Step 1: Understanding the Question:

The question asks which electronic configuration among the given options can exhibit the highest possible oxidation state. The highest oxidation state for a transition metal is generally achieved when all its valence electrons, i.e., electrons from both the (n-1)d and ns subshells, participate in bonding.


Step 2: Analyzing Each Configuration:

We will calculate the maximum possible oxidation state for each configuration by summing the number of electrons in the 3d and 4s orbitals.



(A) \(3d^6 4s^2\):

Total valence electrons = 6 (from 3d) + 2 (from 4s) = 8.

This configuration belongs to Iron (Fe). The highest common oxidation state for Iron is +6 (in \(FeO_4^{2-}\)), though theoretically it could be higher. However, the most stable states are +2 and +3.



(B) \(3d^5 4s^1\):

Total valence electrons = 5 (from 3d) + 1 (from 4s) = 6.

This configuration belongs to Chromium (Cr). Chromium shows a maximum oxidation state of +6, which is common and stable (e.g., in \(CrO_3, Cr_2O_7^{2-}\)).



(C) \(3d^5 4s^2\):

Total valence electrons = 5 (from 3d) + 2 (from 4s) = 7.

This configuration belongs to Manganese (Mn). Manganese can use all 7 valence electrons to exhibit a maximum oxidation state of +7, which is seen in compounds like \(KMnO_4\) and \(Mn_2O_7\).



(D) \(3d^3 4s^2\):

Total valence electrons = 3 (from 3d) + 2 (from 4s) = 5.

This configuration belongs to Vanadium (V). Vanadium shows a maximum oxidation state of +5 (e.g., in \(V_2O_5\)).


Step 3: Conclusion:

Comparing the maximum possible oxidation states:

- (A) \(\rightarrow\) 8 (theoretically), but +6 is common for Fe.

- (B) \(\rightarrow\) +6 for Cr.

- (C) \(\rightarrow\) +7 for Mn.

- (D) \(\rightarrow\) +5 for V.

The highest oxidation state (+7) is shown by the element with the configuration \(3d^5 4s^2\) (Manganese).
Quick Tip: The maximum oxidation state for the 3d transition series elements (from Sc to Mn) is typically the sum of the electrons in the 4s and 3d orbitals. This trend holds up to Manganese (Mn, +7). After Mn, the tendency to exhibit such high oxidation states decreases due to increasing nuclear charge and pairing of d-electrons.

Step 1: Understanding the Question:

The question asks which 1% solution will have the lowest freezing point. The lowering of the freezing point is a colligative property, which depends on the concentration of solute particles in the solution, not their identity. The relevant formula is the depression in freezing point:
\[ \Delta T_f = i \times K_f \times m \]
Where:

- \(\Delta T_f\) is the depression in freezing point. A larger \(\Delta T_f\) means a lower freezing point (\(T_f = T_f^\circ - \Delta T_f\)).

- \(i\) is the van 't Hoff factor, which represents the number of particles the solute dissociates into.

- \(K_f\) is the cryoscopic constant of the solvent (water), which is the same for all solutions.

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

To find the lowest freezing point, we need to find the solution with the largest value of the product \(i \times m\).


Step 2: Calculating Moles and Comparing \(i \times m\):

Let's assume we have 100 g of each solution. This means we have 1 g of solute and 99 g of water. Molality (m) is \(\frac{moles of solute}{kg of solvent}\).

For comparison, since the mass of solute (1 g) and solvent (approx. 99 g) is the same in each case, the molality will be inversely proportional to the molar mass (M) of the solute. So, \(m \propto \frac{1}{M}\). We need to compare the value of \(\frac{i}{M}\) for each solute. The one with the highest \(\frac{i}{M}\) value will have the greatest \(\Delta T_f\).



(A) KCl:

- Dissociation: \(KCl \rightarrow K^+ + Cl^-\). So, \(i = 2\).

- Molar Mass (M) = 39 + 35.5 = 74.5 g/mol.

- \(\frac{i}{M} = \frac{2}{74.5} \approx 0.0268\)



(B) Glucose (\(C_6H_{12}O_6\)):

- Glucose is a non-electrolyte and does not dissociate. So, \(i = 1\).

- Molar Mass (M) = 180 g/mol.

- \(\frac{i}{M} = \frac{1}{180} \approx 0.0055\)



(C) Urea (\(NH_2CONH_2\)):

- Urea is a non-electrolyte and does not dissociate. So, \(i = 1\).

- Molar Mass (M) = 60 g/mol.

- \(\frac{i}{M} = \frac{1}{60} \approx 0.0167\)



(D) CaCl\(_2\):

- Dissociation: \(CaCl_2 \rightarrow Ca^{2+} + 2Cl^-\). So, \(i = 3\).

- Molar Mass (M) = 40 + 2(35.5) = 111 g/mol.

- \(\frac{i}{M} = \frac{3}{111} \approx 0.0270\)


Step 3: Conclusion:

Comparing the \(\frac{i}{M}\) values:

- KCl: 0.0268

- Glucose: 0.0055

- Urea: 0.0167

- CaCl\(_2\): 0.0270

The value of \(\frac{i}{M}\) is highest for CaCl\(_2\). This means it will produce the largest concentration of particles for a given mass percentage, leading to the greatest depression in freezing point and therefore the lowest freezing point.
Quick Tip: For colligative property problems comparing different solutes at the same mass percentage, the key is to compare the ratio of the van 't Hoff factor to the molar mass (\(i/M\)). A higher value of this ratio means a greater effect on the colligative property (greater boiling point elevation, greater freezing point depression, etc.).

Step 1: Understanding the Equation:

The given equation, \(\Lambda_m = \Lambda_m^\circ - A\sqrt{C}\), is the Debye-Hückel-Onsager equation for strong electrolytes. It describes how the molar conductivity (\(\Lambda_m\)) of a strong electrolyte varies with its concentration (C).

- \(\Lambda_m\): Molar conductivity at concentration C.

- \(\Lambda_m^\circ\): Limiting molar conductivity, or molar conductivity at infinite dilution. This is the maximum possible molar conductivity for the electrolyte.

- A: A constant that depends on the nature of the electrolyte and the solvent.

- C: Concentration of the electrolyte.


Step 2: Analyzing the Relationship:

The equation shows that the molar conductivity \(\Lambda_m\) decreases as the concentration C increases. This is because at higher concentrations, inter-ionic attractions are stronger, which retards the motion of ions and reduces conductivity.

The term \(\Lambda_m^\circ\) represents the molar conductivity when the inter-ionic attractions are negligible. This condition is achieved when the ions are infinitely far apart from each other.


Step 3: Finding the Condition where \(\Lambda_m = \Lambda_m^\circ\):

We want to find the condition under which \(\Lambda_m\) becomes equal to \(\Lambda_m^\circ\). Let's look at the equation:
\[ \Lambda_m = \Lambda_m^\circ - A\sqrt{C} \]
For \(\Lambda_m\) to be equal to \(\Lambda_m^\circ\), the term \(A\sqrt{C}\) must be equal to zero.
\[ A\sqrt{C} = 0 \]
Since A is a non-zero constant, this equality holds true only when \(\sqrt{C} = 0\), which means \(C = 0\).

Therefore, the molar conductivity \(\Lambda_m\) approaches the limiting molar conductivity \(\Lambda_m^\circ\) as the concentration C approaches zero (i.e., at infinite dilution).
Quick Tip: The term \(\Lambda_m^\circ\) is called the "limiting molar conductivity" or "molar conductivity at infinite dilution." The name itself tells you the condition. "Infinite dilution" is the theoretical limit where the concentration (C) approaches zero. At this point, ions are so far apart that they don't interact, allowing for maximum conductivity.

Step 1: Understanding the Rate Law:

For a general n-th order reaction, the rate law is given by:

Rate = \(k \times [Concentration]^n\)

Where:

- Rate has units of mol L\(^{-1}\) s\(^{-1}\) (change in concentration per unit time).

- k is the rate constant.

- [Concentration] has units of mol L\(^{-1}\).

- n is the order of the reaction.


Step 2: Deriving the Units for k:

We can rearrange the rate law to solve for the units of k:

Units of k = \(\frac{Units of Rate}{(Units of Concentration)^n}\)

Substituting the standard units:

Units of k = \(\frac{mol L^{-1} s^{-1}}{(mol L^{-1})^n}\)

Units of k = \((mol L^{-1})^{1-n} s^{-1}\)

Units of k = \(mol^{1-n} L^{n-1} s^{-1}\)


Step 3: Applying the Formula for a Second Order Reaction:

For a second order reaction, the order n = 2.

Substitute n = 2 into the general formula for the units of k:

Units of k = \(mol^{1-2} L^{2-1} s^{-1}\)

Units of k = \(mol^{-1} L^{1} s^{-1}\)

So, the unit is mol\(^{-1}\) L s\(^{-1}\).
Quick Tip: Memorize the general formula for the units of the rate constant k: \(mol^{1-n} L^{n-1} s^{-1}\), where n is the reaction order. You can then quickly find the units for any order:
Zero order (n=0): mol L\(^{-1}\) s\(^{-1}\)
First order (n=1): s\(^{-1}\)
Second order (n=2): mol\(^{-1}\) L s\(^{-1}\)

Step 1: Understanding the Concept:

The question relates to the Arrhenius equation and the Maxwell-Boltzmann distribution of molecular energies. The rate of a chemical reaction depends on the number of molecules that possess sufficient energy to overcome the activation energy barrier (E\(_a\)). The fraction of molecules that have this minimum energy is a key factor in determining the reaction rate.


Step 2: The Arrhenius Equation:

The Arrhenius equation relates the rate constant (k) to the activation energy (E\(_a\)) and temperature (T):
\[ k = A e^{-E_a/RT} \]
Where:

- k is the rate constant.

- A is the pre-exponential factor or frequency factor, which relates to the frequency of collisions with the correct orientation.

- E\(_a\) is the activation energy.

- R is the universal gas constant.

- T is the absolute temperature.


Step 3: Identifying the Fraction Term:

In the Arrhenius equation, the term \(e^{-E_a/RT}\) is known as the Boltzmann factor. This term represents the fraction of molecules in a system that have kinetic energy equal to or greater than the activation energy, E\(_a\), at a given temperature T.

- The term is a dimensionless fraction.

- Since E\(_a\), R, and T are all positive, the exponent \(-E_a/RT\) is negative. This means the value of the fraction is always between 0 and 1, which is expected for a fraction.

- As T increases, the exponent becomes less negative, the value of \(e^{-E_a/RT}\) increases, meaning a larger fraction of molecules can overcome the energy barrier, and the reaction rate increases.
Quick Tip: The term \(e^{-E_a/RT}\) from the Arrhenius equation is fundamental. It's the mathematical representation of the fraction of effective collisions (those with sufficient energy). Always remember this term is the answer when asked for the fraction of molecules possessing energy \(\geq E_a\).

Step 1: Understanding Werner's Theory of Coordination Compounds:

According to Werner's theory, coordination compounds have two types of valencies:

- Primary Valency: This corresponds to the oxidation state of the central metal ion and is satisfied by negative ions. These bonds are ionizable.

- Secondary Valency: This corresponds to the coordination number of the central metal ion and is satisfied by ligands (neutral molecules or anions). These bonds are non-ionizable.

The part of the compound written inside the square brackets, \([\ldots]\), is the coordination sphere. The ions and molecules inside the sphere are attached by secondary valencies and do not dissociate in solution. The ions written outside the square brackets are counter-ions, attached by primary valencies, and are ionizable.


Step 2: Analyzing the Given Complex:

The formula of the complex is [Co(NH\(_3\))_4Cl\(_2\)]Cl.

- The coordination sphere is [Co(NH\(_3\))_4Cl\(_2\)]\(^+\). The two chloride ions inside are ligands and are non-ionizable.

- The counter-ion is the single Cl\(^-\) ion outside the brackets. This chloride ion is attached by a primary valency and is ionizable.


Step 3: Predicting the Precipitation Reaction:

When the complex is dissolved in water, it dissociates as follows:
\[ [Co(NH_3)_4Cl_2]Cl(aq) \rightarrow [Co(NH_3)_4Cl_2]^+(aq) + Cl^-(aq) \]
This dissociation produces one mole of free chloride ions (Cl\(^-\)) per mole of the complex.

When excess silver nitrate (AgNO\(_3\)) solution is added, these free chloride ions will react to form a precipitate of silver chloride (AgCl).
\[ Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s) \]
Since one mole of the complex provides one mole of ionizable Cl\(^-\) ions, it will precipitate one mole of AgCl.
Quick Tip: To determine the moles of AgCl precipitated, simply count the number of chloride ions (or any other halide) that are written \textbf{outside} the square brackets of the coordination complex formula. Each counter-ion chloride will precipitate one mole of AgCl. Ligand chlorides inside the brackets will not precipitate.

Step 1: Analyzing the Requirement:

The question asks for the best method to synthesize alkyl fluorides (R-F). Alkyl fluorides cannot be easily prepared by direct fluorination of alkanes due to the extreme reactivity of fluorine. Therefore, indirect methods are used.


Step 2: Evaluating the Options:

(A) Finkelstein reaction: This is a halide exchange reaction used primarily for the synthesis of alkyl iodides. An alkyl chloride or bromide is treated with sodium iodide (NaI) in acetone.
\[ R-Cl + NaI \xrightarrow{acetone} R-I + NaCl(s) \]


(B) Swarts reaction: This is a halide exchange reaction specifically designed for the synthesis of alkyl fluorides. An alkyl chloride or bromide is heated with a heavy metal fluoride, such as AgF, Hg\(_2\)F\(_2\), CoF\(_2\), or SbF\(_3\).
\[ R-Br + AgF \rightarrow R-F + AgBr \]
This is the most common and effective method for preparing alkyl fluorides.



(C) Sandmeyer reaction: This reaction is used to synthesize aryl halides (aryl chlorides and bromides) from diazonium salts, using CuCl/HCl or CuBr/HBr. It is not used for alkyl halides.
\[ ArN_2^+X^- \xrightarrow{CuCl/HCl} Ar-Cl + N_2 \]


(D) Wurtz reaction: This reaction is used to synthesize symmetrical alkanes with an even number of carbon atoms by reacting two molecules of an alkyl halide with sodium metal in dry ether. It does not produce alkyl fluorides.
\[ 2R-X + 2Na \xrightarrow{dry ether} R-R + 2NaX \]

Step 3: Conclusion:

Based on the analysis, the Swarts reaction is the specific named reaction for the synthesis of alkyl fluorides.
Quick Tip: Remember the specific products of these important named reactions:
\textbf{Finkelstein Reaction} \(\rightarrow\) Alkyl Iodides
\textbf{Swarts Reaction} \(\rightarrow\) Alkyl Fluorides
\textbf{Sandmeyer Reaction} \(\rightarrow\) Aryl Halides/Nitriles
\textbf{Wurtz Reaction} \(\rightarrow\) Symmetrical Alkanes

Step 1: Understanding the Reaction:

The reaction shown is the conversion of an alcohol to an alkyl chloride using HCl and a catalyst, anhydrous ZnCl\(_2\). The reagent (conc. HCl + anhyd. ZnCl\(_2\)) is known as the Lucas reagent, and this reaction forms the basis of the Lucas test for distinguishing between primary, secondary, and tertiary alcohols.


Step 2: Analyzing the Reaction Mechanism:

The reaction proceeds through a nucleophilic substitution mechanism.

1. The oxygen of the alcohol's -OH group is protonated by HCl (or coordinates with the Lewis acid ZnCl\(_2\)), converting the poor leaving group (-OH) into a good leaving group (\(-OH_2^+\) or \(-OZnCl_2\)).

2. The good leaving group departs, forming a carbocation intermediate.
\[ R-OH_2^+ \rightarrow R^+ + H_2O \]
3. The nucleophile (Cl\(^-\)) attacks the carbocation to form the final product, R-Cl.


Step 3: Relating Reactivity to Carbocation Stability:

The rate-determining step of this reaction is the formation of the carbocation intermediate. Therefore, the reactivity of the alcohol is directly proportional to the stability of the carbocation it can form.

The stability of carbocations follows the order:
\[ Tertiary (3^\circ) carbocation > Secondary (2^\circ) carbocation > Primary (1^\circ) carbocation \]
This is due to the electron-donating inductive effect (+I effect) and hyperconjugation from the alkyl groups attached to the positively charged carbon, which help to disperse the positive charge and stabilize the carbocation.


Step 4: Conclusion:

Since the reactivity of the alcohol depends on the stability of the carbocation formed, the order of reactivity of alcohols towards the Lucas reagent will be the same as the order of stability of the carbocations.
\[ Tertiary (3^\circ) alcohol > Secondary (2^\circ) alcohol > Primary (1^\circ) alcohol \]
In increasing order, this is 1\(^\circ\) \(<\) 2\(^\circ\) \(<\) 3\(^\circ\).
Quick Tip: The Lucas test is a classic application of this reactivity order.
\textbf{3\(^\circ\) Alcohols:} React instantly to form turbidity (insoluble alkyl chloride).
\textbf{2\(^\circ\) Alcohols:} React in about 5-10 minutes.
\textbf{1\(^\circ\) Alcohols:} Do not react at room temperature (only upon heating).
Remember: Reactivity of alcohols with HX follows a carbocation mechanism, so the order is 3\(^\circ\) \(>\) 2\(^\circ\) \(>\) 1\(^\circ\).

Step 1: Understanding the Reactants:

- Phenol (\(C_6H_5OH\)): The hydroxyl (-OH) group attached to the benzene ring is a very powerful activating group. It strongly increases the electron density on the benzene ring, especially at the ortho and para positions, through the +R (resonance) effect.

- Bromine Water (Br\(_2\)(aq)): This is an aqueous solution of bromine. In a polar solvent like water, bromine gets polarized and acts as a strong electrophile (\(Br^\delta+\)).


Step 2: Analyzing the Reaction (Electrophilic Aromatic Substitution):

The reaction is an electrophilic aromatic substitution. Due to the extremely high activation of the benzene ring by the -OH group, the reaction is very fast and does not require a Lewis acid catalyst (like FeBr\(_3\)).

The high electron density at the two ortho positions (C-2, C-6) and the para position (C-4) makes them highly susceptible to electrophilic attack.


Step 3: Predicting the Product:

Because of the powerful activating effect of the -OH group and the use of a polar medium (water), the bromination does not stop after monosubstitution. The reaction proceeds to completion, with bromine atoms substituting at all available ortho and para positions.

This results in the formation of a white precipitate of 2,4,6-tribromophenol.
\[ C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH(s) + 3HBr(aq) \] Quick Tip: The solvent is crucial in the bromination of phenol.
\textbf{With Bromine Water (polar solvent):} Polysubstitution occurs to give 2,4,6-tribromophenol.
\textbf{With Bromine in a non-polar solvent (like CS\(_2\) or CCl\(_4\)):} The reaction is less vigorous, and monosubstitution occurs, yielding a mixture of o-bromophenol and p-bromophenol.

Step 1: Understanding the Reaction (Ammonolysis):

The reaction of an alkyl halide with ammonia is called ammonolysis. It is a nucleophilic substitution reaction where ammonia acts as the nucleophile. The initial product is a primary amine.
\[ R-I + NH_3 \rightarrow R-NH_2 + HI \]
This is then neutralized to \(R-NH_3^+I^-\).


Step 2: Analyzing the Effect of Excess Alkyl Iodide:

The question specifies that excess alkyl iodide is used. The primary amine (\(R-NH_2\)) formed in the first step is also a nucleophile. It can react further with the excess alkyl iodide present in the reaction mixture.

Further Reactions:

1. The primary amine reacts with another molecule of alkyl iodide to form a secondary amine.
\[ R-NH_2 + R-I \rightarrow R_2NH + HI \]
2. The secondary amine, being nucleophilic, reacts with more alkyl iodide to form a tertiary amine.
\[ R_2NH + R-I \rightarrow R_3N + HI \]
3. The tertiary amine can also react with a final molecule of alkyl iodide to form a quaternary ammonium salt, which is the final product if the reaction goes to completion.
\[ R_3N + R-I \rightarrow R_4N^+I^- \]

Step 3: Conclusion:

Because the amine products at each stage are themselves nucleophilic, they continue to react with the alkyl halide as long as it is available. When an excess of alkyl iodide is used, it is very difficult to stop the reaction at the primary amine stage. The reaction proceeds to give a mixture of the primary amine, secondary amine, tertiary amine, and the quaternary ammonium salt. Therefore, the product is a mixture of amines and the salt.
Quick Tip: The outcome of ammonolysis depends on the relative amounts of reactants.
\textbf{Excess Ammonia:} The reaction favors the formation of the \textbf{primary amine} as the major product. There is a high probability that the alkyl halide will collide with an ammonia molecule rather than an amine molecule.
\textbf{Excess Alkyl Halide:} The reaction leads to a \textbf{mixture} of all possible products, with the \textbf{quaternary ammonium salt} often being the major product if the reaction goes to completion.

Step 1: Understanding the Hinsberg Test:

The Hinsberg test is used to distinguish between primary (1\(^\circ\)), secondary (2\(^\circ\)), and tertiary (3\(^\circ\)) amines. The reagent used is Hinsberg's reagent, which is benzenesulfonyl chloride (\(C_6H_5SO_2Cl\)). The reactivity and the solubility of the product in alkali (like KOH or NaOH) determine the type of amine.


Step 2: Analyzing the Reactivity of Different Amines:

Primary Amine (R-NH\(_2\)):

Reacts with Hinsberg's reagent to form an N-alkylbenzenesulfonamide.
\[ R-NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHR + HCl \]
The resulting sulfonamide still has an acidic hydrogen attached to the nitrogen atom. This acidic hydrogen can be removed by an alkali, making the product soluble in alkali.



Secondary Amine (R\(_2\)NH):

Reacts with Hinsberg's reagent to form an N,N-dialkylbenzenesulfonamide.
\[ R_2NH + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NR_2 + HCl \]
The resulting sulfonamide has no hydrogen atom attached to the nitrogen. Therefore, it is not acidic and is insoluble in alkali.



Tertiary Amine (R\(_3\)N):

Does not have a hydrogen atom attached to the nitrogen. Therefore, it does not react with Hinsberg's reagent.


Step 3: Applying the Conditions to the Options:

The problem states that amine 'X' reacts with the Hinsberg reagent, and the resulting product is insoluble in alkali.
This is the characteristic behavior of a secondary amine.
We will now analyze the given options based on this fact:


[(A)] \((CH_3)_2NH\): This is dimethylamine, a secondary amine.
It reacts to form N,N-dimethylbenzenesulfonamide.
This product lacks an acidic hydrogen on the nitrogen atom and is therefore insoluble in alkali.
This perfectly matches the observations.


[(B)] \((CH_3)_3N\): This is trimethylamine, a tertiary amine.
It does not have a hydrogen atom on the nitrogen, so it does not react with Hinsberg's reagent.


[(C)] \(CH_3CH_2NH_2\): This is ethylamine, a primary amine.
It reacts to form N-ethylbenzenesulfonamide.
This product has an acidic hydrogen on the nitrogen, making it soluble in alkali.


[(D)] \(C_6H_5NH_2\): This is aniline, a primary amine.
It also reacts to form a product that is soluble in alkali due to the presence of an acidic hydrogen on the nitrogen.


Therefore, the amine 'X' that fits the given description must be the secondary amine, \((CH_3)_2NH\).
Quick Tip: Remember the Hinsberg test results:
\textbf{Primary (1\(^\circ\)):} Reacts \(\rightarrow\) Product is \textbf{soluble} in alkali.
\textbf{Secondary (2\(^\circ\)):} Reacts \(\rightarrow\) Product is \textbf{insoluble} in alkali.
\textbf{Tertiary (3\(^\circ\)):} \textbf{No reaction}.
The key is the presence of an acidic hydrogen on the nitrogen in the sulfonamide product.

Step 1: Understanding the Question:

The question asks to identify the vitamin whose deficiency leads to the disease called scurvy.


Step 2: Analyzing the Vitamins and their Deficiency Diseases:

(A) Vitamin A (Retinol): Deficiency of Vitamin A causes night blindness and xerophthalmia (hardening of the cornea of the eye).



(B) Vitamin B\(_2\) (Riboflavin): Deficiency of Vitamin B\(_2\) can cause cheilosis (fissuring at corners of mouth and lips) and digestive disorders.



(C) Vitamin C (Ascorbic Acid): Deficiency of Vitamin C causes scurvy, a disease characterized by bleeding gums, delayed wound healing, and weakness. Vitamin C is essential for the synthesis of collagen, a key protein in connective tissues.



(D) Vitamin D (Calciferol): Deficiency of Vitamin D causes rickets in children (bone deformities) and osteomalacia in adults (soft bones and joint pain).


Step 3: Conclusion:

Based on the known deficiency diseases, scurvy is caused by a lack of Vitamin C.
Quick Tip: It is useful to create a table to memorize the vitamins, their chemical names, and their deficiency diseases for quick recall in exams.

Step 1: Analyze Assertion (A):

The assertion states that a mixture of ortho- and para-nitrophenols can be separated by steam distillation. This is a standard laboratory procedure and is factually correct. The two isomers have significantly different boiling points and volatilities, which allows for this separation technique to be effective. So, Assertion (A) is true.


Step 2: Analyze Reason (R):

The reason provides an explanation for the volatility of o-nitrophenol. It states that o-nitrophenol is steam volatile because of intramolecular hydrogen bonding.

In o-nitrophenol, the -OH group and the -NO\(_2\) group are close to each other (in ortho positions). This proximity allows for a hydrogen bond to form within the same molecule. This is called intramolecular hydrogen bonding. This internal bonding prevents the molecule from forming strong hydrogen bonds with other molecules. As a result, the intermolecular forces of attraction are weak (mainly weak van der Waals forces), making o-nitrophenol have a lower boiling point and higher volatility. It is thus steam volatile. So, Reason (R) is true.


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

The assertion is about the separation of the two isomers. The separation is possible because of a large difference in their volatility.

- o-Nitrophenol: As explained by the reason, it has intramolecular H-bonding, weak intermolecular forces, and is steam volatile.

- p-Nitrophenol: The -OH and -NO\(_2\) groups are far apart. It cannot form intramolecular H-bonds. Instead, it forms strong intermolecular hydrogen bonds with neighboring molecules. This extensive intermolecular association results in a much higher boiling point and lower volatility. It is not steam volatile.

Since one isomer is steam volatile and the other is not, they can be easily separated by steam distillation. The o-nitrophenol will distill over with the steam, while the p-nitrophenol will remain in the distillation flask. Therefore, the reason (intramolecular H-bonding in o-nitrophenol) is the correct explanation for the assertion (their separability by steam distillation).
Quick Tip: Remember the effect of hydrogen bonding on physical properties:
\textbf{Intramolecular H-bonding (within a molecule): Decreases intermolecular forces, leading to lower boiling point, lower water solubility, and higher volatility (steam distillable). Example: o-nitrophenol.
\textbf{Intermolecular H-bonding} (between molecules): Increases intermolecular forces, leading to higher boiling point, higher water solubility, and lower volatility. Example: p-nitrophenol.

Step 1: Analyze Assertion (A):

The assertion states that cooking time is reduced in a pressure cooker. This is the entire purpose of a pressure cooker and is a well-known fact. Food cooks faster at higher temperatures. So, Assertion (A) is true.


Step 2: Analyze Reason (R):

The reason states that the boiling point of water inside the pressure cooker is lowered. The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure. A pressure cooker works by sealing the vessel, which traps the steam produced from boiling water. As more steam is trapped, the pressure inside the cooker builds up to a level higher than the atmospheric pressure.

According to the principles of phase equilibrium, a liquid's boiling point increases as the external pressure on it increases. Because the pressure inside the cooker is higher than atmospheric pressure, the water boils at a temperature significantly higher than 100°C (typically around 121°C). Therefore, the reason's statement that the boiling point is lowered is factually incorrect. So, Reason (R) is false.


Step 3: Conclusion:

The assertion (A) is true because food cooks faster at the higher temperature achieved inside the pressure cooker. The reason (R) is false because the boiling point of water is elevated, not lowered, due to the increased pressure. Since the assertion is true and the reason is false, the correct option is (C).
Quick Tip: Remember the relationship between pressure and boiling point:
\textbf{Higher Pressure \(\rightarrow\) Higher Boiling Point} (e.g., pressure cooker).
\textbf{Lower Pressure \(\rightarrow\) Lower Boiling Point} (e.g., cooking at high altitudes takes longer).
Food cooks faster because chemical reactions (cooking) are faster at higher temperatures.

Step 1: Analyze Assertion (A):

The assertion states that actinoids show irregularities in their electronic configurations. This is true. The filling of electrons in the 5f subshell is not perfectly regular. For example, the configuration of Thorium (Th, Z=90) is \([Rn] 6d^2 7s^2\) instead of the expected \([Rn] 5f^1 6d^1 7s^2\) or \([Rn] 5f^2 7s^2\). Similarly, other elements like Protactinium (Pa), Uranium (U), Neptunium (Np), and Curium (Cm) show deviations from a simple, regular filling pattern. The energies of the 5f and 6d orbitals are very close, allowing electrons to occupy either subshell, which leads to these irregularities. So, Assertion (A) is true.


Step 2: Analyze Reason (R):

The reason attributes these irregularities to the varying stability of f\(^0\) (empty), f\(^7\) (half-filled), and f\(^{14}\) (fully-filled) configurations. This principle of extra stability for empty, half-filled, and fully-filled subshells is a fundamental concept in chemistry that governs electronic configurations across the periodic table, including the f-block. For example, the configuration of Curium (Cm, Z=96) is \([Rn] 5f^7 6d^1 7s^2\), where it achieves a stable half-filled 5f\(^7\) configuration by placing an electron in the 6d orbital. Similarly, Americium (Am, Z=95) has the configuration \([Rn] 5f^7 7s^2\), again prioritizing the stable 5f\(^7\) state. The stability of the f\(^0\) state influences the configuration of early actinoids like Th. So, Reason (R) is true.


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

The irregularities mentioned in the assertion are a direct consequence of the elements adopting configurations that are energetically more favorable. The extra stability associated with empty (f\(^0\)), half-filled (f\(^7\)), and fully-filled (f\(^{14}\)) f-subshells is a major driving force that causes these deviations from the expected aufbau principle filling order. Therefore, the reason provides a correct explanation for the assertion.
Quick Tip: The cause of irregularities in the electronic configurations of both lanthanoids and actinoids is the same: the very small energy gap between the (n-2)f and (n-1)d orbitals, combined with the drive to achieve the extra stability of f\(^0\), f\(^7\), and f\(^{14}\) states.

Step 1: Analyze Assertion (A):

The assertion states that maltose is a reducing sugar. Reducing sugars are carbohydrates that can act as reducing agents because they have a free aldehyde group or a free ketone group, or exist in equilibrium with a form that has one. Maltose is known to give positive tests with Tollens' reagent and Fehling's solution, which confirms that it is a reducing sugar. So, Assertion (A) is true.


Step 2: Analyze Reason (R):

The reason explains why maltose is reducing. Maltose is a disaccharide made of two \(\alpha\)-D-glucose units linked by a C1-C4 glycosidic bond.

Let's label the two glucose units. Unit 1 is linked through its C1 carbon (the anomeric carbon) to the C4 hydroxyl group of Unit 2. The anomeric carbon of Unit 1 is involved in the glycosidic bond and is in an acetal form, so it cannot open.

However, the anomeric carbon (C1) of the second glucose unit (Unit 2) is not involved in the glycosidic linkage. This C1 carbon exists as a hemiacetal. In an aqueous solution, this hemiacetal ring can open up to form the open-chain structure, which contains a free aldehyde group. It is this free aldehyde group that is responsible for the reducing properties of maltose. So, Reason (R) is true.


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

The assertion that maltose is a reducing sugar is directly and correctly explained by the fact that one of its glucose residues has a free hemiacetal group that can open to form a free aldehyde. This aldehyde group can then be oxidized, allowing maltose to act as a reducing agent. Therefore, the reason is the correct explanation for the assertion.
Quick Tip: To determine if a disaccharide is reducing or non-reducing, look at the anomeric carbons (C1 for glucose, C2 for fructose) of all monosaccharide units.
\textbf{Reducing Sugar:} If at least one monosaccharide unit has a free anomeric carbon (i.e., it's a hemiacetal), the ring can open, and the sugar is reducing. Examples: Maltose, Lactose.
\textbf{Non-Reducing Sugar:} If the anomeric carbons of all units are involved in the glycosidic bond (i.e., they are all acetals/ketals), the rings cannot open, and the sugar is non-reducing. Example: Sucrose (C1 of glucose is linked to C2 of fructose).

Step 1: Definition of Negative Deviation:

A solution is said to show negative deviation from Raoult's law when the partial vapour pressure of each component, and hence the total vapour pressure of the solution, is less than that predicted by Raoult's law.

Mathematically, for a binary solution of components A and B:

- \(p_A < p_A^\circ x_A\)

- \(p_B < p_B^\circ x_B\)

- \(P_{Total} = p_A + p_B < p_A^\circ x_A + p_B^\circ x_B\)

This occurs when the intermolecular forces of attraction between the solute (A) and solvent (B) molecules (A-B interactions) are stronger than the intermolecular forces between the pure components (A-A and B-B interactions). Due to these stronger attractions, the escaping tendency of molecules from the solution decreases, resulting in a lower vapour pressure.

Other characteristics of such solutions are:

- Enthalpy of mixing is negative (\(\Delta H_{mix} < 0\)), i.e., the process is exothermic.

- Volume of mixing is negative (\(\Delta V_{mix} < 0\)), i.e., there is a contraction in volume on mixing.


Step 2: Example:

A classic example of a solution showing negative deviation is a mixture of chloroform (\(CHCl_3\)) and acetone (\(CH_3COCH_3\)).

In this mixture, a new hydrogen bond is formed between the hydrogen atom of chloroform and the oxygen atom of acetone, which is a stronger interaction than the dipole-dipole interactions present in pure chloroform and pure acetone.


Step 3: Type of Azeotrope:

Solutions that show a large negative deviation from Raoult's law form a maximum boiling azeotrope.

An azeotrope is a liquid mixture which has a constant boiling point and whose vapour has the same composition as the liquid. For a maximum boiling azeotrope, there is a specific composition at which the boiling point of the mixture is higher than the boiling point of either of the pure components. This occurs because the strong intermolecular forces in the solution make it more difficult for the molecules to escape into the vapour phase, thus requiring a higher temperature to boil.
Quick Tip: To remember the types of deviations and azeotropes:
\textbf{Negative Deviation}: Stronger A-B interactions \(\rightarrow\) Lower vapour pressure \(\rightarrow\) Harder to boil \(\rightarrow\) \textbf{Maximum boiling azeotrope}.
\textbf{Positive Deviation}: Weaker A-B interactions \(\rightarrow\) Higher vapour pressure \(\rightarrow\) Easier to boil \(\rightarrow\) \textbf{Minimum boiling azeotrope}.

Step 1: IUPAC Naming of the Complex:

The complex is \([Pt(en)_2Cl_2]^{2+}\).

1. Identify Ligands: The ligands are 'Cl' (chlorido) and 'en' (ethane-1,2-diamine). They are named in alphabetical order: chlorido before ethane-1,2-diamine.

2. Use Prefixes: There are two 'chlorido' ligands, so we use the prefix 'di-'. There are two 'ethane-1,2-diamine' ligands. Since the ligand name contains a numerical prefix ('di'), we use the special prefix 'bis-'. So we have dichlorido and bis(ethane-1,2-diamine).

3. Name the Central Metal: The central metal is Platinum (Pt). Since the complex ion is a cation, the metal name remains 'platinum'.

4. Determine Oxidation State: Let the oxidation state of Pt be x. 'en' is neutral (charge 0), and 'Cl' has a charge of -1. The overall charge is +2.
\[ x + 2(0) + 2(-1) = +2 \] \[ x - 2 = +2 \implies x = +4 \]
The oxidation state is written in Roman numerals as (IV).

5. Combine the parts: The full IUPAC name is
Dichloridobis(ethane-1,2-diamine)platinum(IV) ion.


Step 2: Geometrical Isomers and Optical Activity:

An octahedral complex of the type \([M(AA)_2X_2]\) (where AA is a bidentate ligand) can exist as two geometrical isomers: cis and \textit{trans.

- cis-isomer: The two 'Cl' ligands are adjacent (90\(^\circ\) apart). This isomer is asymmetric (chiral) and is therefore optically active.

- trans-isomer: The two 'Cl' ligands are opposite (180\(^\circ\) apart). This isomer possesses elements of symmetry (like a plane of symmetry), making it superimposable on its mirror image. Therefore, the trans-isomer is achiral and optically inactive.


Step 3: Structure of the Optically Inactive Isomer:

The question asks for the optically inactive isomer, which is the trans-isomer.



In this structure, the two Cl atoms are on opposite sides of the central Pt atom.
Quick Tip: For octahedral complexes of the type \([M(AA)_2X_2]\), where AA is a symmetric bidentate ligand:
The \textit{cis-isomer is always optically \textbf{active}.
The trans-isomer is always optically \textbf{inactive}.

Step 1: Deconstruct the IUPAC Name:

- Central Metal: Cobalt(III) \(\implies Co^{3+}\).

- Ligands:
- 'Pentaammine' \(\implies\) Five ammine (\(NH_3\)) ligands. \(NH_3\) is neutral.
- 'carbonato' \(\implies\) One carbonate (\(CO_3^{2-}\)) ligand. The charge is -2.

- Counter-ion: 'chloride' \(\implies Cl^-\).


Step 2: Assemble the Coordination Sphere and Determine its Charge:

The coordination sphere contains the central metal and the ligands. We write it as \([Co(NH_3)_5(CO_3)]\).

The charge on this complex ion is calculated as:

Charge = (Charge of Co) + 5 \(\times\) (Charge of \(NH_3\)) + (Charge of \(CO_3\))

Charge = (+3) + 5 \(\times\) (0) + (-2) = +1.

So, the complex ion is \([Co(NH_3)_5(CO_3)]^+\).


Step 3: Write the Final Formula:

To neutralize the +1 charge of the complex ion, one chloride ion (\(Cl^-\)) is needed as the counter-ion.

The final formula is \([Co(NH_3)_5(CO_3)]Cl\).
Quick Tip: When writing a formula from an IUPAC name, work from the inside out. First, assemble the metal and ligands inside the square brackets. Then, calculate the charge of this coordination sphere. Finally, add the necessary number of counter-ions outside the brackets to make the overall compound electrically neutral.

Step 1: Understanding Linkage Isomerism:

Linkage isomerism occurs when a coordination compound contains an ambidentate ligand, which is a ligand that can bond to the central metal ion through two different donor atoms. The given complex has the ligand \(NO_2^-\), which is ambidentate.


Step 2: Identifying the Isomers:

- In the given complex, \([Co(NH_3)_5(NO_2)]Cl_2\), the ligand is written as -NO\(_2\). This denotes that it is bonded through the Nitrogen atom. This form is called nitro.

- The linkage isomer will have the ligand bonded through an Oxygen atom. This form is written as -ONO and is called nitrito.

So, the formula of the linkage isomer is \([Co(NH_3)_5(ONO)]Cl_2\).


Step 3: Naming the Linkage Isomer:

We need to name the complex \([Co(NH_3)_5(ONO)]Cl_2\).

1. Ligands: 'ammine' and 'nitrito'. Alphabetically, ammine comes first.

2. Prefixes: Five ammine ligands \(\implies\) 'pentaammine'. One nitrito ligand.

3. Metal: Cobalt (Co). Since the complex is a cation, the name is 'cobalt'.

4. Oxidation State: Let the oxidation state of Co be x. \(NH_3\) is neutral (0), and ONO\(^-\) has a charge of -1. The two Cl\(^-\) counter-ions mean the complex ion has a +2 charge.
\[ x + 5(0) + 1(-1) = +2 \implies x = +3 \]
The oxidation state is (III).

5. Counter-ion: 'chloride'.

6. Full Name: Pentaamminenitritocobalt(III) chloride. (To be more specific, it can be named Pentaammine(nitrito-O)cobalt(III) chloride).
Quick Tip: Memorize the common ambidentate ligands and their names for both bonding modes:
M-NO\(_2\) (nitro) vs. M-ONO (nitrito)
M-CN (cyano) vs. M-NC (isocyano)
M-SCN (thiocyanato) vs. M-NCS (isothiocyanato)

Condition for Bimolecular Reaction to be First Order:

A bimolecular reaction can behave as a first-order reaction (this type of reaction is called a pseudo-first-order reaction) under a specific condition: one of the two reactants must be present in a large excess compared to the other.

When one reactant is in large excess, its concentration remains practically constant throughout the course of the reaction. The reaction rate then becomes dependent only on the concentration of the reactant present in the smaller amount.

For a reaction: A + B \(\rightarrow\) Products, the rate law is Rate = k[A][B].

If [B] is very large (\([B] \gg [A]\)), then [B] \(\approx\) constant. The rate law can be written as:

Rate = k'[A], where k' = k[B].

The reaction is bimolecular but follows first-order kinetics.


Example:

The acid-catalyzed hydrolysis of an ester, such as the hydrolysis of ethyl acetate.
\[ CH_3COOC_2H_5 (Ethyl Acetate) + H_2O \xrightarrow{H^+} CH_3COOH (Acetic Acid) + C_2H_5OH (Ethanol) \]
In this reaction, water acts as one of the reactants. If the reaction is carried out in an aqueous solution, water is present in such a large excess that its concentration does not change significantly. The rate of reaction thus depends only on the concentration of the ester.

Rate = k[\(CH_3COOC_2H_5\)]


Reactions Where Order and Molecularity are the Same:

The order and molecularity of a reaction have the same value only for elementary reactions.

An elementary reaction is a reaction that occurs in a single step. In such reactions, the rate law can be written directly from the stoichiometry of the balanced equation. The order with respect to each reactant is equal to its stoichiometric coefficient, and the overall order is equal to the molecularity (the number of species colliding in that single step). For complex reactions, which occur via a multi-step mechanism, the overall order is determined by the slowest step (rate-determining step) and is generally not equal to the sum of stoichiometric coefficients.
Quick Tip: \textbf{Pseudo-first-order:} Think of "hiding" a reactant's concentration by using it in large excess (often the solvent). Examples include hydrolysis of esters and inversion of cane sugar.
\textbf{Order vs. Molecularity:} Molecularity is a theoretical concept for elementary steps (can only be small integers 1, 2, or 3). Order is an experimental value (can be zero, fractional, or integer) that applies to the overall reaction. They are equal only for single-step elementary reactions.

Starch is a polymer of \(\alpha\)-D-glucose and consists of two main components, Amylose and Amylopectin. Here are two key differences between them:


1. Structure:

- Amylose: It is a linear (unbranched) polymer. The \(\alpha\)-D-glucose units are linked together in a long, straight chain, which tends to form a helical structure.

- Amylopectin: It is a highly branched polymer. It consists of a main chain of \(\alpha\)-D-glucose units with numerous branches attached to it.


2. Glycosidic Linkages:

- Amylose: The glucose units are joined only by \(\alpha\)-1,4-glycosidic linkages.

- Amylopectin: It contains two types of linkages. The glucose units in the linear chains are connected by \(\alpha\)-1,4-glycosidic linkages, while the branch points are formed by \(\alpha\)-1,6-glycosidic linkages. These branches occur typically every 24-30 glucose units.


(Other possible differences include solubility and reaction with iodine)

- Solubility in Water: Amylose is soluble in hot water, whereas Amylopectin is insoluble.

- Reaction with Iodine: Amylose gives a characteristic deep blue colour with iodine, while Amylopectin gives a reddish-brown or purple colour.
Quick Tip: A simple way to remember the difference: think of \textbf{Amylose as a long, single "Angle" hair (linear) and \textbf{A}mylo\textbf{pectin} as a "Pecten" comb (branched). The branches in amylopectin are formed by the \(\alpha\)-1,6 linkage, which is the key structural difference.

Part 1: Why Haloarenes are Less Reactive

Haloarenes are significantly less reactive than haloalkanes towards nucleophilic substitution reactions due to the following reasons:


Resonance Effect: The lone pair of electrons on the halogen atom is in conjugation with the \(\pi\)-electrons of the benzene ring. This delocalization gives the Carbon-Halogen (C-X) bond a partial double bond character. As a result, the C-X bond in haloarenes is stronger and shorter than in haloalkanes, making it more difficult to break.

Difference in Hybridization of Carbon Atom: The carbon atom of the C-X bond in haloarenes is sp\(^2\)-hybridized, while in haloalkanes it is sp\(^3\)-hybridized. An sp\(^2\)-hybridized orbital has more s-character, is more electronegative, and forms shorter, stronger bonds than an sp\(^3\)-hybridized orbital. This further strengthens the C-X bond.

Instability of Phenyl Cation: In a self-ionization pathway (\(S_N1\)-like), the departure of the halide ion would leave behind a highly unstable phenyl cation. This pathway is not favored.

Repulsion: The incoming nucleophile is electron-rich. It experiences electrostatic repulsion from the electron-rich benzene ring, making the attack difficult.



Part 2: Effect of Nitro Group on Reactivity

The presence of a strong electron-withdrawing group (EWG) like the nitro group (\(-NO_2\)) at the ortho- and \textit{para- positions increases the reactivity of haloarenes towards nucleophilic substitution.

Reason:

The mechanism for this reaction involves the attack of the nucleophile on the carbon atom bearing the halogen, forming a resonance-stabilized carbanion intermediate called a Meisenheimer complex.


The electron-withdrawing nitro group pulls electron density from the benzene ring through both the inductive effect (-I) and the resonance effect (-R).

When the -NO\(_2\) group is at the ortho or para position, it can effectively delocalize the negative charge of the carbanion intermediate through resonance. One of the resonance structures places the negative charge directly on the carbon atom attached to the nitro group, allowing the nitro group to accommodate the charge very effectively.

This extensive delocalization stabilizes the intermediate carbanion, which in turn lowers the activation energy for its formation. A lower activation energy leads to a faster reaction rate.


The effect is not observed if the -NO\(_2\) group is at the meta-position, as the negative charge cannot be delocalized onto the nitro group through resonance from there.
Quick Tip: Remember the mechanism for Nucleophilic Aromatic Substitution (NAS): it's an \textbf{addition-elimination pathway forming a carbanion intermediate.
\textbf{Deactivating} groups for electrophilic substitution (\(e.g., -NO_2\)) are \textbf{activating} for nucleophilic substitution because they stabilize the negatively charged intermediate.
This activation is only effective from the \textbf{ortho and para} positions, where resonance delocalization is possible.

(a) Essential and Non-essential Amino Acids


Quick Tip: To remember these pairs, focus on their core function or structure:
\textbf{Essential vs. Non-essential:} Think about diet. "Essential" means essential to eat.

(b) Peptide Linkage and Glycosidic Linkage



Quick Tip: To remember these pairs, focus on their core function or structure:
\textbf{Peptide vs. Glycosidic:} Peptide links make \textbf{P}roteins. Glycosidic links involve \textbf{G}lucose (sugars).

(c) Fibrous and Globular Proteins



Quick Tip: To remember these pairs, focus on their core function or structure:
\textbf{Fibrous vs. Globular:} Fibrous means \textbf{f}ibre-like (structural, insoluble). Globular means \textbf{g}lobe-like (functional, soluble).

Reasoning:


1. Deactivation of the Benzene Ring: The carboxylic acid group (-COOH) is a strong electron-withdrawing group. It deactivates the benzene ring towards electrophilic aromatic substitution. Friedel-Crafts reaction (both alkylation and acylation) is an electrophilic substitution, which requires an activated or at least a neutral benzene ring to proceed. The deactivated ring in benzoic acid is not nucleophilic enough to attack the carbocation or acylium ion electrophile.


2. Reaction with Catalyst: The catalyst used in Friedel-Crafts reactions is a Lewis acid, typically anhydrous \(AlCl_3\). The -COOH group in benzoic acid has a lone pair of electrons on its oxygen atom, which makes it a Lewis base. The Lewis acid catalyst (\(AlCl_3\)) reacts with the Lewis basic -COOH group to form a salt. This further deactivates the ring by putting a positive charge on the oxygen atom attached to the ring, making it even more strongly deactivating.
\[ C_6H_5COOH + AlCl_3 \rightarrow C_6H_5COO^-AlCl_3H^+ \]
Because of these two reasons, benzoic acid does not undergo the Friedel-Crafts reaction.
Quick Tip: \textbf{Friedel-Crafts Limitations:} Strongly deactivating groups (-NO2, -NR3+, -COOH, -SO3H) and aniline (-NH2) do not undergo Friedel-Crafts reactions.

\(CH_3CHO\) towards addition of HCN.


Reasoning:

The addition of HCN to an aldehyde or ketone is a nucleophilic addition reaction. The reactivity depends on two main factors: electronic factors and steric factors.


1. Electronic Factors: The reaction is initiated by the attack of the nucleophile (\(CN^-\)) on the electrophilic carbonyl carbon. Any group that increases the positive charge (electrophilicity) on the carbonyl carbon will increase reactivity. In acetaldehyde (\(CH_3CHO\)), the methyl group (\(CH_3\)) is an electron-donating group (+I effect). It reduces the positive charge on the carbonyl carbon, making it less electrophilic and thus less reactive. Formaldehyde (HCHO) has only hydrogen atoms attached, which have a negligible electronic effect compared to the methyl group.


2. Steric Factors: The carbonyl carbon in aldehydes is \(sp^2\) hybridized (trigonal planar). During the nucleophilic attack, it changes to \(sp^3\) hybridized (tetrahedral). A bulky group attached to the carbonyl carbon will hinder the approach of the nucleophile. In acetaldehyde, the methyl group is bulkier than the hydrogen atom in formaldehyde. This steric hindrance makes the attack of the \(CN^-\) nucleophile more difficult in acetaldehyde compared to formaldehyde.

Due to both the electron-donating effect of the methyl group and greater steric hindrance, formaldehyde (HCHO) is more reactive than acetaldehyde (\(CH_3CHO\)) towards nucleophilic addition.
Quick Tip: \textbf{Aldehyde/Ketone Reactivity:} Reactivity in nucleophilic addition decreases as steric hindrance and electron-donating effects increase. Order: HCHO > RCHO > R2C=O.

Reasoning:


The question refers to acrylic acid (\(CH_2=CH-COOH\)). The vinyl group (\(CH_2=CH-\)) has a carbon atom that is \(sp^2\) hybridized.

1. Effect of Hybridization: The acidity of a carboxylic acid depends on the stability of its conjugate base (carboxylate ion, \(R-COO^-\)). An \(sp^2\) hybridized carbon is more electronegative than an \(sp^3\) hybridized carbon (like in propanoic acid, \(CH_3CH_2COOH\)) because it has more s-character (33.3% vs 25%).


2. Inductive Effect: This higher electronegativity of the \(sp^2\) carbon causes it to exert an electron-withdrawing inductive effect (-I effect). This -I effect pulls electron density away from the -COOH group, which helps to disperse the negative charge of the carboxylate anion (\(CH_2=CH-COO^-\)) after the proton is lost. This stabilization of the conjugate base increases the acidity of the acid.


3. Resonance Effect: While the vinyl group can participate in resonance, this effect is less dominant than the inductive effect in determining acidity here. The delocalization of the pi electrons of the C=C bond into the C=O group is possible, but the primary factor stabilizing the conjugate base is the electronegativity (inductive effect) of the \(sp^2\) carbon.


Therefore, due to the electron-withdrawing inductive effect of the \(sp^2\) hybridized vinyl carbon, acrylic acid is more acidic than its saturated analogue, propanoic acid.
Quick Tip: \textbf{Acidity Factors:} Acidity is increased by electron-withdrawing groups (-I, -M effects) which stabilize the conjugate base. Remember that \(sp\) > \(sp^2\) > \(sp^3\) in terms of electronegativity.

(a) Anisole reacts with HI

Step 1: Understanding the Reaction:

This is the cleavage of an ether (anisole) with a hydrogen halide (HI). The \(O-CH_3\) bond is an alkyl-aryl ether bond. The oxygen is bonded to a methyl group and a phenyl group.

Step 2: Reaction Mechanism (Cleavage of Ethers):

The reaction proceeds via protonation of the ether oxygen, followed by a nucleophilic attack by the iodide ion (\(I^-\)). The attack can occur at the methyl carbon or the phenyl carbon.

The bond between the oxygen and the \(sp^2\)-hybridized carbon of the benzene ring is very strong due to resonance (partial double bond character). Cleavage of this bond is difficult.

Therefore, the iodide ion attacks the less hindered methyl group via an \(\(S_N2\)\) mechanism.
\[ \text{C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I \]
Products: The products are Phenol and Iodomethane (Methyl iodide).






(b) Phenol reacts with conc. \(HNO_3\)

Step 1: Understanding the Reaction:

This is the nitration of phenol using concentrated nitric acid. Concentrated \(HNO_3\) is a strong nitrating agent.

Step 2: Reaction Mechanism (Electrophilic Aromatic Substitution):

The -OH group in phenol is a strongly activating and ortho-, para-directing group. When a strong nitrating agent like conc. \(HNO_3\) is used, the reaction is vigorous and leads to polysubstitution at all available ortho and para positions.

The electrophile is the nitronium ion (\(NO_2^+\)).
\[ C_6H_5OH + 3HNO_3 (conc.) \xrightarrow{conc. H_2SO_4} C_6H_2(NO_2)_3OH + 3H_2O \]
Product: The product is 2,4,6-Trinitrophenol, which is commonly known as Picric acid.






(c) Cyclohexylmagnesium bromide reacts with HCHO followed by hydrolysis

Step 1: Understanding the Reaction:

This is a two-step reaction. The first step is the reaction of a Grignard reagent (Cyclohexylmagnesium bromide) with an aldehyde (Formaldehyde, HCHO). The second step is the acidic hydrolysis of the intermediate.

Step 2: Reaction Mechanism:

Step I (Nucleophilic Addition): The Grignard reagent acts as a source of a strong nucleophile, the cyclohexyl carbanion (\(C_6H_{11}^-\)). This nucleophile attacks the electrophilic carbonyl carbon of formaldehyde.
\[ C_6H_{11}MgBr + HCHO \rightarrow C_6H_{11}CH_2O^-MgBr^+ \]
This forms an alkoxide-magnesium bromide salt as an intermediate.

Step II (Hydrolysis): Acidic hydrolysis protonates the alkoxide to form an alcohol.
\[ C_6H_{11}CH_2O^-MgBr^+ + H^+/H_2O \rightarrow C_6H_{11}CH_2OH + Mg(OH)Br \]
Product: The final product is Cyclohexylmethanol. Since the reaction is with formaldehyde, a primary alcohol is formed.


Quick Tip: \textbf{Ether Cleavage:} For alkyl-aryl ethers, cleavage with HX always gives phenol and an alkyl halide.
\textbf{Nitration of Phenol:} Dilute \(HNO_3\) gives a mixture of o- and p-nitrophenol. Concentrated \(HNO_3\) gives picric acid (2,4,6-trinitrophenol).
\textbf{Grignard Reactions:} Formaldehyde (\(HCHO\)) with a Grignard reagent gives a primary alcohol. Any other aldehyde gives a secondary alcohol. A ketone gives a tertiary alcohol.

Step 1: Understanding the Question:

We are given data for a first-order gas-phase reaction and asked to calculate the rate constant, k. The data provided is the total pressure of the system at different times, not the partial pressure of the reactant. We need to relate the total pressure to the partial pressure of the reactant \(N_2O_5\).


Step 2: Key Formula or Approach:

The integrated rate law for a first-order reaction is:
\[ k = \frac{2.303}{t} \log \frac{P_0}{P_t} \]
Where:

- \(k\) is the rate constant.

- \(t\) is the time.

- \(P_0\) is the initial partial pressure of the reactant.

- \(P_t\) is the partial pressure of the reactant at time \(t\).



We need to derive an expression for the partial pressure of \(N_2O_5\) (\(P_t\)) in terms of the total pressure (\(P_{total}\)).


Step 3: Detailed Calculation:

Let's analyze the pressure changes based on the stoichiometry of the reaction:
\[ \begin{array}{lcccc} Reaction: & 2N_2O_5(g) & \rightarrow & 2N_2O_4(g) & + & O_2(g)
Initial Pressure (t=0): & P_0 & & 0 & & 0
Pressure at time t: & P_0 - 2x & & 2x & & x
\end{array} \]


From the given data, the initial pressure \(P_0\) is the total pressure at t=0, so \(P_0 = 0.5\) atm.

The total pressure at time \(t\), \(P_{total}\), is the sum of the partial pressures of all gases:
\[ P_{total} = (P_0 - 2x) + (2x) + (x) = P_0 + x \]
So, we can find \(x\) in terms of \(P_{total}\) and \(P_0\):
\[ x = P_{total} - P_0 \]


The partial pressure of \(N_2O_5\) at time \(t\), denoted as \(P_t\), is:
\[ P_t = P_0 - 2x \]
Substitute the value of \(x\):
\[ P_t = P_0 - 2(P_{total} - P_0) = P_0 - 2P_{total} + 2P_0 = 3P_0 - 2P_{total} \]


Now, let's use the given data:

- At \(t = 0\), \(P_0 = 0.5\) atm.

- At \(t = 100\) s, \(P_{total} = 0.625\) atm.



Calculate \(P_t\) at \(t = 100\) s:
\[ P_t (at 100 s) = 3(0.5) - 2(0.625) = 1.5 - 1.25 = 0.25 atm \]


Now, substitute the values into the first-order rate equation:
\[ k = \frac{2.303}{t} \log \frac{P_0}{P_t} \] \[ k = \frac{2.303}{100} \log \frac{0.5}{0.25} \] \[ k = \frac{2.303}{100} \log(2) \]
Given that log(2) = 0.3010:
\[ k = \frac{2.303}{100} \times 0.3010 \] \[ k = \frac{0.6932}{100} \] \[ k = 6.932 \times 10^{-3} s^{-1} \]

Step 4: Final Answer:

The rate constant for the reaction is \(6.932 \times 10^{-3} s^{-1}\).
Quick Tip: For gas-phase reactions where total pressure is given, always set up an ICE (Initial, Change, Equilibrium/End) table using partial pressures. Derive a relationship between the partial pressure of the reactant (\(P_t\)) and the total pressure (\(P_{total}\)). This is a common type of problem in chemical kinetics. The final formula \(P_t = 3P_0 - 2P_{total}\) is specific to this reaction's stoichiometry.

Step 1: Understanding the Question and Deducing Structure of (A):

We are given a series of reactions starting with a primary alkyl halide (A), \(C_4H_9I\).

First, let's identify the possible structures of primary alkyl halides with the formula \(C_4H_9I\).

There are two possibilities:

1. 1-Iodobutane (n-butyl iodide): \(CH_3CH_2CH_2CH_2I\)

2. 1-Iodo-2-methylpropane (isobutyl iodide): \((CH_3)_2CHCH_2I\)



The problem states that when (A) reacts with Na/dry ether (Wurtz reaction) to form (D), \(C_8H_{18}\), the product is different from the one formed with n-butyl iodide. The Wurtz reaction of n-butyl iodide gives n-octane.

This means (A) cannot be n-butyl iodide. Therefore, (A) must be 1-Iodo-2-methylpropane.

Structure of (A): \((CH_3)_2CHCH_2I\)


Step 2: Deducing Structures of (B), (C), and (D):

Reaction 1: (A) + alcoholic KOH \(\rightarrow\) (B)

This is a dehydrohalogenation reaction (E2 elimination). Alcoholic KOH is a strong base that removes H and I to form an alkene.
\[ (CH_3)_2CHCH_2I \xrightarrow{alc. KOH, \Delta} (CH_3)_2C=CH_2 + KI + H_2O \]
So, (B) is 2-Methylpropene.

Structure of (B): \((CH_3)_2C=CH_2\)



Reaction 2: (B) + HI \(\rightarrow\) (C)

This is the addition of HI to an alkene, which follows Markovnikov's rule. The H atom adds to the carbon with more hydrogen atoms, and the I atom adds to the more substituted carbon.
\[ (CH_3)_2C=CH_2 + HI \rightarrow (CH_3)_2C(I)CH_3 \]
So, (C) is 2-Iodo-2-methylpropane (tert-butyl iodide).

Structure of (C): \((CH_3)_3CI\). This is a tertiary alkyl halide and is an isomer of (A). This confirms our deduction.



Reaction 3: (A) + Na/dry ether \(\rightarrow\) (D)

This is the Wurtz reaction, where two molecules of the alkyl halide couple.
\[ 2(CH_3)_2CHCH_2I + 2Na \xrightarrow{dry ether} (CH_3)_2CHCH_2CH_2CH(CH_3)_2 + 2NaI \]
So, (D) is 2,5-Dimethylhexane.

Structure of (D): \((CH_3)_2CHCH_2CH_2CH(CH_3)_2\)


Step 3: Summary of Structures and Required Chemical Equation:

- (A): 1-Iodo-2-methylpropane, \((CH_3)_2CHCH_2I\)

- (B): 2-Methylpropene, \((CH_3)_2C=CH_2\)

- (C): 2-Iodo-2-methylpropane, \((CH_3)_3CI\)

- (D): 2,5-Dimethylhexane, \((CH_3)_2CHCH_2CH_2CH(CH_3)_2\)



The chemical equation for the reaction of compound (A) with alcoholic KOH is:
\[ (CH_3)_2CHCH_2I + KOH (alcoholic) \xrightarrow{\Delta} (CH_3)_2C=CH_2 + KI + H_2O \] Quick Tip: Organic road-map problems are solved step-by-step. Start with the most definitive clue. In this case, the Wurtz reaction product being different from that of n-butyl iodide was the key to identifying the starting material (A) as the branched isomer. Always remember the standard reagents: alcoholic KOH for elimination and aqueous KOH for substitution.

Step 1: Understanding the Question:

The question asks for the name of the battery commonly used in inverters and the electrochemical reactions that occur at its anode and cathode during its operation (discharging).


Step 2: Identifying the Cell:

The cell commonly used in inverters and automobiles is the Lead-storage battery. It is a secondary (rechargeable) cell.


Step 3: Writing the Electrode Reactions:

When the battery is in use (discharging), it acts as a galvanic cell, producing electrical energy. The electrolyte is typically 38% sulfuric acid (\(H_2SO_4\)) by mass.



At the Anode (Negative Electrode):

The anode is made of spongy lead (Pb). Oxidation occurs here. Lead is oxidized to lead(II) ions, which then react with sulfate ions from the electrolyte to form insoluble lead(II) sulfate.

The reaction is:
\[ Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^- \]


At the Cathode (Positive Electrode):

The cathode is a grid of lead packed with lead dioxide (\(PbO_2\)). Reduction occurs here. Lead dioxide is reduced to lead(II) ions in the presence of H+ ions, and these Pb(II) ions also react with sulfate ions to form lead(II) sulfate.

The reaction is:
\[ PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l) \]


Overall Reaction:

Combining the anode and cathode reactions, the overall cell reaction during discharge is:
\[ Pb(s) + PbO_2(s) + 4H^+(aq) + 2SO_4^{2-}(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l) \]
Or, written with sulfuric acid:
\[ Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l) \] Quick Tip: Remember the key components of a lead-storage battery:
\textbf{Anode:} Spongy Lead (Pb)
\textbf{Cathode:} Lead dioxide (\(PbO_2\))
\textbf{Electrolyte:} Sulfuric acid (\(H_2SO_4\))
During discharge, both electrodes get coated with lead sulfate (\(PbSO_4\)), and the concentration of \(H_2SO_4\) decreases. The reverse happens during charging.

Step 1: Understanding the Question:

We need to explain the products formed during the electrolysis of an aqueous NaCl solution by considering the possible reactions at the cathode and anode and their respective electrode potentials.


Step 2: Reactions at the Cathode (Reduction):

In an aqueous solution of NaCl, there are two species that can be reduced at the cathode: \(Na^+\) ions and \(H_2O\) molecules.

The possible reduction reactions are:

(i) \( Na^+(aq) + e^- \rightarrow Na(s) \), with \( E^{\circ} = -2.71 V \)

(ii) \( 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \), with \( E \) (at pH=7) = -0.83 V (or standard E° = -0.83 V, often simplified this way in textbooks).



The reaction with a higher (less negative) reduction potential is preferred. Comparing the potentials, -0.83 V is much higher than -2.71 V.

Therefore, the reduction of water is thermodynamically favored, and \(H_2\) gas is evolved at the cathode.


Step 3: Reactions at the Anode (Oxidation):

Similarly, there are two species that can be oxidized at the anode: \(Cl^-\) ions and \(H_2O\) molecules.

The possible oxidation reactions are:

(i) \( 2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- \), with \( E^{\circ}_{ox} = -1.36 V \) (since \( E^{\circ}_{red} = +1.36 V \))

(ii) \( 2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- \), with \( E^{\circ}_{ox} = -1.23 V \) (since \( E^{\circ}_{red} = +1.23 V \))



Based on standard potentials, the oxidation of water (requiring -1.23 V) seems more favorable than the oxidation of chloride ions (requiring -1.36 V). However, the oxidation of water to produce oxygen is a kinetically slow process and requires a higher voltage than predicted, a phenomenon known as overpotential or overvoltage.

Due to the overpotential of oxygen, the potential required for water oxidation becomes higher than that for chloride ion oxidation. Consequently, \(Cl^-\) ions are preferentially oxidized, and \(Cl_2\) gas is evolved at the anode.


Step 4: Overall Reaction:

We combine the preferred reactions from the cathode and anode.

Cathode: \( 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \)

Anode: \( 2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- \)

The spectator ions are \(Na^+\). Combining them with \(OH^-\) gives NaOH.

The overall reaction is:
\[ 2NaCl(aq) + 2H_2O(l) \xrightarrow{electrolysis} 2NaOH(aq) + H_2(g) + Cl_2(g) \] Quick Tip: For electrolysis in aqueous solutions, always compare the reduction potentials of the cation and water at the cathode, and the oxidation potentials of the anion and water at the anode.
Remember the concept of \textbf{overpotential}, which is crucial for explaining why \(Cl_2\) is produced instead of \(O_2\) during the electrolysis of brine (concentrated NaCl solution).

Step 1: Understanding the Question:

We are given the mass of solute (acetic acid) and solvent (benzene), the observed depression in freezing point (\(\Delta T_f\)), and the molal freezing point depression constant (\(K_f\)) for benzene. We need to calculate the percentage of association of acetic acid, given that it forms a dimer in benzene.


Step 2: Key Formula or Approach:

The depression in freezing point for a solute that undergoes association or dissociation is given by the modified formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:


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

\(K_f\) is the molal depression constant.

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


The van 't Hoff factor \(i\) is related to the degree of association (\(\alpha\)) by the formula:
\[ i = 1 + \left(\frac{1}{n} - 1\right)\alpha \]
Where \(n\) is the number of molecules that associate (for a dimer, \(n=2\)).


Step 3: Detailed Calculation:

Part A: Calculate the theoretical molality (m):

Moles of acetic acid = \(\frac{Mass}{Molar Mass} = \frac{0.3 \, g}{60 \, g/mol} = 0.005 \, mol\)

Mass of solvent (benzene) = 30 g = 0.030 kg

Molality (m) = \(\frac{Moles of solute}{Mass of solvent in kg} = \frac{0.005 \, mol}{0.030 \, kg} = 0.1667 \, mol/kg\)


Part B: Calculate the experimental van 't Hoff factor (i):

Using the depression in freezing point formula: \(\Delta T_f = i \cdot K_f \cdot m\)

Given: \(\Delta T_f = 0.45^\circC = 0.45 \, K\) (since a change in °C is equal to a change in K)
\(K_f = 5.12 \, K kg/mol\)
\[ 0.45 = i \times 5.12 \times 0.1667 \] \[ i = \frac{0.45}{5.12 \times 0.1667} = \frac{0.45}{0.8533} \approx 0.5274 \]

Part C: Calculate the degree of association (\(\alpha\)):

Acetic acid forms a dimer, so \(n=2\).
\[ i = 1 + \left(\frac{1}{n} - 1\right)\alpha \] \[ 0.5274 = 1 + \left(\frac{1}{2} - 1\right)\alpha \] \[ 0.5274 = 1 + (-0.5)\alpha \] \[ 0.5\alpha = 1 - 0.5274 \] \[ 0.5\alpha = 0.4726 \] \[ \alpha = \frac{0.4726}{0.5} = 0.9452 \]

Part D: Calculate the percentage association:

Percentage association = \(\alpha \times 100 = 0.9452 \times 100 = 94.52%\)


Step 4: Final Answer:

The percentage association of acetic acid in benzene is approximately 94.5%.
Quick Tip: For problems involving colligative properties, always first calculate the theoretical value (e.g., molality). Then use the experimental data (\(\Delta T_f\)) to find the van't Hoff factor (\(i\)). Finally, use the formula relating \(i\) to the degree of dissociation/association (\(\alpha\)) to find the answer. Remember, for association \(i < 1\), and for dissociation \(i > 1\).

(i) Ethanoic acid to methanamine

This conversion can be achieved in multiple steps. A common route involves the formation of an amide followed by Hoffmann bromamide degradation, which reduces the carbon chain by one.

Step 1: Conversion of Ethanoic acid to Ethanamide.

Ethanoic acid is first reacted with ammonia (\(NH_3\)) to form an ammonium salt, which upon heating dehydrates to form ethanamide.
\[ CH_3COOH + NH_3 \rightarrow [CH_3COO^-NH_4^+] \xrightarrow{\Delta} CH_3CONH_2 + H_2O \]
Step 2: Hoffmann Bromamide Degradation.

Ethanamide is then treated with bromine in the presence of an aqueous or ethanolic solution of sodium hydroxide. This reaction converts the amide into a primary amine with one carbon atom less than the parent amide.
\[ CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O \]
The final product is Methanamine.


(ii) Propanenitrile to 1-aminopropane

This conversion involves the reduction of the nitrile (-CN) group to a primary amine (\(-CH_2NH_2\)) group.

Step 1: Reduction of the Nitrile.

Propanenitrile can be reduced using strong reducing agents like Lithium Aluminium Hydride (\(LiAlH_4\)) in ether, or by catalytic hydrogenation using hydrogen gas (\(H_2\)) with a Nickel (Ni), Palladium (Pd), or Platinum (Pt) catalyst.

Using Catalytic Hydrogenation:
\[ CH_3CH_2C\equivN + 2H_2 \xrightarrow{Ni} CH_3CH_2CH_2NH_2 \]
The product is 1-Aminopropane (or Propan-1-amine).
Quick Tip: \textbf{Step-down reactions} (decreasing carbon atoms): Hoffmann bromamide degradation is a key reaction for converting amides to amines with one less carbon.
\textbf{Step-up/same-chain reactions}: Reduction of nitriles (\(LiAlH_4\) or \(H_2\)/\(Ni\)) is an excellent method to prepare primary amines without changing the number of carbon atoms.

Step 1: Understanding pK\(_b\) and Basicity:

The pK\(_b\) value is a measure of the basicity of a substance. It is defined as the negative logarithm of the base dissociation constant (\(K_b\)).
\[ pK_b = -\log_{10}(K_b) \]
A stronger base has a larger \(K_b\) value and a smaller pK\(_b\) value. The question asks why aniline has a higher pK\(_b\) value, which means it is a weaker base than methylamine.


Step 2: Analyzing the Structure of Aniline (\(C_6H_5NH_2\)):

In aniline, the lone pair of electrons on the nitrogen atom is not fully available for donation to a proton. This is because the lone pair is delocalized into the benzene ring through resonance. The lone pair participates in the \(\pi\)-electron system of the ring.

The resonance structures of aniline show that the electron density on the nitrogen atom is decreased, making it less available for protonation.


This delocalization stabilizes the aniline molecule but makes it a weaker base.


Step 3: Analyzing the Structure of Methylamine (\(CH_3NH_2\)):

In methylamine, the nitrogen atom is attached to a methyl group (\(-CH_3\)). The methyl group is an electron-donating group due to its positive inductive effect (+I effect).

This +I effect increases the electron density on the nitrogen atom, making the lone pair more readily available for donation to a proton. This enhanced electron density makes methylamine a stronger base.


Step 4: Conclusion:

Because the lone pair in aniline is delocalized by resonance, making it less available, aniline is a much weaker base than methylamine, where the lone pair availability is enhanced by the +I effect of the methyl group. A weaker base has a higher pK\(_b\) value.
Quick Tip: When comparing the basicity of amines:
\textbf{Aromatic amines (like aniline)} are generally much weaker bases than \textbf{aliphatic amines} due to the delocalization of the nitrogen lone pair into the aromatic ring (resonance effect).
\textbf{Alkyl groups} attached to nitrogen increase basicity due to their +I (electron-donating) effect.

Step 1: Understanding Basicity in Aqueous Solution:

The basic strength of alkylamines in an aqueous solution is determined by a combination of three factors:

1. Inductive Effect (+I effect): Alkyl groups are electron-donating, which increases the electron density on the nitrogen atom and enhances basicity. Based on this, the order should be: tertiary \(>\) secondary \(>\) primary. \((CH_3)_3N > (CH_3)_2NH > CH_3NH_2\).

2. Solvation Effect (Hydration): The conjugate acid formed after the amine accepts a proton (\(RNH_3^+, R_2NH_2^+, R_3NH^+\)) is stabilized by hydrogen bonding with water molecules. Greater the number of hydrogen atoms on nitrogen in the cation, the more extensive the hydrogen bonding and the greater the stability of the cation. Based on this, the order of stability of the conjugate acid is: primary \(>\) secondary \(>\) tertiary. This implies the basicity order: primary \(>\) secondary \(>\) tertiary.

3. Steric Hindrance: Bulkier alkyl groups around the nitrogen atom can hinder the approach of a proton and also hinder the solvation of the conjugate acid. This effect increases with the number of alkyl groups and reduces basicity. Based on this, the order should be: primary \(>\) secondary \(>\) tertiary.


Step 2: Combining the Effects for Methylamines:

For methylamines, the three effects combine to give an overall order.

- Inductive effect favors tertiary amine.

- Solvation and Steric effects favor primary amine.

The secondary amine, \((CH_3)_2NH\), represents the best balance between the opposing factors. The +I effect is stronger than in the primary amine, and the steric hindrance/solvation is more favorable than in the tertiary amine. The two +I effects of the methyl groups make the secondary amine more basic than the primary amine. The poor solvation and significant steric hindrance of the conjugate acid of the tertiary amine make it less basic than the secondary amine. In fact, for methyl groups, the tertiary amine becomes even less basic than the primary amine.


Step 3: Final Order:

The combined result of these effects leads to the following experimental order of basic strength in aqueous solution for methylamines:
\[ (CH_3)_2NH > CH_3NH_2 > (CH_3)_3N \]
Therefore, the increasing order of basic strength is:

\((CH_3)_3N < CH_3-NH_2 < (CH_3)_2NH\)
Quick Tip: The order of basicity of alkylamines is different in the gaseous phase and aqueous phase.
\textbf{Gaseous Phase:} Only the inductive effect matters. Order: \(3^\circ > 2^\circ > 1^\circ\).
\textbf{Aqueous Phase:} A combination of effects matters. For \textbf{Methyl} groups: \(2^\circ > 1^\circ > 3^\circ\). For \textbf{Ethyl} groups: \(2^\circ > 3^\circ > 1^\circ\). This is a very important concept to remember for competitive exams.

Step 1: Analyzing the First Reaction (Formation of A):

The starting material is Nitrobenzene (\(C_6H_5NO_2\)).

The reagent is Fe/HCl (Iron scrap in the presence of hydrochloric acid). This is a standard reagent for the reduction of a nitro group (\(-NO_2\)) to a primary amino group (\(-NH_2\)). This is the preferred method for preparing arylamines.

The reaction is:
\[ C_6H_5NO_2 + 6[H] \xrightarrow{Fe/HCl} C_6H_5NH_2 + 2H_2O \]
So, compound A is Aniline.



Step 2: Analyzing the Second Reaction (Formation of B):

Compound A (Aniline) is treated with \(HNO_2\) (nitrous acid) at 273 K (0 °C). Nitrous acid is unstable and is prepared in situ by reacting \(NaNO_2\) with a strong acid like HCl.

This reaction is called diazotization. It converts a primary aromatic amine into a diazonium salt. The low temperature (273-278 K) is crucial to prevent the diazonium salt from decomposing.

The reaction is:
\[ C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273 K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O \]
So, compound B is Benzenediazonium chloride.
Quick Tip: This two-step sequence is fundamental in aromatic chemistry.
1. \textbf{Reduction of Nitrobenzene:} \(Sn/HCl\) or \(Fe/HCl\) are the classic reagents to make aniline.
2. \textbf{Diazotization of Aniline:} \(NaNO_2/HCl\) at \(0-5^\circC\) is the standard condition to form benzenediazonium chloride. This product is a very versatile intermediate for synthesizing a wide variety of aromatic compounds (Sandmeyer reaction, Gattermann reaction, etc.).

Explanation of Colour:

Step 1: Determine the electronic configuration of the metal ion.

In the complex \([Ti(H_2O)_6]Cl_3\), the complex ion is \([Ti(H_2O)_6]^{3+}\).

The oxidation state of Titanium (Ti) is +3.

The atomic number of Ti is 22, so its ground state configuration is \([Ar] 3d^2 4s^2\).

For \(Ti^{3+}\), three electrons are removed, so the configuration is \([Ar] 3d^1\).



Step 2: Apply Crystal Field Theory (CFT).

The complex \([Ti(H_2O)_6]^{3+}\) is an octahedral complex. In the presence of the six water ligands, the five degenerate d-orbitals of the \(Ti^{3+}\) ion split into two sets of different energies: a lower energy \(t_{2g}\) set (\(d_{xy}, d_{yz}, d_{zx}\)) and a higher energy \(e_g\) set (\(d_{x^2-y^2}, d_{z^2}\)).

The single 3d electron of \(Ti^{3+}\) occupies one of the \(t_{2g}\) orbitals in the ground state. The electronic configuration is \(t_{2g}^1 e_g^0\).



Step 3: Explain the origin of colour (d-d transition).

When the complex absorbs light from the visible region, this single electron gets excited from the lower energy \(t_{2g}\) orbital to the higher energy \(e_g\) orbital. This process is called a d-d transition.
\[ t_{2g}^1 e_g^0 \xrightarrow{absorbs light} t_{2g}^0 e_g^1 \]
The energy required for this transition corresponds to the energy of a specific wavelength of visible light. The complex absorbs this wavelength (in this case, greenish-yellow light) and transmits the complementary colour, which is purple. This is why the aqueous solution of \([Ti(H_2O)_6]Cl_3\) appears coloured (purple).



Effect of Heating:

When the complex \([Ti(H_2O)_6]Cl_3\) is heated, the coordinated water ligands (\(H_2O\)) are lost.
\[ [Ti(H_2O)_6]Cl_3 \xrightarrow{\Delta} TiCl_3 + 6H_2O \]
The resulting compound is anhydrous \(TiCl_3\). In the absence of ligands, there is no crystal field splitting of the d-orbitals. Since there is no splitting, d-d transitions cannot occur. Consequently, no light from the visible region is absorbed, and the anhydrous compound becomes colourless.
Quick Tip: For a transition metal complex to be coloured, two conditions must be met:
It must have a partially filled \(d\)-orbital \((d^{1}\) to \(d^{9}\) configuration).
It must have ligands that cause crystal field splitting.
Complexes with \(d^{0}\) \((e.g., Sc^{3+})\) or \(d^{10}\) \((e.g., Zn^{2+}, Cu^{+})\) configurations are generally colourless because \(d\)--\(d\) transitions are not possible.

Definition:

In a coordination compound, the ligands create an electrostatic field (crystal field) that removes the degeneracy of the d-orbitals of the central metal ion. This causes the d-orbitals to split into two or more sets with different energy levels.



Crystal Field Splitting Energy (CFSE), denoted by \(\Delta\), is defined as the energy difference between these sets of d-orbitals that have been split by the ligand field.



For example, in an octahedral complex, the d-orbitals split into a lower energy \(t_{2g}\) set and a higher energy \(e_g\) set. The energy difference between them is called the octahedral crystal field splitting energy, denoted by \(\Delta_o\) or 10 Dq.



In a tetrahedral complex, the splitting is inverted, with a lower energy \(e\) set and a higher energy \(t_2\) set. The energy difference is the tetrahedral crystal field splitting energy, denoted by \(\Delta_t\).
Quick Tip: The magnitude of the crystal field splitting energy (\(\Delta\)) depends on several factors:
\textbf{Nature of the ligand:} Strong field ligands cause larger splitting.
\textbf{Oxidation state of the metal ion:} Higher oxidation states lead to larger splitting.
\textbf{Geometry of the complex:} \(\Delta_o\) is significantly larger than \(\Delta_t\) (\(\Delta_t \approx \frac{4}{9}\Delta_o\)).
\textbf{Size of the metal ion:} Larger d-orbitals (e.g., in 4d or 5d series) experience greater splitting.

Step 1: Understanding the Terms:

- \(\Delta_o\) (Octahedral Crystal Field Splitting Energy): The energy gap between the \(t_{2g}\) and \(e_g\) orbitals in an octahedral complex.

- P (Pairing Energy): The energy required to pair two electrons in the same orbital, overcoming the electrostatic repulsion between them.



Step 2: The Role of Ligands and Energy Comparison:

The distribution of electrons in the d-orbitals (for \(d^4\) to \(d^7\) configurations) depends on the relative magnitudes of \(\Delta_o\) and P. This is determined by the nature of the ligand.



Case 1: Weak Field Ligand

- Condition: A weak field ligand causes a small crystal field splitting. In this case, the splitting energy is less than the pairing energy: \(\Delta_o < P\).

- Electron Filling: It is energetically more favorable for an electron to occupy a higher energy \(e_g\) orbital than to pair up in a lower energy \(t_{2g}\) orbital.

- Result: After the first three electrons occupy the \(t_{2g}\) orbitals singly, the fourth electron will enter an \(e_g\) orbital rather than pairing up. This leads to the formation of high spin complexes.

- Example: For a \(d^4\) ion, the configuration will be \(t_{2g}^3 e_g^1\).



Case 2: Strong Field Ligand

- Condition: A strong field ligand causes a large crystal field splitting. In this case, the splitting energy is greater than the pairing energy: \(\Delta_o > P\).

- Electron Filling: It is energetically more favorable for an electron to pair up in a lower energy \(t_{2g}\) orbital than to jump the large energy gap to an \(e_g\) orbital.

- Result: The fourth electron will pair up with an electron already in a \(t_{2g}\) orbital. This leads to the formation of low spin complexes.

- Example: For a \(d^4\) ion, the configuration will be \(t_{2g}^4 e_g^0\).



Step 3: Differentiation Summary:

- A ligand is a weak field ligand if it creates a complex where \(\Delta_o < P\), resulting in a high spin configuration.

- A ligand is a strong field ligand if it creates a complex where \(\Delta_o > P\), resulting in a low spin configuration.
Quick Tip: A simple way to remember the spin states:
\textbf{Weak field} \(\rightarrow\) Small \(\Delta_o\) \(\rightarrow\) High spin (electrons spread out).
\textbf{Strong field} \(\rightarrow\) Large \(\Delta_o\) \(\rightarrow\) Low spin (electrons pair up).
This decision point is relevant for metal ions with d-electron configurations of \(d^4, d^5, d^6,\) and \(d^7\). For \(d^1, d^2, d^3, d^8, d^9\), there is only one possible ground state electron configuration, regardless of the ligand field strength.

Reasoning:

Low spin complexes are formed when the crystal field splitting energy (\(\Delta\)) is greater than the pairing energy (P), forcing electrons to pair up in lower energy orbitals. Low spin tetrahedral complexes are rare for the following key reason:



Small Crystal Field Splitting Energy (\(\Delta_t\)):

The magnitude of the crystal field splitting in a tetrahedral geometry (\(\Delta_t\)) is inherently much smaller than in an octahedral geometry (\(\Delta_o\)). There are two main reasons for this:

1. Fewer Ligands: A tetrahedral complex has only four ligands, whereas an octahedral complex has six. With fewer ligands, the electrostatic repulsion and the resulting splitting of d-orbitals are weaker.

2. Indirect Ligand Approach: In a tetrahedral arrangement, the ligands do not point directly at any of the d-orbitals. They approach between the axes, leading to less effective repulsion and smaller energy splitting compared to the octahedral case, where the ligands point directly at the \(e_g\) orbitals.



The relationship between the two splitting energies is approximately \(\Delta_t \approx \frac{4}{9} \Delta_o\).



Because \(\Delta_t\) is so small, it is almost never large enough to overcome the pairing energy (P). The condition \(\Delta_t > P\) is very difficult to achieve. It is almost always energetically more favorable for electrons to occupy the higher energy \(t_2\) orbitals rather than pairing up in the lower energy \(e\) orbitals.



Therefore, tetrahedral complexes are almost exclusively high spin, and low spin configurations are not observed.
Quick Tip: Remember the key reason: \(\Delta_t\) is small. Why? Fewer ligands (4 vs 6) and indirect orbital overlap. This small energy gap (\(\Delta_t\)) is almost always less than the pairing energy (P), so electrons follow Hund's rule and occupy higher orbitals before pairing. Hence, tetrahedral complexes are high spin.

Step 1: Write the Overall Cell Reaction and find n.

Anode (Oxidation): \( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \)

Cathode (Reduction): \( 2H^+(aq) + 2e^- \rightarrow H_2(g) \)

Overall: \( Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g) \). The number of electrons transferred, n = 2.

Step 2: Calculate the Standard EMF (\(E^\circ_{cell}\)).
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00 V - (-0.76 V) = +0.76 V \]
Step 3: Apply the Nernst Equation.
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
The reaction quotient Q is:
\[ Q = \frac{[Zn^{2+}] \times P_{H_2}}{[H^+]^2} = \frac{(0.1) \times (1)}{(0.01)^2} = \frac{10^{-1}}{10^{-4}} = 10^3 \]
Step 4: Calculate the Cell EMF (\(E_{cell}\)).
\[ E_{cell} = 0.76 - \frac{0.0591}{2} \log(10^3) \] \[ E_{cell} = 0.76 - \frac{0.0591}{2} \times 3 \] \[ E_{cell} = 0.76 - 0.08865 = 0.67135 V \]
The EMF of the cell is 0.671 V.
Quick Tip: For Nernst equation calculations, follow a systematic approach: 1. Balanced reaction and 'n'. 2. \(E^\circ_{cell}\). 3. Reaction quotient 'Q'. 4. Substitute into the Nernst equation. Pay close attention to stoichiometric coefficients as they become exponents in the Q expression.

Faraday's Second Law of Electrolysis:

Statement: When the same quantity of electricity is passed through solutions of different electrolytes connected in series, the masses of the substances deposited or liberated at the respective electrodes are directly proportional to their chemical equivalent weights.

Calculation of Electricity Required:

Step 1: Write the balanced reduction half-reaction.

The reduction of dichromate ion (\(Cr_2O_7^{2-}\)) to chromium(III) ion (\(Cr^{3+}\)) in an acidic medium is:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
Step 2: Determine moles of electrons.

The balanced equation shows that for every 1 mole of \(Cr_2O_7^{2-}\) that is reduced, 6 moles of electrons are required.

Step 3: Convert moles of electrons to Faradays.

By definition, the charge carried by 1 mole of electrons is 1 Faraday (1 F). Therefore, the charge carried by 6 moles of electrons is 6 Faradays.

Answer: 6 Faradays of electricity are required.
Quick Tip: The key to solving Faraday's laws problems is to use the mole concept. The stoichiometric coefficient of electrons in the balanced half-reaction directly gives the number of Faradays required per mole of substance reacted.

Step 1: Calculate Molar Conductivity (\(\Lambda_m\)).

The formula relating molar conductivity (\(\Lambda_m\)) to specific conductivity (\(\kappa\)) is:
\[ \Lambda_m = \frac{\kappa \times 1000}{M} \]
Given \(\kappa = 2.48 \times 10^{-2} S cm^{-1}\) and M = 0.20 M:
\[ \Lambda_m = \frac{(2.48 \times 10^{-2}) \times 1000}{0.20} = \frac{24.8}{0.20} = 124 S cm^2 mol^{-1} \]
Step 2: Calculate Limiting Molar Conductivity (\(\Lambda_m^\circ\)).

Using Kohlrausch's law for KCl:
\[ \Lambda_m^\circ(KCl) = \lambda^\circ_{(K^+)} + \lambda^\circ_{(Cl^-)} = 73.5 + 76.5 = 150.0 S cm^2 mol^{-1} \]
Step 3: Calculate the Degree of Dissociation (\(\alpha\)).
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{124}{150} = 0.8267 \]
Answer: The molar conductivity is 124 S cm\(^2\) mol\(^{-1}\) and the degree of dissociation is 0.827.
Quick Tip: Remember to use the factor of 1000 in the molar conductivity formula only when conductivity is in S cm\(^{-1}\) and concentration is in mol L\(^{-1}\). For strong electrolytes like KCl, \(\alpha\) is expected to be close to 1, but in concentrated solutions, inter-ionic attractions reduce it.

Step 1: Find n and \(E^\circ_{cell}\).

The half-reactions are \(Mg \rightarrow Mg^{2+} + 2e^-\) (Anode) and \(Cu^{2+} + 2e^- \rightarrow Cu\) (Cathode). The number of electrons transferred, n = 2.
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Mg^{2+}/Mg} \] \[ E^\circ_{cell} = (+0.34 V) - (-2.37 V) = 2.71 V \]
Step 2: Calculate \(\Delta_rG^\circ\).

The relationship between standard Gibbs free energy and standard cell potential is:
\[ \Delta_rG^\circ = -nFE^\circ_{cell} \]
Substitute the values:
\[ \Delta_rG^\circ = -(2 mol) \times (96500 C mol^{-1}) \times (2.71 J C^{-1}) \] \[ \Delta_rG^\circ = -523030 J mol^{-1} \] \[ \Delta_rG^\circ = -523.03 kJ mol^{-1} \]
The standard Gibbs free energy change is -523.03 kJ mol\(^{-1}\).
Quick Tip: The formula \(\Delta_rG^\circ = -nFE^\circ_{cell}\) is a cornerstone of electrochemistry, linking thermodynamics (\(\Delta G^\circ\)) to electrochemistry (\(E^\circ_{cell}\)). A positive \(E^\circ_{cell}\) (spontaneous reaction) will always correspond to a negative \(\Delta_rG^\circ\).

Type of Cell:

A mercury cell is a primary cell, meaning it is non-rechargeable.

Advantage over Dry Cell:

The primary advantage of a mercury cell over a conventional Leclanché (dry) cell is that it provides a constant and stable voltage throughout its entire lifespan.

Reason:

This constant voltage is a result of the overall cell reaction:
\[ Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l) \]
In this reaction, there are no ions in the solution whose concentrations change during the discharge process. Since the activities of all reactants and products remain constant, the cell potential does not decrease as it is used. In contrast, the voltage of a dry cell drops over time as the concentration of ions in the electrolyte changes.
Quick Tip: The constant voltage of a mercury cell makes it ideal for devices requiring a stable power supply, like watches and hearing aids. The key feature to remember is that its overall reaction involves no change in electrolyte ion concentration.

Highly negative \(E^\circ_{Mn^{2+}/Mn}\):

The electronic configuration of Mn is \([Ar] 3d^5 4s^2\). The highly negative \(E^\circ\) value (-1.18 V) for the \(Mn^{2+}/Mn\) couple indicates that Mn is readily oxidized to \(Mn^{2+}\). This is because in forming \(Mn^{2+}\), the atom loses its two 4s electrons to achieve the configuration \([Ar] 3d^5\). This \(3d^5\) configuration has a half-filled d-subshell, which is exceptionally stable. This extra stability provides the driving force for the oxidation.



Highly positive \(E^\circ_{Mn^{3+}/Mn^{2+}}\):

This potential corresponds to the reduction \(Mn^{3+}(aq) + e^- \rightarrow Mn^{2+}(aq)\). The value is highly positive (+1.57 V), meaning the reaction strongly favors the formation of \(Mn^{2+}\). The configuration of \(Mn^{3+}\) is \([Ar] 3d^4\), while \(Mn^{2+}\) is \([Ar] 3d^5\). There is a strong tendency for the \(Mn^{3+}\) ion to gain an electron to achieve the very stable half-filled \(3d^5\) configuration. This makes \(Mn^{3+}\) a powerful oxidizing agent and results in a highly positive reduction potential.
Quick Tip: When explaining trends in electrode potentials for transition metals, always look at the electronic configurations of the ions involved. Exceptional stability of half-filled (\(d^5\)) and fully-filled (\(d^{10}\)) subshells is the most common reason for anomalous values.

The actinoids are the elements in which the 5f subshell is progressively filled. The reason they exhibit a much wider range of oxidation states than their lanthanoid counterparts is the very small energy difference between the 5f, 6d, and 7s subshells.

Because these orbitals have comparable energies, electrons from all three subshells can participate in chemical bonding. This allows for a large number of oxidation states, particularly in the first half of the series. For example, Uranium (U) and Plutonium (Pu) can show oxidation states from +3 up to +6 and +7, respectively. In contrast, for lanthanoids, the 4f electrons are more shielded and lower in energy, making them less available for bonding and resulting in a predominantly +3 oxidation state.
Quick Tip: The key difference between lanthanoid and actinoid chemistry is the relative energies of their valence orbitals. For actinoids, remember that 5f, 6d, and 7s are all close in energy, leading to more complex chemistry and variable oxidation states.

Transition metals generally have very high melting and boiling points. This property is attributed to the strength of the metallic bonding within their crystal lattices.

The metallic bond in these elements is particularly strong due to the involvement of electrons from both the outer ns orbital and the inner (n-1)d orbitals. The presence of a large number of unpaired electrons in the (n-1)d orbitals leads to the formation of strong covalent-like bonds in addition to the standard metallic bonding. The more unpaired electrons available for bonding, the stronger the interatomic forces and the higher the enthalpy of atomization, which directly correlates with high melting points. This explains why elements near the middle of the transition series, which have the maximum number of unpaired d-electrons, tend to have the highest melting points.
Quick Tip: Strong physical properties of d-block elements (high melting point, high density, hardness) are all linked to strong interatomic forces. The primary reason for these strong forces is strong metallic bonding involving both ns and (n-1)d electrons.

Step 1: Identify the Oxidation and Reduction Half-Reactions.

This is a redox reaction in an acidic medium. The permanganate ion (\(MnO_4^-\)) is a strong oxidizing agent, and the sulfite ion (\(SO_3^{2-}\)) is a reducing agent.

- Oxidation: The sulfite ion (\(SO_3^{2-}\), S = +4) is oxidized to the sulfate ion (\(SO_4^{2-}\), S = +6).

- Reduction: In acidic solution, the permanganate ion (\(MnO_4^-\), Mn = +7) is reduced to the manganese(II) ion (\(Mn^{2+}\)).

Step 2: Write the Products and Balance.

The sulfite ions become sulfate ions, and the permanganate ions become manganese(II) ions. The H\(^+\) ions react with the oxygen from the permanganate to form water.

Completed Equation:
\[ 5SO_3^{2-} + 2MnO_4^- + 6H^+ \rightarrow 5SO_4^{2-} + 2Mn^{2+} + 3H_2O \] Quick Tip: Remember the reduction products of \(KMnO_4\) in different media. In an acidic medium (\(H^+\)), \(MnO_4^-\) (purple, Mn=+7) is always reduced to \(Mn^{2+}\) (colourless, Mn=+2), a 5-electron change.

Step 1: Identify the Half-Reactions and Medium.

This is a redox reaction in a neutral or faintly alkaline medium. Permanganate ion (\(MnO_4^-\)) is the oxidizing agent, and iodide ion (\(I^-\)) is the reducing agent.

- Oxidation: Iodide ion (\(I^-\), I = -1) is oxidized to the iodate ion (\(IO_3^-\), I = +5).

- Reduction: In a neutral medium, the permanganate ion (\(MnO_4^-\), Mn = +7) is reduced to manganese dioxide (\(MnO_2\)), a brown precipitate where Mn is +4.

Step 2: Write the Products and Balance.

The reaction produces \(MnO_2\), \(IO_3^-\), and hydroxide ions (\(OH^-\)) are formed to balance the equation.

Completed Equation:
\[ 2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + IO_3^- + 2OH^- \] Quick Tip: In a neutral or weakly alkaline medium, \(MnO_4^-\) (purple, Mn=+7) is reduced to \(MnO_2\) (a brown precipitate, Mn=+4), which is a 3-electron change. This reaction is known as the Baeyer's test for unsaturation.

The third ionization enthalpy (\(IE_3\)) is the energy required for the process \(M^{2+} \rightarrow M^{3+} + e^-\). This value will be exceptionally high if the electron is being removed from a particularly stable electronic configuration.

1. Manganese (Mn): The configuration of \(Mn^{2+}\) is \([Ar] 3d^5\), a stable half-filled d-subshell. Removing a third electron disrupts this stability, requiring a very large amount of energy.

2. Zinc (Zn): The configuration of \(Zn^{2+}\) is \([Ar] 3d^{10}\), a stable fully-filled d-subshell. Removing a third electron from this highly stable configuration requires an enormous amount of energy.

Answer: Manganese (Mn) and Zinc (Zn).
Quick Tip: To identify elements with high ionization enthalpies for a specific step (e.g., \(IE_3\)), first write the configuration of the ion just before that step (e.g., M\(^{2+}\)). If that configuration is particularly stable (\(d^5, d^{10}\), or a noble gas core), the ionization enthalpy will be very high.

Paramagnetism arises from the presence of unpaired electrons. We must determine the electronic configuration of the manganese ion in each compound.

- In \(KMnO_4\) (Potassium permanganate): The oxidation state of Mn is +7. The ground state configuration of Mn is \([Ar] 3d^5 4s^2\). For \(Mn^{7+}\), all valence electrons are lost, giving a configuration of \([Ar] 3d^0\). Since there are no unpaired electrons, \(KMnO_4\) is diamagnetic.

- In \(K_2MnO_4\) (Potassium manganate): The oxidation state of Mn is +6. The electronic configuration of \(Mn^{6+}\) is \([Ar] 3d^1\). This ion has one unpaired electron in its d-orbital.

Answer: \(K_2MnO_4\) is paramagnetic because the \(Mn^{6+}\) ion has one unpaired electron (\(3d^1\)).
Quick Tip: To determine magnetic properties:
1. Find the oxidation state of the transition metal.
2. Write the electronic configuration of the ion.
3. Count the number of unpaired electrons (n). If n > 0, it is paramagnetic. If n = 0, it is diamagnetic.

Lanthanoid contraction is the gradual decrease in atomic and ionic radii across the lanthanoid series (from La to Lu). This is caused by the poor shielding of the nuclear charge by the 4f electrons.

Consequence: A significant consequence is the similarity in atomic radii of the elements of the second (4d) and third (5d) transition series that come after the lanthanoids. For instance, the atomic radius of Zirconium (Zr, 4d series) is 160 pm, which is almost identical to that of Hafnium (Hf, 5d series) at 159 pm. Because of this similarity in size and electronic configuration, these pairs of elements (e.g., Zr/Hf, Nb/Ta) exhibit very similar chemical properties, making their separation from one another extremely difficult.
Quick Tip: The most frequently cited consequence of lanthanoid contraction is the similarity in properties of 4d and 5d elements. This makes them "chemical twins" and is a very important concept in the chemistry of d-block elements.

Pyrolusite ore consists mainly of manganese dioxide (\(MnO_2\)). The preparation of potassium manganate (\(K_2MnO_4\)) involves the fusion of this ore with an alkali (like KOH) in the presence of an oxidizing agent, such as atmospheric oxygen or potassium nitrate (\(KNO_3\)).

The chemical reaction is an oxidation of manganese from the +4 state to the +6 state.

Reaction:
\[ 2MnO_2 (from Pyrolusite) + 4KOH + O_2 \xrightarrow{Heat/Fuse} 2K_2MnO_4 (Potassium Manganate) + 2H_2O \]
The product, potassium manganate, is a dark green solid. This is the first step in the commercial production of potassium permanganate.
Quick Tip: This is a key industrial preparation. Remember the recipe: \(MnO_2\) + Alkali (KOH) + Oxidizing agent (\(O_2\)) \(\xrightarrow{heat}\) Manganate (\(K_2MnO_4\)). The color change from black/brown (\(MnO_2\)) to green (\(K_2MnO_4\)) is a good indicator.

Although fluorine is the most electronegative element, oxygen is more effective at stabilizing the highest oxidation states of transition metals. The primary reason for this is oxygen's ability to form multiple bonds (specifically p\(\pi\)-d\(\pi\) double bonds) with metal atoms.

To achieve a very high oxidation state (e.g., +6 or +7), a metal atom needs to form several bonds with electronegative elements.
- Fluorine can only form single bonds (\(M-F\)). Stabilizing a +7 state would require seven single bonds (e.g., \(MF_7\)), which can lead to significant steric crowding.
- Oxygen can form double bonds (\(M=O\)). This allows it to satisfy the high valency of the metal with fewer bonded atoms, reducing steric hindrance. For example, in the permanganate ion (\(MnO_4^-\)), manganese is in its highest oxidation state of +7 while being bonded to only four oxygen atoms. The highest fluoride of manganese is \(MnF_4\). The formation of stable oxoanions like \(MnO_4^-\), \(Cr_2O_7^{2-}\), and \(VO_4^{3-}\) showcases this ability of oxygen.
Quick Tip: The answer to "Oxygen vs. Fluorine" for stabilizing high oxidation states always comes down to multiple bonding. Oxygen can form double bonds (\(M=O\)), accommodating high oxidation numbers with less steric strain. Fluorine is limited to single bonds.

Condition: The Cannizzaro reaction is a redox disproportionation reaction given by aldehydes that do not have an \(\alpha\)-hydrogen atom. An \(\alpha\)-hydrogen is a hydrogen atom on the carbon adjacent to the aldehyde group.

Structure Derivation: For the formula \(C_5H_{10}O\), we need to find an aldehyde isomer where the \(\alpha\)-carbon has no hydrogen atoms. This is achieved if the \(\alpha\)-carbon is a quaternary carbon, bonded to three other carbon atoms. The only structure that fits this description is 2,2-Dimethylpropanal.

Structure:
In this molecule, the \(\alpha\)-carbon is bonded to three methyl groups and the carbonyl group, hence it has no \(\alpha\)-hydrogen.
Quick Tip: To quickly identify candidates for the Cannizzaro reaction, look for aldehydes where the -CHO group is attached to a carbon with no C-H bonds. Common examples include formaldehyde, benzaldehyde, and pivaldehyde (2,2-dimethylpropanal).

Condition 1: "Reduces Tollens' reagent" implies that the compound (X) must be an aldehyde.

Condition 2: "Has a chiral carbon" means there must be a carbon atom in the molecule that is bonded to four different groups.

Structure Derivation: We need to find an aldehyde isomer of \(C_5H_{10}O\) with a chiral center. Let's examine branched-chain isomers. The structure 2-Methylbutanal fits both criteria.

Structure:



Verification: It is an aldehyde, so it will give a positive Tollens' test. The carbon atom at position 2 is bonded to four different groups: a hydrogen atom (-H), a methyl group (\(-CH_3\)), an ethyl group (\(-CH_2CH_3\)), and an aldehyde group (-CHO). Therefore, C-2 is a chiral carbon.
Quick Tip: When asked to find a structure with a chiral center, systematically look for carbon atoms bonded to four non-identical atoms or groups. It's often helpful to draw out different isomers and check each one.

Condition: The iodoform test (reaction with \(I_2/NaOH\)) is a positive test for compounds containing a methyl ketone group (\(CH_3CO-R\)) or an alcohol group that can be oxidized to a methyl ketone (\(CH_3CH(OH)-R\)).

Structure Derivation: Since the formula is \(C_5H_{10}O\), which corresponds to a saturated acyclic ketone or aldehyde, we look for a ketone structure with a \(CH_3CO-\) group. This requires the carbonyl group to be at position 2.

Structure: Pentan-2-one.




This structure, \(CH_3COCH_2CH_2CH_3\), contains the required methyl ketone group and has the molecular formula \(C_5H_{10}O\). (Note: Another isomer, 3-Methylbutan-2-one, \(CH_3COCH(CH_3)_2\), also gives a positive iodoform test).
Quick Tip: For a positive iodoform test, the molecule must have the \(CH_3-C=O\) unit or the \(CH_3-CH(OH)\) unit. This is one of the most important chemical tests for identifying specific aldehydes, ketones, and alcohols.

Explanation:

The Clemmensen reduction is a reaction used to reduce the carbonyl group of aldehydes or ketones to a methylene group (\(-CH_2-\)), effectively converting them into alkanes.

Reagent: Zinc amalgam (\(Zn-Hg\)) and concentrated hydrochloric acid (\(HCl\)).

General Reaction:
\[ R-CO-R' \xrightarrow{Zn-Hg, conc. HCl} R-CH_2-R' + H_2O \]
(Where R' can be H for an aldehyde or an alkyl/aryl group for a ketone).

Example Reaction (Reduction of Cyclohexanone):

Cyclohexanone is reduced to Cyclohexane.


Quick Tip: The Clemmensen reduction is performed in acidic conditions. It is not suitable for compounds that are sensitive to acid. For acid-sensitive compounds, the Wolff-Kishner reduction (using basic conditions) is the preferred method for reducing a carbonyl to an alkane.

Explanation:

The Etard reaction is a controlled oxidation used to convert a methyl group attached to an aromatic ring directly into an aldehyde group. It is a common method for synthesizing benzaldehyde from toluene.

Reagent: Chromyl chloride (\(CrO_2Cl_2\)) in an inert solvent like \(CS_2\) or \(CCl_4\), followed by aqueous hydrolysis.

Reaction (Toluene to Benzaldehyde):

The reaction proceeds via the formation of a brown chromium complex, which is then hydrolyzed to yield the aldehyde. This intermediate step prevents over-oxidation to carboxylic acid.



Quick Tip: The Etard reaction is a specific named reaction you should memorize for converting toluene to benzaldehyde. Stronger oxidizing agents like \(KMnO_4\) would oxidize toluene all the way to benzoic acid. The use of chromyl chloride allows the oxidation to be stopped at the aldehyde stage.

A hemiacetal is formed from the nucleophilic addition of one molecule of an alcohol to an aldehyde's carbonyl group.

- Aldehyde: Methanal (HCHO)

- Alcohol: Methanol (\(CH_3OH\))

The reaction involves the attack of the methanol oxygen on the carbonyl carbon, and proton transfer to the carbonyl oxygen.
\[ HCHO + CH_3OH \rightleftharpoons HO-CH_2-OCH_3 \]
The resulting structure is methoxymethanol.

Structure:


Quick Tip: Remember the general structures:
- \textbf{Hemiacetal:} A carbon atom bonded to one -OH group and one -OR group (formed from an aldehyde).
- \textbf{Acetal:} A carbon atom bonded to two -OR groups (formed from a hemiacetal and another alcohol molecule).

The structure of semicarbazide is \(H_2N^{(1)}-CO-N^{(2)}H-N^{(3)}H_2\). The reaction with an aldehyde or ketone to form a semicarbazone is a nucleophilic attack by a nitrogen lone pair on the carbonyl carbon.

Reason:

The nitrogen atoms of the \(-NH_2\) group labeled (1) are directly attached to the electron-withdrawing carbonyl group (\(C=O\)). The lone pair of electrons on this nitrogen is delocalized through resonance with the carbonyl group.
\[ H_2N-C(=O)- \leftrightarrow H_2N^+=C(O^-)- \]
This resonance reduces the electron density and nucleophilicity of the nitrogen atom in group (1).

In contrast, the lone pair on the nitrogen atom of the \(-NH_2\) group labeled (3) is not involved in resonance. It is therefore more available to act as a nucleophile and attack the carbonyl carbon of the aldehyde or ketone.
Quick Tip: When a molecule has multiple potential nucleophilic sites, the most reactive site is typically the one that is least deactivated by electron-withdrawing groups or resonance. In ammonia derivatives like semicarbazide, the nitrogen furthest from any resonance-withdrawing group will be the most nucleophilic.

The target molecule, 3-hydroxybutanal, is a \(\beta\)-hydroxy aldehyde, which is the characteristic product of an aldol addition reaction. The starting material for 3-hydroxybutanal is ethanal (acetaldehyde). Therefore, the conversion requires two steps.

Step 1: Oxidation of Ethanol to Ethanal.

Ethanol, a primary alcohol, must first be oxidized to ethanal. A mild oxidizing agent is used to stop the reaction at the aldehyde stage without further oxidation to carboxylic acid. Pyridinium chlorochromate (PCC) is an ideal reagent for this.
\[ CH_3CH_2OH (Ethanol) \xrightarrow{PCC} CH_3CHO (Ethanal) \]
Step 2: Aldol Addition of Ethanal.

Two molecules of ethanal undergo a self-condensation reaction in the presence of a dilute base (like dil. NaOH) to form the aldol product.
\[ 2CH_3CHO \xrightarrow{dil. NaOH} CH_3CH(OH)CH_2CHO (3-Hydroxybutanal) \] Quick Tip: Recognizing the structure of the target molecule is key. A \(\beta\)-hydroxy aldehyde or ketone is a tell-tale sign that an aldol reaction is needed. Work backwards from the product to identify the required carbonyl starting material, and then figure out how to synthesize it from the given reactant.

Reaction Analysis:

- Substrate: Cyclohexanol is a secondary alcohol.

- Reagent: \(CrO_3\) (Chromium trioxide) is a strong oxidizing agent.

The oxidation of a secondary alcohol yields a ketone. The cyclic structure remains intact.

Reaction:

The hydroxyl group (\(-OH\)) on the cyclohexane ring is oxidized to a carbonyl group (\(C=O\)).
\[ C_6H_{11}OH \xrightarrow{CrO_3} C_6H_{10}O \]
The product is Cyclohexanone.


Quick Tip: Memorize the oxidation products of different types of alcohols:
- Primary Alcohols \(\rightarrow\) Aldehydes (with mild agents like PCC) or Carboxylic Acids (with strong agents like \(KMnO_4\)).
- Secondary Alcohols \(\rightarrow\) Ketones (with most oxidizing agents).
- Tertiary Alcohols \(\rightarrow\) No reaction under normal conditions (C-C bond cleavage under harsh conditions).

This conversion occurs in several stages.

Step 1: Acid-Base Reaction. Phthalic acid (benzene-1,2-dicarboxylic acid) is an acid. It reacts with ammonia (\(NH_3\)), a base, to form a salt, diammonium phthalate.

Step 2: Heating (Dehydration). Mild heating of the ammonium salt causes it to lose two molecules of water to form the corresponding diamide, phthalamide.

Step 3: Strong Heating (Cyclization). When phthalamide is heated strongly, it undergoes an intramolecular condensation reaction, losing one molecule of ammonia (\(NH_3\)) to form a stable five-membered cyclic imide.

The final product is Phthalimide.


Overall Reaction:


Quick Tip: The reaction of dicarboxylic acids with ammonia followed by heating is a standard method for preparing cyclic imides, but only if a stable 5- or 6-membered ring can be formed. Phthalic acid (1,2-), succinic acid (1,4-), and glutaric acid (1,5-) all form stable cyclic imides upon heating their amides.