Step 1: Recall the crystal systems.

- Cubic: \(a = b = c\), \(\alpha = \beta = \gamma = 90^\circ\).

- Tetragonal: \(a = b \neq c\), \(\alpha = \beta = \gamma = 90^\circ\).

- Orthorhombic: \(a \neq b \neq c\), \(\alpha = \beta = \gamma = 90^\circ\).

- Monoclinic: \(a \neq b \neq c\), \(\alpha = \gamma = 90^\circ \neq \beta\).

- Triclinic: \(a \neq b \neq c\), \(\alpha \neq \beta \neq \gamma \neq 90^\circ\).

- Hexagonal: \(a = b \neq c\), \(\alpha = \beta = 90^\circ\), \(\gamma = 120^\circ\).


Step 2: Apply given condition.

The problem states: \(a \neq b \neq c\) and \(\alpha = \beta = \gamma = 90^\circ\).


Step 3: Match with system.

This is the definition of the orthorhombic system.


Final Answer: \[ \boxed{(D) Orthorhombic} \] Quick Tip: When all angles are 90° but edge lengths are unequal, the crystal system is orthorhombic.

Step 1: Recall the relationship between concentration units and temperature.

- Molarity (moles of solute per liter of solution) is affected by temperature because it is dependent on volume, which changes with temperature.

- Normality is similarly influenced by volume, meaning it also varies with temperature.

- Mass-volume percentage is based on the solution's volume, and thus, it too changes with temperature.

- Molality (moles of solute per kg of solvent) is dependent on mass, which remains constant with temperature, meaning molality does not change with temperature.


Step 2: Conclusion.

Thus, the concentration unit that remains unaffected by temperature is molality.


Final Answer: \[ \boxed{(C) Molality} \] Quick Tip: Always remember: molality is mass-based (temperature independent), while molarity is volume-based (temperature dependent).

Step 1: General formula for the units of the rate constant.

For a reaction of order \(n\): \[ Unit of rate constant (k) = mol^{1-n} \, L^{n-1} \, s^{-1} \]

Step 2: Apply this for a first-order reaction.

For \(n = 1\): \[ Unit of k = mol^{1-1} \, L^{1-1} \, s^{-1} = s^{-1} \]

Step 3: Conclusion.

Therefore, the unit of the rate constant for a first-order reaction is s\(^{-1}\).


Final Answer: \[ \boxed{s^{-1}} \] Quick Tip: Always use the formula \(Unit of k = mol^{1-n} \, L^{n-1} \, s^{-1}\) to determine units of rate constants.

Step 1: Recall the Cannizzaro reaction.

Aldehydes that lack \(\alpha\)-hydrogens undergo Cannizzaro's reaction in the presence of a strong base. In this reaction, one molecule of aldehyde is reduced to an alcohol, while another is oxidized to a carboxylic acid.


Step 2: Apply to benzaldehyde.

Benzaldehyde has no \(\alpha\)-hydrogens, so it undergoes Cannizzaro’s reaction: \[ 2C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa \]
In this process, one molecule is reduced to benzyl alcohol (C\(_6\)H\(_5\)CH\(_2\)OH), and the other forms sodium benzoate.


Step 3: Analyze the options.

- (A) Aldol condensation: This requires \(\alpha\)-hydrogens, which benzaldehyde does not have. Thus, it is not possible here.

- (B) Cannizzaro’s reaction: Correct, as it leads to the formation of benzyl alcohol.

- (C) Gattermann-Koch reaction: This reaction is used for formylation of benzene, which is not relevant in this case.

- (D) Clemmensen reduction: This reduces the carbonyl group to a methylene group, not yielding benzyl alcohol.


Step 4: Conclusion.

Therefore, benzyl alcohol is formed from benzaldehyde via Cannizzaro's reaction.


Final Answer: \[ \boxed{(B) Cannizzaro’s reaction} \] Quick Tip: In Cannizzaro’s reaction, aldehydes without \(\alpha\)-hydrogens undergo disproportionation to form alcohol and carboxylate salt.

Step 1: Recall Hinsberg’s reagent.

Hinsberg’s reagent is benzene sulphonyl chloride (C\(_6\)H\(_5\)SO\(_2\)Cl). It is used to distinguish between primary, secondary, and tertiary amines.


Step 2: Reasoning.

- Primary amines react with Hinsberg’s reagent to form soluble sulphonamides.

- Secondary amines form insoluble sulphonamides.

- Tertiary amines do not react.


Step 3: Analyzing options.

- (A) Benzene sulphonic acid: Incorrect.

- (B) Benzene sulphonyl chloride: Correct.

- (C) Benzene sulphonamide: This is a product, not the reagent.

- (D) Phenyl isocyanide: Related to carbylamine test, not Hinsberg’s reagent.


Step 4: Conclusion.

Hence, Hinsberg’s reagent is benzene sulphonyl chloride.


Final Answer: \[ \boxed{(B) Benzene sulphonyl chloride} \] Quick Tip: Hinsberg’s reagent (C\(_6\)H\(_5\)SO\(_2\)Cl) is used for distinguishing primary, secondary, and tertiary amines.

Step 1: Recall DNA bases.

DNA contains four nitrogenous bases: Adenine (A), Thymine (T), Guanine (G), and Cytosine (C).


Step 2: RNA bases.

In RNA, uracil (U) replaces thymine and pairs with adenine.


Step 3: Analyzing options.

- (A) Adenine: Present in DNA.

- (B) Thymine: Present in DNA.

- (C) Uracil: Not present in DNA, found in RNA instead.

- (D) Guanine: Present in DNA.


Step 4: Conclusion.

Thus, uracil is not present in DNA.


Final Answer: \[ \boxed{(C) Uracil} \] Quick Tip: DNA bases: A, T, G, C; RNA bases: A, U, G, C. Uracil replaces thymine in RNA.

Step 1: Definition.

Schottky defect is a type of point defect in ionic crystals where equal number of cations and anions are missing from their lattice sites, maintaining electrical neutrality.

Step 2: Effect on density.

Since some ions are missing from the lattice, the mass of the unit cell decreases but the volume remains the same. Hence, the density of the lattice decreases.


Conclusion:

Schottky defect lowers the density of the crystal without disturbing its electrical neutrality. Quick Tip: Schottky defect is common in highly ionic crystals like NaCl, KCl, CsCl, and AgBr.

Step 1: Definition.

Osmotic pressure is the excess pressure applied to a solution to prevent the inward flow of solvent molecules across a semipermeable membrane.


Step 2: Expression for osmotic pressure.
\[ \pi = CRT \]
where, \(\pi\) = osmotic pressure, \(C\) = molar concentration of solute, \(R\) = gas constant, \(T\) = absolute temperature.


Step 3: Proof as a colligative property.

From the equation, osmotic pressure depends only on the number of solute particles (\(C\)) in a given volume of solution, and not on their nature.
Therefore, it is a colligative property.


Conclusion:

Osmotic pressure is directly proportional to solute concentration and depends only on the number of solute particles, hence it is a colligative property. Quick Tip: Colligative properties (vapour pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure) depend only on the number of solute particles, not their type.

Step 1: Statement of Kohlrausch’s law.

Kohlrausch’s law states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the contributions of its individual ions. Each ion migrates independently of the other.

\[ \Lambda_m^\infty = \lambda^\infty_+ + \lambda^\infty_- \]

where,
\(\Lambda_m^\infty\) = molar conductivity at infinite dilution
\(\lambda^\infty_+\) = limiting molar conductivity of cation
\(\lambda^\infty_-\) = limiting molar conductivity of anion


Step 2: Application.

It is used to calculate the molar conductivity of weak electrolytes (like acetic acid) at infinite dilution by using values of strong electrolytes. Example: \[ \Lambda_m^\infty (CH_3COOH) = \Lambda_m^\infty (CH_3COONa) + \Lambda_m^\infty (HCl) - \Lambda_m^\infty (NaCl) \]


Conclusion:

Kohlrausch’s law helps in determining the degree of dissociation and dissociation constant of weak electrolytes. Quick Tip: At infinite dilution, ions migrate independently, hence their conductivities can be simply added.

Step 1: Definition.

- Lyophilic colloids: Colloids in which the dispersed phase has high affinity for the dispersion medium. They are stable and reversible.

Example: Starch sol, Gum, Gelatin.


- Lyophobic colloids: Colloids in which the dispersed phase has little or no affinity for the dispersion medium. They are unstable and irreversible.

Example: Gold sol, Sulphur sol, As\(_2\)S\(_3\) sol.


Step 2: Reason for easy coagulation of lyophobic colloids.

Lyophobic sols are not strongly solvated and lack strong interactions with the solvent. They are stabilized mainly by charges on their particles. When an electrolyte is added, these charges are neutralized, leading to easy coagulation (precipitation).



Conclusion:

Lyophilic sols are stable due to strong solvation, while lyophobic sols are easily coagulated because they rely only on electrostatic stabilization. Quick Tip: Lyophilic = solvent-loving (stable); Lyophobic = solvent-hating (easily coagulated by electrolytes).

Step 1: Relation between atomic radius and edge length in BCC.

In a body centred cubic structure, the body diagonal passes through 2 radii from corner atoms and 1 radius from the central atom. \[ \sqrt{3}a = 4r \] \[ a = \frac{4r}{\sqrt{3}} \]

Step 2: Volume of unit cell.
\[ V_{cell} = a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}} \]

Step 3: Number of atoms in BCC unit cell.

In BCC:
- 8 corner atoms contribute = \( \tfrac{1}{8} \times 8 = 1\) atom

- 1 body-centred atom = 1 atom

Total = 2 atoms per unit cell.

Step 4: Volume occupied by atoms.
\[ V_{atoms} = 2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3 \]

Step 5: Packing efficiency.
\[ Packing Efficiency = \frac{V_{atoms}}{V_{cell}} \times 100 \] \[ = \frac{\tfrac{8}{3}\pi r^3}{\tfrac{64r^3}{3\sqrt{3}}} \times 100 \] \[ = \frac{\pi \sqrt{3}}{8} \times 100 = 68% \]


Conclusion:

The packing efficiency of a BCC unit cell is \(\boxed{68%}\). Quick Tip: BCC packing efficiency (68%) is less than FCC (74%) but greater than simple cubic (52%).

Step 1: Nature of intermolecular forces.

- In \(H_2O\), due to the high electronegativity of oxygen, strong intermolecular hydrogen bonding is present.

- In \(H_2S\), sulphur is less electronegative, and hence hydrogen bonding is absent. Only weak van der Waals forces are present.


Step 2: Effect of intermolecular forces.

- Strong hydrogen bonding in \(H_2O\) holds the molecules together, giving it a liquid state at room temperature.

- Weak forces in \(H_2S\) cannot hold molecules strongly, so it exists as a gas at room temperature.



Conclusion:
\(H_2O\) is liquid due to strong hydrogen bonding, while \(H_2S\) is gas due to weak van der Waals forces. Quick Tip: Hydrogen bonding significantly increases boiling points of molecules like \(H_2O\), \(HF\), and \(NH_3\).

(i) Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride:

- Ligands: Two chloride ions (Cl\(^-\)) and two ethane-1,2-diamine (en, neutral).

- Central metal: Cobalt in +3 oxidation state.

- Outside the coordination sphere: Chloride counter ions.

Formula: \[ [CoCl_2(en)_2]Cl \]



(ii) Iron(III) hexacyanidoferrate(II):

- The compound contains two types of metal centers.

- Iron(II) is coordinated with six cyanide ligands to form \([Fe(CN)_6]^{4-}\).

- Iron(III) balances the charge.

Formula: \[ Fe_4[Fe(CN)_6]_3 \]


Conclusion:

According to IUPAC rules, the systematic formulas are:
1. \([CoCl_2(en)_2]Cl\)
2. \(Fe_4[Fe(CN)_6]_3\) Quick Tip: Always write ligands inside square brackets, metals with oxidation states in Roman numerals, and counter ions outside the coordination sphere.

Step 1: Primary structure.

- The primary structure of proteins refers to the linear sequence of amino acids in a polypeptide chain, joined by peptide bonds.
- This sequence determines the unique characteristics and function of the protein.
- Even a single change in the amino acid sequence can alter protein activity (e.g., sickle-cell anemia caused by a single amino acid substitution in hemoglobin).


Step 2: Secondary structure.

- The secondary structure refers to the regular folding or coiling of the polypeptide chain due to hydrogen bonding between the C=O and N–H groups of peptide bonds.
- Two major forms are:

Alpha-helix: Right-handed coiled structure stabilized by intramolecular hydrogen bonds.
Beta-pleated sheet: Zig-zag structure where polypeptide chains lie side by side, stabilized by intermolecular hydrogen bonds.

- These structures provide stability and define the shape of proteins.


Conclusion:

The primary structure gives the amino acid sequence, while the secondary structure defines local folding patterns such as alpha-helices and beta-sheets, both crucial for protein function. Quick Tip: Remember: Primary = sequence of amino acids, Secondary = hydrogen-bonded folding (alpha-helix, beta-sheet).

Step 1: Write the cell reaction.
\[ Ni (s) + Cu^{2+} (aq) \;\longrightarrow\; Ni^{2+} (aq) + Cu (s) \]


Step 2: Nernst equation.

General Nernst equation: \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
where, \(n = 2\) (electrons transferred)
\(Q = \dfrac{[Ni^{2+}]}{[Cu^{2+}]} = \dfrac{0.01}{0.1} = 0.1\)


Step 3: Substitution.
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log (0.1) \] \[ E_{cell} = E^\circ_{cell} - 0.02955 \times (-1) \] \[ E_{cell} = E^\circ_{cell} + 0.02955 \]


Step 4: Calculate \(E^\circ_{cell}\).

Given: \(E_{cell} = 0.59 \, V\) \[ 0.59 = E^\circ_{cell} + 0.02955 \] \[ E^\circ_{cell} = 0.59 - 0.02955 \] \[ E^\circ_{cell} \approx 0.56 \, V \]


Conclusion:

The standard emf of the cell is: \[ \boxed{0.56 \, V} \] Quick Tip: Remember: \(E_{cell}\) decreases with increasing concentration of products and increases with increasing concentration of reactants.

Step 1: Definition.

Coagulation is the process of aggregation of colloidal particles, resulting in the settling down (precipitation) of colloids. It occurs when the stability of the colloidal system is disturbed.


Step 2: Causes of coagulation.


By addition of electrolytes: Electrolyte ions neutralize the charges on colloidal particles, causing them to aggregate.
By mutual coagulation: Mixing two oppositely charged sols can neutralize charges, causing precipitation.
By boiling: Heating increases particle collisions and reduces stability.
By electrophoresis: Charged particles move towards the oppositely charged electrode and get coagulated.



Conclusion:

Coagulation is the destruction of colloidal stability, leading to precipitation of particles. Quick Tip: Lyophobic colloids are easily coagulated by electrolytes due to weak solvation.

Step 1: Structure of soap molecules.

Soap molecules consist of two parts:
- A long non-polar hydrophobic tail (hydrocarbon chain).
- A polar hydrophilic head (–COO\(^-\) group).


Step 2: Micelle formation.

When soap is added to water, molecules arrange themselves into spherical aggregates called micelles. The hydrophobic tails remain inside, while the hydrophilic heads face water. This allows soap to trap grease and oil.


Step 3: Emulsification.

Grease and oil droplets are surrounded by soap micelles, converting them into small emulsified particles that can be washed away with water.



Conclusion:

The cleansing action of soap is due to micelle formation and emulsification, which removes dirt and oil from clothes and skin. Quick Tip: Soaps and detergents work by emulsifying oily dirt into water-soluble micelles.

Step 1: Gabriel’s Phthalimide Synthesis.

Gabriel’s phthalimide synthesis is a method for preparing primary amines from phthalimide. The reaction involves:
1. Phthalimide is treated with alcoholic KOH to form potassium phthalimide (nucleophile).

2. Potassium phthalimide undergoes nucleophilic substitution with alkyl halide (\(R{-}X\)) to form N-alkylphthalimide.

3. Hydrolysis of N-alkylphthalimide gives a primary amine.
\[ Phthalimide \xrightarrow{KOH} Potassium phthalimide \xrightarrow{R{-}X} N-alkylphthalimide \xrightarrow{Hydrolysis} R{-}NH_2 \]

Step 2: Why preferred for primary amines.

Gabriel’s synthesis gives only primary amines and avoids the formation of secondary and tertiary amines, which are formed in other methods like ammonolysis of alkyl halides. Hence, it is highly selective.


Conclusion:

Gabriel’s phthalimide synthesis is a reliable method to obtain pure primary amines. Quick Tip: Always remember: Gabriel’s synthesis cannot be used for preparing aryl amines due to the inert nature of aryl halides towards nucleophilic substitution.

Step 1: D and L Configurations.

The D and L notations of carbohydrates are based on the configuration of the asymmetric carbon atom farthest from the carbonyl group (the penultimate carbon).

- If the OH group on the penultimate carbon is on the right → D-configuration.

- If the OH group on the penultimate carbon is on the left → L-configuration.


Step 2: Example.

- D-Glucose and D-Fructose both belong to the D-series, meaning that the OH group on the penultimate carbon is on the right-hand side in their Fischer projections.

Step 3: Structures.


D-Glucose (Fischer projection): \[ \begin{array}{c} CHO
[-4pt] |
[-4pt] H - C - OH
[-4pt] |
[-4pt] HO - C - H
[-4pt] |
[-4pt] H - C - OH
[-4pt] |
[-4pt] H - C - OH
[-4pt] |
[-4pt] CH_2OH \end{array} \]

D-Fructose (Fischer projection): \[ \begin{array}{c} CH_2OH
[-4pt] |
[-4pt] C{=}O
[-4pt] |
[-4pt] HO - C - H
[-4pt] |
[-4pt] H - C - OH
[-4pt] |
[-4pt] H - C - OH
[-4pt] |
[-4pt] CH_2OH \end{array} \]


Conclusion:

D and L configurations are based on the arrangement of the OH group on the penultimate carbon atom, and both D-Glucose and D-Fructose belong to the D-series. Quick Tip: Do not confuse D/L notation with optical activity (+ or –). D and L are about configuration, not about dextrorotation or levorotation.

Given:

10% (w/w) solution \(\implies\) 10 g glucose in 100 g solution.

Mass of solution = 100 g.

Mass of solute (glucose) = 10 g.

Mass of solvent (water) = 90 g = 0.090 kg.

Molar mass of glucose = 180 g/mol.

Density = 1.20 g/mL = 1200 g/L.



(i) Molality:
\[ Moles of glucose = \frac{10}{180} = 0.0556 \, mol \] \[ Molality = \frac{Moles of solute}{Mass of solvent (kg)} = \frac{0.0556}{0.090} = 0.617 \, mol/kg \]


(ii) Molarity:
\[ Volume of solution = \frac{Mass of solution}{Density} = \frac{100}{1200} = 0.0833 \, L \] \[ Molarity = \frac{0.0556}{0.0833} = 0.667 \, M \]


(iii) Mole Fraction:
\[ Moles of water = \frac{90}{18} = 5.0 \, mol \] \[ Total moles = 0.0556 + 5.0 = 5.0556 \] \[ X_{glucose} = \frac{0.0556}{5.0556} = 0.011, \quad X_{water} = \frac{5.0}{5.0556} = 0.989 \]


Final Answer:

Molality = 0.617 mol/kg

Molarity = 0.667 M

Mole fraction of glucose = 0.011

Mole fraction of water = 0.989
Quick Tip: Always remember: w/w percentage is based on 100 g of solution. Use density to convert mass into volume when calculating molarity.

Step 1: Definition.

The half-life (\(t_{1/2}\)) of a reaction is the time required for the concentration of the reactant to become half of its initial concentration.


Step 2: Integrated rate law for first order reaction.

For a first order reaction: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \]


Step 3: Apply half-life condition.

At \(t = t_{1/2}\), \([R] = \frac{[R]_0}{2}\). \[ k = \frac{2.303}{t_{1/2}} \log \frac{[R]_0}{[R]_0/2} \] \[ k = \frac{2.303}{t_{1/2}} \log 2 \]


Step 4: Simplification.

Since \(\log 2 = 0.3010\), \[ k = \frac{2.303 \times 0.3010}{t_{1/2}} \] \[ k = \frac{0.693}{t_{1/2}} \] \[ t_{1/2} = \frac{0.693}{k} \]


Step 5: Independence of concentration.

From the equation, \(t_{1/2}\) depends only on the rate constant \(k\) and not on the initial concentration \([R]_0\). Thus, for first-order reactions, half-life is independent of concentration.


Conclusion:

The half-life of a first order reaction is given by: \[ \boxed{t_{1/2} = \dfrac{0.693}{k}} \]
It is independent of initial concentration. Quick Tip: First-order reactions always have constant half-life, while zero and second order reactions depend on initial concentration.

Step 1: Formula.
\[ t_{1/2} = \frac{0.693}{k} \]


Step 2: Substitution.
\[ t_{1/2} = \frac{0.693}{6.93 \times 10^{-3}} \] \[ t_{1/2} = 100 \, min \]


Conclusion:

The half-life of the given reaction is: \[ \boxed{100 \, minutes} \] Quick Tip: For first-order kinetics: higher \(k\) → shorter half-life, lower \(k\) → longer half-life.

Step 1: Electronic configuration.

- Fe atom: [Ar] 3d\(^6\) 4s\(^2\)

- Fe\(^{2+}\): [Ar] 3d\(^6\)

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


Step 2: Count unpaired electrons.

- Fe\(^{2+}\) (3d\(^6\)): 4 unpaired electrons.

- Fe\(^{3+}\) (3d\(^5\)): 5 unpaired electrons.


Step 3: Paramagnetism.

Paramagnetism depends on the number of unpaired electrons. More unpaired electrons → higher paramagnetism.



Conclusion:

Fe\(^{3+}\) is more paramagnetic than Fe\(^{2+}\) because it has 5 unpaired electrons compared to 4 in Fe\(^{2+}\). Quick Tip: Maximum paramagnetism occurs when d-orbitals are half-filled (d\(^5\)).

Step 1: Electronic configuration.

- Zn\(^{2+}\): [Ar] 3d\(^{10}\) (completely filled d-orbitals).

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


Step 2: Colour and d–d transitions.

- In Zn\(^{2+}\), no d–d electronic transitions are possible because d-orbitals are completely filled. Hence, Zn\(^{2+}\) salts are colourless.

- In Ni\(^{2+}\), d–d electronic transitions are possible due to partially filled 3d orbitals. Absorption of visible light during these transitions imparts colour to Ni\(^{2+}\) salts.



Conclusion:

Zn\(^{2+}\) salts are colourless (d\(^{10}\), no transitions) while Ni\(^{2+}\) salts are coloured (d\(^8\), d–d transitions possible). Quick Tip: In transition metals, colour is due to d–d electronic transitions; fully filled d-orbitals give colourless compounds.

Step 1: Introduction.

Alfred Werner, in 1893, proposed a theory to explain the structure and bonding in coordination compounds. His postulates laid the foundation of coordination chemistry.


Step 2: Werner’s Postulates.

1. Dual valencies:

Every metal atom exhibits two types of valencies:
- Primary valency (ionizable): These correspond to the oxidation state of the metal and are satisfied by negative ions. They are ionizable and represented by dotted lines.

- Secondary valency (non-ionizable): These correspond to the coordination number and are satisfied by ligands. They are non-ionizable and represented by solid lines.


2. Fixed coordination number:

Each metal has a fixed number of secondary valencies (coordination number). For example, Co(III) has coordination number 6.


3. Geometry of complexes:

The secondary valencies are directed towards fixed positions in space, leading to definite geometrical shapes (octahedral, tetrahedral, square planar, etc.).


4. Satisfaction of both valencies:

A complex compound is formed when both primary and secondary valencies of a metal are satisfied. Example: \([Co(NH_3)_6]Cl_3\). Here, Co has:
- Primary valency = 3 (satisfied by 3Cl\(^-\))

- Secondary valency = 6 (satisfied by 6 NH\(_3\))


Conclusion:

Werner’s postulates explain the ionizable and non-ionizable nature of ligands, coordination number, and geometry of coordination compounds, forming the basis of modern coordination chemistry. Quick Tip: Remember: Primary valencies = oxidation state (ionizable), Secondary valencies = coordination number (non-ionizable).

Ostwald’s Process for Nitric Acid Manufacture:

The Ostwald process is the commercial method for producing nitric acid. It involves the catalytic oxidation of ammonia in three main steps:

1. Oxidation of ammonia to nitric oxide: \[ 4NH_3 + 5O_2 \; \xrightarrow{Pt/Rh, \; 500K, \; 9 atm} \; 4NO + 6H_2O \]

2. Oxidation of nitric oxide to nitrogen dioxide: \[ 2NO + O_2 \; \longrightarrow \; 2NO_2 \]

3. Absorption of nitrogen dioxide in water to form nitric acid: \[ 4NO_2 + 2H_2O + O_2 \; \longrightarrow \; 4HNO_3 \]

Thus, nitric acid is obtained industrially using the Ostwald process.



(i) Reaction of Zinc metal with conc. HNO\(_3\):

Concentrated nitric acid is an oxidizing acid. With zinc, it forms zinc nitrate and nitrogen dioxide.
\[ Zn + 4HNO_3(conc.) \; \longrightarrow \; Zn(NO_3)_2 + 2NO_2 + 2H_2O \]



(ii) Reaction of Iodine with conc. HNO\(_3\):

Concentrated nitric acid oxidizes iodine to iodic acid.
\[ I_2 + 10HNO_3(conc.) \; \longrightarrow \; 2HIO_3 + 10NO_2 + 4H_2O \]


Conclusion:

- The Ostwald process is the standard industrial route for nitric acid manufacture.

- Conc. HNO\(_3\) acts as a strong oxidizing agent, oxidizing both zinc and iodine to their higher oxidation states while releasing NO\(_2\). Quick Tip: Remember: In the Ostwald process, NH\(_3\) → NO → NO\(_2\) → HNO\(_3\). Also, conc. HNO\(_3\) behaves as an oxidizing acid, producing NO\(_2\) gas in most reactions.

Step 1: Bleaching action of chlorine.

Chlorine reacts with water to form nascent oxygen: \[ Cl_2 + H_2O \;\longrightarrow\; HCl + [O] \]
The nascent oxygen oxidises the coloured substances to colourless products by chemical change. Since oxidation is permanent, the bleaching action of chlorine is permanent.


Step 2: Bleaching action of SO\(_2\).

Sulphur dioxide bleaches substances by reduction in presence of moisture: \[ SO_2 + 2H_2O \;\longrightarrow\; H_2SO_4 + 2[H] \]
The nascent hydrogen reduces coloured substances to colourless forms. On exposure to air, the substances regain colour due to re-oxidation. Hence, bleaching action of SO\(_2\) is temporary.


Conclusion:

Chlorine bleaches by oxidation (permanent), while SO\(_2\) bleaches by reduction (temporary). Quick Tip: Oxidation bleaching (Cl\(_2\)) is permanent, while reduction bleaching (SO\(_2\)) is reversible.

Step 1: Structure of NH\(_3\).

Ammonia has a lone pair of electrons on nitrogen atom.


Step 2: Lewis base behaviour.

Due to this lone pair, NH\(_3\) can donate electrons to electron-deficient species (protons or metal ions). Hence, it acts as a Lewis base.


Step 3: Bronsted-Lowry base behaviour.

According to Bronsted-Lowry concept, a base is a proton acceptor. NH\(_3\) readily accepts a proton to form ammonium ion: \[ NH_3 + H^+ \;\longrightarrow\; NH_4^+ \]


Conclusion:

Ammonia is basic in nature because nitrogen atom has a lone pair which can accept protons, showing both Bronsted-Lowry and Lewis base properties. Quick Tip: The basic nature of NH\(_3\) is due to the lone pair of nitrogen, which readily accepts protons to form NH\(_4^+\).

Step 1: Nature of reaction.

Sulphur dioxide (SO\(_2\)) acts as a strong reducing agent, while potassium permanganate (KMnO\(_4\)) in acidic medium acts as a strong oxidising agent. When they react, SO\(_2\) is oxidised to sulphuric acid (H\(_2\)SO\(_4\)) and KMnO\(_4\) is reduced to Mn\(^{2+}\).


Step 2: Balanced redox reaction.

In acidic medium: \[ 2MnO_4^- + 5SO_2 + 2H_2O \;\longrightarrow\; 2Mn^{2+} + 5SO_4^{2-} + 4H^+ \]


Step 3: Reaction with potassium permanganate.

Since KMnO\(_4\) provides MnO\(_4^-\) ions, the reaction is: \[ 2KMnO_4 + 5SO_2 + 2H_2O \;\longrightarrow\; K_2SO_4 + 2MnSO_4 + 2H_2SO_4 \]


Observation.

The purple colour of KMnO\(_4\) solution is discharged, showing that reduction of MnO\(_4^-\) to Mn\(^{2+}\) has taken place.


Conclusion:

SO\(_2\) reduces KMnO\(_4\) in acidic medium, forming MnSO\(_4\), K\(_2\)SO\(_4\), and H\(_2\)SO\(_4\), while the pink colour of permanganate disappears. Quick Tip: Always balance redox reactions in acidic medium by adding H\(_2\)O, H\(^+\), and electrons properly. KMnO\(_4\) turns colourless as Mn\(^{2+}\) is formed.

Dehydration of ethanol occurs in the presence of concentrated sulphuric acid (H\(_2\)SO\(_4\)) at 443–463 K. The mechanism involves the following steps:


Step 1: Protonation of ethanol.

Ethanol gets protonated by H\(^+\) from conc. H\(_2\)SO\(_4\): \[ CH_3CH_2OH + H^+ \;\longrightarrow\; CH_3CH_2OH_2^+ \]


Step 2: Formation of carbocation.

The protonated alcohol loses a water molecule to form an ethyl carbocation: \[ CH_3CH_2OH_2^+ \;\longrightarrow\; CH_3CH_2^+ + H_2O \]


Step 3: Deprotonation to form alkene.

The ethyl carbocation loses a proton (H\(^+\)) to give ethene: \[ CH_3CH_2^+ \;\longrightarrow\; CH_2=CH_2 + H^+ \]


Overall reaction.
\[ CH_3CH_2OH \;\xrightarrow{conc. H_2SO_4, \, 443-463K}\; CH_2=CH_2 + H_2O \]


Conclusion:

The dehydration of ethanol is an acid-catalysed elimination reaction (E1 mechanism), leading to the formation of ethene. Quick Tip: At 443 K, ethanol undergoes dehydration to form ethene. At 413 K, it forms diethyl ether instead. Temperature controls the product.

(A) Reimer–Tiemann Reaction:

The Reimer–Tiemann reaction is used to introduce a formyl group (\(-CHO\)) at the ortho position of phenol. Phenol reacts with chloroform (CHCl\(_3\)) and sodium hydroxide (NaOH) to give salicylaldehyde.
\[ C_6H_5OH + CHCl_3 + 3NaOH \;\; \longrightarrow \;\; o\!-\!HOC_6H_4CHO + 3NaCl + 2H_2O \]



(B) Kolbe’s Reaction:

The Kolbe reaction is the carboxylation of phenol to produce salicylic acid. Sodium phenoxide reacts with carbon dioxide under high temperature and pressure, followed by acidification.
\[ C_6H_5ONa + CO_2 \;\; \xrightarrow{373K, \; 4-7 atm} \;\; o\!-\!HOC_6H_4COONa \] \[ o\!-\!HOC_6H_4COONa + HCl \;\; \longrightarrow \;\; o\!-\!HOC_6H_4COOH \]
This method is important for synthesizing salicylic acid (a precursor of aspirin).



(C) Williamson Ether Synthesis:

This is a laboratory method for preparing ethers. An alkoxide ion reacts with a primary alkyl halide in a nucleophilic substitution reaction to produce ether.
\[ R\!-\!ONa + R'X \;\; \longrightarrow \;\; R\!-\!O\!-\!R' + NaX \]
Example: \[ C_2H_5ONa + CH_3I \;\; \longrightarrow \;\; C_2H_5OCH_3 + NaI \]
This reaction is widely used for the synthesis of both symmetrical and unsymmetrical ethers.


Conclusion:

- Reimer–Tiemann introduces \(-CHO\) group in phenol.

- Kolbe introduces \(-COOH\) group in phenol.

- Williamson ether synthesis is the standard method to prepare ethers. Quick Tip: Reimer–Tiemann = aldehyde formation, Kolbe = acid formation, Williamson = ether formation.

(i) Diethyl ether from Sodium ethoxide:

This can be prepared by Williamson Ether Synthesis. Sodium ethoxide reacts with ethyl iodide (or ethyl bromide) to form diethyl ether.
\[ C_2H_5ONa + C_2H_5I \;\; \longrightarrow \;\; C_2H_5OC_2H_5 + NaI \]



(ii) Benzyl alcohol from Benzyl chloride:

Benzyl chloride undergoes hydrolysis with aqueous NaOH (or KOH) to give benzyl alcohol.
\[ C_6H_5CH_2Cl + NaOH(aq) \;\; \longrightarrow \;\; C_6H_5CH_2OH + NaCl \]



(iii) Propan-1-ol from Ethyl Magnesium bromide:

Ethyl magnesium bromide (Grignard reagent) reacts with formaldehyde, followed by hydrolysis, to give primary alcohol (propan-1-ol).
\[ C_2H_5MgBr + HCHO \;\; \longrightarrow \;\; C_2H_5CH_2OMgBr \] \[ C_2H_5CH_2OMgBr + H_2O \;\; \longrightarrow \;\; C_2H_5CH_2OH \; (Propan\!-\!1\!-\!ol) + Mg(OH)Br \]


Conclusion:

- Williamson ether synthesis gives diethyl ether.

- Hydrolysis of benzyl halides yields benzyl alcohol.

- Grignard reagents with formaldehyde followed by hydrolysis yield primary alcohols. Quick Tip: Williamson synthesis → ethers, Hydrolysis of alkyl/aryl halides → alcohols, Grignard reagents + formaldehyde → primary alcohols.

Haloalkanes undergo nucleophilic substitution reactions in which a nucleophile replaces the halogen atom. These reactions proceed by two different mechanisms: SN1 (unimolecular) and SN2 (bimolecular).


1. Unimolecular Nucleophilic Substitution (SN1 Mechanism):

The reaction takes place in two steps.
Step 1 (slow step): The C–X bond breaks heterolytically, forming a carbocation.
\[ R–X \;\longrightarrow\; R^+ + X^- \]
Step 2 (fast step): The nucleophile attacks the carbocation to form the product.
\[ R^+ + Nu^- \;\longrightarrow\; R–Nu \]
Since the rate-determining step involves only the haloalkane, the reaction rate is:
\[ Rate = k [R–X] \]
Follows first-order kinetics and usually occurs in tertiary haloalkanes due to stable carbocation formation.



2. Bimolecular Nucleophilic Substitution (SN2 Mechanism):

The reaction takes place in a single step.
The nucleophile directly attacks the carbon atom bonded to the halogen, from the backside, while the leaving group departs simultaneously.
\[ R–X + Nu^- \;\longrightarrow\; [Transition state] \;\longrightarrow\; R–Nu + X^- \]
The transition state has a partial bond between carbon, halogen, and nucleophile.
Since both haloalkane and nucleophile participate in the rate-determining step:
\[ Rate = k [R–X][Nu^-] \]
Follows second-order kinetics and usually occurs in primary haloalkanes.



Conclusion:

SN1 mechanism involves carbocation intermediate and is favored in polar protic solvents with tertiary haloalkanes.
SN2 mechanism is a concerted reaction involving backside attack, favored in polar aprotic solvents with primary haloalkanes. Quick Tip: Remember: SN1 = two steps with carbocation intermediate, rate depends only on substrate. SN2 = one step, backside attack, rate depends on both substrate and nucleophile.

(A) Wurtz–Fittig Reaction:

- The Wurtz–Fittig reaction is used for the synthesis of alkylbenzenes.

- An aryl halide reacts with an alkyl halide in the presence of sodium metal in dry ether.

- This coupling reaction produces alkyl-substituted benzene.
\[ C_6H_5Cl + CH_3Cl + 2Na \;\; \xrightarrow{dry ether} \;\; C_6H_5CH_3 + 2NaCl \]
Example: Chlorobenzene reacts with methyl chloride to give toluene.



(B) Fittig’s Reaction:

- The Fittig reaction is used for the synthesis of biphenyl compounds.

- Two molecules of an aryl halide react with sodium metal in dry ether to produce biphenyl.
\[ 2C_6H_5Cl + 2Na \;\; \xrightarrow{dry ether} \;\; C_6H_5\!-\!C_6H_5 + 2NaCl \]
Example: Chlorobenzene gives biphenyl under this reaction.


Conclusion:

- Wurtz–Fittig = Aryl halide + Alkyl halide \(\rightarrow\) Alkylbenzene.

- Fittig = Aryl halide + Aryl halide \(\rightarrow\) Biphenyl.
Quick Tip: Remember: Both reactions involve sodium in dry ether. Wurtz–Fittig makes alkylbenzenes, while Fittig makes biphenyls.

(i) Etard Reaction:

- In this reaction, an aromatic hydrocarbon containing a \(-CH_3\) group is oxidized to an aldehyde.

- Chromyl chloride (CrO\(_2\)Cl\(_2\)) is used as the oxidizing agent.
\[ C_6H_5CH_3 \; \xrightarrow{CrO_2Cl_2} \; C_6H_5CHO \]
Example: Toluene is converted to benzaldehyde.



(ii) Gattermann–Koch Reaction:

- This reaction is used for formylation of aromatic rings (introduction of \(-CHO\) group).

- Benzene reacts with CO and HCl in the presence of AlCl\(_3\) and CuCl as catalysts to give benzaldehyde.
\[ C_6H_6 + CO + HCl \; \xrightarrow{AlCl_3/CuCl} \; C_6H_5CHO \]



(iii) Cannizzaro’s Reaction:

- Aldehydes without \(\alpha\)-hydrogen undergo self-oxidation and reduction in the presence of concentrated alkali.

- One molecule is reduced to alcohol and the other is oxidized to carboxylate salt.
\[ 2HCHO \; \xrightarrow{Conc. NaOH} \; CH_3OH + HCOONa \]



(iv) Aldol Condensation:

- Aldehydes or ketones containing \(\alpha\)-hydrogen undergo condensation in the presence of dilute alkali to form \(\beta\)-hydroxy aldehydes or ketones.

- On heating, these give \(\alpha,\beta\)-unsaturated carbonyl compounds.
\[ 2CH_3CHO \; \xrightarrow{NaOH} \; CH_3CH(OH)CH_2CHO \; \xrightarrow{\Delta} \; CH_3CH=CHCHO \]


Conclusion:

- Etard → toluene \(\to\) benzaldehyde.

- Gattermann–Koch → benzene \(\to\) benzaldehyde.

- Cannizzaro → aldehyde without \(\alpha\)-H \(\to\) alcohol + salt.

- Aldol → aldehyde/ketone with \(\alpha\)-H \(\to\) \(\alpha,\beta\)-unsaturated compound. Quick Tip: Etard and Gattermann–Koch introduce \(-CHO\), Cannizzaro requires no \(\alpha\)-H, while Aldol requires \(\alpha\)-H for condensation.

(i) Acetaldehyde with Tollen’s reagent:
\[ CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \;\; \longrightarrow \;\; CH_3COO^- + 2Ag \downarrow + 4NH_3 + 2H_2O \]



(ii) Formaldehyde with NaOH (Cannizzaro reaction):
\[ 2HCHO + NaOH \;\; \longrightarrow \;\; HCOONa + CH_3OH \]



(iii) Acidic hydrolysis of Benzamide:
\[ C_6H_5CONH_2 + H_2O + HCl \;\; \longrightarrow \;\; C_6H_5COOH + NH_4Cl \]


Final Answer:

(i) Acetaldehyde \(\to\) Acetate ion + Silver mirror.

(ii) Formaldehyde \(\to\) Sodium formate + Methanol.

(iii) Benzamide \(\to\) Benzoic acid + Ammonium chloride. Quick Tip: - Tollen’s test \(\to\) aldehydes form silver mirror.
- Cannizzaro reaction occurs with aldehydes having no \(\alpha\)-H.
- Amides hydrolyze to acids + ammonium salts in acidic medium.