FYUG Even Semester Exam, 2025

CHEMISTRY (4th Semester) Course No.: CHMDSM-252 Time: 3 Hours | Full Marks: 70 | Pass Marks: 28

UNIT-I

1. (a) What is inorganic benzene? Write its structure. 2

Borazine (B3N3H6) is known as inorganic benzene because its structure is isoelectronic and isostructural with benzene. It consists of a hexagonal ring of alternating Boron and Nitrogen atoms with one Hydrogen atom attached to each.

1. (b) Discuss briefly the catenation property of carbon. 2

Catenation is the ability of an element to form covalent bonds with other atoms of the same element, resulting in long chains or rings. Carbon exhibits the highest catenation property due to its small size and high C-C bond energy, allowing for the vast diversity of organic compounds.

1. (c) Write two structural differences between diamond and graphite. 2

Feature	Diamond	Graphite
Hybridization	sp³	sp²
Structure	3D tetrahedral network	2D hexagonal layered sheets

2. (a) What happens when... 10

(i) Diborane is hydrolyzed: It reacts with water to form Boric acid and Hydrogen gas.

B2H6 + 6H2O → 2H3BO3 + 6H2

(ii) Diborane is treated with NH3:

At low temperatures: Forms an unsymmetrical adduct [H2B(NH3)2]+ [BH4]-.

On heating (1:2 ratio): Forms Borazine (B3N3H6).

(iii) Calcium carbide is treated with water: It produces Ethyne (Acetylene) and Calcium hydroxide.

CaC2 + 2H2O → C2H2 + Ca(OH)2

(iv) Excess CO2 is passed through Ca(OH)2: The initial milkiness (CaCO3) disappears due to the formation of soluble Calcium bicarbonate.

CaCO3 + H2O + CO2 → Ca(HCO3)2

(v) BF3 is treated with NaBH4 in diglyme: It produces Diborane gas.

3NaBH4 + 4BF3 → 2B2H6 + 3NaBF4

2. (b) (ii) Discuss the bonding in B2H6. 3

Diborane (B2H6) features unique 3-center-2-electron (3c-2e) bonds, often called "banana bonds".

There are four terminal B-H bonds which are normal 2-center-2-electron bonds.

The two bridging Hydrogen atoms are bonded to both Boron atoms, where two electrons are shared over three centers (B-H-B).

UNIT-II

3. (a) What are open system and closed system? 2

Open System: A system that can exchange both matter and energy with its surroundings.

Closed System: A system that can exchange energy (heat or work) but not matter with its surroundings.

3. (c) State the first law of thermodynamics. Write its mathematical form. 2

The First Law of Thermodynamics states that energy can neither be created nor destroyed, only transformed from one form to another.

ΔU = q + w

Where ΔU is the change in internal energy, q is heat added to the system, and w is work done on the system.

4. (b) (i) Derive Kirchhoff's equation. 6

Kirchhoff's equation describes how the enthalpy of a reaction changes with temperature.

For a reaction: ΔH = H(products) - H(reactants).

Differentiating with respect to Temperature (T) at constant Pressure (P):

d(ΔH)/dT = dH(prod)/dT - dH(react)/dT.

Since (dH/dT)p = Cp (heat capacity at constant pressure):

d(ΔH)/dT = ΔCp

Integrating from T1 to T2:

ΔH2 - ΔH1 = ΔCp(T2 - T1)

UNIT-III

5. (a) Distinguish between ideal and non-ideal solutions. 2

Ideal Solution	Non-Ideal Solution
Obeys Raoult's law at all concentrations.	Does not obey Raoult's law.
ΔH_mixing = 0.	ΔH_mixing ≠ 0.
ΔV_mixing = 0.	ΔV_mixing ≠ 0.

5. (c) Define Gibbs' phase rule. Explain terms. 2

F = C - P + 2

F (Degrees of Freedom): Number of independent variables (P, T, concentration) that must be specified to define the system.

C (Components): Minimum number of chemically independent constituents.

P (Phases): Homogeneous, physically distinct part of the system.

6. (b) (i) Draw the phase diagram of water system. 6

The water system is a one-component system with three possible phases: ice, water, and vapor.

Triple Point: Point where all three phases coexist in equilibrium (F=0).

Curves: Represent two phases in equilibrium (F=1) e.g., Vaporization curve.

Areas: Represent a single phase (F=2) e.g., Liquid region.

UNIT-IV

7. (b) Propene treated with HBr (Peroxide effect). 2

Absence of Peroxide (Markownikoff's Rule): H attaches to the carbon with more hydrogens. Product: 2-Bromopropane.
Presence of Peroxide (Anti-Markownikoff's Rule): Br attaches to the carbon with more hydrogens. Product: 1-Bromopropane.

CH3-CH=CH2 + HBr → CH3-CHBr-CH3 (No Peroxide)

CH3-CH=CH2 + HBr (Peroxide) → CH3-CH2-CH2Br

8. (a) (ii) Friedel-Crafts Alkylation and Acylation. 6

Alkylation: Introduction of an alkyl group into the benzene ring using an alkyl halide and AlCl3.

C6H6 + CH3Cl (AlCl3) → C6H5-CH3 + HCl

Acylation: Introduction of an acyl group (RCO-) using an acid chloride and AlCl3.

C6H6 + CH3COCl (AlCl3) → C6H5-COCH3 + HCl

UNIT-V

9. (a) Why does CH3X favour SN2 reaction? 2

Methyl halides (CH3X) favour SN2 because they have the least steric hindrance. The nucleophile can easily attack the central carbon atom from the backside. SN1 is not favored because the methyl carbocation is highly unstable.

10. (a) (i) Discuss SN1 mechanism. 6

SN1 (Unimolecular Nucleophilic Substitution) is a two-step process:

Formation of Carbocation: The leaving group departs (slowest step).

Nucleophilic Attack: The nucleophile attacks the flat carbocation from either side.

Racemic Mixture: Since the carbocation intermediate is planar, the nucleophile can attack from the front or back with equal probability, leading to an equal mixture of two enantiomers (racemization).

Exam Focus Enhancements

Exam Tips

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Always write balanced chemical equations for "What happens when" questions to secure full marks[span_49](end_span).

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For mechanisms, clearly show electron movement using curved arrows[span_51](end_span).

In Phase Rule questions, define all variables (F, C, P) before solving.

Common Mistakes

Confusing the peroxide effect (Anti-Markownikoff) with normal addition in propene.

Forgetting to include the "2" in the phase rule formula (F = C - P + 2).

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In SN1/SN2 questions, failing to mention steric hindrance or carbocation stability[span_56](end_span).

Important Formulas List

First Law: ΔU = q + w

Kirchhoff's: ΔH2 - ΔH1 = ΔCp(T2 - T1)

Phase Rule: F = C - P + 2

Raoult's Law: P_total = P°A*XA + P°B*XB

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