Repeated question
Here's your question paper with neatly organized answers included after each question:
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Q1) Solve any five of the following. [5]
a) Define Isomerism.
Isomerism is the phenomenon where compounds have the same molecular formula but different
structural arrangements or spatial orientations.
b) Define inner orbital complex.
Inner orbital complexes use (n–1)d orbitals in hybridization (d²sp³), typically forming low-spin
octahedral structures.
c) Find the number of unpaired electrons in d⁵ weak field octahedral complex.
In weak field: electrons don't pair → configuration: t₂g³e_g² → 5 unpaired electrons.
d) What is silver mirror test?
It is a test to detect aldehydes using Tollens’ reagent (AgNO₃ + NH₃). Aldehydes reduce Ag⁺ to
metallic Ag, forming a shiny silver mirror.
e) How is carboxylic acid prepared using dry ice?
Grignard reagent reacts with solid CO₂ (dry ice) to form carboxylic acid after acid hydrolysis.
R–MgX + CO₂ → RCOOMgX → (H⁺) → RCOOH
f) Explain why amines are basic in nature.
Amines have a lone pair on nitrogen that can accept a proton, making them Lewis bases.
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Q2a) Attempt the following (any two). [6]
i) Draw cis and trans isomers of [Co(NH₃)₄Cl₂]⁺
cis-isomer: Cl⁻ ligands adjacent
trans-isomer: Cl⁻ ligands opposite
(Draw a square planar or octahedral complex showing ligand positions.)
ii) Calculate the CFSE for d⁵ ion in high and low spin octahedral field.
High spin d⁵ → t₂g³e_g² → CFSE = 3(–0.4) + 2(+0.6) = 0 Δ₀
Low spin d⁵ → t₂g⁵e_g⁰ → CFSE = 5(–0.4) = –2.0 Δ₀
iii) Draw Geometrical isomers of [Cr(en)₂Cl₂]⁺
cis-isomer: both Cl⁻ ligands next to each other
trans-isomer: Cl⁻ ligands opposite
(en = ethylenediamine, a bidentate ligand)
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Q2b) Attempt the following. [4]
i) Baeyer’s strain theory:
It states that small ring compounds (like cyclopropane) experience angle strain because their bond
angles deviate from the ideal tetrahedral angle (109.5°), leading to instability.
ii) Locking of conformation:
Bulky groups or ring structures can "lock" a molecule in a particular conformation, preventing ring
flip or rotation, increasing stability in certain cases.
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Q3a) Attempt the following (any two). [6]
i) Explain how n-propylamine is prepared from cycloethane.
Cycloethane → Halogenation → Propyl halide → reacts with NH₃ → n-propylamine
C₃H₇Br + NH₃ → C₃H₇NH₂
ii) Explain Aldol condensation with suitable example.
Two aldehyde/ketone molecules with α-H combine to form β-hydroxy aldehyde, then dehydrate to
form α,β-unsaturated carbonyl compound.
CH₃CHO → CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO
iii) Any two methods for preparation of aldehyde:
1. Oxidation of primary alcohols: RCH₂OH → RCHO
2. Rosenmund reduction: RCOCl + H₂/Pd–BaSO₄ → RCHO
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Q3b) Discuss the factors affecting the magnitude of 10Dq (Δ₀): [4]
1. Nature of metal ion: Higher charge and smaller size → higher Δ₀
2. Nature of ligand: Strong field ligands increase Δ₀ (per spectrochemical series)
3. Geometry: Octahedral > Tetrahedral
4. Oxidation state: Higher oxidation → larger splitting
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Q4a) Attempt any two. [6]
i) Give assumptions of VBT:
1. Metal-ligand bonds involve overlap of orbitals
2. Hybrid orbitals determine geometry (sp³, d²sp³, etc.)
3. Ligands cause pairing depending on strength
4. Magnetism predicted by unpaired electrons
ii) Discuss Reformatsky reaction:
α-Halo ester + aldehyde/ketone + Zn → β-hydroxy ester
Example: BrCH₂COOC₂H₅ + aldehyde → HOCH₂CH(R)COOC₂H₅
iii) Hell-Volhard-Zelinsky reaction:
Halogenation at α-position of carboxylic acids in presence of PCl₃/Br₂:
RCH₂COOH + Br₂ → RCHBrCOOH
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Q4b) Jahn–Teller distortion with example. [4]
In certain configurations like d⁹ ([Cu(H₂O)₆]²⁺), uneven occupancy of eg orbitals leads to distortion
(elongation or compression) to remove degeneracy and lower energy.
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Q5) Attempt any four. [10]
a) Explain the term high spin and low spin complexes:
High spin: Weak field ligands → maximum unpaired electrons
Low spin: Strong field ligands → electron pairing → fewer unpaired electrons
b) Give assumptions of CFT:
1. Ligands are point charges
2. Metal-ligand bonding is purely electrostatic
3. Splitting of d-orbitals due to ligand repulsion
4. Explains color, magnetism, and stability of complexes
c) Draw chair conformation of cyclohexane indicating axial and equatorial H:
(Draw a hexagonal ring with alternating vertical (axial) and slanted (equatorial) hydrogens)
d) Classification of carboxylic acid:
Aliphatic/Aromatic
Saturated/Unsaturated
Monocarboxylic, Dicarboxylic, etc.
e) Identify products ‘A’ and ‘B’ and rewrite the reaction:
CH₃COOH + PCl₅ → CH₃COCl (A)
CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ (B)
A = Acetyl chloride, B = Ethyl acetate
f) Explain bonding, structure, and magnetic properties of [Ni(CN)₄]²⁻:
Ni²⁺ = d⁸, CN⁻ = strong field ligand
Hybridization: dsp²
Geometry: square planar
Magnetism: diamagnetic (all electrons paired)
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