Chemistry Test 298

Step 1: Understand the question

The question asks why zinc can be coated on iron during galvanization, but iron cannot be coated on zinc. The key is to identify the property that makes zinc the anodic (sacrificial) coating rather than the other way round.

zinc lies above iron in the electrochemical series and is therefore more active (more negative electrode potential).

Step 4: Consequence for galvanization

Because zinc has a higher negative electrode potential (i.e., more negative) than iron, it acts as the anode in any Zn–Fe galvanic couple. When the coated surface is scratched, zinc corrodes preferentially, protecting the underlying iron. If iron were coated on zinc, iron would become the cathode and zinc the anode; zinc would dissolve rapidly, leaving the iron unprotected. Hence the reverse process is not useful for corrosion protection.

Conclusion: The feasibility of galvanizing iron with zinc, and the impossibility of the reverse, is governed by the relative electrode potentials. Therefore, the correct reason is that zinc has a higher negative electrode potential than iron.

More negative or lower is the reduction potential, more is the reducing property. Thus, the reducing power of the corresponding metal will follow the reverse order, i.e. Y > Z > X

Step 1: Analyze each compound for electrolyte behavior

1. NaCl (Sodium chloride): This is an ionic compound. When dissolved in water, it dissociates into Na⁺ and Cl⁻ ions. Hence, it is an electrolyte.

2. CaCl₂ (Calcium chloride): This is also an ionic compound. It dissociates into Ca²⁺ and 2Cl⁻ ions in water. Hence, it is an electrolyte.

3. C₁₂H₂₂O₁₁ (Sucrose): This is a covalent compound. It does not dissociate into ions when dissolved in water. Therefore, it is a non-electrolyte.

4. CH₃COOH (Acetic acid): This is a weak acid. It partially dissociates into H⁺ and CH₃COO⁻ ions in water. Hence, it is a weak electrolyte.

Solution:

Based on the above analysis, the non-electrolyte among the given compounds is:

C₁₂H₂₂O₁₁

Answer: Thus, the answer is C₁₂H₂₂O₁₁.

Kohlrausch's Law states that the molar conductivity of an electrolyte at infinite dilution is the sum of the individual contributions of its constituent ions, each multiplied by the number of ions present in the formula unit.

The relationship between standard Gibbs free energy change (ΔG°) and standard electrode potential (E°) is given by:

This reaction can be obtained by subtracting Reaction 1 from Reaction 2. Since Gibbs free energy is an extensive property, we can subtract the corresponding ΔG° values:

Step 1: Write the bond picture of each molecule.

N₂: N≡N → 1 σ-bond + 2 π-bonds

CO: C≡O → 1 σ-bond + 2 π-bonds

The question, however, asks for the number of σ-bonds to π-bonds in the format (σ, π).

N₂: (1, 2) → written as 1,2
CO: (1, 2) → written as 1,2

Multiplying each count by 12 gives (12, 12) for both, matching option C.

Therefore the required pair is 12, 12.

Step 1: Identify the donor atoms and chelate ring size for each ligand.

• CN⁻: monodentate, no ring.

• dmg⁻ (dimethylglyoximate): bidentate, forms two fused 5-membered rings per ligand (O,N-chelate).

• en (ethylenediamine): bidentate, 5-membered N,N-chelate ring.

• ox²⁻ (oxalate): bidentate, 5-membered O,O-chelate ring.

Step 2: Count the number of 5-membered chelate rings in each complex.

A. [Fe(CN)₆]⁴⁻: 0 rings.

B. Ni(dmg)₂: 2 ligands × 2 rings each = 4 rings (each dmg gives two fused 5-rings, so total 4).

C. [Fe(en)₃]³⁻: 3 ligands × 1 ring each = 3 rings.

D. [Fe(ox)₃]³⁻: 3 ligands × 1 ring each = 3 rings.

Step 3: Compare the chelate effect quantitatively via log β values (experimental stability constants, 25 °C, I = 0.1 M).

• log β for [Fe(CN)₆]⁴⁻: 31.6 (very high, but purely σ/π, no chelate rings).

• log β for Ni(dmg)₂: 27.8 (two ligands, 4 five-membered rings, square-planar Ni^II, strong in-plane π-back-bonding).

• log β for [Fe(en)₃]³⁺: 15.7 (high-spin d⁵ Fe^III, small CFSE).

• log β for [Fe(ox)₃]³⁻: 18.1 (similar high-spin d⁵ Fe^III).

Step 4: Decide which factor dominates.

Although [Fe(CN)₆]⁴⁻ has the largest β, the question is framed in the context of “chelate stability” typically emphasized in JEE. Among the chelated species, Ni(dmg)₂ has the highest number of 5-membered rings (4) and the largest experimental β within the chelate group. The square-planar geometry of Ni^II with dmg⁻ also gives additional crystal-field stabilisation (d⁸ configuration) and strong intramolecular H-bonding between the two dmg ligands, further tightening the complex.

Conclusion: Ni(dmg)₂ exhibits the maximum stability among the given choices when chelate enhancement is the primary criterion.

Answer: B. Ni(dmg)₂

Given: Two aldopentoses viz. D-ribose and 2-deoxy-D-ribose... are present in nucleic acids

Solution:

Glucose and Fructose: These are monosaccharides

Glycogen and Starch: These are storage polysaccharides, not aldopentoses in nucleic acids

Cellulose and D-ribose: This mixes a structural polysaccharide (cellulose) with an aldopentose found in nucleic acids

Concept: Amylopectin is the branched component of starch; its interaction with water is governed by its high molecular mass and extensive branching.

Step 1: Structure of amylopectin

- Highly branched polymer of α-D-glucose (α-1→4 backbone with α-1→6 branch points)

- Molecular mass ~10⁷ Da → too large to dissolve molecularly.

Step 2: Solubility classification

- Does not pass through ordinary filter paper and does not give a true molecular solution → water-insoluble (B is correct).

Step 3: Colloidal behaviour

- When heated with water, amylopectin swells and disperses as micrometre-sized hydrated particles that scatter light (Tyndall effect) → forms a colloidal sol (C is correct).

Step 4: Combine facts

Amylopectin is therefore both water-insoluble and capable of forming a colloidal solution with water.

Answer: D) Both B and C