Step 1: Recall the one-letter codes for amino acids.

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

Step 2: Match the pairs.

(A) – IV, (B) – I, (C) – II, (D) – III.

Final Answer: The correct matching is (A) – IV, (B) – I, (C) – II, (D) – III.

Step 1: Understand confirmatory tests for Pb²⁺.

• Lead sulfate, lead chromate, and lead iodide form insoluble precipitates and are confirmatory tests for Pb²⁺ ions.
• Lead nitrate is a soluble, colorless compound and cannot be used as a confirmatory test.

Final Answer: Lead nitrate does not confirm the presence of Pb²⁺ ions.

Step 1: Write the neutralization reaction.

Ba(OH)₂ + 2HBr → BaBr₂ + 2H₂O.

Step 2: Use the equivalent concept.

m.e.q. of HBr = m.e.q. of Ba(OH)₂.

M₁ × n₁ × V₁ = M₂ × n₂ × V₂.

Substitute the values:

0.02 × 1 × V₁ = 0.01 × 2 × 10.

V₁ = (0.01 × 2 × 10) / 0.02 = 10.0 mL.

Final Answer: 10.0 mL of 0.02 M HBr is required.

Step 1: Understand the basicity of oxides.

• As the metallic character increases, the basicity of the oxide increases.
• In Group-13, the order of metallic character is: B₂O₃ < Al₂O₃ < Ga₂O₃ < In₂O₃ < Tl₂O₃.
• Hence, Tl₂O₃ is the most basic.

Final Answer: Tl₂O₃ is the most basic oxide.

Step 1: Identify the ligand and oxidation state.

• The ligand is oxalate (C₂O₄^2-), a bidentate ligand.
• The oxidation state of cobalt is +3.

Step 2: Write the IUPAC name.

• Prefix "trioxalato" indicates three oxalate ligands.
• Suffix "cobaltate" indicates a coordination compound with cobalt in an anionic complex.
• The oxidation state of cobalt is indicated as (III).

Final Answer: Potassium trioxalatocobaltate(III).

Step 1: Use the de Broglie principle.

2πr = nλ, where r is the radius of the orbit, n is the principal quantum number, and λ is the wavelength.

Step 2: Substitute r for the 3rd orbit.

The radius of the nth orbit is r = n²a₀.

For n = 3, r = 9a₀.

Step 3: Solve for λ.

2π(9a₀) = 3λ → λ = 6πa₀.

Final Answer: λ = 6πa₀.

Step 1: Identify the pesticides in the options.

• DDT (Dichlorodiphenyltrichloroethane) and Aldrin are commonly used pesticides.
• The other compounds are not typically classified as pesticides.

Final Answer: DDT and Aldrin are pesticides.

Step 1: Analyze the chromatography column.

The compounds elute in the order of polarity: a > b > c.

Elution order: c > b > a.

Step 2: Identify the incorrect statements.

• (A) Compound ‘c’ is more polar than ‘a’ and ‘b’ – Incorrect (least polar).
• (B) Compound ‘a’ is least polar – Incorrect (most polar).
• (C) Compound ‘b’ comes out of the column before ‘c’ and after ‘a’ – Incorrect.
• (D) Compound ‘a’ spends more time in the column – Correct.

Final Answer: Incorrect statements are (A), (B), and (C).

Step 1: Recall the trend for hydration enthalpy.

Hydration enthalpy is inversely proportional to ionic size:

Hydration enthalpy ∝ 1 / size.

Step 2: Order of ionic size.

Size order: Be²⁺ < Mg²⁺ < Ca²⁺ < Sr²⁺ < Ba²⁺.

Step 3: Determine the order of hydration enthalpy.

Hydration enthalpy: Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺.

Final Answer: Be²⁺ has the highest hydration enthalpy.

Step 1: Analyze the substituents.

• –NO₂ is an electron-withdrawing group (strong –I effect).
• –Cl is an electron-withdrawing group (weaker –I effect).
• –CH₃ is an electron-donating group (+I effect).

Step 2: Determine the effect on acidity.

Electron-withdrawing groups stabilize the conjugate base, increasing acidity. The order of acidity is:

p-Nitrophenol > p-Chlorophenol > Phenol > p-Methylphenol.

Final Answer: p-Nitrophenol is the strongest acid.

Step 1: Analyze the reaction mechanism.

• The reaction involves an alcohol in acidic conditions, typically leading to dehydration.
• Formation of a more stable carbocation intermediate, followed by a rearrangement or reaction, directs the product formation.

Final Answer: The product formation is guided by the stability of the intermediates formed during the reaction.

Step 1: Understand the molecular structure of BeCl₂ in different phases.

• In the solid state, BeCl₂ forms a polymer due to its electron-deficient nature.
• In the vapor phase at moderate temperatures, it exists as a dimer.
• At very high temperatures, it dissociates into monomers.

Final Answer: BeCl₂ exhibits polymeric, dimeric, and monomeric structures in the solid state, vapor phase, and at high temperatures, respectively.

Step 1: Analyze the energy profile diagram.

• The diagram shows three peaks, representing three transition states (activated complexes).
• Two valleys between these peaks represent the intermediates.
• The highest peak corresponds to the rate-determining step.

Final Answer: The reaction has two intermediates and three activated complexes, with the second step being the rate-determining step.

Step 1: Understand the electrolysis of brine.

• At the cathode, hydrogen ions are reduced to form hydrogen gas (H₂).
• At the anode, chloride ions are oxidized to chlorine gas (Cl₂).
• Sodium ions react with hydroxide ions formed at the cathode to produce NaOH.
• HCl is not a direct product of brine electrolysis.

Final Answer: HCl is not produced during the electrolysis of brine.

Step 1: Determine the physical properties of the elements at 100°C.

• Ga (Gallium) melts at 29.76°C and remains liquid at the boiling point of water.
• Li and Cs react with water, while Br evaporates below 100°C.

Final Answer: Gallium remains liquid at the boiling point of water.

Analysis:

• In Ni(CO)₄ and Fe(CO)₅, Ni and Fe indeed have zero oxidation states as the CO ligand is a neutral molecule.
• Reason R incorrectly suggests a π-donor characteristic of CO, which is primarily a σ-donor and a π-acceptor.

Final Answer: While the assertion about the oxidation state is correct, the reason given is not accurate regarding the character of the ligands.

Step 1: Determine the physical properties of the elements at 100°C.

• Ga (Gallium) melts at 29.76°C and remains liquid at the boiling point of water.
• Li and Cs react with water, while Br evaporates below 100°C.

Final Answer: Gallium remains liquid at the boiling point of water.

Reaction Analysis:

• The reaction involves the Grignard reagent reacting with a ketone, followed by protonation.
• The major product results from the addition of the organomagnesium compound to the carbonyl group, forming an alcohol after protonation.

Final Answer: The product formed is primarily due to the addition-elimination mechanism typical of Grignard reactions with ketones.

Reaction Conditions:

• Permanganate is reduced more effectively in neutral or slightly alkaline solutions when reacting with thiosulphate.
• In acidic conditions, permanganate tends to form manganese dioxide, which is not favored in this specific reaction pathway.

Final Answer: A neutral medium best supports the desired oxidation state change in manganese during this reaction.

Composition Analysis:

• Nessler's reagent is formulated as K₂[HgI₄], containing potassium (K), mercury (Hg), and iodine (I).
• Nitrogen (N) is not part of this chemical composition.

Final Answer: Nitrogen is not a component of Nessler's reagent.

Solution:

The reaction for oxidation is:

Metal + NO₃^- → Metal Nitrate

We need to determine which metals have a lower reduction potential than the reduction potential of NO₃^-, which is 0.97 V. Metals with reduction potentials lower than 0.97 V will be oxidized.

The reduction potentials are:

• V²⁺(aq) + 2e^- → V, E° = -1.19 V
• Fe³⁺(aq) + 3e^- → Fe, E° = -0.04 V
• Ag⁺(aq) + e^- → Ag(s), E° = 0.80 V

Since the reduction potentials of V, Fe, and Ag are all less than 0.97 V, they will be oxidized by NO₃^-.

Thus, the metals that will be oxidized are V, Fe, and Ag.

Final Answer: 3 metals.

Solution:

A body-centred unit cell (BCC) is a type of crystal structure where one atom is located at each corner of the unit cell and one atom is at the center of the cell.

BCC can be found in the following crystal systems:

• Cubic
• Tetragonal
• Orthorhombic

Thus, the number of crystal systems where BCC is present is 3.

Solution:

The iodoform reaction occurs with compounds that contain a structure where a methyl ketone (–COCH₃) group is present, or compounds with a hydroxyl group adjacent to a methyl group (–CHOH–CH₃).

Let’s analyze the compounds one by one:

• (a) 1–Phenylbutan–2–one has a methyl ketone group – positive.
• (b) 2–Methylbutan–2–ol has a tertiary alcohol adjacent to a methyl group – negative.
• (c) 3–Methylbutan–2–ol has a secondary alcohol adjacent to a methyl group – positive.
• (d) 1–Phenylethanol does not have a methyl ketone or the appropriate adjacent structure – negative.
• (e) 3,3–dimethylbutan–2–one has a methyl ketone group – positive.
• (f) 1–Phenylpropan–2–ol has a secondary alcohol adjacent to a methyl group – positive.

Thus, the compounds that will give a positive iodoform reaction are 1–Phenylbutan–2–one, 3–Methylbutan–2–ol, 3,3–dimethylbutan–2–one, and 1–Phenylpropan–2–ol.

Therefore, the number of compounds is 4.

Solution:

Using the formula for degrees of unsaturation (Du):

Du = C + 1 - (H - N)/2

Substituting the values:

Du = 8 + 1 - (11 - 1)/2 = 9 - 5 = 4

This indicates that the compound contains a benzene ring, which leads to 5 possible isomers.

Final Answer: 5

Solution:

The number of ions produced in each solution can be used to determine isotonicity. Isotonic solutions have the same effective concentration of particles.

• (A) 1 M aq. NaCl → 2 M ions (Na⁺ and Cl^-), isotonic with 2 M aq. Urea (1 particle per molecule).
• (B) 1 M aq. CaCl₂ → 3 M ions (Ca²⁺ and 2 Cl^-), isotonic with 1.5 M aq. KCl (2 particles per molecule).
• (C) 1.5 M aq. AlCl₃ → 6 M ions (Al³⁺ and 3 Cl^-), isotonic with 2 M aq. Na₂SO₄ (3 particles per molecule).
• (D) 2.5 M aq. KCl → 5 M ions (K⁺ and Cl^-), isotonic with 1 M aq. Al₂(SO₄)₃ (5 particles per molecule).

Thus, the number of isotonic pairs is 4.

Solution:

A liquid dispersion medium is found in:

• Paints
• Milk
• Hair cream
• Froth
• Soap lather

These systems involve a liquid phase as the dispersion medium, whereas other systems like gem stones, cheese, and smoke do not.

Final Answer: 5

Solution:

In an ice crystal, each water molecule forms hydrogen bonds with four neighboring molecules in a tetrahedral arrangement. However, for each individual water molecule, two hydrogen bonds are formed in the basic crystalline structure.

Final Answer: 2

Solution:

Using Hess's Law, we combine the reactions to find the heat of formation of C₂H₅OH(l). The reactions are:

• (1) 2C(s) + O₂ → 2CO₂          (-393.5 × 2) = -787 kJ
• (2) 3H₂ + 1.5O₂ → 3H₂O       (-241.8 × 3) = -725.4 kJ
• (3) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O (-1234.7 kJ)
• (4) 3H₂O + 2CO₂ → C₂H₅OH + 3O₂ (+1234.7 kJ)

Now combine equations (1), (2), and (4):

eq (5) = eq (1) + eq (2) + eq (4)

= (-787) + (-725.4) + (+1234.7) = -277.7 ≈ -278 kJ/mol

Thus, the heat of formation of C₂H₅OH(l) is -278 kJ/mol.

Final Answer: -278 kJ/mol.

Solution:

The equilibrium constant K is calculated as:

K = [PCl₅] / ([PCl₃][Cl₂]) = 0.4 / (0.2 × 0.1) = 20

When 0.2 mol of Cl₂ is added, the system shifts to re-establish equilibrium. Let the change in concentration of PCl₅ be x.

Initial concentrations:

• [PCl₃] = 0.2 mol L^-1
• [Cl₂] = 0.1 mol L^-1
• [PCl₅] = 0.40 mol L^-1

After addition of Cl₂:

• [Cl₂] = 0.1 + 0.2 - x = 0.3 - x
• [PCl₃] = 0.2 - x
• [PCl₅] = 0.40 + x

Substitute into the equilibrium expression:

20 = (0.40 + x) / [(0.2 - x)(0.3 - x)]

Solve the quadratic equation to find x ≈ 0.084 mol L^-1

The new concentration of PCl₅ is:

[PCl₅] = 0.40 + 0.084 = 0.484 mol L^-1 = 48.4 × 10^-2 mol L^-1

Final Answer: 48 × 10^-2 mol L^-1.

Solution:

The species with a square planar geometry are:

• XeF₄
• BrF₄^-
• [Cu(NH₃)₄]²⁺
• [PtCl₄]^2-

These species exhibit a square planar geometry due to the specific coordination and electronic arrangement.

Final Answer: 4