Step 1: Understanding NaCl crystal structure.

The crystal structure of sodium chloride (NaCl) is based on the arrangement of Na\(^+\) and Cl\(^-\) ions. Each Na\(^+\) ion is surrounded by 6 Cl\(^-\) ions and each Cl\(^-\) ion is surrounded by 6 Na\(^+\) ions. This gives it a coordination number of 6:6.


Step 2: Identifying lattice type.

This arrangement corresponds to a face centred cubic (fcc) lattice, also known as the rock-salt structure.


Step 3: Conclusion.

Thus, the correct answer is (ii) face centred cubic (fcc).
Quick Tip: NaCl crystal is an example of a rock-salt structure with fcc arrangement, where each ion is octahedrally surrounded by opposite ions.

Step 1: Nature of ferric oxide sol.

Ferric oxide sol (Fe\(_2\)O\(_3 \cdot xH_2O\)) is a positively charged colloid when dispersed in water. This is because Fe\(^ {3+}\) ions from hydrolysis impart a positive charge to the dispersed particles.


Step 2: Verification of options.

- (i) Negative: Incorrect, ferric oxide sol is not negatively charged.

- (ii) Positive: Correct, because it acquires Fe\(^ {3+}\) ions.

- (iii) No charge: Incorrect, as colloidal particles always carry some charge.

- (iv) None of these: Incorrect, as positive charge is correct.


Step 3: Conclusion.

Hence, the charge on colloidal particles of Fe\(_2\)O\(_3 \cdot xH_2O\) is positive.
Quick Tip: Metal oxides in colloidal form generally acquire charge depending on the medium; Fe\(_2\)O\(_3\) sol is positively charged in water.

Step 1: Physical state of sulphur.

At ordinary temperature, sulphur exists in the solid state. Its most stable form is rhombic sulphur.


Step 2: Molecular structure.

In rhombic sulphur, each molecule consists of 8 sulphur atoms forming a puckered ring (crown shape). Hence, its molecular formula is S\(_8\).


Step 3: Conclusion.

Therefore, the correct molecular formula of sulphur at ordinary temperature is (iv) S\(_8\).
Quick Tip: Sulphur commonly exists as S\(_8\) molecules at room temperature, forming a crown-shaped cyclic structure.

Step 1: Purpose of gas in food packets.

To prevent spoilage and oxidation, an inert atmosphere is required inside sealed food packets.


Step 2: Gas used.

Nitrogen gas (N\(_2\)) is used because it is inert, does not react with food, and displaces oxygen which causes oxidation and spoilage.


Step 3: Conclusion.

Thus, the gas present in food packets is (iii) N\(_2\).
Quick Tip: Nitrogen gas prevents oxidation and rancidity, keeping packed food fresh for longer.

Step 1: Definition.

Specific conductance (or conductivity, \(\kappa\)) is the conductance of 1 cm\(^3\) of a solution placed between two electrodes 1 cm apart.


Step 2: Unit.

Since conductance unit is ohm\(^{-1}\) (or siemens, S), and length factor introduces cm\(^{-1}\), the unit is ohm\(^{-1}\) cm\(^{-1}\).


Step 3: Conclusion.

Therefore, the unit of specific conductance is ohm\(^{-1}\) cm\(^{-1}\).
Quick Tip: Specific conductance is often expressed in S cm\(^{-1}\), where S = siemens = ohm\(^{-1}\).

Step 1: Understanding electrolytes.

Electrolytes are substances that dissociate into ions in aqueous solution and conduct electricity. Non-electrolytes do not dissociate and hence do not conduct electricity.


Step 2: Analysis of options.

- Sodium chloride: Electrolyte, produces Na\(^+\) and Cl\(^-\) ions.

- Urea: Non-electrolyte, dissolves in water but does not produce ions.

- Ammonium nitrate: Electrolyte, dissociates into NH\(_4^+\) and NO\(_3^-\).

- Nitric acid: Strong electrolyte, ionizes completely.


Step 3: Conclusion.

Thus, the non-electrolyte among the options is (ii) Urea.
Quick Tip: Non-electrolytes dissolve in water but do not form ions, hence they do not conduct electricity (e.g., urea, glucose).

Step 1: Understanding lyophilic colloids.

Lyophilic colloids are "liquid-loving" colloids that form readily when mixed with the dispersion medium. They are stable and reversible in nature. Examples: starch sol, gum sol.


Step 2: Understanding lyophobic colloids.

Lyophobic colloids are "liquid-hating" colloids that do not form readily and require special methods for preparation. They are less stable and irreversible. Examples: ferric hydroxide sol, gold sol.


Step 3: Conclusion.

Thus, starch sol and gum sol are lyophilic colloids, while ferric hydroxide sol and gold sol are lyophobic colloids.
Quick Tip: Lyophilic = stable and reversible, Lyophobic = unstable and irreversible.

Step 1: Nature of colloidal particles.

Colloidal particles carry either a positive or negative charge. Their stability depends on the repulsion between similarly charged particles.


Step 2: Role of oppositely charged ions.

When oppositely charged ions are added, they neutralize the charge on the colloidal particles and cause coagulation.


Step 3: Order of coagulating power.

For negatively charged sols: Al\(^ {3+}\) > Ba\(^ {2+}\) > Na\(^+\).
For positively charged sols: [Fe(CN)\(_6\)]\(^{4-}\) > PO\(_4^{3-}\) > SO\(_4^{2-}\) > Cl\(^-\).


Step 4: Conclusion.

Thus, the higher the valency of the oppositely charged ion, the greater its coagulating power.
Quick Tip: Valency matters! Multivalent ions coagulate colloids much faster than monovalent ions.

Step 1: Recall osmosis.

In osmosis, solvent flows naturally from dilute to concentrated solution across a semipermeable membrane.


Step 2: Applying pressure.

If pressure greater than osmotic pressure is applied on the concentrated side, the natural flow is reversed. Solvent moves from concentrated solution to dilute solution. This is called anti-osmosis or reverse osmosis.


Step 3: Example.

Reverse osmosis is used in water purification, such as desalination of seawater where pure water passes through the membrane leaving salts behind.


Step 4: Conclusion.

Thus, anti-osmosis is reverse osmosis caused by applying external pressure greater than osmotic pressure.
Quick Tip: Anti-osmosis is the principle behind RO water purifiers used in homes.

Step 1: Recall nitration mechanism.

Nitration is an electrophilic substitution reaction where an electrophile attacks the benzene ring.


Step 2: Electrophile formation.

In nitration, concentrated HNO\(_3\) reacts with concentrated H\(_2\)SO\(_4\) to form the nitronium ion (NO\(_2^+\)).

\[ HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^- \]

Step 3: Role of electrophile.

This nitronium ion attacks the benzene ring of benzaldehyde to carry out nitration.


Step 4: Conclusion.

Thus, the electrophile is NO\(_2^+\) (nitronium ion).
Quick Tip: Nitration always involves the nitronium ion (NO\(_2^+\)) as the electrophile.

Step 1: Definition of disaccharide.

A disaccharide is a carbohydrate formed when two monosaccharides are joined by a glycosidic bond.


Step 2: Example.

The most common disaccharide is sucrose, formed from glucose + fructose.


Step 3: Molecular formula.

The molecular formula of sucrose is C\(_{12}\)H\(_{22}\)O\(_{11}\).


Step 4: Conclusion.

Thus, one disaccharide example is sucrose with formula C\(_{12}\)H\(_{22}\)O\(_{11}\).
Quick Tip: Other disaccharides include maltose and lactose (same molecular formula: C\(_{12}\)H\(_{22}\)O\(_{11}\)).

Step 1: Recall definitions.

- A coordination compound consists of a central metal atom/ion bonded to ligands forming a complex ion.
- A double salt is formed by crystallization of two salts together.


Step 2: Key distinction.

Coordination compounds retain their identity in solution (complex ion remains intact), while double salts dissociate completely.


Step 3: Conclusion.

Thus, coordination compounds differ from double salts in terms of ion dissociation and stability in solution.
Quick Tip: Coordination compounds keep their complex identity in water, but double salts lose it by dissociation.

Step 1: Oxidation state of Ni.

In [Ni(CN)\(_4\)]\(^{2-}\), Ni is in +2 oxidation state. Electronic configuration of Ni atom = [Ar] 3d\(^8\)4s\(^2\). For Ni\(^{2+}\) = [Ar] 3d\(^8\).


Step 2: Effect of strong ligand (CN\(^-\)).

CN\(^-\) is a strong field ligand (according to spectrochemical series). It causes pairing of 3d electrons. Thus configuration becomes: 3d\(^{10}\).


Step 3: Hybridisation.

Now Ni\(^{2+}\) uses one 3d, one 4s, and two 4p orbitals → dsp\(^2\) hybridisation. This gives a square planar geometry.


Step 4: Conclusion.

Therefore, Ni in [Ni(CN)\(_4\)]\(^{2-}\) undergoes dsp\(^2\) hybridisation with square planar shape.
Quick Tip: Strong field ligands like CN\(^-\) cause pairing → dsp\(^2\) (square planar). Weak field ligands → sp\(^3\) (tetrahedral).

Step 1: General expression for first order kinetics.

For a first order reaction, the integrated rate law is: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \]

Step 2: Condition for half-life.

At half-life, \([R] = \frac{[R]_0}{2}\).


Step 3: Substitution.
\[ k = \frac{2.303}{t_{1/2}} \log \frac{[R]_0}{[R]_0/2} = \frac{2.303}{t_{1/2}} \log 2 \] \[ k = \frac{0.693}{t_{1/2}} \]

Step 4: Rearranging.
\[ t_{1/2} = \frac{0.693}{k} \] Quick Tip: Half-life for a first order reaction is independent of initial concentration.

Step 1: Formula.

The complex is K\(_4\)[Fe(CN)\(_6\)]. Potassium has charge +1. Cyanide ion (CN\(^-\)) has charge –1.


Step 2: Calculation of oxidation state.

Let oxidation state of Fe = \(x\). \[ 4(+1) + x + 6(-1) = 0 \] \[ 4 + x - 6 = 0 \implies x - 2 = 0 \implies x = +2 \]

Step 3: Coordination number.

The number of ligands directly attached to Fe is 6 (from 6 CN\(^-\) ligands). Hence, coordination number = 6.


Step 4: Conclusion.

Oxidation number of Fe = +2, coordination number = 6.
Quick Tip: Oxidation number is found by charge balance; coordination number = number of ligands directly bonded.

Step 1: Recall integrated rate law for first order.
\[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \]

Step 2: Apply given values.

99% dissociated → remaining concentration = 1% of initial. \[ \frac{[R]_0}{[R]} = \frac{100}{1} = 100 \] \(t = 100 \ min\).


Step 3: Substitution.
\[ k = \frac{2.303}{100} \log (100) \] \[ k = \frac{2.303}{100} \times 2 = 0.04606 \ min^{-1} \]

Step 4: Conclusion.

The velocity constant = \(0.046 \ min^{-1}\).
Quick Tip: In first order kinetics, percentage dissociation is directly related to \(\log \frac{[R]_0}{[R]}\).

Step 1: Conductance.

Resistance (\(R\)) opposes current; conductance is ease of current flow. \[ G = \frac{1}{R} \]

Step 2: Cell constant.

In a conductivity cell: \[ Cell constant = \frac{l}{A} \]
where \(l\) = distance between electrodes, \(A\) = electrode area.

Step 3: Importance.

Cell constant helps to convert measured conductance into specific conductance.
Quick Tip: Conductance = \(1/R\). Cell constant = \(l/A\), crucial for conductivity experiments.

Step 1: Recall formula for density.
\[ \rho = \frac{Z \times M}{N_A \times a^3} \]
where \(Z = 4\) (ccp), \(M = 108 \ g \ mol^{-1}\), \(a = 408.6 \ pm\), \(N_A = 6.022 \times 10^{23}\).

Step 2: Convert edge length.
\[ a = 408.6 \times 10^{-10} \ cm = 4.086 \times 10^{-8} \ cm \]

Step 3: Substitution.
\[ \rho = \frac{4 \times 108}{6.022 \times 10^{23} \times (4.086 \times 10^{-8})^3} \] \[ \rho \approx 10.5 \ g \ cm^{-3} \]

Step 4: Conclusion.

Density of silver = \(10.5 \ g \ cm^{-3}\).
Quick Tip: For ccp/fcc structures, always take \(Z=4\).

Step 1: Recall formula.
\[ \pi = C R T \]
where \(C\) = molarity, \(R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}\), \(T = 27+273 = 300 \ K\).

Step 2: Calculate molarity.

5% (w/v) = 5 g urea in 100 mL solution = 50 g in 1 L.
Moles of urea = \(\frac{50}{60} = 0.833 \ mol\).
Molarity = 0.833 M.

Step 3: Substitution.
\[ \pi = 0.833 \times 0.0821 \times 300 \] \[ \pi \approx 20.5 \ atm \]

Step 4: Conclusion.

Osmotic pressure = \(20.5 \ atm\).
Quick Tip: Osmotic pressure depends directly on molarity and temperature (\(\pi \propto C \times T\)).

Step 1: Recall position.

d-block elements are transition metals, lying in groups 3–12. Their valence electrons enter the (n–1)d orbitals.


Step 2: Properties.

- Variable oxidation states due to similar energies of (n–1)d and ns orbitals.
- Coloured compounds due to d–d electronic transitions.
- Catalytic properties as they provide active sites and variable oxidation states.
- Complex formation due to vacant d-orbitals and high charge density.


Step 3: Conclusion.

Thus, d-block elements have distinct properties like variable oxidation states, colour, catalysis, and complex formation.
Quick Tip: Transition metals are best known for colour, variable oxidation states, and catalytic activity.

Step 1: Compare acidity.

Both phenol and ethanol can release a proton (H\(^+\)).


Step 2: Stability of conjugate base.

- Phenoxide ion is resonance stabilized over the aromatic ring.
- Ethoxide ion has no resonance stabilization, only negative charge localized on oxygen.


Step 3: Conclusion.

Thus, phenol is acidic whereas ethanol is neutral because resonance stabilization makes phenol more likely to lose H\(^+\).
Quick Tip: Acidity depends on stability of conjugate base; resonance increases acidity.

Step 1: Identify parent chain.

Longest chain = propane (3 carbons).


Step 2: Identify substituent.

–OCH\(_2\)CH\(_3\) group is an ethoxy group.


Step 3: Numbering.

The ethoxy group is attached to C-1 of propane.


Step 4: Conclusion.

IUPAC name = 1-ethoxypropane.
Quick Tip: In ethers, name smaller alkyl group as alkoxy substituent on main chain.

Step 1: Recall concept of voids.

Voids are empty spaces in close packing of spheres.


Step 2: Tetrahedral void formation.

When three spheres form a triangle in one layer and a sphere from the next layer is placed above it, the resulting void is tetrahedral in shape.


Step 3: Ratio.

Number of tetrahedral voids = 2 × number of spheres.


Step 4: Conclusion.

Thus, tetrahedral voids are small empty spaces formed in close packing with tetrahedral geometry.
Quick Tip: Every atom in close packing is associated with 2 tetrahedral voids.

Step 1: Recall osmotic pressure formula.
\[ \pi = i C R T \]
where \(i\) = van’t Hoff factor.


Step 2: For urea.

Urea is non-electrolyte → \(i=1\).


Step 3: For NaCl.

NaCl dissociates completely → \(i=2\). So effective concentration doubles.


Step 4: Conclusion.

Osmotic pressure of NaCl solution is higher than that of urea.
Quick Tip: Electrolytes give higher colligative properties due to dissociation (\(i > 1\)).

Step 1: Nature of reaction.

Aliphatic primary amines react with nitrous acid to form alcohols, releasing nitrogen gas.


Step 2: Reaction.

Ethanamine + nitrous acid → Ethanol + N\(_2\) + H\(_2\)O.


Step 3: Conclusion.

Hence ethanamine gives ethanol by reaction with NaNO\(_2\)/HCl.
Quick Tip: Primary aliphatic amines liberate N\(_2\) gas on reaction with nitrous acid – useful for identification.

Step 1: Recall Hinsberg test.

Primary amines react with Hinsberg’s reagent (benzenesulphonyl chloride) to form sulphonamide, soluble in alkali.


Step 2: Reaction.

Ethanamine + C\(_6\)H\(_5\)SO\(_2\)Cl → Ethanamide sulphonamide (soluble in alkali).


Step 3: Importance.

This reaction is used to distinguish primary, secondary, and tertiary amines.


Step 4: Conclusion.

Ethanamine forms a soluble sulphonamide with Hinsberg’s reagent.
Quick Tip: Hinsberg’s reagent helps in classification of amines: primary = soluble, secondary = insoluble, tertiary = no reaction.

Step 1: Type of reaction.

When sodium nitrate is heated with concentrated sulphuric acid, nitric acid vapours are liberated.


Step 2: Balanced reaction.
\[ NaNO_3 + H_2SO_4 \xrightarrow{\Delta} NaHSO_4 + HNO_3 \uparrow \]

Step 3: Conclusion.

This is the laboratory method of preparing nitric acid.
Quick Tip: Conc. H\(_2\)SO\(_4\) acts as a strong dehydrating acid to liberate HNO\(_3\).

Step 1: Oxidising nature.

Conc. HNO\(_3\) acts as a powerful oxidising agent.


Step 2: Reaction with iodine.

Iodine is oxidised to iodic acid (HIO\(_3\)) and HNO\(_3\) is reduced to NO\(_2\).


Step 3: Balanced reaction.
\[ 6 HNO_3 + I_2 \rightarrow 2 HIO_3 + 6 NO_2 + 2 H_2O \]

Step 4: Conclusion.

Thus, conc. HNO\(_3\) oxidises iodine to iodic acid.
Quick Tip: Conc. HNO\(_3\) is a strong oxidiser; it readily converts halogens to higher oxo-acids.

Step 1: Nature of reaction.

Nitric acid oxidises zinc to zinc nitrate, liberating nitrogen oxides.


Step 2: Reaction.

With conc. HNO\(_3\): \[ Zn + 4 HNO_3 \rightarrow Zn(NO_3)_2 + 2 NO_2 + 2 H_2O \]

With dilute HNO\(_3\): \[ Zn + 2 HNO_3 \rightarrow Zn(NO_3)_2 + H_2 \]

Step 3: Conclusion.

Thus, nitric acid reacts with zinc producing zinc nitrate and nitrogen oxides (or H\(_2\) with dilute acid).
Quick Tip: Conc. HNO\(_3\) acts as oxidiser, dilute HNO\(_3\) acts more like a normal acid.

Step 1: Principle.

Electric discharge splits O\(_2\) molecules into oxygen atoms, which combine with O\(_2\) to form O\(_3\).


Step 2: Reaction.
\[ 3 O_2 \xrightarrow{electric discharge} 2 O_3 \]

Step 3: Apparatus.

Silent discharge (ozoniser) is used to prevent decomposition of ozone by spark.


Step 4: Conclusion.

Thus, ozone is prepared by passing silent electric discharge through oxygen.
Quick Tip: Always use silent discharge to avoid decomposition of ozone.

Step 1: Nature of reaction.

Ozone is a strong oxidising agent.


Step 2: Reaction.

It oxidises PbS (black) to PbSO\(_4\) (white).


Step 3: Conclusion.

This is used as a test for ozone.
Quick Tip: PbS (black) \(\rightarrow\) PbSO\(_4\) (white): classic ozone test.

Step 1: Reactants.

Nitric oxide (NO) reacts rapidly with ozone.


Step 2: Reaction.
\[ NO + O_3 \rightarrow NO_2 + O_2 \]

Step 3: Conclusion.

Thus ozone oxidises NO to NO\(_2\).
Quick Tip: Ozone readily oxidises NO to NO\(_2\), an important atmospheric reaction.

Step 1: Structural formula.

The open-chain form of D-glucose: \[ CH_2OH - CHOH - CHOH - CHOH - CHOH - CHO \]
It has 6 carbons, 5 hydroxyl groups, and 1 aldehyde group.


Step 2: IUPAC name.

As an aldohexose, the correct name is 2,3,4,5,6-pentahydroxyhexanal. In stereochemical form, it is (2R,3S,4R,5R).


Step 3: Test for aldehyde group.

- Fehling’s Test: On heating with Fehling’s solution, glucose reduces Cu\(^{2+}\) to red precipitate of Cu\(_2\)O.
- Tollen’s Test: On heating with ammoniacal silver nitrate, glucose reduces Ag\(^{+}\) to metallic silver (silver mirror).


Step 4: Conclusion.

Thus, D-glucose contains an aldehyde group confirmed by positive Fehling’s and Tollen’s tests.
Quick Tip: Glucose is an aldohexose; aldehyde group is confirmed by classical silver mirror and Fehling’s tests.

Step 1: Understanding protein structure.

Proteins have four levels of structure: primary (sequence of amino acids), secondary (α-helix, β-sheet), tertiary (3D folding), and quaternary (multiple chains).

Step 2: What is denaturation?

Denaturation is the process in which the secondary, tertiary, and quaternary structures of proteins are destroyed, while the primary structure (amino acid sequence) remains unchanged.

Step 3: Causes.

Denaturation occurs due to heat, acids, alkalis, organic solvents, or heavy metals that disrupt hydrogen bonds, ionic bonds, and hydrophobic interactions.

Step 4: Example.

The most common example is the coagulation of egg white (albumin) when boiled – it turns from soluble to insoluble solid.

Step 5: Conclusion.

Hence, denaturation destroys biological activity of proteins by altering their shape. Quick Tip: Denaturation changes shape of proteins, not their peptide bond sequence.

Step 1: Nature of amino acids.

Amino acids contain both an acidic group (–COOH) and a basic group (–NH\(_2\)).

Step 2: Behaviour in water.

In aqueous solution, –COOH donates a proton forming –COO\(^{-}\), while –NH\(_2\) accepts a proton forming –NH\(_3^+\).

Step 3: Dipolar form.

Thus, the molecule carries both positive and negative charges simultaneously, existing as a zwitterion.

Step 4: Example.

For glycine: \[ H_2N-CH_2-COOH \ \longrightarrow \ ^+H_3N-CH_2-COO^- \]

Step 5: Conclusion.

At isoelectric point, the amino acid exists mainly in zwitterionic form with no net charge. Quick Tip: Zwitter ions are electrically neutral overall, but carry both +ve and –ve charges.

Step 1: Structural role.

Proteins provide support and strength to body structures (collagen in connective tissues, keratin in hair, nails, skin).

Step 2: Catalytic role.

Proteins act as enzymes that catalyse biochemical reactions (e.g., amylase, protease).

Step 3: Transport and storage.

Some proteins carry vital substances – haemoglobin transports oxygen, myoglobin stores oxygen in muscles.

Step 4: Defence mechanism.

Antibodies are protein molecules that protect the body against infections.

Step 5: Hormonal regulation.

Certain hormones like insulin and glucagon are proteins that regulate metabolism.

Step 6: Conclusion.

Proteins are indispensable biomolecules for structural, catalytic, protective, transport, and regulatory functions. Quick Tip: Proteins = body’s building blocks: structure, enzymes, transport, defence, and regulation.

Step 1: Structural formula.

Benzaldehyde has formula C\(_6\)H\(_5\)CHO. \[ \]

Step 2: Reaction with NH\(_2\)NH\(_2\).

Benzaldehyde reacts with hydrazine to form benzaldehyde hydrazone: \[ C_6H_5CHO + NH_2NH_2 \rightarrow C_6H_5CH=NNH_2 + H_2O \]

Step 3: Reaction with Tollen’s reagent.

Tollen’s reagent oxidises the –CHO group to –COOH, producing benzoic acid and metallic silver (silver mirror): \[ C_6H_5CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow C_6H_5COO^- + 2Ag \downarrow + 4NH_3 + 2H_2O \]

Step 4: Reaction with NaOH (Cannizzaro reaction).

Benzaldehyde (without α-H) undergoes disproportionation in presence of conc. NaOH: \[ 2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \]

Step 5: Conclusion.

Thus, benzaldehyde reacts with hydrazine (condensation), Tollen’s reagent (oxidation), and NaOH (Cannizzaro reaction). Quick Tip: Benzaldehyde lacks α-H, hence undergoes Cannizzaro reaction instead of aldol condensation.

Step 1: IUPAC name.

Acetaldehyde = Ethanal (CH\(_3\)CHO).

Step 2: Reaction with NaHSO\(_3\).
\[ CH_3CHO + NaHSO_3 \rightarrow CH_3CH(OH)SO_3Na \]

Step 3: Reaction with NaOH.

Aldol condensation: \[ 2CH_3CHO \xrightarrow{NaOH} CH_3CH(OH)CH_2CHO \ (\beta-hydroxybutanal) \]

Step 4: Reaction with NH\(_2\)NH\(_2\).
\[ CH_3CHO + NH_2NH_2 \rightarrow CH_3CH=NNH_2 + H_2O \]

Step 5: Reaction with HCN.
\[ CH_3CHO + HCN \rightarrow CH_3CH(OH)CN \ (acetaldehyde cyanohydrin) \]

Step 6: Conclusion.

Thus, acetaldehyde shows addition and condensation reactions due to its –CHO group. Quick Tip: Aldehydes readily undergo nucleophilic addition reactions at the carbonyl carbon.

Step 1: Reactivity comparison.

- In chloroethane, the C–Cl bond is a simple polar covalent bond; chlorine can easily undergo nucleophilic substitution.
- In chlorobenzene, the lone pair of chlorine interacts with the benzene ring (\(\pi\)-resonance). This delocalisation gives partial double bond character to the C–Cl bond.

Step 2: Effect of resonance.

Due to resonance:
1. Bond length of C–Cl decreases.
2. Bond strength increases.
3. Nucleophilic substitution becomes difficult.

Step 3: Reaction with Cl\(_2\).

In presence of FeCl\(_3\), chlorobenzene undergoes electrophilic substitution: \[ C_6H_5Cl + Cl_2 \xrightarrow{FeCl_3} C_6H_4Cl_2 + HCl \]
Products: 1,2-dichlorobenzene (ortho) and 1,4-dichlorobenzene (para).

Step 4: Reaction with conc. H\(_2\)SO\(_4\).

Sulphonation occurs: \[ C_6H_5Cl + H_2SO_4 \xrightarrow{\Delta} C_6H_4ClSO_3H + H_2O \]
Product: p-chlorobenzene sulphonic acid (major).

Step 5: Conclusion.

Thus, chlorobenzene is less reactive than chloroethane due to resonance, and it undergoes electrophilic substitution with Cl\(_2\) and H\(_2\)SO\(_4\). Quick Tip: In aryl halides, resonance stabilisation makes C–Cl bond strong and unreactive toward nucleophiles, but benzene ring favours electrophilic substitution.

(i) Electrophilic substitution in halobenzene

Step 1: Nature of halogen substituent.

In halobenzene, halogen atoms are deactivating due to the –I effect (electron withdrawing inductive effect). However, they also exhibit +R effect (electron donating by resonance), which increases electron density at ortho and para positions.

Step 2: Directive influence.

Because of this dual effect, halogens are ortho/para directing but overall deactivate the ring towards electrophilic substitution.

Step 3: Example reaction.

Chlorobenzene + Cl\(_2\) (in presence of FeCl\(_3\)): \[ C_6H_5Cl + Cl_2 \xrightarrow{FeCl_3} o-C_6H_4Cl_2 + p-C_6H_4Cl_2 \]

Step 4: Conclusion.

Thus, halobenzene undergoes electrophilic substitution at ortho and para positions, but at a slower rate than benzene. Quick Tip: Halogen = ortho/para directing but deactivating.