Step 1: Understanding the properties of solids.

- Molecular solids are soft and poor conductors.

- Ionic solids conduct only in molten/aqueous form.

- Metallic solids are good conductors, ductile and malleable.

- Coordinate solids are not ductile or conductive like metals.


Step 2: Conclusion.

Hence, the correct answer is Metallic solid.
Quick Tip: Metallic solids show metallic bonding ("sea of electrons") responsible for conductivity and ductility.

Step 1: Mass percent formula.

Mass percent = \( \frac{Mass of solute}{Mass of solution} \times 100 \).


Step 2: Calculation.
\[ \frac{Mass of solute}{200} \times 100 = 10 \quad \Rightarrow \quad Mass of solute = 20 \, g \]

Step 3: Conclusion.

Therefore, glucose present = 20 g.
Quick Tip: For mass percent: Mass of solute = (Mass percent × Mass of solution)/100.

Step 1: Dimensions of k.

- For zero order: unit = mol L\(^{-1}\) s\(^{-1}\).

- For first order: unit = s\(^{-1}\).

- For second order: unit = L mol\(^{-1}\) s\(^{-1}\).


Step 2: Given unit.

Here, \(k = 2.3 \times 10^{-5} \, L\, mol^{-1}\, s^{-1}\), which matches second order.


Step 3: Conclusion.

Thus, the reaction is second order.
Quick Tip: Always determine reaction order by comparing units of rate constant.

Step 1: Bases in RNA.

RNA contains Adenine (A), Guanine (G), Cytosine (C), and Uracil (U).


Step 2: Absence of Thymine.

Thymine is present only in DNA, where it pairs with Adenine. In RNA, Uracil replaces Thymine.


Step 3: Conclusion.

Therefore, the base not present in RNA is Thymine.
Quick Tip: DNA has Thymine, while RNA has Uracil instead.

Step 1: Understanding Cannizzaro’s reaction.

Cannizzaro’s reaction occurs in aldehydes that lack \(\alpha\)-hydrogen atoms.


Step 2: Application.

- Benzaldehyde has no \(\alpha\)-hydrogen → undergoes Cannizzaro reaction.

- Benzoic acid is not an aldehyde → does not undergo.

- Toluene is hydrocarbon → no reaction.

- Formic acid is carboxylic acid → no reaction.


Step 3: Conclusion.

Therefore, the compound is Benzaldehyde.
Quick Tip: Remember: Aldehydes without \(\alpha\)-hydrogen undergo Cannizzaro’s reaction.

Step 1: Reaction principle.

Carbylamine reaction (isocyanide test) occurs with primary amines.


Step 2: Application.

- Primary amine (CH\(_3\)NH\(_2\)) gives foul-smelling isocyanide.

- Secondary and tertiary amines (B, C, D) do not give the reaction.


Step 3: Conclusion.

Hence, the correct answer is CH\(_3\)NH\(_2\).
Quick Tip: Carbylamine test is a qualitative test to detect primary amines.

Step 1: Atoms per unit cell.

In a simple cubic lattice, each corner atom is shared by 8 unit cells. Thus, number of atoms per unit cell = \(\frac{1}{8} \times 8 = 1\).


Step 2: Relation between atomic radius and edge length.

For simple cubic: \(a = 2r\).


Step 3: Volume of atom and unit cell.

- Volume of one atom = \(\frac{4}{3}\pi r^3\).

- Volume of unit cell = \(a^3 = (2r)^3 = 8r^3\).


Step 4: Packing efficiency.
\[ Packing efficiency = \frac{Volume of atoms in cell}{Volume of unit cell} \times 100 = \frac{\frac{4}{3}\pi r^3}{8r^3} \times 100 \approx 52.4% \]

Step 5: Conclusion.

Thus, the packing efficiency of a simple cubic lattice is 52.4%.
Quick Tip: Remember: Packing efficiency is lowest for simple cubic (52.4%), higher for bcc (68%), and highest for fcc (74%).

Step 1: Relation in bcc structure.

In bcc, body diagonal = \(4r = \sqrt{3}a\).

Here, edge length \(a = 200 \, pm = 2 \times 10^{-8} \, cm\).


Step 2: Volume of unit cell.
\[ a^3 = (2 \times 10^{-8})^3 = 8 \times 10^{-24} \, cm^3 \]

Step 3: Mass of unit cell.
\[ Mass = Density \times Volume = 7 \times 8 \times 10^{-24} = 5.6 \times 10^{-23} \, g \]

Step 4: Atoms per unit cell.

In bcc: 2 atoms/unit cell.

So, molar mass = \(\frac{Mass of unit cell \times N_A}{2}\).
\[ M = \frac{5.6 \times 10^{-23} \times 6.022 \times 10^{23}}{2} \approx 17 \, g/mol \]

Step 5: Number of atoms in 20 g.
\[ Moles = \frac{20}{17} \approx 1.18 \, mol \] \[ Atoms = 1.18 \times 6.022 \times 10^{23} \approx 1.2 \times 10^{23} \]

Step 6: Conclusion.

Number of atoms in 20 g element = \(\approx 1.2 \times 10^{23}\).
Quick Tip: In bcc, 2 atoms per unit cell; always use density formula: \(\rho = \frac{Z \times M}{a^3 \times N_A}\).

(i) Reason.

Conductivity decreases with dilution because the number of ions per unit volume decreases, reducing total current carrying capacity.


(ii) Calculation.

Molar conductivity: \[ \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{0.248 \times 1000}{0.20} = 124 \, S \, cm^2 \, mol^{-1} \]

Step 3: Conclusion.

(i) Fewer ions per volume → lower conductivity.

(ii) Molar conductivity = 124 S cm\(^2\) mol\(^{-1}\).
Quick Tip: Conductivity decreases with dilution, but molar conductivity increases due to increased ion mobility.

Step 1: First-order rate equation.
\[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \]

Step 2: Substitution.
\[ k = \frac{2.303}{30} \log \frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}} = \frac{2.303}{30} \log (6.2) \]

Step 3: Simplification.
\[ k = \frac{2.303}{30} \times 0.7924 \approx 0.044 \, min^{-1} \]

Step 4: Conclusion.

The velocity constant is \(\approx 0.044 \, min^{-1}\).
Quick Tip: Always use \(\log_{10}\) in first-order kinetics formula: \(k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\).

(i) Explanation.

Physical adsorption is exothermic in nature. According to Le Chatelier’s principle, increasing temperature decreases the extent of exothermic processes, hence adsorption decreases.


(ii) Explanation.

Powdered materials have more surface area exposed compared to crystalline solids. Adsorption being a surface phenomenon, greater surface area enhances adsorption.


Conclusion.

(i) Adsorption decreases with temperature.

(ii) Powdered form acts as a better adsorbent due to large surface area.
Quick Tip: Remember: Adsorption is surface phenomenon, and higher surface area leads to higher adsorption.

(i) Explanation.

Nitrogen has only \(2s\) and \(2p\) orbitals, no vacant d-orbitals. Therefore, nitrogen cannot expand its octet to form NCl\(_5\). But phosphorus (in PCl\(_5\)) can, due to vacant 3d orbitals. Hence only NCl\(_3\) exists.


(ii) Explanation.

Halogens have high electronegativity and very high electron affinity. They readily accept electrons to form halide ions, making them strong oxidising agents.
Quick Tip: NCl\(_5\) cannot form as nitrogen lacks d-orbitals; halogens are strong oxidisers due to high electron affinity.

Bidentate ligands.

These have two donor atoms that coordinate simultaneously to the metal ion. Example: Ethylenediamine (en), \(H_2N-CH_2CH_2NH_2\).


Ambidentate ligands.

These have two donor atoms but can bind through only one at a time, leading to linkage isomerism. Example: NO\(_2^-\) (binds via N or O).


Conclusion.

Bidentate ligands → two donor atoms at once; Ambidentate ligands → two possible donor atoms, but only one attaches.
Quick Tip: Bidentate ligands increase stability of complexes (chelation effect); ambidentate ligands cause linkage isomerism.

Step 1: D-Glucose.

It is an aldohexose (6-carbon sugar with an aldehyde group at C-1). Open-chain form: \(CH_2OH-(CHOH)_4-CHO\). In cyclic form, it forms a pyranose ring.


Step 2: D-Fructose.

It is a ketohexose (6-carbon sugar with a ketone group at C-2). Open-chain form: \(CH_2OH-C=O-(CHOH)_3-CH_2OH\). In cyclic form, it forms a furanose ring.


Step 3: Key Difference.

Glucose has an aldehyde (aldohexose), Fructose has a ketone (ketohexose).
Quick Tip: Glucose = aldohexose; Fructose = ketohexose. Both are hexoses but differ in functional groups.

Step 1: Calculate cell constant.

Conductivity (\(\kappa\)) = Cell constant / Resistance.

For 0.1 M KCl: \[ 1.29 = \frac{Cell constant}{100} \quad \Rightarrow \quad Cell constant = 129 \, m^{-1} \]

Step 2: Find conductivity for 0.02 M KCl.
\[ \kappa = \frac{Cell constant}{R} = \frac{129}{500} = 0.258 \, S \, m^{-1} \]

Step 3: Find molar conductivity.
\[ \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{0.258 \times 1000}{0.02 \times 1000} = 12.9 \, S \, cm^2 \, mol^{-1} \]

Step 4: Conclusion.

Conductivity = \(\mathbf{0.258 \, S \, m^{-1}}\), Molar conductivity = \(\mathbf{12.9 \, S \, cm^2 \, mol^{-1}}\).
Quick Tip: Always calculate cell constant from a standard solution first, then apply it to other resistances.

(i) True solution vs Suspension.

- True solution: Homogeneous, particle size < 1 nm, particles invisible, stable (e.g., sugar solution).

- Suspension: Heterogeneous, particle size > 1000 nm, visible particles, unstable (e.g., chalk in water).


(ii) Lyophilic vs Lyophobic colloids.

- Lyophilic: Solvent-loving, stable, easily formed, reversible (e.g., starch sol, gelatin).

- Lyophobic: Solvent-hating, unstable, requires special methods, irreversible (e.g., gold sol).


(iii) Multimolecular vs Macromolecular colloids.

- Multimolecular: Colloidal particles are aggregates of small molecules (e.g., gold sol).

- Macromolecular: Large molecules act as colloids (e.g., starch, proteins).
Quick Tip: Remember: Stability order → True solution > Lyophilic colloid > Lyophobic colloid > Suspension.

(i) Secondary structure of proteins.

Formed by hydrogen bonding between peptide chains. Two main types: \(\alpha\)-helix (spiral) and \(\beta\)-pleated sheet (zig-zag). Gives stability and shape to proteins.


(ii) Peptide bond.

A covalent linkage \(-CO-NH-\) formed between carboxyl group of one amino acid and amino group of another. Responsible for formation of polypeptides and proteins.


(iii) Monosaccharides.

Simplest carbohydrates (C\(_n\)H\(_{2n}\)O\(_n\)). Cannot be hydrolysed further. Examples: glucose, fructose, galactose. They are reducing sugars.
Quick Tip: Proteins → peptides → amino acids; Carbohydrates → monosaccharides are the basic units.

(i) Friedel–Crafts reaction.

Aniline reacts with Lewis acid catalysts (like AlCl\(_3\)) forming insoluble complexes. This deactivates the benzene ring, preventing Friedel–Crafts alkylation/acylation.


(ii) Solubility difference.

Ethyl amine: Small size, strong H-bonding with water → soluble.

Aniline: Bulkier phenyl group decreases polarity and reduces hydrogen bonding with water → very low solubility.


Step 3: Conclusion.

(i) Aniline does not undergo Friedel–Crafts due to catalyst complexation.

(ii) Ethyl amine soluble; Aniline not soluble.
Quick Tip: Remember: Solubility in water depends on hydrogen bonding capacity and molecular size.

Step 1: Definitions.

- Osmosis: Flow of solvent molecules through a semipermeable membrane from lower solute concentration to higher solute concentration.

- Osmotic pressure: The pressure required to stop osmosis.


Step 2: Moles of protein.
\[ n = \frac{Mass}{Molar mass} = \frac{1.26}{61022} \approx 2.07 \times 10^{-5} \, mol \]

Step 3: Concentration of solution.

Volume = 200 cm\(^3\) = 0.200 L.
\[ C = \frac{n}{V} = \frac{2.07 \times 10^{-5}}{0.200} = 1.035 \times 10^{-4} \, mol/L \]

Step 4: Osmotic pressure formula.
\[ \pi = C R T = (1.035 \times 10^{-4})(0.0821)(300) \approx 0.103 \, atm \]

Step 5: Conclusion.

Osmotic pressure = \(\mathbf{0.103 \, atm}\).
Quick Tip: For dilute solutions, osmotic pressure follows \(\pi = CRT\), analogous to ideal gas law.

(i) Velocity of reaction.

The velocity (rate) of a chemical reaction is the change in concentration of reactants or products per unit time. \[ Rate = -\frac{d[Reactant]}{dt} = \frac{d[Product]}{dt} \]

(ii) Raoult’s law.

According to Raoult’s law, for a solution of non-volatile solute: \[ \frac{\Delta p}{p^0} = x_{solute} \]
where \(\Delta p = p^0 - p\), \(p^0\) = vapour pressure of pure solvent, \(x_{solute}\) = mole fraction of solute.


Step 3: Conclusion.

- Reaction velocity = rate of change of concentration.

- Raoult’s law = vapour pressure lowering proportional to solute mole fraction.
Quick Tip: Reaction rate measures speed of reaction; Raoult’s law explains colligative properties.

(i) Coloured compounds.

Transition metals have partially filled d-orbitals. When light falls, electrons undergo d–d transitions, absorbing certain wavelengths, and the complementary colour is observed. Example: \(Ti^{3+}\) (violet), \(Cu^{2+}\) (blue).


(ii) Paramagnetism.

Transition metals and their ions often contain unpaired d-electrons. These unpaired electrons produce magnetic moments, leading to paramagnetic behaviour. Example: Fe\(^{2+}\), Mn\(^{2+}\).
Quick Tip: Colour in transition metal compounds arises from d–d transitions; paramagnetism from unpaired d-electrons.

(i) [CrCl\(_2\)(en)\(_2\)]Cl: Coordination sphere contains 2 Cl\(^-\) ligands and 2 en ligands. Oxidation state of Cr = +3. Name = Dichloridobis(ethane-1,2-diamine)chromium(III) chloride.


(ii) Cs[FeCl\(_4\)]: Complex anion = [FeCl\(_4\)]\(^-\). Oxidation state of Fe = +3. Name = Caesium tetrachloridoferrate(III).


(iii) K\(_3\)[Co(C\(_2\)O\(_4\))\(_3\)]: Oxidation state of Co = +3. Oxalate = bidentate ligand. Name = Potassium tris(oxalato)cobaltate(III).


(iv) [CoCl\(_3\)(NH\(_3\))\(_3\)]: Coordination sphere has 3 Cl\(^-\), 3 NH\(_3\). Oxidation state of Co = +3. Name = Triamminetrichloridocobalt(III).
Quick Tip: Always name ligands alphabetically, specify coordination number, and oxidation state in Roman numerals.

Step 1: Manufacture of SO\(_2\).

SO\(_2\) is produced by burning sulphur or roasting metal sulphides in air/oxygen.


Step 2: Important reactions.
\[ 2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2 \uparrow \] \[ 4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2 \uparrow \]

Step 3: Reactions of H\(_2\)SO\(_4\).
\[ CaF_2 + H_2SO_4 \rightarrow CaSO_4 + 2HF \uparrow \] \[ Cu + 2H_2SO_4 \, (conc.) \rightarrow CuSO_4 + SO_2 \uparrow + 2H_2O \] \[ S + 2H_2SO_4 \, (conc.) \rightarrow 3SO_2 \uparrow + 2H_2O \]

Step 4: Conclusion.

SO\(_2\) is industrially produced by roasting sulphides, and conc. H\(_2\)SO\(_4\) liberates gases from CaF\(_2\), Cu, and S.
Quick Tip: SO\(_2\) is mainly obtained by roasting metal sulphides in industry, and conc. H\(_2\)SO\(_4\) acts as a dehydrating & oxidising agent.

(i) Iodine with HNO\(_3\).

Nitric acid oxidises iodine to iodic acid: \[ I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O \]

(ii) Chlorine with SO\(_2\).

Chlorine oxidises SO\(_2\) to H\(_2\)SO\(_4\): \[ Cl_2 + SO_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl \]

(iii) Chlorine with hot conc. NaOH.

Disproportionation reaction: \[ 3Cl_2 + 6NaOH \, (hot, conc.) \rightarrow 5NaCl + NaClO_3 + 3H_2O \]

(iv) Chlorine with fluorine.

Formation of chlorine trifluoride: \[ Cl_2 + 3F_2 \rightarrow 2ClF_3 \]

(v) Zinc with dilute nitric acid.

Zinc gives zinc nitrate and nitrogen dioxide gas: \[ Zn + 4HNO_3 \, (dil.) \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O \] Quick Tip: Chlorine often shows disproportionation reactions with alkalis, while nitric acid oxidises both iodine and zinc.

(i) 2-chloro-3-methylpentane.

Pentane chain, Cl at C-2, CH\(_3\) at C-3: \[ CH_3-CH(Cl)-CH(CH_3)-CH_2-CH_3 \]

(ii) 1,4-dibromobut-2-ene.

4-carbon chain, double bond at C-2, Br at C-1 and C-4: \[ Br-CH_2-CH=CH-CH_2-Br \]

(iii) 1-chloro-2-methylbenzene.

Benzene ring with Cl at position 1, CH\(_3\) at position 2 (ortho-chlorotoluene).


(iv) 1-chloro-4-ethylcyclohexane.

Cyclohexane ring with Cl at C-1 and –CH\(_2\)CH\(_3\) at C-4.


(v) 3-bromo-2-methylbut-2-ene.

4-carbon chain with double bond at C-2, CH\(_3\) substituent at C-2, Br at C-3: \[ CH_2=C(CH_3)-CH(Br)-CH_3 \] Quick Tip: Always select the longest chain, number to give lowest locants, and place substituents alphabetically in naming.

(i) n-Butyl chloride + alc. KOH.

Dehydrohalogenation occurs, giving 1-butene: \[ C_4H_9Cl + alc. KOH \rightarrow C_4H_8 + HCl \]

(ii) Methyl iodide + Mg (dry ether).

Grignard reagent formation: \[ CH_3I + Mg \rightarrow CH_3MgI \]

(iii) Methyl bromide + Na (dry ether).

Wurtz reaction gives ethane: \[ 2CH_3Br + 2Na \rightarrow C_2H_6 + 2NaBr \]

(iv) Methyl iodide + KCN.

Forms methyl cyanide: \[ CH_3I + KCN \rightarrow CH_3CN + KI \]

(v) Chlorobenzene + aq. NaOH.

At high temp. and pressure forms phenol: \[ C_6H_5Cl + NaOH \, (300^\circ C, 200 \, atm) \rightarrow C_6H_5OH + NaCl \] Quick Tip: Alc. KOH causes elimination, Mg in ether gives Grignard reagents, Na gives Wurtz products, KCN gives nitriles, and chlorobenzene needs drastic conditions for nucleophilic substitution.

Step 1: Industrial manufacture of ethanol.

(i) Fermentation: Glucose is converted into ethanol using enzymes. \[ C_6H_{12}O_6 \xrightarrow{zymase} 2C_2H_5OH + 2CO_2 \]

(ii) From ethene: Ethene reacts with steam at 300°C in presence of phosphoric acid catalyst. \[ C_2H_4 + H_2O \xrightarrow{H_3PO_4, 300^\circ C} C_2H_5OH \]

Step 2: Dehydrogenation of ethanol.

On heating with Cu at 573 K: \[ C_2H_5OH \xrightarrow{Cu, 573K} CH_3CHO + H_2 \]
Thus, ethanol gives acetaldehyde.


Step 3: Dehydration mechanism (acidic).
\[ C_2H_5OH \xrightarrow{H_2SO_4, 443K} C_2H_4 + H_2O \]

\underline{Mechanism:
1. Protonation of OH group → formation of oxonium ion.

2. Loss of water → carbocation formation.

3. Rearrangement → elimination of H\(^+\) to give ethene.


Step 4: Conclusion.

- Ethanol industrially: fermentation/hydration of ethene.

- Dehydrogenation product = acetaldehyde.

- Dehydration (acidic) = ethene.
Quick Tip: Fermentation is biological; industrially, ethene hydration is faster. Cu gives dehydrogenation, H\(_2\)SO\(_4\) gives dehydration.

(i) Phenol + conc. HNO\(_3\).

Forms picric acid (2,4,6-trinitrophenol). \[ C_6H_5OH + 3HNO_3 \rightarrow C_6H_2(NO_2)_3OH + 3H_2O \]

(ii) Phenol + CHCl\(_3\) + NaOH (Reimer–Tiemann).

Forms salicylaldehyde. \[ C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O \]

(iii) Ether cleavage with HI.
\[ CH_3CH_2CH_2-O-C(CH_3)_2CH_2CH_3 + HI \rightarrow CH_3CH_2CH_2OH + (CH_3)_2CHCH_2CH_2I \]
C = Propanol, D = 2-iodo-2-methylbutane.


(iv) Oxidation of isopropanol.
\[ CH_3CH(OH)CH_3 \xrightarrow{CrO_3} CH_3COCH_3 \]
E = Acetone.


(v) Reduction of acetic acid.
\[ CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH \]
F = Ethanol.


Step 3: Conclusion.

- A = Picric acid.

- B = Salicylaldehyde.

- C = Propanol.

- D = 2-iodo-2-methylbutane.

- E = Acetone.

- F = Ethanol.
Quick Tip: Picric acid forms by nitration, salicylaldehyde by Reimer–Tiemann, ethers split with HI, secondary alcohols oxidise to ketones, and acids reduce to alcohols.

Step 1: 2,4-DNP test.

Compound A gives orange-red precipitate → confirms presence of a carbonyl group (aldehyde or ketone).


Step 2: Iodoform test.

Yellow precipitate with iodine + NaOH → confirms presence of CH\(_3\)–CO group. Hence A must be a methyl ketone.


Step 3: Negative tests.

- No Tollen’s/Fehling’s reduction → not an aldehyde.

- No bromine water decolourisation → not an alkene.


Step 4: Oxidation product.

On oxidation with chromic acid, A gives a monocarboxylic acid B (C\(_7\)H\(_6\)O\(_2\)) = benzoic acid.
\[ C_6H_5COCH_3 \xrightarrow[]{[O]} C_6H_5COOH \]

Step 5: Conclusion.

- A = Acetophenone (C\(_6\)H\(_5\)COCH\(_3\))

- B = Benzoic acid (C\(_6\)H\(_5\)COOH)
Quick Tip: Iodoform test always confirms CH\(_3\)–CO group; oxidation of side chain in aromatic ketones gives benzoic acid.

N/A Quick Tip: Aldol condensation gives β-hydroxy carbonyls; oxidation of aldehydes/alcohols always yields acids.