Chemistry 2023 Paper I 50 marks Explain

Paper I — Q8

(a) Sketch the synthesis of [Fe(η⁵-C₅H₅)(η⁵-C₅H₄COCH₃)] and [Fe(η⁵-C₅H₅)(η⁵-C₅H₄COOH)] complexes starting from [Fe(η⁵-C₅H₅)₂]…

(a)

Sketch the synthesis of [Fe(η⁵-C₅H₅)(η⁵-C₅H₄COCH₃)] and [Fe(η⁵-C₅H₅)(η⁵-C₅H₄COOH)] complexes starting from [Fe(η⁵-C₅H₅)₂]. 10 marks

(b)
(i)

Calculate the spin only magnetic moment (μ_s.o.) of the central metal atom in the following complexes: [Fe(H₂O)₅NO]²⁺

(ii)

[Cr(NCS)₆]³⁻

(iii)

[V(H₂O)₆]³⁺

(iv)

[Co(bpy)₃]²⁺ 10 marks

(c)

When an incident light of wavelength 300 nm is passed through a solution in a 1 cm cell, it transmits only 10% of the incident light. What percentage of light would be absorbed by the same solution if taken in a 0.5 cm cell? 10 marks

(d)

Explain the transition-state theory for reaction rates. How is this theory considered superior to collision theory in providing a much more complete interpretation of the pre-exponential factor A in the Arrhenius equation? 20 marks

हिंदी में प्रश्न पढ़ें
(a)

[Fe(η⁵-C₅H₅)₂] से शुरू करते हुए [Fe(η⁵-C₅H₅)(η⁵-C₅H₄COCH₃)] और [Fe(η⁵-C₅H₅)(η⁵-C₅H₄COOH)] संकुलों के संश्लेषण की रूपरेखा बनाइए। (10 अंक)

(b)
(i)

निम्नलिखित संकुलों में केंद्रीय धातु परमाणु के केवल प्रचक्रण चुंबकीय आघूर्ण (μ_s.o.) का परिकलन कीजिए: [Fe(H₂O)₅NO]²⁺

(ii)

[Cr(NCS)₆]³⁻

(iii)

[V(H₂O)₆]³⁺

(iv)

[Co(bpy)₃]²⁺ (10 अंक)

(c)

जब 300 nm तरंगदैर्घ्य का आपतित प्रकाश एक विलयन में से गुजारा जाता है, जो कि 1 cm सेल में है, तो यह आपतित प्रकाश का केवल 10% भाग ही संचारित करता है। 0.5 cm सेल में समान विलयन लेने पर प्रकाश का कितना प्रतिशत अवशोषित होगा? (10 अंक)

(d)

अभिक्रिया दर के लिए संक्रमण-अवस्था सिद्धांत की व्याख्या कीजिए। कैसे इस सिद्धांत को अर्रेनियस समीकरण में पूर्व-चरघातांकी गुणक A की अधिक पूर्ण व्याख्या प्रदान करने में संघट्ट सिद्धांत से बेहतर माना जाता है? (20 अंक)

Q8 of the 2023 UPSC Mains Chemistry Paper I, as printed
The question as printed in the 2023 Chemistry paper

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a) Synthesis of acetylferrocene and ferrocenecarboxylic acid from ferrocene

Step 1, Friedel–Crafts acylation: ferrocene is an electron-rich, aromatic-like sandwich compound and is acylated readily, with acetic anhydride and phosphoric acid (or acetyl chloride and AlCl₃) as the source of the acylium ion CH₃CO⁺. One cyclopentadienyl ring is substituted: Fe(η⁵-C₅H₅)₂ + (CH₃CO)₂O → [Fe(η⁵-C₅H₅)(η⁵-C₅H₄COCH₃)] + CH₃COOH. The acetyl group withdraws electrons and deactivates that ring, so the second acylation goes to the other ring (heteroannular), and mild conditions with limited acylating agent give mainly the monoacetyl compound. Fe remains Fe(II).

Step 2, haloform oxidation: the methyl ketone is oxidised by NaOCl (or NaOBr) in NaOH. The CH₃ group is halogenated to CX₃, and hydroxide cleaves the C–CX₃ bond, giving the carboxylate and CHX₃. Acidification gives [Fe(η⁵-C₅H₅)(η⁵-C₅H₄COOH)].

(b) Spin-only magnetic moments (μ = √[n(n + 2)] BM)

(i) [Fe(H₂O)₅NO]²⁺ (brown-ring complex): written as Fe(I) with NO⁺, d⁷, high-spin, n = 3; μ = √15 = 3.87 BM.

(ii) [Cr(NCS)₆]³⁻: Cr(III), d³, n = 3; μ = √15 = 3.87 BM.

(iii) [V(H₂O)₆]³⁺: V(III), d², n = 2; μ = √8 = 2.83 BM.

(iv) [Co(bpy)₃]²⁺: Co(II), d⁷, high-spin (t₂g⁵e_g²) at room temperature, n = 3; μ = √15 = 3.87 BM. A low-spin d⁷ configuration would give n = 1 and 1.73 BM, but bpy does not produce a field strong enough to force pairing in Co(II) at room temperature.

(c) Beer–Lambert law

For the 1 cm cell, T = 0.10, so A = −log T = 1.0 = εc(1 cm).

For a 0.5 cm cell, concentration and ε are unchanged, so A = 1.0 × 0.5/1 = 0.50. T = 10⁻⁰·⁵ = 0.316 (31.6%).

Light absorbed = 100 − 31.6 = 68.4%.

(d) Transition-state theory and the pre-exponential factor

Theory: reactants A + B are in a rapid equilibrium with an activated complex (transition state) [AB]‡ at the saddle point of the potential-energy surface, and the complex decays to products at the frequency kBT/h along the reaction coordinate:

A + B ⇌ [AB]‡ → products; k = (kBT/h)K‡.

Since K‡ = exp(−ΔG‡/RT) = exp(ΔS‡/R)exp(−ΔH‡/RT), the rate constant is the Eyring equation

k = (kBT/h) exp(ΔS‡/R) exp(−ΔH‡/RT).

Comparing with k = A exp(−Eₐ/RT), and using Eₐ = ΔH‡ + RT for a solution reaction, gives A = e(kBT/h)exp(ΔS‡/R). In statistical terms, A depends on the ratio of partition functions of the complex to those of the reactants.

Superiority over collision theory: collision theory gives A = PZ, where the collision frequency Z treats molecules as hard spheres, and the steric factor P is an empirical adjustment for the orientation requirement and cannot be calculated. In TST, A is given by the entropy of activation, which is worked out from molecular masses, moments of inertia, vibrational frequencies and the geometry of the complex. A tightly ordered transition state has a negative ΔS‡ and a small A (low P), and a loose one has a larger A. So the steric factor is explained, not assumed, and the temperature dependence of A can be predicted. TST also explains solution reactions and solvent effects, through ΔG‡, and gives thermodynamic ideas such as ΔH‡ and ΔS‡ from an Eyring plot of ln(k/T) against 1/T.

How this answer will be evaluated

What earns full marks: Precise mechanisms, correct calculations, and deep theoretical insight.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) (a) Sketch the synthesis of acetylferrocene and ferrocenecarboxylic acid starting from ferrocene (10 marks). 10 marks

    trace— start point → the stages in sequence → end point → what changed

    Must cover

    • Friedel-Crafts acylation with acetic anhydride/H3PO4 or AcCl/AlCl3
    • Monoacetylferrocene; deactivation leads to heteroannular second acylation
    • Haloform oxidation with NaOCl/NaOBr in NaOH
    • Acidification gives the carboxylic acid; Fe stays Fe(II)

    Loses marks

    • Using oxidising agents that destroy ferrocene.
    • Missing the haloform step.

    Earns more

    • Reaction equations.
    • Mild conditions for mono-product.

    Extra mark

    • Alternative lithiation route
  2. (b) (b) Calculate the spin-only magnetic moment for (i) [Fe(H2O)5NO]2+, (ii) [Cr(NCS)6]3-, (iii) [V(H2O)6]3+, (iv) [Co(bpy)3]2+ (10 marks). 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Formula sqrt(n(n+2))
    • (i) and (ii): n = 3, 3.87 BM
    • (iii): n = 2, 2.83 BM
    • (iv): Co(II) d7 with stated spin state and value

    Loses marks

    • Missing oxidation states.
    • Units omitted.

    Earns more

    • Oxidation states for each.
    • Brown-ring Fe(I) with NO+ description.

    Extra mark

    • Low-spin alternative 1.73 BM
  3. (c) (c) Light of 300 nm is 10% transmitted in a 1 cm cell. What percentage is absorbed in a 0.5 cm cell? (10 marks) 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • A = -log T = 1.0 for 1 cm
    • Absorbance proportional to path length: 0.5
    • T = 10^-0.5 = 31.6%
    • Absorbed = 68.4%

    Loses marks

    • Halving the percentage absorbed instead of absorbance.
    • Arithmetic errors.

    Earns more

    • Beer-Lambert statement.

    Extra mark

    • Conceptual reason
  4. (d) (d) Explain transition-state theory. How is it superior to collision theory in interpreting the pre-exponential factor A? (20 marks) 20 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Activated complex in equilibrium with reactants at the saddle point
    • k = (kBT/h)K and the Eyring equation with delta S and delta H
    • A related to entropy of activation and partition functions
    • Collision theory's empirical steric factor P versus TST's explanation

    Loses marks

    • Saying TST gives A = PZ.
    • No explanation of steric factor.

    Earns more

    • Relation Ea = delta H + RT.
    • Eyring plot.

    Extra mark

    • Solvent effects
    • Temperature dependence of A
What "Explain" is asking you to do

Make the working of something clear — what sets it off, what follows from what, and what it produces. Explain is the Commission's mechanism word: it dominates the technical papers and the “explain why” stems, where the marks sit in the causal chain and not in the label.

Structure that answers it

State what it is → the initiating condition → the chain of cause, step by step → an instance where it plays out → what the chain produces

Where marks are lost

Describing what something looks like instead of why it works that way. Naming the stages without linking them reads as description too.

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