Step 1: Understanding stoichiometric defects.

Stoichiometric defects are those that do not alter the stoichiometric ratio of the compound, meaning the proportion of cations to anions remains unchanged. Examples of stoichiometric defects include Schottky defects, Frenkel defects, and interstitial defects.


Step 2: Identifying non-stoichiometric defects.

Non-stoichiometric defects occur when the ratio of cations to anions is disturbed. A typical example of this is the metal excess defect, where an imbalance is created by the presence of extra metal ions.


Step 3: Analyzing the options.

- (A) Interstitial defect: A stoichiometric defect where small ions occupy interstitial sites.

- (B) Frenkel defect: A stoichiometric defect where an ion is displaced to an interstitial position.

- (C) Metal excess defect: A non-stoichiometric defect that changes the ratio of cations to anions.

- (D) Schottky defect: A stoichiometric defect involving equal numbers of cations and anions missing from the crystal.


Step 4: Conclusion.

Therefore, the defect that is not stoichiometric is the metal excess defect.


Final Answer:
The correct answer is (C) Metal excess defect.
Quick Tip: Stoichiometric defects keep the chemical composition unchanged, whereas non-stoichiometric defects alter the cation-anion ratio.

Step 1: Formula for calculating the number of moles.

The number of moles is determined by the formula: \[ Number of moles = \frac{Given mass}{Molar mass} \]

Step 2: Substituting the given values.

The molar mass of water \( H_2O \) is calculated as: \[ M = 2 \times 1 + 16 = 18 \, g/mol \]
The given mass of water is \( 180 \, g \).
\[ Number of moles = \frac{180}{18} = 10 \]

Step 3: Correct option.

Therefore, the number of moles in 180 g of water is 10.

Final Answer: \[ \boxed{10} \] Quick Tip: Number of moles = Mass of substance / Molar mass. Always check the molar mass of the compound before solving.

Step 1: Understanding zero-order reactions.

In zero-order reactions, the rate of reaction does not depend on the concentration of the reactants. These reactions often occur under specific conditions, such as the presence of a catalyst or on metal surfaces.


Step 2: Analyzing the given reactions.

- (A) \(2FeCl_{3} + SnCl_{2} \to 2FeCl_{2} + SnCl_{4}\): This is a redox reaction and is not a zero-order reaction.

- (B) \(H_{2} + Cl_{2} \to 2HCl\): This is a photochemical reaction that occurs under high light intensity. At such intensities, the rate of the reaction becomes independent of the concentrations of the reactants, making it a zero-order reaction.

- (C) \(CH_{3}COOC_{2}H_{5} + NaOH \to CH_{3}COONa + C_{2}H_{5}OH\): This is saponification, which is typically a second-order reaction.

- (D) \(CH_{3}COOCH_{3} + H_{2}O \to CH_{3}COOH + CH_{3}OH\): This is the acid-catalyzed hydrolysis of an ester, a reaction that usually follows pseudo-first-order kinetics.


Step 3: Conclusion.

Thus, the only zero-order reaction from the options is the photochemical reaction of \(H_{2}\) and \(Cl_{2}\) to form \(HCl\).


Final Answer: \[ \boxed{(B) \, H_{2} + Cl_{2} \longrightarrow 2HCl} \] Quick Tip: Zero-order reactions are rare and typically occur in photochemical processes or when a catalyst surface is saturated.

Step 1: Understanding Formalin.

Formalin is a widely used chemical in laboratories, primarily as a disinfectant and preservative. It is an aqueous solution of formaldehyde gas dissolved in water.


Step 2: Composition of Formalin.

Formalin typically contains about 37–40% formaldehyde by weight, dissolved in water. Additionally, a small amount of methanol is often added to prevent polymerization of formaldehyde.


Step 3: Analyzing the options.

- (A) Fluorescein: This is a dye, unrelated to formalin.

- (B) Formaldehyde: Correct. Formalin is simply an aqueous solution of formaldehyde.

- (C) Formic acid: This is a different organic acid and is not the same as formalin.

- (D) Acetic acid: This is the main component of vinegar and is not related to formalin.


Step 4: Conclusion.

Therefore, the correct answer is that formalin is an aqueous solution of formaldehyde.


Final Answer: \[ \boxed{(B) Formaldehyde} \] Quick Tip: Formalin is about 37–40% formaldehyde in water and is widely used as a preservative in biological specimens.

Step 1: Recall the conversion of amides to amines.

Amides can be converted into amines (with one carbon fewer) through the Hofmann bromamide degradation reaction. In this reaction, amides react with bromine and potassium hydroxide (Br\(_2\)/KOH) to produce primary amines.


Step 2: Analyzing the options.

- (A) HCl/ZnCl\(_2\): This is used for Friedel–Crafts type reactions, not for converting amides to amines.

- (B) K\(_2\)Cr\(_2\)O\(_7\)/H\(_2\)SO\(_4\): This is an oxidizing mixture, not suitable for this conversion.

- (C) NaOH/Ca(OH)\(_2\): This combination is used in soda lime decarboxylation, not for converting amides to amines.

- (D) Br\(_2\)/KOH: Correct. This reagent is used in the Hofmann bromamide degradation reaction to convert amides into amines.


Step 3: Conclusion.

Thus, the correct reagent for preparing amines from amides is Br\(_2\)/KOH.


Final Answer: \[ \boxed{(D) Br\(_2\)/KOH} \] Quick Tip: Remember: The Hofmann bromamide reaction converts amides into primary amines with one carbon less than the original amide.

Step 1: Recall the Tollen’s reagent test.

Tollen’s reagent is an ammoniacal solution of silver nitrate, represented as \([Ag(NH_3)_2]^+\). It acts as a mild oxidizing agent, used primarily to differentiate aldehydes from ketones.


Step 2: Reaction with aldehyde or glucose.

When an aldehyde (or a reducing sugar like glucose) reacts with Tollen’s reagent, the aldehyde group is oxidized to a carboxylic acid, while the silver ions (\(Ag^+\)) are reduced to metallic silver (Ag).


Step 3: Observation.

The metallic silver is deposited on the inner walls of the test tube, creating the characteristic “silver mirror” effect.


Step 4: Analyzing the options.

- (A) Ag\(_2\)O: This is not formed in this reaction.

- (B) Ag: Correct. Metallic silver is deposited.

- (C) AgCl: Not formed in this reaction. AgCl forms when Ag\(^+\) reacts with chloride ions.

- (D) Ag(NH\(_3\))Cl: This is part of the reagent and not the product formed during the reaction.


Step 5: Conclusion.

Therefore, when glucose or an aldehyde reacts with Tollen’s reagent, metallic silver is produced.


Final Answer: \[ \boxed{(B) Ag} \] Quick Tip: Tollen’s reagent test is used to detect aldehydes; it gives a characteristic silver mirror due to the deposition of metallic Ag.

Step 1: Understanding the lattice structure.

The atom at the centre touches the corner atoms along the body diagonal. This indicates that the structure is a Body-Centred Cubic (BCC) lattice.


Step 2: Relation between atomic radius and cube edge length.

In a BCC lattice, atoms touch each other along the body diagonal. The length of the body diagonal is: \[ Body diagonal = \sqrt{3}a \]
where \(a\) is the edge length of the cube.

The body diagonal passes through two radii from opposite corner atoms and two radii from the centre atom, i.e., a total of \(4r\). \[ \sqrt{3}a = 4r \quad \Rightarrow \quad a = \frac{4r}{\sqrt{3}} \]


Step 3: Number of atoms per unit cell.

In BCC: \[ N = 2 \quad atoms per unit cell. \]


Step 4: Volume of atoms in the unit cell.
\[ V_{atoms} = N \times \frac{4}{3}\pi r^3 = 2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3 \]


Step 5: Volume of the unit cell.
\[ V_{cell} = a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}} \]


Step 6: Packing efficiency formula.
\[ Packing Efficiency = \frac{V_{atoms}}{V_{cell}} \times 100 \] \[ = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100 \] \[ = \frac{8\pi r^3 \sqrt{3}}{64r^3} \times 100 \] \[ = \frac{\pi \sqrt{3}}{8} \times 100 \approx 68% \]


Conclusion:

The packing efficiency of the cubic lattice in this case is approximately: \[ \boxed{68%} \] Quick Tip: In a BCC structure, atoms touch each other along the body diagonal, not along the edge. Always use \(\sqrt{3}a = 4r\) to relate radius and edge length.

Step 1: Understanding the problem.

We are given a 30% by weight aqueous solution of glycerol. This means that in 100 g of solution, glycerol contributes 30 g and water contributes 70 g.


Step 2: Calculate moles of glycerol.

Molecular mass of glycerol (C\(_3\)H\(_8\)O\(_3\)) = (3 × 12) + (8 × 1) + (3 × 16)

= 36 + 8 + 48

= 92 g/mol.

Moles of glycerol = \(\dfrac{30}{92} \approx 0.326\) mol.


Step 3: Calculate moles of water.

Molecular mass of water (H\(_2\)O) = 18 g/mol.

Mass of water = 70 g.

Moles of water = \(\dfrac{70}{18} \approx 3.889\) mol.


Step 4: Calculate mole fraction of glycerol.

Total moles = 0.326 + 3.889 = 4.215 mol.

Mole fraction of glycerol = \(\dfrac{0.326}{4.215} \approx 0.077\).



Conclusion:

The mole fraction of glycerol in 30% by weight aqueous glycerol solution is approximately: \[ \boxed{0.077} \] Quick Tip: In weight percent solutions, always assume 100 g of solution to simplify calculations. Then convert masses to moles for mole fraction determination.

Step 1: Recall the relationship between Gibbs free energy and cell potential.

The formula is: \[ \Delta G^\circ = -nFE^\circ_{cell} \]
where,
- \(n\) = number of electrons transferred in the reaction = 2

- \(F\) = Faraday constant = \(96500 \, C mol^{-1}\)

- \(E^\circ_{cell}\) = standard electrode potential = \(1.1 \, V\)


Step 2: Substitute the values.
\[ \Delta G^\circ = - (2)(96500)(1.1) \] \[ \Delta G^\circ = -212300 \, J mol^{-1} \] \[ \Delta G^\circ = -212.3 \, kJ mol^{-1} \]


Conclusion:

The standard Gibbs energy for the Daniell cell reaction is: \[ \boxed{-212.3 \, kJ mol^{-1}} \] Quick Tip: Always remember: A negative Gibbs free energy (\(\Delta G^\circ < 0\)) indicates a spontaneous reaction under standard conditions.

Step 1: Definition.

Electrophoresis is the process of movement of charged particles under the influence of an electric field. It is commonly observed in colloidal solutions where particles carry either positive or negative charges.


Step 2: Principle.

When an electric potential is applied across the colloidal solution, the charged colloidal particles migrate towards the electrode of opposite charge:
- Positively charged particles (\( cations \)) move towards the cathode.
- Negatively charged particles (\( anions \)) move towards the anode.


Step 3: Applications.

Electrophoresis is widely used in:

Separation of proteins and nucleic acids: In biochemical and medical laboratories, gel electrophoresis is used to separate DNA, RNA, and proteins.
Determining charge on colloids: The direction of movement helps identify whether the colloid is positively or negatively charged.
Purification processes: Used to remove impurities from colloidal systems.



Conclusion:

Electrophoresis is a fundamental technique to study and separate charged particles in a colloidal system under the influence of an electric field. Quick Tip: Remember: Electrophoresis is always based on the principle of movement of charged particles towards electrodes of opposite charge under an applied electric field.

Step 1: Understanding the Schottky defect.

A Schottky defect occurs when equal numbers of cations and anions are missing from their lattice sites in an ionic crystal.


Step 2: Properties of Schottky defect.


It decreases the density of the crystal because some ions are missing.
It usually occurs in highly ionic compounds with high coordination numbers, such as NaCl, KCl, and CsCl.
Electrical neutrality is maintained because the number of missing cations equals the number of missing anions.


Step 3: Diagram representation.
\[ \begin{array}{|c|c|c|c|} \hline + & - & + & -
\hline - & \square & - & +
\hline + & - & \square & -
\hline - & + & - & +
\hline \end{array} \]
Here, \(\square\) indicates missing ions representing Schottky defects.



Conclusion:

A Schottky defect is a type of vacancy defect in ionic crystals where equal numbers of cations and anions are missing, leading to a decrease in density. Quick Tip: Remember: Schottky defect reduces density, whereas Frenkel defect does not.

Step 1: Deacon’s process.

In Deacon’s process, chlorine is manufactured by the oxidation of hydrogen chloride gas using oxygen in the presence of a catalyst (CuCl\(_2\)).
\[ 4HCl + O_2 \xrightarrow{CuCl_2, 723K} 2Cl_2 + 2H_2O \]

Step 2: Reaction of chlorine with sulphur.

When chlorine reacts with sulphur, sulphur dichloride is formed.
\[ S + Cl_2 \rightarrow SCl_2 \]


Conclusion:

- Deacon’s process is an industrial method for preparing chlorine.

- Chlorine reacts with sulphur to give sulphur dichloride (SCl\(_2\)). Quick Tip: Always mention the catalyst (CuCl\(_2\)) and temperature in Deacon’s process, as it is crucial for industrial application.

Step 1: Definition.

Coordination number is defined as the number of ligand donor atoms that are directly bonded to the central metal atom or ion in a coordination compound.


Step 2: Explanation.

It tells us how many atoms, ions, or molecules are attached to the central atom in a complex. This depends on the size, charge, and electronic configuration of the central atom.


Step 3: Example.

In the complex \([Co(NH_3)_6]^{3+}\):

- The central metal ion is \(Co^{3+}\).

- Six ammonia molecules (\(NH_3\)) are bonded directly to cobalt.

Thus, the coordination number of cobalt is 6.


Conclusion:

The coordination number represents the total number of bonds formed between the central metal ion and the surrounding ligands. Quick Tip: Common coordination numbers are 2, 4, and 6, which lead to geometries like linear, tetrahedral/square planar, and octahedral respectively.

(i) Reaction of glucose with hydroxylamine:

Glucose reacts with hydroxylamine (\(NH_2OH\)) to form an oxime by reacting with the aldehyde group (-CHO) of glucose. \[ C_6H_{12}O_6 + NH_2OH \longrightarrow C_6H_{11}O_6{-}CH{=}NOH + H_2O \]

(ii) Reaction of glucose with bromine water:

Glucose undergoes mild oxidation with bromine water, converting the aldehyde group (-CHO) into a carboxylic acid (-COOH), giving gluconic acid. \[ C_6H_{12}O_6 + Br_2 + H_2O \longrightarrow C_6H_{12}O_7 + 2HBr \]


Conclusion:

- With hydroxylamine, glucose forms an oxime derivative.

- With bromine water, glucose is oxidized to gluconic acid. Quick Tip: Remember: Glucose behaves like an aldehyde, so it reacts with reagents like hydroxylamine and bromine water that test aldehyde functionality.

Step 1: General form of Nernst equation.

The Nernst equation relates the cell potential under non-standard conditions to the standard electrode potential and the reaction quotient: \[ E = E^\circ - \frac{RT}{nF} \ln Q \]
where:
- \( E \) = electrode potential under given conditions

- \( E^\circ \) = standard electrode potential

- \( R \) = gas constant (\(8.314 \, J \, mol^{-1}K^{-1}\))

- \( T \) = absolute temperature (in K)

- \( n \) = number of electrons transferred

- \( F \) = Faraday constant (\(96500 \, C \, mol^{-1}\))

- \( Q \) = reaction quotient

At \( 298 \, K \), the equation becomes: \[ E = E^\circ - \frac{0.0591}{n} \log Q \]


Step 2: Application.

One application of the Nernst equation is in calculating the electrode potential of the hydrogen electrode: \[ E = 0 - 0.0591 \, \log \frac{1}{[H^+]} \]
Thus, the Nernst equation helps determine the pH of a solution using hydrogen electrode.


Conclusion:

The Nernst equation is essential in electrochemistry for predicting cell potentials under non-standard conditions and is widely used in calculating pH values and electrode potentials. Quick Tip: Always remember: At \(25^\circ C\), the simplified form is \(E = E^\circ - \tfrac{0.0591}{n} \log Q\).

Step 1: Understanding heterogeneous catalysis.

Heterogeneous catalysis occurs when the catalyst is in a different phase from the reactants. Usually, solid catalysts act on gaseous or liquid reactants.


Step 2: Adsorption theory.

The adsorption theory explains catalysis in terms of adsorption of reactant molecules on the surface of the solid catalyst:

Reactant molecules are adsorbed on the surface of the catalyst.
Adsorption weakens the chemical bonds in reactant molecules, lowering activation energy.
The adsorbed reactants combine to form products.
The product molecules desorb (leave) from the catalyst surface, making it free for the next reaction.



Step 3: Example.

In the Haber process, \( N_2 \) and \( H_2 \) gases are adsorbed on finely divided Fe catalyst. Their bonds weaken, allowing formation of \( NH_3 \).


Conclusion:

The adsorption theory provides a clear explanation for the mechanism of heterogeneous catalysis, showing how catalysts work by adsorption, bond weakening, and product desorption. Quick Tip: Remember the three steps: Adsorption of reactants → Reaction on surface → Desorption of products.

Step 1: Understanding the classification.

Amines are classified based on the number of alkyl or aryl groups attached to the nitrogen atom.


Step 2: Differences with examples.



Primary amine (1°): Nitrogen atom is attached to one alkyl/aryl group.

General formula: R–NH\(_2\)

Example: Methylamine (CH\(_3\)NH\(_2\))

Reaction:

\[ CH_3NH_2 + HNO_2 \rightarrow CH_3OH + N_2 \uparrow + H_2O \]

Secondary amine (2°): Nitrogen atom is attached to two alkyl/aryl groups.

General formula: R\(_2\)–NH

Example: Dimethylamine ((CH\(_3\))\(_2\)NH)

Reaction:

\[ (CH_3)_2NH + HNO_2 \rightarrow (CH_3)_2NNO + H_2O \]

Tertiary amine (3°): Nitrogen atom is attached to three alkyl/aryl groups.

General formula: R\(_3\)–N

Example: Trimethylamine ((CH\(_3\))\(_3\)N)

Reaction:

\[ (CH_3)_3N + HNO_2 \rightarrow (CH_3)_3NNO \]



Conclusion:

- Primary amines give alcohol and nitrogen gas with nitrous acid.

- Secondary amines form nitrosoamines.

- Tertiary amines form soluble salts or nitrosonium complexes. Quick Tip: Always use nitrous acid test to distinguish between primary, secondary and tertiary amines in organic chemistry.

Step 1: Structure of DNA and RNA.

DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are nucleic acids but differ in sugar, bases, structure, and function.


Step 2: Key differences.


\begin{tabular{|c|c|c|
\hline
Feature & DNA & RNA

\hline
Sugar & Deoxyribose & Ribose

\hline
Bases & A, T, G, C & A, U, G, C (U replaces T)

\hline
Strands & Double-stranded helix & Single-stranded

\hline
Function & Genetic material, stores information & Protein synthesis (mRNA, tRNA, rRNA)

\hline
Location & Mostly in nucleus & Nucleus and cytoplasm

\hline
\end{tabular


Conclusion:

DNA stores and transmits genetic information, while RNA helps in protein synthesis and gene expression. Quick Tip: Remember: DNA = Storage of genetic code; RNA = Expression and translation of genetic code.

Step 1: Definition of Osmotic Pressure.

Osmotic pressure is the pressure required to stop the flow of solvent molecules through a semipermeable membrane from pure solvent to solution. It is a colligative property and depends on the number of solute particles in solution, not their nature.


Step 2: Mathematical Expression.

The osmotic pressure is given by: \[ \pi = CRT \]
where,
- \(\pi\) = osmotic pressure

- \(C\) = molar concentration of solute

- \(R\) = gas constant

- \(T\) = absolute temperature

Step 3: Relation with Molar Mass.

Molar concentration can be written as: \[ C = \frac{n}{V} = \frac{w}{M \cdot V} \]
where,
- \(w\) = mass of solute

- \(M\) = molar mass of solute

- \(V\) = volume of solution (in litres)

Substituting in the osmotic pressure formula: \[ \pi = \frac{wRT}{MV} \]


Conclusion:

The relation between osmotic pressure and molar mass of solute is: \[ M = \frac{wRT}{\pi V} \] Quick Tip: Osmotic pressure measurements are a practical way to determine the molar mass of large biomolecules like proteins and polymers.

Step 1: Average Rate of Reaction.

The average rate of reaction is the change in concentration of reactants or products divided by the time interval during which the change occurs. \[ Average rate = \frac{-\Delta [R]}{\Delta t} = \frac{\Delta [P]}{\Delta t} \]

Step 2: Instantaneous Rate of Reaction.

The instantaneous rate of reaction is the rate at a specific moment of time. It is obtained by finding the slope of the tangent to the concentration-time curve at that instant. \[ Instantaneous rate = -\frac{d[R]}{dt} = \frac{d[P]}{dt} \]

Step 3: Factors Affecting Reaction Rates.

Two important factors are:
1. Concentration of Reactants: Higher concentration increases the frequency of collisions, leading to a faster rate.

2. Temperature: Increasing temperature raises the kinetic energy of molecules, resulting in more effective collisions and faster reaction rate.



Conclusion:

The average rate gives a broad measure over an interval, while instantaneous rate measures the exact speed of reaction at a given time. Both are influenced by factors like concentration and temperature. Quick Tip: Instantaneous rate is always obtained from the tangent to the concentration vs. time curve, while average rate comes from a secant line over a time interval.

Step 1: Electronic configuration of Cr (Z = 24).

The expected configuration is: \[ Cr: [Ar] \, 3d^4 \, 4s^2 \]
But due to extra stability of half-filled orbitals, the actual configuration is: \[ Cr: [Ar] \, 3d^5 \, 4s^1 \]


Step 2: Electronic configuration of Cu (Z = 29).

The expected configuration is: \[ Cu: [Ar] \, 3d^9 \, 4s^2 \]
But due to extra stability of fully filled orbitals, the actual configuration is: \[ Cu: [Ar] \, 3d^{10} \, 4s^1 \]


Step 3: Two main characteristics of transition elements.


Variable oxidation states: Transition metals exhibit multiple oxidation states due to the involvement of both \(3d\) and \(4s\) electrons in bonding.
Formation of coloured compounds: Due to the presence of partially filled \(d\)-orbitals, they show \(d \rightarrow d\) electronic transitions which give rise to characteristic colours.



Conclusion:

Chromium and copper show exceptional electronic configurations due to stability of half-filled and fully-filled \(d\)-orbitals. Transition elements are unique for their variable oxidation states and coloured compounds. Quick Tip: Always check for exceptional cases (Cr, Cu, Mo, Ag, etc.) where half-filled or fully-filled \(d\)-orbitals provide extra stability.

Step 1: Introduction.

Crystal Field Theory (CFT) explains the bonding in coordination compounds in terms of the effect of the electric field produced by ligands on the \(d\)-orbitals of the central metal ion.


Step 2: Splitting of d-orbitals.

In a free metal ion, all five \(d\)-orbitals are degenerate (equal energy).

When ligands approach the central metal ion:
- In an octahedral field, the \(d\)-orbitals split into two sets: \[ t_{2g} (d_{xy}, d_{xz}, d_{yz}) \quad (lower energy) \] \[ e_g (d_{z^2}, d_{x^2-y^2}) \quad (higher energy) \]
- The energy difference is called the crystal field splitting energy (\(\Delta\)).


Step 3: Consequences.


Explains colour of complexes due to \(d \rightarrow d\) transitions.
Explains magnetic properties (high spin or low spin) based on pairing of electrons and magnitude of \(\Delta\).



Step 4: Limitations of CFT.


It treats metal-ligand bonds as purely ionic and ignores covalent character.
Cannot explain the spectra of some complexes accurately.
Does not account for ligand-metal orbital overlap or back bonding.



Conclusion:

Crystal Field Theory successfully explains splitting of \(d\)-orbitals, colour, and magnetic properties of coordination compounds, but fails to explain covalent interactions and detailed spectra. Quick Tip: Remember: Octahedral → \(t_{2g}\) lower, \(e_g\) higher; Tetrahedral → reverse splitting.

(1.) Sodium Azide is heated:

When sodium azide (\(NaN_3\)) is heated, it decomposes to form sodium metal and nitrogen gas.
\[ 2NaN_3 \; \xrightarrow{\Delta} \; 2Na + 3N_2 \uparrow \]
This reaction is used in airbag inflation systems.



(2.) Lithium is heated with Nitrogen:

Lithium reacts directly with nitrogen on heating to form lithium nitride.
\[ 6Li + N_2 \; \xrightarrow{\Delta} \; 2Li_3N \]
Lithium nitride is the only stable alkali metal nitride.



(3.) Ammonium chromate is heated:

Ammonium chromate (\((NH_4)_2CrO_4\)) decomposes on heating to form chromium(III) oxide, nitrogen, and water vapor.
\[ (NH_4)_2CrO_4 \; \xrightarrow{\Delta} \; Cr_2O_3 + N_2 \uparrow + H_2O \uparrow \]



(4.) Nitrogen is heated with Oxygen:

At high temperature, nitrogen combines with oxygen to form nitric oxide (NO).
\[ N_2 + O_2 \; \xrightarrow{\Delta} \; 2NO \]
This is an endothermic reaction and occurs during lightning in the atmosphere.



(5.) The aqueous solutions of Ammonium chloride and Sodium nitrite are allowed to react:

Ammonium chloride reacts with sodium nitrite to produce nitrogen gas, sodium chloride, and water.
\[ NH_4Cl + NaNO_2 \; \longrightarrow \; N_2 \uparrow + NaCl + 2H_2O \]


Conclusion:

These reactions demonstrate thermal decomposition, direct combination, and double decomposition reactions involving nitrogen and its compounds, producing nitrogen gas in several cases. Quick Tip: Heating azides, chromates, or ammonium salts often leads to decomposition with the release of nitrogen gas. This property is widely used in airbags and analytical chemistry.

Step 1: Industrial preparation of Ammonia.

Ammonia is prepared industrially by the Haber process. It involves direct combination of nitrogen and hydrogen gases under specific conditions.


Step 2: Flow diagram of Haber process.
\[ N_2 (g) + 3H_2 (g) \;\xrightleftharpoons[450-500°C]{Fe catalyst, 200 atm}\; 2NH_3 (g) + \Delta H \]

Flow Diagram: \[ N_2 + H_2 \;\longrightarrow\; Compression (200 atm) \;\longrightarrow\; Fe Catalyst Chamber (450-500°C) \;\longrightarrow\; NH_3 formed and condensed. \]


Step 3: Reaction of Ammonia with Copper ion (\(Cu^{2+}\)).

Ammonia reacts with copper(II) ions to form a deep blue complex: \[ Cu^{2+} + 4NH_3 \;\longrightarrow\; [Cu(NH_3)_4]^{2+} \]


Step 4: Reaction of Ammonia with Silver ion (\(Ag^+\)).

Ammonia reacts with silver ions to form a soluble complex: \[ Ag^+ + 2NH_3 \;\longrightarrow\; [Ag(NH_3)_2]^+ \]


Conclusion:

Ammonia is prepared industrially by the Haber process under high temperature, pressure, and Fe catalyst. It also forms characteristic coordination complexes with \(Cu^{2+}\) (deep blue) and \(Ag^+\) (diammine silver complex). Quick Tip: In complex formation reactions, ammonia acts as a Lewis base by donating a lone pair of electrons to the metal ion.

(i) Kolbe Reaction:

Phenol reacts with sodium hydroxide to form sodium phenoxide, which further reacts with carbon dioxide under pressure, followed by acidification to give salicylic acid.
\[ C_6H_5OH + NaOH \; \longrightarrow \; C_6H_5ONa + H_2O \] \[ C_6H_5ONa + CO_2 \; \xrightarrow{373K, 4-7 atm} \; o\!-\!HOC_6H_4COONa \] \[ o\!-\!HOC_6H_4COONa + HCl \; \longrightarrow \; o\!-\!HOC_6H_4COOH \]



(ii) Reimer-Tiemann Reaction:

Phenol reacts with chloroform in the presence of sodium hydroxide to give salicylaldehyde (ortho-hydroxybenzaldehyde).
\[ C_6H_5OH + CHCl_3 + 3NaOH \; \longrightarrow \; o\!-\!HOC_6H_4CHO + 3NaCl + 2H_2O \]



(iii) Oxidation of Phenol:

Phenol undergoes oxidation with neutral ferric chloride (\(FeCl_3\)) or chromic acid to give para-benzoquinone.
\[ C_6H_5OH \; \xrightarrow{[O]} \; C_6H_4O_2 \; (p\!-\!benzoquinone) \]



(iv) Williamson Synthesis:

An alkoxide reacts with an alkyl halide to form an ether.
\[ R\!-\!ONa + R'X \; \longrightarrow \; R\!-\!O\!-\!R' + NaX \]
Example: \[ C_2H_5ONa + CH_3I \; \longrightarrow \; C_2H_5OCH_3 + NaI \]



(v) Industrial Preparation of Methanol:

Methanol is prepared industrially by the catalytic hydrogenation of carbon monoxide.
\[ CO + 2H_2 \; \xrightarrow{ZnO-Cr_2O_3, \; 573K, \; 300 atm} \; CH_3OH \]


Conclusion:

Each reaction illustrates a fundamental organic transformation—carboxylation, formylation, oxidation, ether synthesis, and catalytic hydrogenation—all central to industrial and laboratory organic chemistry. Quick Tip: Remember: Kolbe gives \textbf{salicylic acid}, Reimer–Tiemann gives \textbf{salicylaldehyde}, oxidation of phenol gives \textbf{quinones}, Williamson synthesis produces \textbf{ethers}, and industrial methanol is prepared by \textbf{CO + H\(_2\) hydrogenation}.

Step 1: General chemical equation.

Acid-catalysed hydration of alkenes follows Markovnikov’s rule. Example with ethene: \[ CH_2 = CH_2 + H_2O \;\xrightarrow{H_2SO_4}\; CH_3CH_2OH \]


Step 2: Mechanism.


Protonation of alkene:
\[ CH_2 = CH_2 + H^+ \;\longrightarrow\; CH_3-CH_2^+ \]
(formation of carbocation)

Nucleophilic attack by water:
\[ CH_3-CH_2^+ + H_2O \;\longrightarrow\; CH_3-CH_2OH_2^+ \]

Deprotonation:
\[ CH_3-CH_2OH_2^+ \;\longrightarrow\; CH_3-CH_2OH + H^+ \]



Conclusion:

The final product is an alcohol, and the reaction is catalysed by an acid (usually \(H_2SO_4\)). Quick Tip: Always apply Markovnikov’s rule: “In hydration of alkenes, H attaches to the carbon with more hydrogens.”

Step 1: Oxidation test (using acidified \(K_2Cr_2O_7\)).


Primary alcohols: Oxidised to aldehydes, which further oxidise to carboxylic acids.
\[ R-CH_2OH \;\xrightarrow{[O]}\; R-CHO \;\xrightarrow{[O]}\; R-COOH \]

Secondary alcohols: Oxidised to ketones only.
\[ R-CHOH-R' \;\xrightarrow{[O]}\; R-CO-R' \]



Step 2: Lucas Test (ZnCl\(_2\)/HCl).


Primary alcohols react very slowly (no turbidity at room temp).
Secondary alcohols give turbidity within 5–10 minutes.



Conclusion:

Primary and secondary alcohols can be distinguished by their oxidation behaviour and Lucas test. Quick Tip: Remember: Primary alcohol → oxidises to acid; Secondary alcohol → oxidises to ketone; Tertiary alcohol → generally resistant to oxidation.

Step 1: Nature of chlorine as a substituent.

Chlorine is an electron withdrawing group due to its strong –I (inductive) effect because of high electronegativity. This tends to decrease electron density on the benzene ring.


Step 2: Resonance effect of chlorine.

Chlorine possesses lone pairs of electrons which it can donate to the benzene ring through resonance (+R effect). This resonance increases electron density particularly at the ortho and para positions of the benzene ring.


Step 3: Combined effect.

- The –I effect makes chlorine a deactivating group (reaction rate decreases).

- The +R effect directs the incoming electrophile to ortho and para positions.



Conclusion:

Although chlorine is overall deactivating due to its –I effect, it acts as an ortho-para directing group in electrophilic substitution reactions because of its strong +R effect. Quick Tip: Remember: Halogens are exceptions – they are deactivating but ortho-para directing.

(I) Wurtz–Fittig Reaction:

Definition: The Wurtz–Fittig reaction involves the reaction of an aryl halide with an alkyl halide in the presence of sodium metal in dry ether to give alkyl-substituted aromatic hydrocarbons.


Equation:
\[ C_6H_5Cl + CH_3Cl + 2Na \xrightarrow{dry\ ether} C_6H_5CH_3 + 2NaCl \]

Conclusion: It is used to prepare alkylbenzenes (like toluene).




(II) Sandmeyer Reaction:

Definition: The Sandmeyer reaction involves the replacement of the diazonium group (-N\(_2^+\)) in an aromatic diazonium salt with a halogen (Cl, Br, CN) using copper(I) salts as catalysts.


Equation:
\[ C_6H_5N_2^+Cl^- + CuCl \rightarrow C_6H_5Cl + N_2 \uparrow \]

Conclusion: It is an important method for the preparation of aryl halides and aryl cyanides.
Quick Tip: - Wurtz–Fittig = Aryl halide + Alkyl halide → Alkylbenzene.
- Sandmeyer = Diazonium salt → Aryl halide or cyanide (with Cu(I) salts).

Step 1: Understanding Resonance in Haloarenes.

In haloarenes (like chlorobenzene), the lone pair of electrons on the halogen atom interacts with the \(\pi\)-electrons of the benzene ring. This delocalization leads to resonance structures.


Step 2: Resonance Structures.

The \(p\)-orbital of the halogen atom overlaps with the \(p\)-orbitals of the benzene ring, giving rise to multiple resonance structures. As a result, the C–Cl bond acquires partial double bond character.


Step 3: Effect on Nucleophilic Substitution.

Because of resonance:
- The C–Cl bond in haloarenes is shorter and stronger than a normal C–Cl bond.

- This makes the bond difficult to break, thereby making nucleophilic substitution reactions of haloarenes much less reactive compared to haloalkanes.



Conclusion:

Resonance stabilizes the C–Cl bond in haloarenes, giving it partial double bond character and reducing the tendency for nucleophilic substitution. Quick Tip: Haloarenes are less reactive towards nucleophilic substitution due to both resonance stabilization and the partial double bond character of the C–Cl bond.

1. Reaction with Sodium (Wurtz Reaction):

When haloalkanes react with sodium in dry ether, higher alkanes are formed. \[ 2R{-}X + 2Na \longrightarrow R{-}R + 2NaX \]
Example: \[ 2CH_3Cl + 2Na \longrightarrow C_2H_6 + 2NaCl \]

2. Reaction with Magnesium:

Haloalkanes react with magnesium in dry ether to form Grignard reagents. \[ R{-}X + Mg \xrightarrow{dry \, ether} R{-}MgX \]
Example: \[ CH_3Br + Mg \xrightarrow{dry \, ether} CH_3MgBr \]


Conclusion:

- With sodium, haloalkanes undergo Wurtz reaction to form higher alkanes.

- With magnesium, they form important organometallic compounds known as Grignard reagents. Quick Tip: The Wurtz reaction is useful for the preparation of symmetrical alkanes, while Grignard reagents are versatile intermediates in organic synthesis.

(i) Aldol and Crossed Aldol Condensation in Aldehyde and Ketone:


Step 1: Aldol condensation.

When aldehydes or ketones containing at least one \(\alpha\)-hydrogen are treated with dilute alkali (NaOH, Ba(OH)\(_2\)), they undergo aldol condensation. The \(\alpha\)-hydrogen is removed to form an enolate ion which attacks another carbonyl group, producing \(\beta\)-hydroxy aldehydes or ketones (aldols). On heating, they lose water to give \(\alpha, \beta\)-unsaturated compounds.
\[ 2CH_3CHO \;\xrightarrow{NaOH}\; CH_3CH(OH)CH_2CHO \;\xrightarrow{\Delta}\; CH_3CH=CHCHO \]

Step 2: Crossed Aldol condensation.

When two different carbonyl compounds are used, the reaction is called crossed aldol condensation. For example: \[ CH_3CHO + CH_3COCH_3 \;\xrightarrow{NaOH}\; CH_3CH(OH)CH_2COCH_3 \;\xrightarrow{\Delta}\; CH_3CH=CHCOCH_3 \]


Conclusion:

Aldol condensation gives \(\alpha, \beta\)-unsaturated aldehydes/ketones, while crossed aldol involves two different aldehydes/ketones. Quick Tip: Remember: Aldol condensation requires \(\alpha\)-hydrogen. Compounds lacking \(\alpha\)-hydrogen cannot undergo this reaction.

Carboxylic acids can be prepared by several methods. The five important methods are:


Method 1: Oxidation of Primary Alcohols.
\[ R-CH_2OH \;\xrightarrow{[O]}\; R-COOH \]


Method 2: Oxidation of Aldehydes.
\[ R-CHO \;\xrightarrow{[O]}\; R-COOH \]


Method 3: Hydrolysis of Nitriles.
\[ R-CN + 2H_2O \;\xrightarrow{H^+ / \Delta}\; R-COOH + NH_3 \]


Method 4: Carbonation of Grignard Reagents.
\[ R-MgX + CO_2 \;\xrightarrow{H_3O^+}\; R-COOH \]


Method 5: Hydrolysis of Esters.
\[ R-COOR' + H_2O \;\xrightarrow{H^+ / \Delta}\; R-COOH + R'-OH \]


Summary Table:



\begin{tabular{|c|l|l|
\hline
Method & Reaction & Product

\hline
1 & \(R-CH_2OH \;\xrightarrow{[O]}\; R-COOH\) & From Primary Alcohol

\hline
2 & \(R-CHO \;\xrightarrow{[O]}\; R-COOH\) & From Aldehyde

\hline
3 & \(R-CN + 2H_2O \;\xrightarrow{H^+}\; R-COOH + NH_3\) & From Nitrile

\hline
4 & \(R-MgX + CO_2 \;\xrightarrow{H_3O^+}\; R-COOH\) & From Grignard Reagent

\hline
5 & \(R-COOR' + H_2O \;\xrightarrow{H^+}\; R-COOH + R'-OH\) & From Ester

\hline
\end{tabular



Conclusion:

Carboxylic acids can be synthesised by oxidation, hydrolysis, and carbonation reactions, making them versatile in both laboratory and industry. Quick Tip: Easy trick: Alcohol → Aldehyde → Acid; Nitrile/ester → Hydrolysis → Acid; Grignard + CO\(_2\) → Acid.