CBSE Class 12 Chemistry Question Paper 2024 (Set 2 - 56/4/2) with Answer Key

The limiting molar conductivity (\( \Lambda_m \)) of MgSO\(_4\) is given by the sum of the limiting conductivities of its ions:
\[ \Lambda_m = \lambda_{Mg^{2+}} + \lambda_{SO_4^{2-}} \]

Where:
- \( \lambda_{Mg^{2+}} = 106.0 \, S cm^2 \, mol^{-1} \) (given),

- \( \lambda_{SO_4^{2-}} = 160.0 \, S cm^2 \, mol^{-1} \) (given).


Thus, the limiting molar conductivity of MgSO\(_4\) is:
\[ \Lambda_m = 106.0 + 160.0 = 266.0 \, S cm^2 \, mol^{-1} \] Quick Tip: The limiting molar conductivity of an ionic compound is the sum of the limiting conductivities of its individual ions.

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 \( +7 \).

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.

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.

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.

In effective collisions, the colliding molecules must not only have sufficient energy to overcome the activation energy barrier (threshold energy), but they must also collide with the proper orientation so that the reacting molecules can form new bonds.

Activation energy is required to initiate a reaction.

Proper orientation ensures that the colliding molecules can align in a way that promotes bond formation. Quick Tip: For a reaction to occur, colliding molecules must have sufficient energy (threshold energy) and proper orientation.

A secondary amine has two alkyl or aryl groups attached to the nitrogen atom. Option (B), CH\(_3\)NHCH(CH\(_3\))\(_2\), is the secondary amine because the nitrogen is bonded to one methyl group (CH\(_3\)) and one ethyl group (CH\(_2\)CH\(_3\)), making it a secondary amine.

Step 1: Secondary amines have two substituents on the nitrogen atom.

Step 2: In option (B), the nitrogen is attached to two different alkyl groups, confirming it as a secondary amine. Quick Tip: Secondary amines have two substituents attached to the nitrogen atom, as seen in option (B).

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 general electronic configuration for the d-block elements (transition metals) is:
\[ (n-1) d^{1-10} ns^{1-2} \]

Where \(n\) represents the principal quantum number. The transition elements fill the \(d\)-orbitals after filling the \(s\)-orbitals of the outermost shell.


Step 1: Transition metals are characterized by the filling of the \(d\)-orbitals.

Step 2: The \(d\)-orbitals are filled after the \(ns\)-orbitals, leading to the general configuration \( (n-1) d^{1-10} ns^{1-2} \). Quick Tip: The electronic configuration of d-block elements follows the pattern \( (n-1) d^{1-10} ns^{1-2} \), where \(n\) is the principal quantum number.

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 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 Quick Tip: For oxidation reactions, NaBH\(_4\) is commonly used, and for reduction reactions, phosphoric acid is often the reagent.

The Assertion (A) is true as phenol undergoes bromination even without the presence of a Lewis acid due to the activating effect of the OH group.
The Reason (R) is also true as the OH group in phenol increases the electron density on the ring, making it more reactive toward electrophilic substitution reactions like bromination.
Thus, Reason (R) correctly explains Assertion (A). Quick Tip: Phenol is more reactive in electrophilic aromatic substitution reactions due to the electron-donating nature of the hydroxyl group (-OH).

- 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.

- 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.

(a) Molecularity of reaction:

Molecularity refers to the number of molecules or ions that must collide in order to result in a chemical reaction. It describes the number of reactant molecules involved in the elementary step of a reaction. For example, in a bimolecular reaction, two reactant molecules collide and react to form products. Molecularity can be classified as follows:

- Unimolecular reaction: Involves one molecule.

- Bimolecular reaction: Involves two molecules.

- Termolecular reaction: Involves three molecules.


(b) Complex reaction:

A complex reaction involves multiple elementary steps, and intermediates may form during the reaction. The rate of the complex reaction is governed by the rate-determining step, which is the slowest elementary step in the sequence. Complex reactions often consist of several steps that must occur in sequence.

Step 1: Molecularity is the number of molecules involved in an elementary reaction.

Step 2: Complex reactions involve multiple steps, with intermediates and a rate-determining step.
Quick Tip: Molecularity refers to the number of reactant molecules in the elementary step of a reaction. Complex reactions consist of multiple steps, and the overall rate depends on the slowest step.

The change in boiling point, \(\Delta T_b\), is given by: \[ \Delta T_b = T_b - T_{b0} = 68.04°C - 61.04°C = 7°C \]

Now, using the formula for boiling point elevation: \[ \Delta T_b = K_b \times m \]
where \(m\) is the molality of the solution. Rewriting the equation for molality: \[ m = \frac{K_b \times w_B \times 1000}{\Delta T_b \times W_A} \]
Here, \(w_B\) is the mass of the solute, and \(W_A\) is the mass of the solvent. Substituting the given values: \[ m = \frac{3.63 \times 6.3 \times 1000}{7 \times 27} \]
Simplifying: \[ m = \frac{22839}{189} = 121 \, g mol^{-1} \]
Thus, the molar mass of the compound is \(121 \, g mol^{-1}\) Quick Tip: Boiling point elevation can be used to calculate the molar mass of a solute. The formula is \(\Delta T_b = K_b \times m\), where \(m\) is the molality.

CH\(_3\)–C–Br will react more rapidly because it is a primary alkyl halide, and primary alkyl halides undergo S\(_N\)2 reactions more readily than secondary or tertiary alkyl halides. The size of the alkyl group also affects the rate of the S\(_N\)2 reaction, with smaller alkyl groups allowing for easier backside attack by the nucleophile. The bulky secondary alkyl halide in the second compound will be less reactive due to steric hindrance.

In S\(_N\)2 reactions, the rate is affected by the steric hindrance around the carbon attached to the leaving group. CH\(_3\)–C–Br is a primary alkyl halide, and primary halides undergo S\(_N\)2 reactions faster than secondary or tertiary halides due to less steric hindrance.
Quick Tip: In S\(_N\)2 reactions, primary alkyl halides are more reactive than secondary or tertiary alkyl halides due to less steric hindrance.

The increasing order of boiling points is:

CH\(_3\)CH\(_2\)Br \(<\) CH\(_3\)CH\(_2\)CH\(_2\)Br \(<\) CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)Br
This is because the boiling point increases with the size of the molecule and the presence of more carbon atoms, leading to increased London dispersion forces.

The boiling point of a compound is influenced by its molecular weight and intermolecular forces. The larger the alkyl group, the higher the boiling point due to increased van der Waals forces. Thus, the order of boiling points is based on the length of the carbon chain. Quick Tip: The boiling point of an alkyl halide increases with the length of the carbon chain due to the increased van der Waals forces between the molecules.

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.

The DNA double helix is held together by hydrogen bonds. These bonds form between the nitrogenous bases of the two complementary strands of DNA. Specifically, adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C) via hydrogen bonding. Quick Tip: The two strands of DNA are held together by hydrogen bonds between complementary base pairs (A-T and G-C).

The geometrical isomers of the complex \([Co(en)_2Cl_2]^+\) are cis and trans isomers. In the cis isomer, the two chloride ions are adjacent to each other, while in the trans isomer, the chloride ions are opposite each other. These isomers arise due to the spatial arrangement of the ligands around the central metal ion.


\[ Cis Isomer: \quad [Co(en)_2Cl_2]^+ \quad (chlorides adjacent) \] \[ Trans Isomer: \quad [Co(en)_2Cl_2]^+ \quad (chlorides opposite) \] Quick Tip: In coordination complexes, cis and trans isomerism occurs due to different spatial arrangements of ligands around the central metal ion.

For a d\(^4\) ion with \(\Delta_0 > P\), the electronic configuration would be:
\[ t_{2g}^6 e_g^0 \]

This configuration indicates that all six electrons are placed in the lower-energy \(t_{2g}\) orbitals, leaving the higher-energy \(e_g\) orbitals empty. Quick Tip: When \(\Delta_0 > P\), electrons prefer to pair in the lower-energy \(t_{2g}\) orbitals, leading to low-spin configurations.

A dentate ligand is a ligand that can form multiple bonds with a metal center through more than one donor atom. An example of a dentate ligand is ethylenediamine (en), which can form two bonds to a metal ion due to its two nitrogen donor atoms. Quick Tip: A dentate ligand can bind to a metal center through multiple donor atoms, like ethylenediamine (en) which is a bidentate ligand.

(i) Phenol gives a violet color when treated with neutral FeCl\(_3\), forming a complex, while benzoic acid does not react in the same way. The violet coloration is a characteristic test for phenol.

Test for phenol: Add neutral FeCl\(_3\) → Violet color.

Test for benzoic acid: No reaction with FeCl\(_3\).

(ii) Propanal (an aldehyde) gives a silver mirror when treated with Tollens' reagent, while propanone (a ketone) does not. This is because aldehydes are easily oxidized to carboxylic acids, whereas ketones do not readily undergo oxidation.

Test for propanal: Add Tollens' reagent → Silver mirror formation.
Test for propanone: No reaction with Tollens' reagent. Quick Tip: Phenol reacts with FeCl\(_3\) to form a violet complex, while benzoic acid does not react similarly.
Tollens' reagent gives a silver mirror with aldehydes (propanal), but not with ketones (propanone).

The compound CH\(_2\)FCH\(_2\)CH\(_2\)COOH is a stronger acid than CH\(_3\)CHFCH\(_2\)COOH. This is due to the presence of the fluorine atom in CH\(_2\)FCH\(_2\)CH\(_2\)COOH, which is highly electronegative and withdraws electron density from the carboxyl group through the inductive effect. This increases the positive character of the hydrogen in the carboxyl group, making it more likely to dissociate as a proton (H\(^+\)).

On the other hand, CH\(_3\)CHFCH\(_2\)COOH contains fluorine attached to a carbon in the alkyl chain, which still has some electron-withdrawing effect, but it is less pronounced than the effect in CH\(_2\)FCH\(_2\)CH\(_2\)COOH where the fluorine is directly attached to the carbon adjacent to the carboxyl group. Quick Tip: Fluorine is a highly electronegative atom and when placed close to a functional group like the carboxyl group, it increases the acidity of the compound by withdrawing electron density via the 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\(_{1/2}\)).

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.

(a) Non-essential amino acids are amino acids that the body can synthesize on its own and thus do not need to be obtained from the diet. These amino acids are part of the metabolic pathways that allow the body to create them as required. Examples of non-essential amino acids include alanine, aspartic acid, and glutamic acid. The body is capable of synthesizing them from other compounds in the body.

Non-essential amino acids are not needed through the diet because the body can synthesize them on its own from other molecules.


(b) Monosaccharides are the simplest form of carbohydrates. They cannot be hydrolyzed into smaller carbohydrates. They are the building blocks of more complex carbohydrates like disaccharides and polysaccharides. Common monosaccharides include glucose, fructose, and galactose. They have a general formula of \( C_nH_{2n}O_n \), where \( n \) is typically 3, 4, 5, or 6, corresponding to triose, tetrose, pentose, and hexose sugars, respectively.

Monosaccharides are simple sugars, such as glucose, that cannot be broken down into simpler carbohydrates.

(c) Anomers are a type of stereoisomer that specifically applies to cyclic forms of sugars. Anomers occur when the molecule cyclizes, and the new stereochemistry at the anomeric carbon (the carbon that was part of the carbonyl group) differs. Anomers are of two types: \(\alpha\)-anomer and \(\beta\)-anomer, depending on the orientation of the hydroxyl group at the anomeric carbon. For example, in glucose, the anomeric carbon can either have the -OH group pointing down (\(\alpha\)-anomer) or up (\(\beta\)-anomer) relative to the plane of the ring.
Anomers are isomers that differ in the configuration at the anomeric carbon in cyclic forms of sugars, leading to \(\alpha\)- and \(\beta\)-anomer types.

The given compound is 1-bromo-2-chlorobenzene. In this compound, the positions of the substituents (bromine and chlorine) are denoted by their respective numbers. The benzene ring is numbered such that the bromo group is at position 1 and the chloro group is at position 2. Therefore, the IUPAC name is 1-bromo-2-chlorobenzene.
\[ IUPAC Name: \quad 1-bromo-2-chlorobenzene \] Quick Tip: When naming substituted benzene rings, the positions of the substituents are indicated by numbers, such as 1-bromo-2-chlorobenzene.

Haloalkanes are more reactive towards nucleophilic substitution reactions compared to haloarenes because the carbon-halogen bond in haloalkanes is more polar and weaker than the carbon-halogen bond in haloarenes. In haloalkanes, the halogen (Cl, Br, or I) is attached to a saturated carbon, and the bond is easier to break. In contrast, in haloarenes, the halogen is attached to a carbon in the benzene ring, which has delocalized electrons, making the carbon-halogen bond stronger and less susceptible to nucleophilic attack. This resonance stabilization in haloarenes makes nucleophilic substitution reactions slower. Quick Tip: Haloalkanes undergo nucleophilic substitution more readily than haloarenes because of the weaker carbon-halogen bonds in haloalkanes.

When ethyl chloride (C\(_2\)H\(_5\)Cl) is treated with aqueous KOH, a nucleophilic substitution reaction occurs. The hydroxide ion (OH\(^-\)) from KOH displaces the chlorine atom, replacing it with a hydroxyl group, forming ethanol (C\(_2\)H\(_5\)OH). The reaction is:
\[ C_2H_5Cl + KOH (aq) \xrightarrow{aq} C_2H_5OH + KCl \]

This is an example of an S\(_N\)2 reaction. Quick Tip: When a haloalkane like ethyl chloride reacts with aqueous KOH, a nucleophilic substitution occurs, resulting in the formation of an alcohol.

(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.

(a) 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(II)).

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: 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.

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.

Basic strength of amines is influenced by the electron-donating or electron-withdrawing nature of their substituents. Alkyl groups are electron-donating, which increases the basicity of the amine, whereas an aromatic group (like in aniline) is electron-withdrawing and decreases the basicity.

Order of basicity in the gaseous phase:
\(C_2H_5NH_2\) \(<\) \((C_2H_5)_2NH\) \(<\) \((C_2H_5)_3N\)

- C\(_2\)H\(_5\)NH\(_2\) (Aniline) has the lowest basicity because the phenyl group withdraws electron density from the nitrogen.

- (C\(_2\)H\(_5\))\(_2\)NH (Diethylamine) has higher basicity because two ethyl groups donate electron density to the nitrogen.

- (C\(_2\)H\(_5\))\(_3\)N (Triethylamine) has the highest basicity due to the three electron-donating ethyl groups.
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.


% Solution 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.