Step 1: Understanding the Concepts

This question requires an analysis of the properties of different reaction orders (zero and first) and the concept of molecularity.

- Zero-Order Reaction: A reaction whose rate does not depend on the concentration of the reactants.

- First-Order Reaction: A reaction whose rate is directly proportional to the concentration of one reactant.

- Molecularity: The number of reacting species (atoms, ions, or molecules) that collide simultaneously to bring about a chemical reaction in a single elementary step.


Step 2: Analyzing each statement

(A) Rate of zero order reaction is independent of initial concentration of reactant.

For a zero-order reaction, the rate law is given by: Rate = k[Reactant]\(^0\).

Since any quantity raised to the power of zero is 1, the rate expression becomes: Rate = k.

This shows that the rate is constant and independent of the reactant's concentration. Thus, statement (A) is true.


(B) Half-life of a zero order reaction is inversely proportional to the rate constant.

The formula for the half-life (t\(_{1/2}\)) of a zero-order reaction is:
\[ t_{1/2} = \frac{[R]_0}{2k} \]
Here, [R]\(_0\) is the initial concentration and k is the rate constant.

From the formula, it is clear that t\(_{1/2}\) is inversely proportional to the rate constant k (t\(_{1/2} \propto 1/k\)). Thus, statement (B) is true.


(C) Molecularity of a reaction may be zero.

Molecularity is the count of species participating in an elementary reaction step. For a reaction to occur, at least one molecule must be present to react. Therefore, molecularity must be a non-zero positive integer (e.g., 1, 2, or 3). It cannot be zero or a fraction. Thus, statement (C) is false.


(D) For a first order reaction, t\(_{1/2}\) = 0.693/k.

The half-life (t\(_{1/2}\)) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{k} \]
This is a standard and correct formula for a first-order reaction. Thus, statement (D) is true.


Step 3: Final Answer

The question asks for the false statement. Based on the analysis, statement (C) is the only false statement.
Quick Tip: Remember the key differences: \textbf{Order of a reaction} is an experimental value that can be zero, fractional, or an integer. \textbf{Molecularity} is a theoretical concept for elementary reactions and must be a non-zero positive integer.

Step 1: Understanding the Concept

This problem involves calculating the total charge required for a redox reaction. The total charge is related to the number of moles of electrons transferred, where 1 Faraday (F) is the charge of one mole of electrons (approximately 96500 Coulombs).


Step 2: Determining the Change in Oxidation State

We need to find the number of electrons gained during the reduction of MnO\(_4^-\) to MnO\(_2\). We can do this by calculating the oxidation state of Manganese (Mn) in both species.

In MnO\(_4^-\):

Let the oxidation state of Mn be \(x\). The oxidation state of Oxygen (O) is -2.
\[ x + 4(-2) = -1 \] \[ x - 8 = -1 \] \[ x = +7 \]
So, the oxidation state of Mn in MnO\(_4^-\) is +7.


In MnO\(_2\):

Let the oxidation state of Mn be \(y\). The oxidation state of O is -2.
\[ y + 2(-2) = 0 \] \[ y - 4 = 0 \] \[ y = +4 \]
So, the oxidation state of Mn in MnO\(_2\) is +4.


Step 3: Calculating the Number of Electrons Transferred

The change in oxidation state of Mn is from +7 to +4.

Change = Final Oxidation State - Initial Oxidation State = 4 - 7 = -3.

A change of -3 indicates a gain of 3 electrons per Mn atom.

Therefore, for the reduction of 1 mole of MnO\(_4^-\) to 1 mole of MnO\(_2\), 3 moles of electrons are required.


Step 4: Calculating the Total Charge

The charge of 1 mole of electrons is equal to 1 Faraday (1 F).

Since 3 moles of electrons are required, the total charge needed is 3 F.


Alternative Method: Balancing the Half-Reaction

1. Write the unbalanced reaction: MnO\(_4^- \rightarrow\) MnO\(_2\).

2. Balance atoms other than O and H (Mn is balanced).

3. Balance O atoms by adding H\(_2\)O: MnO\(_4^- \rightarrow\) MnO\(_2\) + 2H\(_2\)O.

4. Balance H atoms by adding H\(^+\) (in acidic medium): MnO\(_4^-\) + 4H\(^+ \rightarrow\) MnO\(_2\) + 2H\(_2\)O.

5. Balance the charge by adding electrons (e\(^-\)):

- Charge on LHS = (-1) + 4(+1) = +3.

- Charge on RHS = 0.

- Add 3e\(^-\) to the LHS: MnO\(_4^-\) + 4H\(^+\) + 3e\(^-\) \(\rightarrow\) MnO\(_2\) + 2H\(_2\)O.

The balanced half-reaction shows that 3 moles of electrons are required for 1 mole of MnO\(_4^-\). Thus, the charge required is 3 F.
Quick Tip: To quickly find the number of Faradays needed for a redox reaction, simply calculate the total change in oxidation number for one mole of the substance. The magnitude of this change is equal to the number of Faradays required.

Step 1: Understanding Electronic Configuration

The electronic configuration describes the distribution of electrons of an atom in atomic orbitals. The given configuration is in noble gas notation, where [Ar] represents the electron configuration of Argon.


Step 2: Calculating the Atomic Number

To identify the element, we need to find its atomic number (Z), which is equal to the total number of electrons in a neutral atom.

- The number of electrons in Argon (Ar) is 18.

- The number of electrons in the 3d orbital is 10.

- The number of electrons in the 4s orbital is 1.

Total number of electrons = Electrons in [Ar] + Electrons in 3d + Electrons in 4s

Total electrons = 18 + 10 + 1 = 29.

The atomic number (Z) of the element is 29.


Step 3: Identifying the Element

We now identify the element with Z = 29 from the given options.

- (A) Copper (Cu) has an atomic number Z = 29.

- (B) Zinc (Zn) has an atomic number Z = 30.

- (C) Chromium (Cr) has an atomic number Z = 24.

- (D) Manganese (Mn) has an atomic number Z = 25.

The element with atomic number 29 is Copper (Cu).


Step 4: Verifying the Configuration

Copper (Cu, Z=29) is a known exception to the Aufbau principle. According to the Aufbau principle, its configuration should be [Ar]3d\(^9\)4s\(^2\). However, to achieve greater stability through a completely filled d-orbital, one electron from the 4s orbital shifts to the 3d orbital. This results in the more stable configuration [Ar]3d\(^{10}\)4s\(^1\). The given configuration matches that of Copper.
Quick Tip: Memorize the exceptional electronic configurations of Chromium (Cr: [Ar]3d\(^5\)4s\(^1\)) and Copper (Cu: [Ar]3d\(^{10}\)4s\(^1\)). These are frequently asked in exams due to their deviation from the standard Aufbau principle to attain the stability of half-filled and fully-filled d-orbitals.

Step 1: Understanding the Definitions

Let's define each term given in the options to find the correct match for the question's description.


(A) Activation energy of the reaction: This is the minimum amount of energy that must be provided to compounds to result in a chemical reaction. It relates to the energy barrier of the reaction, not the number of reacting molecules.


(B) Stoichiometry of the reaction: This refers to the quantitative relationship between the number of moles of reactants and products as represented by a balanced chemical equation. While it gives the ratio of reactants, it applies to the overall reaction, which may consist of multiple elementary steps. It doesn't describe the number of molecules colliding in a single step.


(C) Molecularity of the reaction: This is defined as the number of reacting species (atoms, ions, or molecules) that collide simultaneously in an elementary (single-step) reaction. This perfectly matches the description given in the question. For example, a unimolecular reaction has a molecularity of 1, and a bimolecular reaction has a molecularity of 2.


(D) Order of the reaction: This is the sum of the powers of the concentration terms in the experimentally determined rate law expression. It describes how the rate of reaction is affected by the concentration of reactants. It is an experimental quantity and can be zero, fractional, or an integer, and it may not be equal to the number of reacting molecules.


Step 2: Final Answer

The definition that directly corresponds to "The number of molecules that react with each other in an elementary reaction" is the molecularity of the reaction.
Quick Tip: For an elementary (single-step) reaction, the order of the reaction with respect to a reactant is equal to its stoichiometric coefficient, and the overall order is equal to the molecularity. However, for complex reactions, this is not true. Always remember molecularity is theoretical, while order is experimental.

Step 1: Understanding Diamagnetism

A species is diamagnetic if it has no unpaired electrons in its atomic or molecular orbitals. All electrons are paired. Paramagnetic species, on the other hand, have one or more unpaired electrons. We need to examine the electronic configuration of the central metal ion in each complex.


Step 2: Analyzing each complex

(A) [Ni(CN)\(_4\)]\(^{2-}\)

- Central metal: Nickel (Ni). Atomic number = 28. Ground state config: [Ar] 3d\(^8\) 4s\(^2\).

- Oxidation state of Ni: Let it be \(x\). \(x + 4(-1) = -2 \implies x = +2\).

- Ni\(^{2+}\) configuration: [Ar] 3d\(^8\).

- Ligand: CN\(^-\) is a strong field ligand. It will cause pairing of electrons.

- In the presence of a strong field ligand and with a coordination number of 4, the hybridization is typically dsp\(^2\) (square planar).

- The 8 electrons in the 3d orbitals will pair up, occupying four d-orbitals. This leaves one d-orbital empty for dsp\(^2\) hybridization.
- Orbital diagram for Ni\(^{2+}\) in [Ni(CN)\(_4\)]\(^{2-}\):
\[ \begin{array}{|c|c|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \quad
\hline \end{array} \]
- Since all electrons are paired, [Ni(CN)\(_4\)]\(^{2-}\) is diamagnetic.


(B) [NiCl\(_4\)]\(^{2-}\)

- Central metal: Ni\(^{2+}\) ([Ar] 3d\(^8\)).

- Ligand: Cl\(^-\) is a weak field ligand. It will not cause pairing of electrons.

- With a coordination number of 4 and a weak field ligand, the hybridization is sp\(^3\) (tetrahedral).

- The 8 electrons in the 3d orbitals will be arranged according to Hund's rule.

- Orbital diagram for Ni\(^{2+}\) in [NiCl\(_4\)]\(^{2-}\):
\[ \begin{array}{|c|c|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow & \uparrow
\hline \end{array} \]
- There are 2 unpaired electrons. So, [NiCl\(_4\)]\(^{2-}\) is paramagnetic.


(C) [Fe(CN)\(_6\)]\(^{3-}\)

- Central metal: Iron (Fe). Atomic number = 26. Ground state config: [Ar] 3d\(^6\) 4s\(^2\).

- Oxidation state of Fe: Let it be \(x\). \(x + 6(-1) = -3 \implies x = +3\).

- Fe\(^{3+}\) configuration: [Ar] 3d\(^5\).

- Ligand: CN\(^-\) is a strong field ligand. It causes pairing.

- Hybridization is d\(^2\)sp\(^3\) (octahedral, inner orbital complex). The electrons occupy the t\(_{2g}\) orbitals first.

- The 5 electrons will be arranged as (t\(_{2g}\))\(^5\)(e\(_g\))\(^0\).

- Orbital diagram for Fe\(^{3+}\) in t\(_{2g}\):
\[ \begin{array}{|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow
\hline \end{array} \]
- There is 1 unpaired electron. So, [Fe(CN)\(_6\)]\(^{3-}\) is paramagnetic.


(D) [CoF\(_6\)]\(^{3-}\)

- Central metal: Cobalt (Co). Atomic number = 27. Ground state config: [Ar] 3d\(^7\) 4s\(^2\).

- Oxidation state of Co: Let it be \(x\). \(x + 6(-1) = -3 \implies x = +3\).

- Co\(^{3+}\) configuration: [Ar] 3d\(^6\).

- Ligand: F\(^-\) is a weak field ligand. No pairing occurs.

- Hybridization is sp\(^3\)d\(^2\) (octahedral, outer orbital complex).

- The 6 electrons will be arranged as (t\(_{2g}\))\(^4\)(e\(_g\))\(^2\).

- Orbital diagram:
\[ t_{2g}: \begin{array}{|c|c|c|} \hline \uparrow\downarrow & \uparrow & \uparrow
\hline \end{array} \quad e_g: \begin{array}{|c|c|} \hline \uparrow & \uparrow
\hline \end{array} \]
- There are 4 unpaired electrons. So, [CoF\(_6\)]\(^{3-}\) is paramagnetic.


Step 3: Final Answer

The only complex with zero unpaired electrons is [Ni(CN)\(_4\)]\(^{2-}\). Therefore, it is the diamagnetic species.
Quick Tip: To determine magnetic properties, always check for unpaired electrons. Remember the spectrochemical series to identify strong-field (cause pairing, e.g., CN\(^-\), CO) and weak-field (no pairing, e.g., F\(^-\), Cl\(^-\), H\(_2\)O) ligands.

Step 1: Understanding Isomerism in Coordination Compounds

Isomers are compounds that have the same chemical formula but different arrangements of atoms. In coordination chemistry, there are several types of structural isomerism.

- Ionization Isomers: Differ in the ions present inside and outside the coordination sphere. Example: [Co(NH\(_3\))\(_5\)SO\(_4\)]Br and [Co(NH\(_3\))\(_5\)Br]SO\(_4\).

- Coordination Isomers: Occur when both the cation and anion are complex ions, and they differ by the exchange of ligands between them. Example: [Co(NH\(_3\))\(_6\)][Cr(CN)\(_6\)] and [Cr(NH\(_3\))\(_6\)][Co(CN)\(_6\)].

- Geometrical Isomers (Stereoisomerism): Have the same bonds but different spatial arrangements (cis/trans).

- Linkage Isomers: Occur when a ligand can bond to the central metal ion through more than one atom. Such ligands are called ambidentate ligands.


Step 2: Analyzing the Given Complex Ions

The two complex ions are [Co(NH\(_3\))\(_5\)(NO\(_2\))]\(^{2+}\) and [Co(NH\(_3\))\(_5\)(ONO)]\(^{2+}\).

- Both complexes have the same overall formula: [Co(NH\(_3\))\(_5\)(NO\(_2\))]\(^{2+}\).

- The central metal ion is Co\(^{3+}\) and there are five ammine (NH\(_3\)) ligands in both.

- The difference lies in the sixth ligand, the nitrite ion (NO\(_2^-\)).

- In [Co(NH\(_3\))\(_5\)(NO\(_2\))]\(^{2+}\), the nitrite ligand is bonded to Cobalt through the Nitrogen atom (Co-NO\(_2\), called nitro).

- In [Co(NH\(_3\))\(_5\)(ONO)]\(^{2+}\), the nitrite ligand is bonded to Cobalt through one of the Oxygen atoms (Co-ONO, called nitrito).

- The nitrite ion (NO\(_2^-\)) is an ambidentate ligand because it can coordinate through either the N atom or the O atom.


Step 3: Final Answer

Since the isomerism arises due to the different attachment points of the ambidentate ligand (NO\(_2^-\)), these two complex ions are linkage isomers.
Quick Tip: Look for ambidentate ligands like NO\(_2^-\), SCN\(^-\) (thiocyanate vs. isothiocyanate), and CN\(^-\) (cyanide vs. isocyanide) when asked to identify linkage isomers. The difference is often shown in the formula by writing the donor atom first (e.g., -SCN vs. -NCS).

Step 1: Understanding Oxidation of Alcohols

The product of alcohol oxidation depends on the type of alcohol (primary, secondary, or tertiary) and the strength of the oxidizing agent.

- Primary (1\(^{\circ}\)) alcohol (R-CH\(_2\)OH): Oxidizes first to an aldehyde (R-CHO) and then, with a strong oxidizing agent, further oxidizes to a carboxylic acid (R-COOH).

- Secondary (2\(^{\circ}\)) alcohol (R\(_2\)CHOH): Oxidizes to a ketone (R\(_2\)C=O). Ketones are generally resistant to further oxidation under normal conditions.

- Tertiary (3\(^{\circ}\)) alcohol (R\(_3\)COH): Does not have a hydrogen atom on the carbinol carbon and is resistant to oxidation under normal conditions. Strong conditions can cause cleavage of C-C bonds.


Step 2: Identifying the Reactants

- Reactant: A secondary alcohol. The general structure is R-CH(OH)-R', where R and R' are alkyl or aryl groups.

- Reagent: Chromic anhydride (CrO\(_3\)). This is a chromium(VI) based oxidizing agent, often used in acetone (Jones reagent) or pyridine (Collins reagent), and it is capable of oxidizing primary and secondary alcohols.


Step 3: Predicting the Product

The oxidation of a secondary alcohol involves the removal of two hydrogen atoms: one from the -OH group and one from the carbon atom attached to the -OH group.
\[ R-CH(OH)-R' \xrightarrow{[O]} R-C(=O)-R' + H_2O \]
The functional group formed is a carbonyl group (C=O) bonded to two carbon atoms, which is the definition of a ketone.

Therefore, the oxidation of a secondary alcohol with chromic anhydride yields a ketone.
Quick Tip: A simple way to remember the oxidation products:
1\(^{\circ}\) Alcohol \(\rightarrow\) Aldehyde \(\rightarrow\) Carboxylic Acid
2\(^{\circ}\) Alcohol \(\rightarrow\) Ketone
3\(^{\circ}\) Alcohol \(\rightarrow\) No reaction (under mild conditions)

Step 1: Understanding IUPAC Nomenclature Rules for Benzene Derivatives

For disubstituted benzene rings, the following rules apply:

1. Identify the parent compound. It can be benzene itself or a monosubstituted benzene with a common name (like toluene, phenol, aniline).

2. Number the carbon atoms of the ring to give the substituents the lowest possible locants (positions).

3. If there is a choice in numbering, give the lower number to the substituent that comes first in alphabetical order.

4. List the substituents alphabetically, followed by the name of the parent compound.


Step 2: Applying the Rules to the Given Structure

The structure has a benzene ring with two substituents: a methyl group (-CH\(_3\)) and a chlorine atom (-Cl).


Method 1: Using Benzene as the parent name

- The substituents are 'chloro' and 'methyl'.

- We need to number the ring to give them the lowest locants. The possible sets are (1,4) or (4,1), which are equivalent.

- Now, we apply the alphabetical rule. 'Chloro' comes before 'methyl' alphabetically. Therefore, the chlorine atom gets position 1, and the methyl group gets position 4.

- The name is constructed by listing the substituents alphabetically with their positions, followed by the parent name 'benzene'.

- Name: 1-Chloro-4-methylbenzene.


Method 2: Using Toluene as the parent name

- Toluene is the common name for methylbenzene. If we use toluene as the parent, the methyl group is automatically at position 1.

- The chlorine atom is then at position 4.

- The name would be 4-Chlorotoluene. This is also a correct IUPAC-accepted name, but it is not among the options (A), (C), or (D) in that exact form.


Step 3: Evaluating the Options

(A) Methylchlorobenzene: This name is ambiguous as it doesn't specify the positions of the substituents (ortho, meta, or para).

(B) Toluene: This is the name for methylbenzene, not the given compound.

(C) 1-Chloro-4-Methylbenzene: This name correctly follows the IUPAC rules by treating benzene as the parent and assigning locants based on alphabetical priority.

(D) 1-Methyl-4-Chlorobenzene: This name gives the correct locants but violates the rule of assigning the lower number to the group cited first alphabetically. 'Chloro' should get the lower number.


Therefore, the most correct IUPAC name among the choices is 1-Chloro-4-methylbenzene.
Quick Tip: When numbering a substituted benzene ring, if the lowest locant rule results in a tie (e.g., 1,4 vs. 4,1), always give the lower number to the substituent that comes first alphabetically. In this case, Chloro before Methyl.

Step 1: Understanding Denaturation of Protein

Denaturation is a process in which a protein loses its native three-dimensional structure (quaternary, tertiary, and secondary structures). This is caused by external factors such as heat, change in pH, presence of certain chemicals (like acids, bases, alcohol), or mechanical stress. Denaturation disrupts the weak bonds (like hydrogen bonds, hydrophobic interactions) that hold the protein in its specific shape, leading to the loss of its biological activity. The primary structure (sequence of amino acids) remains intact.


Step 2: Analyzing the Examples

(A) Coagulation of egg white:

Egg white is primarily composed of a protein called albumin. When an egg is heated, the heat energy breaks the weak bonds in the albumin molecules. The protein chains unfold and then aggregate together in a random, tangled mesh. This process is irreversible and is a classic example of denaturation by heat.


(B) Curdling of milk:

Milk contains a protein called casein. When the pH of milk is lowered (e.g., by adding an acid like lemon juice or due to the production of lactic acid by bacteria), the negative charges on the casein micelles are neutralized. This disrupts the electrostatic repulsions that keep them suspended, causing them to clump together, or curdle. This is an example of denaturation by a change in pH.


(C) Clotting of blood:

Blood clotting is a complex, enzyme-controlled physiological process. It involves the conversion of a soluble protein, fibrinogen, into insoluble fibrin fibers to form a clot. While it involves a change in protein structure, it is a highly specific and regulated biological cascade, not a random unfolding like in typical denaturation. Therefore, while related, (A) and (B) are considered more direct and general examples of denaturation.


Step 3: Final Answer

Both the coagulation of egg white (denaturation by heat) and the curdling of milk (denaturation by pH change) are prime examples of protein denaturation. Therefore, option (D) is the most appropriate answer.
Quick Tip: Denaturation generally refers to the loss of the 2\(^\circ\), 3\(^\circ\), and 4\(^\circ\) structures of a protein. Common denaturing agents are heat, acids, bases, organic solvents, and heavy metal ions. Remember that denaturation leads to the loss of the protein's biological function.

Step 1: Understanding the Target Transformation

The reaction is the conversion of Phenol to Salicylic acid.

- Phenol has the structure C\(_6\)H\(_5\)OH.

- Salicylic acid is 2-hydroxybenzoic acid. It has a carboxylic acid group (-COOH) at the ortho position relative to the hydroxyl group (-OH) on the benzene ring.

The reaction involves introducing a -COOH group onto the phenol ring, a process known as carboxylation.


Step 2: Analyzing the Named Reactions

(A) Reimer-Tiemann reaction: This reaction converts phenol to salicylaldehyde (2-hydroxybenzaldehyde) by reacting phenol with chloroform (CHCl\(_3\)) in the presence of a base (like NaOH). It introduces a -CHO group, not a -COOH group.


(B) Friedel-Crafts reaction: This is a method for attaching substituents to an aromatic ring. Friedel-Crafts alkylation and acylation introduce alkyl (-R) and acyl (-COR) groups, respectively. The -OH group of phenol is a strongly activating group, but it can interfere with the Lewis acid catalyst (e.g., AlCl\(_3\)) used in this reaction. This is not the standard method for carboxylation of phenol.


(C) Kolbe reaction (or Kolbe-Schmitt reaction): This is the specific industrial method for synthesizing salicylic acid from phenol. The process involves:

1. Reacting phenol with a strong base (like NaOH) to form the sodium phenoxide ion. The phenoxide ion is more reactive than phenol towards electrophilic aromatic substitution.

2. Reacting the sodium phenoxide with carbon dioxide (CO\(_2\), a weak electrophile) under high pressure and temperature (around 125\(^\circ\)C).

3. Acidification of the resulting sodium salicylate to yield salicylic acid.

This reaction perfectly matches the required transformation.


(D) Coupling reaction: This typically refers to the reaction of a diazonium salt with an activated aromatic compound (like phenol or aniline) to form an azo compound (R-N=N-R'), which are often brightly colored dyes. This does not produce salicylic acid.


Step 3: Final Answer

The correct named reaction for converting phenol to salicylic acid is the Kolbe reaction.
Quick Tip: Associate key transformations with named reactions:
- Phenol \(\rightarrow\) Salicyl\textbf{aldehyde}: \textbf{R}eimer-\textbf{T}iemann (uses CH\textbf{Cl}\(_3\)).
- Phenol \(\rightarrow\) Salicylic \textbf{acid}: \textbf{K}olbe's reaction (uses \textbf{C}O\(_2\)).

Step 1: Understanding Vitamin Deficiency Diseases

Vitamins are essential micronutrients that the body needs for various biochemical functions. A deficiency in any particular vitamin can lead to a specific disease.


Step 2: Identifying the Cause of Scurvy

Scurvy is a well-known disease characterized by symptoms like weakness, fatigue, sore arms and legs, and in advanced stages, gum disease, bleeding from the skin, and poor wound healing. This disease is specifically caused by a lack of Vitamin C in the diet.


Step 3: Matching the Vitamin with its Chemical Name

The chemical name for Vitamin C is Ascorbic acid. Therefore, scurvy is caused by a deficiency of ascorbic acid.


Step 4: Analyzing the Other Options

- (A) Vitamin B1 (Thiamine): Deficiency of Vitamin B1 causes Beriberi, a disease affecting the nervous system and cardiovascular system.

- (B) Vitamin B2 (Riboflavin): Deficiency of Vitamin B2 can lead to skin disorders, sores at the corners of the mouth (cheilosis), and swollen tongue (glossitis).

- (D) Glutamic acid: This is a non-essential amino acid, a building block of proteins. It is not a vitamin.


Step 5: Final Answer

Scurvy is caused by the deficiency of Vitamin C, which is chemically known as ascorbic acid.
Quick Tip: It is crucial for exams to memorize the chemical names of vitamins and the diseases caused by their deficiency. Create a table for quick revision. For example: Vit A (Retinol) - Night blindness; Vit B1 (Thiamine) - Beriberi; Vit C (Ascorbic acid) - Scurvy; Vit D (Calciferol) - Rickets.

Step 1: Understanding the Structure of Nucleic Acids

Nucleic acids (DNA and RNA) are polymers made up of repeating monomer units called nucleotides. Each nucleotide consists of three components: a pentose sugar (deoxyribose in DNA, ribose in RNA), a phosphate group, and a nitrogenous base.


Step 2: Analyzing the Linkages within a Nucleic Acid Polymer

Let's examine the roles of the different types of bonds mentioned in the options in the context of nucleic acids.

- (A) Glycosidic linkage: This bond connects the nitrogenous base to the 1' carbon of the pentose sugar within a single nucleotide. It does not join two separate nucleotides together.

- (B) Peptide linkage (-CO-NH-): This is the covalent bond that joins amino acids together to form proteins. It is not found in nucleic acids.

- (C) Hydrogen bonding: In DNA, hydrogen bonds form between complementary nitrogenous bases (A with T, and G with C) on opposite strands, holding the double helix structure together. They do not form the backbone of the polymer chain.

- (D) Phosphodiester linkage: This is the strong covalent bond that forms the backbone of DNA and RNA strands. It links the 3' carbon of one sugar molecule to the 5' carbon of another through a phosphate group. This linkage is what joins individual nucleotides together to form a polynucleotide chain.


Step 3: Final Answer

The linkage responsible for joining nucleotides together in a sequence is the phosphodiester linkage.
Quick Tip: Visualize the structure of a DNA/RNA strand: The "backbone" is made of alternating sugar and phosphate groups connected by phosphodiester bonds. The "rungs" of the DNA ladder are pairs of bases connected by hydrogen bonds. The base itself is connected to the sugar by a glycosidic bond.

Step 1: Analyze the Assertion (A)

Assertion (A) states that Copper (Cu) cannot liberate Hydrogen (H\(_2\)) gas when reacting with dilute mineral acids (like HCl, H\(_2\)SO\(_4\)). For a metal to liberate H\(_2\) from an acid, it must be able to displace hydrogen. The reaction would be:
\[ M(s) + 2H^+(aq) \rightarrow M^{2+}(aq) + H_2(g) \]
This reaction occurs only if the metal is more reactive than hydrogen. Looking at the electrochemical or activity series, Copper is placed below Hydrogen. This means Copper is less reactive and cannot displace hydrogen from dilute acids. Therefore, Assertion (A) is true.


Step 2: Analyze the Reason (R)

Reason (R) states that Cu has a positive electrode potential. The standard electrode potential (reduction potential) is a measure of a substance's tendency to be reduced.

- The standard reduction potential of the hydrogen electrode is defined as 0.00 V:

\[ 2H^+(aq) + 2e^- \rightarrow H_2(g), \quad E^\circ = 0.00 \, V \]
- The standard reduction potential of Copper is positive:

\[ Cu^{2+}(aq) + 2e^- \rightarrow Cu(s), \quad E^\circ = +0.34 \, V \]
A positive E\(^\circ\) value means that Cu\(^{2+}\) has a greater tendency to be reduced (gain electrons) compared to H\(^+\). Conversely, Cu metal has a lower tendency to be oxidized (lose electrons) compared to H\(_2\) gas. Therefore, Reason (R) is true.


Step 3: Evaluate the Relationship between Assertion (A) and Reason (R)

For the reaction (Cu(s) + 2H\(^+\) \(\rightarrow\) Cu\(^{2+}\) + H\(_2\)) to be spontaneous, the overall cell potential (E\(^\circ_{cell}\)) must be positive.

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

Here, H\(^+\) is reduced (cathode) and Cu is oxidized (anode).
\[ E^\circ_{cell} = E^\circ_{H^+/H_2} - E^\circ_{Cu^{2+}/Cu} \] \[ E^\circ_{cell} = 0.00 \, V - (+0.34 \, V) = -0.34 \, V \]
Since E\(^\circ_{cell}\) is negative, the reaction is non-spontaneous. The reason the reaction is non-spontaneous is precisely because the reduction potential of copper is higher (more positive) than that of hydrogen. Thus, the positive electrode potential of Cu is the correct explanation for why it cannot liberate H\(_2\) from dilute acids.


Step 4: Final Answer

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).
Quick Tip: In the electrochemical series, any metal with a negative standard reduction potential (E\(^\circ \textless 0\)) is placed above hydrogen and can displace H\(_2\) from dilute acids. Metals with a positive E\(^\circ\) (like Cu, Ag, Au, Pt) are below hydrogen and cannot.

Step 1: Analyze the Assertion (A)

Assertion (A) states that for a first-order reaction, doubling the initial concentration of the reactant also doubles its half-life.

The formula for the half-life (t\(_{1/2}\)) of a first-order reaction is:
\[ t_{1/2} = \frac{0.693}{k} \]
where k is the rate constant.

This formula clearly shows that the half-life of a first-order reaction depends only on the rate constant (k) and is independent of the initial concentration of the reactant. Therefore, doubling the initial concentration will have no effect on the half-life. The assertion is false.


Step 2: Analyze the Reason (R)

Reason (R) states that the half-life of a reaction does not depend upon the initial concentration of the reactant in a first-order reaction.

As established in Step 1, the formula t\(_{1/2}\) = 0.693/k does not contain any term for the initial concentration ([R]\(_0\)). This means the statement is a correct property of first-order reactions. Therefore, the Reason (R) is true.


Step 3: Final Answer

The Assertion (A) is false, and the Reason (R) is true. This corresponds to option (D).
Quick Tip: Memorize the half-life dependencies for different reaction orders:
- \textbf{Zero Order:} t\(_{1/2} = \frac{[R]_0}{2k}\) (Directly proportional to initial concentration)
- \textbf{First Order:} t\(_{1/2} = \frac{0.693}{k}\) (Independent of initial concentration)
- \textbf{Second Order:} t\(_{1/2} = \frac{1}{k[R]_0}\) (Inversely proportional to initial concentration)

Step 1: Analyze the Assertion (A)

Assertion (A) states that Vitamin D cannot be stored in our body.

Vitamins are classified into two groups: water-soluble (B-complex and C) and fat-soluble (A, D, E, and K).

Water-soluble vitamins are not stored in the body to a significant extent and must be regularly supplied through the diet. Excess amounts are typically excreted in urine.

Fat-soluble vitamins, including Vitamin D, dissolve in fat and can be stored in the body's fatty tissues and the liver for long periods.

Therefore, the statement that Vitamin D cannot be stored in our body is incorrect. Assertion (A) is false.


Step 2: Analyze the Reason (R)

Reason (R) states that Vitamin D is a fat-soluble vitamin and is not excreted from the body in urine.

As explained above, Vitamin D is indeed a fat-soluble vitamin. Because it is stored in fat, it is not readily excreted through the watery medium of urine. This statement is correct. Therefore, Reason (R) is true.


Step 3: Final Answer

The Assertion (A) is false, and the Reason (R) is true. This corresponds to option (D).

The reason actually contradicts the assertion. If Vitamin D is fat-soluble and not easily excreted (Reason), then it logically follows that it *can* be stored in the body, which makes the Assertion false.
Quick Tip: Remember the two classes of vitamins: \textbf{Fat-Soluble:} A, D, E, K (Can be stored in the body; overdose can be toxic). \textbf{Water-Soluble:} B vitamins and Vitamin C (Not stored; must be consumed regularly).

Step 1: Analyze the Assertion (A)

Assertion (A) states that aromatic primary amines (like aniline) cannot be prepared by the Gabriel Phthalimide synthesis.

The Gabriel synthesis is a classic method for preparing pure primary alkyl amines. The key step in this synthesis is the S\(_N\)2 reaction of an alkyl halide with the potassium salt of phthalimide (phthalimide anion).
This method is indeed limited to the synthesis of primary alkyl amines. It cannot be used for aromatic primary amines. Therefore, Assertion (A) is true.


Step 2: Analyze the Reason (R)

Reason (R) states that aryl halides do not undergo nucleophilic substitution reaction with the anion formed by phthalimide.

The phthalimide anion is a nucleophile. For the synthesis of an aromatic amine, an aryl halide (e.g., chlorobenzene) would be needed as the substrate. However, aryl halides are very unreactive towards nucleophilic substitution (S\(_N\)Ar) reactions under normal conditions. This is due to two main factors:

1. Partial double-bond character: The C-X bond (where X is a halogen) in an aryl halide has partial double-bond character due to resonance with the benzene ring, making the bond stronger and harder to break.

2. Instability of phenyl cation: The S\(_N\)1 mechanism is ruled out as it would require the formation of a highly unstable phenyl cation.

3. Repulsion: The electron-rich nucleophile is repelled by the electron-rich benzene ring.

Because of this low reactivity, the phthalimide anion cannot displace the halide from the aryl halide. Therefore, Reason (R) is true.


Step 3: Evaluate the Relationship between Assertion (A) and Reason (R)

The inability of aryl halides to react with the phthalimide nucleophile (the reason) is the direct cause for the failure of the Gabriel synthesis to produce aromatic primary amines (the assertion). The key step of the reaction simply does not work with aryl halides. Thus, the Reason is the correct explanation for the Assertion.


Step 4: Final Answer

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).
Quick Tip: Remember that Gabriel Phthalimide synthesis is a method specifically for \textbf{primary alkyl amines}. It fails for secondary, tertiary, and aromatic amines. The core of the method is an S\(_N\)2 reaction, which works well with primary alkyl halides but not with aryl or sterically hindered halides.

Step 1: Understanding IUPAC Nomenclature Rules for Coordination Compounds

1. The cation is named before the anion.

2. Within the coordination sphere (the part in square brackets), ligands are named first in alphabetical order, followed by the central metal atom.

3. The oxidation state of the central metal is written in Roman numerals in parentheses.

4. The names of anionic ligands end in '-o' (e.g., chloro), neutral ligands have their usual names (except for special cases like aqua for H\(_2\)O, ammine for NH\(_3\)).

5. Prefixes like di-, tri-, tetra- are used to indicate the number of each ligand. These prefixes are ignored for alphabetical ordering.


Step 2: Applying the Rules to [Co(NH\(_3\))\(_4\)(H\(_2\)O)Cl]Cl\(_2\)

1. Identify Cation and Anion:

The cation is the complex ion [Co(NH\(_3\))\(_4\)(H\(_2\)O)Cl]\(^{2+}\).

The anion is the counter-ion, which is two Cl\(^-\) ions (chloride).


2. Name the Ligands in the Cation:

- NH\(_3\): ammine (four of them, so tetraammine)

- H\(_2\)O: aqua (one of them, so aqua)

- Cl: chloro (one of them, so chloro)


3. Order the Ligands Alphabetically:

Comparing aqua, ammine, and chloro: ammine comes first, then aqua, then chloro.

So the order is: tetraammineaquachloro.


4. Determine the Oxidation State of the Central Metal (Cobalt):

Let the oxidation state of Cobalt (Co) be \(x\).

The charge of NH\(_3\) is 0. The charge of H\(_2\)O is 0. The charge of Cl inside the sphere is -1. The charge of each counter-ion Cl is -1.

The overall charge of the compound is 0.
\[ x + 4(0) + 1(0) + 1(-1) + 2(-1) = 0 \] \[ x - 1 - 2 = 0 \] \[ x = +3 \]
The oxidation state of Cobalt is +3, written as (III).


5. Name the Central Metal:

Since the complex ion is a cation, the metal name is used as is: cobalt.


6. Assemble the Cation Name:

Tetraammineaquachlorocobalt(III)


7. Name the Anion:

The anion is Cl\(^-\), which is named chloride. (Note: we don't use 'di-' for counter-ions).


Step 3: Final Answer

Combining the cation and anion names gives the final IUPAC name: Tetraammineaquachlorocobalt(III) chloride.
Quick Tip: When ordering ligands alphabetically, ignore the numerical prefixes (di, tri, tetra). For example, in 'tetraammine', you consider 'ammine' for alphabetization.

Step 1: Understanding IUPAC Nomenclature Rules

The rules are the same as in the previous question. An additional rule is applied for complex ligands (ligands that already have a numerical prefix in their name, like ethylenediamine). For such ligands, prefixes like bis- (for 2), tris- (for 3), tetrakis- (for 4) are used, and the ligand name is placed in parentheses.


Step 2: Applying the Rules to [CrCl\(_2\)(en)\(_2\)]Cl

1. Identify Cation and Anion:

The cation is [CrCl\(_2\)(en)\(_2\)]\(^+\).

The anion is Cl\(^-\).


2. Name the Ligands in the Cation:

- Cl: chloro (two of them, so dichloro)

- en: ethylenediamine (a neutral, bidentate ligand). There are two of them. Since its name already contains 'di', we use the prefix bis- and enclose the ligand name in parentheses: bis(ethylenediamine).


3. Order the Ligands Alphabetically:

Comparing chloro and ethylenediamine: chloro comes first.

So the order is: dichlorobis(ethylenediamine).


4. Determine the Oxidation State of the Central Metal (Chromium):

Let the oxidation state of Chromium (Cr) be \(x\).

The charge of Cl inside the sphere is -1. The charge of ethylenediamine (en) is 0. The charge of the counter-ion Cl is -1.
\[ x + 2(-1) + 2(0) + 1(-1) = 0 \] \[ x - 2 - 1 = 0 \] \[ x = +3 \]
The oxidation state of Chromium is +3, written as (III).


5. Name the Central Metal:

Since the complex is a cation, the metal name is chromium.


6. Assemble the Cation Name:

Dichlorobis(ethylenediamine)chromium(III)


7. Name the Anion:

The anion is Cl\(^-\), named chloride.


Step 3: Final Answer

The complete IUPAC name is Dichlorobis(ethylenediamine)chromium(III) chloride.
Quick Tip: For complex ligands (e.g., ethylenediamine, EDTA), use prefixes like bis-, tris-, tetrakis- instead of di-, tri-, tetra-, and always enclose the ligand name in parentheses.

Step 1: Definitions

Rate Law:

The Rate Law (or Rate Equation) for a chemical reaction is a mathematical expression that describes how the rate of reaction depends on the concentration of the reactants. For a general reaction A + B \(\rightarrow\) Products, the rate law is typically written as:
\[ Rate = k[A]^x[B]^y \]
Here, [A] and [B] are the molar concentrations of the reactants, \(k\) is the rate constant, and the exponents \(x\) and \(y\) are the orders of the reaction with respect to A and B, respectively. These exponents must be determined experimentally and are not necessarily equal to the stoichiometric coefficients.


Rate Constant (k):

The Rate Constant, denoted by \(k\), is the proportionality constant in the rate law expression. It is a measure of the intrinsic speed of a reaction. The value of \(k\) is constant for a given reaction at a specific temperature and is independent of the concentrations of the reactants. It can be defined as the rate of the reaction when the concentration of each reactant is unity (1 mol L\(^{-1}\)). The units of the rate constant depend on the overall order of the reaction.


Step 2: Identifying the Order from Units of k

Key Formula:

The general formula for the units of the rate constant \(k\) for a reaction of order \(n\) is:
\[ Units of k = (mol L^{-1})^{1-n} s^{-1} \]

(a) Units are s\(^{-1}\)

We compare the given unit with the general formula:
\[ (mol L^{-1})^{1-n} s^{-1} = s^{-1} \]
For this equality to hold, the term \((mol L^{-1})^{1-n}\) must be equal to 1. This happens when the exponent is zero:
\[ 1 - n = 0 \] \[ n = 1 \]
Therefore, the reaction is of the first order.


(b) Units are mol\(^{-1}\) L s\(^{-1}\)

First, we rewrite the given unit to match the form of the general formula:

mol\(^{-1}\) L s\(^{-1}\) = (mol L\(^{-1}\))\(^{-1}\) s\(^{-1}\).

Now, we compare the exponents:
\[ (mol L^{-1})^{1-n} s^{-1} = (mol L^{-1})^{-1} s^{-1} \]
Equating the powers of the concentration term:
\[ 1 - n = -1 \] \[ n = 1 + 1 = 2 \]
Therefore, the reaction is of the second order.
Quick Tip: A quick way to find the order `n` from the units of `k`: If units are s⁻¹, `n=1`. If units involve `mol` and `L`, let `p` be the power of `mol`. Then `n = 1 - p`. For (b), unit is mol⁻¹ L s⁻¹, so `p = -1`. Thus `n = 1 - (-1) = 2`.

Step 1: Understanding the Reaction

This is a redox reaction in a neutral or weakly alkaline medium (indicated by H\(_2\)O as a reactant). We need to identify the oxidation and reduction half-reactions and balance them.
- Permanganate ion (MnO\(_4^-\)) is a strong oxidizing agent. In neutral/alkaline solution, Mn(+7) is reduced to MnO\(_2\) (Mn(+4)).
- Thiosulphate ion (S\(_2\)O\(_3^{2-}\)) will be oxidized. The sulphur atoms in S\(_2\)O\(_3^{2-}\) have an average oxidation state of +2. It gets oxidized to sulphate ion (SO\(_4^{2-}\)), where the oxidation state of sulphur is +6.

Step 2: Balancing the Half-Reactions (in basic/neutral medium)

Reduction Half-Reaction: \[ MnO_4^- \longrightarrow MnO_2 \]
- Balance Mn: Already balanced.
- Balance O atoms by adding H\(_2\)O: MnO\(_4^-\) \(\longrightarrow\) MnO\(_2\) + 2H\(_2\)O
- Balance H atoms by adding H\(^+\) (and then convert to basic medium): MnO\(_4^-\) + 4H\(^+\) \(\longrightarrow\) MnO\(_2\) + 2H\(_2\)O
- Balance charge by adding electrons: MnO\(_4^-\) + 4H\(^+\) + 3e\(^-\) \(\longrightarrow\) MnO\(_2\) + 2H\(_2\)O
- (Alternative for basic medium): Balance O by H₂O, then H by OH⁻ and H₂O.
MnO\(_4^-\) + 2H\(_2\)O \(\longrightarrow\) MnO\(_2\) + 4OH⁻
Balance charge with e⁻: MnO\(_4^-\) + 2H\(_2\)O + 3e\(^-\) \(\longrightarrow\) MnO\(_2\) + 4OH⁻ (This is the balanced half-reaction in basic medium).

Oxidation Half-Reaction: \[ S_2O_3^{2-} \longrightarrow SO_4^{2-} \]
- Balance S atoms: S\(_2\)O\(_3^{2-}\) \(\longrightarrow\) 2SO\(_4^{2-}\)
- Balance O atoms by adding H\(_2\)O: S\(_2\)O\(_3^{2-}\) + 5H\(_2\)O \(\longrightarrow\) 2SO\(_4^{2-}\)
- Balance H atoms by adding H\(^+\): S\(_2\)O\(_3^{2-}\) + 5H\(_2\)O \(\longrightarrow\) 2SO\(_4^{2-}\) + 10H\(^+\)
- Balance charge by adding electrons: S\(_2\)O\(_3^{2-}\) + 5H\(_2\)O \(\longrightarrow\) 2SO\(_4^{2-}\) + 10H\(^+\) + 8e\(^-\)
- (Alternative for basic medium):
S\(_2\)O\(_3^{2-}\) + 10OH⁻ \(\longrightarrow\) 2SO\(_4^{2-}\) + 5H\(_2\)O + 8e⁻

Step 3: Combining the Half-Reactions

We have:
Reduction: MnO\(_4^-\) + 2H\(_2\)O + 3e\(^-\) \(\longrightarrow\) MnO\(_2\) + 4OH⁻ (multiply by 8)
Oxidation: S\(_2\)O\(_3^{2-}\) + 10OH⁻ \(\longrightarrow\) 2SO\(_4^{2-}\) + 5H\(_2\)O + 8e⁻ (multiply by 3)

The number of electrons must be equal. The least common multiple of 3 and 8 is 24. \[ 8(MnO_4^- + 2H_2O + 3e^- \longrightarrow MnO_2 + 4OH^-) \] \[ \Rightarrow 8MnO_4^- + 16H_2O + 24e^- \longrightarrow 8MnO_2 + 32OH^- \]
\[ 3(S_2O_3^{2-} + 10OH^- \longrightarrow 2SO_4^{2-} + 5H_2O + 8e^-) \] \[ \Rightarrow 3S_2O_3^{2-} + 30OH^- \longrightarrow 6SO_4^{2-} + 15H_2O + 24e^- \]

Now, add the two balanced equations and cancel common species: \[ 8MnO_4^- + 16H_2O + 3S_2O_3^{2-} + 30OH^- \longrightarrow 8MnO_2 + 32OH^- + 6SO_4^{2-} + 15H_2O \]

Cancel H\(_2\)O and OH\(^-\) from both sides:
- 15 H\(_2\)O on RHS cancels 15 of 16 H\(_2\)O on LHS, leaving 1 H\(_2\)O on LHS.
- 30 OH\(^-\) on LHS cancels 30 of 32 OH\(^-\) on RHS, leaving 2 OH\(^-\) on RHS.

Step 4: Final Answer

The final balanced equation is: \[ 8MnO_4^- + 3S_2O_3^{2-} + H_2O \longrightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^- \] Quick Tip: When balancing redox reactions, always start by writing the oxidation and reduction half-reactions. Balance atoms other than O and H first, then balance O with H₂O, H with H⁺ (in acid) or H₂O/OH⁻ (in base), and finally balance the charge with electrons.

Step 1: Understanding the Reaction

This is a redox reaction in an acidic medium, as indicated by the presence of H\(^+\) ions.
- Dichromate ion (Cr\(_2\)O\(_7^{2-}\)) is a strong oxidizing agent. Chromium in +6 oxidation state is reduced, typically to Cr\(^{3+}\).
- Tin(II) ion (Sn\(^{2+}\)) is a reducing agent and will be oxidized to Tin(IV) ion (Sn\(^{4+}\)).

Step 2: Balancing the Half-Reactions (in acidic medium)

Reduction Half-Reaction: \[ Cr_2O_7^{2-} \longrightarrow Cr^{3+} \]
- Balance Cr atoms: Cr\(_2\)O\(_7^{2-}\) \(\longrightarrow\) 2Cr\(^{3+}\)
- Balance O atoms by adding H\(_2\)O: Cr\(_2\)O\(_7^{2-}\) \(\longrightarrow\) 2Cr\(^{3+}\) + 7H\(_2\)O
- Balance H atoms by adding H\(^+\): Cr\(_2\)O\(_7^{2-}\) + 14H\(^+\) \(\longrightarrow\) 2Cr\(^{3+}\) + 7H\(_2\)O
- Balance charge by adding electrons: The charge on LHS is (-2) + 14(+1) = +12. The charge on RHS is 2(+3) = +6. Add 6e\(^-\) to LHS. \[ Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O \]

Oxidation Half-Reaction: \[ Sn^{2+} \longrightarrow Sn^{4+} \]
- Balance Sn atoms: Already balanced.
- Balance charge by adding electrons: The charge on LHS is +2, on RHS is +4. Add 2e\(^-\) to RHS. \[ Sn^{2+} \longrightarrow Sn^{4+} + 2e^- \]

Step 3: Combining the Half-Reactions

To make the number of electrons equal in both half-reactions, we need to multiply the oxidation half-reaction by 3.
Reduction: Cr\(_2\)O\(_7^{2-}\) + 14H\(^+\) + 6e\(^-\) \(\longrightarrow\) 2Cr\(^{3+}\) + 7H\(_2\)O
Oxidation: 3(Sn\(^{2+}\) \(\longrightarrow\) Sn\(^{4+}\) + 2e\(^-\)) \(\Rightarrow\) 3Sn\(^{2+}\) \(\longrightarrow\) 3Sn\(^{4+}\) + 6e\(^-\)

Now add the two equations: \[ Cr_2O_7^{2-} + 14H^+ + 6e^- + 3Sn^{2+} \longrightarrow 2Cr^{3+} + 7H_2O + 3Sn^{4+} + 6e^- \]
Cancel the electrons (6e\(^-\)) from both sides.

Step 4: Final Answer

The final balanced equation is: \[ Cr_2O_7^{2-} + 3Sn^{2+} + 14H^+ \longrightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O \] Quick Tip: In acidic medium balancing, remember the sequence: 1. Balance main atoms. 2. Balance Oxygen with H₂O. 3. Balance Hydrogen with H⁺. 4. Balance charge with e⁻. This systematic approach prevents errors.

Step 1: Understanding the Concept

The question asks for the time required for a concentration change in a zero-order reaction, given the rate constant and the initial and final concentrations. We need to use the integrated rate law for a zero-order reaction.


Step 2: Key Formula or Approach

The integrated rate law for a zero-order reaction is given by: \[ [A]_t = -kt + [A]_0 \]
where:
- \([A]_t\) is the concentration of reactant A at time \(t\).
- \([A]_0\) is the initial concentration of reactant A.
- \(k\) is the rate constant.
- \(t\) is the time.

Step 3: Detailed Explanation

We are given the following values:
- Rate constant, \(k = 0.0030 mol L^{-1}s^{-1}\)
- Initial concentration, \([A]_0 = 0.10 M\)
- Final concentration, \([A]_t = 0.075 M\)

We need to find the time, \(t\).

Rearrange the integrated rate law to solve for \(t\): \[ kt = [A]_0 - [A]_t \] \[ t = \frac{[A]_0 - [A]_t}{k} \]
Now, substitute the given values into the formula: \[ t = \frac{0.10 M - 0.075 M}{0.0030 mol L^{-1}s^{-1}} \] \[ t = \frac{0.025 mol L^{-1}}{0.0030 mol L^{-1}s^{-1}} \] \[ t = \frac{25}{3} s \] \[ t \approx 8.33 s \]

Step 4: Final Answer

It will take 8.33 seconds for the concentration of A to fall from 0.10 M to 0.075 M.
Quick Tip: For a zero-order reaction, the rate of reaction is constant (Rate = k). This means the concentration decreases linearly with time. You can think of it as `change in concentration = rate × time`. Here, `Δ[A] = kt`.

Step 1: Understanding the Concept

The problem involves relating the overall rate of a zero-order reaction to the rates of appearance of products. The relationship depends on the stoichiometry of the balanced chemical equation.


Step 2: Key Formula or Approach

First, write the balanced chemical equation for the decomposition of ammonia (NH\(_3\)): \[ 2NH_3(g) \xrightarrow{Pt} N_2(g) + 3H_2(g) \]
The rate of reaction can be expressed in terms of the change in concentration of reactants and products: \[ Rate = -\frac{1}{2} \frac{d[NH_3]}{dt} = +\frac{1}{1} \frac{d[N_2]}{dt} = +\frac{1}{3} \frac{d[H_2]}{dt} \]
For a zero-order reaction, the rate law is: \[ Rate = k \]
where \(k\) is the rate constant.

Step 3: Detailed Explanation

We are given:
- The reaction is zero-order.
- Rate constant, \(k = 2.5 \times 10^{-4} mol L^{-1}s^{-1}\).

From the rate law for a zero-order reaction, the overall rate of the reaction is equal to the rate constant. \[ Rate = k = 2.5 \times 10^{-4} mol L^{-1}s^{-1} \]

Now, we can find the rates of production of N\(_2\) and H\(_2\) using the stoichiometric relationship.


Rate of production of N\(_2\):
The rate of production of N\(_2\) is \(\frac{d[N_2]}{dt}\).
From the relationship, \(Rate = \frac{d[N_2]}{dt}\). \[ \frac{d[N_2]}{dt} = Rate = k \] \[ \frac{d[N_2]}{dt} = 2.5 \times 10^{-4} mol L^{-1}s^{-1} \]

Rate of production of H\(_2\):
The rate of production of H\(_2\) is \(\frac{d[H_2]}{dt}\).
From the relationship, \(Rate = \frac{1}{3} \frac{d[H_2]}{dt}\).
Rearranging for the rate of production of H\(_2\): \[ \frac{d[H_2]}{dt} = 3 \times Rate = 3 \times k \] \[ \frac{d[H_2]}{dt} = 3 \times (2.5 \times 10^{-4} mol L^{-1}s^{-1}) \] \[ \frac{d[H_2]}{dt} = 7.5 \times 10^{-4} mol L^{-1}s^{-1} \]

Step 4: Final Answer

The rate of production of N\(_2\) is \(2.5 \times 10^{-4}\) mol L\(^{-1}\) s\(^{-1}\).

The rate of production of H\(_2\) is \(7.5 \times 10^{-4}\) mol L\(^{-1}\) s\(^{-1}\).
Quick Tip: For any reaction `aA + bB → cC + dD`, the rate expression is `Rate = -1/a d[A]/dt = -1/b d[B]/dt = +1/c d[C]/dt = +1/d d[D]/dt`. For a zero-order reaction, this entire expression is simply equal to `k`.

Step 1: Understanding the Concept

This question addresses the reactivity of aryl halides towards nucleophilic aromatic substitution (S\(_N\)Ar). Generally, aryl halides like chlorobenzene are unreactive towards this type of reaction because of the strong C-Cl bond (due to partial double bond character from resonance) and repulsion between the incoming nucleophile and the electron-rich benzene ring. However, the presence of certain groups on the ring can drastically alter this reactivity.


Step 2: Analyzing the Role of the Nitro Group

The reaction is the substitution of -Cl by the nucleophile -OH from NaOH.

1. Chlorobenzene: It has no activating groups. For the reaction to occur, extremely harsh conditions are required (Dow's Process: 623 K and 300 atm pressure). This indicates a very high activation energy for the reaction.

2. p-Chloronitrobenzene: This molecule has a strong electron-withdrawing group (-NO\(_2\)) at the para position relative to the chlorine atom. This group significantly increases the reactivity of the aryl halide towards nucleophilic substitution. The reaction occurs under much milder conditions (443 K).


Step 3: Detailed Explanation (Mechanism)

The nucleophilic aromatic substitution proceeds via a two-step addition-elimination mechanism, forming a resonance-stabilized carbanion intermediate known as a Meisenheimer complex.


- When the OH\(^-\) nucleophile attacks the carbon atom bonded to chlorine in p-chloronitrobenzene, a negative charge develops on the ring.

- This negative charge is delocalized over the ring through resonance. Crucially, the -NO\(_2\) group at the para-position participates in this delocalization. One of the resonance structures places the negative charge directly on the oxygen atom of the nitro group.


- This particular resonance structure is highly stable because the negative charge is on a very electronegative oxygen atom. This extensive delocalization and stabilization of the intermediate carbanion lowers the activation energy of the reaction, making it proceed much faster and under milder conditions.

- In contrast, for chlorobenzene, the intermediate carbanion is only stabilized by delocalization within the benzene ring itself. It lacks the extra stabilization provided by an electron-withdrawing group. Consequently, the activation energy is much higher, and drastic conditions are needed to force the reaction.
Quick Tip: Remember that electron-withdrawing groups (like -NO₂, -CN, -SO₃H) at ortho and para positions activate aryl halides for nucleophilic substitution, while electron-donating groups deactivate them. The meta position has little to no effect.

Step 1: Understanding the Concept

This observation is a classic example of the reactivity of an ambidentate nucleophile. The cyanide ion (CN\(^-\)) is an ambidentate nucleophile because it has two potential nucleophilic centers: the carbon atom and the nitrogen atom. The product formed depends on the nature of the attacking reagent (KCN or AgCN) and the reaction conditions.


Step 2: Reaction with Potassium Cyanide (KCN)

Nature of KCN: Potassium cyanide is predominantly ionic in nature. In a polar solvent, it readily dissociates to provide K\(^+\) and free cyanide ions, CN\(^-\).

Nucleophilic Attack: The cyanide ion ([:C\(\equiv\)N:]\(^-\)) can attack the electrophilic carbon of chloroethane. Although both C and N can donate electrons, the attack preferentially occurs through the carbon atom. This is because the resulting Carbon-Carbon (C-C) bond in the nitrile is stronger and more stable than the Carbon-Nitrogen (C-N) bond that would be formed if the attack occurred through nitrogen.

Reaction: \[ CH_3CH_2-Cl + K^+[:C\equivN:]^- \longrightarrow CH_3CH_2-C\equivN + KCl \]
The product is propane nitrile (or ethyl cyanide).


Step 3: Reaction with Silver Cyanide (AgCN)

Nature of AgCN: Silver cyanide is predominantly covalent in nature. The bond between silver and carbon (Ag-C) is strong and does not easily dissociate to give free CN\(^-\) ions.

Nucleophilic Attack: Since the carbon atom is covalently bonded to silver, it is not freely available to form a new bond. However, the nitrogen atom has a lone pair of electrons, which is available for donation. Therefore, the nucleophilic attack on the electrophilic carbon of chloroethane occurs through the nitrogen atom.

Reaction: \[ CH_3CH_2-Cl + Ag-C\equivN: \longrightarrow CH_3CH_2-N^+\equivC^- + AgCl \]
The product is ethyl isocyanide (or ethyl isonitrile).


Step 4: Final Answer

The difference in products is due to the ionic nature of KCN versus the covalent nature of AgCN. KCN provides free CN\(^-\) ions for C-attack, forming nitriles. In AgCN, the carbon is blocked by the covalent bond to Ag, forcing the N-atom's lone pair to act as the nucleophile, forming isocyanides.
Quick Tip: A simple way to remember this: \textbf{I}onic KCN gives N\textbf{i}trile. \textbf{C}ovalent AgCN gives Isocyanide (think of the 'C' in covalent and isocyanide being different from the 'N' in nitrile). A similar logic applies to KNO₂ (ionic, gives R-ONO, nitrite) and AgNO₂ (covalent, gives R-NO₂, nitro).

Step 1: Understanding the Concept

This problem requires the application of Henry's Law, which relates the partial pressure of a gas over a solution to the mole fraction of the gas dissolved in the solution. We will use this law to find the mole fraction of CO\(_2\) and then use the amount of solvent (water) to calculate the moles of CO\(_2\).


Step 2: Key Formula or Approach

Henry's Law is given by the formula: \[ p = K_H \times x \]
where:
- \(p\) is the partial pressure of the gas (CO\(_2\)).
- \(K_H\) is Henry's law constant.
- \(x\) is the mole fraction of the gas in the solution.

The mole fraction of CO\(_2\), \(x_{CO_2}\), is defined as: \[ x_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \]
where \(n_{CO_2}\) and \(n_{H_2O}\) are the number of moles of CO\(_2\) and water, respectively.


Step 3: Detailed Calculation

Given values:

- Partial pressure of CO\(_2\), \(p = 3.34 \times 10^5\) Pa.

- Henry's law constant, \(K_H = 1.67 \times 10^8\) Pa.

- Mass of water (soda water is mostly water), \(m_{H_2O} = 540\) g.

- Molar mass of water, \(M_{H_2O} = 18.015\) g/mol.


1. Calculate the mole fraction of CO\(_2\) (\(x_{CO_2}\)):

Rearranging Henry's Law: \[ x_{CO_2} = \frac{p}{K_H} \] \[ x_{CO_2} = \frac{3.34 \times 10^5 Pa}{1.67 \times 10^8 Pa} = 2 \times 10^{-3} = 0.002 \]

2. Calculate the number of moles of water (\(n_{H_2O}\)):
\[ n_{H_2O} = \frac{Mass of water}{Molar mass of water} = \frac{540 g}{18 g/mol} = 30 mol \]

3. Calculate the number of moles of CO\(_2\) (\(n_{CO_2}\)):

From the definition of mole fraction: \[ x_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \]
Since the solubility of CO\(_2\) is low, the number of moles of CO\(_2\) is much smaller than the number of moles of water (\(n_{CO_2} \ll n_{H_2O}\)). Therefore, we can approximate the denominator: \[ n_{CO_2} + n_{H_2O} \approx n_{H_2O} \]
The formula simplifies to: \[ x_{CO_2} \approx \frac{n_{CO_2}}{n_{H_2O}} \]
Rearranging to solve for \(n_{CO_2}\): \[ n_{CO_2} \approx x_{CO_2} \times n_{H_2O} \] \[ n_{CO_2} \approx 0.002 \times 30 mol = 0.06 mol \]

Step 4: Final Answer

The number of moles of CO\(_2\) in 540 g of soda water is 0.06 mol.
Quick Tip: In Henry's Law problems for gases with low solubility, you can almost always approximate the term `(moles of gas + moles of solvent)` in the mole fraction calculation to just `moles of solvent`. This simplifies the calculation significantly.

Step 1: Understanding the Technologies

- Thermal Power Plants: Generate electricity by burning fossil fuels (like coal) to produce heat, which boils water to create steam. The steam drives a turbine, which in turn spins a generator to produce electricity. This involves multiple energy conversions (Chemical \(\rightarrow\) Heat \(\rightarrow\) Mechanical \(\rightarrow\) Electrical).

- Fuel Cells: Generate electricity through an electrochemical reaction, typically by reacting hydrogen and oxygen. They convert chemical energy directly into electrical energy without combustion.


Step 2: Key Reasons for Preference

There are two primary reasons why fuel cells are preferred over thermal plants:

1. Higher Efficiency:

In a thermal plant, each energy conversion step involves some energy loss, primarily as heat, according to the second law of thermodynamics. The overall efficiency of a modern thermal plant is typically around 30-40%.

Fuel cells, on the other hand, convert chemical energy directly into electrical energy. This direct conversion process is much more efficient, with efficiencies reaching 60-70%. If the waste heat is also utilized (cogeneration), the overall efficiency can exceed 85%.


2. Low Pollution:

Thermal plants burn fossil fuels, which releases harmful pollutants into the atmosphere, including greenhouse gases (like CO\(_2\)), sulfur oxides (SOx), nitrogen oxides (NOx), and particulate matter. These contribute to acid rain, smog, and global warming.

The primary product of a hydrogen-oxygen fuel cell is water. This makes them essentially pollution-free at the point of operation, which is a significant environmental advantage.
Quick Tip: When comparing energy sources, always consider two main factors: \textbf{efficiency} (how much useful energy you get out) and \textbf{environmental impact} (what pollutants are produced). Fuel cells win on both counts against traditional thermal plants.

Step 1: Understanding Galvanization and Corrosion

Galvanization is the process of applying a protective zinc coating to iron or steel to prevent rusting. Rusting (corrosion of iron) is an electrochemical process that requires both oxygen and water. It involves the oxidation of iron.


Step 2: Electrochemical Principle

The reason for the protection lies in the relative positions of zinc and iron in the electrochemical series. Zinc is more reactive (more easily oxidized) than iron. Their standard reduction potentials are: \[ E^\circ_{Zn^{2+}/Zn} = -0.76 \, V \] \[ E^\circ_{Fe^{2+}/Fe} = -0.44 \, V \]
Since zinc has a more negative reduction potential, it has a greater tendency to be oxidized than iron.


Step 3: Detailed Explanation

When the zinc coating on a galvanized pipe is broken and both metals are exposed to oxygen and moisture, an electrochemical cell (or a galvanic cell) is formed.

- Anode (Oxidation): The more reactive metal, zinc, acts as the anode. It gets oxidized, meaning it corrodes and loses electrons.
\[ Zn(s) \longrightarrow Zn^{2+}(aq) + 2e^- \]
- Cathode (Reduction): The less reactive metal, iron, acts as the cathode. The electrons released by the zinc travel to the iron surface, where they are used to reduce oxygen.
\[ O_2(g) + 4H^+(aq) + 4e^- \longrightarrow 2H_2O(l) \]
Because zinc is continuously being oxidized in place of iron, the iron itself is protected from rusting. This method of protection is called sacrificial protection, as the zinc "sacrifices" itself to protect the iron. This protection continues as long as there is some zinc present.
Quick Tip: For sacrificial protection, the coating metal must be \textbf{more reactive} (have a more negative E\(^\circ\)) than the metal being protected. For example, zinc protects iron, but a tin coating on iron (used in tin cans) would cause the iron to rust faster if scratched, because iron is more reactive than tin.

Step 1: Understanding Electrolytic Conductance Measurement

The measurement of electrolytic conductance (or its reciprocal, resistance) is typically done using a Wheatstone bridge setup with a conductivity cell. The goal is to measure the resistance of the electrolyte solution itself.


Step 2: Problems with Using Direct Current (DC)

If a Direct Current (DC) source is used, two main problems arise:

1. Electrolysis and Change in Concentration:

DC causes a continuous, one-way flow of ions towards the electrodes. Cations move to the cathode and anions move to the anode. At the electrodes, these ions undergo redox reactions (electrolysis), getting discharged. For example, in a CuSO\(_4\) solution, Cu\(^{2+}\) would plate onto the cathode and water might be oxidized at the anode. This process changes the chemical composition and the concentration of the electrolyte in the vicinity of the electrodes. Since conductance is dependent on concentration, the resistance of the solution will change during the measurement, leading to fluctuating and incorrect readings.


2. Polarization of Electrodes:

The accumulation of products of electrolysis on the surface of the electrodes creates a "back EMF" or polarization effect. This opposing potential interferes with the applied voltage, effectively increasing the resistance of the cell and leading to erroneous measurements.


Step 3: The Solution - Using Alternating Current (AC)

To overcome these problems, an Alternating Current (AC) source is used.

- AC rapidly reverses the direction of the current flow. This means the ions just oscillate back and forth in the solution without significant net movement towards either electrode.

- As a result, electrolysis does not occur to any significant extent, and the concentration of the solution remains uniform.

- The polarization effect is also minimized because any products that might form during one half-cycle are decomposed in the next.

This ensures that the measured resistance is solely due to the bulk properties of the electrolyte solution, providing an accurate value for its conductance.
Quick Tip: Remember the simple rule: DC for \textbf{electrolysis} (when you want a reaction), AC for \textbf{conductance measurement} (when you want to avoid a reaction).

Step 1: Understanding Standard Electrode Potential (E\(^\circ\))

The standard electrode potential (E\(^\circ_{M^{2+}/M}\)) reflects the tendency for the reduction M\(^{2+}\) + 2e\(^-\) \(\rightarrow\) M to occur. A more negative E\(^\circ\) value indicates a lower tendency for the ion to be reduced, or conversely, a higher tendency for the metal (M) to be oxidized to the ion (M\(^{2+}\)). The value of E\(^\circ\) depends on the net energy change of three processes: Atomization enthalpy (\(\Delta_a H\)), Ionization enthalpy (IE\(_1\) + IE\(_2\)), and Hydration enthalpy (\(\Delta_{hyd} H\)).


Step 2: Analyzing the Electronic Configuration of Manganese

- Manganese atom (Mn): Atomic number Z = 25. Electronic configuration: [Ar] 3d\(^5\) 4s\(^2\).
- Manganese ion (Mn\(^{2+}\)): When the atom is oxidized, it loses the two 4s electrons, resulting in the electronic configuration: [Ar] 3d\(^5\).


Step 3: Explanation for the Highly Negative E\(^\circ\) Value

The key to the unusually negative E\(^\circ\) value for Mn lies in the exceptional stability of the Mn\(^{2+}\) ion.

The [Ar] 3d\(^5\) configuration is a half-filled d-subshell. According to Hund's rule of maximum multiplicity, electronic configurations with half-filled or completely filled orbitals are particularly stable. This stability arises from two factors:

1. Symmetrical distribution of electrons: The five d-orbitals are each occupied by one electron, leading to a spherically symmetrical and stable arrangement.

2. High exchange energy: The number of possible exchanges between electrons with the same spin is maximized, which leads to a large release of energy, further stabilizing the configuration.


Because the resulting Mn\(^{2+}\) ion is so stable, the oxidation of Mn metal to Mn\(^{2+}\) is more favorable than for its neighbours (like Fe and Cr), whose M\(^{2+}\) ions do not have this special half-filled d-orbital stability. This greater tendency for Mn to be oxidized is reflected in its more negative standard electrode potential.
Quick Tip: Look for electronic stability when explaining anomalies in trends for d-block elements. The stability of d⁰, d⁵ (half-filled), and d¹⁰ (fully-filled) configurations is a very common reason for unusual values of ionization enthalpy, electrode potential, and stability of oxidation states.

Step 1: Definition

Lanthanoid contraction is the steady and gradual decrease in the size (atomic and ionic radii) of the lanthanoid elements as the atomic number increases from Lanthanum (La, Z=57) to Lutetium (Lu, Z=71).


Step 2: Cause of Lanthanoid Contraction

The cause of this phenomenon lies in the electronic structure of the lanthanoids.

1. Filling of 4f Orbitals: As we move across the lanthanoid series, each successive element adds one proton to the nucleus and one electron to the inner 4f subshell.

2. Poor Shielding Effect of 4f Electrons: The electrons in the 4f orbitals have a very diffuse (spread out) shape. Because of this shape, they are very poor at shielding the outer shell electrons (in the 5th and 6th shells) from the increasing positive charge of the nucleus.

3. Increase in Effective Nuclear Charge: As the nuclear charge increases by +1 for each element, the poor shielding by the 4f electrons means that the effective nuclear charge experienced by the outer electrons increases significantly.

4. Contraction in Size: This increased effective nuclear charge pulls the outer electron shells closer to the nucleus, resulting in a gradual decrease in both atomic and ionic radii across the series.


Step 3: Consequences of Lanthanoid Contraction

The lanthanoid contraction has important consequences for the chemistry of the elements that follow them in the periodic table. For example:

- Similarity of 2nd and 3rd Transition Series: The atomic radii of the elements of the second transition series (e.g., Zr, Zirconium) and the third transition series (e.g., Hf, Hafnium) are almost identical. This makes their separation very difficult and their chemical properties very similar.

- Basicity of Lanthanoid Hydroxides: The basic character of the hydroxides M(OH)\(_3\) decreases across the series as the ionic size decreases and the covalent character of the M-OH bond increases.
Quick Tip: Remember the cause and effect: The cause is the \textbf{poor shielding by 4f electrons}. The effect is a \textbf{decrease in size} across the lanthanoids and the \textbf{remarkable similarity} in size and properties of elements in the subsequent periods (e.g., Zr/Hf, Nb/Ta).

Step 1: Understanding Metallic Bonding in Transition Metals

The characteristic properties of transition metals, such as high melting points, high boiling points, and hardness, are attributed to the strength of their metallic bonds. This strong bonding arises from the participation of not only the outermost s-electrons but also the inner (n-1)d-electrons in the metallic lattice. The presence of a large number of unpaired d-electrons generally leads to stronger interatomic bonding.


Step 2: Analyzing the Electronic Configuration of Zn, Cd, and Hg

The elements Zinc (Zn), Cadmium (Cd), and Mercury (Hg) belong to Group 12 of the d-block. Their electronic configurations are:

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

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

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

In all three elements, the d-subshell is completely filled (d\(^{10}\)).


Step 3: Explanation for their Softness

Because the d-orbitals in Zn, Cd, and Hg are completely filled, the d-electrons are held tightly by the nucleus and do not participate in forming metallic bonds. The metallic bonding in these elements is solely due to the involvement of the outermost s-electrons (two electrons per atom).

The resulting metallic bonds are much weaker compared to those in other transition metals where both s and d-electrons contribute. This weakness in the interatomic forces has several consequences:

- They are soft metals.

- They have low melting and boiling points (Mercury is a liquid at room temperature).

- They have a lower enthalpy of atomization.

For these reasons, they are sometimes considered not to be true transition elements, as they do not show the typical characteristics of transition metals.
Quick Tip: The strength of metallic bonding in d-block elements is roughly proportional to the number of unpaired electrons. The maximum is in the middle (e.g., Cr, Mo, W), and the minimum is at the end (Zn, Cd, Hg), where there are no unpaired d-electrons.

Step 1: Determine the Oxidation State of Cobalt

The complex is [Co(NH\(_3\))\(_6\)]Cl\(_3\). The complex ion is [Co(NH\(_3\))\(_6\)]\(^{3+}\).

Let the oxidation state of Co be \(x\). Ammonia (NH\(_3\)) is a neutral ligand (charge = 0).
\(x + 6(0) = +3 \implies x = +3\).

So, we have Cobalt in the +3 oxidation state (Co\(^{3+}\)).


Step 2: Write the Electronic Configurations

- Atomic Cobalt (Co, Z=27): [Ar] 3d\(^7\) 4s\(^2\).

- Cobalt(III) ion (Co\(^{3+}\)): The atom loses three electrons (two from 4s, one from 3d). Configuration is [Ar] 3d\(^6\).


Step 3: Analyze the Ligand and Apply VBT

- The ligand is ammonia (NH\(_3\)), which is a strong field ligand.

- According to VBT, in the presence of a strong field ligand, the electrons in the d-orbitals of the central metal ion will pair up if possible, to make inner d-orbitals available for hybridization.


Step 4: Determine Hybridization and Geometry

- The ground state orbital diagram for Co\(^{3+}\) (3d\(^6\)) is:
\[ \begin{array}{cc} 3d & \begin{array}{|c|c|c|c|c|} \hline \uparrow\downarrow & \uparrow & \uparrow & \uparrow & \uparrow
\hline \end{array}
\end{array} \]
- Due to the strong field NH\(_3\) ligands, the six 3d electrons are forced to pair up:
\[ \begin{array}{cc} 3d (after pairing) & \begin{array}{|c|c|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \quad & \quad
\hline \end{array}
\end{array} \]
- Now, the complex needs six empty orbitals to accommodate the six lone pairs from the six NH\(_3\) ligands.

- The available empty orbitals are two 3d, one 4s, and three 4p orbitals.

- These orbitals hybridize to form six equivalent d\(^2\)sp\(^3\) hybrid orbitals.

\[ \begin{array}{cccccc} d\(^2\)sp\(^3\) hybridization & \underbrace{\begin{array}{|c|c|} \hline \quad & \quad
\hline \end{array}}_{3d} & \underbrace{\begin{array}{|c|} \hline \quad
\hline \end{array}}_{4s} & \underbrace{\begin{array}{|c|c|c|} \hline \quad & \quad & \quad
\hline \end{array}}_{4p}
\end{array} \]
- Each of these six hybrid orbitals accepts a pair of electrons from an NH\(_3\) ligand.

- The geometry corresponding to d\(^2\)sp\(^3\) hybridization is octahedral. Because it uses inner (3d) orbitals, it is called an inner orbital complex.


Step 5: Determine Magnetic Behaviour

- Looking at the orbital diagram of Co\(^{3+}\) after the electrons have paired up, we can see that there are no unpaired electrons.

- A species with no unpaired electrons is diamagnetic (repelled by a magnetic field).


Final Answer Summary:

- Hybridization: d\(^2\)sp\(^3\)

- Magnetic Behaviour: Diamagnetic
Quick Tip: For VBT, the key is to identify the ligand as strong field or weak field. Strong field ligands (like NH₃, CN⁻, en) cause pairing of electrons in the d-orbitals, often leading to inner orbital complexes (e.g., d²sp³) and low spin states.

Step 1: Determine the Oxidation State of Nickel

The complex is K\(_2\)[NiCl\(_4\)]. The complex ion is [NiCl\(_4\)]\(^{2-}\).

Let the oxidation state of Ni be \(x\). The chloride ligand (Cl\(^-\)) has a charge of -1.
\(x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2\).

So, we have Nickel in the +2 oxidation state (Ni\(^{2+}\)).


Step 2: Write the Electronic Configurations

- Atomic Nickel (Ni, Z=28): [Ar] 3d\(^8\) 4s\(^2\).

- Nickel(II) ion (Ni\(^{2+}\)): The atom loses the two 4s electrons. Configuration is [Ar] 3d\(^8\).


Step 3: Analyze the Ligand and Apply VBT

- The ligand is chloride (Cl\(^-\)), which is a weak field ligand.

- According to VBT, weak field ligands do not have enough energy to force the pairing of electrons in the d-orbitals of the central metal ion.


Step 4: Determine Hybridization and Geometry

- The ground state orbital diagram for Ni\(^{2+}\) (3d\(^8\)) is (following Hund's rule):
\[ \begin{array}{cc} 3d & \begin{array}{|c|c|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow & \uparrow
\hline \end{array}
\end{array} \]
- Since Cl\(^-\) is a weak field ligand, no pairing of electrons will occur.

- The complex needs four empty orbitals to accommodate the four lone pairs from the four Cl\(^-\) ligands.

- The inner 3d orbitals are not available for hybridization. The complex must use the outer orbitals. The available empty orbitals are one 4s and three 4p orbitals.

- These orbitals hybridize to form four equivalent sp\(^3\) hybrid orbitals.

\[ \begin{array}{cccc} sp\(^3\) hybridization & \underbrace{\begin{array}{|c|} \hline \quad
\hline \end{array}}_{4s} & \underbrace{\begin{array}{|c|c|c|} \hline \quad & \quad & \quad
\hline \end{array}}_{4p}
\end{array} \]
- Each of these four hybrid orbitals accepts a pair of electrons from a Cl\(^-\) ligand.

- The geometry corresponding to sp\(^3\) hybridization is tetrahedral. Because it uses outer (4s, 4p) orbitals, it is called an outer orbital complex.


Step 5: Determine Magnetic Behaviour

- Looking at the orbital diagram of Ni\(^{2+}\), we can see that there are two unpaired electrons in the 3d orbitals.

- A species with unpaired electrons is paramagnetic (attracted by a magnetic field).


Final Answer Summary:

- Hybridization: sp\(^3\)

- Magnetic Behaviour: Paramagnetic
Quick Tip: Weak field ligands (halides like Cl⁻, F⁻; H₂O) generally do not cause electron pairing. This often leads to outer orbital complexes (e.g., sp³, sp³d²) and high spin states.

Step 1: Understanding Crystal Field Theory Concepts

This question is based on Crystal Field Theory (CFT) for an octahedral complex.

- In an octahedral field, the five degenerate d-orbitals of the central metal ion split into two sets of different energy levels: a lower-energy set of three orbitals called t\(_{2g}\) (\(d_{xy}, d_{yz}, d_{zx}\)) and a higher-energy set of two orbitals called e\(_g\) (\(d_{x^2-y^2}, d_{z^2}\)).

- \(\Delta_o\) (Crystal Field Splitting Energy): This is the energy difference between the t\(_{2g}\) and e\(_g\) sets.

- P (Pairing Energy): This is the energy required to place two electrons in the same orbital, overcoming the electron-electron repulsion.


Step 2: Analyzing the Condition \(\Delta_o >\) P

The condition \(\Delta_o > P\) means that the energy gap between the t\(_{2g}\) and e\(_g\) orbitals is larger than the energy required to pair electrons in a t\(_{2g}\) orbital.

- This condition arises in the presence of strong field ligands.

- It is energetically more favorable for an electron to pair up in a lower-energy t\(_{2g}\) orbital than to jump up to a higher-energy e\(_g\) orbital. This results in a low spin complex.


Step 3: Filling the Orbitals for a d\(^5\) Ion

We need to fill 5 electrons into the d-orbitals according to the low spin rule (Aufbau principle and Hund's rule, but with the pairing priority).

1. The first three electrons will go into the three t\(_{2g}\) orbitals singly, with parallel spins (Hund's rule).
\[ t_{2g} \begin{array}{|c|c|c|} \hline \uparrow & \uparrow & \uparrow
\hline \end{array} \quad e_g \begin{array}{|c|c|} \hline \quad & \quad
\hline \end{array} \]
2. The fourth electron, instead of going into the high-energy e\(_g\) orbital (which would cost energy \(\Delta_o\)), will pair up with an electron in one of the t\(_{2g}\) orbitals (costing energy P). Since \(\Delta_o > P\), pairing is preferred.
\[ t_{2g} \begin{array}{|c|c|c|} \hline \uparrow\downarrow & \uparrow & \uparrow
\hline \end{array} \quad e_g \begin{array}{|c|c|} \hline \quad & \quad
\hline \end{array} \]
3. The fifth electron will also pair up in another t\(_{2g}\) orbital.
\[ t_{2g} \begin{array}{|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow
\hline \end{array} \quad e_g \begin{array}{|c|c|} \hline \quad & \quad
\hline \end{array} \]

Step 4: Final Answer

The final distribution of electrons shows five electrons in the t\(_{2g}\) set and zero electrons in the e\(_g\) set.

The electronic configuration is therefore t\(_{2g}^5\) e\(_g^0\).
Quick Tip: Remember the two conditions for octahedral complexes: - \textbf{Strong field (\(\Delta_o > P\))} \(\rightarrow\) Low Spin \(\rightarrow\) Fill t\(_{2g}\) completely before e\(_g\). - \textbf{Weak field (\(\Delta_o \textless P\))} \(\rightarrow\) High Spin \(\rightarrow\) Fill all orbitals singly before pairing.

Step 1: Definition

Enantiomers are a specific type of stereoisomer. They are defined as molecules that are non-superimposable mirror images of each other.


Step 2: Key Properties

- Chirality: For a molecule to have an enantiomer, it must be chiral. A common feature of chiral molecules is the presence of a chiral center (an atom, usually carbon, bonded to four different groups).

- Physical Properties: Enantiomers have identical physical properties (such as melting point, boiling point, density, and solubility in achiral solvents).

- Optical Activity: Their defining difference is their interaction with plane-polarized light. One enantiomer will rotate the plane of polarized light in a clockwise direction (dextrorotatory, denoted by (+) or d-), while its mirror image will rotate it by an equal magnitude in the counter-clockwise direction (levorotatory, denoted by (-) or l-).

- Chemical Properties: They have identical chemical properties when reacting with achiral reagents, but they may react at different rates and give different products when reacting with other chiral reagents (like enzymes in biological systems).


Example: The two enantiomers of 2-butanol.

The mirror image of (R)-2-butanol is (S)-2-butanol. No matter how you rotate the (S)-2-butanol molecule, you cannot make it identical to the (R)-2-butanol molecule, hence they are non-superimposable.
Quick Tip: A simple analogy for enantiomers is your left and right hands. They are mirror images, but you cannot superimpose them (you can't fit your left hand into a right-handed glove). This property of "handedness" is called chirality.

Step 1: Definition

A racemic mixture (or racemate) is a mixture that contains equal amounts (an equimolar mixture) of a pair of enantiomers (the left- and right-handed forms of a chiral molecule).


Step 2: Key Property - Optical Inactivity

The most important characteristic of a racemic mixture is that it is optically inactive.

This happens because the two enantiomers in the mixture rotate plane-polarized light in equal but opposite directions.

- The dextrorotatory (+) enantiomer causes a clockwise rotation.

- The levorotatory (-) enantiomer causes a counter-clockwise rotation of the exact same magnitude.

In a 50:50 mixture, the rotation caused by one type of molecule is perfectly cancelled out by the opposite rotation caused by the other type. The net rotation is zero. This phenomenon is called external compensation.


Notation:

A racemic mixture is often denoted by the prefix (±)- or (dl)- before the name of the compound, for example, (±)-2-butanol.


Formation:

Racemic mixtures are often formed in chemical reactions when a chiral product is synthesized from achiral reactants without the use of a chiral catalyst or reagent. The probability of forming the R enantiomer is the same as forming the S enantiomer, resulting in a 50:50 mixture.
Quick Tip: Distinguish between optical inactivity due to \textbf{external compensation} (in a racemic mixture) and \textbf{internal compensation} (in a meso compound). Meso compounds are achiral molecules that contain chiral centers, but are optically inactive due to an internal plane of symmetry.

Chlorobenzene is significantly less reactive than alkyl halides (like chloroethane) towards nucleophilic substitution reactions. This resistance is due to several factors:


1. Resonance Effect:

The lone pair of electrons on the chlorine atom can participate in resonance with the \(\pi\)-electrons of the benzene ring.

This delocalization gives the carbon-chlorine (C-Cl) bond a partial double-bond character. A double bond is stronger and shorter than a single bond, making it much more difficult for a nucleophile to break the C-Cl bond compared to the pure single C-Cl bond in an alkyl halide.


2. Difference in Hybridization of Carbon Atom:

In chlorobenzene, the carbon atom attached to the chlorine is sp\(^2\) hybridized. In an alkyl halide like chloroethane, the carbon is sp\(^3\) hybridized.

An sp\(^2\) orbital has more s-character (33.3%) than an sp\(^3\) orbital (25%). This makes the sp\(^2\) hybridized carbon more electronegative. As a result, it holds the electrons of the C-Cl bond more tightly, making the bond shorter and stronger, and thus harder to break.


3. Instability of the Phenyl Cation:

If the reaction were to proceed via an S\(_N\)1 mechanism, it would require the departure of the Cl\(^-\) ion to form a phenyl cation (C\(_6\)H\(_5^+\)). The phenyl cation is highly unstable because the positive charge is localized on an sp\(^2\) orbital and cannot be stabilized by resonance. Therefore, the S\(_N\)1 pathway is not feasible.


4. Repulsion between Nucleophile and Benzene Ring:

The benzene ring is an electron-rich system due to its delocalized \(\pi\)-electron cloud. An incoming nucleophile (e.g., OH\(^-\), CN\(^-\)) is also electron-rich. There is a strong electrostatic repulsion between the incoming nucleophile and the \(\pi\)-electron cloud of the ring, which makes it difficult for the nucleophile to approach the carbon atom.
Quick Tip: The most important and frequently cited reason for the low reactivity of haloarenes is the \textbf{partial double-bond character of the C-X bond due to resonance}. Always mention this as the primary factor.

Step 1: Explanation

The Wolff-Kishner reduction is a chemical reaction used to convert the carbonyl group (C=O) of an aldehyde or a ketone into a methylene group (-CH\(_2\)-), effectively reducing the carbonyl compound to an alkane. This reaction is particularly useful for carbonyl compounds that are sensitive to acid (for which the Clemmensen reduction cannot be used). The reaction is carried out in two steps:
1. Formation of a hydrazone by reacting the carbonyl compound with hydrazine (H\(_2\)N-NH\(_2\)).
2. Treatment of the hydrazone with a strong base (like KOH or potassium tert-butoxide) at high temperatures, usually in a high-boiling solvent like ethylene glycol. This step eliminates nitrogen gas (N\(_2\)) and forms the alkane.

Step 2: Chemical Equation

The general reaction can be represented as: \[ \underset{Ketone/Aldehyde}{R-CO-R'} \xrightarrow[-H_2O]{H_2N-NH_2} \underset{Hydrazone}{R-C(=NNH_2)-R'} \xrightarrow[\Delta]{KOH, Ethylene Glycol} \underset{Alkane}{R-CH_2-R'} + N_2 \]

For example, the reduction of Acetone to Propane: \[ \underset{Acetone}{CH_3-CO-CH_3} \xrightarrow{H_2N-NH_2, KOH, Ethylene Glycol, \Delta} \underset{Propane}{CH_3-CH_2-CH_3} + N_2 \] Quick Tip: Remember the key reagents for carbonyl-to-alkane reductions: - \textbf{Wolff-Kishner:} Hydrazine + Strong Base (Basic conditions). - \textbf{Clemmensen:} Zn(Hg) + Conc. HCl (Acidic conditions). Choose the reaction based on the stability of other functional groups in the molecule.

Step 1: Explanation

The Etard reaction is a specific method for the controlled oxidation of a methyl group attached to an aromatic ring to an aldehyde group. The most common example is the conversion of toluene to benzaldehyde. The reagent used is chromyl chloride (CrO\(_2\)Cl\(_2\)) in an inert, non-polar solvent such as carbon tetrachloride (CCl\(_4\)) or carbon disulfide (CS\(_2\)). The reaction proceeds via the formation of a brown, insoluble intermediate complex (the Etard complex), which precipitates from the solution. This complex is then carefully hydrolyzed with water (H\(_3\)O\(^+\)) to yield the aldehyde. The formation of the intermediate complex is crucial as it prevents the over-oxidation of the aldehyde to a carboxylic acid, which would occur with stronger oxidizing agents like KMnO\(_4\).

Step 2: Chemical Equation

The reaction for the conversion of Toluene to Benzaldehyde is: \[ \underset{Toluene}{C_6H_5-CH_3} + 2CrO_2Cl_2 \xrightarrow{CS_2} \underset{Etard Complex (brown)}{C_6H_5-CH(OCrOHCl_2)_2} \xrightarrow{H_3O^+} \underset{Benzaldehyde}{C_6H_5-CHO} \] Quick Tip: The Etard reaction is a key named reaction for preparing aromatic aldehydes from alkylbenzenes. Remember the specific reagent: Chromyl Chloride (CrO₂Cl₂). If you use a strong oxidizing agent like KMnO₄, you will get a carboxylic acid instead of an aldehyde.

Step 1: Explanation

The Cannizzaro reaction is a base-induced disproportionation reaction. The key requirement for this reaction is that the aldehyde substrate must not have any hydrogen atoms on the alpha-carbon (the carbon atom adjacent to the carbonyl group). Examples of such aldehydes include formaldehyde (HCHO) and benzaldehyde (C\(_6\)H\(_5\)CHO).
When treated with a concentrated strong base (like 50% NaOH or KOH), two molecules of the aldehyde react. One molecule is oxidized to the corresponding carboxylic acid (which is immediately deprotonated by the base to form a salt), and the other molecule is reduced to the corresponding primary alcohol. The reaction is thus a self-oxidation-reduction (redox) process.

Step 2: Chemical Equation

Using formaldehyde as an example: \[ \underset{Formaldehyde}{2 HCHO} + Conc. NaOH \xrightarrow{\Delta} \underset{Methanol}{CH_3OH} + \underset{Sodium formate}{HCOONa} \]
Using benzaldehyde as an example: \[ \underset{Benzaldehyde}{2 C_6H_5CHO} + Conc. KOH \xrightarrow{\Delta} \underset{Benzyl alcohol}{C_6H_5CH_2OH} + \underset{Potassium benzoate}{C_6H_5COOK} \] Quick Tip: To decide if an aldehyde will undergo an Aldol reaction or a Cannizzaro reaction in the presence of a base, check for \(\alpha\)-hydrogens. - \textbf{Has \(\alpha\)-H} + Dilute base \(\rightarrow\) Aldol reaction. - \textbf{No \(\alpha\)-H} + Concentrated base \(\rightarrow\) Cannizzaro reaction.

Step 1: Identify A

The starting material is acetic acid (CH\(_3\)COOH). It is treated with thionyl chloride (SOCl\(_2\)). This is a standard laboratory method to convert a carboxylic acid into an acyl chloride by replacing the -OH group with a -Cl group. \[ CH_3COOH + SOCl_2 \longrightarrow \underset{\textbf{A: Acetyl chloride}}{CH_3COCl} + SO_2 \uparrow + HCl \uparrow \]
So, A is Acetyl chloride (CH\(_3\)COCl).

Step 2: Identify B

Compound A (Acetyl chloride) is subjected to catalytic hydrogenation (H\(_2\)) with a palladium catalyst poisoned with barium sulfate (Pd-BaSO\(_4\)). This is the Rosenmund reduction. This specific set of reagents reduces an acyl chloride to an aldehyde. The BaSO\(_4\) acts as a poison to the catalyst, preventing the further reduction of the aldehyde (B) to a primary alcohol. \[ \underset{\textbf{A: Acetyl chloride}}{CH_3COCl} + H_2 \xrightarrow{Pd/BaSO_4} \underset{\textbf{B: Acetaldehyde}}{CH_3CHO} + HCl \]
So, B is Acetaldehyde (CH\(_3\)CHO).

Step 3: Identify C

Compound B (Acetaldehyde) is treated with hydrazine (H\(_2\)N-NH\(_2\)). The carbonyl group of an aldehyde reacts with hydrazine in a condensation reaction (loss of a water molecule) to form a hydrazone. \[ \underset{\textbf{B: Acetaldehyde}}{CH_3CHO} + H_2N-NH_2 \longrightarrow \underset{\textbf{C: Acetaldehyde hydrazone}}{CH_3CH=N-NH_2} + H_2O \]
So, C is Acetaldehyde hydrazone (CH\(_3\)CH=N-NH\(_2\)). Quick Tip: The Rosenmund reduction (H₂/Pd-BaSO₄) is a crucial reaction for converting acyl chlorides to aldehydes. Recognize the "poisoned catalyst" as the key to stopping the reduction at the aldehyde stage.

Step 1: Identify A

The starting material is acetonitrile (CH\(_3\)CN). It is treated with DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis (H\(_2\)O). DIBAL-H is a reducing agent that reduces nitriles to an intermediate imine, which is then hydrolyzed to an aldehyde upon workup with water. \[ CH_3C\equivN \xrightarrow{1. DIBAL-H} [CH_3CH=NH] \xrightarrow{2. H_2O} \underset{\textbf{A: Acetaldehyde}}{CH_3CHO} \]
So, A is Acetaldehyde (CH\(_3\)CHO).

Step 2: Identify B

Compound A (Acetaldehyde) is treated with dilute sodium hydroxide (Dil. NaOH). Acetaldehyde has \(\alpha\)-hydrogens, and this is the classic condition for the Aldol addition reaction. The enolate of one acetaldehyde molecule attacks the carbonyl carbon of another acetaldehyde molecule. \[ 2CH_3CHO \xrightarrow{Dil. NaOH} \underset{\textbf{B: 3-Hydroxybutanal}}{CH_3-CH(OH)-CH_2-CHO} \]
So, B is 3-Hydroxybutanal.

Step 3: Identify C

Compound B (3-Hydroxybutanal), which is a \(\beta\)-hydroxy aldehyde, is heated (\(\Delta\)). Heating an aldol addition product causes dehydration (loss of a water molecule) to form an \(\alpha,\beta\)-unsaturated carbonyl compound. \[ \underset{\textbf{B: 3-Hydroxybutanal}}{CH_3-CH(OH)-CH_2-CHO} \xrightarrow{\Delta} \underset{\textbf{C: But-2-enal (Crotonaldehyde)}}{CH_3-CH=CH-CHO} + H_2O \]
So, C is But-2-enal (commonly known as Crotonaldehyde). Quick Tip: Recognize DIBAL-H as a versatile reducing agent. It reduces nitriles and esters to aldehydes. Also, remember the two-stage aldol process: `Aldehyde with α-H + Dil. Base → Aldol (β-hydroxy aldehyde)`, and `Aldol + Heat → α,β-Unsaturated Aldehyde`.

A native protein is a protein in its fully folded and biologically functional conformation. Every protein has a unique and specific three-dimensional structure, including its secondary (alpha-helices, beta-sheets), tertiary (overall 3D folding), and sometimes quaternary (arrangement of multiple subunits) structures. This precise architecture is held together by various interactions like hydrogen bonds, disulfide bridges, hydrophobic interactions, and ionic bonds. The native state is essential for the protein's biological activity, whether it's acting as an enzyme, a structural component, or a signaling molecule. Any process (like heating or changing pH) that disrupts this native structure is called denaturation, which leads to the loss of biological function. Quick Tip: Think of "native" as meaning "natural and functional". When a protein is in its native state, it's correctly folded and able to do its job. Denaturation is the process of losing that native state.

A nucleotide is the fundamental building block of nucleic acids, such as DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). It is an organic molecule composed of three distinct components joined by covalent bonds:

A Nitrogenous Base: A nitrogen-containing ring structure. These can be purines (Adenine - A, Guanine - G) or pyrimidines (Cytosine - C, Thymine - T in DNA, Uracil - U in RNA).
A Pentose Sugar: A five-carbon sugar molecule. It is deoxyribose in DNA and ribose in RNA.
A Phosphate Group: At least one phosphate group (PO\(_4^{3-}\)) is attached to the 5' carbon of the pentose sugar.

Nucleotides are joined together by phosphodiester bonds between the sugar of one nucleotide and the phosphate of the next to form polynucleotide chains. Nucleotides like ATP (adenosine triphosphate) also serve as primary carriers of chemical energy in cells. Quick Tip: Remember the difference: \textbf{Nucleoside = Base + Sugar}. \textbf{Nucleotide = Base + Sugar + Phosphate}. You can think of it as the "tide" adding the phosphate to the "side".

An essential amino acid is an amino acid that an organism cannot synthesize from scratch in sufficient quantities to meet its physiological needs. Therefore, it must be supplied in its diet. For humans, there are nine essential amino acids:

Histidine
Isoleucine
Leucine
Lysine
Methionine
Phenylalanine
Threonine
Tryptophan
Valine

These amino acids are crucial for protein synthesis and various metabolic functions. A dietary deficiency in any of the essential amino acids can lead to severe health problems, including protein-energy malnutrition (e.g., kwashiorkor). In contrast, non-essential amino acids can be synthesized by the body from other compounds. Quick Tip: A useful mnemonic to remember the nine essential amino acids is "PVT TIM HALL": Phenylalanine, Valine, Threonine, Tryptophan, Isoleucine, Methionine, Histidine, Arginine (conditionally essential), Leucine, Lysine. (Note: Arginine is often considered semi-essential).

Step 1: Definition of Reverse Osmosis (RO)

Osmosis is the natural, spontaneous flow of solvent molecules across a semipermeable membrane (SPM) from a dilute solution (lower solute concentration) to a concentrated solution (higher solute concentration). The pressure required to just stop this flow is called osmotic pressure (\(\pi\)).

Reverse Osmosis is a process in which the direction of this natural flow is reversed. This is achieved by applying an external pressure to the concentrated solution side that is greater than the osmotic pressure (\(P_{ext} > \pi\)). This high pressure forces the solvent molecules to move from the concentrated solution to the dilute (or pure solvent) side, leaving the solute behind. This process is widely used for the desalination of seawater to produce fresh drinking water.


Step 2: Naming a Semipermeable Membrane (SPM)

A semipermeable membrane is a membrane that allows the passage of solvent molecules but blocks the passage of larger solute particles. For the high pressures used in reverse osmosis, a robust membrane is required. A commonly used synthetic SPM is cellulose acetate. Other examples include thin-film composite membranes made from polyamide.
Quick Tip: Remember the pressure conditions: - \textbf{Osmosis}: No external pressure, natural flow. - \textbf{Stopping Osmosis}: \(P_{ext} = \pi\). - \textbf{Reverse Osmosis}: \(P_{ext} > \pi\).

Step 1: Understanding Tonicity

The behavior of cells like RBCs in a solution depends on the relative concentration of solutes inside the cell compared to the surrounding solution.
- The fluid inside red blood cells (cytoplasm) has a salt concentration that is osmotically equivalent to a 0.9% (mass/volume) NaCl solution. This is called an isotonic solution.
- A solution with a higher concentration (\(>\) 0.9% NaCl) is hypertonic.
- A solution with a lower concentration (\(\textless\) 0.9% NaCl) is hypotonic.

Step 2: Applying the Concept to the Problem

The RBCs are placed in a 0.5% NaCl solution. Since 0.5% is less than 0.9%, the external solution is hypotonic relative to the fluid inside the RBCs.

Due to osmosis, water (the solvent) will move from the region of lower solute concentration (the 0.5% NaCl solution) to the region of higher solute concentration (inside the RBCs) through the cell membrane, which acts as a semipermeable membrane.


Step 3: Predicting the Outcome

As water flows into the RBCs, they will begin to swell up. Unlike plant cells, animal cells like RBCs do not have a rigid cell wall. The cell membrane can only stretch so much. If the influx of water is significant, the cells will swell to a point where the membrane can no longer withstand the internal pressure and will rupture or burst. This bursting of red blood cells is called hemolysis.
Quick Tip: Remember the rule: Water moves towards the higher solute concentration. - In \textbf{hypo}tonic solutions, cells swell (like a hippo). - In \textbf{hyper}tonic solutions, cells shrink (crenation).

Step 1: Understanding Osmotic Pressure and the van't Hoff Factor

Osmotic pressure (\(\pi\)) is a colligative property, meaning it depends on the number of solute particles in the solution, not their chemical nature. The formula for osmotic pressure is: \[ \pi = i \times C \times R \times T \]
where:
- \(i\) is the van't Hoff factor, which represents the number of particles the solute dissociates into in the solution.
- \(C\) is the molar concentration of the solution.
- \(R\) is the ideal gas constant.
- \(T\) is the temperature in Kelvin.

Step 2: Determining the van't Hoff Factor (i) for each solute

- Urea (NH\(_2\)CONH\(_2\)): Urea is a non-electrolyte. It dissolves in water but does not dissociate into ions. Therefore, one molecule of urea remains as one particle in the solution. So, for urea, \(i=1\).
- Potassium Chloride (KCl): KCl is a strong electrolyte. It completely dissociates in water into two ions: one K\(^+\) ion and one Cl\(^-\) ion.
\[ KCl(s) \xrightarrow{Water} K^+(aq) + Cl^-(aq) \]
Therefore, one formula unit of KCl produces two particles (ions) in the solution. So, for KCl, \(i=2\).

Step 3: Comparing the Osmotic Pressures

Both solutions have the same molar concentration (C = 1 M) and are presumably at the same temperature (T). Since R is a constant, the osmotic pressure is directly proportional to the van't Hoff factor (\(i\)).
- For 1 M urea: \(\pi_{urea} = 1 \times C \times R \times T\)
- For 1 M KCl: \(\pi_{KCl} = 2 \times C \times R \times T\)

Since the van't Hoff factor for KCl (2) is greater than that for urea (1), the effective concentration of particles in the KCl solution is double that of the urea solution. Consequently, the 1 M KCl solution will exert a higher osmotic pressure.
Quick Tip: When comparing colligative properties of solutions with the same molarity, always check if the solutes are electrolytes or non-electrolytes. Electrolytes will always have a larger effect due to dissociation (i > 1).

Step 1: Definition of Colligative Properties

Colligative properties of solutions are properties that depend upon the concentration of solute molecules or ions, but not upon the identity of the solute. In simple terms, they depend on the number of solute particles, not on their size, mass, or chemical nature. The four main colligative properties are:

Relative lowering of vapor pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure


Step 2: Relating Osmotic Pressure to this Definition

Osmosis is the movement of solvent molecules from a region of high solvent concentration to a region of low solvent concentration. The presence of solute particles lowers the effective concentration (or chemical potential) of the solvent. The greater the number of solute particles, the lower the solvent concentration, and the greater the driving force for osmosis.
Osmotic pressure (\(\pi\)) is the pressure required to counteract this flow. Therefore, the magnitude of the osmotic pressure is directly proportional to the extent to which the solute has lowered the solvent's concentration. This, in turn, is directly proportional to the total concentration of solute particles.
The formula \(\pi = iCRT\) shows this dependence clearly: \(\pi\) is directly proportional to the effective molar concentration of particles (\(i \times C\)). It does not depend on what the solute particles are (e.g., glucose, urea, or Na\(^+\) ions), only on how many of them are present per unit volume of solution.
Because its value is determined by the number of solute particles and not their identity, osmotic pressure is classified as a colligative property. Quick Tip: The key phrase for defining any colligative property is that it "depends on the \textbf{number of solute particles, not their \textbf{nature}."

Step 1: Understanding Basicity of Amines

The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom to donate to a proton (Lewis base concept). Any factor that increases the electron density on the nitrogen atom increases its basicity. Conversely, any factor that decreases the electron density on the nitrogen atom decreases its basicity.


Step 2: Analyzing the Structures

The three compounds are substituted anilines. We need to analyze the electronic effect of the substituent at the para position on the -NH\(_2\) group.


- Aniline (C\(_6\)H\(_5\)NH\(_2\)): This is our reference compound. The lone pair on the nitrogen is delocalized into the benzene ring via resonance, which makes it less available for protonation compared to aliphatic amines.

- p-Nitroaniline: This has a nitro group (-NO\(_2\)) at the para position. The nitro group is a powerful electron-withdrawing group (EWG) due to both the -I (inductive) effect and the strong -R (resonance or mesomeric) effect. It pulls electron density away from the ring and from the -NH\(_2\) group, making the lone pair on nitrogen significantly less available for donation.

- p-Toluidine (p-Methylaniline): This has a methyl group (-CH\(_3\)) at the para position. The methyl group is an electron-donating group (EDG) due to its +I (inductive) effect and hyperconjugation. It pushes electron density into the ring, which in turn increases the electron density on the nitrogen atom, making the lone pair more available for donation.


Step 3: Ordering the Compounds

- The electron-withdrawing -NO\(_2\) group makes p-nitroaniline the least basic.

- The electron-donating -CH\(_3\) group makes p-toluidine the most basic.

- Aniline, with no substituent, is of intermediate basicity.


Therefore, the increasing order of basic character is: \[ p-Nitroaniline \textless Aniline \textless p-Toluidine \] Quick Tip: For substituted anilines: Electron Donating Groups (EDGs like -CH₃, -OCH₃, -NH₂) increase basicity. Electron Withdrawing Groups (EWGs like -NO₂, -CN, -X) decrease basicity. The effect is strongest at the ortho and para positions.

Step 1: Understanding the Relationship between 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, due to the negative logarithm, a smaller pK\(_b\) value. Therefore, the question "Why is the pK\(_b\) of aniline more than that of methylamine?" is equivalent to asking "Why is aniline a weaker base than methylamine?".


Step 2: Comparing the Structures of Aniline and Methylamine

- Aniline (C\(_6\)H\(_5\)NH\(_2\)): This is an aromatic amine. The nitrogen atom is attached to an sp\(^2\)-hybridized carbon of the benzene ring.

- Methylamine (CH\(_3\)NH\(_2\)): This is a primary aliphatic amine. The nitrogen atom is attached to an sp\(^3\)-hybridized carbon of the methyl group.


Step 3: Explaining the Difference in Basicity

There are two main reasons why aniline is a much weaker base than methylamine:

1. Resonance Effect in Aniline:

In aniline, the lone pair of electrons on the nitrogen atom is not localized. It is involved in resonance with the \(\pi\)-electron system of the benzene ring. This delocalization spreads the electron density over the entire molecule, making the lone pair less available to be donated to a proton.

Furthermore, when the anilinium ion (C\(_6\)H\(_5\)NH\(_3^+\)) is formed after protonation, it has only two resonance structures (Kekulé structures of the ring), whereas aniline itself has five. This means the anilinium ion is less resonance-stabilized than aniline, so its formation is less favorable.


2. Inductive Effect in Methylamine:

In methylamine, the nitrogen atom is attached to a methyl group (-CH\(_3\)). The methyl group is an electron-donating group due to its +I (positive inductive) effect. It pushes electron density towards the nitrogen atom, thereby increasing the electron density of the lone pair and making it more readily available for donation to a proton. The resulting methylammonium ion is also stabilized by the +I effect.


Conclusion:

The electron-withdrawing resonance effect in aniline greatly reduces the availability of its lone pair, making it a weak base. The electron-donating inductive effect in methylamine enhances the availability of its lone pair, making it a stronger base. A weaker base has a higher pK\(_b\), so pK\(_b\) (aniline) > pK\(_b\) (methylamine). (Typical values: pK\(_b\) of aniline is ~9.4, while pK\(_b\) of methylamine is ~3.4). Quick Tip: As a general rule, aliphatic amines are significantly stronger bases than aromatic amines. The resonance delocalization in aromatic amines is the dominant factor that reduces their basicity.

Step 1: Understanding Basicity of Amines in Aqueous Solution

The basic character of amines in an aqueous solution is a complex property determined by the interplay of three factors:

Inductive Effect (+I effect): Alkyl groups (like -CH\(_3\)) are electron-donating. They increase the electron density on the nitrogen atom, making its lone pair more available for donation and thus increasing basicity. Based on this effect alone, the order would be: Tertiary (3\(^\circ\)) > Secondary (2\(^\circ\)) > Primary (1\(^\circ\)) > Ammonia.
Solvation Effect (Hydration): After accepting a proton, the amine forms a substituted ammonium cation (R-NH\(_3^+\)). This cation is stabilized by hydrogen bonding with water molecules. Greater the number of hydrogen atoms on the nitrogen, stronger the hydrogen bonding and greater the stability of the conjugate acid. This makes the parent amine more basic. Based on this effect alone, the order would be: Ammonia > Primary (1\(^\circ\)) > Secondary (2\(^\circ\)) > Tertiary (3\(^\circ\)).
Steric Hindrance: Bulky alkyl groups around the nitrogen atom can physically block the lone pair, making it difficult for a proton to approach and bond. This steric hindrance decreases basicity. Based on this effect alone, the order would be: Ammonia > Primary (1\(^\circ\)) > Secondary (2\(^\circ\)) > Tertiary (3\(^\circ\)).


Step 2: Analyzing the Combined Effect for Methylamines

For methylamines, the combined result of these opposing effects determines the final order of basicity in water.
- (CH\(_3\))\(_2\)NH (Secondary): It has the best balance between the strong +I effect of two methyl groups and good stabilization of its conjugate acid by solvation (two H-atoms for H-bonding). This makes it the strongest base in the series.
- CH\(_3\)NH\(_2\) (Primary): It has a moderate +I effect and excellent solvation of its conjugate acid (three H-atoms for H-bonding). It is a strong base, but less so than the secondary amine.
- (CH\(_3\))\(_3\)N (Tertiary): It has the strongest +I effect from three methyl groups. However, its conjugate acid, [(CH\(_3\))\(_3\)NH]\(^+\), has only one hydrogen atom available for H-bonding, leading to very poor solvation. Significant steric hindrance also reduces its basicity. These factors make it a weaker base than both primary and secondary methylamines.
- NH\(_3\) (Ammonia): It lacks any electron-donating inductive effect from alkyl groups, making it the weakest base in the series.

Step 3: Final Order

Based on the combined effects, the increasing order of basic strength in aqueous solution is: \[ NH_3 \textless (CH_3)_3N \textless CH_3NH_2 \textless (CH_3)_2NH \] Quick Tip: The order of basicity for amines is different in the gas phase and in aqueous solution. In the gas phase, only the inductive effect matters, so the order is 3° > 2° > 1° > NH₃. In aqueous solution, solvation and steric effects become important, leading to the irregular order (for methyl groups: 2° > 1° > 3° > NH₃).

Step 1: Understanding Ammonolysis

Ammonolysis is the reaction of an alkyl halide (R-X) with ammonia (NH\(_3\)). It is a nucleophilic substitution reaction where the ammonia molecule acts as a nucleophile, displacing the halide ion to form a primary amine. \[ R-X + NH_3 \longrightarrow \underset{Primary amine}{R-NH_2} + HX \]

Step 2: The Problem of Over-alkylation

The primary amine (R-NH\(_2\)) formed in the first step is also a nucleophile. In fact, due to the electron-donating (+I) effect of the alkyl group (R), the primary amine is a stronger nucleophile than the initial ammonia molecule.

Consequently, this primary amine can react further with another molecule of the alkyl halide: \[ R-X + R-NH_2 \longrightarrow \underset{Secondary amine}{R_2NH} + HX \]
The process does not stop here. The secondary amine (R\(_2\)NH) is also nucleophilic and can react with more alkyl halide to form a tertiary amine (R\(_3\)N). \[ R-X + R_2NH \longrightarrow \underset{Tertiary amine}{R_3N} + HX \]
Finally, the tertiary amine can react with a final molecule of alkyl halide to form a quaternary ammonium salt ([R\(_4\)N]\(^+\)X\(^-\)). \[ R-X + R_3N \longrightarrow \underset{Quaternary ammonium salt}{[R_4N]^+X^-} \]

Step 3: Final Conclusion

Because the product amine at each stage is also a reactant for the next stage, the reaction is difficult to control. It inevitably leads to a mixture of primary, secondary, and tertiary amines, along with the quaternary ammonium salt. Separating these products is often difficult due to their similar physical properties (like boiling points). Therefore, ammonolysis of alkyl halides is not a good method for the synthesis of pure amines, especially for primary amines. To favor the formation of the primary amine, a very large excess of ammonia must be used. Quick Tip: For the clean synthesis of pure primary amines, better methods like the \textbf{Gabriel Phthalimide Synthesis} or the reduction of nitriles and amides are preferred, as they avoid the problem of over-alkylation.

Step 1: Understanding the Concept

We need to calculate the standard Gibbs free energy change (\(\Delta_r G^\circ\)) for the given electrochemical reaction and predict its spontaneity. The relationship between standard Gibbs energy and standard cell potential (E\(^\circ_{cell}\)) is used for this. A reaction is spontaneous if its \(\Delta_r G^\circ\) is negative (or its E\(^\circ_{cell}\) is positive).


Step 2: Key Formula or Approach

The formula relating standard Gibbs energy and standard cell potential is: \[ \Delta_r G^\circ = -n F E^\circ_{cell} \]
where:
- \(n\) is the number of moles of electrons transferred in the balanced reaction.
- \(F\) is the Faraday constant (96500 C mol\(^{-1}\)).
- \(E^\circ_{cell}\) is the standard cell potential.

The standard cell potential is calculated as: \[ E^\circ_{cell} = E^\circ_{cathode} (reduction) - E^\circ_{anode} (oxidation) \]

Step 3: Detailed Calculation

1. Identify Half-Reactions and Electrodes:

From the given overall reaction: Au(s) + Ca\(^{2+}\)(1M) \(\rightarrow\) Au\(^{3+}\)(1M) + Ca(s)
- Gold is oxidized: Au(s) \(\rightarrow\) Au\(^{3+}\)(aq) + 3e\(^-\) (Anode)
- Calcium is reduced: Ca\(^{2+}\)(aq) + 2e\(^-\) \(\rightarrow\) Ca(s) (Cathode)

2. Calculate E\(^\circ_{cell}\):
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Ca^{2+}/Ca} - E^\circ_{Au^{3+}/Au} \] \[ E^\circ_{cell} = (-2.87 V) - (+1.5 V) = -4.37 V \]

3. Determine the value of 'n':

To balance the electrons in the half-reactions, we multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3:
- 2Au(s) \(\rightarrow\) 2Au\(^{3+}\)(aq) + 6e\(^-\)
- 3Ca\(^{2+}\)(aq) + 6e\(^-\) \(\rightarrow\) 3Ca(s)
The total number of electrons transferred in the balanced reaction is n = 6.

4. Calculate \(\Delta_r G^\circ\):
\[ \Delta_r G^\circ = -n F E^\circ_{cell} \] \[ \Delta_r G^\circ = -(6 mol) \times (96500 C mol^{-1}) \times (-4.37 V) \] \[ \Delta_r G^\circ = + (6 \times 96500 \times 4.37) J \quad (since C \times V = J) \] \[ \Delta_r G^\circ = + 2530230 J \]
To convert to kJ, divide by 1000: \[ \Delta_r G^\circ = +2530.23 kJ mol^{-1} \]

Step 4: Predict Spontaneity

A reaction is spontaneous under standard conditions if \(\Delta_r G^\circ \textless 0\).

Since our calculated \(\Delta_r G^\circ\) is positive (+2530.23 kJ mol\(^{-1}\)), the reaction is non-spontaneous as written. This is also confirmed by the negative value of E\(^\circ_{cell}\). Quick Tip: A quick way to check spontaneity: if E\(^\circ_{cell}\) is positive, the reaction is spontaneous (\(\Delta G^\circ\) is negative). If E\(^\circ_{cell}\) is negative, the reaction is non-spontaneous (\(\Delta G^\circ\) is positive).

Step 1: Understanding the Concept

The setup described creates a galvanic (electrochemical) cell. For the tarnish (Ag\(_2\)S) to be removed, it must be reduced back to metallic silver (Ag). This reduction requires a source of electrons. The aluminum pan can provide these electrons if it is oxidized to Al\(^{3+}\) ions. We need to determine if this overall redox process is spontaneous by calculating the standard cell potential (E\(^\circ_{cell}\)). A positive E\(^\circ_{cell}\) indicates a spontaneous reaction.


Step 2: Identifying Half-Reactions and Electrodes

- Tarnish Removal (Reduction): The silver sulfide must be reduced. This will be the cathode reaction.
\[ Ag_2S(s) + 2e^- \longrightarrow 2Ag(s) + S^{2-}(aq) \quad E^\circ_{cathode} = -0.71 V \]
- Aluminum Pan (Oxidation): The aluminum pan will act as the anode, getting oxidized. The oxidation half-reaction is the reverse of the given reduction reaction. The potential for the oxidation half-reaction is the negative of the reduction potential.
\[ Al(s) \longrightarrow Al^{3+}(aq) + 3e^- \quad E^\circ_{anode} = -E^\circ_{Al^{3+}/Al} = -(-1.66 V) = +1.66 V \]

Step 3: Calculating the Standard Cell Potential (E\(^\circ_{cell}\))

The overall cell potential is the sum of the potentials for the reduction and oxidation half-reactions. \[ E^\circ_{cell} = E^\circ_{cathode} (reduction) + E^\circ_{anode} (oxidation) \] \[ E^\circ_{cell} = (-0.71 V) + (+1.66 V) \] \[ E^\circ_{cell} = +0.95 V \]

Step 4: Final Conclusion

Since the calculated standard cell potential (\(E^\circ_{cell}\)) is positive (+0.95 V), the overall reaction is spontaneous. The aluminum pan will act as a sacrificial anode, corroding to supply electrons that reduce the silver sulfide tarnish back to pure silver. The NaCl solution acts as an electrolyte (salt bridge) to complete the circuit by allowing ions to flow between the electrodes. Therefore, yes, this method will remove the tarnish. Quick Tip: This is a practical application of sacrificial protection. The more reactive metal (aluminum, with a more negative reduction potential) is oxidized to protect the less reactive metal compound (silver sulfide).

The cell potential, also known as the electromotive force (EMF), is the difference in electrical potential between the anode and the cathode of a galvanic (voltaic) cell. It is a measure of the energy per unit charge available from a redox reaction and represents the driving force that causes electrons to flow through the external circuit.
Key points:

It is measured in Volts (V).
By convention, it is calculated as the potential of the cathode minus the potential of the anode: \(E_{cell} = E_{cathode} - E_{anode}\).
The cell potential is an intensive property, meaning it does not depend on the size of the electrodes or the amount of reactants.
When the cell potential is positive (\(E_{cell} > 0\)), the cell reaction is spontaneous. Quick Tip: Think of cell potential as the "electrical pressure" that pushes electrons from the site of oxidation (anode) to the site of reduction (cathode). A higher cell potential means a stronger push.

A fuel cell is a type of electrochemical cell that generates electricity directly from a chemical reaction between a fuel and an oxidizing agent. Unlike a conventional battery, a fuel cell is an open system that requires a continuous supply of fuel (e.g., hydrogen, methane, methanol) and an oxidant (e.g., oxygen from the air) from external sources to operate.
Key features:

It converts chemical energy directly into electrical energy without combustion.
The reactants are supplied continuously, allowing it to produce power for as long as fuel is available.
Fuel cells are highly efficient (typically 40-60% efficiency, or higher with cogeneration).
They are environmentally friendly, as the by-products are often harmless. For example, a hydrogen-oxygen fuel cell produces only water as its by-product. Quick Tip: The key difference between a battery and a fuel cell: a battery stores energy (a closed system), while a fuel cell generates energy from an external fuel supply (an open system).

Step 1: Understanding the Concept

We need to calculate the electromotive force (emf) of a non-standard galvanic cell. Since the concentrations are not 1 M, we must use the Nernst equation, which relates the cell potential under non-standard conditions (\(E_{cell}\)) to the standard cell potential (\(E^\circ_{cell}\)) and the concentrations of the reactants and products.


Step 2: Key Formula or Approach

The Nernst equation at 25 \(^\circ\)C (298 K) is: \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
where:
- \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\)
- \(n\) is the number of moles of electrons transferred.
- \(Q\) is the reaction quotient.

Step 3: Detailed Calculation

1. Identify Electrodes and Calculate E\(^\circ_{cell}\):

From the cell notation:
- Anode (Oxidation, left side): Zn(s) \(\rightarrow\) Zn\(^{2+}\)(aq) + 2e\(^-\)
- Cathode (Reduction, right side): Cd\(^{2+}\)(aq) + 2e\(^-\) \(\rightarrow\) Cd(s) \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Cd^{2+}/Cd} - E^\circ_{Zn^{2+}/Zn} \] \[ E^\circ_{cell} = (-0.40 V) - (-0.76 V) = +0.36 V \]

2. Write the Overall Reaction and Determine 'n' and 'Q':

The overall cell reaction is: \[ Zn(s) + Cd^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cd(s) \]
The number of electrons transferred is n = 2.
The reaction quotient is: \[ Q = \frac{[Products]}{[Reactants]} = \frac{[Zn^{2+}]}{[Cd^{2+}]} \]
Given concentrations: [Zn\(^{2+}\)] = 0.1 M and [Cd\(^{2+}\)] = 0.01 M. \[ Q = \frac{0.1}{0.01} = 10 \]

3. Apply the Nernst Equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \] \[ E_{cell} = 0.36 V - \frac{0.0591}{2} \log(10) \]
Using the given value log 10 = 1: \[ E_{cell} = 0.36 - \frac{0.0591}{2} \times 1 \] \[ E_{cell} = 0.36 - 0.02955 \] \[ E_{cell} = 0.33045 V \]

Step 4: Final Answer

The emf of the cell at 25 \(^\circ\)C is approximately 0.33 V. Quick Tip: Remember the structure of the reaction quotient Q for a cell `A|A⁺||B⁺|B`: It's always `[Anode ion]/[Cathode ion]`. In this case, `[Zn²⁺]/[Cd²⁺]`.

Step 1: Identify Compound 'A'

- The molecular formula is C\(_2\)H\(_6\)O. The possible isomers are ethanol (CH\(_3\)CH\(_2\)OH) and dimethyl ether (CH\(_3\)OCH\(_3\)).
- The compound 'A' undergoes oxidation with CrO\(_3\). Ethers are generally resistant to oxidation, while primary alcohols are oxidized. This suggests that 'A' is ethanol.
- A = Ethanol (CH\(_3\)CH\(_2\)OH)


Step 2: Identify Compound 'B'

- Compound 'A' (ethanol, a primary alcohol) is oxidized by CrO\(_3\) (a mild oxidizing agent) to form 'B'. Primary alcohols are oxidized to aldehydes by CrO\(_3\).
- B = Acetaldehyde (CH\(_3\)CHO)

- Reaction:
\[ \underset{(A) Ethanol}{CH_3CH_2OH} \xrightarrow{CrO_3} \underset{(B) Acetaldehyde}{CH_3CHO} \]

Step 3: Identify Compound 'C'

- Compound 'B' (acetaldehyde) is warmed with iodine and aqueous NaOH, giving a yellow precipitate 'C'. This is the Iodoform Test.
- The test is positive for compounds containing a CH\(_3\)CO- group or a CH\(_3\)CH(OH)- group. Acetaldehyde has a CH\(_3\)CO- group.
- The yellow precipitate 'C' is iodoform.
- C = Iodoform (CHI\(_3\))

- Reaction:
\[ \underset{(B) Acetaldehyde}{CH_3CHO} + 3I_2 + 4NaOH \longrightarrow \underset{(C) Iodoform}{CHI_3 \downarrow} + HCOONa + 3NaI + 3H_2O \]

Step 4: Identify Compound 'D'

- Compound 'A' (ethanol) is heated with concentrated H\(_2\)SO\(_4\) at 413 K. This is the acid-catalyzed intermolecular dehydration of an alcohol to form an ether.
- D = Diethyl ether (CH\(_3\)CH\(_2\)OCH\(_2\)CH\(_3\))

- Reaction:
\[ 2\underset{(A) Ethanol}{CH_3CH_2OH} \xrightarrow{Conc. H_2SO_4, 413 K} \underset{(D) Diethyl ether}{CH_3CH_2OCH_2CH_3} + H_2O \]

Step 5: Identify Compound 'E'

- Compound 'D' (diethyl ether) reacts with excess HI. This reaction is the cleavage of ethers by strong hydrohalic acids. With excess HI, both C-O bonds are cleaved to form two molecules of the corresponding alkyl iodide.
- E = Ethyl iodide (CH\(_3\)CH\(_2\)I)

- Reaction:
\[ \underset{(D) Diethyl ether}{CH_3CH_2OCH_2CH_3} + 2HI (excess) \xrightarrow{\Delta} 2\underset{(E) Ethyl iodide}{CH_3CH_2I} + H_2O \] Quick Tip: Remember the temperature dependence of ethanol dehydration with conc. H₂SO₄: - At 413 K (140°C), intermolecular dehydration occurs, forming an ether. - At 443 K (170°C), intramolecular dehydration occurs, forming an alkene (ethene).

The -OH group of phenol is a strongly activating group, making the benzene ring highly susceptible to electrophilic substitution. When phenol is treated with a strong nitrating agent like concentrated nitric acid (in the presence of concentrated sulfuric acid), nitration occurs at all available ortho and para positions. This leads to the formation of a yellow crystalline solid, 2,4,6-trinitrophenol, commonly known as Picric acid. \[ \underset{Phenol}{C_6H_5OH} + 3HNO_3 (conc.) \xrightarrow{Conc. H_2SO_4} \underset{2,4,6-Trinitrophenol (Picric acid)}{C_6H_2(NO_2)_3OH} + 3H_2O \] Quick Tip: To get mononitrophenols (ortho- and para-nitrophenol), a milder nitrating agent must be used, such as dilute nitric acid at a low temperature. Concentrated acid leads to polysubstitution.

This two-step reaction is the hydroboration-oxidation of an alkene. It achieves the hydration of an alkene with anti-Markovnikov regioselectivity.
Step 1: Hydroboration. Propene reacts with diborane (B\(_2\)H\(_6\), often written as BH\(_3\)) to form a trialkylborane intermediate. The boron atom adds to the less substituted carbon of the double bond, and the hydrogen adds to the more substituted carbon.
Step 2: Oxidation. The trialkylborane is then oxidized with hydrogen peroxide (H\(_2\)O\(_2\)) in the presence of a base (OH\(^-\)). The boron group is replaced by a hydroxyl (-OH) group with retention of stereochemistry.
The net result is the addition of H and OH across the double bond, with the -OH group ending up on the terminal carbon. \[ \underset{Propene}{CH_3-CH=CH_2} \xrightarrow{1. B_2H_6 / THF} \xrightarrow{2. H_2O_2, OH^-} \underset{Propan-1-ol}{CH_3-CH_2-CH_2OH} \] Quick Tip: Remember the regioselectivity of alkene hydration: - \textbf{Acid-catalyzed hydration (dil. H₂SO₄):} Markovnikov addition (OH on more substituted carbon). - \textbf{Hydroboration-oxidation (B₂H₆ then H₂O₂/OH⁻):} Anti-Markovnikov addition (OH on less substituted carbon).

This reaction is an example of the Williamson ether synthesis. It involves a nucleophilic substitution (S\(_N\)2) reaction between an alkoxide ion and an alkyl halide.
- Sodium t-butoxide [(CH\(_3\))\(_3\)CO\(^-\)Na\(^+\)]: Provides the strong nucleophile, the tert-butoxide ion.
- Methyl chloride (CH\(_3\)Cl): A primary alkyl halide, which is an excellent substrate for S\(_N\)2 reactions.
The tert-butoxide ion attacks the methyl chloride, displacing the chloride ion to form an ether. \[ \underset{Sodium t-butoxide}{(CH_3)_3CO^-Na^+} + \underset{Methyl chloride}{CH_3Cl} \longrightarrow \underset{Methyl tert-butyl ether}{(CH_3)_3C-O-CH_3} + NaCl \] Quick Tip: The Williamson ether synthesis works best when the alkyl halide is primary. If a secondary or tertiary alkyl halide is used with a strong base like t-butoxide, the elimination (E2) reaction will dominate, forming an alkene instead of an ether.

A simple and effective test to distinguish between butan-1-ol and butan-2-ol is the Iodoform Test.
- Principle: The iodoform test gives a positive result (a yellow precipitate of iodoform, CHI\(_3\)) for compounds containing a methyl ketone (CH\(_3\)CO-) group or alcohols that can be oxidized to a methyl ketone, i.e., alcohols with a CH\(_3\)CH(OH)- group.
- Butan-2-ol (CH\(_3\)-CH(OH)-CH\(_2\)-CH\(_3\)): This is a secondary alcohol that has the required CH\(_3\)CH(OH)- structural unit. It will be oxidized by the reagent to butan-2-one (a methyl ketone), which then reacts further to give the yellow precipitate of iodoform.
- Butan-1-ol (CH\(_3\)-CH\(_2\)-CH\(_2\)-CH\(_2\)OH): This is a primary alcohol and does not possess the CH\(_3\)CH(OH)- group. Therefore, it does not give a positive iodoform test.

Procedure and Observations:

1. Add a few drops of the alcohol to a test tube containing aqueous sodium hydroxide.
2. Warm the mixture gently.
3. Add iodine solution dropwise until a faint yellow color persists.

With Butan-2-ol: A pale yellow precipitate of iodoform (CHI\(_3\)) with a characteristic antiseptic smell will be formed.
With Butan-1-ol: No yellow precipitate will be formed.

Reaction for Butan-2-ol: \[ CH_3CH(OH)CH_2CH_3 + 4I_2 + 6NaOH \longrightarrow \underset{Iodoform (yellow ppt)}{CHI_3 \downarrow} + CH_3CH_2COONa + 5NaI + 5H_2O \] Quick Tip: Another valid test is the \textbf{Lucas Test} (anhydrous ZnCl₂/conc. HCl). Secondary alcohols like butan-2-ol give turbidity in about 5-10 minutes, while primary alcohols like butan-1-ol show no reaction at room temperature.

Step 1: Understanding Acidity

The acidic strength of a compound is its ability to donate a proton (H\(^+\)). The acidity is determined by the stability of the conjugate base formed after the proton is donated. A more stable conjugate base corresponds to a stronger acid.

Step 2: Analyzing the Conjugate Bases

- Ethanol (CH\(_3\)CH\(_2\)OH): When ethanol loses a proton, it forms the ethoxide ion (CH\(_3\)CH\(_2\)O\(^-\)). The ethyl group (-C\(_2\)H\(_5\)) is an electron-donating group (+I effect). It pushes electron density onto the oxygen atom, intensifying the negative charge and destabilizing the ethoxide ion. This makes it less favorable for the proton to leave.
- Water (H\(_2\)O): When water loses a proton, it forms the hydroxide ion (OH\(^-\)). There are no inductive effects to either stabilize or destabilize the ion significantly, so it serves as a neutral reference point.
- Phenol (C\(_6\)H\(_5\)OH): When phenol loses a proton, it forms the phenoxide ion (C\(_6\)H\(_5\)O\(^-\)). This ion is highly stabilized because the negative charge on the oxygen atom is delocalized over the entire benzene ring through resonance. Spreading the charge over several atoms makes the ion much more stable.

Step 3: Comparing the Stabilities and Ordering

- The ethoxide ion is destabilized by the +I effect of the ethyl group, making it less stable than the hydroxide ion. Therefore, ethanol is a weaker acid than water.
- The phenoxide ion is highly stabilized by resonance. This makes it much more stable than the hydroxide ion. Therefore, phenol is a stronger acid than water.
- Combining these findings gives the increasing order of acid strength.

Final Order: \[ Ethanol \textless Water \textless Phenol \] Quick Tip: A key principle: Resonance stabilization of a conjugate base has a much stronger effect on acidity than inductive effects. This is why phenols are acidic, while alcohols are neutral or very weakly acidic.

Step 1: Identify the Principal Functional Group

The molecule contains a -CHO group, which is an aldehyde. The aldehyde group has the highest priority here, so the suffix of the name will be '-al'. The carbon of the aldehyde group is always numbered as C-1.


Step 2: Identify the Parent Chain

The longest carbon chain containing the principal functional group has four carbon atoms. Therefore, the parent alkane is butane. The root name is 'but'.


Step 3: Number the Carbon Chain

Numbering starts from the aldehyde carbon: \[ CH_3-CH=CH-CHO \] \[ 4 \quad 3 \quad 2 \quad 1 \]

Step 4: Locate Substituents and Multiple Bonds

There is a carbon-carbon double bond starting at C-2. The infix for a double bond is '-en-'. Its position is indicated as '2-en'.


Step 5: Assemble the Full IUPAC Name

Combining the parts: Root (But) + position of double bond (2-en) + suffix (al).
The name is But-2-enal.
(Note: The 'e' from 'ene' is dropped before the vowel 'a' of 'al').
Quick Tip: For polyfunctional compounds, remember the priority order of functional groups. The highest priority group determines the suffix, and its carbon gets the lowest possible number.

A simple chemical test to distinguish between an aldehyde (propanal) and a ketone (propanone) is Tollens' Test.
- Principle: Aldehydes are easily oxidized, whereas ketones are resistant to oxidation by mild oxidizing agents. Tollens' reagent is a mild oxidizing agent, consisting of an ammoniacal silver nitrate solution [Ag(NH\(_3\))\(_2\)]\(^+\). It oxidizes aldehydes to the corresponding carboxylate ion and is itself reduced to metallic silver.
- Propanal (CH\(_3\)CH\(_2\)CHO): As an aldehyde, it will be oxidized by Tollens' reagent.
- Propanone (CH\(_3\)COCH\(_3\)): As a ketone, it will not react with Tollens' reagent.

Procedure and Observations:

1. Prepare fresh Tollens' reagent in a clean test tube.
2. Add a few drops of the sample (propanal or propanone) to the reagent.
3. Warm the test tube gently in a water bath for a few minutes.

With Propanal: A shiny silver mirror will be deposited on the inner wall of the test tube, or a black precipitate of silver will form.
With Propanone: The solution will remain clear and colorless. No reaction will occur.

Reaction for Propanal: \[ CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \longrightarrow \underset{Propanoate ion}{CH_3CH_2COO^-} + \underset{Silver mirror}{2Ag \downarrow} + 4NH_3 + 2H_2O \] Quick Tip: Other tests for aldehydes include Fehling's test (red precipitate with aliphatic aldehydes) and Benedict's test. Tollens' test is the most general test for aldehydes as it works for both aliphatic and aromatic aldehydes.

The methyl group of toluene can be oxidized to a carboxylic acid group using a strong oxidizing agent. A common and effective reagent for this is potassium permanganate (KMnO\(_4\)) in a basic (alkaline) medium, followed by acidification.
Step 1: Oxidation. Toluene is heated under reflux with alkaline KMnO\(_4\). The methyl side chain is oxidized to a carboxylate salt (potassium benzoate).
Step 2: Acidification. The resulting solution is acidified with a dilute mineral acid (like HCl or H\(_2\)SO\(_4\)) to protonate the carboxylate salt, yielding the free benzoic acid, which precipitates out as it is sparingly soluble in cold water. \[ \underset{Toluene}{C_6H_5CH_3} \xrightarrow[\Delta]{KMnO_4, KOH} \underset{Potassium benzoate}{C_6H_5COO^-K^+} \xrightarrow{H_3O^+} \underset{Benzoic acid}{C_6H_5COOH} \] Quick Tip: The oxidation of alkyl side chains on a benzene ring using strong oxidizing agents like KMnO₄ or K₂Cr₂O₇ will always produce benzoic acid, regardless of the length of the alkyl chain, as long as there is at least one benzylic hydrogen.

This conversion involves increasing the carbon chain length from two to three and converting a primary alcohol to a secondary alcohol. This can be achieved using a Grignard reagent.
Step 1: Oxidation of Ethanol. Ethanol is oxidized to acetaldehyde (ethanal) using a mild oxidizing agent like Pyridinium chlorochromate (PCC) or by passing its vapors over heated copper at 573 K. \[ \underset{Ethanol}{CH_3CH_2OH} \xrightarrow{PCC} \underset{Acetaldehyde}{CH_3CHO} \]
Step 2: Grignard Reaction. Acetaldehyde is treated with a Grignard reagent, methyl magnesium bromide (CH\(_3\)MgBr), in dry ether. The nucleophilic methyl group from the Grignard reagent attacks the electrophilic carbonyl carbon. \[ \underset{Acetaldehyde}{CH_3CHO} + CH_3MgBr \xrightarrow{Dry Ether} \underset{Adduct}{CH_3-CH(OMgBr)-CH_3} \]
Step 3: Hydrolysis. The resulting magnesium adduct is hydrolyzed with dilute acid (e.g., H\(_3\)O\(^+\)) to yield the final product, propan-2-ol. \[ CH_3-CH(OMgBr)-CH_3 \xrightarrow{H_3O^+} \underset{Propan-2-ol}{CH_3-CH(OH)-CH_3} + Mg(OH)Br \] Quick Tip: Grignard reactions are extremely versatile for C-C bond formation. Remember the general patterns: - Formaldehyde + Grignard → 1° alcohol - Other Aldehydes + Grignard → 2° alcohol - Ketones + Grignard → 3° alcohol

(Note: There appears to be a typo in the question. The conversion of Propanal (3 carbons) to 2-hydroxy propanoic acid (3 carbons) is not a standard, direct synthesis. A more likely intended question is the conversion of Propanal to 2-hydroxybutanoic acid, which increases the carbon chain by one, a characteristic of cyanohydrin synthesis. The solution below is for this likely intended conversion.)

The synthesis of an \(\alpha\)-hydroxy carboxylic acid from an aldehyde is a classic two-step process involving the formation and subsequent hydrolysis of a cyanohydrin.
Step 1: Cyanohydrin Formation. Propanal is treated with hydrogen cyanide (HCN). The nucleophilic cyanide ion (CN\(^-\)) attacks the carbonyl carbon of propanal, followed by protonation of the oxygen atom to form 2-hydroxybutanenitrile (propanal cyanohydrin). \[ \underset{Propanal}{CH_3CH_2CHO} + HCN \longrightarrow \underset{2-Hydroxybutanenitrile}{CH_3CH_2-CH(OH)-CN} \]
Step 2: Acid Hydrolysis. The nitrile group (-CN) of the cyanohydrin is hydrolyzed by heating with an aqueous mineral acid (e.g., H\(_3\)O\(^+\)). The carbon-nitrogen triple bond is converted into a carboxylic acid group (-COOH). \[ CH_3CH_2-CH(OH)-CN \xrightarrow{H_3O^+, \Delta} \underset{2-Hydroxybutanoic acid}{CH_3CH_2-CH(OH)-COOH} + NH_4^+ \] Quick Tip: The cyanohydrin reaction is an excellent method for converting an aldehyde or ketone into an \(\alpha\)-hydroxy carboxylic acid. It is a step-up reaction, as it increases the number of carbon atoms in the molecule by one.

This synthesis shows the conversion of benzene to acetophenone (methyl phenyl ketone). This is a classic example of Friedel-Crafts Acylation.
This reaction introduces an acyl group (-COR) onto an aromatic ring. It is an electrophilic aromatic substitution reaction.
The reagents required are:
1. An acylating agent, which for introducing an acetyl group (CH\(_3\)CO-) would be acetyl chloride (CH\(_3\)COCl) or acetic anhydride ((CH\(_3\)CO)\(_2\)O).
2. A Lewis acid catalyst, most commonly anhydrous aluminum chloride (AlCl\(_3\)).
The catalyst activates the acylating agent by forming a highly electrophilic acylium ion ([CH\(_3\)C=O]\(^+\)), which then attacks the benzene ring.
The reaction is: \[ \underset{Benzene}{C_6H_6} + \underset{Acetyl chloride}{CH_3COCl} \xrightarrow{Anhyd. AlCl_3} \underset{Acetophenone}{C_6H_5COCH_3} + HCl \] Quick Tip: Friedel-Crafts acylation is a superior method to Friedel-Crafts alkylation because it does not suffer from rearrangement of the electrophile and the product is deactivated, preventing poly-acylation.