Step 1: Understanding the Question:

The question asks for the by-product formed during the commercial preparation of phenol from cumene (isopropylbenzene). This process is known as the Cumene process or Hock process.


Step 2: Key Formula or Approach:

The Cumene process involves two main steps:

1. Oxidation of cumene with air to form cumene hydroperoxide.

2. Acid-catalyzed cleavage of cumene hydroperoxide to form phenol and a by-product.


Step 3: Detailed Explanation:

Step 2a: Formation of Cumene Hydroperoxide

Cumene (isopropylbenzene) is oxidized in the presence of air (oxygen) to form cumene hydroperoxide.



Step 2b: Cleavage of Cumene Hydroperoxide

The cumene hydroperoxide is then treated with a dilute acid (like H\(_2\)SO\(_4\)), which causes it to rearrange and cleave into phenol and acetone.



Step 4: Final Answer:

As shown in the reaction, the two products are phenol and acetone. Since phenol is the main product, acetone is the by-product. Therefore, acetone is obtained as a by-product in this process.
Quick Tip: The Cumene process is a very important industrial method for synthesizing both phenol and acetone. Remembering this single reaction provides the answer to two important manufacturing processes. The name "cumene" itself gives a hint, as the structure is related to both the final products.

Step 1: Understanding the Question:

The question asks for the set of reagents that can convert chlorobenzene (C\(_6\)H\(_5\)Cl) to toluene (C\(_6\)H\(_5\)CH\(_3\)) in a single step. This involves replacing the -Cl atom with a -CH\(_3\) group.


Step 2: Analyzing the Options:

(A) NaNO\(_2\) + HCl: These reagents are used to form nitrous acid (HNO\(_2\)), which is used in diazotization of primary aromatic amines, not for this conversion.


(B) CH\(_3\)Cl/Anhydrous AlCl\(_3\): This is the reagent for the Friedel-Crafts alkylation reaction. This reaction introduces an alkyl group onto an activated benzene ring. However, chlorobenzene is deactivated towards Friedel-Crafts reaction due to the -I effect of chlorine, and the reaction does not proceed easily. It is not considered a standard one-step conversion.


(C) CH\(_3\)Cl/Fe, Dark: These are reagents for the electrophilic halogenation of an aromatic ring, not for alkylation.


(D) CH\(_3\)Cl, Na, Dry ether: This set of reagents corresponds to the Wurtz-Fittig reaction.


Step 3: The Wurtz-Fittig Reaction:

The Wurtz-Fittig reaction is a coupling reaction between an aryl halide (like chlorobenzene) and an alkyl halide (like methyl chloride) in the presence of sodium metal and dry ether as a solvent.

The reaction proceeds as follows:
\[ \underset{Chlorobenzene}{C_6H_5Cl} + \underset{Sodium}{2Na} + \underset{Methyl chloride}{ClCH_3} \xrightarrow{Dry ether} \underset{Toluene}{C_6H_5CH_3} + 2NaCl \]

Step 4: Final Answer:

The correct set of reagents for the one-step conversion of chlorobenzene to toluene is CH\(_3\)Cl, Na, and Dry ether, which corresponds to the Wurtz-Fittig reaction.
Quick Tip: Remember the names of coupling reactions involving sodium metal:
- \textbf{Wurtz Reaction:} Alkyl halide + Alkyl halide \(\xrightarrow{Na, ether}\) Alkane.
- \textbf{Fittig Reaction:} Aryl halide + Aryl halide \(\xrightarrow{Na, ether}\) Biphenyl derivative.
- \textbf{Wurtz-Fittig Reaction:} Aryl halide + Alkyl halide \(\xrightarrow{Na, ether}\) Alkylbenzene.

Step 1: Understanding the Question:

We need to determine the coordination number of the central metal ion, Iron (Fe), in the complex ion [Fe(C\(_2\)O\(_4\))\(_3\)]\(^{3-}\).


Step 2: Key Formula or Approach:

The coordination number is defined as the total number of coordinate bonds formed by the central metal atom/ion with the ligands.

Coordination Number = (Number of ligands) \(\times\) (Denticity of each ligand)


Step 3: Detailed Explanation:

- The central metal ion is Iron (Fe).

- The ligand is C\(_2\)O\(_4^{2-}\), which is the oxalate ion.


- The oxalate ion has two oxygen atoms that can donate a lone pair of electrons to the central metal ion simultaneously. Therefore, it is a bidentate ligand (its denticity is 2).

- In the complex [Fe(C\(_2\)O\(_4\))\(_3\)]\(^{3-}\), there are three oxalate ligands attached to the iron ion.


Step 4: Calculation:

Using the formula:

Coordination Number = (Number of oxalate ligands) \(\times\) (Denticity of oxalate)

Coordination Number = 3 \(\times\) 2

Coordination Number = 6


Final Answer:

The coordination number of Fe in the complex is 6.
Quick Tip: It is crucial to know the denticity of common ligands. Always be careful with polydentate ligands like oxalate (C\(_2\)O\(_4^{2-}\)) and ethylenediamine (en), where the coordination number is not simply the number of ligand molecules.

Step 1: Understanding the Question:

The question asks to identify which of the given octahedral complexes can exhibit geometrical isomerism (cis-trans isomerism).


Step 2: Conditions for Geometrical Isomerism in Octahedral Complexes:

Geometrical isomerism arises when ligands can occupy different spatial positions around the central metal ion. For octahedral complexes, this is commonly seen in complexes of the type:

- MA\(_4\)B\(_2\)

- MA\(_3\)B\(_3\)

- M(AA)\(_2\)B\(_2\) (where AA is a symmetrical bidentate ligand)


Complexes of the type MA\(_6\) and MA\(_5\)B do not show geometrical isomerism because all possible arrangements of the ligands are identical.


Step 3: Analyzing the Options:

(A) [Co(NH\(_3\))\(_6\)]\(^{3+}\): This is of the type MA\(_6\). All six ligands are identical (ammonia). No geometrical isomerism is possible.


(B) [Co(NH\(_3\))\(_5\)Cl]\(^{2+}\): This is of the type MA\(_5\)B. The single 'B' ligand (Cl) can be placed in any of the six positions, and all resulting structures are superimposable. No geometrical isomerism is possible.


(C) [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^{+}\): This is of the type MA\(_4\)B\(_2\). The two 'B' ligands (Cl) can be arranged in two different ways:

- cis-isomer: The two Cl ligands are adjacent to each other (at a 90\(^\circ\) angle).

- trans-isomer: The two Cl ligands are opposite to each other (at a 180\(^\circ\) angle).

Since two different spatial arrangements are possible, this complex shows geometrical isomerism.


(D) [Co(NH\(_3\))\(_5\)(ONO)]\(^{2+}\): This is of the type MA\(_5\)B. The ligand ONO\(^{-}\) (nitrito-O) is a monodentate ligand. This complex does not show geometrical isomerism. However, it can show linkage isomerism with [Co(NH\(_3\))\(_5\)(NO\(_2\))]\(^{2+}\) where the ligand is nitrito-N.


Step 4: Final Answer:

The complex [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^{+}\) is the only one that can exhibit geometrical isomerism.
Quick Tip: For octahedral complexes, a quick check for geometrical isomerism is to look for at least two different types of ligands, and the formula should not be MA\(_6\) or MA\(_5\)B. The simplest types that show it are MA\(_4\)B\(_2\) and MA\(_3\)B\(_3\).

Step 1: Understanding the Question:

The question asks which of the given d-block elements does not exhibit variable oxidation states. Variable oxidation states are a characteristic property of transition elements.


Step 2: Analyzing the Electronic Configurations and Oxidation States of the Options:

(A) Scandium (Sc): Atomic number Z = 21. Electronic configuration: [Ar] 3d\(^1\) 4s\(^2\).

Scandium can lose its two 4s electrons and one 3d electron to achieve a stable noble gas configuration ([Ar]). It exclusively shows the +3 oxidation state in its compounds. It does not show variable oxidation states.


(B) Manganese (Mn): Atomic number Z = 25. Electronic configuration: [Ar] 3d\(^5\) 4s\(^2\).

Manganese has a large number of unpaired electrons and can lose a variable number of electrons from both 4s and 3d subshells. It shows a wide range of oxidation states from +2 to +7 (e.g., +2 in MnCl\(_2\), +4 in MnO\(_2\), +7 in KMnO\(_4\)).


(C) Chromium (Cr): Atomic number Z = 24. Electronic configuration: [Ar] 3d\(^5\) 4s\(^1\).

Chromium also shows variable oxidation states, with the most common being +2, +3, and +6.


(D) Copper (Cu): Atomic number Z = 29. Electronic configuration: [Ar] 3d\(^{10}\) 4s\(^1\).

Copper shows two common oxidation states: +1 (e.g., Cu\(_2\)O) and +2 (e.g., CuSO\(_4\)).


Step 3: Final Answer:

Among the given options, only Scandium (Sc) does not show variable oxidation states; it consistently shows an oxidation state of +3.
Quick Tip: Scandium and Zinc are two important exceptions to remember in the first transition series. Scandium only shows a +3 state, while Zinc only shows a +2 state. All other elements in the 3d series show variable oxidation states.

Step 1: Understanding the Question:

The question asks for a reagent to carry out the conversion of benzenediazonium chloride (C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\)) to benzene (C\(_6\)H\(_6\)). This is a reduction reaction where the diazonium group (-N\(_2^+\)Cl\(^-\)) is replaced by a hydrogen atom.


Step 2: Analyzing the Options:

(A) Cu/HCl: This reagent combination is used in the Gattermann reaction to replace the diazonium group with -Cl.


(B) H\(_2\)O: Warming an aqueous solution of benzenediazonium chloride results in the formation of phenol (C\(_6\)H\(_5\)OH).


(C) CH\(_3\)CH\(_2\)OH (Ethanol): Ethanol acts as a mild reducing agent. When benzenediazonium chloride is warmed with ethanol, the diazonium group is reduced to give benzene. Ethanol itself is oxidized to ethanal (acetaldehyde).
\[ C_6H_5N_2^+Cl^- + CH_3CH_2OH \xrightarrow{\Delta} C_6H_6 + CH_3CHO + N_2 + HCl \]
Another important reducing agent for this conversion is hypophosphorous acid (phosphinic acid, H\(_3\)PO\(_2\)) in the presence of Cu\(^+\) ions.


(D) CuCN: This reagent is used in the Sandmeyer reaction to replace the diazonium group with a cyano group (-CN).


Step 3: Final Answer:

The correct reagent for converting benzenediazonium chloride to benzene among the given options is ethanol (CH\(_3\)CH\(_2\)OH).
Quick Tip: Diazonium salts are extremely versatile intermediates. It's essential to memorize the specific reagents for replacing the diazonium group with various other groups (-H, -OH, -Cl, -Br, -I, -CN, -F, -NO\(_2\)). For replacement by -H (reduction), the two main reagents are H\(_3\)PO\(_2\) and CH\(_3\)CH\(_2\)OH.

Step 1: Understanding the Question:

The question asks to identify the gas evolved from the reaction between methylamine (CH\(_3\)NH\(_2\)) and nitrous acid (HNO\(_2\)).


Step 2: Reaction of Primary Amines with Nitrous Acid:

Methylamine is a primary aliphatic amine. Primary aliphatic amines react with nitrous acid (prepared in situ from NaNO\(_2\) and a mineral acid like HCl) to form an unstable aliphatic diazonium salt. This salt immediately decomposes, liberating nitrogen gas and forming an alcohol.


Step 3: Writing the Reaction:

First, nitrous acid is formed:
\[ NaNO_2 + HCl \rightarrow HNO_2 + NaCl \]
Then, methylamine reacts with nitrous acid:
\[ \underset{Methylamine}{CH_3NH_2} + \underset{Nitrous acid}{HNO_2} \rightarrow [CH_3N_2^+Cl^-]_{(unstable)} + H_2O \]
The unstable methyldiazonium chloride decomposes in the aqueous solution:
\[ [CH_3N_2^+Cl^-] + H_2O \rightarrow \underset{Methanol}{CH_3OH} + \underset{Nitrogen gas}{N_2(g)} + HCl \]

Step 4: Final Answer:

The gas evolved in the reaction is dinitrogen gas (N\(_2\)). This reaction is used as a quantitative test for primary amino groups, as the volume of N\(_2\) evolved can be measured.
Quick Tip: Remember the different outcomes for reactions of amines with nitrous acid:
- \textbf{Primary Aliphatic Amine:} Forms alcohol + N\(_2\) gas.
- \textbf{Primary Aromatic Amine:} Forms a stable diazonium salt (at 0-5\(^\circ\)C).
- \textbf{Secondary Amine (aliphatic or aromatic):} Forms N-nitrosamine (a yellow oily liquid).
- \textbf{Tertiary Aliphatic Amine:} Forms a soluble nitrite salt.
- \textbf{Tertiary Aromatic Amine:} Undergoes electrophilic substitution on the ring.

Step 1: Understanding the Question:

The question asks about the nature of a solution formed by mixing acetone and chloroform with respect to Raoult's law.


Step 2: Analyzing the Intermolecular Interactions:

We need to compare the intermolecular forces in the pure components with the forces in the mixture. Let A = Acetone and B = Chloroform.

- In pure acetone (A-A): The forces are dipole-dipole interactions.

- In pure chloroform (B-B): The forces are dipole-dipole interactions.

- In the mixture (A-B): A new, stronger interaction forms. The oxygen atom in acetone (CH\(_3\)COCH\(_3\)) has lone pairs and is partially negative. The hydrogen atom in chloroform (CHCl\(_3\)) is acidic (partially positive) due to the strong electron-withdrawing effect of the three chlorine atoms. This allows for the formation of an intermolecular hydrogen bond between the acetone and chloroform molecules.


Step 3: Relating Interactions to Deviation from Raoult's Law:

- The A-B interactions (hydrogen bonds) are stronger than the original A-A and B-B interactions (dipole-dipole).

- Because of these stronger attractive forces in the solution, the escaping tendency of both acetone and chloroform molecules from the solution phase to the vapour phase is reduced.

- This means the partial vapour pressure of each component in the solution will be lower than that predicted by Raoult's law for an ideal solution.

- A lower-than-expected vapour pressure corresponds to a negative deviation from Raoult's law.


Step 4: Final Answer:

A solution of acetone in chloroform shows a negative deviation from Raoult's law due to the formation of hydrogen bonds between the two components, leading to stronger intermolecular forces in the solution than in the pure components.
Quick Tip: To predict deviation from Raoult's Law, compare the strength of intermolecular forces:
- \textbf{Negative Deviation:} Solute-solvent forces are STRONGER than solute-solute and solvent-solvent forces (e.g., Acetone + Chloroform, Acid + Water). \(\Delta H_{mix}\) is negative, \(\Delta V_{mix}\) is negative. Forms maximum boiling azeotrope.
- \textbf{Positive Deviation:} Solute-solvent forces are WEAKER than solute-solute and solvent-solvent forces (e.g., Ethanol + Acetone, Ethanol + Water). \(\Delta H_{mix}\) is positive, \(\Delta V_{mix}\) is positive. Forms minimum boiling azeotrope.

Step 1: Understanding the Question:

We need to calculate the freezing point of a 1 molal aqueous solution of KCl. KCl is an electrolyte that dissociates in water.


Step 2: Key Formula or Approach:

The depression in freezing point (\(\Delta T_f\)) for an electrolyte solution is given by the formula:
\[ \Delta T_f = i \times K_f \times m \]
Where:

- \(\Delta T_f\) is the depression in freezing point.

- \(i\) is the van't Hoff factor.

- \(K_f\) is the molal freezing point depression constant (cryoscopic constant).

- \(m\) is the molality of the solution.


The freezing point of the solution (\(T_f\)) is then calculated as:
\[ T_f = T_f^\circ - \Delta T_f \]
where \(T_f^\circ\) is the freezing point of the pure solvent (water), which is 0\(^\circ\)C.


Step 3: Detailed Explanation and Calculation:

- Molality (m): Given as 1 molal.

- K\(_f\) for water: Given as 1.86 K kg mol\(^{-1}\) (which is equivalent to 1.86 \(^\circ\)C kg mol\(^{-1}\)).

- van't Hoff factor (i): The problem states that KCl is completely dissociated. The dissociation of KCl is:

\[ KCl(s) \rightarrow K^+(aq) + Cl^-(aq) \]
One formula unit of KCl produces two ions in solution. Therefore, for complete dissociation, the van't Hoff factor \(i = 2\).


Now, calculate the depression in freezing point:
\[ \Delta T_f = i \times K_f \times m \] \[ \Delta T_f = 2 \times 1.86 \, ^\circC kg mol^{-1} \times 1 \, mol kg^{-1} \] \[ \Delta T_f = 3.72 \, ^\circC \]

Finally, calculate the freezing point of the solution:
\[ T_f = T_f^\circ - \Delta T_f \] \[ T_f = 0^\circC - 3.72^\circC \] \[ T_f = -3.72^\circC \]

Step 4: Final Answer:

The freezing point of the one molal KCl solution is -3.72\(^\circ\)C.
Quick Tip: For problems involving colligative properties of electrolytes, never forget to include the van't Hoff factor (\(i\)). For strong electrolytes assuming 100% dissociation, \(i\) is simply the number of ions produced per formula unit (e.g., \(i=2\) for NaCl, KCl; \(i=3\) for CaCl\(_2\), MgSO\(_4\) gives i=2).

Step 1: Understanding the Difference between Galvanic and Electrolytic Cells:

- Galvanic (or Voltaic) Cell: A device that converts chemical energy into electrical energy through a spontaneous redox reaction. The cell potential (E\(_{cell}\)) is positive. Electrons flow from the anode to the cathode.

- Electrolytic Cell: A device that uses external electrical energy to drive a non-spontaneous redox reaction. This process is called electrolysis. The anode is positive, and the cathode is negative.


Step 2: Analyzing the Effect of an External Potential (E\(_{ext}\)):

Consider a galvanic cell with its own cell potential, E\(_{cell}\). Now, let's apply an external potential (E\(_{ext}\)) from an external source in opposition to the galvanic cell's potential.


- Case 1: E\(_{ext}\) \(<\) E\(_{cell}\): The galvanic cell continues to function normally, but the net cell potential is reduced. The reaction is still spontaneous, and current flows from the galvanic cell.

- Case 2: E\(_{ext}\) = E\(_{cell}\): The external potential perfectly balances the cell's potential. There is no net flow of current, and the cell reaction stops. This is the principle behind a potentiometer.

- Case 3: E\(_{ext}\) \(>\) E\(_{cell}\): The external potential is now stronger than the cell's own potential. It forces the electrons to flow in the opposite direction. The original spontaneous reaction is reversed, and a non-spontaneous reaction occurs. The cell now consumes electrical energy to perform a chemical reaction.


Step 3: Final Answer:

This condition, where an external voltage greater than the cell potential forces the reaction to reverse, is precisely the definition of an electrolytic cell's operation. Therefore, a galvanic cell behaves like an electrolytic cell when the opposing external potential (E\(_{ext}\)) is greater than the cell's potential (E\(_{cell}\)).
Quick Tip: Think of it as a tug-of-war. The galvanic cell wants to push electrons one way with force E\(_{cell}\). The external source pushes the other way with force E\(_{ext}\).
- If E\(_{cell}\) wins (E\(_{cell}\) > E\(_{ext}\)), it's a galvanic cell.
- If it's a tie (E\(_{cell}\) = E\(_{ext}\)), nothing happens.
- If the external source wins (E\(_{ext}\) > E\(_{cell}\)), the process is forced backward, and it becomes an electrolytic cell.

Step 1: Understanding Pseudo First Order Reactions:

A pseudo first order reaction is a bimolecular reaction that is made to behave like a first-order reaction. This occurs when one of the reactants is present in a very large excess compared to the other.


Step 2: Key Formula or Approach:

Consider a reaction: A + B \(\rightarrow\) Products

The true rate law is: Rate = k[A][B], where k is the true rate constant.

If reactant B is in large excess, its concentration [B] remains practically constant throughout the reaction.

So, we can combine the true rate constant k and the constant concentration [B] into a new constant, k', called the pseudo first order rate constant.

Rate = (k[B])[A]

Rate = k'[A]

Here, k' = k[B].


Step 3: Analyzing the Dependence of the Rate Constant:

- The true rate constant, k, depends only on temperature (as per the Arrhenius equation).

- The pseudo first order rate constant, k', depends on both the true rate constant (k) and the concentration of the reactant present in large excess ([B]).

The question asks for the "value of rate constant for a pseudo first order reaction". This is best interpreted as the pseudo rate constant (k'), which is the constant measured experimentally in such a reaction.


Step 4: Evaluating the Options:

(A) depends only on temperature: This is true for the true rate constant k, but not the pseudo constant k'.

(B) depends on the concentration of reactants present in small amount: The rate of reaction depends on this, but the pseudo rate constant k' does not.

(C) depends on the concentration of reactants present in large excess: This is correct, as k' = k[B], where B is the reactant in excess.

(D) is not dependent on the concentration of reactants: This is incorrect.


Final Answer:

The value of the pseudo first order rate constant depends on the concentration of the reactant that is present in large excess.
Quick Tip: A classic example is the hydrolysis of ethyl acetate: CH\(_3\)COOC\(_2\)H\(_5\) + H\(_2\)O \(\rightarrow\) CH\(_3\)COOH + C\(_2\)H\(_5\)OH. If water is the solvent, it is in huge excess. The rate law becomes Rate = k'[CH\(_3\)COOC\(_2\)H\(_5\)], where the pseudo rate constant k' = k[H\(_2\)O].

Step 1: Understanding the Structure of Fructose and Ring Types:

- Fructose is a ketohexose. Its open-chain structure has a ketone group at the C\(_2\) position and hydroxyl groups on the other carbons.

- A furanose ring is a five-membered heterocyclic ring containing one oxygen atom.

- A pyranose ring is a six-membered heterocyclic ring containing one oxygen atom.


Step 2: Mechanism of Ring Formation (Hemiketal Formation):

Cyclic structures of monosaccharides are formed by an intramolecular reaction between a carbonyl group (aldehyde or ketone) and a hydroxyl group within the same molecule. This forms a hemiacetal (from an aldehyde) or a hemiketal (from a ketone).


Step 3: Applying to Fructose for Furanose Ring:

- The carbonyl group in fructose is the ketone group at the C\(_2\) position.

- To form a stable five-membered furanose ring, this C\(_2\) ketone group must react with a hydroxyl group. A five-membered ring involves four carbon atoms and one oxygen atom.

- If the C\(_2\) ketone reacts with the -OH group on C\(_5\), the resulting ring will consist of atoms C\(_2\), C\(_3\), C\(_4\), C\(_5\), and the oxygen from the C\(_5\)-OH group. This is a five-membered ring.


Step 4: Final Answer:

The furanose ring structure of fructose (fructofuranose) is formed by the intramolecular reaction between the ketone group at C\(_2\) and the hydroxyl group at C\(_5\).

(Note: Fructose can also form a six-membered pyranose ring by reaction between C\(_2\) and C\(_6\)-OH, but the furanose form is specifically asked for).
Quick Tip: Remember the ring types and the atoms involved:
- \textbf{Pyranose} = 6-membered ring (think Hexagon). For aldoses like glucose, it's C1-OH and C5-OH. For ketoses like fructose, it's C2=O and C6-OH.
- \textbf{Furanose} = 5-membered ring (think Pentagon). For aldoses, it's C1-OH and C4-OH. For ketoses like fructose, it's C2=O and C5-OH.

Step 1: Analyze the Assertion (A):

Assertion (A) states that actinoids show a wide range of oxidation states. This is a well-known characteristic of actinoids. For example, uranium shows +3, +4, +5, +6 oxidation states, and plutonium shows states from +3 to +7. The assertion is true.


Step 2: Analyze the Reason (R):

Reason (R) states that this wide range of oxidation states is due to the comparable energies of 5f, 6d, and 7s orbitals. The energy difference between these orbitals in the actinoid series is very small. Because of this small energy gap, electrons from all three of these subshells can participate in chemical bonding. This allows for a variable number of electrons to be lost or shared, leading to a wide range of oxidation states. The reason is true.


Step 3: Evaluate if Reason (R) explains Assertion (A):

The ability to use electrons from multiple, closely-spaced energy orbitals (5f, 6d, 7s) is the direct cause of the variable and wide range of oxidation states observed in actinoids. Therefore, the reason correctly explains the assertion.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Quick Tip: The similarity in energy between the valence orbitals (5f, 6d, 7s for actinoids; 4f, 5d, 6s for lanthanoids) is a key concept for f-block elements. For lanthanoids, the energy gap is larger, making it harder for 4f electrons to participate, which is why they show a more limited range of oxidation states (mostly +3).

Step 1: Analyze the Assertion (A):

Assertion (A) states that the order of a reaction can be zero or fractional. The order of a reaction is the sum of the powers of the concentration terms in the experimentally determined rate law. It can be a positive integer, zero, or a fraction. For example, the decomposition of acetaldehyde has an order of 1.5. The catalytic decomposition of ammonia on a platinum surface is a zero-order reaction. So, the assertion is true.


Step 2: Analyze the Reason (R):

Reason (R) states that we cannot determine the order from the balanced chemical equation. The order of a reaction is an experimental quantity that depends on the reaction mechanism. The stoichiometric coefficients in a balanced equation represent the overall reaction but do not necessarily reflect the individual steps of the mechanism that determine the rate. Therefore, the order must be determined experimentally and cannot be simply deduced from the balanced equation (except for elementary reactions). So, the reason is also true.


Step 3: Evaluate if Reason (R) explains Assertion (A):

While both statements are correct, the reason doesn't fully explain why the order can be zero or fractional. The reason explains \textit{how the order is determined (experimentally, not from stoichiometry). The fact that the order can be zero or fractional is a consequence of the complex nature of reaction mechanisms, where the rate might depend on concentrations in complex ways, or not at all (in the case of zero order). The reason states a fact about determining the order, but it is not the fundamental explanation for the existence of zero or fractional orders. Therefore, the reason is not the correct explanation for the assertion.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Quick Tip: Always remember the distinction: \textbf{Stoichiometry tells you 'how much' reacts in the overall equation. \textbf{Order of Reaction} tells you 'how' the rate depends on the concentration of reactants, which is determined by the slowest step (rate-determining step) in the reaction mechanism.

Step 1: Analyze the Assertion (A):

The assertion compares the boiling points of two isomeric amines: diethylamine ((C\(_2\)H\(_5\))\(_2\)NH), a secondary amine, and n-butylamine (n-C\(_4\)H\(_9\)NH\(_2\)), a primary amine. Both have the same molecular formula (C\(_4\)H\(_{11}\)N) and similar molecular masses. The boiling point of n-butylamine is 78\(^\circ\)C, while that of diethylamine is 56\(^\circ\)C. Thus, the boiling point of diethylamine is indeed lower than that of n-butylamine. The assertion is true.


Step 2: Analyze the Reason (R):

The reason attributes the difference in boiling points to the extent of hydrogen bonding. The boiling point of amines is primarily determined by intermolecular hydrogen bonding.
- In primary amines (R-NH\(_2\)), like n-butylamine, there are two hydrogen atoms attached to the nitrogen, allowing for extensive intermolecular hydrogen bonding.
- In secondary amines (R\(_2\)NH), like diethylamine, there is only one hydrogen atom attached to the nitrogen, leading to less extensive hydrogen bonding compared to primary amines.
(Tertiary amines, R\(_3\)N, have no N-H bond and cannot form hydrogen bonds with each other).
Thus, hydrogen bonding is indeed more extensive in the primary amine n-C\(_4\)H\(_9\)NH\(_2\). The reason is true.


Step 3: Evaluate if Reason (R) explains Assertion (A):

The extent of intermolecular hydrogen bonding is the primary factor that determines the boiling points of isomeric amines. Since n-butylamine can form more extensive hydrogen bonds, more energy is required to separate its molecules, resulting in a higher boiling point. The reason correctly and directly explains the assertion.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Quick Tip: For isomeric amines, the order of boiling points is always: \textbf{Primary (1\(^\circ\)) > Secondary (2\(^\circ\)) > Tertiary (3\(^\circ\))}. This is a direct consequence of the decreasing ability to form intermolecular hydrogen bonds as the number of hydrogen atoms on the nitrogen decreases.

Step 1: Analyze the Assertion (A):

Assertion (A) states that phenol is less acidic than alcohol (e.g., ethanol). Acidity is the tendency to donate a proton. Phenols are significantly more acidic than alcohols. For example, the pK\(_a\) of phenol is about 10, while the pK\(_a\) of ethanol is about 16. A lower pK\(_a\) value indicates a stronger acid. Therefore, phenol is much more acidic than alcohol. The assertion is false.


Step 2: Analyze the Reason (R):

Reason (R) states that the phenoxide ion is more stable than the alkoxide ion. This is the key to understanding the difference in acidity.
- When a phenol loses a proton, it forms a phenoxide ion (C\(_6\)H\(_5\)O\(^-\)). In this ion, the negative charge on the oxygen is delocalized over the entire benzene ring through resonance. This delocalization stabilizes the ion.
- When an alcohol (R-OH) loses a proton, it forms an alkoxide ion (R-O\(^-\)). In this ion, the negative charge is localized on the oxygen atom. There is no resonance stabilization. In fact, the alkyl group (R) has a +I (electron-donating) effect, which further destabilizes the anion by intensifying the negative charge.
Thus, the phenoxide ion is much more stable than the alkoxide ion. The reason is true.


Step 3: Final Answer:

Since Assertion (A) is false and Reason (R) is true, the correct option is (D).
Quick Tip: When comparing the acidity of organic compounds, always look at the stability of the conjugate base formed after removing a proton. A more stable conjugate base means a stronger parent acid. Resonance is a very powerful stabilizing factor for conjugate bases.

Part 1: Raoult's Law for Volatile Components

Raoult's law states that for an ideal solution containing two or more volatile liquids, the partial vapour pressure of each component at a given temperature is equal to the product of the vapour pressure of that component in its pure state and its mole fraction in the solution.


Mathematically, for a component 'A' in the solution:
\[ P_A = P_A^\circ \times x_A \]
Where:

- \(P_A\) is the partial vapour pressure of component A over the solution.

- \(P_A^\circ\) is the vapour pressure of pure component A.

- \(x_A\) is the mole fraction of component A in the solution.


According to Dalton's law of partial pressures, the total vapour pressure (P\(_{total}\)) above the solution is the sum of the partial pressures of all components:
\[ P_{total} = P_A + P_B + ... = (P_A^\circ x_A) + (P_B^\circ x_B) + ... \]

Part 2: Vapour Pressure of Glucose Solution

This phenomenon is known as the "lowering of vapour pressure," which is a colligative property. The reason for this is as follows:


1. Nature of Solute: Glucose is a non-volatile solute. This means it has a negligible tendency to escape into the vapour phase. Water is a volatile solvent.

2. Surface Phenomenon: Evaporation is a surface phenomenon. In pure water, the entire surface is occupied by volatile water molecules, which can escape into the vapour phase, creating a certain vapour pressure.

3. Effect of Adding Solute: When glucose is dissolved in water, the non-volatile glucose molecules occupy some of the sites on the surface of the solution. This reduces the fraction of the surface area that is available for the volatile water molecules.

4. Reduced Escaping Tendency: With a smaller surface area available for them, fewer water molecules can escape from the liquid phase to the vapour phase per unit time. This results in a lower equilibrium vapour pressure for the solution compared to that of the pure solvent (water).
Quick Tip: A simple way to remember the reason for vapour pressure lowering is: "Non-volatile solutes act like obstacles on the surface, making it harder for the solvent molecules to escape." This applies to all solutions with non-volatile solutes.

(i) [CoCl\(_2\)(en)\(_2\)]SO\(_4\)


Step 1: Identify Cation and Anion. The complex ion [CoCl\(_2\)(en)\(_2\)]\(^+\) is the cation and SO\(_4^{2-}\) is the anion. We name the cation first.


Step 2: Name the Ligands. There are two types of ligands: 'Cl' (chloro or chlorido) and 'en' (ethane-1,2-diamine). Ligands are named in alphabetical order.
- Chloro comes before ethane-1,2-diamine.
- There are two chloro ligands, so we use the prefix 'di-'. \(\rightarrow\) dichlorido
- There are two 'en' ligands. Since the ligand name already contains a numerical prefix ('di' in diamine), we use the prefixes bis, tris, tetrakis. So, we use 'bis-'. \(\rightarrow\) bis(ethane-1,2-diamine)


Step 3: Name the Central Metal. The central metal is Cobalt (Co). Since the complex is a cation, the name remains cobalt.


Step 4: Determine the Oxidation State. Let the oxidation state of Co be x. Cl has a charge of -1, 'en' is neutral (0), and the sulfate ion (SO\(_4\)) has a charge of -2.
x + 2(-1) + 2(0) = +2 (to balance the -2 of sulfate)
x - 2 = +2 \(\implies\) x = +4. Wait, there is a mistake in balancing. Let's recheck. The complex ion must balance the SO4(2-) charge, so the charge on the complex ion is +1.
Let's check the oxidation state based on the formula: [CoCl\(_2\)(en)\(_2\)]SO\(_4\). The sulfate is SO\(_4^{2-}\). So the complex ion must be [CoCl\(_2\)(en)\(_2\)]\(^{+}\).
x + 2(-1) + 2(0) = +1
x - 2 = +1 \(\implies\) x = +3.

The oxidation state is +3, written as (III).

Step 5: Assemble the Name. Cation: Dichloridobis(ethane-1,2-diamine)cobalt(III). Anion: sulfate.
Final Name: Dichloridobis(ethane-1,2-diamine)cobalt(III) sulfate


(ii) K\(_3\)[Fe(C\(_2\)O\(_4\))\(_3\)]


Step 1: Identify Cation and Anion. K\(^+\) is the cation and [Fe(C\(_2\)O\(_4\))\(_3\)]\(^{3-}\) is the complex anion. We name the cation first. Cation: Potassium.


Step 2: Name the Ligands. The ligand is C\(_2\)O\(_4^{2-}\) (oxalate or oxalato). There are three of them, so we use the prefix 'tri-'. \(\rightarrow\) trioxalato


Step 3: Name the Central Metal. The central metal is Iron (Fe). Since the complex is an anion, the name ends in '-ate'. So, iron becomes ferrate.


Step 4: Determine the Oxidation State. Let the oxidation state of Fe be x. Potassium (K) has a charge of +1, and oxalate has a charge of -2.
3(+1) + x + 3(-2) = 0
3 + x - 6 = 0 \(\implies\) x = +3.
The oxidation state is +3, written as (III).


Step 5: Assemble the Name. Cation: Potassium. Anion: trioxalatoferrate(III).
Final Name: Potassium trioxalatoferrate(III) Quick Tip: Key IUPAC Naming Rules: 1. Cation is named before the anion. 2. Inside the complex, name ligands alphabetically, then the metal. 3. Use prefixes di-, tri- for simple ligands and bis-, tris- for complex ligands. 4. If the complex is an anion, the metal name ends in '-ate' (e.g., ferrate, cuprate, cobaltate). 5. The oxidation state of the metal is written in Roman numerals in parentheses.

Step 1: Understanding the Question:

The question asks for the key differences between a double salt and a complex compound.


Step 2: Detailed Explanation:

The main points of differentiation are as follows:


\begin{tabular{|p{3.5cm|p{6.5cm|p{6.5cm|
\hline
Property & Double Salt & Complex Compound

\hline
Definition & Formed by the combination of two or more stable salts in stoichiometric ratio. & Contains a central metal atom/ion bonded to a number of ions or neutral molecules (ligands).

\hline
Identity in Solution & Loses its identity when dissolved in water. It dissociates completely into its constituent ions. & Retains its identity in solution. The complex ion does not dissociate into its constituent parts.

\hline
Dissociation & Undergoes complete ionic dissociation. & Undergoes partial dissociation; the coordination sphere remains intact.

\hline
Test for Ions & Gives positive tests for all of its constituent ions. & Does not give tests for all constituent ions, specifically those inside the coordination sphere.

\hline
Example & Mohr's salt: FeSO\(_{4}\).(NH\(_{4}\))\(_{2}\)SO\(_{4}\).6H\(_{2}\)O. In water, it gives tests for Fe\(^{2+}\), NH\(_{4}^{+}\), and SO\(_{4}^{2-}\) ions. & Potassium ferrocyanide: K\(_{4}\)[Fe(CN)\(_{6}\)]. In water, it gives test for K\(^{+}\) but not for Fe\(^{2+}\) or CN\(^{-}\) ions, as [Fe(CN)\(_{6}\)]\(^{4-}\) remains as a single unit.

\hline
Bonding & The bonding is ionic between the constituent ions. & The bonding between the central metal and ligands is coordinate covalent.

\hline
\end{tabular
Quick Tip: The simplest way to distinguish them is the "water test":
If you dissolve the substance in water and can detect every single ion from the original salts, it's a double salt.
If some ions are "hidden" inside a complex and cannot be detected, it's a complex compound.

Step 1: Understanding the Core Concepts

The question asks to distinguish between two types of ligands based on their binding behavior with a central metal ion. The key difference lies in how many donor atoms are used at once and the consequences of that binding mode.


Step 2: Detailed Differentiation

Here is a point-by-point comparison between didentate and ambidentate ligands:


\begin{tabular{|l|l|l|
\hline
Feature & Didentate Ligand & Ambidentate Ligand

\hline
Definition & A ligand that can bind to the central & A monodentate ligand that can bind to

& metal ion through two donor atoms & the central metal ion through two

& simultaneously, forming a chelate ring. & different donor atoms, but only one at a time.

\hline
Denticity & Denticity is 2. & Denticity is 1 (it is monodentate).

\hline
Bonding & Forms two coordinate bonds at the & Forms only one coordinate bond at a time,

& same time. & using either of its available donor sites.

\hline
Isomerism & Can be involved in optical isomerism. & Gives rise to linkage isomerism.

\hline
Example & Ethane-1,2-diamine (en): & Thiocyanate ion (SCN\(^-\)):

& H\(_2\)N-CH\(_2\)-CH\(_2\)-NH\(_2\). & Can bind through sulfur (thiocyanato) or

& Binds through both Nitrogen atoms. & nitrogen (isothiocyanato).

& Oxalate ion (C\(_2\)O\(_4^{2-}\)). & Nitrite ion (NO\(_2^-\)): Can bind through N or O.

\hline
\end{tabular
Quick Tip: To remember the difference, focus on the prefixes:
- \textbf{Di-} in "didentate" means two. It uses \textbf{two} binding sites at once.
- \textbf{Ambi-} in "ambidentate" relates to "ambiguous" or "ambidextrous". It has \textbf{two choices} but can only use \textbf{one} at a time, just like an ambidextrous person chooses to write with either their left or right hand.

Step 1: Understanding the Question:

The question asks to predict the order of a reaction based on a given graph. The graph plots log([R]\(_{0}\)/[R]) on the y-axis against Time on the x-axis, and the result is a straight line passing through the origin.


Step 2: Key Formula or Approach:

We need to compare the given graphical representation with the integrated rate laws for different reaction orders.

- Zero Order: \([R] = -kt + [R]_{0}\). A plot of \([R]\) vs. time is a straight line.

- First Order: \(\ln[R] = -kt + \ln[R]_{0}\). This can be rearranged as \(\ln([R]_{0}/[R]) = kt\). Converting to base-10 logarithm: \(2.303 \log([R]_{0}/[R]) = kt\), which gives \(\log([R]_{0}/[R]) = (k/2.303)t\). A plot of \(\log([R]_{0}/[R])\) vs. time is a straight line.

- Second Order: \(1/[R] = kt + 1/[R]_{0}\). A plot of \(1/[R]\) vs. time is a straight line.


Step 3: Detailed Explanation:

The given graph plots \(\log([R]_{0}/[R])\) versus time (\(t\)).

The plot is a straight line that passes through the origin.

This matches the integrated rate law for a first-order reaction in the form:
\[ \log\left(\frac{[R]_{0}}{[R]}\right) = \left(\frac{k}{2.303}\right)t \]

This equation is in the form of a straight line, \(y = mx\), where:

- \(y = \log([R]_{0}/[R])\)

- \(x = t\) (time)

- \(m = k/2.303\) (the slope)

Since the graph of \(\log([R]_{0}/[R])\) vs. \(t\) is a straight line, the reaction must be of the first order.


Step 4: Final Answer:

The linear relationship shown in the graph is characteristic of a first-order reaction. Therefore, the order of the reaction is 1.
Quick Tip: Memorize the linear plots for different reaction orders:
- \textbf{Zero Order}: [A] vs. t (slope = -k)
- \textbf{First Order}: ln[A] vs. t (slope = -k) OR log([A]\(_{0}\)/[A]) vs. t (slope = k/2.303)
- \textbf{Second Order}: 1/[A] vs. t (slope = k)

Step 1: Understanding the Question:

The question asks for the value of the slope of the given curve, which is a plot of log([R]\(_{0}\)/[R]) versus time for a chemical reaction.


Step 2: Key Formula or Approach:

From the previous part, we identified the reaction as first-order. The integrated rate law for a first-order reaction is:
\[ k = \frac{2.303}{t} \log\left(\frac{[R]_{0}}{[R]}\right) \]

We need to rearrange this equation into the form of a straight line, \(y = mx + c\), to identify the slope.


Step 3: Detailed Explanation:

Rearranging the first-order integrated rate law:
\[ \log\left(\frac{[R]_{0}}{[R]}\right) = \frac{k}{2.303} \times t \]

This equation relates the variables plotted on the graph:

- The y-axis variable is \(y = \log([R]_{0}/[R])\).

- The x-axis variable is \(x = t\).

Comparing this equation to the standard equation of a straight line passing through the origin, \(y = mx\):
\[ \underbrace{\log\left(\frac{[R]_{0}}{[R]}\right)}_{y} = \underbrace{\left(\frac{k}{2.303}\right)}_{m} \underbrace{t}_{x} \]

We can clearly see that the slope of the line, \(m\), is equal to \(k/2.303\).


Step 4: Final Answer:

The slope of the curve is equal to \(k/2.303\), where \(k\) is the rate constant of the first-order reaction.
Quick Tip: Be careful with the logarithm base. If the plot is \(\ln([R]_{0}/[R])\) vs. \(t\), the slope is simply \(k\). If the plot is \(\log([R]_{0}/[R])\) vs. \(t\) (using base-10 log), the slope is \(k/2.303\). Always check the y-axis label carefully in such graphical questions.

Step 1: Understanding the Question:

The question asks for a chemical test or reaction that proves the presence of an aldehyde (-CHO) functional group in the structure of glucose.


Step 2: Key Formula or Approach:

Aldehydes are easily oxidized to carboxylic acids by mild oxidizing agents, whereas ketones are resistant to oxidation under mild conditions. We can use a reaction that is specific to aldehydes.


Step 3: Detailed Explanation:

The presence of an aldehydic group in glucose can be explained by the following chemical evidence:

Reaction with Bromine Water:

When glucose is treated with a mild oxidizing agent like bromine water (Br\(_{2}\) in H\(_{2}\)O), the aldehydic group (-CHO) is oxidized to a carboxylic acid group (-COOH), while the other hydroxyl groups remain unaffected. The product formed is gluconic acid, which is a six-carbon carboxylic acid.





This reaction is a characteristic test for aldehydes. Since glucose gives a positive result, it confirms the presence of an aldehydic functional group.

Other confirmatory tests include:

- Tollens' Test: Glucose reduces Tollens' reagent ([Ag(NH\(_{3}\))\(_{2}\)]\(^{+}\)) to metallic silver, forming a silver mirror.

- Fehling's Test: Glucose reduces Fehling's solution (Cu\(^{2+}\)) to give a red precipitate of copper(I) oxide (Cu\(_{2}\)O).


Step 4: Final Answer:

A key piece of evidence for the presence of an aldehydic group in glucose is its reaction with bromine water. Glucose gets oxidized to gluconic acid, confirming the presence of a -CHO group.
Quick Tip: For "presence of" questions in biomolecules, you need to recall specific chemical tests.
- Aldehyde in glucose \(\rightarrow\) Bromine water oxidation.
- Five -OH groups in glucose \(\rightarrow\) Acetylation with acetic anhydride.
- Carbonyl group (general) \(\rightarrow\) Reaction with HCN or NH\(_{2}\)OH.

Step 1: Understanding the Question:

The question asks for a chemical reaction that proves the presence of five hydroxyl (-OH) functional groups in the structure of glucose.


Step 2: Key Formula or Approach:

Alcohols (compounds with -OH groups) react with acetic anhydride in a process called acetylation to form esters. By quantifying the amount of acetic anhydride consumed or analyzing the product, we can determine the number of -OH groups present.


Step 3: Detailed Explanation:

The presence of five hydroxyl groups in glucose is confirmed by its acetylation reaction.

When glucose is treated with acetic anhydride ((CH\(_{3}\)CO)\(_{2}\)O) in the presence of a catalyst like pyridine or zinc chloride, it undergoes acylation. All five of the hydroxyl groups react to form a penta-ester derivative called glucose pentaacetate.






The molecular formula of the product is C\(_{6}\)H\(_{7}\)O(OCOCH\(_{3}\))\(_{5}\). The fact that a stable pentaacetate derivative is formed confirms two things:

1. There are exactly five hydroxyl groups in one molecule of glucose.

2. All five -OH groups are on different carbon atoms, because if two were on the same carbon (a gem-diol), the molecule would be unstable and readily lose a molecule of water.


Step 4: Final Answer:

The formation of glucose pentaacetate upon reaction with excess acetic anhydride provides conclusive evidence for the presence of five hydroxyl groups in the glucose molecule.
Quick Tip: Acetylation is the standard test for confirming the number of alcoholic hydroxyl groups in a molecule. The number of acetyl groups incorporated into the product directly corresponds to the number of -OH groups present in the starting material.

Step 1: Understanding the Question:

We need to find the major product when p-hydroxybenzyl alcohol reacts with thionyl chloride (SOCl\(_2\)).


Step 2: Identifying the Functional Groups and Reagent:

- The reactant has two types of hydroxyl groups: an alcoholic -OH group (attached to the -CH\(_2\)- group) and a phenolic -OH group (directly attached to the benzene ring).

- The reagent SOCl\(_2\) (thionyl chloride) is a standard reagent used to convert primary and secondary alcohols into alkyl chlorides (Darzens process).


Step 3: Comparing the Reactivity of the -OH Groups:

- Alcoholic -OH groups react readily with SOCl\(_2\) to be replaced by a -Cl atom.

- Phenolic -OH groups do not react with SOCl\(_2\) under these conditions. The C-O bond in phenols has partial double bond character due to resonance, making it much stronger and harder to break than the C-O single bond in alcohols.


Step 4: Final Answer:

Therefore, only the alcoholic -OH group of the -CH\(_2\)OH side chain will be substituted by chlorine, while the phenolic -OH group remains unchanged. The major product is p-hydroxybenzyl chloride.
Quick Tip: Remember the reactivity difference: SOCl\(_2\) and PCl\(_5\) readily convert alcoholic -OH to -Cl, but they do not affect the phenolic -OH group. This selectivity is a key concept in organic synthesis.

Step 1: Understanding the Question:

We need to find the major product of the reaction between allylbenzene and HBr in the presence of a peroxide.


Step 2: Identifying the Reaction Type and Rule:

- The reaction is the addition of HBr across the double bond of the allyl group (-CH\(_2\)-CH=CH\(_2\)).

- The presence of peroxide indicates that the reaction proceeds via a free-radical mechanism.

- The addition of HBr in the presence of peroxide follows the Anti-Markovnikov's rule. This is also known as the peroxide effect or Kharasch effect.


Step 3: Applying the Anti-Markovnikov's Rule:

- Anti-Markovnikov's rule states that in the addition of HBr to an unsymmetrical alkene, the negative part of the addendum (Br atom) gets attached to the carbon atom of the double bond that has the greater number of hydrogen atoms.

- In allylbenzene (C\(_6\)H\(_5\)-CH\(_2\)-CH=CH\(_2\)), the double bond is between a CH carbon (one H) and a CH\(_2\) carbon (two H's).

- According to the rule, the Br atom will add to the CH\(_2\) carbon (the one with more hydrogens), and the H atom will add to the CH carbon.


Step 4: Final Answer:

The product formed is 3-bromo-1-phenylpropane.
\[ C_6H_5-CH_2-CH=CH_2 + HBr \xrightarrow{Peroxide} C_6H_5-CH_2-CH_2-CH_2Br \] Quick Tip: Remember the addition rules for HBr:
- \textbf{HBr alone:} Markovnikov's rule (electrophilic addition). Br adds to the more substituted carbon.
- \textbf{HBr + Peroxide:} Anti-Markovnikov's rule (free-radical addition). Br adds to the less substituted carbon.
This peroxide effect is specific to HBr and does not work for HCl or HI.

(a) p-nitroaniline from aniline

Direct nitration of aniline is not suitable as it leads to oxidation and formation of meta-product. Therefore, the amino group is first protected by acetylation.


Step 1: Acetylation


Aniline reacts with acetic anhydride to form acetanilide.





Step 2: Nitration

Acetanilide is nitrated with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\). The acetamido group is ortho, para-directing. The para isomer is the major product.





Step 3: Hydrolysis

The p-nitroacetanilide is hydrolyzed with acid or base to give p-nitroaniline.





(b) Chlorobenzene from aniline

This conversion is done via the Sandmeyer reaction.

Step 1: Diazotization

Aniline is treated with nitrous acid (NaNO\(_2\) + HCl) at low temperature (0-5 \(^\circ\)C or 273-278 K) to form benzenediazonium chloride.
\[ \underset{Aniline}{C_6H_5NH_2} + NaNO_2 + 2HCl \xrightarrow{273-278 K} \underset{Benzenediazonium chloride}{C_6H_5N_2^+Cl^-} + NaCl + 2H_2O \]
Step 2: Sandmeyer Reaction

The diazonium salt solution is treated with cuprous chloride (CuCl) dissolved in HCl. The diazonium group is replaced by -Cl.
\[ \underset{Benzenediazonium chloride}{C_6H_5N_2^+Cl^-} \xrightarrow{CuCl/HCl} \underset{Chlorobenzene}{C_6H_5Cl} + N_2 \]

(c) Phenol from aniline

This conversion also proceeds via the diazonium salt.

Step 1: Diazotization

Aniline is converted to benzenediazonium chloride as shown above.
\[ C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 273-278 K} C_6H_5N_2^+Cl^- \]
Step 2: Hydrolysis

The aqueous solution of benzenediazonium chloride is warmed. The diazonium group is replaced by a hydroxyl group from water.
\[ C_6H_5N_2^+Cl^- + H_2O \xrightarrow{Warm} \underset{Phenol}{C_6H_5OH} + N_2 + HCl \] Quick Tip: Benzenediazonium chloride is a key intermediate in aromatic chemistry. Mastering its preparation (diazotization) and its various substitution reactions (Sandmeyer, Gattermann, coupling, etc.) is essential for solving many conversion problems involving aniline.

Step 1: Decoding the properties of Compound (B):

1. Positive Tollens' test: This indicates that compound (B) is an aldehyde.

2. Does not give iodoform test: This means compound (B) does not have a methyl ketone (CH\(_3\)-C=O) group or a CH\(_3\)-CH(OH)- group.

3. Formation from ethanal: (B) is formed when ethanal (CH\(_3\)CHO) is treated with dilute NaOH followed by heating. This is a classic Aldol Condensation reaction.


Step 2: Identifying Compound (B) from the Aldol Condensation:

The reaction of ethanal with dilute NaOH is an aldol addition, followed by dehydration upon heating.

- Aldol Addition:

\[ 2CH_3CHO \xrightarrow{dil. NaOH} \underset{3-Hydroxybutanal}{CH_3-CH(OH)-CH_2-CHO} \]
- Dehydration (Heating): The aldol product loses a molecule of water.

\[ CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta} \underset{But-2-enal}{CH_3-CH=CH-CHO} + H_2O \]
The product is But-2-enal, commonly known as Crotonaldehyde. Let's verify this structure with the given tests. It is an aldehyde (gives Tollens' test) and does not have a methyl ketone group (no iodoform test).

Therefore, Compound (B) is But-2-enal (CH\(_3\)-CH=CH-CHO).


Step 3: Identifying Compound (A):

We are told that compound (A) has the molecular formula C\(_4\)H\(_5\)N.

Compound (A) upon reduction with DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis gives compound (B), which is an aldehyde. DIBAL-H is a specific reducing agent for the partial reduction of nitriles (-CN) to aldehydes.

Since (A) is C\(_4\)H\(_5\)N and reduces to the aldehyde (B) But-2-enal (C\(_4\)H\(_6\)O), it must be the corresponding nitrile.

The nitrile corresponding to But-2-enal (CH\(_3\)-CH=CH-CHO) is But-2-enenitrile (CH\(_3\)-CH=CH-CN).

Let's check the molecular formula for But-2-enenitrile: It has 4 carbons, (3+1+1) = 5 hydrogens, and 1 nitrogen. The formula is C\(_4\)H\(_5\)N. This matches.

Therefore, Compound (A) is But-2-enenitrile.


Step 4: Writing the Reaction of (A) with DIBAL-H:

The reaction involves the reduction of the nitrile group to an imine intermediate by DIBAL-H, which is then hydrolyzed to form the aldehyde.
\[ \underset{(A) But-2-enenitrile}{CH_3-CH=CH-C \equiv N} \xrightarrow[(ii) H_2O]{(i) DIBAL-H} \underset{(B) But-2-enal}{CH_3-CH=CH-CHO} \] Quick Tip: In organic synthesis problems, work backwards from the known product or the reaction that gives a clear identification. Here, the "ethanal + dil. NaOH" clue firmly identifies B as the product of an aldol condensation. Then, identifying the reagent (DIBAL-H) that converts A to B helps deduce the functional group and structure of A.

Part 1: Differences between S\(_N\)1 and S\(_N\)2 Reactions


% Corrected table with paragraph columns (p{width) to allow text wrapping
\begin{tabular{|l|p{4.2cm|p{4.5cm|
\hline
Feature & S\(_N\)1 Reaction & S\(_N\)2 Reaction

\hline
1. Mechanism & Two-step mechanism. Involves a carbocation intermediate. & One-step (concerted) mechanism. Involves a pentavalent transition state.

\hline
2. Kinetics & Unimolecular, first-order kinetics. Rate = k[Substrate] & Bimolecular, second-order kinetics. Rate = k[Substrate][Nucleophile]

\hline
3. Reactivity & Reactivity order: 3\(^\circ\) \(>\) 2\(^\circ\) \(>\) 1\(^\circ\) \(>\) CH\(_3\)X. Favoured by stable carbocations. & Reactivity order: CH\(_3\)X \(>\) 1\(^\circ\) \(>\) 2\(^\circ\) \(>\) 3\(^\circ\). Favoured by less sterically hindered substrates.

\hline
4. Stereochemistry & Leads to racemization (formation of both enantiomers). & Leads to complete inversion of configuration (Walden Inversion).

\hline
\end{tabular


Part 2: Faster S\(_N\)1 Reaction


Step 1: Understanding the Requirement for S\(_N\)1 Reactions:

The rate-determining step of an S\(_N\)1 reaction is the formation of a carbocation intermediate. Therefore, the rate of an S\(_N\)1 reaction is directly proportional to the stability of the carbocation formed after the leaving group departs.


Step 2: Analyzing the Given Compounds:

The two compounds given are:

1. (Chloromethyl)cyclohexane:


2. Benzyl chloride:


Step 3: Comparing the Stability of Carbocation Intermediates:

- When (Chloromethyl)cyclohexane loses Cl\(^{-}\), it forms the cyclohexylmethyl carbocation. This is a primary (1\(^\circ\)) carbocation, which is relatively unstable.


- When Benzyl chloride loses Cl\(^{-}\), it forms the benzyl carbocation. This carbocation is highly stable because the positive charge on the benzylic carbon can be delocalized over the entire benzene ring through resonance.


The resonance structures of the benzyl carbocation are:





Step 4: Conclusion:

Because the benzyl carbocation is significantly more stable than the primary cyclohexylmethyl carbocation due to resonance, Benzyl chloride will form its carbocation intermediate much more readily. This lowers the activation energy of the rate-determining step, making the S\(_N\)1 reaction much faster.


Final Answer:

Benzyl chloride will undergo the S\(_N\)1 reaction faster. This is because the intermediate benzyl carbocation formed from it is highly stabilized by resonance.
Quick Tip: For S\(_N\)1 reactivity, always think "\textbf{carbocation stability}". Tertiary, allylic, and benzylic carbocations are the most stable due to hyperconjugation and/or resonance. For S\(_N\)2 reactivity, think "\textbf{steric hindrance}". Less crowded (primary) substrates react fastest.

Step 1: Understanding the Question:

We are given the freezing point of an aqueous solution and the vapour pressure of pure water. We need to calculate the lowering in vapour pressure (\(\Delta P\)) for this solution. This requires connecting two different colligative properties: depression in freezing point and lowering of vapour pressure.


Step 2: Key Formula or Approach:

1. Use the depression in freezing point formula to find the effective molality of the solution.

\[ \Delta T_{f} = i \cdot K_{f} \cdot m \]

2. Use the molality to find the mole fraction of the solute (\(X_{solute}\)).

\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{water}} = \frac{i \cdot m}{i \cdot m + \frac{1000}{M_{water}}} \]

where \(M_{water}\) is the molar mass of water (18.015 g/mol).

3. Use Raoult's law to calculate the lowering of vapour pressure.

\[ \Delta P = P^{o} \cdot X_{solute} \]

Here, 'i' is the van't Hoff factor and 'm' is the molality. We can calculate the product (\(i \cdot m\)) directly.


Step 3: Detailed Explanation:

Given data:

- Vapour pressure of pure water, \(P^{o}\) = 24.8 mm Hg

- Freezing point of solution, \(T_{f}\) = -0.3\(^{\circ}\)C

- Freezing point of pure water, \(T_{f}^{o}\) = 0\(^{\circ}\)C

- Molal freezing point depression constant for water, \(K_{f}\) = 1.86 K kg mol\(^{-1}\)


Calculation of effective molality (\(i \cdot m\)):

First, calculate the depression in freezing point, \(\Delta T_{f}\).
\[ \Delta T_{f} = T_{f}^{o} - T_{f} = 0^{\circ}C - (-0.3^{\circ}C) = 0.3^{\circ}C = 0.3 K \]

Now, use the formula \(\Delta T_{f} = i \cdot K_{f} \cdot m\).
\[ 0.3 = i \cdot m \cdot (1.86) \] \[ i \cdot m = \frac{0.3}{1.86} \approx 0.1613 \ mol \ kg^{-1} \]

This value represents the total molality of all solute particles in the solution.


Calculation of mole fraction of solute (\(X_{solute}\)):

A molality of 0.1613 mol kg\(^{-1}\) means there are 0.1613 moles of solute particles in 1 kg (1000 g) of water.

- Moles of solute particles, \(n_{solute, total} = i \cdot m = 0.1613\) mol

- Moles of water, \(n_{water} = \frac{Mass}{Molar \ Mass} = \frac{1000 \ g}{18.015 \ g/mol} \approx 55.51\) mol

Now, calculate the total mole fraction of solute particles.
\[ X_{solute} = \frac{n_{solute, total}}{n_{solute, total} + n_{water}} = \frac{0.1613}{0.1613 + 55.51} = \frac{0.1613}{55.6713} \approx 0.002897 \]


Calculation of lowering in vapour pressure (\(\Delta P\)):

Using Raoult's Law:
\[ \Delta P = P^{o} \cdot X_{solute} \] \[ \Delta P = 24.8 \ mm \ Hg \times 0.002897 \] \[ \Delta P \approx 0.07185 \ mm \ Hg \]

Step 4: Final Answer:

The lowering in vapour pressure of the aqueous solution is approximately 0.072 mm Hg.
Quick Tip: This problem links two colligative properties. The key is to realize that all colligative properties depend on the concentration of solute particles.
You can use one property (like freezing point depression) to find the effective concentration (like \(i \cdot m\)) and then use that concentration to calculate another property (like vapour pressure lowering).

Step 1: Understanding the Question:


We need to analyze two coordination complexes of iron, [FeF\(_6\)]\(^{3-}\) and [Fe(CN)\(_6\)]\(^{4-}\), based on Valence Bond Theory (VBT) to determine their hybridization, orbital type (inner/outer), and magnetic properties.


Step 2: Detailed Explanation for [FeF\(_6\)]\(^{3-}\):


a. Oxidation State of Fe: Let the oxidation state of Fe be x.

x + 6(-1) = -3 \(\implies\) x = +3. So, we have Fe\(^{3+}\).


b. Electronic Configuration:
Fe (Z=26): [Ar] 3d\(^6\) 4s\(^2\).


Fe\(^{3+}\): [Ar] 3d\(^5\). The orbital diagram for Fe\(^{3+}\) is:


\begin{tabular{|c|c|c|c|c|
\hline \(\uparrow\) & \(\uparrow\) & \(\uparrow\) & \(\uparrow\) & \(\uparrow\)

\hline
\end{tabular (3d) \quad
\begin{tabular{|c|
\hline
\phantom{\(\uparrow\downarrow\)

\hline
\end{tabular (4s) \quad
\begin{tabular{|c|c|c|
\hline
\phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\)

\hline
\end{tabular (4p) \quad
\begin{tabular{|c|c|c|c|c|
\hline
\phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\)

\hline
\end{tabular (4d)


c. Ligand Type: F\(^{-}\) is a weak-field ligand. It does not cause the pairing of electrons in the 3d orbitals.


d. Hybridization: For the formation of six coordinate bonds with six F\(^{-}\) ligands, the Fe\(^{3+}\) ion needs six empty orbitals. Since the 3d orbitals are singly occupied and pairing does not occur, the vacant outer orbitals (one 4s, three 4p, and two 4d) are used for hybridization.


Hybridization is sp\(^3\)d\(^2\).


e. Orbital Type: Since the outer 4d orbitals are used, it is an outer orbital complex (or high-spin complex).


f. Magnetic Behaviour: It has five unpaired electrons in the 3d orbitals. Therefore, it is strongly paramagnetic.


Step 3: Detailed Explanation for [Fe(CN)\(_6\)]\(^{4-}\):


a. Oxidation State of Fe: Let the oxidation state of Fe be y.

y + 6(-1) = -4 \(\implies\) y = +2. So, we have Fe\(^{2+}\).


b. Electronic Configuration:
Fe (Z=26): [Ar] 3d\(^6\) 4s\(^2\).


Fe\(^{2+}\): [Ar] 3d\(^6\). The orbital diagram for Fe\(^{2+}\) is:

\begin{tabular{|c|c|c|c|c|
\hline \(\uparrow\downarrow\) & \(\uparrow\) & \(\uparrow\) & \(\uparrow\) & \(\uparrow\)

\hline
\end{tabular (3d) \quad
\begin{tabular{|c|
\hline
\phantom{\(\uparrow\downarrow\)

\hline
\end{tabular (4s) \quad
\begin{tabular{|c|c|c|
\hline
\phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\)

\hline
\end{tabular (4p)


c. Ligand Type: CN\(^{-}\) is a strong-field ligand. It forces the pairing of electrons in the 3d orbitals.


The 3d electrons rearrange as:


\begin{tabular{|c|c|c|c|c|
\hline \(\uparrow\downarrow\) & \(\uparrow\downarrow\) & \(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\)

\hline
\end{tabular (3d)


d. Hybridization: After pairing, two inner 3d orbitals become vacant. These two 3d orbitals, along with one 4s and three 4p orbitals, hybridize to form six equivalent orbitals for bonding with six CN\(^{-}\) ligands.


Hybridization is d\(^2\)sp\(^3\).


e. Orbital Type: Since the inner 3d orbitals are used, it is an inner orbital complex (or low-spin complex).


f. Magnetic Behaviour: After pairing, there are no unpaired electrons. Therefore, the complex is diamagnetic.
Quick Tip: The key to solving such problems is to identify the ligand type. Spectrochemical series helps: strong-field ligands (like CN\(^{-}\), CO) cause pairing (low-spin, inner orbital), while weak-field ligands (like F\(^{-}\), Cl\(^{-}\), H\(_2\)O) do not (high-spin, outer orbital). The number of unpaired electrons determines the magnetic character (paramagnetic if unpaired e\(^{-}\) are present, diamagnetic if not).

(i) What happens to the colour of complex [Ti(H\(_2\)O)\(_6\)]\(^{3+}\) when heated gradually ?


Step 1: Explanation of Colour:


In [Ti(H\(_2\)O)\(_6\)]\(^{3+}\), the oxidation state of Ti is +3. The electronic configuration of Ti\(^{3+}\) is [Ar] 3d\(^1\).


The complex is violet in colour. This colour is due to the d-d transition. The single d-electron in the lower energy t\(_{2g}\) orbital absorbs light from the visible region (yellow-green light) and gets promoted to the higher energy e\(_g\) orbital. The transmitted light appears complementary, which is violet.


Step 2: Effect of Heating:


When the complex is heated gradually, the water ligands (H\(_2\)O), which are coordinated to the central metal ion, are lost.

\[ [Ti(H_2O)_6]^{3+} \xrightarrow{Heat} Ti^{3+} + 6H_2O \]
The resulting anhydrous Ti\(^{3+}\) ion has no ligands surrounding it. Without ligands, there is no crystal field splitting of the d-orbitals.


Step 3: Final Answer:


Since there is no splitting, d-d transitions are not possible. Consequently, the substance does not absorb light from the visible region and becomes colourless.



(ii) Write the electronic configuration for d\(^5\) ion if \(\Delta_o < P\).


Step 1: Understanding the Condition:


The condition \(\Delta_o < P\) means that the crystal field splitting energy (\(\Delta_o\)) is less than the pairing energy (P). This situation occurs with weak-field ligands.


Step 2: Electron Filling:


When \(\Delta_o < P\), it is energetically more favourable for electrons to occupy the higher energy e\(_g\) orbitals than to pair up in the lower energy t\(_{2g}\) orbitals.


For a d\(^5\) configuration, the electrons will be filled according to Hund's rule of maximum multiplicity. The first three electrons will go into the t\(_{2g}\) orbitals singly. The next two electrons will go into the e\(_g\) orbitals singly, rather than pairing in t\(_{2g}\).


Step 3: Final Answer:


The electronic configuration will be t\(_{2g}^3\) e\(_g^2\). This corresponds to a high-spin complex.



(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)\(_4\)].


Step 1: Oxidation State and Configuration:


The complex is [Ni(CO)\(_4\)], which is tetracarbonylnickel(0). CO is a neutral ligand, so the oxidation state of Nickel (Ni) is 0.


The atomic number of Ni is 28. Its ground state electronic configuration is [Ar] 3d\(^8\) 4s\(^2\).


Step 2: Effect of Ligand:


CO is a very strong-field ligand. In its presence, the electrons from the 4s orbital are pushed into the 3d orbitals to pair up with the existing d-electrons.


So, the configuration of Ni in the complex becomes [Ar] 3d\(^{10}\) 4s\(^0\).


The orbital diagram for Ni in the complex is:


\begin{tabular{|c|c|c|c|c|
\hline \(\uparrow\downarrow\) & \(\uparrow\downarrow\) & \(\uparrow\downarrow\) & \(\uparrow\downarrow\) & \(\uparrow\downarrow\)

\hline
\end{tabular (3d) \quad
\begin{tabular{|c|
\hline
\phantom{\(\uparrow\downarrow\)

\hline
\end{tabular (4s) \quad
\begin{tabular{|c|c|c|
\hline
\phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\) & \phantom{\(\uparrow\downarrow\)

\hline
\end{tabular (4p)


Step 3: Hybridization and Magnetic Behaviour:


For bonding with four CO ligands, Ni uses its empty valence orbitals. The empty 4s orbital and the three empty 4p orbitals hybridize to form four sp\(^3\) hybrid orbitals. These orbitals are then used to accept electron pairs from the four CO ligands.


The hybridization is sp\(^3\), which corresponds to a tetrahedral geometry.


Since the 3d orbitals are completely filled (3d\(^{10}\)), there are no unpaired electrons. Therefore, the complex is diamagnetic.
Quick Tip: For carbonyl complexes like [Ni(CO)\(_4\)] and [Fe(CO)\(_5\)], the metal is in a zero oxidation state. CO is a strong ligand that forces all valence electrons (from both s and d subshells) to pair up in the d-orbitals, leading to diamagnetic character. The hybridization then involves the empty s and p orbitals.

Step 1: Understanding the Question:

The question asks for the reason why the pH of an aqueous solution of sodium chloride (NaCl) increases during electrolysis. An increase in pH means the solution becomes more alkaline (basic).


Step 2: Detailed Explanation:

Aqueous NaCl solution contains Na\(^+\), Cl\(^{-}\) ions from NaCl and H\(^+\) and OH\(^{-}\) ions from the dissociation of water.

During electrolysis, reactions occur at two electrodes: the cathode (negative electrode) and the anode (positive electrode).


At the Cathode (Reduction):

There are two possible reduction reactions:

1. Na\(^+\)(aq) + e\(^{-}\) \(\rightarrow\) Na(s) \quad (\(E^\circ = -2.71\) V)

2. 2H\(_2\)O(l) + 2e\(^{-}\) \(\rightarrow\) H\(_2\)(g) + 2OH\(^{-}\)(aq) \quad (\(E^\circ = -0.83\) V at pH 7)

Since the reduction potential of water is higher (less negative) than that of Na\(^+\), water will be preferentially reduced at the cathode.


At the Anode (Oxidation):

There are two possible oxidation reactions:

1. 2Cl\(^{-}\)(aq) \(\rightarrow\) Cl\(_2\)(g) + 2e\(^{-}\) \quad (\(E^\circ = -1.36\) V)

2. 2H\(_2\)O(l) \(\rightarrow\) O\(_2\)(g) + 4H\(^+\)(aq) + 4e\(^{-}\) \quad (\(E^\circ = -1.23\) V)

Due to overpotential, the oxidation of Cl\(^{-}\) is preferred over the oxidation of water.


Overall Effect:

The key reaction causing the pH change is the one at the cathode:

2H\(_2\)O(l) + 2e\(^{-}\) \(\rightarrow\) H\(_2\)(g) + 2OH\(^{-}\)(aq)

This reaction produces hydroxide ions (OH\(^{-}\)) in the solution.

The accumulation of OH\(^{-}\) ions increases the basicity of the solution.


Step 3: Final Answer:

The increase in the concentration of hydroxide ions (OH\(^{-}\)) makes the solution alkaline. According to the definition of pH (pH = -log[H\(^+\)] or pOH = -log[OH\(^{-}\)], with pH + pOH = 14), an increase in [OH\(^{-}\)] leads to a decrease in pOH and consequently an increase in pH above 7.
Quick Tip: In the electrolysis of aqueous solutions of salts, always compare the standard electrode potentials of the cation/anion with that of water to determine what gets reduced at the cathode and oxidized at the anode. Remember that H\(_2\)O reduction produces OH\(^-\) (basic) and H\(_2\)O oxidation produces H\(^+\) (acidic).

Step 1: Understanding the Question:

The question asks why a mercury cell maintains a constant voltage throughout its operational life, which is a characteristic that distinguishes it from a common dry cell (Leclanché cell).


Step 2: Detailed Explanation:

The reactions in a mercury cell are as follows:

Anode: Zinc amalgam (Zn(Hg)) is oxidized.

Zn(Hg) + 2OH\(^{-}\)(aq) \(\rightarrow\) ZnO(s) + H\(_2\)O(l) + 2e\(^{-}\)

Cathode: Mercuric oxide (HgO) is reduced.

HgO(s) + H\(_2\)O(l) + 2e\(^{-}\) \(\rightarrow\) Hg(l) + 2OH\(^{-}\)(aq)


Overall Reaction:

By adding the anode and cathode half-reactions, we get the net reaction:

Zn(Hg) + HgO(s) \(\rightarrow\) ZnO(s) + Hg(l)


Reason for Constant Potential:

1. No Change in Ion Concentration: Look at the overall cell reaction. The reactants (Zn, HgO) and products (ZnO, Hg) are all either solids or liquids. There are no ions from the electrolyte (like OH\(^{-}\)) present in the net equation.

2. Constant Activity: The concentrations (or more accurately, activities) of pure solids and liquids are considered to be constant (unity).

3. Nernst Equation: The cell potential is given by the Nernst equation: \(E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q\). The reaction quotient, Q, for this reaction would be \(Q = \frac{[ZnO][Hg]}{[Zn][HgO]}\). Since all components are in their pure solid or liquid states, their activities are constant and equal to 1. Thus, Q=1.

4. Conclusion: Because the concentrations of the species involved in the overall reaction do not change as the cell discharges, the cell potential (\(E_{cell}\)) remains constant. In contrast, in a dry cell, the concentration of ions like Zn\(^{2+}\) and NH\(_4^+\) changes, causing the voltage to drop over time.


Step 3: Final Answer:

The mercury cell provides a constant cell potential because the overall cell reaction does not involve any ions in the solution whose concentrations change during the cell's lifetime. The reactants and products are all solids or liquids with constant activities.
Quick Tip: When asked about constant cell potential, check the overall reaction. If the reaction involves only pure solids and liquids, and no ions from the electrolyte, the potential will be constant. This is because the reaction quotient Q in the Nernst equation will be constant (usually 1).

Step 1: Understanding the Question:

The question asks to explain why the conductivity (\(\kappa\)) of an electrolyte solution decreases when it is diluted (i.e., when more solvent is added).


Step 2: Detailed Explanation:

1. Definition of Conductivity: Conductivity, also known as specific conductance, is a measure of a solution's ability to conduct electricity. It is specifically defined as the conductance of 1 cubic centimeter (or 1 cubic meter) of the solution. It essentially measures the concentration of effective charge carriers.


2. Effect of Dilution: When we dilute a solution by adding more solvent (e.g., water), two things happen:

a) The total number of ions in the solution may increase (in the case of a weak electrolyte due to increased dissociation) or remain the same (in the case of a strong electrolyte which is already fully dissociated).

b) The total volume of the solution increases significantly.


3. Ions per Unit Volume: The crucial factor for conductivity is the number of ions per unit volume. Although the total number of ions might increase or stay the same, the volume increases much more substantially. As a result, the number of ions present in any given unit volume (like 1 cm\(^3\)) of the solution decreases.


4. Conclusion: Since conductivity is directly proportional to the number of current-carrying ions per unit volume, a decrease in this number upon dilution leads to a decrease in the conductivity of the solution.


Distinction from Molar Conductivity: It is important not to confuse conductivity (\(\kappa\)) with molar conductivity (\(\Lambda_m\)). Molar conductivity is the conducting power of all the ions produced by dissolving one mole of an electrolyte. Molar conductivity increases with dilution because the increased volume allows ions to move more freely with less inter-ionic attraction, and for weak electrolytes, the degree of dissociation increases.


Step 3: Final Answer:

The conductivity of a solution decreases with dilution because the number of ions that carry the current per unit volume of the solution decreases.
Quick Tip: Remember the key difference:
- \textbf{Conductivity (\(\kappa\)): Depends on ions per \textbf{unit volume}. Decreases on dilution.
- \textbf{Molar Conductivity (\(\Lambda_m\)):} Depends on ions from \textbf{one mole}. Increases on dilution.
This is a very common point of confusion in exams.

Step 1: Understanding the Question:

We are given experimental data for the reaction A + B \(\rightarrow\) product.

We need to find the order of the reaction with respect to reactants A and B, and then calculate the rate constant (k).


Step 2: Key Formula or Approach:

The rate law for the reaction can be written as:

Rate = k[A]\(^x\)[B]\(^y\)

where x is the order of reaction with respect to A, and y is the order of reaction with respect to B.

We will use the given data to find the values of x and y by comparing the rates of different experiments.


Step 3: Detailed Explanation:

To find the order with respect to A (x):

Let's compare Experiment 1 and Experiment 2. In these experiments, the concentration of B is kept constant ([B] = 0.01 mol L\(^{-1}\)).

Rate\(_1\) = k(0.01)\(^x\)(0.01)\(^y\) = \(5 \times 10^{-3}\)

Rate\(_2\) = k(0.02)\(^x\)(0.01)\(^y\) = \(1 \times 10^{-2}\)

Divide Rate\(_2\) by Rate\(_1\):
\[ \frac{Rate_2}{Rate_1} = \frac{k(0.02)^x(0.01)^y}{k(0.01)^x(0.01)^y} = \frac{1 \times 10^{-2}}{5 \times 10^{-3}} \] \[ \left(\frac{0.02}{0.01}\right)^x = \frac{10 \times 10^{-3}}{5 \times 10^{-3}} \] \[ (2)^x = 2 \] \[ x = 1 \]
So, the order of the reaction with respect to A is 1.


To find the order with respect to B (y):

Let's compare Experiment 1 and Experiment 3. In these experiments, the concentration of A is kept constant ([A] = 0.01 mol L\(^{-1}\)).

Rate\(_1\) = k(0.01)\(^x\)(0.01)\(^y\) = \(5 \times 10^{-3}\)

Rate\(_3\) = k(0.01)\(^x\)(0.02)\(^y\) = \(5 \times 10^{-3}\)

Divide Rate\(_3\) by Rate\(_1\):
\[ \frac{Rate_3}{Rate_1} = \frac{k(0.01)^x(0.02)^y}{k(0.01)^x(0.01)^y} = \frac{5 \times 10^{-3}}{5 \times 10^{-3}} \] \[ \left(\frac{0.02}{0.01}\right)^y = 1 \] \[ (2)^y = 1 \]
Any number raised to the power of 0 is 1. Therefore, y = 0.

So, the order of the reaction with respect to B is 0.


To determine the rate constant (k):

The rate law is Rate = k[A]\(^1\)[B]\(^0\) = k[A].

We can use the data from any experiment to calculate k. Let's use Experiment 1.

Rate\(_1\) = k[A]\(_1\)
\(5 \times 10^{-3}\) mol L\(^{-1}\) min\(^{-1}\) = k (0.01 mol L\(^{-1}\))
\[ k = \frac{5 \times 10^{-3}}{0.01} = \frac{5 \times 10^{-3}}{1 \times 10^{-2}} = 0.5 min^{-1} \]

Step 4: Final Answer:

The order with respect to A is 1.

The order with respect to B is 0.

The rate constant (k) is 0.5 min\(^{-1}\).
Quick Tip: To find the order of a reaction with respect to a specific reactant, always compare two experiments where the concentration of only that reactant changes, while the concentrations of all other reactants are kept constant. This isolates the effect of that single reactant on the rate.

(i) Peptide linkage

A peptide linkage or peptide bond is a covalent chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amino group of the other molecule, releasing a molecule of water (H\(_2\)O). This is a dehydration synthesis reaction (also known as a condensation reaction) and usually occurs between amino acids. The resulting C(O)NH bond is called a peptide bond, and the resulting molecule is an amide.


The linkage -CO-NH- is the peptide linkage.


(ii) Denatured protein

Denaturation is a process in which a protein loses its native shape due to the disruption of weak chemical bonds and interactions, thereby becoming biologically inactive. The native conformation of a protein is its unique three-dimensional structure, including secondary, tertiary, and quaternary structures.

Denaturation can be caused by external stress such as:

- Heat: Breaks hydrogen bonds.

- Acids or Bases: Disrupt salt bridges by changing the state of protonation.

- Organic solvents, urea, or detergents.

During denaturation, the primary structure (the sequence of amino acids) remains the same. A common example is the coagulation of egg white (albumin) when it is cooked.
Quick Tip: Remember that denaturation is the loss of 2\(^\circ\), 3\(^\circ\), and 4\(^\circ\) structures, which causes the loss of biological function. The primary structure (amino acid sequence) is not affected by denaturation.

Step 1: Understanding Amphoteric Behaviour:

A substance is called amphoteric if it can react as both an acid and a base.


Step 2: Structure of an Amino Acid:

An \(\alpha\)-amino acid has a central carbon atom (the \(\alpha\)-carbon) bonded to:

- An amino group (-NH\(_2\))

- A carboxyl group (-COOH)

- A hydrogen atom (-H)

- A variable side chain (-R group)


Step 3: Dual Functionality:

- The carboxyl group (-COOH) is acidic and can donate a proton (H\(^+\)).

- The amino group (-NH\(_2\)) is basic and can accept a proton (H\(^+\)).

Since a single amino acid molecule contains both an acidic and a basic functional group, it has the ability to act as either an acid or a base depending on the pH of the surrounding medium.


Step 4: Zwitterion Formation:

In neutral aqueous solution, the acidic carboxyl group donates its proton to the basic amino group within the same molecule. This forms a dipolar ion called a zwitterion, which has both a positive charge (-NH\(_3^+\)) and a negative charge (-COO\(^-\)).


This zwitterionic form can then:

- React with an acid (H\(^+\)): The -COO\(^-\) group accepts a proton, and the amino acid acts as a base.

- React with a base (OH\(^-\)): The -NH\(_3^+\) group donates a proton, and the amino acid acts as an acid.


Step 5: Final Answer:

Amino acids are amphoteric because their structure contains both an acidic carboxyl group (-COOH) and a basic amino group (-NH\(_2\)). This dual functionality allows them to react with both acids and bases.
Quick Tip: The concept of the zwitterion is key to understanding the properties of amino acids, including their high melting points, solubility in water, and amphoteric nature. Remember that at its isoelectric point (pI), an amino acid exists predominantly as a zwitterion.

Fibrous and globular proteins can be differentiated based on their structure, solubility, and function.


\begin{tabular{|l|l|l|
\hline
Property & Fibrous Proteins & Globular Proteins

\hline
Shape & Long, narrow, thread-like or & Spherical, ovoid, or ellipsoidal in

& sheet-like structure. Polypeptide & shape. Polypeptide chains are

& chains are arranged in parallel. & tightly folded into a compact form.

\hline
Solubility & Generally insoluble in water and & Generally soluble in water and

& aqueous solutions of acids and bases. & aqueous solutions.

\hline
Function & Primarily have a structural or & Primarily involved in metabolic

& protective role in organisms. & and functional roles like catalysis

& Provide strength and elasticity. & (enzymes), transport, and regulation.

\hline
Stability & More stable to changes in & Less stable; sensitive to changes

& temperature and pH. & in temperature and pH (easily denatured).

\hline
Examples & Keratin (in hair, nails), Collagen & Insulin, Haemoglobin, Albumin,

& (in connective tissue), Myosin (in muscle). & and all enzymes.

\hline
\end{tabular
Quick Tip: A simple way to remember is: \textbf{Fibrous = Fiber = Structural} (like threads in a rope) and are insoluble. \textbf{Globular = Globe = Functional} (like compact balls that do jobs) and are soluble.

The secondary structure of a protein refers to the local, regular, folded structures that form within a polypeptide chain due to hydrogen bonding between the atoms of the polypeptide backbone (not the side chains).

The two most common and stable types of secondary structures are:


1. \(\alpha\)-Helix:

- This structure resembles a coiled spring or a spiral staircase.

- The polypeptide chain is twisted into a right-handed helix.

- It is stabilized by intramolecular hydrogen bonds between the C=O group of one amino acid and the N-H group of the amino acid that is four residues ahead in the chain.

- An example of a protein rich in \(\alpha\)-helices is keratin, found in hair and nails.


2. \(\beta\)-Pleated Sheet:

- This structure consists of polypeptide chains (called \(\beta\)-strands) lying side-by-side.

- The structure is stabilized by intermolecular or intramolecular hydrogen bonds between the C=O groups of one strand and the N-H groups of an adjacent strand.

- The sheet has a pleated or folded appearance.

- The strands can run in the same direction (parallel \(\beta\)-sheet) or in opposite directions (antiparallel \(\beta\)-sheet).

- An example is fibroin, the protein in silk.
Quick Tip: Remember that secondary structures are all about hydrogen bonds in the \textbf{backbone} of the polypeptide chain. The \(\alpha\)-helix involves H-bonds within a single chain, while the \(\beta\)-sheet involves H-bonds between chains (or distant parts of the same chain).

(i) Reaction with Br\(_2\) water:

The -OH group in phenol is a strongly activating group. It activates the benzene ring for electrophilic substitution, particularly at the ortho and para positions. Bromine water is a polar medium which facilitates the ionization of Br\(_2\) to Br\(^+\).

Due to the high activation of the ring, the reaction is very fast and all three available ortho and para positions are substituted by bromine atoms.

Reaction: Phenol reacts with excess bromine water at room temperature to give a white precipitate of 2,4,6-tribromophenol.

Equation:





(ii) Reaction with Conc. HNO\(_3\):

Concentrated nitric acid, in the presence of concentrated sulfuric acid (which acts as a catalyst), is a strong nitrating agent.

Again, due to the highly activating nature of the -OH group, nitration occurs at all three ortho and para positions.


Reaction: Phenol reacts with concentrated nitric acid to yield 2,4,6-trinitrophenol, which is commonly known as Picric acid.


Equation:



Quick Tip: The -OH group in phenol is so strongly activating that controlling the substitution to get a mono-substituted product is difficult. To get mono-bromophenol, the reaction must be carried out in a non-polar solvent like CS\(_2\) or CCl\(_4\) at low temperatures. To get mono-nitrophenol, dilute HNO\(_3\) is used.

Step 1: Understanding the Question:

The question asks for the reaction mechanism when an alcohol (R-OH) acts as a nucleophile and attacks an electrophile, which is a methyl carbocation (CH\(_3^+\)). A nucleophile is an electron-rich species that donates an electron pair, and an electrophile is an electron-deficient species that accepts an electron pair. The lone pairs on the oxygen atom of the alcohol make it a good nucleophile.


Step 2: Detailed Mechanism:

The reaction proceeds in two steps:


Step 2a: Nucleophilic Attack of Alcohol on the Carbocation

The lone pair of electrons on the oxygen atom of the alcohol attacks the electron-deficient carbon of the methyl carbocation. This results in the formation of a new carbon-oxygen bond and an intermediate called a protonated ether or an oxonium ion.
\[ \underset{Alcohol (Nucleophile)}{R-\ddot{O}-H} + \underset{Carbocation (Electrophile)}{CH_3^{\oplus}} \longrightarrow \underset{Protonated Ether (Oxonium ion)}{R-\underset{\displaystyle\oplus}{\overset{\displaystyle H}{\ddot{O}}}-CH_3} \]

Step 2b: Deprotonation to form Ether

The oxonium ion formed in the first step is unstable because of the positive charge on the highly electronegative oxygen atom. It readily loses a proton to a weak base (B:), which could be another alcohol molecule or water, to form a stable ether.
\[ \underset{Protonated Ether}{R-\underset{\displaystyle\oplus}{\overset{\displaystyle H}{\ddot{O}}}-CH_3} + \underset{Base}{:B} \longrightarrow \underset{Ether}{R-\ddot{O}-CH_3} + \underset{Protonated Base}{H-B^{\oplus}} \]

Step 3: Final Answer:

The overall reaction is the formation of an ether from an alcohol and a carbocation. The mechanism involves a nucleophilic attack followed by a deprotonation step.
Quick Tip: This mechanism is fundamental to understanding reactions like the S\(_N\)1 reaction of alkyl halides with alcohol as the solvent/nucleophile, or the acid-catalyzed dehydration of alcohols to form ethers. The key is that the oxygen atom in alcohols and water can act as a nucleophile due to its lone pairs.

Step 1: Understanding the Bond in Phenol:

The question asks why the bond between the benzene ring's carbon atom and the oxygen atom (C-OH) is difficult to break in phenols, unlike in alcohols.


Step 2: The Role of Resonance:

The key reason lies in the electronic structure of phenol. The oxygen atom of the -OH group has lone pairs of electrons. One of these lone pairs can participate in resonance with the \(\pi\)-electron system of the benzene ring.

The resonance structures of phenol are:






Step 3: Effect of Resonance on the C-O Bond:

As seen in the resonance structures II, III, and IV, there is a double bond between the carbon of the ring and the oxygen atom.

This means that the actual C-O bond in phenol is not a pure single bond but a resonance hybrid that has a significant partial double bond character.

A double bond is stronger and shorter than a single bond.


Step 4: Comparison with Alcohols:

In alcohols (R-OH), the C-O bond is a pure single bond. There is no resonance to strengthen it.

For example, the C-O bond length in methanol is 142 pm, whereas in phenol it is 136 pm. The shorter bond length in phenol indicates a stronger bond.


Step 5: Final Answer:

Due to the partial double bond character acquired through resonance, the C-O bond in phenols is much stronger than the C-O single bond in alcohols. Therefore, reactions that require the cleavage of this C-O bond do not occur easily in phenols.
Quick Tip: Whenever comparing the reactivity of functional groups attached to a benzene ring versus an alkyl chain (e.g., phenol vs alcohol, or chlorobenzene vs chloroalkane), always consider the effect of resonance. Resonance usually strengthens the bond between the ring and the atom of the functional group.

Step 1: Identifying the Reagent and Test:

The reagent described, a solution of concentrated HCl in the presence of anhydrous ZnCl\(_2\), is known as the Lucas reagent. The test performed using this reagent to distinguish between primary, secondary, and tertiary alcohols is called the Lucas test.


Step 2: Principle of the Lucas Test:

The Lucas test is based on the difference in the rate of reaction of primary, secondary, and tertiary alcohols with hydrogen halides. The reaction follows an S\(_N\)1 mechanism, where the rate-determining step is the formation of a carbocation. The reaction is:
\[ R-OH + HCl \xrightarrow{anhy. ZnCl_2} R-Cl + H_2O \]
The product, an alkyl chloride (R-Cl), is insoluble in the reagent and appears as a cloudy suspension or turbidity.

The reactivity order is: Tertiary > Secondary > Primary, because the stability of the corresponding carbocation follows the same order.


Step 3: Classifying the Given Alcohols:

1. Butan-1-ol: CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)OH. The -OH group is attached to a primary carbon. This is a primary (1\(^\circ\)) alcohol.

2. 2-Methylpropan-2-ol: (CH\(_3\))\(_3\)COH. The -OH group is attached to a tertiary carbon. This is a tertiary (3\(^\circ\)) alcohol.


Step 4: Predicting the Observations:

- With 2-Methylpropan-2-ol (tertiary alcohol): It will react very rapidly with the Lucas reagent. A tertiary carbocation ((CH\(_3\))\(_3\)C\(^+\)) is formed, which is very stable. This leads to the immediate formation of 2-chloro-2-methylpropane, causing immediate turbidity to appear in the solution.

- With Butan-1-ol (primary alcohol): It will react very slowly. Primary carbocations are highly unstable. At room temperature, there will be no visible reaction or turbidity. Turbidity will only appear if the mixture is heated for a significant amount of time.


Step 5: Final Answer:

To distinguish between the two alcohols, add Lucas reagent to both test tubes at room temperature.

- The test tube containing 2-Methylpropan-2-ol will show immediate turbidity.

- The test tube containing Butan-1-ol will remain clear.
Quick Tip: Remember the Lucas test results by the speed of reaction:
- \textbf{Tertiary (3\(^\circ\)):} Turbidity is immediate (Fastest).
- \textbf{Secondary (2\(^\circ\)):} Turbidity appears in 5-10 minutes.
- \textbf{Primary (1\(^\circ\)):} No turbidity at room temperature; appears only on heating (Slowest).

Step 1: Identify the first reaction (Toluene to A and then B)


The starting material is Toluene (methylbenzene). It is treated with CrO\(_3\) in acetic anhydride ((CH\(_3\)CO)\(_2\)O). This is a method for the oxidation of the methyl group of toluene.


- The reagent combination CrO\(_3\) / (CH\(_3\)CO)\(_2\)O first oxidizes the methyl group to a gem-diacetate, which is stable and resists further oxidation to carboxylic acid. This intermediate is Benzylidene diacetate. So, A = C\(_6\)H\(_5\)CH(OCOCH\(_3\))\(_2\).


- This gem-diacetate (A) is then subjected to acidic hydrolysis (H\(_3\)O\(^+\)). Hydrolysis of the diacetate yields an aldehyde. So, B = Benzaldehyde (C\(_6\)H\(_5\)CHO).


This two-step conversion is an alternative to the Etard reaction for preparing benzaldehyde from toluene.


Step 2: Identify the second reaction (B to C and Sodium Benzoate)


Compound B (Benzaldehyde) is treated with concentrated NaOH.


- Identify the reaction type: Benzaldehyde (C\(_6\)H\(_5\)CHO) is an aldehyde with no \(\alpha\)-hydrogen atoms. When such aldehydes are treated with a strong base (like concentrated NaOH), they undergo a self-oxidation-reduction reaction known as the Cannizzaro reaction.


- Predict the products: In the Cannizzaro reaction, one molecule of the aldehyde is reduced to the corresponding primary alcohol, and another molecule is oxidized to the salt of the corresponding carboxylic acid.


- Reduction of Benzaldehyde gives Benzyl alcohol (C\(_6\)H\(_5\)CH\(_2\)OH). This is compound C.


- Oxidation of Benzaldehyde gives Benzoic acid, which in the basic medium forms the salt, Sodium Benzoate (C\(_6\)H\(_5\)COONa). This is the other product shown.


The reaction is:

\[ \underset{(B) Benzaldehyde}{2C_6H_5CHO} + conc. NaOH \rightarrow \underset{(C) Benzyl alcohol}{C_6H_5CH_2OH} + \underset{Sodium benzoate}{C_6H_5COONa} \]

Step 3: Final Answer Summary:


- A is Benzylidene diacetate, C\(_6\)H\(_5\)CH(OCOCH\(_3\))\(_2\).


- B is Benzaldehyde, C\(_6\)H\(_5\)CHO.


- C is Benzyl alcohol, C\(_6\)H\(_5\)CH\(_2\)OH.
Quick Tip: Recognizing named reactions is crucial. Here, the final step is a classic Cannizzaro reaction (aldehyde with no \(\alpha\)-H + conc. base). The first step is a variation of the Etard reaction. Knowing the reagents and products for these named reactions can solve multi-step synthesis problems quickly.

The characteristic reactions of the carbonyl group in aldehydes and ketones are nucleophilic addition reactions. In these compounds, the carbonyl carbon is highly electrophilic.


However, in carboxylic acids (-COOH), the carbonyl group is attached to a hydroxyl (-OH) group. The lone pair of electrons on the oxygen of the -OH group participates in resonance with the C=O double bond.


Resonance in Carboxylic Acid:





Due to this resonance:


1. The C=O bond acquires some single bond character.


2. The carbonyl carbon's positive charge (electrophilicity) is reduced because of the delocalization of the lone pair from the adjacent oxygen atom.


Since the carbonyl carbon is less electrophilic in carboxylic acids, it does not readily undergo nucleophilic addition reactions that are characteristic of aldehydes and ketones (e.g., reaction with HCN, NaHSO\(_3\)).
Quick Tip: The reactivity of a carbonyl group is significantly altered by the group attached to it. An adjacent atom with a lone pair (like in acids, esters, amides) will always reduce the electrophilicity of the carbonyl carbon via resonance, making it less reactive than aldehydes or ketones.

The strength of an acid is determined by the stability of its conjugate base formed after donating a proton (H\(^+\)).


- Ethanoic Acid (CH\(_3\)COOH):


When it loses a proton, it forms the ethanoate (acetate) ion (CH\(_3\)COO\(^-\)).


\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]

The ethanoate ion is highly stabilized by resonance. The negative charge is delocalized equally over both oxygen atoms.


\includegraphics[width=0.5\linewidth{image_acetate_resonance.png

This delocalization makes the ethanoate ion very stable.


- Ethanol (CH\(_3\)CH\(_2\)OH):


When it loses a proton, it forms the ethoxide ion (CH\(_3\)CH\(_2\)O\(^-\)).


\[ CH_3CH_2OH \rightleftharpoons CH_3CH_2O^- + H^+ \]

In the ethoxide ion, the negative charge is localized on the single oxygen atom. There is no resonance stabilization. In fact, the ethyl group (+I effect) slightly destabilizes the anion by increasing the electron density on the oxygen.


Conclusion:


Because the conjugate base of ethanoic acid is much more stable than the conjugate base of ethanol, the equilibrium for the dissociation of ethanoic acid lies further to the right. This means ethanoic acid donates its proton more readily and is therefore a much stronger acid than ethanol.
Quick Tip: When comparing acidity, always analyze the stability of the conjugate base. Factors that stabilize the conjugate base (like resonance, inductive effect of electron-withdrawing groups) will increase the acidity of the parent acid. Resonance is a very powerful stabilizing effect.

- Reagents: Acetic acid (CH\(_3\)COOH) is heated with phosphorus pentoxide (P\(_4\)O\(_{10}\)).


- Reaction Type: P\(_4\)O\(_{10}\) is a very powerful dehydrating agent. It removes one molecule of water from two molecules of carboxylic acid to form an acid anhydride.


- Reaction:


\[ 2CH_3COOH \xrightarrow{P_4O_{10}, \Delta} \underset{Acetic anhydride}{(CH_3CO)_2O} + H_2O \]

- Product: Acetic anhydride.
Quick Tip: Recognize the function of key reagents. P\(_4\)O\(_{10}\) is a classic dehydrating agent used to prepare acid anhydrides from carboxylic acids and nitriles from amides.

- Reagents: Benzoyl chloride (an acid chloride) reacts with dimethyl cadmium (an organocadmium reagent).


- Reaction Type: Organocadmium reagents are milder than Grignard reagents and are specifically used to synthesize ketones from acid chlorides. They react with acid chlorides but do not react further with the ketone product. The reaction proceeds via nucleophilic acyl substitution.


- Reaction: (The stoichiometry is usually 2 moles of acid chloride per mole of dialkylcadmium)


\[ 2\underset{Benzoyl chloride}{C_6H_5COCl} + \underset{Dimethylcadmium}{(CH_3)_2Cd} \rightarrow 2\underset{Acetophenone}{C_6H_5COCH_3} + CdCl_2 \]

- Product: Acetophenone.
Quick Tip: While Grignard reagents (R-MgX) react with acid chlorides to produce tertiary alcohols (after reacting twice), the less reactive organocadmium reagents (R\(_2\)Cd) are perfect for stopping the reaction at the ketone stage.

- Reagent: The starting material is phthalamide, which is benzene-1,2-dicarboxamide.


- Reaction Type: This is an intramolecular dehydration reaction (or more accurately, deammoniation). On strong heating, the two adjacent amide groups lose a molecule of ammonia (NH\(_3\)).


- Product: Phthalimide. This is a cyclic imide.
Quick Tip: Heating dicarboxylic acids or their derivatives (like diamides) that are positioned to form a 5- or 6-membered ring often leads to cyclization with the elimination of a small stable molecule like water or ammonia.

- Purpose: This reaction is used for the complete reduction of the carbonyl group (C=O) of aldehydes and ketones to a methylene group (-CH\(_2\)-). It is particularly useful for compounds that are sensitive to acid (for which the Clemmensen reduction cannot be used).


- Reagents: Hydrazine (NH\(_2\)NH\(_2\)) followed by a strong base like potassium hydroxide (KOH) or potassium tert-butoxide, usually in a high-boiling polar solvent like ethylene glycol.


- General Reaction:


\[ \underset{Aldehyde or Ketone}{R-CO-R'} \xrightarrow{NH_2NH_2, KOH, ethylene glycol, \Delta} \underset{Alkane}{R-CH_2-R'} + N_2(g) \]

- Mechanism Outline: The carbonyl compound first reacts with hydrazine to form a hydrazone. Then, in the presence of a strong base, the hydrazone is deprotonated and rearranges to eliminate nitrogen gas, leaving behind the alkane.
Quick Tip: To choose between Wolff-Kishner and Clemmensen reduction (which both convert C=O to CH\(_2\)), check the stability of the rest of the molecule. Use Wolff-Kishner (basic conditions) if the molecule has acid-sensitive groups. Use Clemmensen (acidic conditions) for base-sensitive groups.

- Purpose: This reaction involves the removal of a carboxyl group (-COOH) from a molecule, which is released as carbon dioxide (CO\(_2\)).


- Reagents and Method: There are several methods. A very common one is the Soda-Lime Decarboxylation. In this method, the sodium salt of a carboxylic acid is heated with soda-lime (a mixture of NaOH and CaO).


- General Reaction (Soda-Lime):


\[ \underset{Sodium carboxylate}{R-COONa} + \underset{from Soda-lime}{NaOH} \xrightarrow{CaO, \Delta} \underset{Alkane}{R-H} + Na_2CO_3 \]

- Note: Carboxylic acids having a keto group at the \(\beta\)-position ( \(\beta\)-keto acids) are particularly easy to decarboxylate and often do so just upon gentle heating, without the need for soda-lime.


\[ R-CO-CH_2-COOH \xrightarrow{\Delta} R-CO-CH_3 + CO_2 \] Quick Tip: The easiest decarboxylation occurs with \(\beta\)-keto acids because they can form a stable, cyclic, six-membered transition state that facilitates the loss of CO\(_2\). This is a very common reaction in biochemical pathways and organic synthesis.

Step 1: Understanding the Question:

We need to calculate the cell potential (E\(_{cell}\)) for a non-standard galvanic cell using the Nernst equation. The temperature is 25\(^\circ\)C (298 K).


Step 2: Identify Anode and Cathode and Calculate E\(^\circ_{cell}\):

From the overall reaction: Zn(s) \(\rightarrow\) Zn\(^{2+}\)(aq) and Pb\(^{2+}\)(aq) \(\rightarrow\) Pb(s).

- Oxidation occurs for Zinc: Zn is the Anode.

- Reduction occurs for Lead: Pb is the Cathode.


The standard cell potential (E\(^\circ_{cell}\)) is calculated as:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] \[ E^\circ_{cell} = E^\circ_{Pb^{2+}/Pb} - E^\circ_{Zn^{2+}/Zn} \] \[ E^\circ_{cell} = (-0.13 V) - (-0.76 V) \] \[ E^\circ_{cell} = -0.13 + 0.76 = 0.63 V \]

Step 3: Apply the Nernst Equation:

The Nernst equation at 25\(^\circ\)C (298 K) is:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
Where:

- n is the number of moles of electrons transferred in the balanced equation.

Zn \(\rightarrow\) Zn\(^{2+}\) + 2e\(^{-}\)

Pb\(^{2+}\) + 2e\(^{-}\) \(\rightarrow\) Pb

So, n = 2.

- Q is the reaction quotient.

\[ Q = \frac{[Products]}{[Reactants]} = \frac{[Zn^{2+}]}{[Pb^{2+}]} \]
(Activities of pure solids Zn and Pb are taken as 1).


Step 4: Calculate Q and E\(_{cell}\):

Given concentrations are [Zn\(^{2+}\)] = 0.1 M and [Pb\(^{2+}\)] = 0.02 M.
\[ Q = \frac{0.1}{0.02} = 5 \]

Now, substitute all values into the Nernst equation:
\[ E_{cell} = 0.63 - \frac{0.0591}{2} \log(5) \]
Given log(5) = 0.6990.
\[ E_{cell} = 0.63 - (0.02955) \times (0.6990) \] \[ E_{cell} = 0.63 - 0.02065545 \] \[ E_{cell} \approx 0.6093 V \]

Step 5: Final Answer:

The calculated E\(_{cell}\) for the galvanic cell is 0.609 V.
Quick Tip: Always start by calculating the standard cell potential, \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\). Remember that the cathode is where reduction happens (higher reduction potential) and the anode is where oxidation happens (lower reduction potential). Then, carefully set up the reaction quotient Q for the Nernst equation, making sure to exclude solids and liquids.

Part 1: Faraday's First Law of Electrolysis

The law states that the amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

Mathematically, if 'm' is the mass of the substance deposited or liberated and 'Q' is the quantity of electricity passed, then:
\[ m \propto Q \]
Since Q = I \(\times\) t (where I is current in amperes and t is time in seconds),
\[ m \propto I \times t \] \[ m = Z \times I \times t \]
Where Z is a constant of proportionality known as the electrochemical equivalent of the substance.


Part 2: Electricity Required for Reduction of MnO\(_4^-\)

Step 1: Determine the half-reaction and change in oxidation state.

We need to find the electricity required to reduce one mole of MnO\(_4^-\) to Mn\(^{2+}\).

- First, find the oxidation state of Mn in MnO\(_4^-\). Let it be x.

x + 4(-2) = -1 \(\implies\) x = +7.

- The oxidation state of Mn in Mn\(^{2+}\) is +2.

- The change in oxidation state is from +7 to +2.

Change = 7 - 2 = 5.


Step 2: Relate the change in oxidation state to moles of electrons.

A change in oxidation state of 5 means that 5 moles of electrons are gained for every one mole of MnO\(_4^-\) that is reduced.

The balanced half-reaction in acidic medium is:
\[ MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l) \]
This confirms that 5 moles of electrons are required to reduce 1 mole of MnO\(_4^-\).


Step 3: Convert moles of electrons to Faradays.

By definition, the charge carried by one mole of electrons is equal to one Faraday (1 F).

1 F \(\approx\) 96500 C/mol.

Since 5 moles of electrons are required, the total quantity of electricity needed is 5 Faradays.


Step 4: Final Answer:

The amount of electricity required to reduce one mole of MnO\(_4^-\) to Mn\(^{2+}\) is 5 Faradays.
Quick Tip: To quickly find the Faradays needed for a redox reaction, simply calculate the total change in oxidation number for one mole of the substance. This number is equal to the number of moles of electrons transferred, which is also the number of Faradays required.

Step 1: Understanding the Question:

We are given the resistance (R) of a KCl solution in a conductivity cell and the conductivity (\(\kappa\)) of that solution. We need to find the cell constant (G*).


Step 2: Key Formula or Approach:

The relationship between conductivity (\(\kappa\)), resistance (R), and the cell constant (G*) is given by the formula:
\[ \kappa = \frac{1}{R} \times G^* \]
Where:

- \(\kappa\) (kappa) is the conductivity in S cm\(^{-1}\).

- R is the resistance in ohms (\(\Omega\)).

- G* is the cell constant in cm\(^{-1}\). The cell constant is the ratio of the distance between the electrodes (l) to their area of cross-section (A), i.e., G* = l/A.


Step 3: Rearrange the Formula and Substitute the Values:

We can rearrange the formula to solve for the cell constant (G*):
\[ G^* = \kappa \times R \]
We are given:

- R = 1000 \(\Omega\)

- \(\kappa\) = 0.125 \(\times\) 10\(^{-3}\) S cm\(^{-1}\)


Now, substitute these values into the rearranged formula:
\[ G^* = (0.125 \times 10^{-3} S cm^{-1}) \times (1000 \Omega) \]
Since S (Siemens) is the reciprocal of ohm (\(\Omega^{-1}\)), the units S and \(\Omega\) will cancel out.
\[ G^* = 0.125 \times 10^{-3} \times 10^3 cm^{-1} \] \[ G^* = 0.125 cm^{-1} \]

Step 4: Final Answer:

The cell constant is 0.125 cm\(^{-1}\).
Quick Tip: Remember the fundamental formulas for conductance and conductivity.
- Conductance (G) = 1/Resistance (R)
- Conductivity (\(\kappa\)) = Conductance (G) \(\times\) Cell Constant (G*)
Combining these gives \(\kappa = (1/R) \times G^*\). The cell constant is a property of the cell itself and does not change with the solution used.

Note: The standard reduction potential for Mg\(^{2+}\)/Mg is -2.36 V. The value given in the question, E\(^\circ_{Mg^{2+}/Mg}\) = +2.36 V, is the standard oxidation potential (E\(^\circ_{Mg/Mg^{2+}}\)). We will solve the problem using the data as given in the question.


Step 1: Understanding the Question:

We need to calculate the non-standard reduction potential (E) for the Mg\(^{2+}\)/Mg half-cell using the Nernst equation.


Step 2: Write the Half-Reaction and Nernst Equation:

The reduction half-reaction is:
\[ Mg^{2+}(aq) + 2e^- \rightarrow Mg(s) \]
The Nernst equation for this half-reaction at 25\(^\circ\)C is:
\[ E_{Mg^{2+}/Mg} = E^\circ_{Mg^{2+}/Mg} - \frac{0.0591}{n} \log \frac{[Products]}{[Reactants]} \] \[ E_{Mg^{2+}/Mg} = E^\circ_{Mg^{2+}/Mg} - \frac{0.0591}{n} \log \frac{1}{[Mg^{2+}]} \]
Here, n = 2 (number of electrons transferred).


Step 3: Substitute the Given Values:

We are given:

- [Mg\(^{2+}\)] = 1 \(\times\) 10\(^{-4}\) M

- E\(^\circ_{Mg^{2+}/Mg}\) = +2.36 V (as per the question)

- n = 2


Substitute these values into the equation:
\[ E_{Mg^{2+}/Mg} = 2.36 - \frac{0.0591}{2} \log \frac{1}{1 \times 10^{-4}} \] \[ E_{Mg^{2+}/Mg} = 2.36 - 0.02955 \log(10^4) \]
Using the property log(a\(^b\)) = b log(a):
\[ E_{Mg^{2+}/Mg} = 2.36 - 0.02955 \times (4 \log 10) \]
Given log 10 = 1.
\[ E_{Mg^{2+}/Mg} = 2.36 - 0.02955 \times 4 \] \[ E_{Mg^{2+}/Mg} = 2.36 - 0.1182 \] \[ E_{Mg^{2+}/Mg} = +2.2418 V \]

Step 4: Final Answer:

Based on the data provided in the question, the potential for the half-cell is +2.2418 V.
Quick Tip: Be very careful with the signs of standard electrode potentials. The standard convention is to use reduction potentials. If an oxidation potential is given (like in this question, where \(E^\circ_{Mg^{2+}/Mg}\) is positive), you must still use it as provided in the problem statement, even if it contradicts the standard data book values. Always state any assumptions or note potential typos if you suspect them.

Step 1: Understanding Metallic Conduction:

In metallic conductors, the flow of electricity is due to the movement of delocalized electrons through a fixed lattice of positive metal ions (kernels).


Step 2: Effect of Temperature:

When the temperature of a metal is increased, the metal ions in the lattice gain kinetic energy and begin to vibrate more vigorously about their mean positions.


Step 3: Resistance to Electron Flow:

These increased vibrations of the positive ions create a greater obstruction or hindrance to the flow of electrons. The electrons collide more frequently with the vibrating ions, which impedes their smooth passage through the conductor.


Step 4: Conclusion on Resistance and Conductance:

- This increased hindrance to electron flow means that the electrical resistance of the metal increases with an increase in temperature.

- Electrical conductance is the reciprocal of resistance (Conductance = 1/Resistance).

- Therefore, since resistance increases with temperature, the electrical conductance of a metallic conductor must decrease with an increase in temperature.


Step 5: Final Answer:

An increase in temperature causes the electrical conductance of a metallic conductor to decrease. This is because the increased thermal vibrations of the metal ions in the lattice obstruct the flow of electrons, increasing the resistance.
Quick Tip: Remember the opposite behavior for electrolytic conductors. For electrolytic solutions, conductance increases with temperature. This is because the increased temperature increases the kinetic energy of the ions, making them move faster, and also decreases the viscosity of the solvent, reducing the friction on the moving ions.

In aqueous solution, the dichromate ion (Cr\(_2\)O\(_7^{2-}\)), which is orange, and the chromate ion (CrO\(_4^{2-}\)), which is yellow, exist in a pH-dependent equilibrium.


The equilibrium can be represented as:

\[ \underset{(Orange)}{Cr_2O_7^{2-}(aq)} + 2OH^-(aq) \rightleftharpoons \underset{(Yellow)}{2CrO_4^{2-}(aq)} + H_2O(l) \]

When an alkali (a source of OH\(^-\) ions) is added, according to Le Chatelier's principle, the equilibrium shifts to the right to consume the added OH\(^-\).


This results in the formation of the yellow chromate ion (CrO\(_4^{2-}\)). Therefore, the colour of the solution changes from orange to yellow.
Quick Tip: Remember this key equilibrium: Dichromate (Cr\(_2\)O\(_7^{2-}\)) is stable in acidic solution, while Chromate (CrO\(_4^{2-}\)) is stable in alkaline solution. Adding acid to yellow chromate turns it orange, and adding base to orange dichromate turns it yellow.

The definition of a transition element is an element that has an incompletely filled d-subshell in its ground state or in any of its common oxidation states.


Let's examine Zinc (Zn), Cadmium (Cd), and Mercury (Hg):


- Ground State Electronic Configuration:


Zn (Z=30): [Ar] 3d\(^{10}\) 4s\(^2\)


Cd (Z=48): [Kr] 4d\(^{10}\) 5s\(^2\)


Hg (Z=80): [Xe] 4f\(^{14}\) 5d\(^{10}\) 6s\(^2\)


In their ground state, all three have a completely filled d-subshell (d\(^{10}\)).


- Common Oxidation State: Their most common and stable oxidation state is +2.


Zn\(^{2+}\): [Ar] 3d\(^{10}\)


Cd\(^{2+}\): [Kr] 4d\(^{10}\)


Hg\(^{2+}\): [Xe] 4f\(^{14}\) 5d\(^{10}\)


In their +2 oxidation state, they still have a completely filled d-subshell.


Since they do not have a partially filled d-orbital in either their elemental form or their common ionic form, they are not considered typical transition elements and are often referred to as pseudo-transition elements or Group 12 elements.
Quick Tip: The key to identifying a transition element is to look for a partially filled d-orbital (d\(^1\) to d\(^9\)) in either the neutral atom or any of its common ions. If all possible states are d\(^0\) or d\(^{10}\), it is not a transition element.

The standard electrode potential (E\(^\circ\)) value indicates the tendency for a reduction to occur. A high positive value means the reduction is highly favourable.


- For Manganese (Mn): The reduction is Mn\(^{3+}\) + e\(^-\) \(\rightarrow\) Mn\(^{2+}\).


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


Electronic configuration of Mn\(^{2+}\) is [Ar] 3d\(^5\).


The 3d\(^5\) configuration is a half-filled d-subshell, which is an exceptionally stable electronic arrangement due to symmetry and high exchange energy. The strong tendency to achieve this stable configuration makes the reduction of Mn\(^{3+}\) to Mn\(^{2+}\) very favourable, resulting in a large positive E\(^\circ\) value.


- For Chromium (Cr): The reduction is Cr\(^{3+}\) + e\(^-\) \(\rightarrow\) Cr\(^{2+}\).


Electronic configuration of Cr\(^{3+}\) is [Ar] 3d\(^3\). This is also stable as it has a half-filled t\(_{2g}\) level in an octahedral field.


Electronic configuration of Cr\(^{2+}\) is [Ar] 3d\(^4\).


The reduction from the stable Cr\(^{3+}\) (d\(^3\)) to the less stable Cr\(^{2+}\) (d\(^4\)) is not as favourable. In fact, the reverse reaction (oxidation of Cr\(^{2+}\) to Cr\(^{3+}\)) is favoured. This is reflected in its negative E\(^\circ\) value (E\(^\circ_{Cr^{3+}/Cr^{2+}}\) = -0.41 V).


Thus, the E\(^\circ\) for Mn\(^{3+}\)/Mn\(^{2+}\) is highly positive due to the extra stability of the d\(^5\) configuration of Mn\(^{2+}\).
Quick Tip: When explaining trends in E\(^\circ\) values for transition metals, always look at the electronic configurations of the ions involved. The exceptional stability of half-filled (d\(^5\)) and fully-filled (d\(^{10}\)) configurations is a very common explanation for unusually high or low values.

A disproportionation reaction is a redox reaction in which a species is simultaneously oxidized and reduced.


The manganate ion (MnO\(_4^{2-}\)), in which Mn is in the +6 oxidation state, is only stable in strongly alkaline solutions. In neutral or acidic medium, it is unstable and undergoes disproportionation.


- Oxidation: Mn\(^{6+}\) (in MnO\(_4^{2-}\)) is oxidized to Mn\(^{7+}\) (in MnO\(_4^-\)).


- Reduction: Mn\(^{6+}\) (in MnO\(_4^{2-}\)) is reduced to Mn\(^{4+}\) (in MnO\(_2\)).


Balanced Chemical Equation:

\[ \underset{(Manganate, green)}{3MnO_4^{2-}(aq)} + 4H^+(aq) \rightarrow \underset{(Permanganate, purple)}{2MnO_4^-(aq)} + \underset{(Manganese dioxide, brown ppt)}{MnO_2(s)} + 2H_2O(l) \]

So, when manganate ion is in an acidic medium, it disproportionates to form permanganate ion and manganese dioxide.
Quick Tip: A key feature of disproportionation reactions is that an element in an intermediate oxidation state converts to species with both higher and lower oxidation states. For manganese, the +6 state (manganate) is intermediate and unstable in acidic conditions.

Potassium permanganate (KMnO\(_4\)) is thermally unstable. When it is heated strongly (to about 513 K or 240\(^\circ\)C), it undergoes decomposition.


In this reaction, the manganese in KMnO\(_4\) (oxidation state +7) is reduced to both +6 (in K\(_2\)MnO\(_4\)) and +4 (in MnO\(_2\)).


Balanced Chemical Equation:

\[ \underset{(Potassium permanganate, purple)}{2KMnO_4(s)} \xrightarrow{\Delta} \underset{(Potassium manganate, green)}{K_2MnO_4(s)} + \underset{(Manganese dioxide, black)}{MnO_2(s)} + \underset{(Oxygen gas)}{O_2(g)} \]

This reaction is a common laboratory method for the preparation of small amounts of pure oxygen gas.
Quick Tip: The chemistry of manganese is rich in different oxidation states, each with a characteristic colour. Memorizing these can be helpful:
- Mn\(^{2+}\): Pale pink
- MnO\(_2\) (Mn\(^{4+}\)): Black/Brown solid
- K\(_2\)MnO\(_4\) (Mn\(^{6+}\)): Green
- KMnO\(_4\) (Mn\(^{7+}\)): Dark purple

- Definition: Misch metal is an alloy which consists predominantly of lanthanoid metals (about 95%) and iron (about 5%), with small traces of S, C, Ca, and Al. The typical composition of the lanthanoid part is ~50% Cerium, ~25% Lanthanum, and other rare-earth metals.


- Use: It is pyrophoric (sparks when struck). A major use is in the manufacture of flints for cigarette lighters and gas lighters. It is also used in magnesium-based alloys to produce bullets, shells and tracer bullets.
Quick Tip: The name 'Mischmetal' comes from the German word 'Mischmetall', which literally means "mixed metal". This is a good way to remember that it's an alloy made of a mixture of lanthanoid metals.

- Chromium (Cr) is in Group 6 of the periodic table.


- We need an oxoanion where Cr has an oxidation state of +6.


- Two common examples are:


1. Chromate ion (CrO\(_4^{2-}\)): Let oxidation state of Cr be x. x + 4(-2) = -2 \(\implies\) x = +6.


2. Dichromate ion (Cr\(_2\)O\(_7^{2-}\)): Let oxidation state of Cr be y. 2y + 7(-2) = -2 \(\implies\) 2y = 12 \(\implies\) y = +6.
Quick Tip: For many transition metals in Groups 4 through 8, the highest possible oxidation state is equal to their group number (e.g., Ti in Group 4 is +4, V in Group 5 is +5, Cr in Group 6 is +6, Mn in Group 7 is +7).

Vanadium pentoxide acts as a catalyst primarily because of the ability of vanadium to exhibit variable oxidation states (from +2 to +5).


This allows it to participate in redox reactions by forming unstable intermediate compounds. It provides an alternative reaction pathway with a lower activation energy.


For example, in the Contact Process for manufacturing H\(_2\)SO\(_4\), V\(_2\)O\(_5\) oxidizes SO\(_2\) to SO\(_3\):

\[ 2SO_2 + V_2O_5 \rightarrow 2SO_3 + 2VO_2 \]

The catalyst is then regenerated by oxygen:

\[ 2VO_2 + \frac{1}{2}O_2 \rightarrow V_2O_5 \]

This cyclic change in oxidation state (V\(^{5+}\) \(\leftrightarrow\) V\(^{4+}\)) is key to its catalytic activity.
Quick Tip: The ability to show variable oxidation states is a hallmark of transition metals and is the primary reason why many of them and their compounds (like V\(_2\)O\(_5\), Ni, Pt) are excellent catalysts.

Enthalpy of atomisation is the energy required to convert one mole of a substance from its standard state into gaseous atoms. For metals, it is a measure of the strength of the metallic bond.


Transition elements have high enthalpies of atomisation because they have strong interatomic attractions. This is due to:


1. Large number of valence electrons: They have a large number of valence electrons in both ns and (n-1)d orbitals.


2. Participation of d-electrons: The (n-1)d electrons, in addition to the ns electrons, participate in forming strong metallic bonds. The more unpaired d-electrons, the stronger the bonding (generally).


The combination of these factors results in very strong metallic bonding, which requires a large amount of energy to break, leading to high enthalpies of atomisation.
Quick Tip: The strength of metallic bonding in d-block elements generally increases across the period up to the middle (due to increasing unpaired d-electrons) and then decreases. This trend is directly reflected in properties like melting point and enthalpy of atomisation.

Sodium dichromate (Na\(_2\)Cr\(_2\)O\(_7\)) is prepared from sodium chromate (Na\(_2\)CrO\(_4\)) by acidification of its aqueous solution.


The yellow solution of sodium chromate is treated with an acid, typically sulfuric acid (H\(_2\)SO\(_4\)). This shifts the chromate-dichromate equilibrium towards the dichromate side.


Equation:

\[ \underset{(Sodium Chromate, yellow solution)}{2Na_2CrO_4} + H_2SO_4 \rightarrow \underset{(Sodium Dichromate, orange solution)}{Na_2Cr_2O_7} + Na_2SO_4 + H_2O \]

The less soluble sodium sulfate can be crystallized out, leaving sodium dichromate in the solution.
Quick Tip: This is the reverse of the reaction in question 32(a)(i)(I). The chromate/dichromate system is a classic example of a pH-controlled equilibrium. Base favors chromate (yellow), while acid favors dichromate (orange).