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Chemistry 2021 Paper I 50 marks Calculate

Paper I — Q2

(a) Hydrogen atoms are observed to have radiative transitions from n = 101 to n = 100 to occur. (i) What are the frequency and…

(a)

Hydrogen atoms are observed to have radiative transitions from n = 101 to n = 100 to occur.

(i)

What are the frequency and wavelength of the radiation emitted in this transition?

(ii)

Why is it difficult to observe this transition? 10 marks

(b)

Draw the geometrical arrangements for the following hybridized systems and identify the type of d-orbitals involved in each system :

sp³d, sp³d², dsp², sd³ 20 marks

(c)

Iron crystallizes in a b.c.c. unit cell at room temperature (ρ = 7·86 g/cm³). Calculate the radius of an iron atom in this crystal. At temperatures more than 910 °C, iron prefers to be in f.c.c. If we neglect the temperature dependence of radius of iron on the grounds that it is negligible, use this information to determine whether iron expands or contracts when it undergoes transformation from b.c.c. to f.c.c. structure. The atomic mass of iron is 55·845 u. 20 marks

हिंदी में प्रश्न पढ़ें
(a)

हाइड्रोजन परमाणु में विकिरणी संक्रमण n = 101 से n = 100 पाए जाने का अवलोकन किया गया।

(i)

इस संक्रमण में उत्सर्जित विकिरण की आवृत्ति और तरंगदैर्घ्य क्या है?

(ii)

इस संक्रमण का अवलोकन करना क्यों मुश्किल है? (10 अंक)

(b)

निम्नलिखित संकरित समुदायों के लिए ज्यामितीय व्यवस्थाओं को खींचिए और अभिनिर्धारित कीजिए कि प्रत्येक समुदाय में किस प्रकार का d-क्षक सम्मिलित है :

sp³d, sp³d², dsp², sd³ (20 अंक)

(c)

लोहा कक्ष ताप पर b.c.c. एकक सेल में क्रिस्टलित होता है (ρ = 7·86 g/cm³)| इस क्रिस्टल में लोहा परमाणु की त्रिज्या का परिकलन कीजिए। 910 °C तापमान से ऊपर लोहा f.c.c. को प्राथमिकता/तर्जीह देता है। अगर हम लोहे की त्रिज्या की तापमान पर निर्भरता को इस आधार पर छोड़ दें कि वह उपेक्षणीय है, इस जानकारी का प्रयोग करके निर्धारित कीजिए कि लोहा जब b.c.c. से f.c.c. संरचना में रूपांतरण करेगा, तो वह प्रसारित होगा या आकुंचित। लोहे का परमाणविक द्रव्यमान 55·845 u है। (20 अंक)

Q2 of the 2021 UPSC Mains Chemistry Paper I, as printed
The question as printed in the 2021 Chemistry paper
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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) Hydrogen transition n = 101 → 100 (i) 1/λ = R_H(1/n₁² − 1/n₂²) with n₁ = 100, n₂ = 101 and R_H = 109,677 cm⁻¹: 1/100² − 1/101² = 1.970 × 10⁻⁶, so ν̃ = 0.2161 cm⁻¹. λ = 1/0.2161 = 4.63 cm. ν = c/λ = 3 × 10¹⁰ cm s⁻¹ / 4.63 cm ≈ 6.5 × 10⁹ Hz (6.5 GHz), in the radio/microwave region. (ii) The energy hν is only about 2.7 × 10⁻⁵ eV, so the photons are very weak, and the spontaneous emission rate (∝ ν³) is extremely small. Such a highly excited atom is huge (r ≈ n²a₀ ≈ 0.5 μm), and is readily ionised or perturbed by collisions and stray fields, which broaden the line. Laboratory detection is therefore hard, though radio recombination lines of this type are seen from interstellar gas.

(b) Hybridisation geometries and d-orbitals sp³d: trigonal bipyramidal (three equatorial bonds at 120°, two axial at 90°); d-orbital: d_z² (as in PCl₅). sp³d²: octahedral (all angles 90°); d-orbitals: d_z² and d_x²−y² (the e_g set, outer shell, as in SF₆). dsp²: square planar (90° angles in one plane); d-orbital: d_x²−y², with s, p_x and p_y (as in [PtCl₄]²⁻, [Ni(CN)₄]²⁻). sd³: tetrahedral (109.5°); d-orbitals: d_xy, d_yz and d_zx (the t₂ set, which point between the axes), with s (as in MnO₄⁻, CrO₄²⁻). In each drawing the bonds point along the lobes of the hybrid orbitals, and the axes are labelled so that the participating d-orbital lobes are identifiable. Inner (n−1)d involvement gives low-spin d²sp³ octahedral complexes, unlike outer sp³d².

Diagram: Geometry of sp3d, sp3d2, dsp2 and sd3 hybridisation

(c) Radius of iron and b.c.c. → f.c.c. change For b.c.c., Z = 2: a³ = ZM/(ρN_A) = (2 × 55.845)/(7.86 × 6.022 × 10²³) = 2.360 × 10⁻²³ cm³, so a = 2.867 × 10⁻⁸ cm = 286.7 pm. Atoms touch along the body diagonal: √3a = 4r, so r = √3a/4 = 124.1 pm (1.24 Å). Taking r constant in the f.c.c. form (Z = 4, 4r = √2a′): a′ = 4(124.1)/1.414 = 351.0 pm. Volume per atom: b.c.c. = a³/2 = 1.180 × 10⁻²³ cm³; f.c.c. = a′³/4 = 1.081 × 10⁻²³ cm³. The ratio is 0.916 (matching packing efficiencies, 68% vs 74%). Hence iron contracts by about 8% in volume on transformation to f.c.c. at 910 °C.

How this answer will be evaluated

Framework: Concept > Structure or mechanism > Reasoning > Result

What earns full marks: Accurate calculations, clear diagrams, and correct reasoning for all parts.

Evaluation rubric

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

  1. (a) Determine frequency and wavelength for n=101 to n=100 transition and explain observation difficulty. 10 marks

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

    Must cover

    • Apply Rydberg formula for frequency calculation
    • Calculate wavelength using c = νλ
    • Identify radiation as radio/microwave region
    • Explain difficulty due to low energy/long wavelength

    Loses marks

    • Using visible light formulas for radio transition
    • Ignoring the high n-value context

    Earns more

    • Mention specific values (approx 6.5 GHz, 4.6 cm)
    • Reference to radio astronomy techniques

    Extra mark

    • Mention specific telescope types (e.g., radio interferometry)
  2. (b) Draw geometrical arrangements for sp³d, sp³d², dsp², sd³ and identify d-orbitals. 20 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • Draw correct geometry for each hybridization
    • Identify specific d-orbitals (e.g., dxy, dz2)
    • Label axes and orbital lobes correctly
    • Distinguish between inner and outer d-orbitals

    Loses marks

    • Confusing dsp² (square planar) with sp³d (trigonal bipyramidal)
    • Incorrect identification of d-orbital types

    Earns more

    • Clear, labelled diagrams for all four systems
    • Correct association of geometry with hybridization

    Extra mark

    • Mentioning specific molecular examples for each geometry
  3. (c) Calculate iron atom radius in b.c.c. and determine expansion/contraction in f.c.c. transformation. 20 marks

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

    Must cover

    • Calculate radius using b.c.c. density formula
    • Use correct atomic mass and unit cell parameters
    • Compare packing efficiency of b.c.c. vs f.c.c.
    • Determine volume change based on radius constancy

    Loses marks

    • Using wrong unit cell parameters for b.c.c.
    • Ignoring the assumption of constant radius

    Earns more

    • Step-by-step calculation of radius (approx 1.24 Å)
    • Correct application of packing efficiency (68% vs 74%)

    Extra mark

    • Mentioning specific temperature of transformation (910 °C)
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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