CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 1 - 56/1/1)

\textcolor{red{Step 1: Concept
For a first-order reaction, the reaction rate depends linearly on the concentration of a single reactant. The integrated rate law yields a specific half-life formula given by \(t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}\), where \(k\) is the rate constant.

\textcolor{red{Step 2: Analysis

Looking closely at this formula, the initial concentration term, \([R]_0\), is completely absent. This mathematically demonstrates that the time required for half of the reactant to be consumed does not depend on how much starting material is present.

For a first-order reaction, the half-life is given by:
\[ t_{1/2}=\frac{0.693}{k} \]
where \(k\) is the rate constant.

Notice that the expression contains only the rate constant \(k\) and does not include the initial concentration \([R]_0\).

This means that regardless of whether the reaction starts with a low concentration or a very high concentration of the reactant, the time required for the concentration to decrease to half of its initial value remains exactly the same.

For example, if a sample initially contains \(1.0\,\mathrm{mol\,L^{-1}}\), it will decrease to \(0.5\,\mathrm{mol\,L^{-1}}\) in one half-life. Similarly, if another sample starts at \(10.0\,\mathrm{mol\,L^{-1}}\), it will decrease to \(5.0\,\mathrm{mol\,L^{-1}}\) in the same amount of time.

Thus, the half-life of a first-order reaction is independent of the initial concentration and depends solely on the magnitude of the rate constant. A larger value of \(k\) results in a shorter half-life, while a smaller value of \(k\) leads to a longer half-life.



\textcolor{red{Step 3: Conclusion
Since \(t_{1/2}\) remains perfectly constant regardless of any variations or changes in the initial concentration \([R]_0\), plotting \(t_{1/2}\) on the y-axis against \([R]_0\) on the x-axis will yield a straight horizontal line parallel to the x-axis.


\textcolor{red{Final Answer: (B) Quick Tip: For first order, \(t_{1/2}\) is independent of \([R]_0\). For zero order, \(t_{1/2} \propto [R]_0\).

\textcolor{red{Step 1: Concept
Freezing point depression (\(\Delta T_f = i \cdot K_f \cdot m\)) is a colligative property. This means the magnitude of the freezing point decrease depends entirely on the total number of dissolved solute particles (represented by the van't Hoff factor, \(i\)) rather than their specific chemical identity.

\textcolor{red{Step 2: Analysis

Let us determine the \(i\) values by assuming complete dissociation for the electrolytes: Glucose is a non-electrolyte (\(i=1\)), \(CaCl_{2}\) dissociates into \(Ca^{2+}\) and \(2Cl^{-}\) (\(i=3\)), KCl dissociates into \(K^{+}\) and \(Cl^{-}\) (\(i=2\)), and Urea is a non-electrolyte (\(i=1\)).

The van't Hoff factor (\(i\)) represents the number of particles produced in solution by one formula unit of a solute.

Non-electrolytes do not ionize or dissociate in water. Therefore, each molecule remains intact and contributes only one particle to the solution.

Glucose (\(C_6H_{12}O_6\)) is a non-electrolyte:
\[ C_6H_{12}O_6(aq) \rightarrow C_6H_{12}O_6(aq) \]
Hence,
\[ i=1. \]

Calcium chloride is a strong electrolyte and dissociates completely as:
\[ CaCl_2(aq)\rightarrow Ca^{2+}(aq)+2Cl^{-}(aq) \]
One formula unit produces three ions, so
\[ i=3. \]

Potassium chloride is also a strong electrolyte:
\[ KCl(aq)\rightarrow K^{+}(aq)+Cl^{-}(aq) \]
Since two ions are produced,
\[ i=2. \]

Urea (\(NH_2CONH_2\)) is a non-electrolyte and does not dissociate in water:
\[ NH_2CONH_2(aq)\rightarrow NH_2CONH_2(aq) \]
Therefore,
\[ i=1. \]

These \(i\) values are important because colligative properties such as boiling point elevation, freezing point depression, osmotic pressure, and lowering of vapour pressure depend on the total number of dissolved particles rather than the chemical identity of the solute.


\textcolor{red{Step 3: Conclusion
Because \(CaCl_{2}\) provides the maximum number of particles per mole of solute dissolved (\(i=3\)), it will cause the greatest depression in the freezing point, resulting in the lowest overall freezing temperature for the solution.


\textcolor{red{Final Answer: (B) Quick Tip: Higher \(i\) value leads to a greater decrease in freezing point.

\textcolor{red{Step 1: Concept
According to IUPAC definitions, a transition metal is an element that has partially filled d-orbitals either in its neutral ground state or in any of its common oxidation states.

\textcolor{red{Step 2: Analysis

Let us examine the electronic configuration of Mercury (Hg, atomic number 80). In its ground state, Hg has a configuration of \([Xe]\,4f^{14}5d^{10}6s^2\). When it forms its most common \(Hg^{2+}\) ion, it loses the two \(6s\) electrons, leaving a strictly \([Xe]\,4f^{14}5d^{10}\) configuration.

Mercury has atomic number \(80\), meaning it possesses \(80\) electrons in its neutral state.

The electronic configuration of neutral mercury is:
\[ [Xe]\,4f^{14}5d^{10}6s^2. \]

During ion formation, electrons are removed first from the outermost shell, which in mercury is the \(6s\) orbital.

Removal of the two \(6s\) electrons gives the stable ion:
\[ Hg^{2+}:[Xe]\,4f^{14}5d^{10}. \]

The \(5d\) subshell remains completely filled with ten electrons. A completely filled \(d^{10}\) configuration is especially stable due to its symmetrical electron distribution and lower energy.

Since all the electrons in the \(5d\) subshell are paired, there are no unpaired electrons present in the \(Hg^{2+}\) ion.

Consequently, \(Hg^{2+}\) is diamagnetic and is not attracted by an external magnetic field.


\textcolor{red{Step 3: Conclusion
Because the 5d subshell remains completely filled with 10 electrons in both its elemental state and its stable ionic state, Mercury lacks partially filled d-orbitals and is therefore excluded from being classified as a transition metal.


\textcolor{red{Final Answer: (C) Quick Tip: Zn, Cd, and Hg are d-block elements but not transition metals.

\textcolor{red{Step 1: Concept
The Arrhenius equation (\(k = A e^{-E_{a}/RT}\)) relates the rate constant of a reaction to temperature and activation energy. This formula is fundamentally rooted in the Boltzmann distribution of molecular kinetic energies.

\textcolor{red{Step 2: Analysis

In the context of the Boltzmann distribution, the dimensionless exponential term \(e^{-E_a/RT}\) calculates the precise mathematical fraction of the total molecules that possess kinetic energy equal to or greater than the activation energy barrier (\(E_a\)) at a specific absolute temperature (\(T\)).

According to the Maxwell--Boltzmann distribution, molecules in a sample possess a wide range of kinetic energies rather than all having the same energy.

For a chemical reaction to occur, the reacting molecules must possess energy equal to or greater than the activation energy, \(E_a\).

The exponential term in the Arrhenius equation,
\[ e^{-E_a/RT}, \]
represents the fraction of molecules that have sufficient energy to overcome the activation energy barrier.

Here,

\(E_a\) is the activation energy,
\(R\) is the universal gas constant,
\(T\) is the absolute temperature in kelvin.


Since this term is exponential, even a small increase in temperature significantly increases the value of \(e^{-E_a/RT}\).

As a result, a much larger fraction of molecules acquires sufficient energy to undergo successful collisions, causing the reaction rate to increase rapidly with temperature.

This explains why many chemical reactions proceed much faster when heated, even though the average molecular energy increases only slightly.


\textcolor{red{Step 3: Conclusion
Therefore, this standard exponential term directly quantifies the proportion of molecules capable of having effective collisions that result in a chemical reaction.


\textcolor{red{Final Answer: (B) Quick Tip: This is known as the Arrhenius factor or Boltzmann factor.

\textcolor{red{Step 1: Concept
Electrolysis in aqueous solutions involves a competition between the dissolved ions and water molecules at the electrodes. The species that actually discharges depends on the relative standard reduction and oxidation potentials.

\textcolor{red{Step 2: Analysis
In an aqueous \(CuCl_2\) solution, \(Cu^{2+}\) has a higher reduction potential than water, so copper metal safely deposits at the cathode. At the anode, \(Cl^{-}\) and water compete for oxidation. Although water has a slightly lower standard oxidation potential, the kinetic ``overvoltage'' required to produce oxygen gas makes the oxidation of \(Cl^{-}\) more favorable.

During electrolysis of aqueous \(CuCl_2\), the solution contains \(Cu^{2+}\), \(Cl^{-}\), and water molecules.

At the cathode (negative electrode), reduction occurs. Two possible reduction reactions are:
\[ Cu^{2+}+2e^{-}\rightarrow Cu(s) \]
and
\[ 2H_2O+2e^{-}\rightarrow H_2+2OH^{-}. \]

Since the standard reduction potential of \(Cu^{2+}\) is higher than that of water, \(Cu^{2+}\) ions are reduced more readily.

Therefore, metallic copper is deposited on the cathode:
\[ Cu^{2+}+2e^{-}\rightarrow Cu(s). \]

At the anode (positive electrode), oxidation occurs. The competing reactions are:
\[ 2Cl^{-}\rightarrow Cl_2+2e^{-} \]
and
\[ 2H_2O\rightarrow O_2+4H^{+}+4e^{-}. \]

Although water appears slightly more favourable based on standard electrode potentials, oxygen evolution requires a significant kinetic overpotential (overvoltage). This extra energy barrier slows the oxidation of water.

The oxidation of chloride ions has a lower kinetic barrier under these conditions, making chlorine gas evolution the preferred process.

Hence, the overall products of electrolysis are:

Cathode: Copper metal is deposited.
Anode: Chlorine gas is evolved.



\textcolor{red{Step 3: Conclusion
As a consequence of this overpotential, chloride ions are preferentially oxidized to form diatomic chlorine, meaning \(Cl_{2}\) gas is actively released at the platinum anode.


\textcolor{red{Final Answer: (D) Quick Tip: In aqueous \(CuCl_{2}\), chloride ions are discharged at the anode.

\textcolor{red{Step 1: Concept
Acetylation is a chemical substitution reaction where an active hydrogen atom—typically from a hydroxyl (\(-OH\)) or amino (\(-NH_2\)) group—is replaced by an acetyl functional group (\(-COCH_3\)), usually using acetic anhydride.

\textcolor{red{Step 2: Analysis

The commercial synthesis of aspirin involves the esterification of the phenolic hydroxyl group found on salicylic acid. The systematic IUPAC name for salicylic acid is 2-hydroxybenzoic acid.

Aspirin is prepared commercially by reacting salicylic acid with acetic anhydride in the presence of a suitable acid catalyst, such as concentrated sulfuric acid or phosphoric acid.

The IUPAC name of salicylic acid is:
\[ \boxed{2-hydroxybenzoic acid} \]
indicating that a hydroxyl (\(-OH\)) group is attached at the second carbon (ortho position) relative to the carboxylic acid (\(-COOH\)) group.

Salicylic acid contains two functional groups:

A phenolic hydroxyl group (\(-OH\)).
A carboxylic acid group (\(-COOH\)).


During aspirin synthesis, only the phenolic hydroxyl group undergoes esterification with acetic anhydride, while the carboxylic acid group remains unchanged.

The reaction can be represented as:
\[ Salicylic acid+Acetic anhydride \rightarrow Aspirin+Acetic acid \]

The product formed is acetylsalicylic acid (aspirin), in which the hydrogen atom of the phenolic \(-OH\) group is replaced by an acetyl (\(-COCH_3\)) group.

This acetylation reduces the acidity and irritation associated with the phenolic hydroxyl group while retaining the medicinal properties of the compound.



\textcolor{red{Step 3: Conclusion
When 2-hydroxybenzoic acid is treated with acetic anhydride in the presence of an acid catalyst, the phenol group successfully undergoes acetylation to form acetylsalicylic acid, which is the active pharmaceutical ingredient in Aspirin.


\textcolor{red{Final Answer: (C) Quick Tip: Aspirin is also known as acetylsalicylic acid.

\textcolor{red{Step 1: Concept
According to Werner's coordination theory, secondary valency specifically refers to the coordination number of the central metal ion. This is the total number of ligand donor atoms that are directly bonded to the metal via coordinate covalent bonds.

\textcolor{red{Step 2: Analysis

Inspecting the coordination sphere of \([Co(NH_{3})_{5}(NO_{2})]^{2+}\), we can see that the central cobalt ion is directly bonded to five neutral ammonia (\(NH_{3}\)) ligands and one anionic nitrite (\(NO_{2}^{-}\)) ligand. All of these act as monodentate ligands.

The coordination sphere consists of the central cobalt ion surrounded by six ligands.

The complex contains:

Five ammonia (\(NH_3\)) ligands.
One nitrite (\(NO_2^{-}\)) ligand.


Ammonia is a neutral ligand that donates one lone pair of electrons through its nitrogen atom to form a coordinate bond with the metal ion.

The nitrite ion also donates only one lone pair to the metal centre, coordinating through either the nitrogen atom (nitro form) or an oxygen atom (nitrito form).

Since each ligand donates only one pair of electrons through a single donor atom, both \(NH_3\) and \(NO_2^{-}\) are classified as monodentate ligands.

The total number of donor atoms attached to the cobalt ion is therefore:
\[ 5+1=6. \]

Hence, the coordination number of cobalt in
\[ [Co(NH_3)_5(NO_2)]^{2+} \]
is
\[ \boxed{6}. \]

A coordination number of six generally corresponds to an octahedral geometry around the central metal ion.



\textcolor{red{Step 3: Conclusion
Adding the number of coordinating donor atoms together gives a total of \(5 + 1 = 6\). Therefore, the secondary valency, or the coordination number, for cobalt in this complex is 6.


\textcolor{red{Final Answer: (D) Quick Tip: Secondary valency equals the number of coordinate bonds formed.

\textcolor{red{Step 1: Concept
The nitration of phenol is an electrophilic aromatic substitution reaction. The hydroxyl (\(-OH\)) group attached to the benzene ring is highly activating and strongly directs incoming electrophiles toward the ortho and para positions.

\textcolor{red{Step 2: Analysis

When dilute nitric acid (\(HNO_{3}\)) is utilized at low temperatures (around 298 K), the concentration of the nitronium ion (\(NO_{2}^{+}\)) electrophile is kept relatively low. These mild conditions prevent the complete, exhaustive nitration of the highly reactive phenol ring.

The hydroxyl group (\(-OH\)) present in phenol is a strongly electron-donating group due to resonance.

As a result, the benzene ring becomes highly activated toward electrophilic substitution reactions and directs incoming electrophiles to the ortho and para positions.

The active electrophile responsible for nitration is the nitronium ion:
\[ NO_2^{+}. \]

When dilute nitric acid is used at room temperature (approximately \(298\,\mathrm{K}\)), only a small concentration of \(NO_2^{+}\) ions is produced.

Under these mild conditions, nitration occurs only once, producing mainly ortho-nitrophenol and para-nitrophenol.

If concentrated nitric acid or a nitrating mixture of concentrated \(HNO_3\) and concentrated \(H_2SO_4\) is used, the concentration of \(NO_2^{+}\) becomes much higher.

Since the phenol ring is already highly activated, repeated nitration takes place readily, producing 2,4,6-trinitrophenol (picric acid).

Therefore, dilute nitric acid and low temperature are used to achieve controlled mononitration and prevent exhaustive nitration of the aromatic ring.



\textcolor{red{Step 3: Conclusion
As a result of these controlled conditions, the reaction primarily yields a mixture of mono-nitrated isomers, specifically ortho-nitrophenol and para-nitrophenol, which can subsequently be separated via steam distillation.


\textcolor{red{Final Answer: (D) Quick Tip: Concentrated \(HNO_{3}\) would produce 2,4,6-trinitrophenol (picric acid).

\textcolor{red{Step 1: Concept
Enantiomers are pairs of optical stereoisomers that are non-superimposable mirror images of one another. Because their internal atomic distances and bond angles are identical, they share all typical achiral physical and chemical properties.

\textcolor{red{Step 2: Analysis

This symmetrical relationship means enantiomers will have identical melting points, boiling points, and refractive indices, and will react identically with achiral reagents. However, they interact inversely with chiral entities, such as plane-polarized light.

Enantiomers are stereoisomers that are non-superimposable mirror images of each other.

Since they possess the same molecular formula, bonding pattern, and intermolecular forces, they exhibit identical physical properties such as:

Melting point.
Boiling point.
Density.
Refractive index.
Solubility in achiral solvents.


Enantiomers also undergo identical chemical reactions with achiral reagents because an achiral environment cannot distinguish between the two mirror-image molecules.

Their only significant difference arises when they interact with chiral substances or chiral environments.

One of the most important chiral interactions involves plane-polarized light.

One enantiomer rotates the plane of polarized light in the clockwise direction (dextrorotatory, \(+\)), whereas the other rotates it by exactly the same magnitude in the anticlockwise direction (levorotatory, \(-\)).

Thus, although enantiomers are identical in almost all physical and chemical properties, they exhibit opposite optical activities and may also react differently with chiral reagents, enzymes, or biological systems.


\textcolor{red{Step 3: Conclusion
While the magnitude (degree) of their optical rotation is exactly the same, the direction is entirely opposite—one rotates light dextrorotatory (\(+\)), and the other levorotatory (\(-\)). Therefore, claiming they have the identical specific rotation is factually incorrect.


\textcolor{red{Final Answer: (B) Quick Tip: Enantiomers share all physical constants except for the direction of light rotation.

\textcolor{red{Step 1: Concept
Direct nitration of aniline is typically carried out in a strongly acidic medium using a nitrating mixture. Under normal non-acidic conditions, the amino group is a strong ortho/para-directing activator.

\textcolor{red{Step 2: Analysis

However, in the presence of strong acids, the basic \(-NH_{2}\) group gets readily protonated to form the anilinium ion (\(-NH_{3}^{+}\)). This positively charged ion behaves entirely differently; it acts as a strongly electron-withdrawing and meta-directing deactivating group.

In neutral conditions, the amino group (\(-NH_2\)) possesses a lone pair of electrons that can be donated to the benzene ring through resonance.

This resonance increases the electron density of the ring, making it highly reactive toward electrophilic substitution reactions.

Consequently, the amino group behaves as an activating, ortho-directing, and para-directing substituent.

In the presence of strong acids, such as concentrated hydrochloric acid or sulfuric acid, the amino group becomes protonated:
\[ -NH_2+H^{+}\rightarrow -NH_3^{+}. \]

After protonation, the nitrogen atom no longer possesses a lone pair available for resonance with the aromatic ring.

The positively charged anilinium ion withdraws electron density from the benzene ring through the strong inductive (\(-I\)) effect.

As a result, the aromatic ring becomes less reactive toward electrophilic substitution, making the substituent a deactivating group.

Furthermore, the electron deficiency is greatest at the ortho and para positions, making electrophilic substitution comparatively more favourable at the meta position.

Therefore, the protonated amino group (\(-NH_3^{+}\)) behaves as a strongly electron-withdrawing, meta-directing, and deactivating substituent, unlike the neutral amino group.



\textcolor{red{Step 3: Conclusion
Because a significant fraction of the reactant exists as the anilinium ion during the reaction, nitration at the meta position becomes highly favorable. This dual behavior results in a mixed product distribution where the meta derivative forms a major proportion of the yield.


\textcolor{red{Final Answer: (B) Quick Tip: The anilinium ion is meta-directing, while aniline itself is ortho/para-directing.

\textcolor{red{Step 1: Concept
The \(pK_{b}\) value is a logarithmic measure of the base dissociation constant. It is strictly inversely proportional to basic strength; meaning that a stronger, more reactive base will always possess a much lower \(pK_{b}\) value.

\textcolor{red{Step 2: Analysis

Basic strength in amines depends largely on the availability of the lone pair of electrons on the nitrogen atom. Alkyl groups (like methyl) exhibit an inductive electron-donating effect (+I effect), which increases electron density on nitrogen. Conversely, groups like nitro (\(-NO_{2}\)) are strongly electron-withdrawing.

The basic nature of an amine depends on how readily the nitrogen atom can donate its lone pair of electrons to a proton (\(H^+\)).

The greater the availability of the lone pair, the stronger is the basic character of the amine.

Alkyl groups such as methyl (\(-CH_3\)) exhibit a positive inductive effect (\(+I\) effect), which pushes electron density toward the nitrogen atom.

This increased electron density makes the lone pair more available for protonation:
\[ RNH_2 + H^+ \rightarrow RNH_3^+. \]

Consequently, alkyl-substituted amines are generally more basic than ammonia because the alkyl groups stabilize the positively charged ammonium ion formed after protonation.

In contrast, strongly electron-withdrawing groups such as the nitro group (\(-NO_2\)) exert a negative inductive effect (\(-I\) effect), pulling electron density away from the nitrogen atom.

As a result, the lone pair becomes less available for donation, reducing the tendency of the amine to accept a proton.

Therefore, electron-donating substituents increase the basic strength of amines, whereas electron-withdrawing substituents decrease their basicity.


\textcolor{red{Step 3: Conclusion
N,N-Dimethylaniline (\(C_{6}H_{5}-N(CH_{3})_{2}\)) features two electron-donating methyl groups attached directly to the nitrogen. This localized electron enrichment makes its lone pair highly available for protonation, rendering it the strongest base among the choices, and thus giving it the lowest \(pK_{b}\).


\textcolor{red{Final Answer: (A) Quick Tip: Electron donating groups increase basicity and lower the \(pK_{b}\).

\textcolor{red{Step 1: Concept
Vitamins are essential organic micronutrients that perform specific metabolic and regulatory functions. A lack or poor absorption of these specific vitamins inevitably leads to distinct clinical deficiency diseases and physical symptoms.

\textcolor{red{Step 2: Analysis

Vitamin E (tocopherol) acts as a highly crucial fat-soluble antioxidant within the body. Its primary biological role is to protect delicate cellular membranes from oxidative degradation caused by reactive oxygen species, especially in environments rich in oxygen like red blood cells and muscle tissue.

Vitamin E is a fat-soluble vitamin belonging to the group of compounds known as tocopherols.

Since it is fat-soluble, it is mainly present within the lipid bilayers of cell membranes.

During normal metabolism, reactive oxygen species (ROS) such as free radicals are produced. These highly reactive species can attack membrane lipids and initiate lipid peroxidation.

Lipid peroxidation damages cell membranes, resulting in loss of membrane integrity and impaired cellular function.

Vitamin E interrupts this chain reaction by donating a hydrogen atom to free radicals, thereby converting them into more stable and less harmful species.

In doing so, Vitamin E protects polyunsaturated fatty acids present in cell membranes from oxidative degradation.

This protective role is especially important in tissues exposed to high oxygen concentrations, such as red blood cells, muscle tissue, nervous tissue, and other highly active cells.

Therefore, Vitamin E functions as one of the body's major lipid-phase antioxidants, helping maintain healthy cell membranes and preventing oxidative damage.


\textcolor{red{Step 3: Conclusion
When a diet is deficient in Vitamin E, these cell membranes become susceptible to rapid lipid peroxidation. This damage directly results in the increased physical fragility of red blood cells (leading to hemolysis) as well as noticeable muscular weakness.


\textcolor{red{Final Answer: (C) Quick Tip: Vitamin E deficiency is linked to neuromuscular and hematological symptoms.

\textcolor{red{Step 1: Concept
Azeotropes are constant boiling liquid mixtures.

\textcolor{red{Step 2: Meaning
This means the mixture boils at a specific temperature without any change in its overall composition during the phase change from liquid to gas.

\textcolor{red{Step 3: Analysis

Because the liquid and vapor phases possess identical compositions, boiling and condensing them repeatedly (which is the fundamental mechanism of fractional distillation) does not enrich one component over the other.

Fractional distillation separates liquids based on differences in their volatilities.

During distillation, the vapour formed is richer in the more volatile component than the original liquid.

Repeated cycles of vaporization and condensation inside the fractionating column progressively increase the concentration of the more volatile component in the vapour phase.

However, in the case of an azeotrope, the liquid and vapour phases have exactly the same composition at the boiling point.

Since the vapour composition is identical to that of the liquid,
\[ Composition of liquid=Composition of vapour. \]

Consequently, repeated boiling and condensation do not produce any enrichment of one component over the other.

As a result, fractional distillation becomes ineffective in separating the components beyond the azeotropic composition.

Therefore, azeotropic mixtures cannot be separated completely by ordinary fractional distillation and require special techniques such as azeotropic distillation, extractive distillation, or the use of suitable separating agents.


\textcolor{red{Step 4: Conclusion
Therefore, both statements are factually true, and the identical composition in both phases perfectly explains why fractional distillation fails to separate them.


\textcolor{red{Final Answer: (A) Quick Tip: An azeotrope behaves like a pure liquid in terms of boiling point and distillation.

\textcolor{red{Step 1: Concept
The hydroxyl (\(-\)OH) group attached to a benzene ring has lone pairs of electrons on the oxygen atom.

\textcolor{red{Step 2: Meaning
Through resonance (+R effect), it donates electron density into the ring, which significantly activates the ring towards incoming electrophiles.

\textcolor{red{Step 3: Analysis

This increased electron density is concentrated specifically at the ortho and para positions, not the meta position. Therefore, the Assertion claiming it is a meta-directing group is entirely incorrect. The Reason, however, correctly identifies \(-OH\) as a ring-activating group.

The hydroxyl group (\(-OH\)) contains a lone pair of electrons on the oxygen atom.

Through resonance, this lone pair is donated into the benzene ring, increasing the electron density of the aromatic system.

The resonance structures show that the additional electron density is concentrated mainly at the ortho and para positions relative to the hydroxyl group.

Because these positions become more electron-rich, they are the preferred sites for attack by electrophiles during electrophilic aromatic substitution reactions.

Hence, the \(-OH\) group acts as an ortho- and para-directing substituent.

The increased electron density also activates the benzene ring, making phenol considerably more reactive than benzene toward electrophilic substitution.

Therefore, the Assertion stating that \(-OH\) is a meta-directing group is incorrect.

The Reason is correct because the hydroxyl group is indeed a strongly activating substituent due to its electron-donating resonance effect.

Thus, the Assertion is false, whereas the Reason is true.


\textcolor{red{Step 4: Conclusion
The Assertion statement is false, while the Reason statement presents a true chemical fact.


\textcolor{red{Final Answer: (D) Quick Tip: Groups with lone pairs directly attached to the ring (like \(-\)OH, \(-\)NH2) are generally strongly activating and ortho/para directing.

\textcolor{red{Step 1: Concept
Actinoids are f-block elements found in the 7th period of the periodic table.

\textcolor{red{Step 2: Meaning
They exhibit a large variety of oxidation states because the energy levels of their 5f, 6d, and 7s orbitals are highly comparable.

\textcolor{red{Step 3: Analysis

Because these energy gaps are small, electrons from all three subshells can easily participate in chemical bonding, validating the Assertion. The Reason states that actinoids are radioactive, which is also a true fact. However, radioactivity has no relation to valence electron participation or oxidation states.

The electronic configuration of actinoids generally involves the filling of the \(5f\), \(6d\), and \(7s\) subshells.

The energies of these three subshells are very close to one another.

Owing to these small energy differences, electrons from the \(5f\), \(6d\), and \(7s\) orbitals can all participate in chemical bonding.

As a result, actinoids exhibit a wide range of oxidation states, with the most common being \(+3\), although higher oxidation states such as \(+4\), \(+5\), \(+6\), and even \(+7\) are also observed.

Therefore, the Assertion stating that electrons from multiple subshells participate in bonding is correct.

The Reason states that actinoids are radioactive, which is also true because all actinoid elements possess unstable nuclei and undergo radioactive decay.

However, radioactivity is a nuclear property, whereas the participation of electrons in bonding depends on electronic structure and orbital energies.

Thus, although both the Assertion and the Reason are true, the Reason is not the correct explanation of the Assertion.


\textcolor{red{Step 4: Conclusion
Both individual statements are correct, but the radioactive nature of actinoids is not the cause for their wide range of oxidation states.


\textcolor{red{Final Answer: (B) Quick Tip: Oxidation states depend on electronic configuration and comparable orbital energies, not nuclear instability (radioactivity).

\textcolor{red{Step 1: Concept
Free aldehyde (\(-\)CHO) or ketone carbonyl groups react with hydroxylamine (\(H_2N-OH\)) to form derivatives called oximes.

\textcolor{red{Step 2: Meaning
When D-glucose is acetylated to form glucose pentaacetate, the cyclic hemiacetal ring structure is rigidly locked into place.

\textcolor{red{Step 3: Analysis

Because the structure is locked, the straight-chain form containing the free aldehyde group can no longer be established. Consequently, it fails to react with hydroxylamine. Thus, the Assertion is a true statement. The Reason claims this failure indicates a "free" \(-CHO\) group, which is contradictory; it actually proves the exact opposite—the absence of a free \(-CHO\) group.

In carbohydrates, the open-chain form contains a free aldehyde (\(-CHO\)) group that can undergo characteristic aldehyde reactions.

Hydroxylamine (\(NH_2OH\)) reacts readily with free aldehyde groups to form oximes:
\[ RCHO + NH_2OH \rightarrow RCH=NOH + H_2O. \]

However, when the aldehyde group participates in the formation of a cyclic acetal (glycoside), the ring structure becomes locked.

In this locked cyclic form, the molecule cannot readily open to regenerate the free aldehyde group.

Since no free \(-CHO\) group is available, the compound does not react with hydroxylamine.

Therefore, the failure to form an oxime confirms the absence of a free aldehyde group rather than its presence.

Hence, the Assertion is true because the locked structure prevents regeneration of the free aldehyde group.

The Reason is false because it incorrectly states that the absence of reaction indicates the presence of a free \(-CHO\) group, whereas it actually proves the absence of a free aldehyde group.


\textcolor{red{Step 4: Conclusion
The Assertion provides a correct chemical observation, but the Reason draws a completely false conclusion from it.


\textcolor{red{Final Answer: (C) Quick Tip: The inability to form an oxime provides strong chemical evidence for the cyclic structure of glucose.

\textcolor{red{Step 1: Concept
Raoult's law and non-ideal solutions in liquid mixtures.

\textcolor{red{Step 2: Meaning
Deviations occur based on the comparative strengths of intermolecular forces between the components.

\textcolor{red{Step 3: Analysis

Phenol and aniline form strong intermolecular hydrogen bonds with each other. These new A-B interactions (phenol-aniline) are stronger than the individual A-A (phenol-phenol) and B-B (aniline-aniline) interactions. This stronger bonding decreases the escaping tendency of the molecules, resulting in a lower vapor pressure than predicted by Raoult's law. Because the vapor pressure is lowered, a higher temperature is required to make the vapor pressure equal to the atmospheric pressure.

Raoult's law is obeyed only when the intermolecular forces between unlike molecules (A-B) are nearly equal to those between like molecules (A-A and B-B).

Phenol contains a hydroxyl (\(-OH\)) group, while aniline contains an amino (\(-NH_2\)) group.

These functional groups readily form strong intermolecular hydrogen bonds with each other.

The phenol-aniline (A-B) hydrogen bonds are stronger than the interactions present between phenol molecules themselves or between aniline molecules themselves.

Due to these stronger attractive forces, the molecules are held together more firmly in the liquid phase.

As a result, fewer molecules escape into the vapour phase, causing the vapour pressure of the solution to become lower than that predicted by Raoult's law.

This behaviour is known as a negative deviation from Raoult's law.

Since the vapour pressure is reduced, the solution must be heated to a higher temperature before its vapour pressure becomes equal to the external atmospheric pressure.

Consequently, the solution exhibits a higher boiling point than expected and may even form a maximum-boiling azeotrope.


\textcolor{red{Step 4: Conclusion
The mixture exhibits a negative deviation from Raoult's law, which directly results in an elevation of the solution's boiling point.



\textcolor{red{Final Answer: Negative deviation; The boiling point of the solution will increase. Quick Tip: Stronger intermolecular hydrogen bonding between mixed components leads to a Negative deviation and a Higher boiling point.

\textcolor{red{Step 1: Concept
Reactivity of haloarenes (aryl halides) in nucleophilic substitution reactions compared to haloalkanes.

\textcolor{red{Step 2: Meaning
Nucleophilic substitution requires the cleavage of the carbon-halogen (C-X) bond.

\textcolor{red{Step 3: Analysis

The first major factor is the resonance effect. The lone pairs of electrons on the halogen atom are in conjugation with the pi-electrons of the benzene ring, imparting a partial double bond character to the C--X bond, making it shorter and much stronger to break. The second factor is hybridization. The carbon atom attached to the halogen in a haloarene is \(sp^2\) hybridized (which is more electronegative), whereas in haloalkanes it is \(sp^3\) hybridized. The \(sp^2\) carbon holds the electron pair of the C--X bond more tightly.

Haloarenes are much less reactive toward nucleophilic substitution reactions than haloalkanes.

One important reason is the resonance effect between the halogen atom and the aromatic ring.

The lone pair of electrons on the halogen overlaps with the \(\pi\)-electron cloud of the benzene ring through resonance.

As a result, the carbon-halogen bond acquires partial double-bond character:
\[ \mathrm{C-X} \;\longrightarrow\; \mathrm{C=X\ (partial)}. \]

This partial double-bond character shortens and strengthens the C--X bond, making it more difficult to break.

A second reason involves hybridization.

In haloarenes, the carbon bonded to the halogen is \(sp^2\) hybridized, whereas in haloalkanes it is \(sp^3\) hybridized.

Since an \(sp^2\) hybridized carbon has greater \(s\)-character, it is more electronegative and attracts the bonding electron pair more strongly.

Consequently, the C--X bond in haloarenes is shorter, stronger, and less susceptible to nucleophilic substitution than the corresponding bond in haloalkanes.

These two factors together explain the remarkably lower reactivity of haloarenes toward nucleophilic substitution reactions.


\textcolor{red{Step 4: Conclusion
The combination of partial double bond character and the higher electronegativity of the \(sp^2\) carbon makes bond cleavage very difficult under normal conditions.



\textcolor{red{Final Answer: 1. Resonance effect causing partial double bond character. 2. Difference in hybridization (\(sp^2\) carbon is more electronegative and holds the bond tightly). Quick Tip: Remember the two R's for low reactivity: Resonance and (sp2) Repulsion against incoming nucleophiles.

\textcolor{red{Step 1: Concept
IUPAC nomenclature of coordination compounds containing both complex cations and complex anions.

\textcolor{red{Step 2: Meaning
The cationic part is named first, followed by the anionic part, and the oxidation states of both metal centers must balance out.

\textcolor{red{Step 3: Analysis

Silver typically exhibits an oxidation state of \(+1\). In the cation \([Ag(NH_{3})_{2}]^{+}\), the ligand is ammine, so it is diamminesilver(I). In the anion \([Ag(CN)_{2}]^{-}\), the ligand is cyanido, and since it is an anionic complex, the metal takes the suffix ``-ate'' (argentate), yielding dicyanidoargentate(I).

The rules of IUPAC nomenclature require naming the ligands before the central metal atom.

In the complex
\[ [Ag(NH_3)_2]^+, \]
ammonia (\(NH_3\)) is a neutral ligand known as ammine.

Since there are two ammine ligands, the prefix ``di'' is used.

Silver has an oxidation state of \(+1\), which is indicated by the Roman numeral (I).

Therefore, the correct IUPAC name is:
\[ \boxed{Diamminesilver(I). \]

In the complex
\[ [Ag(CN)_2]^-, \]
cyanide (\(CN^-\)) is named as the ligand cyanido.

Since the complex ion carries an overall negative charge, the name of the metal changes from silver to \textit{argentate.

The oxidation state of silver remains \(+1\) because:
\[ x+2(-1)=-1 \quad\Rightarrow\quad x=+1. \]

Hence, the correct IUPAC name is:
\[ \boxed{Dicyanidoargentate(I). \]



\textcolor{red{Step 4: Conclusion
Combining the two parts yields the full systematic name.



\textcolor{red{Final Answer: diamminesilver(I) dicyanidoargentate(I) Quick Tip: For silver in an anionic complex, always use its Latin root "argentate".

\textcolor{red{Step 1: Concept
IUPAC nomenclature of coordination compounds with simple cations and complex anions.

\textcolor{red{Step 2: Meaning
The positive counter ion is named first. The ligands inside the coordination sphere are named alphabetically before the metal.

\textcolor{red{Step 3: Analysis

The counter ion is potassium. The complex sphere has three oxalate ligands, termed ``oxalato''. The total charge of three oxalates is \(-6\). To balance three \(K^{+}\) ions \((+3)\), the iron must have an oxidation state of \(+3\). Because it sits in an anionic complex, iron becomes ``ferrate''.

Consider the coordination compound:
\[ K_3[Fe(C_2O_4)_3]. \]

The counter ion outside the coordination sphere is potassium (\(K^+\)).

The ligand \(C_2O_4^{2-}\) is called oxalato.

Since three oxalato ligands are present, the prefix ``tris'' is used.

The total charge contributed by the three oxalato ligands is:
\[ 3\times(-2)=-6. \]

The three potassium ions contribute a total charge of:
\[ 3\times(+1)=+3. \]

Therefore, the complex ion must carry a charge of \(-3\), and the oxidation state of iron is calculated as:
\[ x+3(-2)=-3, \]
\[ x=+3. \]

Since the complex ion is anionic, the metal name changes from iron to \textit{ferrate.

Hence, the complete IUPAC name of the compound is:
\[ \boxed{Potassium tris(oxalato)ferrate(III). \]


\textcolor{red{Step 4: Conclusion
Assembling the parts gives the name potassium trioxalatoferrate(III).



\textcolor{red{Final Answer: potassium trioxalatoferrate(III) Quick Tip: Do not specify the number of simple counter ions (it's "potassium", not "tripotassium").

\textcolor{red{Step 1: Concept
Ionisation isomerism in coordination chemistry.

\textcolor{red{Step 2: Meaning
Ionisation isomers yield different ions when dissolved in an aqueous solution due to the exchange of ligands between the coordination sphere and the ionization sphere.

\textcolor{red{Step 3: Analysis

The complex \([Co(NH_{3})_{5}SO_{4}]Cl\) will release \(Cl^{-}\) ions in solution. When treated with aqueous silver nitrate \((AgNO_{3})\), it will form a white precipitate of \(AgCl\). Conversely, \([Co(NH_{3})_{5}Cl]SO_{4}\) releases \(SO_{4}^{2-}\) ions in solution. When treated with aqueous barium chloride \((BaCl_{2})\), it will form a white precipitate of \(BaSO_{4}\).

The ions present outside the coordination sphere behave as counter ions and dissociate freely in aqueous solution.

In the complex
\[ [Co(NH_3)_5SO_4]Cl, \]
the chloride ion lies outside the coordination sphere.

On dissolving in water, the complex ionizes as:
\[ [Co(NH_3)_5SO_4]Cl \rightarrow [Co(NH_3)_5SO_4]^+ + Cl^-. \]

Since free chloride ions are present in solution, addition of aqueous silver nitrate produces a white precipitate of silver chloride:
\[ Ag^+ + Cl^- \rightarrow AgCl(s). \]

In the complex
\[ [Co(NH_3)_5Cl]SO_4, \]
the sulfate ion is present outside the coordination sphere.

The complex dissociates in water as:
\[ [Co(NH_3)_5Cl]SO_4 \rightarrow [Co(NH_3)_5Cl]^{2+}+SO_4^{2-}. \]

The free sulfate ions react with aqueous barium chloride to give a white precipitate of barium sulfate:
\[ Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s). \]

Thus, these precipitation reactions help distinguish whether chloride or sulfate is present inside or outside the coordination sphere of the complex.


\textcolor{red{Step 4: Conclusion
Applying these specific precipitation reagents acts as a definitive chemical test to distinguish between the two isomers.



\textcolor{red{Final Answer: Add aqueous \(AgNO_{3}\) to both. \([Co(NH_{3})_{5}SO_{4}]Cl\) yields a white precipitate of \(AgCl\), whereas the other does not. Add aqueous \(BaCl_{2}\) to both. \([Co(NH_{3})_{5}Cl]SO_{4}\) yields a white precipitate of \(BaSO_{4}\), whereas the other does not. Quick Tip: \(Ag^{+}\) ions specifically test for external halides, while \(Ba^{2+}\) ions specifically test for external sulphates.

\textcolor{red{Step 1: Concept
Thermodynamic stability of coordination complexes.

\textcolor{red{Step 2: Meaning
When a bidentate or polydentate ligand binds to a central metal ion using two or more donor atoms, it forms a cyclic or ring structure known as a chelate ring.

\textcolor{red{Step 3: Analysis

Complexes containing these chelate rings are significantly more stable than their analogous complexes containing only monodentate ligands. This enhanced stability is primarily an entropy-driven phenomenon and is termed the chelate effect.

A chelating ligand is a polydentate ligand that can coordinate to the same metal ion through two or more donor atoms.

When such a ligand binds to the metal ion, one or more ring structures called \textit{chelate rings are formed.

Complexes containing chelating ligands are considerably more stable than similar complexes containing only monodentate ligands.

This increased stability is known as the chelate effect.

The chelate effect is mainly explained by entropy considerations.

For example, when one bidentate ligand replaces two monodentate ligands, the total number of particles in the solution increases.

An increase in the number of particles leads to an increase in entropy (\(\Delta S>0\)), making the formation of the chelate complex thermodynamically more favourable.

In addition, multiple coordinate bonds between the ligand and the metal make dissociation of the ligand more difficult.

Consequently, chelate complexes are generally more stable, less easily dissociated, and more resistant to ligand substitution than corresponding complexes containing only monodentate ligands.


\textcolor{red{Step 4: Conclusion
Providing a standard example like ethylenediamine clarifies the definition.



\textcolor{red{Final Answer: The chelate effect is the enhanced stability observed in complexes containing bidentate or polydentate ligands that form ring structures with the central metal, compared to similar complexes with monodentate ligands. Example: \([Co(en)_{3]^{3+}\) is much more stable than \([Co(NH_{3})_{6}]^{3+}\). Quick Tip: "Chelate" means claw. Ligands that grab the metal in multiple places form highly stable rings.

\textcolor{red{Step 1: Concept
Chemical bonds that join monomers to form biological polymers.

\textcolor{red{Step 2: Meaning
Peptide linkages build proteins, whereas glycosidic linkages build carbohydrates.

\textcolor{red{Step 3: Analysis

A peptide linkage is an amide bond \((-CO-NH-)\) formed by the condensation reaction between the carboxyl group of one amino acid and the amino group of another, accompanied by the loss of a water molecule. A glycosidic linkage is an ether-type oxide bond \((-O-)\) formed by the condensation of the hydroxyl groups of two monosaccharides, also eliminating a water molecule.

A peptide linkage is formed when the carboxyl group (\(-COOH\)) of one amino acid reacts with the amino group (\(-NH_2\)) of another amino acid.

During this condensation reaction, one molecule of water is eliminated:
\[ -COOH + H_2N- \rightarrow -CO-NH- + H_2O. \]

The resulting amide bond,
\[ -CO-NH-, \]
is called a peptide linkage.

Repeated formation of peptide linkages joins amino acids together to produce peptides and proteins.

A glycosidic linkage is formed when the hydroxyl groups of two monosaccharides undergo a condensation reaction.

During this reaction, one molecule of water is removed and an oxygen atom bridges the two sugar units:
\[ -OH + HO- \rightarrow -O- + H_2O. \]

The resulting ether-like linkage,
\[ -O-, \]
is known as a glycosidic linkage.

Glycosidic linkages join monosaccharides together to form disaccharides, oligosaccharides, and polysaccharides such as starch, cellulose, and glycogen.


\textcolor{red{Step 4: Conclusion
The primary difference lies in the specific functional groups involved and the macromolecules they create.



\textcolor{red{Final Answer: A peptide linkage is an amide bond connecting amino acids to form proteins. A glycosidic linkage is an ether bond (oxide linkage) connecting monosaccharides to form complex carbohydrates. Quick Tip: Peptide = Proteins (amide bond). Glycosidic = Sugars/Carbohydrates (ether bond).

\textcolor{red{Step 1: Concept
Nutritional classification of amino acids required by the human body.

\textcolor{red{Step 2: Meaning
The categorization is based entirely on the body's capability to synthesize them internally.

\textcolor{red{Step 3: Analysis

Non-essential amino acids are those that the human body can readily synthesize in sufficient quantities from other intermediates (e.g., Glycine, Alanine). Essential amino acids are those that the human metabolic pathways cannot synthesize; therefore, they must be strictly obtained through external dietary sources to prevent nutritional deficiency (e.g., Valine, Leucine).

Amino acids are classified as essential or non-essential depending on whether the human body can synthesize them.

Non-essential amino acids are those that can be produced by normal metabolic pathways from other biochemical intermediates.

Since the body synthesizes these amino acids in adequate amounts, they do not necessarily need to be supplied through the diet.

Common examples of non-essential amino acids include:
\[ Glycine, Alanine, Aspartic acid, and Glutamic acid. \]

Essential amino acids cannot be synthesized by the human body, or are not produced in sufficient quantities to meet physiological needs.

Therefore, they must be obtained through food sources such as milk, eggs, pulses, meat, fish, and soy products.

Common examples include:
\[ Valine, Leucine, Lysine, Methionine, and Tryptophan. \]

A deficiency of essential amino acids can impair normal growth, tissue repair, enzyme synthesis, and overall metabolic functions.


\textcolor{red{Step 4: Conclusion
The difference is whether dietary intake is physiologically mandatory.



\textcolor{red{Final Answer: Essential amino acids cannot be synthesized by the human body and must be acquired through the diet (e.g., Valine). Non-essential amino acids can be synthesized internally by the body (e.g., Glycine). Quick Tip: "Essential" implies it is essential for you to eat them because your body cannot make them.

\textcolor{red{Step 1: Concept
Reaction kinetics, elementary rate laws, and the relationship between volume, concentration, and reaction rate.

\textcolor{red{Step 2: Meaning
Since it is a one-step (elementary) reaction, the molecularity equals the order. The rate law is written directly from the stoichiometry: \(Rate = k[A]^{2}[B]^{1}\).

\textcolor{red{Step 3: Analysis

Concentration is inversely proportional to volume (\(C=n/V\)). If the volume is decreased to \(\frac{1}{3}\) of the original, the concentration of both reactants A and B increases by a factor of 3. Substituting these new concentrations into the rate law gives: \[ New Rate=k(3[A])^2(3[B])=27k[A]^2[B]. \]
Therefore, the new rate is 27 times the original rate. The order of the reaction is an experimentally determined constant corresponding to the reaction mechanism.

The concentration of a solution is given by:
\[ C=\frac{n}{V}, \]
where \(n\) is the number of moles and \(V\) is the volume.

If the volume of the reaction mixture is reduced to one-third while the number of moles remains constant, the concentration of every reactant becomes three times its original value.

Thus,
\[ [A]\rightarrow 3[A] \qquadand\qquad [B]\rightarrow 3[B]. \]

The given rate law is:
\[ Rate=k[A]^2[B]. \]

Substituting the new concentrations,
\[ New Rate =k(3[A])^2(3[B]). \]

Simplifying,
\[ New Rate =k(9[A]^2)(3[B]) =27k[A]^2[B]. \]

Therefore,
\[ \boxed{New Rate=27\timesOriginal Rate.} \]

The overall order of the reaction remains unchanged because reaction order depends only on the exponents in the experimentally determined rate law and is independent of changes in concentration or volume.

Hence, decreasing the volume increases the reaction rate by a factor of \(27\), while the reaction order remains the same.



\textcolor{red{Step 4: Conclusion
Changing the physical volume alters the rate but never the fundamental order of the chemical reaction.



\textcolor{red{Final Answer: The rate of the reaction will become 27 times its original rate. No, the overall order of the reaction (which is 3) remains unchanged. Quick Tip: Decreasing volume increases concentration. For an overall order of \(n\), if volume scales by factor \(x\), rate scales by \((1/x)^{n}\). Here, \((3)^{3} = 27\).

\textcolor{red{Step 1: Concept
The integrated rate law for a first-order gas-phase reaction relates the rate constant (\(k\)) to the initial pressure of the reactant and its partial pressure at time \(t\).

\textcolor{red{Step 2: Meaning
For the reaction \(A(g) \rightarrow B(g) + C(g)\), if the initial pressure is \(P_i\) and the pressure decreases by \(x\) at time \(t\), the total pressure is \(P_t = (P_i - x) + x + x = P_i + x\). Therefore, the decrease in pressure \(x = P_t - P_i\). The partial pressure of the reactant at time \(t\) becomes \(P_A = P_i - x = P_i - (P_t - P_i) = 2P_i - P_t\).

\textcolor{red{Step 3: Analysis
From the given data: Initial pressure \(P_i = 0.30\) atm. Total pressure at \(t = 30\) s is \(P_t = 0.50\) atm.

Calculating \(P_A\):
\(P_A = 2(0.30) - 0.50 = 0.60 - 0.50 = 0.10\) atm.

Using the first-order rate equation:
\(k = \frac{2.303}{t} \log\left(\frac{P_i}{P_A}\right)\)
\(k = \frac{2.303}{30} \log\left(\frac{0.30}{0.10}\right) = \frac{2.303}{30} \log(3)\)
Substitute \(\log 3 = 0.48\):
\(k = \frac{2.303 \times 0.48}{30} = \frac{1.10544}{30} = 0.036848\) s\(^{-1}\).


\textcolor{red{Step 4: Conclusion

The precise evaluation using the total pressure variations yields the rate constant.




\textcolor{red{Final Answer: \(3.68 \times 10^{-2}\) s\(^{-1}\) Quick Tip: For \(A \rightarrow B + C\) gas phase reactions, partial pressure of \(A\) at time \(t\) is always \(P_A = 2P_i - P_t\).

\textcolor{red{Step 1: Concept
The relationship between electromotive force (EMF) and the reaction quotient/equilibrium constant is given by the Nernst equation.

\textcolor{red{Step 2: Meaning
\(E_{cell}\) represents the cell potential at any given concentration during the reaction, while \(E^\circ_{cell}\) is the standard cell potential measured at standard conditions (\(1~M\), \(1~atm\), \(298~K\)).

\textcolor{red{Step 3: Analysis


As a galvanic cell operates, the reactant concentrations decrease and product concentrations increase, causing \(E_{cell}\) to continuously drop. When the reaction finally reaches chemical equilibrium, the cell stops functioning and \(E_{cell}\) becomes strictly zero. Applying this to the Nernst equation, \[ E_{cell}=E^\circ_{cell}-\frac{RT}{nF}\ln Q, \]
setting \(E_{cell}=0\) and \(Q=K_c\) yields the fundamental relation: \[ E^\circ_{cell}=\frac{RT}{nF}\ln K_c. \]


A galvanic cell converts the chemical energy of a spontaneous redox reaction into electrical energy.

As the cell operates, the concentration of reactants gradually decreases, while the concentration of products increases.

Consequently, the reaction quotient (\(Q\)) continuously increases during the course of the reaction.

According to the Nernst equation,
\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln Q, \]
an increase in \(Q\) causes the cell potential to decrease.

Eventually, the reaction reaches chemical equilibrium, where there is no net tendency for the reaction to proceed in either direction.

At equilibrium,
\[ E_{cell}=0 \]
because the cell can no longer produce electrical energy.

At the same time,
\[ Q=K_c, \]
where \(K_c\) is the equilibrium constant of the reaction.

Substituting these equilibrium conditions into the Nernst equation,
\[ 0 = E^\circ_{cell} - \frac{RT}{nF}\ln K_c, \]
gives
\[ E^\circ_{cell} = \frac{RT}{nF}\ln K_c. \]

This equation establishes a direct relationship between the standard cell potential and the equilibrium constant. A larger positive value of \(E^\circ_{cell}\) corresponds to a larger value of \(K_c\), indicating that the forward reaction is highly spontaneous.



\textcolor{red{Step 4: Conclusion
Because \(E_{cell}\) is zero at equilibrium, \(K_c\) can only be mathematically related to the constant standard potential, \(E^\circ_{cell}\).



\textcolor{red{Final Answer: At equilibrium, the cell potential \(E_{cell}\) becomes zero. Therefore, substituting this into the Nernst equation relates the equilibrium constant exclusively to the standard cell potential \(E^\circ_{cell}\). Quick Tip: A dead battery has reached equilibrium; its \(E_{cell} = 0\).

\textcolor{red{Step 1: Concept
The standard electrochemical series dictates the displacement reactivity of metals with acids.

\textcolor{red{Step 2: Meaning
Metals possessing a negative standard reduction potential (\(E^\circ < 0\)) sit above hydrogen in the reactivity series.

\textcolor{red{Step 3: Analysis

The standard reduction potential of hydrogen (\(H^+/H_2\)) is exactly \(0.00\) V. For a metal to actively reduce \(H^+\) ions into \(H_2\) gas, it must act as a stronger reducing agent than hydrogen itself. This means its standard reduction potential must be algebraically lower (more negative) than zero. Metal A has a negative potential (\(-0.24\) V), making it capable of displacing hydrogen. Metal B has a positive potential (\(+0.80\) V), meaning it is less reactive and cannot displace hydrogen.


The standard hydrogen electrode (SHE) is assigned a standard reduction potential of:
\[ E^\circ = 0.00\;V. \]

The SHE serves as the reference electrode for measuring the standard reduction potentials of all other electrodes.

A metal can liberate hydrogen gas from dilute acids only if it is more easily oxidized than hydrogen.

This means the metal must have a standard reduction potential less than that of hydrogen, that is,
\[ E^\circ < 0.00\;V. \]

Metal A has
\[ E^\circ=-0.24\;V, \]
which is more negative than the hydrogen electrode.

Therefore, Metal A is a stronger reducing agent and readily displaces hydrogen from dilute acids:
\[ M+2H^+\rightarrow M^{2+}+H_2. \]

Metal B has
\[ E^\circ=+0.80\;V, \]
which is greater than zero.

Hence, Metal B is less readily oxidized than hydrogen and cannot reduce \(H^+\) ions to hydrogen gas.

Therefore, only metals having negative standard reduction potentials relative to the SHE can displace hydrogen from dilute acids.



\textcolor{red{Step 4: Conclusion
Only Metal 'A' meets the thermodynamic requirement for hydrogen displacement.



\textcolor{red{Final Answer: Metal 'A' Quick Tip: Only metals with negative \(E^\circ\) values can liberate \(H_2\) gas from dilute acids.

\textcolor{red{Step 1: Concept
Reversibility of secondary batteries (accumulators).

\textcolor{red{Step 2: Meaning
During the charging phase, an external electrical source forces current through the battery in the opposite direction, reversing the normal spontaneous discharge reactions.

\textcolor{red{Step 3: Analysis
While discharging, lead (\(Pb\)) and lead dioxide (\(PbO_2\)) convert into lead sulfate (\(PbSO_4\)). During the charging process, this reaction is forced backward through electrolysis. Solid lead sulfate on the cathode is reduced back to spongy lead (\(Pb\)), and the lead sulfate on the anode is oxidized back to lead dioxide (\(PbO_2\)), simultaneously regenerating aqueous sulfuric acid.

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

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


A lead storage battery is a secondary cell because its chemical reactions are reversible.

During discharge, the battery supplies electrical energy by converting both electrodes into lead sulfate.

The overall discharge reaction is:
\[ Pb+PbO_2+2H_2SO_4 \rightarrow 2PbSO_4+2H_2O. \]

During charging, an external electrical source forces the reverse reaction to occur through electrolysis.

At the cathode, lead sulfate gains electrons and is reduced back to metallic lead:
\[ PbSO_4(s)+2e^- \rightarrow Pb(s)+SO_4^{2-}(aq). \]

The porous metallic lead produced is known as spongy lead.

At the anode, lead sulfate undergoes oxidation to regenerate lead dioxide:
\[ PbSO_4(s)+2H_2O(l) \rightarrow PbO_2(s)+SO_4^{2-}(aq)+4H^+(aq)+2e^-. \]

The sulfate ions and hydrogen ions formed recombine to regenerate sulfuric acid in the electrolyte.

Thus, charging restores both the original electrode materials and the concentration of sulfuric acid, making the battery ready for repeated use.


\textcolor{red{Step 4: Conclusion
Combining these two half-reactions gives the overall forced chemical reaction.



\textcolor{red{Final Answer: Overall charging reaction: \(2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)\) Quick Tip: Charging a lead storage battery consumes water and regenerates concentrated sulfuric acid.

\textcolor{red{Step 1: Concept
Classification and operating principles of commercial primary batteries.

\textcolor{red{Step 2: Meaning
Primary cells are non-rechargeable. The mercury cell is specifically designed for low-current, long-life applications.

\textcolor{red{Step 3: Analysis

A mercury cell falls under the category of a primary cell. Its principal advantage over a standard Leclanche (dry) cell is its remarkably steady voltage output (\(1.35\) V) throughout its entire functional lifespan. This stability occurs because its overall cell reaction exclusively involves solid substances and no dissolved ions whose concentration might change over time and alter the potential.

The reaction utilizes a zinc-mercury amalgam anode and a mercury(II) oxide cathode.

Anode: \[ Zn(Hg)+2OH^- \rightarrow ZnO(s)+H_2O+2e^-. \]

Cathode: \[ HgO+H_2O+2e^- \rightarrow Hg(l)+2OH^-. \]


A mercury cell is a primary cell and therefore cannot be recharged after use.

It consists of a zinc-mercury amalgam as the anode, mercury(II) oxide as the cathode, and an alkaline electrolyte such as potassium hydroxide.

During discharge, oxidation occurs at the anode:
\[ Zn(Hg)+2OH^- \rightarrow ZnO+H_2O+2e^-. \]

Simultaneously, reduction occurs at the cathode:
\[ HgO+H_2O+2e^- \rightarrow Hg+2OH^-. \]

The overall cell potential remains nearly constant at approximately
\[ 1.35\;V \]
throughout the life of the cell.

This constant voltage is obtained because the cell reaction mainly involves solids and liquids whose activities remain essentially constant.

Since the concentrations of dissolved ionic species do not change significantly, the cell potential predicted by the Nernst equation also remains nearly unchanged.

Owing to its stable voltage, the mercury cell was widely used in precision instruments such as watches, hearing aids, cameras, and scientific equipment.


\textcolor{red{Step 4: Conclusion
Adding the half-reactions cancels the mobile ions, leaving only solid/liquid phases in the net equation.



\textcolor{red{Final Answer: It is a primary cell. Advantage: It provides a constant voltage throughout its life because the overall reaction does not involve any ions in solution. Overall reaction: \(Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l)\) Quick Tip: No aqueous ions in the overall reaction = Constant voltage output over time.

\textcolor{red{Step 1: Concept
Elevation of boiling point (\(\Delta T_b\)), which is a colligative property modified by the van't Hoff factor (\(i\)) for associating solutes.

\textcolor{red{Step 2: Meaning
Benzoic acid molecules pair up (dimerize) through hydrogen bonding in non-polar solvents like \(CS_2\). This reduces the total number of effective solute particles.

\textcolor{red{Step 3: Analysis

1. Calculate the molality (\(m\)):

Moles of benzoic acid (\(n\)) = \(\frac{Mass}{Molar Mass} = \frac{0.61 g}{122 g mol^{-1}} = 0.005\) mol.

Mass of solvent \(CS_2\) (\(w_1\)) = \(5 g = 0.005 kg\).

Molality (\(m\)) = \(\frac{n}{w_1} = \frac{0.005}{0.005} = 1.0 mol kg^{-1}\).


2. Calculate the van't Hoff factor (\(i\)):

For dimerization, 2 molecules associate into 1 (\(n=2\)).

Degree of association (\(\alpha\)) = 88% = 0.88.

Formula: \(i = 1 - \left(1 - \frac{1}{n}\right)\alpha = 1 - \left(1 - \frac{1}{2}\right)(0.88) = 1 - \frac{0.88}{2} = 1 - 0.44 = 0.56\).


3. Calculate Elevation in Boiling Point (\(\Delta T_b\)):
\(\Delta T_b = i \times K_b \times m = 0.56 \times 2.3 \times 1.0 = 1.288^\circ C\).


4. Calculate Final Boiling Point of Solution (\(T_b\)):
\(T_b = T_b^\circ + \Delta T_b = 46.2^\circ C + 1.288^\circ C = 47.488^\circ C\).


\textcolor{red{Step 4: Conclusion
The association of particles successfully lowers the expected colligative effect, generating the final adjusted boiling temperature.



\textcolor{red{Final Answer: \(47.488^\circ C\) Quick Tip: For dimerization, the van't Hoff factor formula simplifies directly to \(i = 1 - \frac{\alpha}{2}\).

\textcolor{red{Step 1: Concept
Prolonged heating of glucose with hydrogen iodide (HI) is a severe reduction reaction used for structure elucidation.

\textcolor{red{Step 2: Meaning
Hydrogen iodide is a strong reducing agent that removes all oxygen atoms from the carbohydrate backbone, fully saturating the carbon chain with hydrogen atoms.

\textcolor{red{Step 3: Analysis

When D-glucose is heated with HI and red phosphorus for a prolonged period, all the hydroxyl (\(-OH\)) groups and the aldehyde (\(-CHO\)) group are completely reduced. The resulting product is a straight, unbranched chain of six carbon atoms.


Hydriodic acid (HI) in the presence of red phosphorus is a powerful reducing agent.

On prolonged heating, it reduces every oxygen-containing functional group present in D-glucose.

The aldehyde group (\(-CHO\)) is reduced to a methyl group, while all the hydroxyl groups (\(-OH\)) are replaced by hydrogen atoms.

Consequently, every oxygen atom present in the glucose molecule is removed during the reduction process.

The final product obtained is:
\[ \boxed{n-hexane} \]
which is a straight-chain hydrocarbon containing six carbon atoms.

Since no carbon-carbon bonds are broken during the reaction, the carbon skeleton of glucose remains intact.

The formation of normal hexane proves that all six carbon atoms in D-glucose are arranged in one continuous, unbranched chain.

Thus, this reaction provides important experimental evidence for the straight-chain six-carbon structure of D-glucose.



\textcolor{red{Step 4: Conclusion
This reaction definitively proves that all six carbon atoms in a glucose molecule are linked linearly in a straight chain without any carbon branching.



\textcolor{red{Final Answer: \(CHO-(CHOH)_4-CH_2OH \xrightarrow{HI, \Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3 (n-hexane)\) Quick Tip: HI converts all functional groups in glucose to alkanes. It proves the 6-carbon straight chain.

\textcolor{red{Step 1: Concept
Bromine water (\(Br_2\) in water) is a mild oxidizing agent.

\textcolor{red{Step 2: Meaning
Because it is mild, it selectively oxidizes the easily oxidizable aldehyde group (\(-CHO\)) at the end of the chain, but it is not strong enough to oxidize the primary or secondary alcohol groups.

\textcolor{red{Step 3: Analysis

When D-glucose is heated with HI and red phosphorus for a prolonged period, all the hydroxyl (\(-OH\)) groups and the aldehyde (\(-CHO\)) group are completely reduced. The resulting product is a straight, unbranched chain of six carbon atoms.


Hydriodic acid (HI) in the presence of red phosphorus is a powerful reducing agent.

On prolonged heating, it reduces every oxygen-containing functional group present in D-glucose.

The aldehyde group (\(-CHO\)) is reduced to a methyl group, while all the hydroxyl groups (\(-OH\)) are replaced by hydrogen atoms.

Consequently, every oxygen atom present in the glucose molecule is removed during the reduction process.

The final product obtained is:
\[ \boxed{n-hexane} \]
which is a straight-chain hydrocarbon containing six carbon atoms.

Since no carbon-carbon bonds are broken during the reaction, the carbon skeleton of glucose remains intact.

The formation of normal hexane proves that all six carbon atoms in D-glucose are arranged in one continuous, unbranched chain.

Thus, this reaction provides important experimental evidence for the straight-chain six-carbon structure of D-glucose.



\textcolor{red{Step 4: Conclusion
The formation of a six-carbon monocarboxylic acid indicates that the carbonyl group present in glucose is exclusively an aldehyde, not a ketone.



\textcolor{red{Final Answer: \(CHO-(CHOH)_4-CH_2OH \xrightarrow{Br_2 water} COOH-(CHOH)_4-CH_2OH (Gluconic acid)\) Quick Tip: Mild oxidation with Bromine water yields Gluconic acid, proving the presence of an aldehyde group.

\textcolor{red{Step 1: Concept
Concentrated nitric acid (\(HNO_3\)) is a powerful oxidizing agent.

\textcolor{red{Step 2: Meaning
It is strong enough to simultaneously oxidize both the terminal aldehyde group and the terminal primary alcohol group into carboxylic acids.

\textcolor{red{Step 3: Analysis

Upon treatment with concentrated \(HNO_3\), D-glucose undergoes oxidation at both the top (C1, aldehyde) and the bottom (C6, primary alcohol) of the carbon chain. Both terminal ends are transformed into \(-COOH\) groups, resulting in a dicarboxylic acid.

Concentrated nitric acid (\(HNO_3\)) is a strong oxidizing agent capable of oxidizing both aldehyde and primary alcohol functional groups.

In D-glucose, the carbon atom at position C1 contains an aldehyde group (\(-CHO\)), while the carbon atom at position C6 contains a primary alcohol group (\(-CH_2OH\)).

During oxidation, the aldehyde group at C1 is converted into a carboxylic acid group:
\[ -CHO \;\longrightarrow\; -COOH. \]

Simultaneously, the primary alcohol group at C6 is also oxidized to a carboxylic acid:
\[ -CH_2OH \;\longrightarrow\; -COOH. \]

Since oxidation occurs at both ends of the carbon chain, the product contains two carboxylic acid groups.

The product formed is
\[ \boxed{Saccharic acid (Glucaric acid)}, \]
which is a dicarboxylic acid.

This reaction provides experimental evidence that D-glucose contains an aldehyde group at one end and a primary alcohol group at the other end of its six-carbon chain.



\textcolor{red{Step 4: Conclusion
This reaction provides direct chemical evidence for the presence of a primary alcoholic group (\(-CH_2OH\)) at the end of the glucose chain.



\textcolor{red{Final Answer: \(CHO-(CHOH)_4-CH_2OH \xrightarrow{Conc. HNO_3} COOH-(CHOH)_4-COOH (Saccharic acid)\) Quick Tip: Strong oxidation yields Saccharic (Glucaric) acid, proving the presence of a primary alcohol group.

\textcolor{red{Step 1: Concept
The boiling point of organic compounds depends directly on the strength of their intermolecular forces.

\textcolor{red{Step 2: Meaning
Both carboxylic acids and alcohols exhibit intermolecular hydrogen bonding due to their \(-OH\) groups.

\textcolor{red{Step 3: Analysis

The \(O-H\) bond in a carboxylic acid is more strongly polarized than in an alcohol because of the electron-withdrawing adjacent carbonyl (\(>C=O\)) group. This results in significantly stronger intermolecular hydrogen bonds. Furthermore, carboxylic acids uniquely associate to form stable cyclic dimers in the vapor phase or in non-polar solvents, effectively doubling their apparent molecular mass.

Carboxylic acids contain both a hydroxyl (\(-OH\)) group and a carbonyl (\(>C=O\)) group within the same functional group.

The highly electronegative carbonyl oxygen withdraws electron density through the inductive (\(-I\)) effect.

As a result, the \(O-H\) bond becomes more strongly polarized than the corresponding bond in alcohols.

This greater polarity enables carboxylic acids to form stronger intermolecular hydrogen bonds.

In addition, each carboxylic acid molecule can simultaneously act as both a hydrogen bond donor and a hydrogen bond acceptor.

Consequently, two carboxylic acid molecules associate through two hydrogen bonds to form a stable cyclic dimer:
\[ 2RCOOH \rightleftharpoons (RCOOH)_2. \]

This dimerization commonly occurs in the vapour phase and in non-polar solvents.

Since the molecules exist as dimers, their apparent molecular mass becomes nearly twice that of a single molecule.

Therefore, carboxylic acids possess unusually high boiling points compared with alcohols and other organic compounds of similar molecular mass.


\textcolor{red{Step 4: Conclusion
Breaking these highly extensive and stronger hydrogen bonds, especially the dimeric structures, requires a much greater input of thermal energy compared to the simple hydrogen bonds in alcohols.



\textcolor{red{Final Answer: Carboxylic acids have more extensive and stronger intermolecular hydrogen bonding than alcohols due to higher polarization. They also form highly stable dimeric structures, which require significantly more energy to break, leading to a higher boiling point. Quick Tip: Remember the "Dimer effect": Carboxylic acids pair up, making them heavier and harder to boil.

\textcolor{red{Step 1: Concept
Acidity in organic molecules is determined by the stability of the conjugate base formed after a proton (\(H^+\)) is removed.

\textcolor{red{Step 2: Meaning
The \(\alpha\)-hydrogen is the hydrogen attached to the carbon atom immediately adjacent to the carbonyl carbon.

\textcolor{red{Step 3: Analysis

Two main factors contribute to this acidity. First, the strongly electronegative carbonyl oxygen exerts a powerful electron-withdrawing inductive effect (\(-I\) effect), which pulls electron density away from the \(\alpha\)-carbon, weakening the \(C_{\alpha}-H\) bond. Second, when the proton is extracted by a base, the resulting carbanion (enolate ion) is highly stabilized by resonance.

The hydrogen atom attached to the \(\alpha\)-carbon of aldehydes and ketones is relatively acidic compared with that of ordinary hydrocarbons.

One reason is the strong electron-withdrawing inductive (\(-I\)) effect of the adjacent carbonyl group.

The carbonyl oxygen attracts electron density toward itself, reducing the electron density around the \(\alpha\)-carbon.

This weakens the \(C_{\alpha}-H\) bond, making removal of the proton easier.

When a base removes the \(\alpha\)-hydrogen, an enolate ion is formed:
\[ RCOCH_2R' + B^- \rightarrow RCOCHR'^- + HB. \]

The negative charge of the enolate ion is not confined to a single atom.

Instead, it is delocalized between the \(\alpha\)-carbon and the oxygen atom through resonance.

Resonance stabilization lowers the energy of the conjugate base, thereby increasing the acidity of the \(\alpha\)-hydrogen.

Thus, the combined effects of the electron-withdrawing carbonyl group and resonance stabilization of the enolate ion account for the relatively acidic nature of \(\alpha\)-hydrogen atoms.


\textcolor{red{Step 4: Conclusion
The negative charge of the enolate ion delocalizes onto the electronegative oxygen atom, making the conjugate base exceptionally stable and facilitating the release of the \(H^+\) ion.



\textcolor{red{Final Answer: The strong electron-withdrawing inductive effect (\(-I\)) of the carbonyl group weakens the \(C_\alpha-H\) bond. Furthermore, the conjugate base (enolate anion) formed after losing the proton is highly stabilized by resonance delocalization of the negative charge onto the oxygen atom. Quick Tip: Acidity of \(\alpha\)-hydrogens = \(-I\) effect of carbonyl + Resonance stabilization of the enolate.

\textcolor{red{Step 1: Concept
Nucleophilic addition reactions require an active nucleophile carrying a lone pair of electrons to attack the carbonyl carbon.

\textcolor{red{Step 2: Meaning
Ammonia (\(NH_3\)) and its derivatives (\(NH_2-Z\)) act as strong nucleophiles precisely because of the available lone pair of electrons located on the nitrogen atom.

\textcolor{red{Step 3: Analysis

A strongly acidic medium contains a high concentration of free protons (\(H^+\)). Because ammonia derivatives are basic, they readily react with these protons. The basic nitrogen atom accepts an \(H^+\), forming a stable protonated ammonium salt (\(^+NH_3-Z\)).

Amines and ammonia derivatives contain a lone pair of electrons on the nitrogen atom.

This lone pair makes them basic because it can readily accept a proton.

In a strongly acidic medium, the concentration of hydrogen ions (\(H^+\)) is very high.

The nitrogen atom accepts a proton according to the reaction:
\[ RNH_2 + H^+ \rightarrow RNH_3^+. \]

The resulting species is a protonated ammonium ion or ammonium salt.

Protonation removes the availability of the nitrogen lone pair for further reactions.

Consequently, the nucleophilic character of the amine is greatly reduced.

This is the reason why reactions involving free amines are often suppressed in strongly acidic media unless the amine is first protected or the reaction conditions are modified.


\textcolor{red{Step 4: Conclusion
Once the nitrogen is protonated, its lone pair of electrons is entirely tied up in the new \(N-H\) bond. Devoid of its lone pair, the molecule instantly loses all of its nucleophilic character and cannot attack the electrophilic carbonyl carbon.



\textcolor{red{Final Answer: In a strongly acidic medium, the basic ammonia derivatives undergo rapid protonation to form ammonium salts (\(H^+ + :NH_2-Z \rightarrow ^+NH_3-Z\)). Because the nitrogen's lone pair is consumed in this bond, the molecule completely loses its nucleophilic character and fails to attack the carbonyl group. Quick Tip: Acid protonates the nucleophile. No lone pair = No nucleophilic attack.

\textcolor{red{Step 1: Concept
The Reimer-Tiemann reaction is a classic method for the ortho-formylation of phenols.

\textcolor{red{Step 2: Meaning
It introduces an aldehyde group (\(-CHO\)) onto the aromatic ring of phenol, primarily at the ortho position relative to the hydroxyl group.

\textcolor{red{Step 3: Analysis

The reaction involves treating phenol with chloroform (\(CHCl_3\)) in the presence of an aqueous base like sodium hydroxide (\(NaOH\)) at approximately \(340\,\mathrm{K}\). The intermediate formed is a substituted benzal chloride, which undergoes rapid alkaline hydrolysis to form sodium salicylate.

This reaction is known as the Reimer--Tiemann reaction, which is used for the ortho-formylation of phenols.

Phenol is treated with chloroform (\(CHCl_3\)) and aqueous sodium hydroxide at about
\[ 340\,\mathrm{K}. \]

Under these strongly basic conditions, chloroform generates the highly reactive electrophile dichlorocarbene:
\[ CHCl_3 + OH^- \rightarrow :CCl_2 + Cl^- + H_2O. \]

The activated phenoxide ion attacks the dichlorocarbene, producing an intermediate substituted benzal chloride derivative.

This intermediate undergoes alkaline hydrolysis, replacing the chlorine atoms with oxygen-containing groups to form sodium salicylate.

Upon acidification, sodium salicylate is converted into salicylaldehyde:
\[ \boxed{o-Hydroxybenzaldehyde (Salicylaldehyde)}. \]

Since the hydroxyl group is an ortho- and para-directing group, the major product is the ortho-substituted aldehyde, while only a small amount of the para isomer is formed.

Thus, the Reimer--Tiemann reaction provides an important method for introducing an aldehyde (\(-CHO\)) group into the ortho position of phenol.


\textcolor{red{Step 4: Conclusion
Subsequent acidification of the reaction mixture yields the final functional product, 2-hydroxybenzaldehyde.



\textcolor{red{Final Answer:
Phenol + \(CHCl_3\) + \(3NaOH \xrightarrow{340K}\) Sodium 2-hydroxybenzaldehyde (intermediate) \(\xrightarrow{H^+}\) Salicylaldehyde (2-Hydroxybenzaldehyde) Quick Tip: Reimer-Tiemann uses Chloroform (\(CHCl_3\)) and base to add an aldehyde (\(-CHO\)) group to phenol.

\textcolor{red{Step 1: Concept
Kolbe's reaction (or Kolbe-Schmitt reaction) is used for the carboxylation of phenols.

\textcolor{red{Step 2: Meaning
It involves the electrophilic aromatic substitution of a phenoxide ion with carbon dioxide gas to yield a phenolic acid.

\textcolor{red{Step 3: Analysis

First, phenol is reacted with sodium hydroxide to form highly reactive sodium phenoxide. This phenoxide is then subjected to carbon dioxide gas at elevated temperature (\(400\,\mathrm{K}\)) and high pressure (4--7 atm). The \(CO_2\) acts as a weak electrophile, attacking the ortho position.

This reaction is known as the Kolbe--Schmitt reaction, which is used for the preparation of salicylic acid from phenol.

In the first step, phenol reacts with sodium hydroxide to form sodium phenoxide:
\[ C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O. \]

Sodium phenoxide is much more reactive than phenol because the negative charge on the oxygen atom increases the electron density of the aromatic ring.

The phenoxide ion is then treated with carbon dioxide under high pressure (4--7 atm) and at about
\[ 400\,\mathrm{K}. \]

Carbon dioxide behaves as a weak electrophile and is attacked mainly at the ortho position of the activated benzene ring.

This produces sodium salicylate as the intermediate product:
\[ C_6H_5ONa + CO_2 \rightarrow o-HOC_6H_4COONa. \]

Finally, acidification with a dilute mineral acid converts sodium salicylate into salicylic acid:
\[ o-HOC_6H_4COONa + HCl \rightarrow o-HOC_6H_4COOH + NaCl. \]

Thus, the Kolbe--Schmitt reaction provides an industrially important method for preparing salicylic acid, which is the starting material for the manufacture of aspirin.


\textcolor{red{Step 4: Conclusion
The resulting sodium salt is treated with dilute acid to precipitate the final carboxylic acid compound.



\textcolor{red{Final Answer:
Phenol \(\xrightarrow{NaOH}\) Sodium phenoxide \(\xrightarrow{CO_2, 400K, 4-7 atm}\) Sodium salicylate \(\xrightarrow{H^+}\) Salicylic acid (2-Hydroxybenzoic acid) Quick Tip: Kolbe's reaction uses Carbon dioxide (\(CO_2\)) and base to add a carboxylic acid (\(-COOH\)) group to phenol.

\textcolor{red{Step 1: Concept
Friedel-Crafts acylation introduces an acyl group (\(-COR\)) onto an aromatic ring via electrophilic substitution.

\textcolor{red{Step 2: Meaning
Anisole (methoxybenzene) has an electron-donating methoxy group (\(-OCH_3\)), which activates the ring and directs incoming electrophiles to the ortho and para positions.

\textcolor{red{Step 3: Analysis

Anisole is reacted with acetyl chloride (\(CH_3COCl\)) in the presence of anhydrous aluminum chloride (\(AlCl_3\)), a Lewis acid catalyst. The catalyst generates an active acylium ion electrophile (\(CH_3CO^+\)). The reaction yields a mixture of ortho and para substituted products.

This reaction is an example of the Friedel--Crafts acylation reaction.

Anhydrous aluminum chloride (\(AlCl_3\)) acts as a Lewis acid catalyst by accepting an electron pair from acetyl chloride.

This interaction generates the highly reactive acylium ion:
\[ CH_3COCl + AlCl_3 \rightarrow CH_3CO^+ + AlCl_4^-. \]

The acylium ion serves as the electrophile in the electrophilic aromatic substitution reaction.

In anisole, the methoxy group (\(-OCH_3\)) donates electron density to the benzene ring through resonance.

Consequently, the ortho and para positions become more electron-rich and are the preferred sites for electrophilic attack.

The acylium ion therefore substitutes mainly at the ortho and para positions, producing a mixture of:

o-Methoxyacetophenone.
\textit{p-Methoxyacetophenone (major product).


The para product is usually formed in greater amount because it experiences less steric hindrance than the ortho product.

Thus, anisole undergoes Friedel--Crafts acylation to give predominantly para-substituted products along with a smaller amount of the ortho isomer.


\textcolor{red{Step 4: Conclusion
The para isomer dominates as the major product because the bulky methoxy group causes significant steric hindrance at the ortho positions.



\textcolor{red{Final Answer:
Anisole (\(C_6H_5OCH_3\)) + \(CH_3COCl \xrightarrow{Anhydrous AlCl_3\) 2-Methoxyacetophenone (Minor) + 4-Methoxyacetophenone (Major) Quick Tip: Acylation uses acetyl chloride (\(CH_3COCl\)) to attach a ketone (\(-COCH_3\)) group to the ring. Para is major due to steric reasons.

\textcolor{red{Step 1: Concept
The kinetics of nucleophilic substitution determine the underlying mechanism (\(S_N1\) vs \(S_N2\)) and structure of alkyl halides.

\textcolor{red{Step 2: Meaning
A rate depending only on the substrate (\(Rate = k[X]\)) indicates a unimolecular \(S_N1\) mechanism, which requires a highly stable carbocation (typically tertiary). A rate depending on both substrate and nucleophile (\(Rate = k[Y][OH^-]\)) indicates a bimolecular \(S_N2\) mechanism, which prefers unhindered primary or secondary halides.

\textcolor{red{Step 3: Analysis

For \(C_4H_9Br\), X must be a tertiary alkyl halide to favor the \(S_N1\) pathway. Thus, X is tert-butyl bromide. Y undergoes \(S_N2\) and is specified as optically active, meaning it must have a chiral center. The only chiral isomer of \(C_4H_9Br\) is 2-bromobutane.

Four structural isomers of \(C_4H_9Br\) are possible:

1-Bromobutane.
2-Bromobutane.
1-Bromo-2-methylpropane.
2-Bromo-2-methylpropane (tert-butyl bromide).


The \(S_N1\) mechanism proceeds through the formation of a carbocation intermediate.

Since tertiary carbocations are the most stable, tertiary alkyl halides react most readily by the \(S_N1\) mechanism.

Therefore,
\[ X=\boxed{tert-butyl bromide (2-bromo-2-methylpropane)}. \]

Compound Y is stated to undergo an \(S_N2\) reaction and is optically active.

Optical activity requires the presence of a chiral carbon atom bonded to four different groups.

Among the four isomers of \(C_4H_9Br\), only 2-bromobutane possesses such a chiral centre.

Therefore,
\[ Y=\boxed{2-bromobutane}. \]

Thus, tert-butyl bromide is identified as the compound favoring the \(S_N1\) mechanism, whereas 2-bromobutane is the optically active compound capable of undergoing an \(S_N2\) reaction.



\textcolor{red{Step 4: Conclusion
The kinetic dependencies perfectly map to the specific structural isomers.



\textcolor{red{Final Answer:
Compound 'X' (tert-butyl bromide): \(CH_3-C(CH_3)(Br)-CH_3\)

Compound 'Y' (2-bromobutane): \(CH_3-CH_2-CH(Br)-CH_3\) Quick Tip: Rate = \(k[RX]\) means \(S_N1\) (tertiary preferred). Rate = \(k[RX][Nu]\) means \(S_N2\) (needs chiral center if optically active).

\textcolor{red{Step 1: Concept
Stereochemistry of the \(S_N1\) substitution mechanism.

\textcolor{red{Step 2: Meaning
Racemisation is the formation of a 50:50 mixture of two enantiomers, resulting in an optically inactive product mixture.

\textcolor{red{Step 3: Analysis

Compound X reacts via the \(S_N1\) pathway. The first, rate-determining step is the departure of the bromide ion, which creates a carbocation intermediate. The central carbon in a carbocation is \(sp^2\) hybridized, making its geometry perfectly trigonal planar.

The \(S_N1\) mechanism proceeds in two distinct steps.

The first and slowest step is the ionization of the alkyl halide:
\[ R-Br \rightarrow R^+ + Br^-. \]

This step is the rate-determining step because it involves breaking the carbon--bromine bond to form a carbocation.

The carbon atom in the carbocation possesses only three sigma bonds and no lone pair of electrons.

Therefore, it is
\[ sp^2 \]
hybridized.

The three \(sp^2\) hybrid orbitals arrange themselves at angles of approximately
\[ 120^\circ, \]
giving the carbocation a trigonal planar geometry.

An empty unhybridized \(p\) orbital remains perpendicular to the plane of the molecule.

Because the carbocation is planar, the nucleophile can attack from either side with nearly equal probability, often leading to racemization if the reaction centre is chiral.



\textcolor{red{Step 4: Conclusion
Because the intermediate is completely flat, the incoming hydroxide nucleophile (\(OH^-\)) has an equal probability of attacking from the front face or the back face, yielding a racemic mixture of the product alcohols.



\textcolor{red{Final Answer: Compound 'X' will undergo racemisation. This is because it proceeds via an \(S_N1\) mechanism, forming a planar carbocation intermediate. The nucleophile (\(OH^-\)) can attack this planar intermediate from either side with equal probability, generating a racemic mixture. Quick Tip: \(S_N1\) = Planar Carbocation = Front & Back Attack = Racemisation.

\textcolor{red{Step 1: Concept
Stereochemistry of the \(S_N2\) substitution mechanism.

\textcolor{red{Step 2: Meaning
Inversion of configuration refers to the stereochemical reversal of spatial arrangement around the chiral carbon atom, much like an umbrella flipping inside out in strong wind.

\textcolor{red{Step 3: Analysis

Compound Y reacts exclusively via the \(S_N2\) pathway. This is a concerted, single-step reaction where bond-breaking and bond-forming occur simultaneously. The bulky leaving group (\(Br^-\)) blocks the front side.

The \(S_N2\) mechanism occurs in a single elementary step.

The nucleophile attacks the carbon atom at the same time as the leaving group departs:
\[ Nu^- + R-Br \rightarrow R-Nu + Br^-. \]

Since bond formation and bond breaking occur simultaneously, no carbocation intermediate is formed.

The leaving group partially blocks the front side of the carbon atom.

Therefore, the nucleophile approaches only from the side opposite to the leaving group, a process known as backside attack.

During the reaction, a transition state is formed in which the carbon atom is partially bonded to both the nucleophile and the leaving group.

Completion of the reaction results in inversion of the spatial arrangement around the chiral carbon atom.

This stereochemical inversion is called the Walden inversion.

Thus, \(S_N2\) reactions are characterized by a concerted mechanism, backside attack, and complete inversion of configuration.


\textcolor{red{Step 4: Conclusion
Therefore, the incoming nucleophile (\(OH^-\)) is forced to attack the chiral carbon exclusively from the backside, exactly 180 degrees opposite to the leaving group, resulting in complete stereochemical inversion (Walden inversion).



\textcolor{red{Final Answer: Compound 'Y' will form a product with inversion of configuration. It proceeds via a concerted \(S_N2\) mechanism where the incoming nucleophile (\(OH^-\)) attacks strictly from the side opposite to the leaving group (\(Br^-\)) to avoid steric repulsion, leading to 100% inversion. Quick Tip: \(S_N2\) = Concerted Single Step = Backside Attack = Inversion of Configuration.

\textcolor{red{Step 1: Concept
The basicity of amines in an aqueous solution depends on a delicate balance of the inductive effect (\(+I\)), steric hindrance, and solvation (hydration) of the resulting conjugate acid.

\textcolor{red{Step 2: Meaning
A stronger base forms a more stable conjugate acid in water.

\textcolor{red{Step 3: Analysis

While trimethylamine, \((CH_3)_3N\), has three electron-donating methyl groups maximizing the \(+I\) effect, its bulky nature creates severe steric hindrance. Furthermore, its conjugate acid, \((CH_3)_3NH^+\), has only one hydrogen atom available for hydrogen bonding with water. In contrast, the conjugate acid of methylamine, \(CH_3NH_3^+\), is less sterically hindered and has three hydrogens, allowing for extensive and stabilizing hydrogen bonding with water.

The basic strength of amines in aqueous solution depends on two important factors:

The availability of the lone pair of electrons on the nitrogen atom.
The stability of the protonated ammonium ion formed after accepting a proton.


Trimethylamine,
\[ (CH_3)_3N, \]
contains three methyl groups that exert a strong positive inductive (\(+I\)) effect.

This electron-donating effect increases the electron density on the nitrogen atom, making the lone pair more available for protonation.

However, the three bulky methyl groups create considerable steric hindrance around the nitrogen atom.

After protonation,
\[ (CH_3)_3N+H^+ \rightarrow (CH_3)_3NH^+, \]
the resulting conjugate acid possesses only one N--H bond and therefore forms comparatively fewer hydrogen bonds with water.

In contrast, methylamine,
\[ CH_3NH_2, \]
produces the methylammonium ion:
\[ CH_3NH_2+H^+ \rightarrow CH_3NH_3^+. \]

This ion is much less sterically hindered and contains three N--H bonds, enabling it to form extensive hydrogen bonding with surrounding water molecules.

Greater solvation stabilizes the conjugate acid, thereby increasing the basic strength of methylamine in aqueous solution.

Hence, despite the stronger \(+I\) effect in trimethylamine, methylamine is more basic than trimethylamine in aqueous medium because of better solvation and lower steric hindrance.


\textcolor{red{Step 4: Conclusion
The dominant stabilization from solvation (hydrogen bonding) outweighs the inductive effect, making the primary amine more basic in water.



\textcolor{red{Final Answer: In aqueous solution, the conjugate acid of \(CH_3-NH_2\) is much more stabilized by hydrogen bonding (solvation) and experiences less steric hindrance compared to the conjugate acid of \((CH_3)_3N\). This solvation effect overcomes the greater \(+I\) effect of the tertiary amine. Quick Tip: In water, Solvation \& Sterics \textgreater\ Inductive effect for methyl amines. Order: \(2^\circ > 1^\circ > 3^\circ\).

\textcolor{red{Step 1: Concept
Reaction 1 is the Hoffmann bromamide degradation. Reaction 2 is a nucleophilic acyl substitution (Benzoylation/Schotten-Baumann reaction).

\textcolor{red{Step 2: Meaning
Hoffmann degradation removes the carbonyl carbon from an amide to form a primary amine with one less carbon. Benzoylation attaches a benzoyl group to the amine nitrogen.

\textcolor{red{Step 3: Analysis

Acetamide (\(CH_3CONH_2\)) reacts with \(NaOBr\) to lose its carbonyl group, forming methanamine (\(CH_3NH_2\)), which is compound A. Methanamine then reacts with benzoyl chloride (\(C_6H_5COCl\)) in the presence of a base to eliminate \(HCl\) and form an amide linkage, resulting in N-methylbenzamide, which is compound B.

Acetamide undergoes the Hofmann bromamide degradation reaction when treated with sodium hypobromite (\(NaOBr\)).

In this reaction, the amide loses one carbon atom as carbon dioxide, producing a primary amine containing one carbon atom fewer than the original amide.

The reaction is represented as:
\[ CH_3CONH_2 \xrightarrow{NaOBr} CH_3NH_2. \]

Therefore, compound A is:
\[ \boxed{Methanamine (Methylamine)}. \]

Methanamine then reacts with benzoyl chloride in the presence of a base such as sodium hydroxide.

This reaction is known as the Schotten--Baumann reaction, in which an amine reacts with an acid chloride to form an amide.

During the reaction, hydrogen chloride is eliminated:
\[ CH_3NH_2+C_6H_5COCl \rightarrow C_6H_5CONHCH_3+HCl. \]

The base present in the reaction neutralizes the hydrogen chloride formed.

Hence, compound B is:
\[ \boxed{N-methylbenzamide}. \]


\textcolor{red{Step 4: Conclusion
The sequential transformations map directly to the intermediate and final structures.



\textcolor{red{Final Answer:

Compound A: \(CH_3-NH_2\) (Methanamine)

Compound B: \(C_6H_5-CO-NH-CH_3\) (N-Methylbenzamide) Quick Tip: \(NaOBr\) chops off the \(C=O\) from the amide. Benzoylation replaces an \(N-H\) hydrogen with a \(C_6H_5-CO-\) group.

\textcolor{red{Step 1: Concept
Hinsberg's test is used to distinguish primary, secondary, and tertiary amines using benzenesulfonyl chloride.

\textcolor{red{Step 2: Meaning
Primary amines form a sulfonamide soluble in alkali. Secondary amines form a sulfonamide insoluble in alkali. Tertiary amines do not react.

\textcolor{red{Step 3: Analysis

Since the product formed by compound X is explicitly stated to be insoluble in alkali, X must definitely be a secondary amine. The given molecular formula is \(C_3H_9N\). The only possible secondary amine structure that can be drawn with three carbon atoms is N-methylethanamine.

Primary, secondary, and tertiary amines can be distinguished using the Hinsberg test.

Primary amines react with benzenesulfonyl chloride to form sulfonamides that are soluble in alkali because they contain an acidic hydrogen atom attached to nitrogen.

Secondary amines also react with benzenesulfonyl chloride, but the resulting sulfonamides contain no acidic hydrogen.

Consequently, these sulfonamides are insoluble in aqueous alkali.

Tertiary amines do not react with benzenesulfonyl chloride under ordinary conditions.

Since the product obtained from compound X is insoluble in alkali, X must be a secondary amine.

The molecular formula is:
\[ C_3H_9N. \]

Among all possible isomers with this molecular formula, the only secondary amine is:
\[ \boxed{CH_3NHCH_2CH_3}, \]
which is called N-methylethanamine (ethylmethylamine).

Therefore, compound X is N-methylethanamine.


\textcolor{red{Step 4: Conclusion
Matching the kinetic observation with the molecular formula isolates the exact isomer.



\textcolor{red{Final Answer: 'X' is N-Methylethanamine. Structure: \(CH_3-NH-CH_2-CH_3\) Quick Tip: Hinsberg product insoluble in alkali = Secondary (\(2^\circ\)) Amine.

\textcolor{red{Step 1: Concept
Diazotization followed by the Sandmeyer reaction.

\textcolor{red{Step 2: Meaning
Direct substitution of the amino group on a benzene ring is difficult, so it is first converted into a highly reactive diazonium leaving group.

\textcolor{red{Step 3: Analysis

First, aniline is treated with a mixture of sodium nitrite and hydrochloric acid (\(NaNO_2+HCl\)) at cold temperatures (\(0{-}5^\circ C\)) to form benzenediazonium chloride. Second, this diazonium salt is treated with cuprous cyanide (\(CuCN\)) and potassium cyanide (\(KCN\)). The diazonium group is expelled as \(N_2\) gas and replaced by the cyanide group.

The first step is the diazotization of aniline.

Sodium nitrite reacts with hydrochloric acid to generate nitrous acid \((HNO_2)\) in situ.

At low temperatures
\[ 0-5^\circ\mathrm{C}, \]
aniline reacts with nitrous acid to form benzenediazonium chloride:
\[ C_6H_5NH_2 \xrightarrow{NaNO_2/HCl} C_6H_5N_2^+Cl^-. \]

The reaction is carried out at low temperature because diazonium salts are unstable at higher temperatures.

In the second step, the diazonium salt is treated with cuprous cyanide (\(CuCN\)), often in the presence of potassium cyanide.

This reaction is known as the Sandmeyer reaction.

The diazonium group is replaced by the cyanide group with the evolution of nitrogen gas:
\[ C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN+N_2. \]

The product formed is:
\[ \boxed{Benzonitrile}. \]

Nitrogen gas is an exceptionally stable molecule, and its evolution provides the driving force for the reaction.


\textcolor{red{Step 4: Conclusion
This two-step pathway cleanly yields benzonitrile.



\textcolor{red{Final Answer:

Step 1: Aniline \(\xrightarrow{NaNO_2 + HCl , 0-5^\circ C}\) Benzenediazonium chloride

Step 2: Benzenediazonium chloride \(\xrightarrow{CuCN/KCN , \Delta}\) Benzonitrile (\(C_6H_5-CN\)) Quick Tip: Always convert aniline to a diazonium salt first when you need to substitute the ring with a new functional group like \(-CN\) or halogens.

\textcolor{red{Step 1: Concept
Protection of highly activating groups during electrophilic aromatic substitution.

\textcolor{red{Step 2: Meaning
The \(-NH_2\) group is a powerful activator and makes the benzene ring highly susceptible to oxidation.

\textcolor{red{Step 3: Analysis

Direct nitration involves a strongly acidic nitrating mixture consisting of concentrated nitric acid and concentrated sulfuric acid. If aniline is nitrated directly, the amino group becomes protonated to form the anilinium ion (\(-NH_3^+\)), which is a meta-directing and deactivating group. In addition, oxidation and side reactions produce tarry by-products. Acetylation converts the amino group into an amide (\(-NHCOCH_3\)), which is less activating because the lone pair on nitrogen is partially delocalized toward the carbonyl group.

Direct nitration of aniline is not suitable because the reaction is carried out in a strongly acidic medium containing concentrated \(HNO_3\) and concentrated \(H_2SO_4\).

Under these conditions, the amino group is protonated:
\[ -NH_2+H^+ \rightarrow -NH_3^+. \]

The protonated amino group is strongly electron-withdrawing and behaves as a meta-directing, deactivating substituent.

Consequently, direct nitration gives poor yields and may produce unwanted meta-substituted products.

The strongly oxidizing nitrating mixture may also oxidize aniline, producing dark, resinous (tarry) by-products.

To overcome these problems, the amino group is first protected by acetylation:
\[ C_6H_5NH_2 \rightarrow C_6H_5NHCOCH_3. \]

The acetyl group reduces the activating effect of the amino group because the nitrogen lone pair is partially delocalized toward the carbonyl group by resonance.

The protected amide group remains ortho- and para-directing but is much less reactive than the free amino group, allowing controlled nitration.

After nitration, the acetyl protecting group is removed by hydrolysis to regenerate the amino group and obtain the desired nitroaniline.


\textcolor{red{Step 4: Conclusion
This protection prevents both oxidation and protonation, ensuring the reaction selectively yields ortho and para products.



\textcolor{red{Final Answer: Acetylation protects the aniline from destructive oxidation by the strongly oxidizing nitrating mixture. It also lowers the activating power of the amino group and prevents its protonation into a meta-directing anilinium ion, thereby ensuring the selective formation of ortho and para-nitro derivatives. Quick Tip: Acetylation acts like a "shield", lowering reactivity to prevent ring oxidation and stopping acid from turning the ortho/para director into a meta director.

\textcolor{red{Step 1: Concept
Crystal Field Splitting in octahedral coordination complexes.

\textcolor{red{Step 2: Meaning
Degenerate d-orbitals split into two sets (\(t_{2g}\) and \(e_g\)) based on the electrostatic repulsion from approaching ligands.

\textcolor{red{Step 3: Analysis

In an octahedral geometry, the six ligands approach the central metal ion directly along the Cartesian axes (\(x\), \(y\), and \(z\)). The \(d_{x^2-y^2}\) and \(d_{z^2}\) orbitals (the \(e_g\) set) have their electron lobes pointing directly along these exact axes. Consequently, they experience maximum electrostatic repulsion from the incoming ligand electrons.

In an octahedral complex, six ligands surround the central metal ion and approach it along the positive and negative directions of the three Cartesian axes (\(x\), \(y\), and \(z\)).

The five \(d\) orbitals of the metal ion do not have identical orientations in space.

The orbitals
\[ d_{x^2-y^2}\quadand\quad d_{z^2} \]
have their lobes directed exactly along the coordinate axes.

Since the ligands also approach along these axes, the electrons present in these orbitals experience maximum electrostatic repulsion from the ligand electron pairs.

As a result, these two orbitals increase in energy and form the higher-energy
\[ e_g \]
set.

The remaining three orbitals,
\[ d_{xy},\; d_{yz},\; d_{xz}, \]
have their lobes directed between the coordinate axes.

These orbitals experience comparatively less repulsion because they are not oriented directly toward the approaching ligands.

Therefore, they become lower in energy and constitute the
\[ t_{2g} \]
set.

The energy difference between the \(e_g\) and \(t_{2g}\) sets is called the octahedral crystal field splitting energy,
\[ \Delta_o. \]



\textcolor{red{Step 4: Conclusion
Due to this direct head-on repulsion, their energy levels are elevated above the average energy level.



\textcolor{red{Final Answer: The energies of the \(d_{x^2-y^2}\) and \(d_{z^2}\) orbitals (\(e_g\) set) will be raised. This is because their lobes point directly along the axes where the ligands are approaching, causing maximum electrostatic repulsion. Quick Tip: Axial orbitals (\(d_{x^2-y^2}\), \(d_{z^2}\)) face the ligands directly in an octahedral field, so their energy goes up.

\textcolor{red{Step 1: Concept
Electronic configuration under Crystal Field Theory (CFT).

\textcolor{red{Step 2: Meaning
Electron filling in \(t_{2g}\) and \(e_g\) levels depends on the ligand field strength relative to the pairing energy (\(P\)).

\textcolor{red{Step 3: Analysis

The oxidation state of cobalt is \(+3\), which means it has a \(3d^6\) configuration. Fluoride (\(F^-\)) is a weak field ligand, meaning the crystal field splitting energy is less than the pairing energy (\(\Delta_o
In the complex, cobalt is present in the
\[ +3 \]
oxidation state.

The electronic configuration of
\[ Co^{3+} \]
is:
\[ 3d^6. \]

Fluoride ion (\(F^-\)) is a weak field ligand according to the spectrochemical series.

Weak field ligands produce only a small crystal field splitting:
\[ \Delta_o where \(P\) is the electron pairing energy.

Since pairing requires more energy than occupying higher-energy orbitals, electrons prefer to remain unpaired according to Hund's rule.

Therefore, the six electrons are distributed as:
\[ t_{2g}^{4}e_g^{2}. \]

This arrangement contains
\[ \boxed{4\ unpaired electrons}. \]

Such a complex is called a
\[ \boxed{high-spin octahedral complex} \]
and exhibits strong paramagnetic behaviour because of its large number of unpaired electrons.



\textcolor{red{Step 4: Conclusion
According to Hund's rule, the electrons will occupy all available orbitals singly before any pairing occurs. Thus, 3 electrons go to \(t_{2g}\), 2 go to \(e_g\), and the 6th electron pairs in \(t_{2g}\).



\textcolor{red{Final Answer: \(t_{2g}^4 e_g^2\) Quick Tip: Weak field ligand = High spin complex (\(\Delta_o < P\)). Electrons fill both levels before pairing.

\textcolor{red{Step 1: Concept
Electronic configuration under Crystal Field Theory (CFT).

\textcolor{red{Step 2: Meaning
Strong field ligands force electron pairing in the lower energy level before occupying the higher energy level.

\textcolor{red{Step 3: Analysis

Cobalt is again in the \(+3\) oxidation state (\(3d^6\)). However, ammonia (\(NH_3\)) acts as a strong field ligand for \(Co^{3+}\), meaning the splitting energy is greater than the pairing energy (\(\Delta_o>P\)).

In this complex, cobalt again has the oxidation state:
\[ +3. \]

Therefore, its electronic configuration remains:
\[ 3d^6. \]

Ammonia (\(NH_3\)) behaves as a strong field ligand for
\[ Co^{3+}. \]

Strong field ligands produce a large crystal field splitting such that:
\[ \Delta_o>P. \]

Since the splitting energy exceeds the pairing energy, electrons pair within the lower-energy
\[ t_{2g} \]
orbitals instead of occupying the higher-energy
\[ e_g \]
orbitals.

The electronic configuration therefore becomes:
\[ t_{2g}^{6}e_g^{0}. \]

All six electrons are paired, leaving
\[ \boxed{0\ unpaired electrons}. \]

Such complexes are called
\[ \boxed{low-spin octahedral complexes} \]
and are diamagnetic because no unpaired electrons are present.


\textcolor{red{Step 4: Conclusion
It is more energetically favorable for the electrons to pair up in the lower \(t_{2g}\) level rather than jump the large gap to the \(e_g\) level. All 6 electrons fill the \(t_{2g}\) orbitals.



\textcolor{red{Final Answer: \(t_{2g}^6 e_g^0\) Quick Tip: Strong field ligand = Low spin complex (\(\Delta_o > P\)). Electrons pair up in the \(t_{2g}\) level first.

\textcolor{red{Step 1: Concept
Magnetic behavior is dictated by the presence or absence of unpaired electrons in the d-orbitals.

\textcolor{red{Step 2: Meaning
Paramagnetic complexes have unpaired electrons; diamagnetic complexes have all electrons paired.

\textcolor{red{Step 3: Analysis

In \([NiCl_4]^{2-}\), the oxidation state of Ni is \(+2\), giving a \(3d^8\) configuration. Chloride (\(Cl^-\)) is a weak field ligand and cannot force pairing against Hund's rule. The \(sp^3\) hybridization leaves two unpaired electrons in the \(3d\) subshell. In \([Ni(CO)_4]\), the oxidation state of Ni is \(0\), giving a \(3d^8\,4s^2\) configuration. Carbon monoxide (\(CO\)) is a very strong field ligand, forcing the two \(4s\) electrons to pair up in the \(3d\) subshell, shifting the configuration to a fully filled \(3d^{10}\).

Consider the complex
\[ [NiCl_4]^{2-}. \]

The oxidation state of nickel is:
\[ +2, \]
giving the electronic configuration:
\[ Ni^{2+}:3d^8. \]

Chloride ion is a weak field ligand and cannot pair the \(3d\) electrons.

Therefore, the complex utilizes the outer orbitals to undergo:
\[ sp^3 \]
hybridization.

The complex has tetrahedral geometry and contains
\[ \boxed{2\ unpaired electrons}, \]
making it paramagnetic.

Now consider
\[ [Ni(CO)_4]. \]

Here nickel is in the oxidation state:
\[ 0, \]
with the electronic configuration:
\[ 3d^8\,4s^2. \]

Carbon monoxide (\(CO\)) is a very strong field ligand.

It causes the two \(4s\) electrons to pair in the \(3d\) orbitals, producing:
\[ 3d^{10}. \]

The metal then undergoes:
\[ sp^3 \]
hybridization using one \(4s\) and three \(4p\) orbitals.

Since all ten \(3d\) electrons are paired, the complex contains no unpaired electrons and is therefore
\[ \boxed{diamagnetic}. \]


\textcolor{red{Step 4: Conclusion
The filled \(3d^{10}\) subshell has zero unpaired electrons, resulting in diamagnetism.



\textcolor{red{Final Answer: In \([NiCl_4]^{2-}\), \(Ni^{2+}\) has a \(3d^8\) configuration with a weak field \(Cl^-\) ligand, leaving 2 unpaired electrons (paramagnetic). In \([Ni(CO)_4]\), Ni is in the \(0\) oxidation state (\(3d^8 4s^2\)). The strong \(CO\) ligand forces 4s electrons into the 3d orbitals to form \(3d^{10}\), leaving zero unpaired electrons (diamagnetic). Quick Tip: \(Ni(0)\) with a strong ligand forces 4s electrons to pair into 3d, creating a diamagnetic \(d^{10}\) state.

\textcolor{red{Step 1: Concept
Valence Bond Theory (VBT) applied to coordination complexes.

\textcolor{red{Step 2: Meaning
The nature of the ligand determines whether inner (3d) or outer (4d) orbitals are used for hybridization.

\textcolor{red{Step 3: Analysis

Iron is in the \(+3\) oxidation state, yielding a \(3d^5\) configuration. Cyanide (\(CN^-\)) is a strong field ligand. It forces the pairing of the five \(3d\) electrons, resulting in two fully paired orbitals and one singly occupied orbital. This pairing vacates two inner \(3d\) orbitals. These two \(3d\) orbitals hybridize with one \(4s\) and three \(4p\) orbitals to form an octahedral geometry.

In the complex, iron has the oxidation state:
\[ +3. \]

Therefore, its electronic configuration is:
\[ Fe^{3+}:3d^5. \]

Cyanide ion (\(CN^-\)) is a strong field ligand and produces a large crystal field splitting.

Since
\[ \Delta_o>P, \]
the five electrons pair within the lower-energy orbitals.

The electron distribution becomes:
\[ t_{2g}^{5}e_g^{0}, \]
leaving only
\[ \boxed{1\ unpaired electron}. \]

Pairing of the electrons creates two vacant inner \(3d\) orbitals.

These two vacant \(3d\) orbitals combine with one \(4s\) and three \(4p\) orbitals to produce:
\[ d^2sp^3 \]
hybridization.

The resulting complex possesses an octahedral geometry.

Since inner \(3d\) orbitals participate in hybridization, the complex is called an
\[ \boxed{inner orbital (low-spin) octahedral complex}. \]

\textcolor{red{Step 4: Conclusion
The single remaining unpaired electron makes the complex paramagnetic.



\textcolor{red{Final Answer:

Hybridization: \(d^2sp^3\) (Inner orbital complex)

Magnetic behaviour: Paramagnetic (due to 1 unpaired electron) Quick Tip: Strong field \(CN^-\) forces pairing, vacating inner d-orbitals for \(d^2sp^3\) hybridization.

\textcolor{red{Step 1: Concept
Distinct chemical tests are used to identify the specific nature of carbonyl compounds (aldehydes vs. ketones, and methyl ketones).

\textcolor{red{Step 2: Meaning
A negative Tollen's test indicates the compound is a ketone. A positive iodoform test confirms the presence of a methyl ketone group (\(-COCH_3\)). An Aldol condensation requires \(\alpha\)-hydrogens. A Cannizzaro reaction requires the complete absence of \(\alpha\)-hydrogens.

\textcolor{red{Step 3: Analysis

(I) A five-carbon ketone that is a methyl ketone is pentan-2-one.

(II) A five-carbon ketone that is not a methyl ketone but possesses \(\alpha\)-hydrogen atoms is pentan-3-one.

(III) A five-carbon aldehyde possessing no \(\alpha\)-hydrogens must have a highly branched adjacent carbon. This compound is 2,2-dimethylpropanal.

A methyl ketone is a ketone containing the functional group:
\[ -COCH_3. \]

Among the ketones having the molecular formula
\[ C_5H_{10}O, \]
pentan-2-one contains the
\[ -COCH_3 \]
group and therefore qualifies as a methyl ketone:
\[ CH_3COCH_2CH_2CH_3. \]

Pentan-3-one has the structure:
\[ CH_3CH_2COCH_2CH_3. \]

Since the carbonyl carbon is bonded to two ethyl groups instead of a methyl group, it is not a methyl ketone.

However, both carbon atoms adjacent to the carbonyl group contain hydrogen atoms, so pentan-3-one possesses \(\alpha\)-hydrogens.

A five-carbon aldehyde without any \(\alpha\)-hydrogen must have no hydrogen attached to the carbon adjacent to the aldehyde group.

This is possible only when the \(\alpha\)-carbon is fully substituted by carbon atoms.

The required compound is:
\[ (CH_3)_3CCHO, \]
whose IUPAC name is
\[ \boxed{2,2-dimethylpropanal}. \]

Since the \(\alpha\)-carbon carries no hydrogen atom, this aldehyde cannot undergo reactions that require \(\alpha\)-hydrogen atoms, such as aldol condensation.


\textcolor{red{Step 4: Conclusion
The structural formulas correspond exactly to the kinetic and qualitative test constraints provided.



\textcolor{red{Final Answer:

(I) Pentan-2-one: \(CH_3-CO-CH_2-CH_2-CH_3\)

(II) Pentan-3-one: \(CH_3-CH_2-CO-CH_2-CH_3\)

(III) 2,2-Dimethylpropanal: \(CH_3-C(CH_3)_2-CHO\) Quick Tip: Positive Iodoform = Methyl Ketone. Cannizzaro = Zero \(\alpha\)-hydrogens.

\textcolor{red{Step 1: Concept
Vigorous oxidation of alkyl or acyl side chains attached to an aromatic ring.

\textcolor{red{Step 2: Meaning
Any side chain containing at least one benzylic hydrogen is completely oxidized to a carboxyl group (\(-COOH\)) upon treatment with strong oxidizing agents.

\textcolor{red{Step 3: Analysis

Both acetophenone (which contains a \(-COCH_3\) group) and ethylbenzene (which contains a \(-CH_2CH_3\) group) possess benzylic hydrogen atoms. When refluxed with alkaline potassium permanganate (\(KMnO_4/KOH\)), the entire side chain is oxidized to form potassium benzoate.

Alkaline potassium permanganate is a powerful oxidizing agent.

Any alkyl side chain attached to a benzene ring can be completely oxidized provided it contains at least one benzylic hydrogen atom.

Acetophenone contains the side chain:
\[ -COCH_3, \]
in which the methyl group provides benzylic hydrogen atoms.

Ethylbenzene contains the side chain:
\[ -CH_2CH_3, \]
where the benzylic carbon also possesses hydrogen atoms.

During oxidation, the entire side chain is removed irrespective of its length.

The carbon directly attached to the benzene ring is ultimately converted into a carboxylate group.

In alkaline medium, the product formed is:
\[ \boxed{C_6H_5COOK} \]
(potassium benzoate).

On subsequent acidification, potassium benzoate is converted into benzoic acid:
\[ C_6H_5COOK+HCl \rightarrow C_6H_5COOH+KCl. \]

Thus, both acetophenone and ethylbenzene yield the same oxidation product because both possess benzylic hydrogen atoms.


\textcolor{red{Step 4: Conclusion
Subsequent acidification of this intermediate yields benzoic acid. The identical reagent works for both starting materials.



\textcolor{red{Final Answer:

(I) Acetophenone \(\xrightarrow{KMnO_4, KOH, \Delta} Potassium benzoate \xrightarrow{H_3O^+} Benzoic acid\)

(II) Ethylbenzene \(\xrightarrow{KMnO_4, KOH, \Delta} Potassium benzoate \xrightarrow{H_3O^+} Benzoic acid\) Quick Tip: Alkaline \(KMnO_4\) is the universal "eraser" for aromatic side chains, reducing them all down to a benzoic acid group as long as one benzylic H exists.

\textcolor{red{Step 1: Concept
Nucleophilic addition-elimination reaction between a carbonyl compound and an ammonia derivative.

\textcolor{red{Step 2: Meaning
Benzaldehyde reacts with 2,4-dinitrophenylhydrazine (2,4-DNP) to eliminate a molecule of water and form a colored hydrazone derivative.

\textcolor{red{Step 3: Analysis

The reaction involves the carbonyl oxygen (\(=O\)) of benzaldehyde (\(C_6H_5CHO\)) and the two hydrogen atoms of hydrazine (\(NH_2NH_2\)). Removal of one molecule of water links the carbonyl carbon directly to nitrogen through a double bond, producing a hydrazone.

Benzaldehyde reacts with hydrazine in a condensation reaction.

The nucleophilic nitrogen atom of hydrazine attacks the electrophilic carbonyl carbon of benzaldehyde.

An unstable addition intermediate is first formed.

This intermediate subsequently eliminates one molecule of water:
\[ C_6H_5CHO+NH_2NH_2 \rightarrow C_6H_5CH=NNH_2+H_2O. \]

The product formed contains the characteristic functional group:
\[ -CH=N-NH_2, \]
which is known as a hydrazone.

The carbonyl oxygen atom and two hydrogen atoms from hydrazine together constitute the water molecule removed during the condensation.

Hydrazone formation is widely used for the identification and characterization of aldehydes and ketones.


\textcolor{red{Step 4: Conclusion
This yields the specific imine structure known as a hydrazone.



\textcolor{red{Final Answer: \(C_6H_5-CH=N-NH-C_6H_3(NO_2)_2\) Quick Tip: To draw hydrazones, simply remove the \(=O\) from the carbonyl and \(H_2\) from the hydrazine, then connect the remaining fragments with a double bond (\(C=N\)).

\textcolor{red{Step 1: Concept
Reactivity of carbonyl compounds towards nucleophilic addition reactions.

\textcolor{red{Step 2: Meaning
Reactivity is primarily governed by steric hindrance (crowding around the carbonyl carbon) and the inductive effect (electron donation by alkyl groups).

\textcolor{red{Step 3: Analysis

Aldehydes are generally more reactive than ketones because they possess only one alkyl group, resulting in less steric hindrance and a smaller positive inductive (\(+I\)) effect. Consequently, the carbonyl carbon is more electrophilic. Acetaldehyde is therefore more reactive than ketones. Among the ketones, di-tert-butyl ketone contains extremely bulky substituents that severely hinder the approach of the nucleophile compared with acetone.

Nucleophilic addition reactions occur at the electrophilic carbonyl carbon atom.

The reactivity of the carbonyl group depends mainly on:

The electrophilic character of the carbonyl carbon.
Steric hindrance around the carbonyl group.


Aldehydes possess only one alkyl group, whereas ketones possess two alkyl groups.

Therefore, aldehydes experience less steric hindrance and a weaker electron-donating (\(+I\)) effect than ketones.

As a result, the carbonyl carbon in aldehydes carries a larger partial positive charge and is attacked more readily by nucleophiles.

Among the aldehydes, acetaldehyde is highly reactive because it contains only one small methyl group.

Acetone contains two methyl groups, which reduce its reactivity slightly through both steric effects and the \(+I\) effect.

Di-\textit{tert-butyl ketone possesses two bulky \textit{tert-butyl groups surrounding the carbonyl carbon.

These bulky groups strongly hinder the approach of nucleophiles such as
\[ CN^-, \]
making nucleophilic addition extremely difficult.

Hence, the order of reactivity is:
\[ \boxed{Acetaldehyde>Acetone>Di-\textit{tert-butyl ketone.} \]


\textcolor{red{Step 4: Conclusion
Therefore, di-tert-butyl ketone is the least reactive.



\textcolor{red{Final Answer: Di-tert. butyl ketone \(<\) Acetone \(<\) Acetaldehyde Quick Tip: Less steric crowding and less \(+I\) effect = Higher reactivity towards nucleophiles. Aldehydes \textgreater\ Ketones.

\textcolor{red{Step 1: Concept
Identification of the carboxylic acid functional group using mild bases.

\textcolor{red{Step 2: Meaning
Carboxylic acids are sufficiently acidic to decompose sodium bicarbonate, whereas esters (like ethyl benzoate) are neutral and do not react.

\textcolor{red{Step 3: Analysis

When an aqueous solution of sodium bicarbonate (\(NaHCO_3\)) is added to benzoic acid, an acid--base reaction occurs, rapidly liberating carbon dioxide gas in the form of brisk effervescence. Ethyl benzoate, lacking an acidic proton, shows no such reaction.

Benzoic acid contains the acidic carboxyl group:
\[ -COOH. \]

Sodium bicarbonate is a weak base that reacts readily with carboxylic acids.

The reaction is:
\[ C_6H_5COOH+NaHCO_3 \rightarrow C_6H_5COONa+CO_2+H_2O. \]

Carbon dioxide gas is evolved rapidly, producing brisk effervescence.

This test is commonly used to distinguish carboxylic acids from many other organic compounds.

Ethyl benzoate is an ester and does not contain a replaceable acidic hydrogen atom.

Therefore, it does not react with sodium bicarbonate and no effervescence is observed.

Hence, the bicarbonate test provides a simple method to distinguish benzoic acid from ethyl benzoate.


\textcolor{red{Step 4: Conclusion
The visual confirmation of gas bubbles makes this an effective distinguishing test.



\textcolor{red{Final Answer: Sodium bicarbonate (\(NaHCO_3\)) test. Add aqueous \(NaHCO_3\) to both compounds. Benzoic acid will produce a brisk effervescence of \(CO_2\) gas. Ethyl benzoate will not react. Quick Tip: The \(NaHCO_3\) test is the standard qualitative chemical test for the \(-COOH\) group.

\textcolor{red{Step 1: Concept
Controlled partial reduction of a nitrile group (\(-CN\)) to an aldehyde group (\(-CHO\)).

\textcolor{red{Step 2: Meaning
A strong reducing agent would reduce the nitrile all the way to an amine. A specialized, milder reagent is required to stop the reduction at the intermediate imine stage.

\textcolor{red{Step 3: Analysis

Two standard chemical methods convert nitriles into aldehydes. The Stephen reaction utilizes stannous chloride and hydrochloric acid (\(SnCl_2/HCl\)) to form an imine hydrochloride, which is subsequently hydrolysed with water. Alternatively, DIBAL-H (diisobutylaluminium hydride) selectively reduces the nitrile to an imine, followed by aqueous hydrolysis.

Nitriles can be selectively converted into aldehydes without complete reduction to primary amines.

One important method is the Stephen reduction.

In this reaction, the nitrile is treated with stannous chloride and concentrated hydrochloric acid:
\[ RCN \xrightarrow{SnCl_2/HCl} RCH=NH\cdot HCl. \]

The intermediate imine hydrochloride is then hydrolysed with water:
\[ RCH=NH\cdot HCl+H_2O \rightarrow RCHO+NH_4Cl. \]

Another useful method employs
\[ \boxed{DIBAL-H} \]
(diisobutylaluminium hydride).

DIBAL-H selectively reduces the nitrile to an imine intermediate under controlled low-temperature conditions.

Subsequent aqueous hydrolysis converts the imine into the corresponding aldehyde:
\[ RCN \xrightarrow{DIBAL-H} RCH=NH \xrightarrow{H_2O} RCHO. \]

Both methods stop the reduction at the aldehyde stage and prevent complete reduction to the corresponding primary amine.


\textcolor{red{Step 4: Conclusion
Either reagent accurately satisfies the conversion requirement.



\textcolor{red{Final Answer: DIBAL-H (Diisobutylaluminium hydride) followed by \(H_2O\). (Alternatively, \(SnCl_2\) and \(HCl\) followed by \(H_3O^+\), known as Stephen reduction). Quick Tip: DIBAL-H is an excellent reagent for selectively stopping the reduction of nitriles and esters exactly at the aldehyde stage.

\textcolor{red{Step 1: Concept
Oxidation of secondary alcohols.

\textcolor{red{Step 2: Meaning
Chromic anhydride (\(CrO_3\)) in an acidic medium (Jones reagent) acts as a strong oxidizing agent.

\textcolor{red{Step 3: Analysis

The provided compound is cyclohexanol, a saturated six-membered cyclic alcohol containing a secondary hydroxyl (\(-OH\)) group. When a secondary alcohol is oxidized with chromium trioxide (\(CrO_3\)), it loses two hydrogen atoms (one from the hydroxyl group and one from the \(\alpha\)-carbon) to form the corresponding ketone.

Cyclohexanol is a cyclic alcohol having the molecular formula:
\[ C_6H_{11}OH. \]

The carbon atom bearing the hydroxyl group is attached to two other carbon atoms.

Therefore, cyclohexanol is classified as a
\[ \boxed{secondary alcohol}. \]

Chromium trioxide (\(CrO_3\)) is a powerful oxidizing agent commonly used for the oxidation of alcohols.

During oxidation, one hydrogen atom is removed from the hydroxyl group and another hydrogen atom is removed from the carbon bearing the hydroxyl group.

The removal of these two hydrogen atoms forms a carbonyl group:
\[ -CHOH- \longrightarrow -CO-. \]

Thus, cyclohexanol is oxidized to cyclohexanone:
\[ \boxed{Cyclohexanol \xrightarrow{CrO_3} Cyclohexanone.} \]

Unlike primary alcohols, secondary alcohols are normally oxidized only up to ketones because further oxidation would require breaking a carbon--carbon bond.


\textcolor{red{Step 4: Conclusion
The saturated ring remains fully intact, and the secondary \(-OH\) transforms into a carbonyl \(=O\), yielding cyclohexanone.



\textcolor{red{Final Answer: 'X' is Cyclohexanone. (A six-membered saturated carbon ring with a double-bonded oxygen attached to one of the carbons). Quick Tip: \(CrO_3\) efficiently oxidizes \(1^\circ\) alcohols to carboxylic acids, and \(2^\circ\) alcohols exclusively to ketones.

\textcolor{red{Step 1: Concept
Standard electrode potentials (\(E^\circ\)) of 3d transition metals.

\textcolor{red{Step 2: Meaning
The overall \(E^\circ\) value is determined by the net sum of enthalpy of sublimation, ionization enthalpy, and hydration enthalpy.

\textcolor{red{Step 3: Analysis

(I) The irregular trend across the first transition series arises from the irregular variation in the sum of the first and second ionization enthalpies together with the sublimation enthalpies.

(II) For copper, the high energy required for atomization and ionization is not sufficiently compensated by its hydration enthalpy, resulting in a positive standard reduction potential.

(III) Manganese attains the exceptionally stable half-filled \(3d^5\) configuration in the \(+2\) oxidation state, making the formation of \(Mn^{2+}\) highly favourable.

The standard electrode potentials of transition elements do not show a regular increasing or decreasing trend across the series.

This irregularity arises because several energy factors contribute simultaneously, including:

Sublimation enthalpy.
First and second ionization enthalpies.
Hydration enthalpy of the metal ion.


The combined variation of these quantities produces the observed irregular pattern of standard electrode potentials.

In the case of copper, considerable energy is required to convert the metal atom into
\[ Cu^{2+}, \]
owing to its relatively high sublimation and ionization enthalpies.

Although hydration releases energy, it is insufficient to compensate for these large energy requirements.

Consequently, copper possesses a positive standard reduction potential:
\[ E^\circ(Cu^{2+}/Cu)=+0.34\;V. \]

Manganese, on the other hand, forms the particularly stable
\[ Mn^{2+}(3d^5) \]
ion.

The half-filled \(3d^5\) configuration possesses extra stability due to exchange energy and symmetrical electron distribution.

Therefore, formation of
\[ Mn^{2+} \]
is highly favourable, giving manganese a comparatively large tendency to undergo oxidation.


\textcolor{red{Step 4: Conclusion
The delicate balance of these specific thermodynamic energy terms dictates the observed standard potentials.



\textcolor{red{Final Answer:

(I) Due to irregular variations in ionization enthalpies and sublimation enthalpies across the series.

(II) Its high enthalpies of atomization and ionization are not fully compensated by its hydration enthalpy.

(III) Due to the extra thermodynamic stability of the exactly half-filled \(d^5\) configuration of the \(Mn^{2+}\) ion. Quick Tip: Positive \(E^\circ\) for Cu means it cannot displace \(H_2\) gas from acids. Half-filled \(d^5\) always grants extra stability.

\textcolor{red{Step 1: Concept
Redox behavior of the permanganate ion (\(MnO_4^-\)) is strictly dependent on the pH of the medium.

\textcolor{red{Step 2: Meaning
It oxidizes iodide (\(I^-\)) to different products based entirely on whether the environment is acidic or alkaline.

\textcolor{red{Step 3: Analysis

In an acidic medium, iodide ions (\(I^-\)) are oxidized to iodine (\(I_2\)), while permanganate ions (\(MnO_4^-\)) are reduced to \(Mn^{2+}\). In a neutral or faintly alkaline medium, iodide ions are oxidized further to iodate (\(IO_3^-\)), while permanganate ions are reduced to manganese dioxide (\(MnO_2\)).

Potassium permanganate is a powerful oxidizing agent whose reduction products depend upon the reaction medium.

In acidic solution, permanganate ions are reduced according to:
\[ MnO_4^-+8H^++5e^- \rightarrow Mn^{2+}+4H_2O. \]

Simultaneously, iodide ions are oxidized:
\[ 2I^- \rightarrow I_2+2e^-. \]

Hence, iodine is liberated in acidic medium.

In neutral or faintly alkaline solution, the reduction half-reaction becomes:
\[ MnO_4^-+2H_2O+3e^- \rightarrow MnO_2+4OH^-. \]

Under these conditions, iodide ions undergo stronger oxidation and are converted into iodate ions:
\[ I^- \rightarrow IO_3^-. \]

Thus, the oxidation products of iodide and the reduction products of permanganate depend strongly upon the pH of the reaction medium.


\textcolor{red{Step 4: Conclusion
Balancing the specific electron transfers in both media yields the respective balanced ionic equations.



\textcolor{red{Final Answer:

Acidic solution: \(2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O\)

Alkaline solution: \(2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-\) Quick Tip: Acidic medium yields \(I_2\) (iodine). Alkaline/neutral medium yields \(IO_3^-\) (iodate).

\textcolor{red{Step 1: Concept
Distinct physical and chemical properties of f-block elements based on electronic configuration.

\textcolor{red{Step 2: Meaning
Stability of empty (\(f^0\)), half-filled (\(f^7\)), or fully filled (\(f^{14}\)) subshells dictates the unusual oxidation states.

\textcolor{red{Step 3: Analysis

Cerium (Ce) loses four electrons to attain the highly stable noble-gas configuration with an empty \(4f\) shell (\(4f^0\)), while europium (Eu) loses two electrons to achieve the particularly stable half-filled \(4f^7\) configuration.

Lanthanoids generally exhibit the oxidation state:
\[ +3. \]

However, certain lanthanoids display exceptional oxidation states because of the extra stability associated with specific \(4f\) electron configurations.

Cerium readily loses four electrons to form:
\[ Ce^{4+}. \]

This produces the stable electronic configuration:
\[ 4f^0, \]
corresponding to an empty \(4f\) subshell.

Europium preferentially forms:
\[ Eu^{2+}, \]
by losing only two electrons.

The resulting electronic configuration is:
\[ 4f^7, \]
which represents a completely half-filled \(4f\) subshell.

Half-filled and completely empty subshells possess extra stability due to symmetrical electron distribution and exchange energy.

Therefore, cerium commonly exhibits the
\[ +4 \]
oxidation state, whereas europium commonly exhibits the
\[ +2 \]
oxidation state.



\textcolor{red{Step 4: Conclusion
These specific electronic configurations provide direct reasoning for these observed oxidation states.



\textcolor{red{Final Answer:

(I) Cerium (Ce)

(II) Europium (Eu) Quick Tip: Cerium likes +4 to empty its f-shell (\(f^0\)). Europium likes +2 to half-fill its f-shell (\(f^7\)).

\textcolor{red{Step 1: Concept
Catalytic properties of d-block elements.

\textcolor{red{Step 2: Meaning
Catalysts lower the activation energy of a reaction by providing an alternative pathway.

\textcolor{red{Step 3: Analysis

Transition metals act as excellent catalysts because they possess variable oxidation states and provide large surface areas for adsorption of reactant molecules.

Transition metals exhibit several oxidation states because both the \((n-1)d\) and \(ns\) electrons can participate in bonding.

This enables them to form unstable intermediate oxidation states during chemical reactions.

These intermediate species provide an alternative reaction pathway with a lower activation energy.

Consequently, the reaction proceeds at a much faster rate.

In addition, transition metals possess large surface areas, particularly when finely divided.

Reactant molecules become adsorbed on the metal surface.

Adsorption weakens existing chemical bonds and brings reactant molecules into close proximity, increasing the frequency of effective collisions.

Since the activation energy decreases while the catalyst itself remains chemically unchanged, transition metals are highly effective catalysts in numerous industrial processes.


\textcolor{red{Step 4: Conclusion
These features facilitate the breaking and forming of bonds during chemical reactions.



\textcolor{red{Final Answer: Because they exhibit multiple variable oxidation states to form unstable intermediates and provide a large solid surface area for the adsorption of reactant molecules. Quick Tip: Variable valency (for intermediates) + Large surface area (for adsorption) = Good Catalyst.

\textcolor{red{Step 1: Concept
Melting points and metallic bonding in transition metals.

\textcolor{red{Step 2: Meaning
Stronger metallic bonds require more thermal energy to break, resulting in a higher melting point.

\textcolor{red{Step 3: Analysis

Chromium (\(Cr\)) has the electronic configuration \(3d^5\,4s^1\), containing six unpaired electrons that contribute to strong metallic bonding. Manganese (\(Mn\)) has the configuration \(3d^5\,4s^2\); its \(3d\) electrons are more localized and participate less effectively in metallic bonding.

The strength of metallic bonding depends upon the number of electrons available for delocalization.

Chromium has the electronic configuration:
\[ 3d^5\,4s^1. \]

All six valence electrons participate effectively in metallic bonding.

The presence of many bonding electrons results in exceptionally strong metallic bonds.

Consequently, chromium possesses a very high melting point and high enthalpy of atomization.

Manganese has the electronic configuration:
\[ 3d^5\,4s^2. \]

The half-filled \(3d^5\) subshell is particularly stable and its electrons are relatively less available for metallic bonding.

Therefore, manganese forms comparatively weaker metallic bonds than chromium.

This explains why chromium has a higher melting point and greater hardness than manganese.



\textcolor{red{Step 4: Conclusion
The stronger metallic bonding in Cr directly leads to its higher melting point compared to Mn.



\textcolor{red{Final Answer: Cr has more unpaired electrons (\(3d^5 4s^1\)) participating in strong interatomic metallic bonding compared to Mn (\(3d^5 4s^2\)), resulting in a higher melting point. Quick Tip: Strong metallic bonds require a high number of unpaired electrons. Cr has the maximum (6) in the 3d series.

\textcolor{red{Step 1: Concept
Thermodynamic stability of the permanganate ion in acidic media.

\textcolor{red{Step 2: Meaning
\(KMnO_4\) is a strong oxidizing agent that can slowly react with the solvent (water) if left standing.

\textcolor{red{Step 3: Analysis

Acidified potassium permanganate is thermodynamically unstable and slowly oxidizes water to oxygen gas on standing. During this process, permanganate ions are reduced to manganese dioxide while water is oxidized to oxygen.

Potassium permanganate is a very powerful oxidizing agent.

Even in the absence of another reducing agent, acidified potassium permanganate slowly decomposes on standing.

In this process, water molecules undergo oxidation:
\[ 2H_2O \rightarrow O_2+4H^++4e^-. \]

Simultaneously, permanganate ions undergo reduction to manganese dioxide:
\[ MnO_4^- \rightarrow MnO_2. \]

As a result, oxygen gas is gradually evolved.

The formation of brown manganese dioxide causes the purple colour of the permanganate solution to fade with time.

Therefore, acidified potassium permanganate solutions are not stable for prolonged storage and are generally prepared fresh before use in analytical and laboratory work.



\textcolor{red{Step 4: Conclusion
This slow decomposition is an example of a redox reaction.



\textcolor{red{Final Answer: It slowly decomposes to release oxygen gas. Equation: \(4MnO_4^- + 4H^+ \rightarrow 4MnO_2 + 2H_2O + 3O_2\). This is a redox (decomposition) reaction. Quick Tip: Acidic \(KMnO_4\) slowly oxidizes water to \(O_2\) over time.

\textcolor{red{Step 1: Concept
This electrochemical problem requires the application of the Nernst equation to find the non-standard electromotive force (emf or \(E_{cell}\)) and the thermodynamic relationship \(\Delta G = -nFE_{cell}\) to find the Gibbs free energy change.

\textcolor{red{Step 2: Meaning

Standard Cell Potential: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\)

Nernst Equation: \(E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q\)

Reaction Quotient (\(Q\)) depends on the balanced cell reaction: \(Mg(s) + 2Ag^+(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)\). Here, \(n = 2\) electrons are transferred.

\textcolor{red{Step 3: Analysis

Part A: Calculate \(E^\circ_{cell}\)

The anode (oxidation) is Mg and the cathode (reduction) is Ag.
\(E^\circ_{cell} = E^\circ_{Ag^{+}/Ag} - E^\circ_{Mg^{2+}/Mg}\)
\(E^\circ_{cell} = 0.80~V - (-2.37~V) = 3.17~V\)


Part B: Calculate \(E_{cell}\) (emf)

From the balanced reaction, \(Q = \frac{[Mg^{2+}]}{[Ag^+]^2}\)

Given: \([Mg^{2+}] = 0.01~M = 10^{-2}~M\), \([Ag^+] = 0.001~M = 10^{-3}~M\)
\(Q = \frac{10^{-2}}{(10^{-3})^2} = \frac{10^{-2}}{10^{-6}} = 10^4\)

Applying Nernst equation at 298 K:
\(E_{cell} = 3.17 - \frac{0.059}{2} \log(10^4)\)
\(E_{cell} = 3.17 - 0.0295 \times 4\)
\(E_{cell} = 3.17 - 0.118 = 3.052~V\)


Part C: Calculate \(\Delta G\)
\(\Delta G = -n F E_{cell}\)
\(\Delta G = -2 \times 96500~C~mol^{-1} \times 3.052~V\)
\(\Delta G = -193000 \times 3.052~J~mol^{-1}\)
\(\Delta G = -589036~J~mol^{-1} = -589.036~kJ~mol^{-1}\)


\textcolor{red{Step 4: Conclusion

The cell operates spontaneously with a high positive potential and a corresponding large negative Gibbs free energy.



\textcolor{red{Final Answer:

emf (\(E_{cell}\)) = \(3.052~V\)
\(\Delta G\) = \(-589.036~kJ~mol^{-1}\) (or \(-589036~J~mol^{-1}\)) Quick Tip: Always double-check the stoichiometry for the Nernst equation quotient \(Q\). Because 2 \(Ag^+\) ions are reduced, its concentration must be squared in the denominator.

\textcolor{red{Step 1: Concept
This problem requires linking the standard Gibbs free energy (\(\Delta G^\circ\)) to the standard cell potential (\(E^\circ_{cell}\)), and then using the Nernst equation to find the non-standard emf (\(E_{cell}\)) for a cell involving a gas and sparingly soluble salt.

\textcolor{red{Step 2: Meaning

Formula 1: \(\Delta G^\circ = -n F E^\circ_{cell}\)

Formula 2: \(E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q\)

For the reaction \(2AgCl + H_2 \rightarrow 2Ag + 2H^+ + 2Cl^-\), the number of electrons exchanged is \(n = 2\) (since \(H_2 \rightarrow 2H^+ + 2e^-\)).


\textcolor{red{Step 3: Analysis

Part A: Calculate \(E^\circ_{cell}\)
\(E^\circ_{cell} = \frac{-\Delta G^\circ}{nF}\)
\(E^\circ_{cell} = \frac{-(-43500~J~mol^{-1})}{2 \times 96500~C~mol^{-1}}\)
\(E^\circ_{cell} = \frac{43500}{193000} \approx 0.225~V\)


Part B: Calculate Reaction Quotient (Q)

Solids (\(AgCl\), \(Ag\)) are excluded from \(Q\). Gases use partial pressure.
\(Q = \frac{[H^+]^2 [Cl^-]^2}{P_{H_2}}\)

Given: \([H^+] = 0.1~M = 10^{-1}\), \([Cl^-] = 0.2~M = 2 \times 10^{-1}\), \(P_{H_2} = 0.4~atm\)
\(Q = \frac{(0.1)^2 \times (0.2)^2}{0.4} = \frac{0.01 \times 0.04}{0.4}\)
\(Q = \frac{0.0004}{0.4} = 0.001 = 10^{-3}\)


Part C: Calculate \(E_{cell}\) (emf)

Using the Nernst equation at \(25^\circ C\) (298 K):
\(E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q\)
\(E_{cell} = 0.225 - \frac{0.059}{2} \log(10^{-3})\)
\(E_{cell} = 0.225 - 0.0295 \times (-3)\)
\(E_{cell} = 0.225 + 0.0885 = 0.3135~V\)


\textcolor{red{Step 4: Conclusion

The non-standard potential is higher than the standard potential because the reactant gas pressure is relatively high compared to the extremely low product ion concentrations.



\textcolor{red{Final Answer: emf of the cell (\(E_{cell}\)) = \(0.3135~V\) Quick Tip: When solids or pure liquids are present in a cell reaction, they are omitted (assigned an activity of 1) in the reaction quotient \(Q\). Only aqueous ions and gases are included.