Chemistry 2024 Paper I 50 marks Compulsory Distinguish

Paper I — Q5

(a) Indicating 'Fluorescence' and 'Phosphorescence' processes in Jablonski diagram, distinguish between both processes. (10…

(a)

Indicating 'Fluorescence' and 'Phosphorescence' processes in Jablonski diagram, distinguish between both processes. 10 marks

(b)
(i)

How does water condense onto glass? 5 marks

(ii)

How long will it take for about one-half of a monolayer to form on the surface if the surface is exposed to a pressure of 2·0 × 10⁻¹¹ torr? Assume that a monolayer formation needs an exposure of 1 Langmuir. Also give reason for using such low pressure in cleaning the surface. 5 marks

(c)
(i)

Write the IUPAC nomenclature of the following complexes: (10 marks) Na₂[ZnCl₄]

(ii)

[PtCl₆]²⁻

(iii)

[Pt(py)₄] [PtCl₄]

(iv)

[(H₃N)₄Co Co(NH₃)₄]⁴⁺ H₂N \ O / H

(v)

Ni(CO)₄

(d)

What is Zeise's salt? Outline its synthesis method. Describe its structure. 10 marks

(e)
(i)

Explain lanthanide contraction. 5 marks

(ii)

Hf is placed below Zr in the periodic table. Justify the statement. 5 marks

हिंदी में प्रश्न पढ़ें
(a)

'प्रतिदीप्ति' और 'स्फुरदीप्ति' प्रक्रमों को जाब्लोन्स्की आरेख में सूचित करते हुए दोनों प्रक्रमों में भेद कीजिए। (10 अंक)

(b)
(i)

जल काँच पर संघनित कैसे होता है? (5 अंक)

(ii)

यदि एक पृष्ठ को 2·0 × 10⁻¹¹ torr के दाब से उद्वाषित किया जाए, तो उस पृष्ठ पर एकल परत (मोनोलेयर) की लगभग आधी परत बनाने के लिए कितना समय लगेगा? मान लीजिए एकल परत बनाने के लिए 1 लैंगम्यूर उद्वाषन की आवश्यकता है। उस पृष्ठ को साफ करने के लिए इतने कम दाब का प्रयोग क्यों किया जाता है, कारण भी दीजिए। (5 अंक)

(c)
(i)

निम्नलिखित संकुलों का IUPAC नामपद्धति के अनुसार नाम लिखिए: (10 अंक) Na₂[ZnCl₄]

(ii)

[PtCl₆]²⁻

(iii)

[Pt(py)₄] [PtCl₄]

(iv)

[(H₃N)₄Co Co(NH₃)₄]⁴⁺ H₂ N \ O / H

(v)

Ni(CO)₄

(d)

जाइसे लवण क्या है? इसकी संश्लेषण विधि को प्रस्तुत कीजिए। इसकी संरचना का वर्णन कीजिए। (10 अंक)

(e)
(i)

लैन्थेनाइड आकुंचन की व्याख्या कीजिए। (5 अंक)

(ii)

आवर्त सारणी में Hf को Zr के नीचे रखा गया है। इस कथन को उचित सिद्ध कीजिए। (5 अंक)

Q5 of the 2024 UPSC Mains Chemistry Paper I, as printed
The question as printed in the 2024 Chemistry paper

The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(c) Part (c)(iv) shows a chemical structure for a complex ion with the formula [(H3N)4Co Co(NH3)4]4+. The structure depicts two central Cobalt (Co) atoms. The left Cobalt is bonded to four Ammonia (H3N) groups. The right Cobalt is bonded to four Ammonia (NH3) groups. The two Cobalt atoms are bridged by two ligands: an Amine bridge (labeled H2N) and an Oxygen bridge (labeled O). The H2N bridge connects the top of the left Co to the top of the right Co. The O bridge connects the bottom of the left Co to the bottom of the right Co. The entire complex carries a 4+ charge.

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) Fluorescence and phosphorescence on a Jablonski diagram

After absorption of a photon (S₀ → S₁ or a higher singlet, about 10⁻¹⁵ s), the molecule loses its excess vibrational energy by collisions (vibrational relaxation, 10⁻¹²–10⁻¹⁰ s) and reaches the lowest vibrational level of S₁. From there it can:

  • fluoresce: emit a photon, S₁ → S₀;
  • undergo internal conversion: return to S₀ without radiation;
  • undergo intersystem crossing (ISC) to T₁: a spin flip made possible by spin–orbit coupling. After relaxing to the bottom of T₁, the molecule can phosphoresce (emit a photon, T₁ → S₀) or return to S₀ by a second ISC.
Diagram: Jablonski diagram

Distinctions

  1. Spin multiplicity:
  • Fluorescence is singlet → singlet (ΔS = 0, spin-allowed).
  • Phosphorescence is triplet → singlet (ΔS ≠ 0, spin-forbidden). It occurs only because spin–orbit coupling mixes singlet and triplet character, which is why heavy atoms (Br, I, metal centres) enhance it.
  1. Lifetime:
  • Fluorescence: about 10⁻⁹–10⁻⁷ s; it stops almost as soon as the exciting light is switched off.
  • Phosphorescence: about 10⁻⁴ s to many seconds; it gives an afterglow.
  1. Route: phosphorescence needs an intermediate ISC step; fluorescence does not.
  2. Wavelength: T₁ lies below S₁ (Hund's rule), so phosphorescence appears at longer wavelength, with a larger Stokes shift than fluorescence. Fluorescence is usually a mirror image of the S₀ → S₁ absorption band.
  3. Environment:
  • The long-lived triplet is easily deactivated by collisions, by heat and by paramagnetic O₂ (itself a triplet). Phosphorescence of organic molecules is therefore normally seen only in rigid media or frozen glasses at 77 K, or in solids.
  • Fluorescence is readily observed in fluid solution at room temperature.
  1. Examples:
  • Fluorescence: fluorescein, quinine sulfate, the green fluorescent protein.
  • Phosphorescence: benzophenone at 77 K; ZnS:Cu and SrAl₂O₄:Eu,Dy "glow-in-the-dark" phosphors (in the inorganic phosphors, trap states rather than a molecular triplet produce the long afterglow).

(b)(i) How water condenses on glass

  1. Surface chemistry: a clean glass surface is covered with silanol (Si–OH) groups, roughly 4–5 per nm², and adsorbed water. It is therefore a high-energy, hydrophilic surface.
  2. Adsorption: water molecules from the vapour are first held by hydrogen bonds to the Si–OH groups (physisorption). As the relative humidity rises, further layers build up on this first layer, since water hydrogen-bonds strongly to adsorbed water (multilayer, BET-type adsorption).
  3. Condensation: when the glass is cooled below the dew point of the surrounding air (a cold drinking glass, a car windscreen), the adsorbed film grows into bulk liquid. The surface acts as a site for heterogeneous nucleation, so condensation occurs at much smaller supersaturation than in free air. In fine cracks and pores it starts even below saturation (capillary condensation, Kelvin equation).
  4. Film or droplets:
  • On clean glass the contact angle is nearly zero, so water spreads into a continuous transparent film.
  • On glass contaminated by grease or organic films (a higher contact angle) it condenses as myriad tiny droplets that scatter light: the "fogging" of spectacles and mirrors.
  • Anti-fog coatings work by making the surface strongly hydrophilic, so that a film rather than droplets forms.

(b)(ii) Time to form half a monolayer at 2.0 × 10⁻¹¹ torr

1 langmuir (L) = an exposure of 10⁻⁶ torr × 1 s. If 1 L gives a monolayer (sticking probability ≈ 1), half a monolayer needs 0.5 L:

t = exposure/pressure = (0.5 × 10⁻⁶ torr s)/(2.0 × 10⁻¹¹ torr) = 2.5 × 10⁴ s ≈ 6.9 h

Why such low pressures are used

  • At 10⁻⁶ torr a monolayer of residual gas (H₂O, CO, H₂) builds up in about 1 s. A surface cleaned by ion sputtering, heating (flashing) or cleaving would be recontaminated almost at once.
  • Only in ultra-high vacuum (UHV, below 10⁻⁹ torr) does a clean surface stay clean for hours. That is long enough to characterise it by LEED, XPS, AES or STM, or to study controlled adsorption.
  • UHV also gives the electrons and ions used as probes a long mean free path, so that they reach the surface and the detector unscattered.

(c) IUPAC names

  1. Na₂[ZnCl₄]: sodium tetrachloridozincate(II). The anion is tetrahedral and Zn is +2.
  2. [PtCl₆]²⁻: hexachloridoplatinate(IV) ion (Pt +4; 6Cl⁻ − 4 = 2−).
  3. [Pt(py)₄][PtCl₄]: tetrakis(pyridine)platinum(II) tetrachloridoplatinate(II).
  • Both Pt centres are +2. "Tetrapyridine" is also seen in textbooks.
  • This is a Magnus-type salt. It shows coordination isomerism, e.g. [PtCl(py)₃][PtCl₃(py)], and is a polymerisation isomer of [PtCl₂(py)₂].
  1. [(H₃N)₄Co(μ-NH₂)(μ-OH)Co(NH₃)₄]⁴⁺: μ-amido-μ-hydroxido-bis[tetraamminecobalt(III)] ion, equivalently octaammine-μ-amido-μ-hydroxido-dicobalt(III) ion.
  • Charge check: 2 Co(III) = +6, and NH₂⁻ + OH⁻ = −2, giving 4+.
  • The prefix μ marks a ligand bridging two metal centres.
  1. Ni(CO)₄: tetracarbonylnickel(0), a tetrahedral, 18-electron complex.

(Since the 2005 IUPAC recommendations, anionic ligands take the "-ido" ending, so chloride is chlorido and hydroxide is hydroxido. The older "chloro" and "hydroxo" forms remain common in textbooks.)

(d) Zeise's salt

Definition. Zeise's salt is potassium trichlorido(η²-ethene)platinate(II) monohydrate, K[PtCl₃(C₂H₄)]·H₂O, a lemon-yellow crystalline solid. W. C. Zeise prepared it in 1827. It was the first transition-metal–alkene complex and one of the earliest organometallic compounds known, but its structure was not understood until the Dewar–Chatt–Duncanson model (1951–53).

Synthesis

  • Modern method: ethylene is bubbled at room temperature through a solution of K₂[PtCl₄] in dilute HCl, with a little SnCl₂ as catalyst. Ethylene replaces one chloride:

K₂[PtCl₄] + C₂H₄ → K[PtCl₃(C₂H₄)] + KCl

On concentration the monohydrate crystallises.

  • Zeise's original route: PtCl₄ (or H₂PtCl₆) was boiled with ethanol, which supplies the ethylene, and KCl was then added.
Diagram: Zeise salt structure and bonding

Structure (from X-ray and neutron diffraction)

  • The [PtCl₃(C₂H₄)]⁻ anion is square planar about Pt(II) (d⁸). Three positions are occupied by Cl and the fourth by the midpoint of the C=C bond. Ethylene is bound side-on (η²).
  • The C=C axis is perpendicular to the PtCl₃ coordination plane. This orientation gives the best π back-bonding overlap and the least steric clash with the cis chlorides.
  • The C–C bond lengthens from 134 pm in free ethylene to about 137.5 pm, and the four H atoms are bent back away from the metal. The carbons have acquired some sp³ character.
  • The Pt–Cl bond trans to ethylene (≈ 234 pm) is longer than the cis bonds (≈ 230 pm). This reflects ethylene's strong trans influence, the same property behind the trans effect.

Bonding: the Dewar–Chatt–Duncanson model

  1. σ-donation: the filled π orbital of C=C donates electron density into an empty dsp² hybrid on Pt.
  2. π back-donation: a filled Pt d orbital (dxz) donates into the empty π* orbital of C=C.

Both effects reduce the C=C bond order. As a result the bond lengthens, and the C=C stretching frequency falls from 1623 cm⁻¹ in free ethylene to about 1526 cm⁻¹ in the complex. The two effects are synergic: donation makes the metal richer in electrons, which strengthens back-donation.

(e)(i) Lanthanide contraction

Definition. Lanthanide contraction is the steady decrease in the atomic and ionic radii of the lanthanides with increasing atomic number, from La to Lu. For example, Ln³⁺ falls from 103 pm (La³⁺) to 86 pm (Lu³⁺) for six-coordination.

Cause.

  • Across the series each step adds one proton to the nucleus and one electron to the 4f subshell.
  • The 4f orbitals are diffuse and poorly penetrating, so they shield one another and the outer 5s, 5p, 5d and 6s electrons very imperfectly.
  • The effective nuclear charge felt by the outer electrons therefore rises steadily and pulls the whole electron cloud inward.
  • Relativistic effects make a smaller contribution, estimated at about 10–30%.

Consequences:

  • The lanthanides are very alike and hard to separate (ion exchange and solvent extraction are needed).
  • The basicity of Ln(OH)₃ decreases from La to Lu.
  • Y³⁺ (90 pm) is similar in size to Ho³⁺, so yttrium occurs with the heavy lanthanides.
  • The 5d (third) transition series has almost the same atomic sizes as the 4d series, as in Zr/Hf, Nb/Ta and Mo/W.
  • The 5d metals are very dense (Os and Ir are the densest elements) and have high ionisation energies, which contributes to the nobility of Pt and Au.

(e)(ii) Why Hf is placed below Zr

Electronic basis. Hf ([Xe]4f¹⁴5d²6s²) has the same outer valence configuration, (n−1)d²ns², as Zr ([Kr]4d²5s²) and Ti. It is the first element after the lanthanides to begin filling the 5d subshell beyond one electron, so it belongs to Group 4, directly below Zr.

Size. Moving down a group from the 4d to the 5d period would normally increase the radius substantially. But the 14 intervening lanthanide elements add 32 units of nuclear charge (Z = 40 for Zr, 72 for Hf), including the poorly shielded 4f electrons. The lanthanide contraction cancels the expected increase almost exactly:

  • metallic radii: Zr 160 pm, Hf 159 pm;
  • ionic radii: Zr⁴⁺ 72 pm, Hf⁴⁺ 71 pm.

Chemical twins. Because Zr and Hf have the same charge, the same valence configuration and virtually the same size, they are chemically almost identical:

  • both are stable in the +4 state;
  • they form isostructural oxides (ZrO₂/HfO₂) and halides (ZrCl₄/HfCl₄), and similar complex fluorides K₂[MF₆];
  • they always occur together in nature (zircon, ZrSiO₄, contains 1–3% Hf).

Hafnium was discovered only in 1923 (by Coster and von Hevesy, using X-ray spectroscopy), precisely because it hides inside zirconium minerals. Their separation still requires multistage solvent extraction or ion exchange. Hf is almost twice as dense as Zr (13.3 against 6.5 g cm⁻³): nearly the same atomic volume holds almost double the mass.

Practical importance: nuclear-grade Zr (used for fuel cladding because of its low neutron-capture cross-section) must be freed of Hf, whose capture cross-section is very high and which is therefore used in control rods.

How this answer will be evaluated

What earns full marks: Accurate diagrams, calculations, and nomenclature with clear reasoning.

Evaluation rubric

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

  1. (a) Jablonski diagram showing both processes and a point-wise distinction. 10 marks

    compare— paired headings or point-wise contrast → key differences → significance → conclusion

    Must cover

    • Jablonski diagram with S1, T1, S0 levels
    • Fluorescence shown as S1 to S0 transition
    • Phosphorescence shown as T1 to S0 transition
    • Distinction based on spin multiplicity and lifetime

    Loses marks

    • Missing Jablonski diagram
    • Confusing singlet and triplet states
    • No distinction between the two processes

    Earns more

    • Mention of intersystem crossing
    • Note on 'forbidden' nature of phosphorescence
    • Indication of vibrational relaxation

    Extra mark

    • Quantitative values for typical lifetimes
  2. (b)(i) Mechanism of water condensation on glass surfaces. 5 marks

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

    Must cover

    • Role of surface energy and wettability
    • Nucleation of water droplets
    • Interaction with surface hydroxyl groups

    Loses marks

    • Vague description without mechanism
    • Ignoring surface properties of glass

    Earns more

    • Mention of contact angle
    • Distinction between clean and dirty glass

    Extra mark

    • Reference to specific surface chemistry of glass
  3. (b)(ii) Time for 0.5 monolayer formation at 2.0 x 10^-11 torr and reason for low pressure. 5 marks

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

    Must cover

    • Calculation of time using 1 Langmuir = 10^-6 torr-s
    • Correct unit conversion and substitution
    • Reason: low pressure prevents re-contamination
    • Reason: in UHV a monolayer takes hours to form (vs about 1 s at 10⁻⁶ torr), so the cleaned surface stays clean long enough to be studied

    Loses marks

    • Incorrect calculation of time
    • Missing reason for low pressure
    • Wrong definition of Langmuir

    Earns more

    • Explicit definition of Langmuir
    • Mention of UHV requirements

    Extra mark

    • Reference to specific surface cleaning techniques
  4. (c) IUPAC names for five coordination complexes. 10 marks

    enumerate— list the items in order → one line each → no commentary

    Must cover

    • Correct name for Na2[ZnCl4]
    • Correct name for [PtCl6]2-
    • Correct name for [Pt(py)4][PtCl4]
    • Correct name for the Co-bridged complex

    Loses marks

    • Wrong oxidation state in name
    • Incorrect ligand names
    • Missing counter-ion in name

    Earns more

    • Correct name for Ni(CO)4
    • Correct oxidation states in names
    • Correct ligand nomenclature

    Extra mark

    • Mention of specific isomerism if applicable
  5. (d) Definition, synthesis, and structure of Zeise's salt. 10 marks

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

    Must cover

    • Definition of Zeise's salt
    • Synthesis from K2PtCl4 and ethylene
    • Description of the Pt-C2H4 bond
    • Square-planar Pt(II) with the C=C axis perpendicular to the PtCl₃ plane (η², side-on; not a sandwich structure)

    Loses marks

    • Missing synthesis method
    • Incorrect description of the structure
    • Confusing with other platinum complexes

    Earns more

    • Details of the reaction conditions
    • Explanation of the Dewar-Chatt-Duncanson model
    • Mention of the historical significance

    Extra mark

    • Note that it (1827) was the first transition-metal olefin complex, among the earliest organometallic compounds
  6. (e)(i) Explanation of lanthanide contraction. 5 marks

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

    Must cover

    • Definition of lanthanide contraction
    • Cause: poor shielding by 4f electrons
    • Effect: steady decrease in ionic radius

    Loses marks

    • Vague definition
    • Missing cause of the contraction
    • Confusing with other periodic trends

    Earns more

    • Mention of the effective nuclear charge
    • Comparison with transition metals

    Extra mark

    • Reference to specific ionic radii values
  7. (e)(ii) Justification for placing Hf below Zr in the periodic table. 5 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Similarity in ionic radii of Hf and Zr
    • Role of lanthanide contraction
    • Similar chemical properties

    Loses marks

    • Missing link to lanthanide contraction
    • Vague statement without justification
    • Confusing with other group 4 elements

    Earns more

    • Mention of the 'lanthanide effect'
    • Comparison of oxidation states

    Extra mark

    • Reference to specific chemical reactions
What "Distinguish" is asking you to do

Name the property that separates the items and say which side holds it. Distinguish is marked exactly as differentiate is, with no difference in expectation, but its stems more often line up three terms rather than two — gender equality, gender equity and empowerment — and every pair in the set has to be separated.

Structure that answers it

The category they all sit in → the property dividing the first pair → the second pair → the third → why the boundary matters in practice

Where marks are lost

Separating the two obviously different items and leaving the middle term unplaced. A description of each side from which the line must be inferred is marked as description, not as a distinction.

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