Chemistry DSC (NEP) IV - SEM B.Sc Regular Solved Question Paper
1. Answer any SIX questions. (6 × 2 = 12)
a) Write Van Deemter’s equation and its significance.
Van Deemter equation:
Where:
H: Plate height
A: Eddy diffusion term
B: Longitudinal diffusion term
C: Mass transfer term
u: Linear velocity of mobile phase
Significance: It describes how different factors contribute to band broadening in chromatography and helps to determine the optimal flow rate.
b) Mention two factors affecting the distribution coefficient.
Temperature
Nature of solute and solvents (polarity and miscibility)
c) Mention the hybridization and shape of PCl₅ molecule.
Hybridization: sp³d
Shape: Trigonal bipyramidal
d) State Bent’s rule.
Bent’s rule states that: “Atomic s-character concentrates in orbitals directed toward electropositive substituents, and p-character concentrates toward more electronegative substituents.”
e) Write the Aldol condensation reaction of acetaldehyde.
- Aldol product: 3-hydroxybutanal
- Condensation product: Crotonaldehyde
f) What are arynes? Give an example.
Arynes are reactive intermediates derived from aromatic compounds by elimination of two substituents, usually a hydrogen and a leaving group, from adjacent carbon atoms.
Example: Benzyne.
g) The energy of activation of a reaction is zero and rate constant at 300 K is 3.2×10⁻⁶ s⁻¹. Calculate the rate constant at 310 K.
Given:
Activation energy (Ea) = 0 (since the reaction has no energy barrier)
Rate constant at 300 K (k1) = 3.2 × 10⁻⁶ s⁻¹
Temperature T1=300 K, T2=310 K
h) Define molar conductance and mention its variation with dilution.
Molar Conductance (Λₘ) is the conducting power of all ions from 1 mole of an electrolyte in solution. It is given by:
Variation:
Strong electrolytes (e.g., NaCl, HCl):
Λₘ increases slowly with dilution (more ion mobility).Weak electrolytes (e.g., CH₃COOH):
Λₘ increases sharply with dilution (more dissociation).At infinite dilution, Λₘ reaches maximum (Λₘ°).
2. Answer any THREE questions. (3 × 4 = 12)
a) Write about the following:
i) Stationary phase and mobile phase.
- Stationary phase: Phase that does not move; responsible for selective adsorption (e.g., silica, alumina).
- Mobile phase: Phase that moves through the stationary phase carrying components (e.g., solvents like hexane, ethanol).
ii) Rf value and factors affecting it.
The Rf value (Retention factor) is a ratio used in thin-layer chromatography (TLC) and paper chromatography to identify compounds. It is calculated as
Factors:
Nature of solute
Nature of solvent
Type of stationary phase
Temperature
b) Write a note on column chromatography.
Column chromatography is a separation technique used to purify individual chemical compounds from mixtures. It works on the principle of adsorption and differential migration of components through a stationary phase under the flow of a mobile phase.
Uses:
Purification of compounds.
Isolation of natural products (e.g., plant pigments).
Separation of complex mixtures in organic chemistry.
Advantage:
Handles larger quantities than TLC.
Highly versatile for various compounds.
Applications:
Purification of organic compounds.
Separation of plant pigments.
Used in pharmaceuticals and biochemistry for compound isolation.
Helps in analyzing complex mixtures in laboratories.
Working Principle:
Different substances in the mixture interact differently with the stationary phase.
Components with stronger adsorption move slower, while others move faster.
As a result, they get separated into bands or zones.
c) What is solvent extraction? Explain the batch solvent extraction.
Solvent extraction (also called liquid-liquid extraction) is a separation technique used to isolate a compound from a mixture by transferring it from one solvent (usually aqueous) to another immiscible solvent (organic) based on differential solubility.
Batch Solvent Extraction:
Principle:
A solute is partitioned between two immiscible liquids (e.g., water and organic solvent) until equilibrium is reached.
Procedure (Steps):
Mixing: The mixture is shaken vigorously with an immiscible solvent in a separating funnel.
Settling: The mixture is allowed to separate into two distinct layers.
Separation: The layer containing the desired compound is collected.
Repetition: The process is repeated to improve extraction efficiency.
Example:
Extraction of caffeine from water using dichloromethane.
Advantages:
- Simple and cost-effective.
- Suitable for small-scale extractions.
Limitations:
- Time-consuming for multiple batches.
- Requires large solvent volumes.
Applications:
Purification of organic compounds.
Isolation of pharmaceuticals and natural products.
Environmental analysis (e.g., pollutant extraction).
d) Explain the application of ion-exchange chromatography in the separation of Lanthanides.
Ion-exchange chromatography separates lanthanides based on their differences in affinity for charged resin beads. Lanthanides (Ln³⁺) bind to the resin and are selectively eluted using a complexing agent (e.g., EDTA or α-HIBA).
Process:
Column Packing: A cation-exchange resin (e.g., sulfonated polystyrene) is used.
Loading: Lanthanide mixture (Ln³⁺) in acidic solution is passed through the column.
Elution: A pH gradient or complexing agent (e.g., α-hydroxyisobutyric acid, α-HIBA) is applied.
Smaller Ln³⁺ ions (e.g., Lu³⁺) bind more strongly due to higher charge density.
Larger Ln³⁺ ions (e.g., La³⁺) elute first because of weaker resin interaction.
Collection: Fractions are collected and analyzed (e.g., via spectrophotometry).
Why It Works:
Lanthanide Contraction: Gradual decrease in ionic radii across the series causes slight differences in resin binding.
Complexation: Agents like EDTA enhance separation by forming varying stability complexes.
Applications:
- Purification of individual lanthanides (e.g., Nd, Sm for magnets).
- Nuclear fuel processing (U/Pu/Ln separations).
- Analytical chemistry (trace metal analysis).
Advantages:
High selectivity for +3 ions.
Scalable for industrial use.
Limitations:
Slow process (requires gradient optimization).
Overlapping peaks for adjacent lanthanides.
Example:
Separation of Europium (Eu³⁺) from other lanthanides in nuclear waste.
3. Answer any THREE questions (3×4=12)
a) Explain the sp³ hybridisation of BF₄⁻ molecule by VBT.
BF₄⁻ (tetrafluoroborate ion) is formed when boron trifluoride (BF₃) accepts a fluoride ion (F⁻).
In BF₄⁻, the central atom boron (B) forms four sigma bonds with four fluorine atoms.
Hybridisation:
Boron has 3 valence electrons (2s²2p¹).
Upon accepting one more electron from F⁻, it becomes 4 valence electrons.
Boron undergoes sp³ hybridisation, mixing one 2s and three 2p orbitals to form four sp³ hybrid orbitals.
Each hybrid orbital overlaps with a fluorine atom’s p-orbital to form a sigma bond.
Structure:
The resulting shape is tetrahedral.
Bond angle = 109.5°
All B–F bonds are of equal length.
b) Give the molecular orbital energy level diagram of N₂ and write its magnetic property.
c) Calculate the bond order of O₂ and O₂⁻, and predict their stability, bond length and bond energy.
d) Write in brief the Free Electron Theory of metallic bonding and how it explains the thermal conductivity of metals.
Free Electron Theory:
Proposed to explain the electrical and thermal conductivity of metals.
Assumes that valence electrons are not bound to individual atoms but move freely throughout the metallic lattice.
These electrons are called free electrons or electron gas.
Key Points:
The metal is seen as a lattice of positive ions immersed in a sea of free electrons.
The movement of these electrons explains metallic properties like:
High electrical conductivity
High thermal conductivity
Luster and malleability
Thermal Conductivity:
As temperature increases, free electrons gain kinetic energy and move faster.
These electrons transfer heat energy throughout the lattice by collisions, resulting in high thermal conductivity.