \(Ni^{2+}\) with \(3d^8\).

Step 2: The \(Ni^{2+}\) ion uses its \(3d\), \(4s\), and \(4p\) orbitals to form six hybrid orbitals, leading to an octahedral shape.

Step 3: The hybridization of the \(Ni^{2+}\) ion is sp\(^3\)d\(^2\), and the complex adopts an octahedral shape.
Quick Tip: For a coordination number of 6, the complex adopts an octahedral shape with sp\(^3\)d\(^2\) hybridization.

Step 1: Identify the structure of ‘A’ (C\(_7\)H\(_7\)ON):

The molecular formula C\(_7\)H\(_7\)ON suggests the compound is a derivative of an amide. A common structure that fits this molecular formula is benzamide (C\(_6\)H\(_5\)CONH\(_2\)), which contains a benzene ring attached to a carboxamide group.

Step 2: Hoffmann Bromamide degradation reaction:

In this reaction, benzamide (A) undergoes a degradation when treated with bromine in the presence of alkali (usually NaOH) to give a primary amine. The reaction leads to the loss of one carbon atom from the amide.

- The product ‘B’ is aniline (C\(_6\)H\(_5\)NH\(_2\)), formed after the elimination of CO\(_2\) from the amide group.
\[ C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2 \quad (Aniline) \]

Step 3: Reaction of ‘B’ with nitrous acid:

When aniline (B) reacts with nitrous acid (HNO\(_2\)) at 273-278 K, it undergoes diazotization followed by deamination to form a phenol derivative.

- The product ‘C’ is phenol (C\(_6\)H\(_5\)OH).
\[ C_6H_5NH_2 + HNO_2 \rightarrow C_6H_5OH + N_2 (gas) \]

Step 4: Reaction of ‘B’ with chloroform and ethanolic potassium hydroxide:

When aniline (B) is treated with chloroform (CHCl\(_3\)) and ethanolic potassium hydroxide (KOH), it undergoes the Carbylamine reaction, which produces an isocyanide (or isothiocyanate).

- The product ‘D’ is phenyl isocyanide (C\(_6\)H\(_5\)NC).
\[ C_6H_5NH_2 + CHCl_3 + KOH \rightarrow C_6H_5NC + KCl + H_2O \]

Step 5: Reaction of ‘C’ with ethanol:

When phenol (C) reacts with ethanol, it undergoes etherification to form an ethoxy group attached to the benzene ring.

- The product ‘E’ is ethyl phenyl ether (C\(_6\)H\(_5\)OCH\(_2\)CH\(_3\)).
\[ C_6H_5OH + C_2H_5OH \xrightarrow{HCl} C_6H_5OCH_2CH_3 \] Quick Tip: Hoffmann Bromamide degradation removes the carbonyl group from amides, forming amines. Diazotization of amines with nitrous acid yields phenol derivatives. Carbylamine reaction with chloroform forms isocyanides.

Hinsberg’s reagent is benzenesulfonyl chloride (C\(_6\)H\(_5\)SO\(_2\)Cl), which is used to differentiate between primary, secondary, and tertiary amines. The reagent reacts with primary amines to form sulfonamides, with secondary amines to form N-alkyl sulfonamides, and with tertiary amines, no reaction occurs due to the lack of an available hydrogen atom for reaction. This test helps distinguish between the different classes of amines. Quick Tip: Hinsberg’s reagent is used to differentiate between primary, secondary, and tertiary amines based on their reactions with the sulfonyl chloride group.

The basic strength of amines in the gaseous phase is determined by the availability of the lone pair of electrons on nitrogen for protonation. The steric and inductive effects of the alkyl groups also influence the basicity.

1. Ethylamine (C\(_2\)H\(_5\)NH\(_2\)):
This is a simple primary amine where the lone pair on nitrogen is available for protonation.

2. Diethylamine (C\(_2\)H\(_5\))\(_2\)NH:
This is a secondary amine. The basicity of secondary amines is lower than primary amines because the nitrogen's lone pair is partially sterically hindered by the two ethyl groups, making it less available for protonation.

3. Triethylamine (C\(_2\)H\(_5\))\(_3\)N:
This is a tertiary amine, and the nitrogen atom in tertiary amines is even more sterically hindered than in secondary amines. Furthermore, the inductive effect of the three ethyl groups reduces the availability of the nitrogen lone pair, lowering the basicity further.

Hence, the order of basicity in the gaseous phase is:
\[ C_2H_5NH_2 > (C_2H_5)_2NH > (C_2H_5)_3N \]

This indicates that ethylamine is the strongest base, followed by diethylamine, and triethylamine being the weakest. Quick Tip: Aromatic groups withdraw electron density, reducing basicity, while alkyl groups donate electron density, increasing basicity.

Methylamine (CH\(_3\)NH\(_2\)) is more basic than aniline (C\(_6\)H\(_5\)NH\(_2\)) because the methyl group is an electron-donating group. It increases the electron density on the nitrogen atom, making it more available to accept a proton. In contrast, the phenyl group in aniline is electron-withdrawing through resonance, decreasing the electron density on the nitrogen atom and thus reducing its basicity.

Step 1: Methyl group donates electrons, increasing basicity.

Step 2: Phenyl group withdraws electrons, decreasing basicity. Quick Tip: Electron-donating groups (like methyl) increase basicity, while electron-withdrawing groups (like phenyl) decrease basicity.

Aniline reacts with bromine water in an electrophilic aromatic substitution reaction. The amino group (-NH\(_2\)) in aniline is an electron-donating group, which activates the benzene ring towards electrophilic substitution. The bromine molecules attack the ortho and para positions relative to the amino group, resulting in the formation of 2, 4, 6-tribromoaniline.

Step 1: Amino group donates electrons to the ring, making it more reactive.

Step 2: Bromine undergoes electrophilic substitution at the activated positions. Quick Tip: Electron-donating groups like -NH\(_2\) activate the aromatic ring towards electrophilic substitution, leading to products like 2, 4, 6-tribromoaniline.

Primary amines have higher boiling points than tertiary amines due to the ability to form hydrogen bonds. In primary amines, the nitrogen atom has a hydrogen atom attached, allowing for hydrogen bonding between molecules. Tertiary amines, on the other hand, do not have a hydrogen atom on the nitrogen, so they cannot form hydrogen bonds as efficiently, leading to lower boiling points.

Step 1: Primary amines form hydrogen bonds, increasing boiling point.

Step 2: Tertiary amines cannot form hydrogen bonds, leading to lower boiling points. Quick Tip: Hydrogen bonding increases boiling point. Primary amines have higher boiling points than tertiary amines due to the presence of hydrogen atoms on the nitrogen.

Zinc is not regarded as a transition element because it does not have an incomplete d-orbital in its ground state or any of its common oxidation states. The electronic configuration of zinc is \( [Ar]\, 3d^{10}4s^2 \), and in its most stable oxidation state of \( +2 \), the configuration becomes \( [Ar]\, 3d^{10} \), which is a fully filled d-orbital. For an element to be considered a transition metal, it must have at least one incomplete d-orbital in either its ground state or common oxidation states. Zinc, therefore, does not exhibit the typical characteristics of transition metals such as variable oxidation states and colored complexes.
Quick Tip: Transition metals typically have an incomplete d-orbital in their ground or common oxidation states, which is not the case for zinc.

Lanthanoid contraction refers to the gradual decrease in the size of the lanthanoid ions (from La\(^{3+}\) to Lu\(^{3+}\)) across the lanthanide series. Despite the increasing nuclear charge, the addition of electrons occurs mainly in the 4f orbitals, which do not effectively shield the increased positive charge. As a result, the effective nuclear charge experienced by the electrons increases, leading to a gradual contraction of the ionic radius. This contraction is particularly significant between elements such as La and Lu and affects the properties of the elements and their compounds, including the stability of their coordination complexes and their atomic size.
Quick Tip: Lanthanoid contraction occurs due to poor shielding by the 4f-electrons, causing a decrease in the size of the ions across the series.

The first ionization enthalpy of chromium is lower than that of zinc because of the electron configuration and the stability of the orbitals. Chromium has an electron configuration of \( [Ar]\, 3d^5 4s^1 \), where one electron from the 4s orbital is promoted to the 3d orbital, leading to a half-filled stable d-subshell. This arrangement provides extra stability to chromium, making it easier to lose an electron compared to zinc, which has a completely filled \( 3d^{10} 4s^2 \) configuration. The completely filled d-subshell in zinc results in higher stability, which makes it harder to ionize. Hence, chromium’s first ionization enthalpy is lower.
Quick Tip: Half-filled d-subshells, like in chromium, provide extra stability, making ionization easier compared to elements with completely filled orbitals like zinc.

Transition elements are good catalysts due to several reasons:

1. Variable oxidation states: Transition metals can easily change oxidation states, which helps in facilitating redox reactions by acting as both electron donors and acceptors.

2. Ability to form complexes: Transition metals can coordinate with various ligands, forming complexes that are essential intermediates in catalytic processes.

3. Partial filling of d-orbitals: The presence of partially filled d-orbitals allows transition metals to interact with reactant molecules, lowering activation energies.

4. Surface area: Many transition metals have high surface area, providing more active sites for reactions to occur.

These factors combined make transition metals effective in speeding up reactions without being consumed in the process.

Quick Tip: Transition metals are good catalysts due to their variable oxidation states, ability to form complexes, and partially filled d-orbitals that facilitate reaction mechanisms.

Compounds of transition metals are generally colored because of d-d transitions. In these compounds, the d-orbitals are split into two energy levels due to the ligand field. When electrons in the lower energy d-orbitals absorb visible light, they get excited to the higher energy d-orbitals. The wavelength of light absorbed corresponds to the energy difference between the split d-orbitals, and the remaining transmitted light gives the color of the compound. This phenomenon is common in transition metal complexes, particularly when the metal has an incomplete d-orbital configuration. The presence of ligands and the metal’s oxidation state also affect the color.

Quick Tip: Transition metal compounds are colored because of d-d transitions, where electrons absorb visible light and move between split d-orbitals.

KMnO\(_4\) has Mn in the +7 oxidation state, meaning that there are no unpaired electrons and it is diamagnetic.
However, in K\(_2\)MnO\(_4\), Mn is in the +6 oxidation state, and it has unpaired electrons in the d-orbital, making it paramagnetic. Quick Tip: The magnetic behavior of a compound depends on the number of unpaired electrons. If a compound has unpaired electrons, it is paramagnetic; if all electrons are paired, it is diamagnetic.

The reaction is a reduction half-reaction, where Cr(VI) is reduced to Cr(III). The complete ionic equation is: \[ Cr_2O_7^{2-} + 14 H^+ + 6e^- \longrightarrow 2 Cr^{3+} + 7 H_2O \] Quick Tip: In redox reactions, always balance the charges and atoms by adding electrons and water molecules when necessary. Here, Cr(VI) is reduced to Cr(III) with water molecules added to balance the oxygens.

Step 1: Reverse osmosis is the process in which solvent molecules pass through a semipermeable membrane from a region of low solute concentration to a region of high solute concentration when pressure greater than the osmotic pressure is applied. The solvent moves against the natural osmotic gradient.

Step 2: This process is widely used in water purification and desalination, where solvent (typically water) is purified by removing solutes like salts and other impurities. Quick Tip: In reverse osmosis, pressure is applied to push the solvent from low concentration to high concentration across a semipermeable membrane.

Step 1: Cold water has a higher solubility for gases such as oxygen. Since oxygen is crucial for respiration in aquatic species, cold water allows more oxygen to dissolve, making it more available for aquatic species.

Step 2: As temperature increases, the solubility of gases in water decreases. Therefore, warmer water contains less dissolved oxygen, making it less suitable for aquatic life. Quick Tip: Cold water can hold more dissolved oxygen, making it more suitable for aquatic species, as they rely on dissolved oxygen for respiration.

We can calculate the vapour pressure of the solution using Raoult’s Law, which states that the vapour pressure of the solvent in the solution is proportional to the mole fraction of the solvent.
\[ P = P_0 \times X_{solvent} \]

Where:
- \( P_0 = 32.8 \, mm Hg \) is the vapour pressure of pure water,

- \( X_{solvent} \) is the mole fraction of water.


Step 1: Calculate the moles of glucose: \[ Moles of glucose = \frac{2}{180} = 0.0111 \, mol \]

Step 2: Calculate the moles of water: \[ Moles of water = \frac{100}{18} = 5.56 \, mol \]

Step 3: Calculate the mole fraction of water: \[ X_{water} = \frac{moles of water}{moles of water + moles of glucose} = \frac{5.56}{5.56 + 0.0111} = 0.998 \]

Step 4: Calculate the vapour pressure of the solution: \[ P = 32.8 \times 0.998 = 32.734 \, mm Hg \]

Thus, the vapour pressure of the solution is 32.734 mm Hg. Quick Tip: Raoult's Law states that the vapour pressure of a solvent in a solution is proportional to the mole fraction of the solvent. The presence of a solute lowers the vapour pressure.

Ethanoic acid (CH\(_3\)COOH) is a weak acid and it forms dimers in benzene due to hydrogen bonding between two molecules of ethanoic acid. The formation of dimers reduces the number of particles in the solution. Since Van’t Hoff factor \(i\) represents the number of particles in solution, for ethanoic acid in benzene, \( i \) will be less than 1 due to the dimerization process.


Step 1: Ethanoic acid dissociates into dimers, reducing the total number of particles in the solution.

Step 2: Therefore, the Van’t Hoff factor \( i \) will be less than 1. Quick Tip: Van’t Hoff factor (\(i\)) is less than 1 when solute molecules form complexes or dimers in solution.

An ideal solution is one that obeys Raoult’s Law completely over the entire range of concentrations. In an ideal solution, the enthalpy of mixing is zero, and the intermolecular forces between the components (solvent-solvent, solute-solute, and solvent-solute) are similar. There is no excess enthalpy change upon mixing, and the vapor pressure of the solution is proportional to the mole fraction of each component. Real solutions that do not follow Raoult's law due to differences in intermolecular interactions are termed non-ideal solutions. Quick Tip: Ideal solutions obey Raoult’s law at all concentrations and have no enthalpy change during mixing, implying similar intermolecular forces between components.

We can use the formula for freezing point depression:
\[ \Delta T_f = i \times K_f \times m \]

Where:

- \( \Delta T_f = 2 \, K \),

- \( K_f = 1.86 \, K kg mol^{-1} \) (given),

- \( i = 3 \) (for CaCl\(_2\) as it dissociates into 3 ions: Ca\(^{2+}\) and 2 Cl\(^{-}\)),
- \( m \) is the molality of the solution.


Step 1: Rearrange the equation to solve for molality: \[ m = \frac{\Delta T_f}{i \times K_f} = \frac{2}{3 \times 1.86} = 0.358 \, mol/kg \]

Step 2: Calculate the moles of CaCl\(_2\): \[ moles of CaCl_2 = m \times mass of solvent = 0.358 \times 0.5 = 0.179 \, mol \]

Step 3: Calculate the mass of CaCl\(_2\): \[ mass of CaCl_2 = moles \times molar mass = 0.179 \times 111 = 19.89 \, g \]

Thus, the mass of CaCl\(_2\) required is 19.89 g. Quick Tip: Freezing point depression depends on the molality and the Van’t Hoff factor. For ionic compounds like CaCl\(_2\), dissociation increases the number of particles and the effect on the freezing point.