CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 3 - 56/2/3)

\textcolor{red{Step 1: Concept
Molecularity is defined theoretically as the number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction, which must collide simultaneously.

\textcolor{red{Step 2: Meaning
Since complex reactions proceed via multiple elementary steps, there is no single simultaneous collision involving all reactants.

\textcolor{red{Step 3: Analysis
Option (A) is incorrect because order is experimental, while molecularity is theoretical.

Option (C) is incorrect because molecularity must be a whole number (at least 1).

Option (D) is incorrect because simultaneous collisions of more than three molecules are statistically highly improbable.

Option (B) is mathematically and theoretically correct because molecularity is only defined for individual elementary steps, not the overall complex reaction.

\textcolor{red{Step 4: Conclusion
The statement regarding the lack of meaning of molecularity for complex reactions is the only correct one.


\textcolor{red{Final Answer: (B) Quick Tip: Order is experimental and applies to the overall reaction. Molecularity is theoretical and applies only to single elementary steps.

\textcolor{red{Step 1: Concept
The standard Gibbs free energy change (\(\Delta_r G^\circ\)) of an electrochemical cell is related to its standard cell potential (\(E^\circ_{cell}\)) by the equation \(\Delta_r G^\circ = -nFE^\circ_{cell}\).

\textcolor{red{Step 2: Meaning
Here, \(n\) is the number of moles of electrons transferred in the balanced redox reaction, and \(F\) is Faraday's constant.

\textcolor{red{Step 3: Analysis

1. Identify oxidation and reduction: Zinc goes from \(0\) to \(+2\) (oxidation, anode). Silver goes from \(+1\) to \(0\) (reduction, cathode).

2. Determine \(n\): \(Zn \rightarrow Zn^{2+} + 2e^-\), so \(n = 2\).

3. Calculate \(E^\circ_{cell}\): \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80~V - (-0.76~V) = 1.56~V\).

4. Calculate \(\Delta_r G^\circ\): \(\Delta_r G^\circ = -2 \times 96500~C~mol^{-1} \times 1.56~V = -301080~J~mol^{-1}\).

5. Convert to kJ: \(-301080~J~mol^{-1} = -301.080~kJ~mol^{-1}\).

\textcolor{red{Step 4: Conclusion
The calculated value perfectly matches option (A).


\textcolor{red{Final Answer: (A) Quick Tip: Always ensure you convert the final answer from Joules to kiloJoules if the options demand it (\(1~kJ = 1000~J\)).

\textcolor{red{Step 1: Concept
Chromium (Cr) has an atomic number of 24. It is an exceptional case in the Aufbau principle of electron filling.

\textcolor{red{Step 2: Meaning
Half-filled and fully filled d-subshells offer extra stability due to symmetry and higher exchange energy.

\textcolor{red{Step 3: Analysis
The expected configuration for \(Z=24\) based strictly on the Aufbau principle is \([Ar]~3d^4 4s^2\).

However, by shifting one electron from the 4s orbital to the 3d orbital, chromium achieves a \(3d^5\) configuration.

This exactly half-filled d-subshell is far more thermodynamically stable.

\textcolor{red{Step 4: Conclusion
Therefore, the actual ground-state electronic configuration becomes \([Ar]~3d^5 4s^1\).


\textcolor{red{Final Answer: (C) Quick Tip: Chromium (\(3d^5 4s^1\)) and Copper (\(3d^{10} 4s^1\)) are the two classic exceptions in the 3d series due to half-filled and fully-filled stability.

\textcolor{red{Step 1: Concept
The overall order of a chemical reaction is defined strictly by the experimentally determined rate law.

\textcolor{red{Step 2: Meaning
It is mathematically calculated as the sum of the exponents (powers) of the concentration terms present in the rate law equation.

\textcolor{red{Step 3: Analysis

The experimentally determined rate law is: \[ Rate=k[A]^{1/2}[B]. \]
The exponent of \([A]\) is \(\frac{1}{2}\) and the exponent of \([B]\) is \(1\). Therefore, the overall order of the reaction is: \[ Overall Order =\frac{1}{2}+1 =\frac{3}{2}=1.5. \]


The order of a reaction is determined from the experimentally obtained rate law.

The given rate law is:
\[ Rate=k[A]^{1/2}[B]. \]

The order with respect to reactant \(A\) is equal to the exponent of its concentration:
\[ \boxed{\frac{1}{2}}. \]

Similarly, the order with respect to reactant \(B\) is:
\[ \boxed{1}. \]

The overall order of the reaction is obtained by adding the exponents of all reactants:
\[ Overall Order =\frac{1}{2}+1 =\frac{3}{2}. \]

Hence,
\[ \boxed{Overall Order=1.5.} \]

Since the reaction order is fractional, it indicates that the reaction mechanism is more complex and cannot be inferred directly from the balanced chemical equation.


\textcolor{red{Step 4: Conclusion
The stoichiometric coefficients in the balanced chemical equation (\(A + 2B\)) are entirely irrelevant; only the rate law exponents dictate the order.


\textcolor{red{Final Answer: (A) Quick Tip: Order = Sum of powers in the rate law. Do not use the coefficients from the balanced reaction equation.

\textcolor{red{Step 1: Concept
Coordination complexes are classified based on the types of ligands attached to the central metal ion.

\textcolor{red{Step 2: Meaning
A homoleptic complex contains only one specific type of ligand. A heteroleptic complex contains more than one distinct type of ligand in its coordination sphere.

\textcolor{red{Step 3: Analysis

Options (A), (B), and (C) contain only one type of ligand (\(NH_3\), \(NH_3\), and \(H_2O\), respectively). Since each complex contains only one kind of donor ligand, they are homoleptic complexes. Option (D) contains both ammonia (\(NH_3\)) and chloride (\(Cl^-\)) ligands coordinated to the central cobalt ion.


Coordination compounds are classified as homoleptic or heteroleptic depending upon the number of different ligands attached to the central metal ion.

A homoleptic complex contains only one type of ligand coordinated to the metal ion.

Therefore, complexes containing only:
\[ NH_3 \quador\quad H_2O \]
ligands are homoleptic complexes.

Examples include:
\[ [Co(NH_3)_6]^{3+} \quadand\quad [Co(H_2O)_6]^{3+}. \]

A heteroleptic complex contains two or more different kinds of ligands.

In option (D), the coordination sphere contains both:
\[ NH_3 \quadand\quad Cl^-. \]

Since more than one type of ligand is present, the complex is classified as a
\[ \boxed{heteroleptic complex}. \]

Thus, options (A), (B), and (C) are homoleptic, whereas option (D) is heteroleptic.


\textcolor{red{Step 4: Conclusion
Because there are two different types of ligands, \([Co(NH_3)_4Cl_2]^+\) is classified as a heteroleptic complex.


\textcolor{red{Final Answer: (D) Quick Tip: "Homo" = Same (one type of ligand). "Hetero" = Different (multiple types of ligands).

\textcolor{red{Step 1: Concept
IUPAC nomenclature rules for naming neutral coordination complexes.

\textcolor{red{Step 2: Meaning
Ligands are named first in strictly alphabetical order, followed by the central metal, and then its oxidation state is placed in Roman numerals inside parentheses.

\textcolor{red{Step 3: Analysis
The complex has two ammonia ligands ("ammine", starts with 'a') and two chloride ligands ("chlorido", starts with 'c').

Alphabetically, 'ammine' precedes 'chlorido', giving "diamminedichlorido".

The central metal is Platinum. Let its oxidation state be \(x\). Since the overall complex is neutral:
\(x + 2(0) + 2(-1) = 0 \implies x = +2\).

\textcolor{red{Step 4: Conclusion
Combining these components yields the name diamminedichloridoplatinum(II).


\textcolor{red{Final Answer: (B) Quick Tip: Note the spelling: "ammine" has two 'm's for the \(NH_3\) ligand, distinguishing it from organic amines.

\textcolor{red{Step 1: Concept
The boiling point of alcohols depends on two primary factors: the molar mass (length of the carbon chain) and the extent of molecular branching.

\textcolor{red{Step 2: Meaning
Boiling point increases with an increase in molar mass due to stronger van der Waals dispersion forces. For isomeric alcohols, branching decreases the boiling point by reducing the effective surface area available for intermolecular contact.

\textcolor{red{Step 3: Analysis

1. Molar Mass: Propan-1-ol (3 carbons) < Butanols (4 carbons) < Pentan-1-ol (5 carbons). This establishes the extremes.

2. Branching: Butan-1-ol is a straight chain, whereas butan-2-ol is branched. Therefore, butan-2-ol has a lower boiling point than butan-1-ol.

3. Assembling the order: Propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol.

\textcolor{red{Step 4: Conclusion
This logical deduction matches the order provided in option (C).


\textcolor{red{Final Answer: (C) Quick Tip: More carbons = Higher BP. More branching = Lower BP.

\textcolor{red{Step 1: Concept
A molecule is chiral if it lacks a plane of symmetry and contains at least one stereocenter (an asymmetric chiral carbon).

\textcolor{red{Step 2: Meaning
A chiral carbon must be \(sp^3\) hybridized and bonded to four distinctly different chemical groups or atoms.

\textcolor{red{Step 3: Analysis

- (A) Propan-2-ol: C2 is attached to \(-OH, -H,\) and two identical \(-CH_3\) groups (achiral).

- (C) 1-Bromobutane: C1 is attached to \(-Br,\) a propyl group, and two identical \(-H\) atoms (achiral).

- (D) 2-Bromopropane: C2 is attached to \(-Br, -H,\) and two identical \(-CH_3\) groups (achiral).

- (B) Butan-2-ol: The C2 atom is attached to four distinct groups:

a methyl group (\(-CH_3\)), an ethyl group (\(-CH_2CH_3\)), a hydroxyl group (\(-OH\)), and a hydrogen atom (\(-H\)).

\textcolor{red{Step 4: Conclusion
Having four different groups attached to the central carbon makes Butan-2-ol a chiral molecule.


\textcolor{red{Final Answer: (B) Quick Tip: Draw the structure and check for a carbon with 4 different branches. If any two branches are identical, it is achiral.

\textcolor{red{Step 1: Concept
Nucleic acids (DNA and RNA) are built from nucleotides containing specific nitrogenous bases.

\textcolor{red{Step 2: Meaning
There are purines (Adenine, Guanine) and pyrimidines (Cytosine, Thymine, Uracil).

\textcolor{red{Step 3: Analysis

Adenine (A), Guanine (G), and Cytosine (C) are present in both DNA and RNA molecules. The distinguishing nitrogenous base is the fourth base: DNA contains thymine (T), whereas RNA contains uracil (U).


Nucleic acids are polymers composed of nucleotide units.

Each nucleotide contains:

A pentose sugar.
A phosphate group.
A nitrogenous base.


Both DNA and RNA contain the purine bases:
\[ \boxed{Adenine (A) and Guanine (G)}. \]

Both nucleic acids also contain the pyrimidine base:
\[ \boxed{Cytosine (C)}. \]

The difference lies in the fourth nitrogenous base.

DNA contains:
\[ \boxed{Thymine (T)}. \]

RNA does not contain thymine; instead, it contains:
\[ \boxed{Uracil (U)}. \]

Therefore,
\[ \boxed{DNA: A, G, C, T} \]
and
\[ \boxed{RNA: A, G, C, U.} \]


\textcolor{red{Step 4: Conclusion
Therefore, Thymine is the unique base found only in DNA.


\textcolor{red{Final Answer: (C) Quick Tip: Remember: DNA has ATGC, RNA has AUGC. T replaces U in DNA.

\textcolor{red{Step 1: Concept
Nitriles (alkyl cyanides) undergo strong reduction when treated with Lithium aluminium hydride (\(LiAlH_4\)).

\textcolor{red{Step 2: Meaning
This severe reduction converts the carbon-nitrogen triple bond (\(-C \equiv N\)) completely into a primary amine group (\(-CH_2-NH_2\)).

\textcolor{red{Step 3: Analysis
The starting material is Ethanenitrile, which has the formula \(CH_3-C \equiv N\).

It contains exactly two carbon atoms. Reduction by \(LiAlH_4\) adds two hydrogen atoms to the carbon and two to the nitrogen, maintaining the two-carbon chain.

Reaction: \(CH_3-C \equiv N \xrightarrow{LiAlH_4} CH_3-CH_2-NH_2\).

The IUPAC name of the product is Ethanamine, and its common name is Ethylamine.

\textcolor{red{Step 4: Conclusion
The resulting two-carbon amine matches option (A).


\textcolor{red{Final Answer: (A) Quick Tip: \(LiAlH_4\) reduction of nitriles yields primary amines without losing any carbon atoms.

\textcolor{red{Step 1: Concept
The open-chain Fischer projection of D-glucose correctly predicts many of its chemical properties, but fails to explain certain anomalous behaviors.

\textcolor{red{Step 2: Meaning
Reactions that prove the straight carbon chain, the aldehyde group, and the alcohol groups are all supported by the open-chain structure.

\textcolor{red{Step 3: Analysis
Option (A) confirms the straight 6-carbon chain.

Options (B) and (C) confirm the presence of an active, free aldehyde group.

However, the existence of D-glucose in two distinct crystalline isomeric forms (\(\alpha\)-D-glucose
and \(\beta\)-D-glucose, called anomers) stems entirely from the creation of a new chiral center at C-1 when the molecule folds into a closed, cyclic hemiacetal ring structure.

\textcolor{red{Step 4: Conclusion
The open-chain structure cannot possibly account for the existence of these two separate anomeric forms.


\textcolor{red{Final Answer: (D) Quick Tip: Anomers (alpha and beta forms) and the lack of reaction with Schiff's reagent strongly necessitate the cyclic hemiacetal structure of glucose.

\textcolor{red{Step 1: Concept
Macromolecular polymers in biology are synthesized by linking specific monomer units together through specialized condensation bonds.

\textcolor{red{Step 2: Meaning
Proteins are biological polyamides constructed entirely from sequences of \(\alpha\)-amino acid monomers.

\textcolor{red{Step 3: Analysis
When two \(\alpha\)-amino acids condense, the carboxyl group (\(-COOH\)) of the first amino acid reacts with the amino group (\(-NH_2\)) of the second amino acid.

A molecule of water is eliminated, creating an amide linkage (\(-CO-NH-\)).

While this is technically a type of covalent bond, in the specific context of biochemistry

and protein structure, this specialized amide bond linking amino acids is uniquely defined as a "peptide bond."

\textcolor{red{Step 4: Conclusion
Therefore, the most precise and biologically accurate term is the peptide bond.


\textcolor{red{Final Answer: (B) Quick Tip: Peptide bonds link amino acids (Proteins), Glycosidic bonds link sugars (Carbohydrates), and Phosphodiester bonds link nucleotides (DNA/RNA).

\textcolor{red{Step 1: Concept
IUPAC defines transition metals as elements whose atoms have an incomplete d-subshell, or which can give rise to cations with an incomplete d-subshell.

\textcolor{red{Step 2: Meaning
Elements of Group 12 (Zn, Cd, Hg) must be evaluated based on their d-orbital electron configurations.

\textcolor{red{Step 3: Analysis

Zinc (\(3d^{10}4s^2\)), cadmium (\(4d^{10}5s^2\)), and mercury (\(5d^{10}6s^2\)) possess completely filled \(d\)-subshells in both their atomic states and their common \(+2\) oxidation states.


Zinc, cadmium, and mercury belong to Group 12 of the periodic table.

Their electronic configurations are:
\[ Zn:3d^{10}4s^2, \]
\[ Cd:4d^{10}5s^2, \]
\[ Hg:5d^{10}6s^2. \]

During ion formation, the two outermost \(s\) electrons are removed first.

Consequently, the common ions have the configurations:
\[ Zn^{2+}:3d^{10}, \]
\[ Cd^{2+}:4d^{10}, \]
\[ Hg^{2+}:5d^{10}. \]

Since the \(d\)-subshell remains completely filled, these ions contain no partially filled \(d\) orbitals.

As a result, Group 12 elements differ from typical transition elements in many of their physical and chemical properties.

Therefore, zinc, cadmium, and mercury are generally not regarded as typical transition metals because neither their atoms nor their common ions possess incompletely filled \(d\) orbitals.



\textcolor{red{Step 4: Conclusion
Because they lack partially filled d-orbitals entirely, they are not considered true transition elements. The Reason is factually true and perfectly explains the Assertion.


\textcolor{red{Final Answer: (A) Quick Tip: Group 12 elements (Zn, Cd, Hg) are d-block elements but NOT transition elements because their d-orbitals are completely full (\(d^{10}\)).

\textcolor{red{Step 1: Concept
Molecularity represents the total number of reacting species taking part in an elementary chemical reaction.

\textcolor{red{Step 2: Meaning
These species must collide simultaneously to bring about the chemical change.

\textcolor{red{Step 3: Analysis

The given elementary reaction involves two molecules of hydrogen iodide: \[ 2HI \rightarrow H_2+I_2. \]
For this elementary step to occur, two \(HI\) molecules must collide simultaneously.


Molecularity is defined as the number of reacting species participating in a single elementary reaction step.

The given elementary reaction is:
\[ 2HI \rightarrow H_2+I_2. \]

Since two molecules of hydrogen iodide participate directly in the elementary step, the reaction requires the simultaneous collision of two molecules.

Therefore, the molecularity of the reaction is:
\[ \boxed{2}. \]

A reaction involving two colliding molecules is called a
\[ \boxed{bimolecular reaction}. \]

Molecularity is always a whole positive integer because it represents the actual number of reacting particles involved in a single elementary step.

Unlike reaction order, molecularity cannot be fractional or zero.



\textcolor{red{Step 4: Conclusion
Therefore, the molecularity is exactly 2 (a bimolecular reaction). The Reason accurately describes the physical requirement for the Assertion to be true.


\textcolor{red{Final Answer: (A) Quick Tip: Molecularity is determined by counting the stoichiometric coefficients of reactants in a single-step elementary reaction.

\textcolor{red{Step 1: Concept
Chemical properties and structural elucidation of D-glucose.

\textcolor{red{Step 2: Meaning
Mild oxidizing agents test for the presence of specific functional groups, such as aldehydes.

\textcolor{red{Step 3: Analysis

Bromine water is a mild oxidizing agent. When glucose is treated with bromine water, it is oxidized to gluconic acid, demonstrating the presence of an aldehyde group in its open-chain structure.


Bromine water is a mild oxidizing agent that selectively oxidizes aldehydes without affecting primary alcohol groups.

D-Glucose exists in equilibrium between cyclic forms and a small amount of the open-chain form.

The open-chain structure contains a free aldehyde group at the C1 position:
\[ -CHO. \]

Bromine water oxidizes this aldehyde group into a carboxylic acid:
\[ -CHO \longrightarrow -COOH. \]

Consequently, glucose is converted into:
\[ \boxed{Gluconic acid.} \]

This reaction demonstrates that glucose behaves as a reducing sugar.

The oxidation occurs only because the open-chain form contains a free aldehyde group that is available for oxidation.

Therefore, the successful oxidation of glucose by bromine water provides important experimental evidence for the presence of an aldehyde group in the open-chain structure of glucose.

Hence, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.


\textcolor{red{Step 4: Conclusion
The Reason states that the carbonyl group is absent, which is a completely false chemical statement.


\textcolor{red{Final Answer: (C) Quick Tip: Bromine water oxidizes the terminal aldehyde (\(-CHO\)) of glucose to a carboxylic acid (\(-COOH\)), proving the aldehyde group exists.

\textcolor{red{Step 1: Concept
The Gabriel Phthalimide synthesis is a specific method for preparing pure aliphatic primary amines.

\textcolor{red{Step 2: Meaning
The key step in this synthesis is an \(S_N2\) nucleophilic substitution where a phthalimide anion attacks an alkyl halide.

\textcolor{red{Step 3: Analysis

To prepare an aromatic primary amine (such as aniline), an aryl halide (such as chlorobenzene) would be required. However, aryl halides are extremely unreactive toward nucleophilic substitution because the carbon--halogen bond possesses partial double-bond character due to resonance. Therefore, the phthalimide anion cannot displace the halide ion.

The Gabriel phthalimide synthesis is commonly used for the preparation of primary aliphatic amines.

In this method, the phthalimide anion acts as a nucleophile and attacks an alkyl halide through an
\[ S_N2 \]
mechanism.

However, this method cannot be used to prepare aromatic primary amines such as aniline.

Aromatic halides (aryl halides) like chlorobenzene are highly resistant to nucleophilic substitution.

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

The lone pair of electrons on the halogen overlaps with the \(\pi\)-electron cloud of the benzene ring, giving the carbon--halogen bond partial double-bond character.

Consequently, the C--Cl bond becomes shorter and much stronger than that in alkyl halides.

Moreover, the carbon atom bonded to chlorine is
\[ sp^2 \]
hybridized, making the bond even stronger.

Therefore, the phthalimide anion cannot replace the halide ion, and the Gabriel synthesis is unsuitable for preparing aromatic primary amines.


\textcolor{red{Step 4: Conclusion
Aromatic primary amines cannot be prepared this way, making the Assertion false. The Reason provides the correct chemical justification for why the reaction fails.


\textcolor{red{Final Answer: (D) Quick Tip: Gabriel Phthalimide synthesis works ONLY for aliphatic primary amines, never for aromatic amines.

\textcolor{red{Step 1: Concept
The coordination number in a coordination complex refers to the secondary valency of the central metal atom or ion.

\textcolor{red{Step 2: Meaning
It is an integer that quantifies how many coordinate bonds are formed directly between the central metal and its surrounding ligands.

\textcolor{red{Step 3: Analysis

When calculating the coordination number, the total number of donor atoms directly bonded to the central metal ion is counted. Thus, a bidentate ligand contributes two donor atoms, while a hexadentate ligand such as EDTA contributes six donor atoms.

The coordination number of a complex is defined as the total number of donor atoms directly bonded to the central metal ion.

It is important to note that the coordination number depends on the number of donor atoms rather than the number of ligand molecules.

A monodentate ligand donates only one lone pair through one donor atom.

Therefore, each monodentate ligand contributes:
\[ \boxed{1} \]
to the coordination number.

A bidentate ligand such as ethylenediamine (en) contains two donor nitrogen atoms.

Hence, each ethylenediamine ligand contributes:
\[ \boxed{2} \]
to the coordination number.

Ethylenediaminetetraacetate (EDTA) is a hexadentate ligand containing six donor atoms.

Therefore, a single EDTA molecule contributes:
\[ \boxed{6} \]
to the coordination number.

Thus, the coordination number always represents the total number of coordinate bonds formed between the ligands and the central metal ion.



\textcolor{red{Step 4: Conclusion
The formal definition directly reflects this principle.



\textcolor{red{Final Answer: The coordination number of a central metal ion in a complex is defined as the total number of ligand donor atoms to which it is directly bonded via coordinate covalent bonds. Quick Tip: Coordination Number = Number of Ligands \(\times\) Denticity of each ligand.

\textcolor{red{Step 1: Concept
Stereoisomerism in octahedral coordination complexes containing bidentate ligands.

\textcolor{red{Step 2: Meaning
A complex exhibits optical isomerism if it lacks a plane of symmetry and forms non-superimposable mirror images (enantiomers).

\textcolor{red{Step 3: Analysis

The complex \[ [Co(en)_3]^{3+} \]
contains a central cobalt ion octahedrally surrounded by three bidentate ethylenediamine ligands. The three chelate rings wrap around the metal ion in a propeller-like arrangement, making the complex chiral.

In this complex, cobalt is coordinated by three ethylenediamine (en) ligands.

Each ethylenediamine molecule is a bidentate ligand and donates two nitrogen atoms to the metal ion.

Consequently, the coordination number of cobalt is:
\[ 3\times2=6. \]

A coordination number of six gives the complex an octahedral geometry.

The three bidentate ligands wrap around the metal ion to form three chelate rings.

These chelate rings can twist in two different ways, producing right-handed and left-handed arrangements.

As a result, the complex possesses neither a plane of symmetry nor a centre of inversion.

Therefore, the complex exists as two non-superimposable mirror images known as optical isomers or enantiomers.

These are commonly designated as:
\[ \Delta \quadand\quad \Lambda \]
forms.

Hence,
\[ [Co(en)_3]^{3+} \]
is an optically active coordination compound.


\textcolor{red{Step 4: Conclusion
Molecules lacking a plane of symmetry are chiral and consequently exhibit optical isomerism, existing as dextro (d) and laevo (l) forms.



\textcolor{red{Final Answer: Optical Isomerism. Quick Tip: Any octahedral complex of the type \([M(AA)_3]^{n\pm}\) (where AA is a symmetrical bidentate ligand) is always chiral and shows optical isomerism.

\textcolor{red{Step 1: Concept
Collision theory of chemical kinetics.

\textcolor{red{Step 2: Meaning
Not all collisions between reactant molecules lead to the formation of products. Only a specific fraction of collisions are successful.

\textcolor{red{Step 3: Analysis
For a collision to be considered "effective," two strict conditions must be met simultaneously.

First, the colliding molecules must possess kinetic energy equal to or greater than the activation energy (threshold energy).

Second, the molecules must have the proper spatial orientation at the moment of impact so that the correct bonds can break and form.

\textcolor{red{Step 4: Conclusion
Collisions fulfilling both energetic and spatial criteria result in chemical reactions.



\textcolor{red{Final Answer: Effective collisions are those collisions between reactant molecules that result in a chemical reaction, which only occur when the molecules possess sufficient kinetic energy (activation energy) and collide with the proper spatial orientation. Quick Tip: Effective Collision = Sufficient Energy + Proper Orientation.

\textcolor{red{Step 1: Concept
Dimensional analysis of the rate constant (\(k\)) for different orders of reaction.

\textcolor{red{Step 2: Meaning
The general formula for the unit of a rate constant for an \(n^{th}\) order reaction is \((mol~L^{-1})^{1-n} s^{-1}\).

\textcolor{red{Step 3: Analysis

(i) For a second order reaction, \(n = 2\). Substituting this into the formula gives: \((mol~L^{-1})^{1-2} s^{-1} = (mol~L^{-1})^{-1} s^{-1} = L~mol^{-1}~s^{-1}\).

(ii) For a zero order reaction, \(n = 0\). Substituting this into the formula gives: \((mol~L^{-1})^{1-0} s^{-1} = mol~L^{-1}~s^{-1}\).

\textcolor{red{Step 4: Conclusion
The units dynamically scale based on the concentration dependence of the rate law.



\textcolor{red{Final Answer:

(i) Second order reaction unit: \(L~mol^{-1}~s^{-1}\)

(ii) Zero order reaction unit: \(mol~L^{-1}~s^{-1}\) Quick Tip: Master formula for \(k\) units: \(M^{1-n} \cdot Time^{-1}\) where M is Molarity (\(mol~L^{-1}\)) and \(n\) is the order.

\textcolor{red{Step 1: Concept
Ionization enthalpy depends heavily on the thermodynamic stability of the specific electronic configuration from which the electron is being removed.

\textcolor{red{Step 2: Meaning
Exactly half-filled (\(d^5\)) and fully filled (\(d^{10}\)) subshells possess immense exchange energy and spherical symmetry, making them exceptionally stable.

\textcolor{red{Step 3: Analysis
The ground state configurations are \(Cr\) (\(3d^5 4s^1\)) and \(Cu\) (\(3d^{10} 4s^1\)).
When they lose their first electron (the 4s electron),
they form the univalent ions \(Cr^+\) (\(3d^5\)) and \(Cu^+\) (\(3d^{10}\)).
The second ionization enthalpy (\(IE_2\)) involves removing an electron from these resulting \(Cr^+\) and \(Cu^+\) ions.

\textcolor{red{Step 4: Conclusion
Because the second electron must be extracted from a highly stable, symmetrical half-filled (\(3d^5\)) or completely filled (\(3d^{10}\)) core, an exceptionally large amount of energy is required compared to neighboring elements lacking this extra stability.



\textcolor{red{Final Answer: After losing the first electron, Cr and Cu form \(Cr^+\) (\(3d^5\)) and \(Cu^+\) (\(3d^{10}\)), which have highly stable half-filled and fully filled d-subshells, respectively. Removing a second electron disrupts this immense stability, requiring exceptionally high energy. Quick Tip: Breaking a \(d^5\) or \(d^{10}\) stable core always results in a massive spike in ionization energy.

\textcolor{red{Step 1: Concept
The crystal lattice structure of transition metals contains empty spaces or voids.

\textcolor{red{Step 2: Meaning
Interstitial compounds are formed when small atoms get physically trapped inside these metallic lattices without forming typical ionic or covalent chemical bonds.

\textcolor{red{Step 3: Analysis
Transition metals possess closely packed crystal lattices (like FCC or HCP).
Between the large metal atoms, there are numerous small tetrahedral and octahedral voids (interstices).
Because elements like Hydrogen (H), Carbon (C), and Nitrogen (N) have very small atomic radii,
they can easily slip into and perfectly fit within these empty voids.

\textcolor{red{Step 4: Conclusion
This trapping mechanism creates stable, non-stoichiometric interstitial compounds with enhanced hardness and modified properties.



\textcolor{red{Final Answer: Transition metal crystal lattices contain empty spaces or voids (interstices). Because atoms of non-metals like H, C, and N are very small, they easily slip into and get trapped within these voids to form non-stoichiometric interstitial compounds. Quick Tip: Interstitial = "In the spaces". Small atoms (H, C, N) hide in the gaps between the big transition metal atoms.

\textcolor{red{Step 1: Concept
The integrated rate equation for a first-order chemical reaction relates the rate constant (\(k\)) to time (\(t\)) and reactant concentrations.

\textcolor{red{Step 2: Meaning
The formula is: \(k = \frac{2.303}{t} \log\left(\frac{[R_0]}{[R]}\right)\), where \([R_0]\) is initial concentration and \([R]\) is the concentration at time \(t\).

\textcolor{red{Step 3: Analysis

1. Given values: Initial concentration \([R_0] = 0.6~mol~L^{-1}\). Final concentration \([R] = 0.2~mol~L^{-1}\). Time \(t = 5\) minutes.

2. Substitute the values into the integrated rate equation:
\(k = \frac{2.303}{5} \log\left(\frac{0.6}{0.2}\right)\)
\(k = \frac{2.303}{5} \log(3)\)

3. Substitute the given logarithmic value (\(\log 3 = 0.48\)):
\(k = \frac{2.303 \times 0.48}{5}\)
\(k = \frac{1.10544}{5} = 0.221088 min^{-1}\)


\textcolor{red{Step 4: Conclusion
Rounding to three significant figures, the calculated rate constant is \(0.221 min^{-1}\).



\textcolor{red{Final Answer: \(0.221 min^{-1}\) Quick Tip: Ensure you state the unit of \(k\) correctly. Since time was in minutes, the unit of \(k\) for this 1st order reaction is min\(^{-1}\).

\textcolor{red{Step 1: Concept
The integrated rate law is used to find the time (\(t\)) needed to reach a specific fraction of completion in a first-order reaction.

\textcolor{red{Step 2: Meaning
If three-fourths (\(3/4\)) of the reactant has decomposed, it means exactly one-fourth (\(1/4\)) of the initial amount remains unreacted.

\textcolor{red{Step 3: Analysis

1. The remaining concentration \([R] = [R_0] - \frac{3}{4}[R_0] = \frac{1}{4}[R_0]\).

2. Rearranging the first-order rate equation for time \(t\):
\(t = \frac{2.303}{k} \log\left(\frac{[R_0]}{[R]}\right)\)

3. Substitute \([R]\) and the given rate constant \(k = 2.54 \times 10^{-3} s^{-1}\):
\(t = \frac{2.303}{2.54 \times 10^{-3}} \log\left(\frac{[R_0]}{[R_0]/4}\right) = \frac{2.303}{2.54 \times 10^{-3}} \log(4)\)

4. Substitute \(\log 4 = 0.60\):
\(t = \frac{2.303 \times 0.60}{2.54 \times 10^{-3}} = \frac{1.3818}{0.00254} \approx 544~s\)

\textcolor{red{Step 4: Conclusion
The calculation precisely yields the time taken for \(75%\) completion.



\textcolor{red{Final Answer: \(544 s\) Quick Tip: "Decomposed" means reacted. In the formula, \([R]\) represents the amount left remaining, not the amount reacted.

\textcolor{red{Step 1: Concept
This is the Wurtz-Fittig reaction.

\textcolor{red{Step 2: Meaning
It is a coupling reaction between an aryl halide and an alkyl halide in the presence of sodium metal and dry ether to form an alkylarene.

\textcolor{red{Step 3: Analysis

Sodium metal removes the chlorine atoms from chlorobenzene (\(C_6H_5Cl\)) and chloromethane (\(CH_3Cl\)), producing sodium chloride. The resulting phenyl and methyl radicals then combine to form a new carbon--carbon bond, yielding toluene.

This reaction is known as the
\[ \boxed{Wurtz--Fittig reaction}. \]

It is used for the preparation of alkyl-substituted aromatic hydrocarbons.

The reaction is carried out using sodium metal in dry ether.

Sodium removes the chlorine atoms from both chlorobenzene and chloromethane:
\[ C_6H_5Cl+CH_3Cl+2Na \rightarrow C_6H_5CH_3+2NaCl. \]

During the reaction, highly reactive phenyl and methyl radicals (or equivalent organosodium intermediates) are generated.

These reactive species combine to form a new carbon--carbon bond.

The final product obtained is:
\[ \boxed{Toluene} \]
in which the methyl group becomes directly attached to the benzene ring.

Thus, the Wurtz--Fittig reaction provides a convenient method for synthesizing alkylbenzenes from aryl and alkyl halides.


\textcolor{red{Step 4: Conclusion
The resulting major organic product is methylbenzene, widely known by its common name, Toluene.



\textcolor{red{Final Answer: Toluene (Methylbenzene) Quick Tip: Wurtz = Two alkyl halides. Fittig = Two aryl halides. Wurtz-Fittig = One of each (Aryl + Alkyl).

\textcolor{red{Step 1: Concept
This involves a dehydrohalogenation reaction (\(\beta\)-elimination).

\textcolor{red{Step 2: Meaning
Alcoholic KOH acts as a strong base, removing a hydrogen atom from a \(\beta\)-carbon and the bromine atom from the \(\alpha\)-carbon, resulting in the formation of a double bond (alkene).

\textcolor{red{Step 3: Analysis
2-Bromobutane (\(CH_3-CH_2-CH(Br)-CH_3\)) has two different sets of \(\beta\)-hydrogens.
Elimination can yield either But-1-ene or But-2-ene. According to Zaitsev's rule,
the highly substituted alkene is thermodynamically more stable and is formed as the major product.
Elimination of the internal \(\beta\)-hydrogen forms But-2-ene, which has two alkyl groups attached to the double bond.

\textcolor{red{Step 4: Conclusion
Therefore, the reaction predominantly produces But-2-ene, with a minor amount of But-1-ene.



\textcolor{red{Final Answer: A dehydrohalogenation (\(\beta\)-elimination) reaction occurs. According to Zaitsev's rule, But-2-ene is formed as the major product because it is the more highly substituted, stable alkene. (A small amount of But-1-ene is also formed as a minor product). Quick Tip: Alcoholic KOH = Elimination (Alkene). Aqueous KOH = Substitution (Alcohol). Remember Zaitsev's rule for the major alkene.

\textcolor{red{Step 1: Concept
Depression in freezing point formula: \(\Delta T_f = K_f \times m\).

\textcolor{red{Step 2: Meaning
\(m\) is molality (moles of solute per kg of solvent).

\textcolor{red{Step 3: Analysis

Moles of glucose (\(n\)) = \(\frac{60~g}{180~g~mol^{-1}} = \frac{1}{3}~mol\).

Mass of water (\(W\)) = \(250~g = 0.25~kg\).

Molality (\(m\)) = \(\frac{1/3}{0.25} = \frac{4}{3} = 1.333~m\).
\(\Delta T_f = 1.86 \times \frac{4}{3} = 0.62 \times 4 = 2.48~K\).

\textcolor{red{Step 4: Conclusion

Freezing point of solution = Freezing point of pure water - \(\Delta T_f\)
\(T_f = 273.15~K - 2.48~K = 270.67~K\) (or \(-2.48^\circ C\)).



\textcolor{red{Final Answer: \(270.67~K\) (or \(-2.48^\circ C\)) Quick Tip: Always convert the mass of the solvent to kilograms before calculating molality.

\textcolor{red{Step 1: Concept
Nernst equation for non-standard cell potential.

\textcolor{red{Step 2: Meaning
\(E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q\).

\textcolor{red{Step 3: Analysis

Reaction: \(Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)\). Here, \(n = 2\).
\(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00~V - (-0.14~V) = +0.14~V\).
\(Q = \frac{[Sn^{2+}][P_{H_2}]}{[H^+]^2} = \frac{(0.001)(1)}{(0.01)^2} = \frac{10^{-3}}{10^{-4}} = 10\).
\(E_{cell} = 0.14 - \frac{0.0591}{2} \log(10) = 0.14 - 0.0295(1)\).

\textcolor{red{Step 4: Conclusion
\(E_{cell} = 0.14 - 0.0295 = 0.1105~V\).



\textcolor{red{Final Answer: \(0.1105~V\) Quick Tip: Remember to square the \([H^+]\) concentration in the reaction quotient \(Q\) due to the stoichiometry.

\textcolor{red{Step 1: Concept
Redox reactions of d-block transition metal compounds in acidic medium.

\textcolor{red{Step 2: Meaning
(i) Chromate to dichromate conversion. (ii) Oxidation of sulfide by permanganate.

\textcolor{red{Step 3: Analysis

(i) In acidic medium, yellow chromate ions convert to orange dichromate ions.
\(2Na_2CrO_4 + 2H^+ \longrightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O\).

(ii) Permanganate reduces to \(Mn^{2+}\) in acidic medium, oxidizing sulfide to elemental sulfur.
\(5S^{2-} + 2MnO_4^- + 16H^+ \longrightarrow 2Mn^{2+} + 5S + 8H_2O\).

\textcolor{red{Step 4: Conclusion
The balanced equations follow the conservation of mass and charge.



\textcolor{red{Final Answer:

(i) \(2Na_2CrO_4 + 2H^+ \longrightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O\)

(ii) \(5S^{2-} + 2MnO_4^- + 16H^+ \longrightarrow 2Mn^{2+} + 5S + 8H_2O\) Quick Tip: Chromate (\(CrO_4^{2-}\)) and Dichromate (\(Cr_2O_7^{2-}\)) are interconvertible depending strictly on pH.

\textcolor{red{Step 1: Concept
Electronic configuration and energy levels of f-block elements.

\textcolor{red{Step 2: Meaning
Actinoids show a wider range of chemical behaviors than lanthanoids.

\textcolor{red{Step 3: Analysis

The \(5f\), \(6d\), and \(7s\) orbitals of the actinoids have very similar energies, whereas the energy difference between the corresponding \(4f\), \(5d\), and \(6s\) orbitals in the lanthanoids is comparatively larger.

In actinoids, the energies of the
\[ 5f,\;6d,\;and\;7s \]
orbitals are very close to one another.

Because the energy differences are small, electrons can be removed from or promoted into any of these orbitals with relatively little energy.

Consequently, both the \(5f\) and \(6d\) electrons participate in chemical bonding.

This leads to the occurrence of a large number of oxidation states in the actinoid series.

In contrast, the corresponding orbitals of the lanthanoids,
\[ 4f,\;5d,\;and\;6s, \]
are separated by comparatively larger energy differences.

As a result, the \(4f\) electrons participate only to a limited extent in bonding.

Therefore, lanthanoids predominantly exhibit the
\[ \boxed{+3} \]
oxidation state, whereas actinoids commonly exhibit oxidation states ranging from
\[ \boxed{+3\ to\ +7.} \]

Thus, the close energy spacing of the \(5f\), \(6d\), and \(7s\) orbitals is the principal reason for the variable oxidation states and greater chemical complexity of the actinoids.


\textcolor{red{Step 4: Conclusion
Due to these comparable energies, electrons from all three subshells can participate in bonding, leading to a large number of variable oxidation states. Additionally, all actinoids are radioactive.



\textcolor{red{Final Answer: The chemistry of actinoids is complicated because the 5f, 6d, and 7s orbitals have comparable energies, allowing them to exhibit a wide range of variable oxidation states. Furthermore, all actinoids are radioactive in nature. Quick Tip: Actinoids = Comparable 5f, 6d, 7s energies + Radioactivity.

\textcolor{red{Step 1: Concept
Crystal Field Theory (CFT) and ligand strength.

\textcolor{red{Step 2: Meaning
\(Fe^{3+}\) has a \(3d^5\) configuration. Unpaired electrons determine paramagnetism.

\textcolor{red{Step 3: Analysis

Water (\(H_2O\)) is a weak field ligand, so the crystal field splitting energy is smaller than the electron pairing energy \[ (\Delta_oTherefore, electrons do not pair in the lower-energy orbitals and instead occupy the higher-energy orbitals according to Hund's rule. The electronic configuration becomes \[ t_{2g}^{3}e_g^{2}, \]
containing five unpaired electrons. Cyanide (\(CN^-\)), on the other hand, is a strong field ligand \[ (\Delta_o>P), \]
which forces electron pairing. The resulting configuration is \[ t_{2g}^{5}e_g^{0}, \]
containing only one unpaired electron.


Consider a metal ion having the electronic configuration:
\[ d^5. \]

In an octahedral crystal field, the five \(d\) orbitals split into:
\[ t_{2g} \quadand\quad e_g. \]

When the ligand is water (\(H_2O\)), which is a weak field ligand,
\[ \Delta_o
Since the splitting energy is smaller than the pairing energy, electrons prefer to occupy all five orbitals singly rather than pair up.

The electronic configuration therefore becomes:
\[ t_{2g}^{3}e_g^{2}. \]

This arrangement contains:
\[ \boxed{5\ unpaired electrons}, \]
making the complex strongly paramagnetic.

Cyanide (\(CN^-\)) is a strong field ligand and produces a large crystal field splitting:
\[ \Delta_o>P. \]

Consequently, electrons pair in the lower-energy
\[ t_{2g} \]
orbitals before occupying the higher-energy
\[ e_g \]
orbitals.

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

Thus, weak field ligands form high-spin complexes, whereas strong field ligands form low-spin complexes.



\textcolor{red{Step 4: Conclusion
More unpaired electrons result in a stronger paramagnetic character.



\textcolor{red{Final Answer: \([Fe(H_2O)_6]^{3+}\) has a weak field ligand (\(H_2O\)), resulting in 5 unpaired electrons. \([Fe(CN)_6]^{3-}\) has a strong field ligand (\(CN^-\)), forcing pairing and resulting in only 1 unpaired electron. Therefore, the former is strongly paramagnetic while the latter is weakly paramagnetic. Quick Tip: Weak ligand = High spin (more unpaired \(e^-\)). Strong ligand = Low spin (fewer unpaired \(e^-\)).

\textcolor{red{Step 1: Concept
Crystal field splitting and electron filling rules.

\textcolor{red{Step 2: Meaning
\(\Delta_o < P\) indicates a weak ligand field where pairing energy is high.

\textcolor{red{Step 3: Analysis

Since the crystal field splitting energy \[ (\Delta_o) \]
is smaller than the pairing energy \[ (P), \]
it is energetically more favourable for electrons to occupy the higher-energy \(e_g\) orbitals than to pair within the lower-energy \(t_{2g}\) orbitals. Consequently, all five electrons remain unpaired.


According to crystal field theory, the behaviour of electrons depends on the relative magnitudes of:
\[ \Delta_o \quadand\quad P. \]

If
\[ \Delta_o pairing electrons in the lower-energy orbitals requires more energy than promoting an electron to the higher-energy orbitals.

Therefore, electrons obey Hund's rule and occupy each orbital singly before pairing.

For a
\[ d^5 \]
electronic configuration, the electrons are distributed as:
\[ t_{2g}^{3}e_g^{2}. \]

Every \(d\) orbital contains one electron.

Thus, all five electrons remain unpaired.

Such complexes are called:
\[ \boxed{high-spin complexes}. \]

High-spin complexes exhibit strong paramagnetism because of the large number of unpaired electrons.


\textcolor{red{Step 4: Conclusion
Three electrons enter \(t_{2g}\) and two enter \(e_g\).



\textcolor{red{Final Answer: \(t_{2g}^3 e_g^2\) Quick Tip: \(\Delta_o < P\) means "Pairing is prevented". Fill all orbitals singly first.

\textcolor{red{Step 1: Concept
IUPAC nomenclature rules for branched haloalkanes.

\textcolor{red{Step 2: Meaning
Identify the longest continuous carbon chain and number it to give the substituents the lowest possible locants.

\textcolor{red{Step 3: Analysis
The longest carbon chain contains 4 carbon atoms (butane).
Numbering from the right side gives the bromo group the position 2 and the methyl group the position 3.
Numbering from the left would give methyl position 2 and bromo position 3. Alphabetical priority dictates
that Bromo gets precedence for the lower number when possible, making 2-bromo-3-methylbutane the correct designation.

\textcolor{red{Step 4: Conclusion
Combining the substituents alphabetically gives the final IUPAC name.



\textcolor{red{Final Answer: 2-Bromo-3-methylbutane Quick Tip: Always arrange substituent prefixes alphabetically (Bromo before Methyl) and assign the lowest possible numbers.

\textcolor{red{Step 1: Concept
Stereochemistry and optical activity.

\textcolor{red{Step 2: Meaning
Enantiomers are chiral molecules that rotate plane-polarized light in opposite directions (dextro and laevo).

\textcolor{red{Step 3: Analysis

When a pure enantiomer undergoes a reaction such as an \(S_N1\) reaction that produces equal amounts of both enantiomers, the resulting mixture becomes optically inactive because the optical rotations produced by the two enantiomers exactly cancel each other.


Enantiomers are non-superimposable mirror images of one another.

A pure sample containing only one enantiomer rotates plane-polarized light and is therefore optically active.

In an \(S_N1\) reaction, the leaving group first departs to form a planar carbocation intermediate.

Since the carbocation is trigonal planar, the nucleophile can attack from either side with nearly equal probability.

Consequently, both enantiomers are formed in approximately equal amounts.

A mixture containing equal amounts of the two enantiomers is called a:
\[ \boxed{racemic mixture (racemate)}. \]

One enantiomer rotates plane-polarized light clockwise, whereas the other rotates it by exactly the same magnitude in the opposite direction.

Therefore, the two optical rotations cancel each other completely.

Hence, a racemic mixture is optically inactive despite containing optically active molecules.



\textcolor{red{Step 4: Conclusion
This process of forming a racemic mixture from a pure enantiomer is termed racemisation.



\textcolor{red{Final Answer: Racemisation is the process of conversion of an optically active enantiomer into an optically inactive racemic mixture (an equimolar 50:50 mixture of dextro and laevo enantiomers). Quick Tip: Racemisation = 1 pure enantiomer \(\rightarrow\) 50:50 mixture of two enantiomers (zero net optical rotation).

\textcolor{red{Step 1: Concept
Intermolecular forces governing physical properties of haloalkanes.

\textcolor{red{Step 2: Meaning
Boiling point depends on the magnitude of van der Waals (London dispersion) forces between molecules.

\textcolor{red{Step 3: Analysis
As we move down the halogen group from Fluorine to Iodine,
the atomic size and mass of the halogen atom increase significantly.
A larger atomic size provides a larger surface area and higher electron cloud polarizability.
This leads to much stronger van der Waals forces of attraction between \(RI\) molecules compared to \(RF\) molecules.

\textcolor{red{Step 4: Conclusion
Stronger intermolecular forces require more thermal energy to break, hence the boiling point is highest for alkyl iodides and lowest for alkyl fluorides.



\textcolor{red{Final Answer: As the size and mass of the halogen atom increase from F to I, the magnitude of van der Waals forces of attraction increases due to larger surface area and higher polarizability. Consequently, the boiling point increases in the order \(RF < RCl < RBr < RI\). Quick Tip: Larger halogen atom = Greater polarizability = Stronger van der Waals forces = Higher boiling point.

\textcolor{red{Step 1: Concept
Preparation of aldehydes from acid chlorides.

\textcolor{red{Step 2: Meaning
It is a catalytic hydrogenation reaction.

\textcolor{red{Step 3: Analysis

An acyl chloride (acid chloride) is hydrogenated in the presence of palladium deposited on barium sulfate (\(Pd/BaSO_4\)). The barium sulfate acts as a catalyst poison, reducing the activity of palladium and preventing further reduction of the aldehyde to a primary alcohol.


This selective reduction is known as the
\[ \boxed{Rosenmund reduction}. \]

In this reaction, an acyl chloride is treated with hydrogen gas over a palladium catalyst supported on barium sulfate:
\[ RCOCl+H_2 \xrightarrow{Pd/BaSO_4} RCHO+HCl. \]

Palladium is an efficient hydrogenation catalyst.

However, if ordinary palladium is used, the aldehyde formed would undergo further reduction to a primary alcohol.

To prevent this, palladium is deposited on barium sulfate.

Barium sulfate acts as a catalyst poison, decreasing the catalytic activity of palladium.

Additional catalyst poisons such as sulfur or quinoline are often employed to further suppress over-reduction.

As a result, the reduction stops at the aldehyde stage.

Therefore, Rosenmund reduction provides a convenient method for preparing aldehydes selectively from acid chlorides.


\textcolor{red{Step 4: Conclusion
The controlled reduction successfully stops at the aldehyde stage.



\textcolor{red{Final Answer:
\(R-COCl + H_2 \xrightarrow{Pd / BaSO_4} R-CHO + HCl\)

(Example: Benzoyl chloride to Benzaldehyde) Quick Tip: Rosenmund catalyst (\(Pd/BaSO_4\)) is specifically "poisoned" to stop reduction at the aldehyde stage.

\textcolor{red{Step 1: Concept
Disproportionation (self-oxidation-reduction) of specific aldehydes.

\textcolor{red{Step 2: Meaning
Only aldehydes lacking \(\alpha\)-hydrogens undergo this reaction.

\textcolor{red{Step 3: Analysis

When an aldehyde such as formaldehyde (\(HCHO\)) or benzaldehyde (\(C_6H_5CHO\)) is heated with concentrated aqueous or alcoholic alkali, one molecule is oxidized to the corresponding carboxylate salt while another molecule is simultaneously reduced to the corresponding primary alcohol.


This reaction is known as the
\[ \boxed{Cannizzaro reaction}. \]

It is exhibited by aldehydes that do not possess any \(\alpha\)-hydrogen atoms.

Typical examples include:
\[ HCHO \quadand\quad C_6H_5CHO. \]

The reaction is carried out using concentrated aqueous or alcoholic sodium hydroxide or potassium hydroxide.

During the reaction, one molecule of aldehyde undergoes oxidation to form the salt of the corresponding carboxylic acid.

Simultaneously, another molecule of aldehyde undergoes reduction to form the corresponding primary alcohol.

For benzaldehyde, the reaction is:
\[ 2C_6H_5CHO+NaOH \rightarrow C_6H_5COONa+C_6H_5CH_2OH. \]

Since oxidation and reduction occur simultaneously between two molecules of the same compound, the Cannizzaro reaction is an example of a
\[ \boxed{disproportionation (self-redox) reaction}. \]

Thus, aldehydes lacking \(\alpha\)-hydrogen atoms undergo Cannizzaro reaction to produce an alcohol and a carboxylate salt.


\textcolor{red{Step 4: Conclusion
This yields two different functional groups from one starting reactant.



\textcolor{red{Final Answer:
\(2 HCHO + conc. NaOH \xrightarrow{\Delta} CH_3OH (Methanol) + HCOONa (Sodium formate)\) Quick Tip: Cannizzaro = No \(\alpha\)-H + conc. Base \(\rightarrow\) Alcohol (reduction) + Acid salt (oxidation).

\textcolor{red{Step 1: Concept
\(\alpha\)-halogenation of aliphatic carboxylic acids.

\textcolor{red{Step 2: Meaning
Carboxylic acids possessing at least one \(\alpha\)-hydrogen are specifically halogenated at the \(\alpha\)-position.

\textcolor{red{Step 3: Analysis

A carboxylic acid is treated with chlorine or bromine in the presence of a small amount of red phosphorus, followed by hydrolysis. This selectively replaces an \(\alpha\)-hydrogen atom by a halogen atom to form an \(\alpha\)-halocarboxylic acid.


This reaction is known as the
\[ \boxed{Hell--Volhard--Zelinsky (HVZ) reaction}. \]

It is used for the halogenation of the carbon atom adjacent to the carboxyl group (the \(\alpha\)-carbon).

A carboxylic acid containing at least one \(\alpha\)-hydrogen is treated with chlorine or bromine in the presence of a small amount of red phosphorus.

Red phosphorus reacts with the halogen to generate phosphorus trihalide \((PCl_3\) or \(PBr_3)\) in situ, which converts the carboxylic acid into the corresponding acyl halide.

The acyl halide readily undergoes enolization, allowing halogenation at the \(\alpha\)-carbon.

Finally, hydrolysis converts the acyl halide back into the carboxylic acid.

Thus, an \(\alpha\)-hydrogen atom is replaced by a halogen atom:
\[ RCH_2COOH \xrightarrow{X_2,\;P RCHXCOOH, \]
where
\[ X=Cl\ or\ Br. \]

Therefore, the HVZ reaction provides an important method for preparing \(\alpha\)-halocarboxylic acids.


\textcolor{red{Step 4: Conclusion
The reaction highlights the unique reactivity of the carbon adjacent to the carboxyl group.



\textcolor{red{Final Answer:
\(R-CH_2-COOH + X_2 \xrightarrow{(i) Red P, (ii) H_2O} R-CH(X)-COOH + HX\) \quad (where \(X = Cl, Br\)) Quick Tip: HVZ Reaction targets exclusively the \(\alpha\)-carbon using Red P and a halogen.

\textcolor{red{Step 1: Concept
Classification and properties of carbohydrates.

\textcolor{red{Step 2: Meaning
Terminology defines specific carbohydrate mixtures and polymer sizes.

\textcolor{red{Step 3: Analysis

(i) Sucrose is naturally dextrorotatory (+).
When hydrolyzed, it breaks down into an equimolar mixture of D-(+)-glucose and D-(-)-fructose.
Because the laevorotation of fructose (\(-92.4^\circ\)) is stronger than the dextrorotation of glucose (\(+52.5^\circ\)),
the net rotation of the resulting mixture becomes laevorotatory.
This inversion of optical rotation gives the mixture its name.

(ii) Polysaccharides are complex macromolecular carbohydrates. Upon complete acid or enzymatic hydrolysis,
they break down into a very large number (hundreds or thousands) of monosaccharide monomer units.

\textcolor{red{Step 4: Conclusion
Precise definitions capture these chemical behaviors.



\textcolor{red{Final Answer:

(i) Invert Sugar: The equimolar mixture of glucose and fructose obtained by the hydrolysis of sucrose is called invert sugar because the process brings about an inversion in the sign of specific optical rotation (from dextro to laevo).

(ii) Polysaccharides: These are high molecular mass carbohydrates that yield a large number of monosaccharide units upon hydrolysis (e.g., starch, cellulose). Quick Tip: "Invert" refers to the flip in optical rotation sign (+ to -) during sucrose hydrolysis.

\textcolor{red{Step 1: Concept
Molecular shape and folding patterns of polypeptide chains.

\textcolor{red{Step 2: Meaning
Structural orientation strictly dictates physical properties like solubility.

\textcolor{red{Step 3: Analysis
In fibrous proteins, the polypeptide chains run parallel to each other
and are held tightly together by hydrogen bonds and disulfide bonds, forming long, thread-like structures.
These are robust and generally insoluble in water (e.g., keratin, myosin).
In globular proteins, the polypeptide chain folds intricately around itself,
burying hydrophobic groups inside and exposing hydrophilic groups, to yield a highly compact, spherical 3D shape.

\textcolor{red{Step 4: Conclusion
This spherical shape allows globular proteins to be soluble in water (e.g., insulin, albumin).



\textcolor{red{Final Answer:

1. Structure: Fibrous proteins consist of linear polypeptide chains running parallel to form thread-like structures. Globular proteins consist of polypeptide chains folded around themselves to form compact, spherical structures.

2. Solubility: Fibrous proteins are generally insoluble in water, whereas globular proteins are usually soluble in water. Quick Tip: Fibrous = Thread-like and insoluble (Hair/Keratin). Globular = Spherical and soluble (Blood/Hemoglobin).

\textcolor{red{Step 1: Concept
Energy storage polysaccharides in different biological systems.

\textcolor{red{Step 2: Meaning
Starch serves as the primary energy storage in plants, while glycogen serves the same role in animals.

\textcolor{red{Step 3: Analysis
Structurally, glycogen is a highly branched polymer of \(\alpha\)-D-glucose.
Its chemical structure is remarkably similar to amylopectin, which is the major highly branched component of plant starch.
Because its structure closely mirrors starch and it functions identically as the primary carbohydrate energy reserve stored in the animal body
(specifically in liver, muscles, and brain), it earns its common moniker.

\textcolor{red{Step 4: Conclusion
The structural and functional analogies justify the name.



\textcolor{red{Final Answer: Glycogen is known as animal starch because its structure is highly similar to amylopectin (a key component of plant starch), and it serves identically as the primary carbohydrate energy storage molecule in animals. Quick Tip: Glycogen is the animal equivalent of plant starch (amylopectin), just much more highly branched.

\textcolor{red{Step 1: Concept
Nutritional classification of amino acids for human biology.

\textcolor{red{Step 2: Meaning
The categorization depends purely on the body's internal biochemical synthesis capabilities.

\textcolor{red{Step 3: Analysis

Non-essential amino acids are those that the human body can synthesize in sufficient quantities, whereas essential amino acids cannot be synthesized by normal metabolic pathways and must be supplied through the diet.


Amino acids are classified into two groups based on the body's ability to synthesize them.

Non-essential amino acids are produced by the body from other metabolic intermediates.

Therefore, they do not necessarily need to be obtained from food.

Common examples include:
\[ \boxed{Glycine, Alanine, Aspartic acid, and Glutamic acid.} \]

Essential amino acids cannot be synthesized, or cannot be synthesized in sufficient quantities, by the human body.

Hence, they must be obtained from dietary sources such as milk, eggs, pulses, meat, fish, and soy products.

Common examples are:
\[ \boxed{Valine, Leucine, Lysine, Methionine, and Tryptophan.} \]

A deficiency of essential amino acids can impair protein synthesis, growth, tissue repair, and several important metabolic processes.


\textcolor{red{Step 4: Conclusion
Because they cannot be produced internally, essential amino acids must be strictly supplied through external dietary intake to maintain health (e.g., Valine, Leucine).



\textcolor{red{Final Answer:

1. Essential Amino Acids: These cannot be synthesized by the human body and must necessarily be obtained through the diet (e.g., Valine, Leucine).

2. Non-essential Amino Acids: These can be readily synthesized by the human body internally, so their dietary intake is not strictly mandatory (e.g., Glycine, Alanine). Quick Tip: "Essential" means it is essential to include them in your food because your body can't make them.

\textcolor{red{Step 1: Concept
Protein denaturation caused by extreme physical or chemical stresses (heat, pH changes).

\textcolor{red{Step 2: Meaning
Denaturation involves the disruption of the native, biologically active 3D conformation of a protein.

\textcolor{red{Step 3: Analysis
When a protein is denatured, the weak forces
(like hydrogen bonds, hydrophobic interactions, and disulfide bridges)
that stabilize its higher-order structures are broken.
This causes the complex globular structures to uncoil and the helices to unwind.
Consequently, the secondary and tertiary structures are completely destroyed,
resulting in a loss of biological activity.

\textcolor{red{Step 4: Conclusion
However, the robust covalent peptide bonds holding the specific sequence of amino acids together remain completely unaffected, meaning the primary structure is preserved intact.



\textcolor{red{Final Answer: During denaturation, the secondary and tertiary structures of the protein are destroyed (unfolded) due to the breaking of weak stabilizing bonds like hydrogen bonds. However, the primary structure (the sequence of amino acids linked by peptide bonds) remains completely intact and unaffected. Quick Tip: Denaturation destroys the 2D and 3D shape, but leaves the basic 1D chain (primary structure) untouched.

\textcolor{red{Step 1: Concept
Osmotic pressure (\(\pi\)) formula: \(\pi = iCRT\).

\textcolor{red{Step 2: Meaning
Express molarity \(C\) as \(\frac{w}{M \times V}\), giving \(\pi = \frac{i \times w \times R \times T}{M \times V}\).

\textcolor{red{Step 3: Analysis

Rearrange for mass (\(w\)): \(w = \frac{\pi \times M \times V}{i \times R \times T}\).

Given: \(\pi = 0.70~atm\), \(M = 111~g/mol\), \(V = 2.46~L\), \(i = 2.59\), \(R = 0.082\), \(T = 27 + 273 = 300~K\).
\(w = \frac{0.70 \times 111 \times 2.46}{2.59 \times 0.082 \times 300} = \frac{191.142}{63.714} = 3.0~g\).

\textcolor{red{Step 4: Conclusion
The required mass calculates exactly to 3.0 g.



\textcolor{red{Final Answer: \(3.0~g\) Quick Tip: Always convert Temperature to Kelvin (\(^\circ C + 273\)) before using the ideal gas constant \(R\).

\textcolor{red{Step 1: Concept
Movement of solvent across a semi-permeable membrane.

\textcolor{red{Step 2: Meaning
Water moves from an area of lower solute concentration (outside) to higher solute concentration (inside the raisin).

\textcolor{red{Step 3: Analysis

Water enters the raisin cells by osmosis, causing the cells to swell and increasing the size of the raisin.


A raisin contains a concentrated solution of sugars and other dissolved substances inside its cells.

When a raisin is placed in pure water or a dilute solution, the surrounding solution is hypotonic relative to the cell sap.

The cell membrane behaves as a selectively permeable membrane.

Water molecules therefore move into the raisin cells through the process of osmosis.

Osmosis is the movement of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semipermeable membrane.

As water enters the cells, the internal pressure (turgor pressure) increases.

Consequently, the cells expand and the raisin swells.

Thus, the increase in the size of the raisin is a direct consequence of endosmosis.



\textcolor{red{Step 4: Conclusion
This specific biological process is called endosmosis.



\textcolor{red{Final Answer: Endosmosis (or Osmosis) Quick Tip: "Endo" means inward. Water moving into a cell is Endosmosis.

\textcolor{red{Step 1: Concept
Measurement of molar masses of polymers and macromolecules.

\textcolor{red{Step 2: Meaning
Biomolecules are unstable at high temperatures and have poor solubility.

\textcolor{red{Step 3: Analysis

Osmotic pressure is usually measured at ordinary temperatures, depends upon the molarity of the solution, and can be determined accurately even for very dilute solutions.


Osmotic pressure is the minimum pressure that must be applied to a solution to prevent the flow of solvent through a semipermeable membrane.

It is represented by the symbol:
\[ \pi. \]

Osmotic pressure is related to concentration by the van't Hoff equation:
\[ \pi=iCRT, \]
where

\(i\) is the van't Hoff factor,
\(C\) is the molarity,
\(R\) is the universal gas constant,
\(T\) is the absolute temperature.


Unlike boiling point elevation and freezing point depression, osmotic pressure depends upon molarity rather than molality.

It can be measured conveniently at room temperature because no heating or cooling of the solution is required.

Even very dilute solutions produce measurable osmotic pressures.

Therefore, osmotic pressure is one of the most accurate colligative properties for determining the molar masses of high-molecular-mass substances such as proteins and polymers.


\textcolor{red{Step 4: Conclusion
These conditions make it the most suitable colligative property for large molecules.



\textcolor{red{Final Answer: It is measured at room temperature, relies on molarity instead of molality, and yields a large, measurable magnitude even for very dilute solutions. Quick Tip: Osmotic pressure is the best method for determining the molar mass of proteins and polymers.

\textcolor{red{Step 1: Concept
Reversing the natural direction of osmosis.

\textcolor{red{Step 2: Meaning
Applying an external pressure greater than the osmotic pressure to the solution side.

\textcolor{red{Step 3: Analysis

During reverse osmosis, pressure greater than the osmotic pressure is applied to seawater, forcing pure water to pass through a semipermeable membrane while dissolved salts remain behind.


In ordinary osmosis, water naturally flows from pure water to a concentrated salt solution through a semipermeable membrane.

Reverse osmosis is the reverse of this natural process.

A pressure greater than the osmotic pressure is applied to the concentrated solution, such as seawater.

This external pressure forces water molecules to move in the opposite direction.

The semipermeable membrane allows only water molecules to pass through while preventing dissolved salts and other impurities from crossing.

As a result, pure water is collected on one side of the membrane.

The dissolved salts remain in the concentrated solution.

Reverse osmosis is widely used for:

Desalination of seawater.
Purification of drinking water.
Industrial water treatment.



\textcolor{red{Step 4: Conclusion
This process is termed reverse osmosis.



\textcolor{red{Final Answer: Reverse Osmosis (RO) Quick Tip: Applied Pressure > Osmotic Pressure (\(\pi\)) triggers Reverse Osmosis.

\textcolor{red{Step 1: Concept
Reduction of amides to primary amines.

\textcolor{red{Step 2: Meaning
Lithium aluminium hydride (\(LiAlH_4\)) is a strong reducing agent that fully reduces the carbonyl group (\(>C=O\)) of an amide into a methylene group (\(-CH_2-\)).

\textcolor{red{Step 3: Analysis

Acetamide (\(CH_3CONH_2\)) contains two carbon atoms. Upon reduction with lithium aluminium hydride (\(LiAlH_4\)), followed by hydrolysis, the carbonyl oxygen is removed and the amide is converted into the corresponding primary amine without changing the carbon chain length.


Lithium aluminium hydride (\(LiAlH_4\)) is a powerful reducing agent.

It reduces amides into primary amines.

During the reduction, the carbonyl oxygen atom is completely removed.

The carbonyl carbon remains attached to the nitrogen atom, so the number of carbon atoms remains unchanged.

Acetamide undergoes the reaction:
\[ CH_3CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2NH_2. \]

The product formed is:
\[ \boxed{Ethanamine (Ethylamine)}. \]

Unlike the Hofmann bromamide reaction, this reduction does not shorten the carbon chain.

Therefore, reduction of amides with \(LiAlH_4\) provides a convenient method for preparing primary amines containing the same number of carbon atoms as the original amide.


\textcolor{red{Step 4: Conclusion
The resulting product is a primary amine containing two carbons.



\textcolor{red{Final Answer: \(CH_3CH_2NH_2\) (Ethanamine) Quick Tip: \(LiAlH_4\) reduction of an amide (\(-CONH_2\)) replaces the \(=O\) with two hydrogens to give an amine (\(-CH_2NH_2\)).

\textcolor{red{Step 1: Concept
Reduction of aromatic nitro compounds.

\textcolor{red{Step 2: Meaning
Metals in an acidic medium (such as Fe/HCl or Sn/HCl) act as excellent reducing agents to convert nitro groups (\(-NO_2\)) directly into primary amino groups (\(-NH_2\)).

\textcolor{red{Step 3: Analysis

The reactant is p-nitrotoluene. The mixture of iron and hydrochloric acid selectively reduces the nitro group to an amino group without affecting either the aromatic ring or the methyl substituent.


\textit{p-Nitrotoluene contains two substituents on the benzene ring:

A nitro group (\(-NO_2\)).
A methyl group (\(-CH_3\)).


Iron filings in the presence of hydrochloric acid act as a reducing system.

This reagent selectively reduces the nitro group through a series of intermediate steps to an amino group:
\[ -NO_2 \longrightarrow -NH_2. \]

The aromatic benzene ring remains unaffected under these reaction conditions.

The methyl group is also not altered during the reduction.

The overall reaction is:
\[ p-CH_3C_6H_4NO_2 \xrightarrow{Fe/HCl p-CH_3C_6H_4NH_2. \]

The product obtained is:
\[ \boxed{p-Toluidine (4-methylaniline)}. \]

Thus, iron and hydrochloric acid provide a convenient method for converting aromatic nitro compounds into the corresponding aromatic amines.


\textcolor{red{Step 4: Conclusion
The \(-NO_2\) group becomes an \(-NH_2\) group, yielding p-toluidine.



\textcolor{red{Final Answer: p-Toluidine (or 4-Methylaniline). The \(-NO_2\) group is replaced by an \(-NH_2\) group on the ring. Quick Tip: \(Fe + HCl\) is the preferred industrial reagent for reducing \(-NO_2\) to \(-NH_2\) because the \(FeCl_2\) formed gets hydrolyzed to release \(HCl\), requiring only a small initial amount of acid.

\textcolor{red{Step 1: Concept
Intermolecular forces and their effect on boiling points.

\textcolor{red{Step 2: Meaning
Stronger intermolecular forces (like hydrogen bonding) require more thermal energy to overcome, leading to higher boiling points.

\textcolor{red{Step 3: Analysis
Primary amines (\(R-NH_2\)) possess two hydrogen atoms directly bonded to the highly electronegative nitrogen atom.
This allows them to form extensive intermolecular hydrogen bonds with other primary amine molecules.
Tertiary amines (\(R_3N\)), however, lack any hydrogen atoms directly attached to the nitrogen,
completely preventing them from forming intermolecular hydrogen bonds with each other.

\textcolor{red{Step 4: Conclusion
The presence of strong hydrogen bonding network in primary amines dictates their higher boiling point.



\textcolor{red{Final Answer: Primary amines have two hydrogen atoms directly attached to the nitrogen, allowing them to form strong intermolecular hydrogen bonds. Tertiary amines have no hydrogen atoms attached to the nitrogen and cannot form intermolecular hydrogen bonds, resulting in a lower boiling point. Quick Tip: More N-H bonds = More hydrogen bonding = Higher boiling point. Order of boiling points: \(1^\circ > 2^\circ > 3^\circ\).

\textcolor{red{Step 1: Concept
Classification of amines based on the degree of nitrogen substitution.

\textcolor{red{Step 2: Meaning
A primary (\(1^\circ\)) amine has one carbon attached to nitrogen. A secondary (\(2^\circ\)) amine has two. A tertiary (\(3^\circ\)) amine has three.

\textcolor{red{Step 3: Analysis

(i) In N,N-dimethyl-2-naphthylamine, the nitrogen atom is bonded to three carbon atoms: two from the distinct methyl groups and one from the naphthyl ring. It has zero N-H bonds. Thus, it is a tertiary amine.

(ii) In 2-naphthylamine, the nitrogen atom is bonded to only one carbon atom (from the naphthyl ring) and has two N-H bonds. Thus, it is a primary amine.

\textcolor{red{Step 4: Conclusion
The number of C-N bonds directly defines the class of the amine.



\textcolor{red{Final Answer:

(i) Tertiary (\(3^\circ\)) amine

(ii) Primary (\(1^\circ\)) amine Quick Tip: Look strictly at the nitrogen atom: 2 Hydrogens = Primary. 1 Hydrogen = Secondary. 0 Hydrogens = Tertiary.

\textcolor{red{Step 1: Concept
Solubility of organic compounds in water is driven primarily by their ability to form hydrogen bonds with water molecules.

\textcolor{red{Step 2: Meaning
The stronger the hydrogen bonds formed with water, the greater the degree of solubility.

\textcolor{red{Step 3: Analysis
Butan-1-ol features a hydroxyl (\(-OH\)) group,
while butan-1-amine features an amino (\(-NH_2\)) group.
Oxygen is significantly more electronegative (3.5) than nitrogen (3.0).
Because of this higher electronegativity difference,
the \(O-H\) bond in alcohols is much more strongly polarized than the \(N-H\) bond in amines.

\textcolor{red{Step 4: Conclusion
This greater bond polarity enables alcohols to form much stronger hydrogen bonds with water molecules compared to amines of comparable molecular mass.



\textcolor{red{Final Answer: Butan-1-ol is more soluble in water. This is because oxygen is more electronegative than nitrogen, making the hydrogen bonds formed between the alcohol (\(-OH\)) and water stronger than those formed between the amine (\(-NH_2\)) and water. Quick Tip: Oxygen is more electronegative than Nitrogen \(\rightarrow\) Stronger polarity \(\rightarrow\) Stronger H-bonds \(\rightarrow\) Higher solubility for alcohols over amines.

\textcolor{red{Step 1: Concept
Molar conductivity (\(\Lambda_m\)) relates conductivity to concentration. Degree of dissociation (\(\alpha\)) relates \(\Lambda_m\) to limiting molar conductivity (\(\Lambda^\circ_m\)).

\textcolor{red{Step 2: Meaning
Formulas: \(\Lambda_m = \frac{\kappa \times 1000}{C}\) and \(\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}\).

\textcolor{red{Step 3: Analysis

1. Calculate \(\Lambda_m\):
\(\Lambda_m = \frac{1.06 \times 10^{-2} \times 1000}{0.1} = \frac{10.6}{0.1} = 106~S~cm^2~mol^{-1}\).

2. Calculate \(\Lambda^\circ_m\):
\(\Lambda^\circ_m (NaCl) = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-} = 50.1 + 76.5 = 126.6~S~cm^2~mol^{-1}\).

3. Calculate \(\alpha\):
\(\alpha = \frac{106}{126.6} = 0.837\).


\textcolor{red{Step 4: Conclusion
The values are derived directly from Kohlrausch's law and basic electrochemistry formulas.



\textcolor{red{Final Answer: Molar conductivity = \(106~S~cm^2~mol^{-1}\); Degree of dissociation (\(\alpha\)) = \(0.837\) (or \(83.7%\)). Quick Tip: Always ensure units are consistent (\(cm^3\) to \(L\) conversion requires the 1000 factor in the numerator).

\textcolor{red{Step 1: Concept
Cell operation principles and battery classifications.

\textcolor{red{Step 2: Meaning
Electrons flow from anode (oxidation) to cathode (reduction). Current flows in the opposite direction.

\textcolor{red{Step 3: Analysis
(i) Zn oxidizes to \(Zn^{2+}\) (anode), while \(Ag^+\) reduces to Ag (cathode).
Electrons flow from Zn to Ag. Therefore, conventional current flows from Ag to Zn.
(ii) Primary batteries cannot be recharged because their cell reactions are irreversible.
Secondary batteries can be recharged by passing current through them in the opposite direction.

\textcolor{red{Step 4: Conclusion
Directionality and reversibility characterize these electrochemical systems.



\textcolor{red{Final Answer:
(i) Current flows from Silver (Ag) to Zinc (Zn).
(ii) Primary batteries are non-rechargeable (irreversible cell reaction). Secondary batteries are rechargeable (reversible cell reaction). Quick Tip: Current always flows opposite to electron flow. Anode = Loss of electrons.

\textcolor{red{Step 1: Concept
Cell constant (\(G^*\)) remains constant for the same physical cell.

\textcolor{red{Step 2: Meaning
\(G^* = \kappa \times R\). Once found, use it to find the new \(\kappa = \frac{G^*}{R}\).

\textcolor{red{Step 3: Analysis

1. Find Cell Constant (\(G^*\)):
\(G^* = \kappa_1 \times R_1 = (1.29 \times 10^{-2}~S~cm^{-1}) \times 100~\Omega = 1.29~cm^{-1}\).

2. Find Conductivity (\(\kappa_2\)) for \(0.01~M\):
\(\kappa_2 = \frac{G^*}{R_2} = \frac{1.29}{300} = 0.0043 = 4.3 \times 10^{-3}~S~cm^{-1}\).

3. Find Molar Conductivity (\(\Lambda_m\)):
\(\Lambda_m = \frac{\kappa_2 \times 1000}{C_2} =
\frac{4.3 \times 10^{-3} \times 1000}{0.01} =
\frac{4.3}{0.01} = 430~S~cm^2~mol^{-1}\).

\textcolor{red{Step 4: Conclusion
Successive application of cell constant formulas provides the required values.



\textcolor{red{Final Answer: Conductivity = \(4.3 \times 10^{-3}~S~cm^{-1}\); Molar conductivity = \(430~S~cm^2~mol^{-1}\). Quick Tip: Cell constant (\(l/a\)) is a property of the physical cell and does not change when you change the solution inside it.

\textcolor{red{Step 1: Concept
Properties of commercial cells and fuel cells.

\textcolor{red{Step 2: Meaning
Fuel cells provide continuous energy, and mercury cells offer stable voltage.

\textcolor{red{Step 3: Analysis
(i) \(H_2-O_2\) fuel cells are highly efficient (around 70%) and are pollution-free (produce only water).
(ii) The overall reaction of a mercury cell involves only solid and liquid phases (\(Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l)\)).
It does not involve any ions in solution whose concentration can change during its lifetime.

\textcolor{red{Step 4: Conclusion
These specific characteristics make them highly advantageous for specialized applications.



\textcolor{red{Final Answer:
(i) Advantages: Highly efficient and completely pollution-free.
(ii) The overall cell reaction does not involve any aqueous ions whose concentration can change over time. Quick Tip: No aqueous ions in the overall net reaction = Constant voltage output over time.

\textcolor{red{Step 1: Concept
Reactions of alcohols and ethers.

\textcolor{red{Step 2: Meaning
Reactivity with Na indicates an alcohol. Iodoform test confirms a \(CH_3CH(OH)-\) group.

\textcolor{red{Step 3: Analysis

1. Formula \(C_2H_6O\) releasing \(H_2\) with Na is Ethanol ('A').

Reaction: \(2CH_3CH_2OH
(A) + 2Na \rightarrow 2CH_3CH_2ONa
(B) + H_2\). So 'B' is Sodium ethoxide.

2. Ethanol reacts with \(I_2/NaOH\) (iodoform reaction) to give a yellow precipitate of Iodoform ('C').

Reaction: \(CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3
(C) + HCOONa + 5NaI + 5H_2O\).

3. Ethanol heated with \(H_2SO_4\) at \(413~K\) undergoes intermolecular dehydration to form an ether ('D').

Reaction: \(2CH_3CH_2OH \xrightarrow{H_2SO_4, 413~K} C_2H_5-O-C_2H_5
(D) + H_2O\). 'D' is Diethyl ether (Ethoxyethane).

4. Cleavage of ether 'D' with excess HI yields alkyl iodides.

Reaction: \(C_2H_5-O-C_2H_5 + 2HI \rightarrow 2C_2H_5I
(E) + H_2O\). 'E' is Ethyl iodide (Iodoethane).

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



\textcolor{red{Final Answer:

A: Ethanol (\(CH_3CH_2OH\))

B: Sodium ethoxide (\(CH_3CH_2ONa\))

C: Iodoform (\(CHI_3\))

D: Diethyl ether or Ethoxyethane (\(C_2H_5-O-C_2H_5\))

E: Ethyl iodide or Iodoethane (\(C_2H_5I\))

Reactions are detailed in Step 3. Quick Tip: Temperature matters: \(H_2SO_4\) at \(413~K\) gives ethers, but at \(443~K\) gives alkenes.

\textcolor{red{Step 1: Concept
Selection of specific reagents for organic transformations.

\textcolor{red{Step 2: Meaning
Halogenation of phenols, anti-Markovnikov hydration, and ketone reduction.

\textcolor{red{Step 3: Analysis

(i) Aqueous bromine readily brominates all ortho/para positions of phenol due to strong activation.

(ii) Hydroboration-oxidation gives anti-Markovnikov alcohol from an alkene.

(iii) Weak or strong reducing agents convert ketones to secondary alcohols.


\textcolor{red{Step 4: Conclusion
The standard reagents for these conversions are uniquely suited for the desired products.



\textcolor{red{Final Answer:

(i) Bromine water (\(Br_2(aq)\))

(ii) \(B_2H_6\) followed by \(H_2O_2/OH^-\) (Hydroboration-Oxidation)

(iii) \(NaBH_4\) or \(LiAlH_4\) or \(H_2/Ni\) Quick Tip: For anti-Markovnikov hydration of an alkene to a primary alcohol, always use Hydroboration-Oxidation.

\textcolor{red{Step 1: Concept
Electrophilic addition reaction mechanism.

\textcolor{red{Step 2: Meaning
An alkene is converted to an alcohol following Markovnikov's rule.

\textcolor{red{Step 3: Analysis

Step 1: Protonation of alkene to form a carbocation by electrophilic attack of \(H_3O^+\).
\(>C=C< + H_3O^+ \rightleftharpoons >C^+-C(H)< + H_2O\).

Step 2: Nucleophilic attack of water on the carbocation.
\(>C^+-C(H)< + H_2O \rightleftharpoons >C(O^+H_2)-C(H)<\).

Step 3: Deprotonation to form an alcohol.
\(>C(O^+H_2)-C(H)< + H_2O \rightleftharpoons >C(OH)-C(H)< + H_3O^+\).

\textcolor{red{Step 4: Conclusion
The acid acts as a catalyst, consumed in step 1 and regenerated in step 3.



\textcolor{red{Final Answer:
Step 1: Protonation of alkene to form a carbocation.
Step 2: Nucleophilic attack of \(H_2O\) on the carbocation.
Step 3: Deprotonation to yield the alcohol. Quick Tip: Carbocations can rearrange during acid-catalyzed hydration to form a more stable intermediate.

\textcolor{red{Step 1: Concept
Reduction of a ketone followed by dehydration of the resulting alcohol.

\textcolor{red{Step 2: Meaning
Propanone is first converted to propan-2-ol, which is then dehydrated to form an alkene.

\textcolor{red{Step 3: Analysis

Sodium borohydride (\(NaBH_4\)) reduces propanone to propan-2-ol. Heating this secondary alcohol with concentrated sulfuric acid at \(443\,\mathrm{K}\) removes one molecule of water to produce propene.


Sodium borohydride (\(NaBH_4\)) is a mild reducing agent that readily reduces aldehydes and ketones to alcohols.

Propanone undergoes reduction according to:
\[ CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CHOHCH_3. \]

The product formed is:
\[ \boxed{Propan-2-ol}. \]

On heating propan-2-ol with concentrated sulfuric acid at
\[ 443\,\mathrm{K}, \]
dehydration takes place.

During dehydration, one molecule of water is eliminated:
\[ CH_3CHOHCH_3 \xrightarrow{conc. H_2SO_4,\;443\,K} CH_3CH=CH_2+H_2O. \]

The final product is:
\[ \boxed{Propene}. \]

Thus, the overall sequence converts propanone into propene through reduction followed by dehydration.


\textcolor{red{Step 4: Conclusion
The two-step chemical sequence completes the conversion.



\textcolor{red{Final Answer:
\(CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH_3 \xrightarrow{conc. H_2SO_4, \Delta} CH_3CH=CH_2\) Quick Tip: Ketone \(\rightarrow\) \(2^\circ\) Alcohol \(\rightarrow\) Alkene.

\textcolor{red{Step 1: Concept
Conversion of a carboxylic acid to an acid chloride, followed by Rosenmund reduction.

\textcolor{red{Step 2: Meaning
Direct reduction is difficult, so the acid is activated into a chloride first.

\textcolor{red{Step 3: Analysis

Benzoic acid reacts with thionyl chloride (\(SOCl_2\)) to form benzoyl chloride. The acid chloride is then selectively reduced with hydrogen over palladium supported on barium sulfate (\(Pd/BaSO_4\)) to give benzaldehyde.


Benzoic acid first reacts with thionyl chloride:
\[ C_6H_5COOH+SOCl_2 \rightarrow C_6H_5COCl+SO_2+HCl. \]

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

Benzoyl chloride is then subjected to catalytic hydrogenation.

Hydrogen gas is passed over palladium deposited on barium sulfate:
\[ C_6H_5COCl+H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO+HCl. \]

This reaction is called the
\[ \boxed{Rosenmund reduction}. \]

Barium sulfate acts as a catalyst poison, reducing the activity of palladium.

Consequently, the reduction stops at the aldehyde stage and does not proceed further to benzyl alcohol.

Hence, benzoic acid is converted into benzaldehyde through the sequence:
\[ \boxed{Benzoic acid \rightarrow Benzoyl chloride \rightarrow Benzaldehyde.} \]


\textcolor{red{Step 4: Conclusion
This specific catalytic reduction prevents over-reduction to an alcohol.



\textcolor{red{Final Answer:
\(C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl \xrightarrow{H_2, Pd/BaSO_4} C_6H_5CHO\) Quick Tip: Rosenmund catalyst (\(Pd/BaSO_4\)) selectively reduces acid chlorides to aldehydes.

\textcolor{red{Step 1: Concept
Friedel-Crafts acylation followed by electrophilic aromatic nitration.

\textcolor{red{Step 2: Meaning
An acetyl group must be attached first because it acts as a meta-directing group for the subsequent nitration.

\textcolor{red{Step 3: Analysis

Benzene reacts with acetyl chloride (\(CH_3COCl\)) in the presence of anhydrous aluminum chloride to produce acetophenone. The acetyl group is electron-withdrawing and meta-directing. Subsequent nitration with concentrated nitric acid and concentrated sulfuric acid gives m-nitroacetophenone.


Benzene undergoes Friedel--Crafts acylation with acetyl chloride in the presence of anhydrous aluminum chloride:
\[ C_6H_6+CH_3COCl \xrightarrow{AlCl_3 C_6H_5COCH_3. \]

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

The acetyl group
\[ (-COCH_3) \]
withdraws electron density from the benzene ring through both the inductive and resonance effects.

Consequently, it behaves as a deactivating and meta-directing substituent.

Nitration is carried out using a nitrating mixture of concentrated nitric acid and concentrated sulfuric acid.

The nitro group therefore enters predominantly at the meta position:
\[ C_6H_5COCH_3 \xrightarrow{HNO_3/H_2SO_4} m-NO_2C_6H_4COCH_3. \]

The major product obtained is:
\[ \boxed{m-Nitroacetophenone}. \]


\textcolor{red{Step 4: Conclusion
The sequence order ensures the correct substitution position.



\textcolor{red{Final Answer:
\(C_6H_6 \xrightarrow{CH_3COCl, anh. AlCl_3} C_6H_5COCH_3 \xrightarrow{conc. HNO_3 / conc. H_2SO_4} m-NO_2-C_6H_4-COCH_3\) Quick Tip: Always place the meta-directing group on the ring first to ensure the second group goes to the meta position.

\textcolor{red{Step 1: Concept
Nucleophilic addition reactivity depends on steric hindrance and inductive effects.

\textcolor{red{Step 2: Meaning
Less bulky groups and lower electron donation to the carbonyl carbon increase reactivity.

\textcolor{red{Step 3: Analysis

Aldehydes are generally more reactive toward nucleophilic addition than ketones because they experience less steric hindrance and possess a more electrophilic carbonyl carbon. Among ketones, di-tert-butyl ketone is the least reactive because of the extremely bulky \textit{tert-butyl groups.


Nucleophilic addition reactions occur at the positively polarized carbonyl carbon.

Aldehydes contain only one alkyl group attached to the carbonyl carbon.

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

Consequently, the carbonyl carbon in aldehydes is more electrophilic and reacts more readily with nucleophiles.

Acetaldehyde is therefore more reactive than ketones such as acetone.

Acetone contains two methyl groups, which reduce the electrophilic character of the carbonyl carbon.

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

These bulky groups severely hinder the approach of nucleophiles.

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


\textcolor{red{Step 4: Conclusion
The reactivity strictly follows the inverse of steric bulk.



\textcolor{red{Final Answer: Di-tert. butyl ketone \(<\) Acetone \(<\) Acetaldehyde Quick Tip: Higher steric hindrance = Lower nucleophilic reactivity.

\textcolor{red{Step 1: Concept
Structural requirement for Aldol condensation.

\textcolor{red{Step 2: Meaning
Only carbonyl compounds possessing at least one \(\alpha\)-hydrogen atom can undergo this reaction.

\textcolor{red{Step 3: Analysis

Formaldehyde (\(HCHO\)) and benzaldehyde (\(C_6H_5CHO\)) do not contain any \(\alpha\)-hydrogen atoms, whereas acetaldehyde (\(CH_3CHO\)) contains three \(\alpha\)-hydrogen atoms.


An \(\alpha\)-hydrogen is a hydrogen atom attached to the carbon atom adjacent to the carbonyl group.

Formaldehyde has the structure:
\[ HCHO. \]

Since there is no carbon atom adjacent to the carbonyl carbon, formaldehyde possesses:
\[ \boxed{0\ \alpha-hydrogen atoms.} \]

Benzaldehyde has the structure:
\[ C_6H_5CHO. \]

The carbon adjacent to the carbonyl group is part of the benzene ring and therefore carries no hydrogen atom.

Hence, benzaldehyde also contains:
\[ \boxed{0\ \alpha-hydrogen atoms.} \]

Acetaldehyde has the structure:
\[ CH_3CHO. \]

The methyl group adjacent to the carbonyl carbon contains three hydrogen atoms.

Therefore, acetaldehyde possesses:
\[ \boxed{3\ \alpha-hydrogen atoms.} \]

Consequently, acetaldehyde undergoes reactions requiring \(\alpha\)-hydrogen atoms, such as aldol condensation, whereas formaldehyde and benzaldehyde undergo the Cannizzaro reaction instead.


\textcolor{red{Step 4: Conclusion
Only acetaldehyde satisfies the essential structural requirement.



\textcolor{red{Final Answer: \(CH_3CHO\) (Acetaldehyde) Quick Tip: Aldol condensation strictly requires an \(\alpha\)-hydrogen.

\textcolor{red{Step 1: Concept
Chemical identification of methyl ketones.

\textcolor{red{Step 2: Meaning
The Iodoform test is specific for compounds containing a \(CH_3-CO-\) group.

\textcolor{red{Step 3: Analysis

Acetophenone (\(C_6H_5COCH_3\)) contains the characteristic methyl ketone group (\(-COCH_3\)) and therefore gives a positive iodoform test. Benzophenone (\(C_6H_5COC_6H_5\)) does not contain this group and consequently gives a negative iodoform test.


The iodoform test is used to identify compounds containing the functional group:
\[ \boxed{-COCH_3} \]
(methyl ketone group).

The test is carried out using iodine in the presence of aqueous sodium hydroxide:
\[ I_2/NaOH. \]

Acetophenone has the structure:
\[ C_6H_5COCH_3. \]

Since it contains a methyl group directly attached to the carbonyl carbon, it is a methyl ketone.

Therefore, acetophenone undergoes successive iodination of the methyl group followed by cleavage to produce iodoform:
\[ \boxed{CHI_3}. \]

Iodoform appears as a yellow crystalline precipitate with a characteristic antiseptic smell.

Benzophenone has the structure:
\[ C_6H_5COC_6H_5. \]

In benzophenone, the carbonyl carbon is attached to two phenyl groups and no methyl group is present.

Hence, benzophenone does not undergo the iodoform reaction and no yellow precipitate is formed.

Thus, the iodoform test can be used to distinguish acetophenone from benzophenone.


\textcolor{red{Step 4: Conclusion
The formation of a yellow precipitate effectively distinguishes the two.



\textcolor{red{Final Answer: Iodoform Test. Acetophenone gives a yellow precipitate of iodoform when treated with \(I_2\) and \(NaOH\). Benzophenone does not. Quick Tip: Iodoform test specifically identifies the \(-COCH_3\) (methyl ketone) functional group.

\textcolor{red{Step 1: Concept
Chemical distinction between an aldehyde and a ketone.

\textcolor{red{Step 2: Meaning
Tollen's reagent is a mild oxidizing agent that oxidizes aldehydes but not ketones.

\textcolor{red{Step 3: Analysis

Propanal, being an aldehyde, reduces Tollen's reagent to metallic silver and produces a bright silver mirror. Propanone, being a ketone, does not reduce Tollen's reagent and therefore shows no reaction.


Tollen's reagent is an ammoniacal solution of silver nitrate containing the complex ion:
\[ [Ag(NH_3)_2]^+. \]

It acts as a mild oxidizing agent and is commonly used to distinguish aldehydes from ketones.

Propanal has the structure:
\[ CH_3CH_2CHO. \]

The aldehyde group is readily oxidized to the corresponding carboxylate ion:
\[ CH_3CH_2CHO \longrightarrow CH_3CH_2COOH. \]

Simultaneously, silver ions are reduced to metallic silver:
\[ Ag^+ \longrightarrow Ag. \]

The deposited silver forms a bright reflective coating on the inner walls of the test tube, known as the
\[ \boxed{silver mirror.} \]

Propanone has the structure:
\[ CH_3COCH_3. \]

Since ketones are resistant to oxidation by mild oxidizing agents such as Tollen's reagent, propanone does not react.

Therefore:
\[ \boxed{Propanal gives a positive Tollen's test, whereas propanone gives a negative test.} \]


\textcolor{red{Step 4: Conclusion
The silver mirror provides visual confirmation.



\textcolor{red{Final Answer: Tollen's Test. Propanal forms a silver mirror when heated with Tollen's reagent. Propanone does not. Quick Tip: Tollen's test is the definitive test for distinguishing aldehydes from ketones.

\textcolor{red{Step 1: Concept
Positional identification of the carbonyl group.

\textcolor{red{Step 2: Meaning
The Iodoform test identifies the presence of a terminal methyl group attached to the carbonyl carbon.

\textcolor{red{Step 3: Analysis

Pentan-2-one (\(CH_3COCH_2CH_2CH_3\)) contains a methyl ketone group and therefore gives a positive iodoform test. Pentan-3-one (\(CH_3CH_2COCH_2CH_3\)) contains ethyl groups on both sides of the carbonyl carbon and gives a negative iodoform test.


The iodoform test is specific for compounds containing:
\[ \boxed{-COCH_3} \]
or compounds that can be oxidized to this group.

Pentan-2-one has the structure:
\[ CH_3COCH_2CH_2CH_3. \]

Since a methyl group is directly attached to the carbonyl carbon, pentan-2-one is a methyl ketone.

Consequently, it reacts with iodine and sodium hydroxide to form iodoform:
\[ CHI_3. \]

A yellow precipitate of iodoform is therefore obtained.

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

In this compound, the carbonyl carbon is bonded to two ethyl groups.

No methyl group is directly attached to the carbonyl carbon.

Therefore, pentan-3-one cannot undergo the iodoform reaction.

Hence:


Pentan-2-one gives a positive iodoform test,
whereas pentan-3-one gives a negative iodoform test.



\textcolor{red{Step 4: Conclusion
The Iodoform test distinguishes the structural isomers.



\textcolor{red{Final Answer: Iodoform Test. Pentan-2-one gives a yellow precipitate of iodoform with \(I_2\) and \(NaOH\). Pentan-3-one does not. Quick Tip: Only the "2-one" ketones are methyl ketones and thus respond positively to the Iodoform test.

\textcolor{red{Step 1: Concept
Inductive effect on acid strength.

\textcolor{red{Step 2: Meaning
Electron-withdrawing groups stabilize the conjugate base, increasing acidity.

\textcolor{red{Step 3: Analysis
Fluorine is highly electronegative and exerts a strong \(-I\) (electron-withdrawing) effect.
This effect disperses the negative charge on the carboxylate ion formed after donating a proton,
stabilizing it significantly compared to the acetate ion,
which is destabilized by the \(+I\) effect of the methyl group.

\textcolor{red{Step 4: Conclusion
Therefore, fluoroacetic acid is the stronger acid.



\textcolor{red{Final Answer: \(CH_2FCOOH\) is stronger. The highly electronegative fluorine atom exerts a strong \(-I\) effect, which stabilizes the resulting carboxylate conjugate base. Quick Tip: \(-I\) effect (halogens) increases acidity. \(+I\) effect (alkyl groups) decreases acidity.

\textcolor{red{Step 1: Concept
Magnitude of intermolecular forces determining boiling points.

\textcolor{red{Step 2: Meaning
Stronger intermolecular forces lead to higher boiling points.

\textcolor{red{Step 3: Analysis
Propane (\(CH_3CH_2CH_3\)) has only weak van der Waals forces.
Dimethyl ether (\(CH_3OCH_3\)) has weak dipole-dipole interactions.
Acetaldehyde (\(CH_3CHO\)) has stronger dipole-dipole interactions.
Ethanol (\(CH_3CH_2OH\)) forms strong intermolecular hydrogen bonds.

\textcolor{red{Step 4: Conclusion
The boiling points increase directly with the strength of these intermolecular forces.



\textcolor{red{Final Answer: \(CH_3CH_2CH_3 < CH_3OCH_3 < CH_3CHO < CH_3CH_2OH\) Quick Tip: IMF Strength: van der Waals \(<\) Weak Dipole \(<\) Strong Dipole \(<\) Hydrogen Bonding.