The one-letter codes for the given amino acids are:

• Arginine: R
• Aspartic acid: D
• Asparagine: N
• Alanine: A

Matching these with the options:

(A) Arginine - (IV) R
(B) Aspartic acid - (I) D
(C) Asparagine - (II) N
(D) Alanine - (III) A

Lead nitrate, Pb(NO₃)₂, is a soluble colorless compound. Confirmatory tests for ions usually involve the formation of a precipitate or a distinctly colored complex. Lead sulphate (white precipitate), lead chromate (yellow precipitate), and lead iodide (yellow precipitate) are all used as confirmatory tests for lead(II) ions. Since lead nitrate is colorless and soluble, it’s not used as a confirmatory test.

For neutralization, the milliequivalents of acid must equal the milliequivalents of base.

Milliequivalents = Molarity × n-factor × Volume

For HBr, n-factor = 1. For Ba(OH)₂, n-factor = 2.

Let V be the volume of HBr required.

0.02 × 1 × V = 0.01 × 2 × 10

0.02V = 0.2
V = 0.2 ÷ 0.02 = 10 mL.

As we move down Group 13, the metallic character increases, and hence the electropositive character increases. As electropositivity increases, the basicity of the oxides increases.

The order of basicity for Group 13 oxides is:

B₂O₃ < Al₂O₃ < Ga₂O₃ < In₂O₃ < Tl₂O₃.

Therefore, Tl₂O₃ is the most basic oxide.

The oxalate anion (C₂O₄²⁻) is a bidentate ligand, and the prefix “tris” indicates there are three oxalate ligands.

The complex anion is [Co(C₂O₄)₃]³⁻, and since the potassium ion has a +1 charge, there must be three potassium ions to balance the charge. The oxidation state of cobalt is +3, indicated by the Roman numeral (III).

Step 1: Recall the de Broglie principle
The circumference of an electron's orbit in an atom is an integer multiple of the electron's wavelength:
2πr = nλ, where r is the radius of the orbit, n is the principal quantum number, and λ is the de Broglie wavelength.

Step 2: Use the formula for the radius of the nth orbit
r_n = n²a₀/Z, where Z is the atomic number. For hydrogen (Z = 1) and n = 3, r₃ = 9a₀.

Step 3: Substitute into the de Broglie equation
2π(9a₀) = 3λ
λ = 18πa₀/3 = 6πa₀.

DDT (Dichlorodiphenyltrichloroethane) and Aldrin are well-known pesticides.
PAN (Peroxyacetyl nitrate) is a component of photochemical smog and not a pesticide.
Sodium chlorate is a herbicide, while Sodium arsinite has been used as an insecticide and rodenticide. Tetrachloroethene is used in dry cleaning.
Thus, the correct pair is DDT and Aldrin.

Key Observations:

• In column chromatography, the polarity of the compound determines the retention time.
• More polar compounds elute later, spending more time in the column.

Order of polarity: a > b > c.
Order of elution: c > b > a.
Incorrect statements: A, B, and C.

Step 1: Understand hydration enthalpy
Hydration enthalpy depends on the charge density (charge/size) of the ion. Smaller ions with higher charge have greater hydration enthalpy.

Step 2: Analyze the given ions
Be²⁺ is the smallest ion among the given alkaline earth metals, so it has the highest charge density and hence the highest hydration enthalpy.

Step 1: Stability of the conjugate base
The strength of an acid is determined by the stability of its conjugate base. Electron-withdrawing groups stabilize the conjugate base, increasing acidity.

Step 2: Analyze substituents
- NO₂ is a strong electron-withdrawing group (-I and -M effects), stabilizing the conjugate base the most.
- CH₃ is electron-donating, reducing acidity.
Thus, nitrophenol is the strongest acid.

Step 1: Protonation of the alcohol
The hydroxyl group is protonated by H₃O⁺, making it a better leaving group.

Step 2: Carbocation formation
Water leaves, generating a secondary carbocation.

Step 3: Cyclization
The double bond attacks the carbocation, forming a six-membered ring. This results in a tertiary carbocation adjacent to the oxygen in the newly formed ring.

Step 4: Deprotonation
A proton is lost to give the final cyclic ether product.

Thus, the correct product is Option 4.

Solid state: BeCl₂ exists as a polymeric structure with each beryllium atom bonded to four chlorine atoms, forming a chain-like structure.

Vapor phase: Below 1200 K, BeCl₂ exists as a dimer (Be₂Cl₄) with chloro-bridging.

High temperature: Above 1200 K, BeCl₂ breaks into monomeric units with a linear structure.

Step 1: Analyze intermediates and activated complexes
Intermediates are represented by valleys in the energy diagram (2 valleys). Activated complexes correspond to peaks (3 peaks).

Step 2: Identify the rate-determining step
The slowest step corresponds to the highest energy barrier, which is Step II.

Electrolysis reactions:

At cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻
At anode: 2Cl⁻ → Cl₂ + 2e⁻

Products formed: H₂, Cl₂, and NaOH. HCl is not formed during electrolysis.

Step 1: Analyze melting and boiling points
Boiling water temperature = 100°C.
- Li and Cs react with water.
- Br₂: Boiling point is 58.8°C (gas at 100°C).
- Ga: Melting point = 29.8°C, Boiling point = 2400°C (liquid at 100°C).

Step 1: Analyze Assertion A
In Ni(CO)₄ and Fe(CO)₅, the carbonyl ligand (CO) is neutral. Thus, the oxidation state of Ni and Fe is zero. Assertion A is correct.

Step 2: Analyze Reason R
Low oxidation states are stabilized by synergistic bonding, involving σ-donation from the ligand and π-backdonation from the metal. CO is a π-acceptor ligand, not a π-donor. Therefore, Reason R is incorrect.

Step 1: Analyze Statement I
Morphine is a narcotic analgesic that relieves pain but induces sleep and drowsiness. Hence, Statement I is false.

Step 2: Analyze Statement II
Morphine and its derivatives are indeed obtained from the opium poppy, making Statement II true.

Step 1: Perform 1,4-addition
The organocuprate reagent (MeMgBr/CuI) adds to the α, β-unsaturated ketone in a 1,4-conjugate addition manner.

Step 2: Alkylate the product
The resulting ketone reacts with n-propyl iodide (nPrI) via an alkylation reaction at the α-carbon.

The final product is Option 3.

Step 1: Reaction in a neutral medium
In a neutral medium, permanganate is reduced to manganese(IV) oxide (MnO₂), with a change in oxidation state of 3.

Step 2: Reaction in an acidic medium
In an acidic medium, permanganate is reduced to Mn²⁺, with a change in oxidation state of 5. The condition specified in the question aligns with a neutral medium.

Step 1: Composition of Nessler’s reagent
Nessler’s reagent is an alkaline solution of potassium tetraiodomercurate(II), K₂[HgI₄].

Step 2: Identify absent element
The reagent contains potassium (K), mercury (Hg), and iodine (I) but not nitrogen (N).

Step 1: Understand oxidation and reduction potentials
For a metal to be oxidized, the reduction potential of NO^-₃ (0.97 V) must be greater than the metal's oxidation potential.

Step 2: Reverse reduction potentials for oxidation potentials
- V: E⁰ (oxidation) = 1.19 V
- Fe: E⁰ (oxidation) = 0.04 V
- Ag: E⁰ (oxidation) = -0.80 V
- Au: E⁰ (oxidation) = -1.40 V

Step 3: Compare with NO^-₃
Metals oxidized: Fe, Ag, Au (E⁰ < 0.97 V).

Step 1: Define body-centered cubic (BCC)
A BCC structure has atoms at each corner of a cube and one atom at the center.

Step 2: Identify systems
BCC unit cells are found in cubic, tetragonal, and orthorhombic systems.

Step 3: Exclude others
Hexagonal, rhombohedral, monoclinic, and triclinic do not have BCC arrangements.

Step 1: Understand the iodoform test
The test is positive for methyl ketones (R-CO-CH₃) and secondary alcohols with CH₃CH(OH)- groups.

Step 2: Analyze compounds
Positive test: 1, 3, 4, 6 (methyl ketones or secondary alcohols with required groups).

Step 1: Analyze the molecular formula
C₈H₁₁N implies a benzene ring (degree of unsaturation = 4) with an amine group.

Step 2: Consider positions of substituents
Possible isomers: Aniline, ortho-, meta-, para-toluidine, and their derivatives.

Step 1: Define isotonic solutions
Solutions with the same osmotic pressure have the same number of solute particles.

Step 2: Analyze dissociation
Each pair has equal particle concentrations after dissociation.

Step 1: Identify colloidal systems
A colloid consists of a dispersed phase and a dispersion medium. For this question, focus on systems where the dispersion medium is liquid.

Step 2: Analyze each system
- Paints: Liquid medium
- Milk: Liquid medium
- Hair cream: Liquid medium
- Froth: Liquid medium
- Soap lather: Liquid medium
Others (gemstones, smoke, cheese, insecticide sprays) do not have a liquid dispersion medium.

Step 3: Count the systems
There are 5 systems with liquid as the dispersion medium: paints, milk, hair cream, froth, and soap lather.

Step 1: Understand hydrogen bonding in ice
Each water molecule in ice has two hydrogen atoms and two lone pairs of electrons. These allow it to form hydrogen bonds with four neighboring water molecules.

Step 2: Analyze the structure
- Each hydrogen atom forms one hydrogen bond as a donor.
- Each lone pair forms one hydrogen bond as an acceptor.
Thus, each water molecule is hydrogen bonded to four others, forming a tetrahedral arrangement.

Step 1: Write the target equation
The target reaction is: 2C(s) + 3H₂(g) + 0.5O₂(g) → C₂H₅OH(l).

Step 2: Use Hess's Law
Combine the given reactions:
- Combustion of carbon: 2C(s) + 2O₂(g) → 2CO₂(g), ΔH = -787 kJ
- Combustion of hydrogen: 3H₂(g) + 1.5O₂(g) → 3H₂O(l), ΔH = -725.4 kJ
- Combustion of ethanol: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH = -1234.7 kJ
Reverse the ethanol combustion reaction to form ethanol.

Step 3: Calculate heat of formation
ΔH_formation = (-787) + (-725.4) + (+1234.7) ≈ -278 kJ mol^-1.

Step 1: Write the equilibrium expression
K_c = [PCl₅]/([PCl₃][Cl₂]) = 20.

Step 2: Update concentrations
After adding 0.2 mol of Cl₂:
[PCl₃] = 0.2 - x, [Cl₂] = 0.3 - x, [PCl₅] = 0.4 + x.

Step 3: Solve for x
20 = (0.4 + x)/[(0.2 - x)(0.3 - x)]. Solve for x to find x ≈ 0.086 mol.

Step 4: Calculate [PCl₅]
[PCl₅] = 0.4 + 0.086 ≈ 0.486 mol L^-1 = 49 × 10^-2 mol L^-1.

Step 1: Identify species with square planar geometry
- XeF₄: Square planar due to lone pairs in axial positions.
- BrF^-₄: Square planar for similar reasons.
- [Cu(NH₃)₄]²⁺: Square planar due to Jahn-Teller distortion.
- [PtCl₄]^2-: Square planar due to dsp² hybridization.
Others (SF₄, SiF₄, [FeCl₄]^2-) do not have square planar geometry.

Step 2: Count the species
There are 4 species with square planar geometry.