Complete Detailed Solutions · 2024-25 (BSCH201)
(i) Write the reagents of nitration of benzene.
Answer: Concentrated Nitric Acid (HNO₃) and Concentrated Sulfuric Acid (H₂SO₄).
Explanation: This combination is known as the "nitrating mixture." H₂SO₄ acts as a strong acid to protonate HNO₃, which then loses water to generate the powerful electrophile, the nitronium ion (NO₂⁺).
(ii) Write the value of quantum numbers n, l and m for 2s orbital.
Answer: n = 2, l = 0, m = 0.
Explanation: Principal quantum number (n) is 2. The azimuthal quantum number (l) for an s-orbital is always 0. The magnetic quantum number (m) ranges from -l to +l, so it can only be 0.
(iii) MRI is the application of which spectroscopic method?
Answer: Nuclear Magnetic Resonance (NMR) Spectroscopy.
Explanation: MRI (Magnetic Resonance Imaging) uses the principles of NMR to image the protons (hydrogen nuclei) in the water and fat of the human body.
(iv) London forces or dispersion forces operates between _________.
Answer: Non-polar molecules (and all other molecules/atoms).
Explanation: They arise from temporary, fluctuating dipoles.
(v) How the entropy of the system changes when water is frozen?
Answer: Entropy decreases (ΔS < 0).
Explanation: Liquid water molecules are relatively free to move. In solid ice, the molecules are locked into a highly ordered crystalline lattice, significantly reducing their randomness (entropy).
(vi) What is Hard-soft acid base principle?
Answer: Hard acids prefer to bind with hard bases, and soft acids prefer to bind with soft bases.
Explanation: This is Pearson's HSAB principle. Hard-hard interactions are more ionic, while soft-soft interactions are more covalent.
(vii) If two stereoisomers of tartaric acid have 2S, 3R and 2S, 3S configurations, then they are __________.
Answer: Diastereomers.
Explanation: The configuration at carbon-2 is the same (S in both), but the configuration at carbon-3 is inverted (R vs S). Since they are not perfect mirror images, they are diastereomers.
(viii) What are the conditions of Cannizzaro reaction?
Answer: The aldehyde must lack alpha-hydrogens, and the reagent must be a concentrated strong base (e.g., 50% NaOH or KOH).
Explanation: Without α-hydrogens (like in formaldehyde or benzaldehyde), the aldol condensation cannot occur, leading to a disproportionation (auto-oxidation-reduction) reaction instead.
(ix) State whether naphthalene is aromatic, antiaromatic or nonaromatic and give reason.
Answer: Aromatic.
Explanation: Naphthalene is a fully conjugated, planar bicyclic ring system. It contains 10 π electrons, which completely satisfies Hückel's Rule (4n + 2) where n = 2.
(x) Write the equation of rotational constant B of a molecule.
Answer: B = h / (8π²Ic) (in cm⁻¹)
Explanation: 'h' is Planck's constant, 'I' is the moment of inertia, and 'c' is the speed of light.
(xi) What is the value of critical coefficient?
Answer: RT_c / (P_c V_c) = 8/3 (or ~2.67).
Explanation: This is a universal constant for all gases that follow the van der Waals equation.
(xii) Write the half-cell representation of saturated calomel electrode.
Answer: Pt | Hg(l) | Hg₂Cl₂(s) | KCl(sat)
Role of Lewis acid in halogenation of benzene:
Benzene is an electron-rich aromatic ring that reacts with electrophiles. However, halogens like Cl₂ or Br₂ are not strong enough electrophiles on their own to disrupt the stable aromaticity of benzene.
A Lewis acid catalyst (like FeCl₃, FeBr₃, or AlCl₃) acts as an electron-pair acceptor. It reacts with the halogen molecule (e.g., Br₂) to polarize the bond and create a highly reactive, positive halonium ion (e.g., Br⁺, or a strongly polarized Br_δ+ --- Br-FeBr₃_δ- complex). This incredibly strong electrophile is what successfully attacks the benzene ring in Electrophilic Aromatic Substitution (EAS).
Substrate for ozonolysis yielding only acetone:
Ozonolysis cleaves a carbon-carbon double bond (C=C), placing a double-bonded oxygen (=O) on each of the originally double-bonded carbons.
Acetone is a 3-carbon ketone: (CH₃)₂C=O.
If ozonolysis produces acetone as the only product, the original molecule must be symmetrical, consisting of two acetone halves joined at the carbonyl carbons.
Structure: (CH₃)₂C=C(CH₃)₂
IUPAC Name: 2,3-dimethyl-2-butene.
Significance of van der Waals Constants 'a' and 'b':
The van der Waals equation for real gases is: (P + an²/V²)(V - nb) = nRT
1. Constant 'a' (Pressure Correction Factor):
It is a measure of the magnitude of intermolecular attractive forces between the gas molecules.
- A larger value of 'a' indicates stronger intermolecular forces (like in NH₃ or CO₂) compared to gases with small 'a' (like He or H₂).
- Gases with higher 'a' values are more easily liquefied because the attractive forces can more readily pull the molecules together.
2. Constant 'b' (Volume Correction Factor):
It represents the effective volume occupied by the gas molecules themselves (also called co-volume or excluded volume).
- It accounts for the fact that real gas molecules have a finite, non-zero size, reducing the total free volume available for them to move.
- A larger value of 'b' indicates a larger molecular size (e.g., CCl₄ has a much larger 'b' than H₂).
Conformational Isomers:
Conformational isomers (or conformers) are different spatial arrangements of the atoms in a molecule that arise strictly from the free rotation around single (sigma) bonds. Unlike structural or configurational isomers, conformers interconvert rapidly at room temperature and usually cannot be isolated.
Ethane Example:
Ethane (CH₃-CH₃) consists of two methyl groups connected by a C-C single bond. As one methyl group rotates relative to the other, it creates an infinite number of conformations. The two most extreme conformations are:
- Staggered Conformation: The C-H bonds on the front carbon are exactly positioned between the C-H bonds on the back carbon (dihedral angle = 60°). This minimizes electron repulsion between the bonds, making it the most stable, lowest-energy conformation.
- Eclipsed Conformation: The C-H bonds on the front carbon perfectly align with (eclipse) the C-H bonds on the back carbon (dihedral angle = 0°). This causes maximum torsional strain due to electron repulsion between the aligned bonds, making it the least stable, highest-energy conformation.
Potential Energy Diagram of Ethane:
If we plot Potential Energy vs. Dihedral Angle (Angle of Rotation):
- The curve is a smooth wave with a periodicity of 120°.
- The minima (valleys) occur at 60°, 180°, and 300°, corresponding to the stable staggered conformations.
- The maxima (peaks) occur at 0°, 120°, and 240°, corresponding to the unstable eclipsed conformations.
- The energy barrier to rotation (difference between max and min) is relatively small, about 12.5 kJ/mol (3 kcal/mol), allowing rapid rotation at room temperature.
From the First Law of Thermodynamics, the change in internal energy (ΔU) is:
ΔU = q + w
If the work is only mechanical pressure-volume (P-V) expansion work, then:
w = -PΔV (where P is external pressure)
Substituting this into the first law equation:
ΔU = q - PΔV
For a process occurring at constant pressure, the heat absorbed is denoted as q_p. Thus:
ΔU = q_p - PΔV
q_p = ΔU + PΔV
Expanding the change terms from initial state (1) to final state (2):
q_p = (U₂ - U₁) + P(V₂ - V₁)
q_p = (U₂ + PV₂) - (U₁ + PV₁)
By definition, Enthalpy (H) is defined as H = U + PV. Therefore:
H₂ = U₂ + PV₂ and H₁ = U₁ + PV₁
Substitute H into the q_p equation:
q_p = H₂ - H₁
q_p = ΔH (Proved)
Electronic Configuration of NO (Nitric Oxide):
Total valence electrons = 5 (from N) + 6 (from O) = 11 electrons.
Using Molecular Orbital Theory, the configuration is:
σ(2s)² σ*(2s)² σ(2pz)² π(2px)² = π(2py)² π*(2px)¹
Justification for Magnetic Behavior:
- NO is Paramagnetic: Looking at the MO configuration of NO, the highest occupied molecular orbital (HOMO) is the π*(2px) anti-bonding orbital, which contains exactly one unpaired electron. The presence of this unpaired electron makes NO paramagnetic.
- NO⁺ is Diamagnetic: When NO loses an electron to form the nitrosonium ion (NO⁺), that electron is removed from the highest energy level, which is the π*(2px) orbital. The new configuration for NO⁺ (10 valence electrons) is: σ(2s)² σ*(2s)² σ(2pz)² π(2px)² = π(2py)². Now, all the electrons are fully paired in bonding orbitals. The absence of any unpaired electrons makes NO⁺ diamagnetic.
(a) Define 'Standard cell' with example:
A standard cell is an electrochemical cell whose electromotive force (EMF) is accurately known, highly stable, and reproducible over a long period. It must have a very small temperature coefficient (its voltage should not change much with temperature) and must not be susceptible to polarization when a small current is drawn. It is used as a reference to calibrate instruments like potentiometers.
Example: The Weston Cadmium Cell (produces exactly 1.0183 V at 20°C).
(b) Relation between ΔH and ΔU for an ideal gas undergoing P-V type mechanical work:
By definition, Enthalpy (H) is given by: H = U + PV
For a macroscopic change at constant temperature and pressure: ΔH = ΔU + Δ(PV)
For an ideal gas, PV = nRT. Therefore, Δ(PV) = Δ(nRT).
Since temperature (T) and the gas constant (R) are constant, only the number of moles (n) changes during a chemical reaction. Thus: Δ(PV) = (Δn)RT.
Substituting this back into the enthalpy equation yields:
ΔH = ΔU + ΔnRT
(where Δn is the difference between the number of moles of gaseous products and gaseous reactants).
(c) First law of thermodynamics (Statement and Mathematical explanation):
Statement: Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant.
Mathematical explanation: If a system absorbs a certain amount of heat (q) from the surroundings, this energy can be used in two ways: (1) to increase the internal energy of the system (ΔU), and (2) to do work (w) on the surroundings. Thus: q = ΔU + w, which is often rearranged as ΔU = q + w (where w is work done on the system).
(d) Explain 'Galvanic cell corrosion':
Galvanic corrosion occurs when two different (dissimilar) metals are in physical or electrical contact while immersed in a common conducting electrolyte (like saltwater).
- The more chemically active (less noble) metal becomes the anode and undergoes accelerated oxidation (corrosion), losing electrons and dissolving into ions.
- The less active (more noble) metal becomes the cathode and is protected from corrosion; reduction reactions (like oxygen reduction or hydrogen evolution) occur on its surface.
Example: A steel pipe connected to a copper fitting. The steel (more reactive) acts as the anode and rusts rapidly, while the copper (less reactive) is protected.
(a) Shape of the following compounds (using VSEPR Theory):
(i) ClF₃ (Chlorine trifluoride):
Central atom Cl has 7 valence electrons. It forms 3 bonds with F and has 2 lone pairs. Total domains = 5 (sp³d hybridization). The 2 lone pairs occupy equatorial positions to minimize repulsion. Shape: T-shaped.
(ii) XeF₂ (Xenon difluoride):
Central atom Xe has 8 valence electrons. It forms 2 bonds with F and has 3 lone pairs. Total domains = 5 (sp³d hybridization). The 3 lone pairs occupy all three equatorial positions. The 2 F atoms are axial. Shape: Linear.
(iii) BrF₅ (Bromine pentafluoride):
Central atom Br has 7 valence electrons. It forms 5 bonds with F and has 1 lone pair. Total domains = 6 (sp³d² hybridization). The lone pair occupies one of the positions in the octahedron. Shape: Square Pyramidal.
(iv) SCl₆ (Sulfur hexachloride - theoretically, practically SF₆ is used as SCl₆ is sterically hindered):
Central atom S has 6 valence electrons. It forms 6 bonds with Cl and has 0 lone pairs. Total domains = 6 (sp³d² hybridization). Shape: Octahedral.
(b) Arrange NH₃, H₂O, CH₄ according to increasing bond angle stating the reason:
Order: H₂O (104.5°) < NH₃ (107°) < CH₄ (109.5°)
Reason: All three central atoms (O, N, C) are sp³ hybridized, which gives a basic tetrahedral geometry with an ideal bond angle of 109.5°.
- CH₄ has 4 bonding pairs and 0 lone pairs. The repulsions are equal, so the angle remains exactly 109.5°.
- NH₃ has 3 bonding pairs and 1 lone pair. According to VSEPR theory, Lone Pair-Bond Pair repulsion is greater than Bond Pair-Bond Pair repulsion. The lone pair pushes the N-H bonds closer together, reducing the angle to 107°.
- H₂O has 2 bonding pairs and 2 lone pairs. The Lone Pair-Lone Pair repulsion is the strongest, pushing the two O-H bonds even closer together, reducing the angle further to 104.5°.
(c) Dipole moment of BF₃ is zero while that of NF₃ is 0.24D. Justify:
- BF₃ (Boron trifluoride): Boron is sp² hybridized. It forms 3 B-F bonds and has no lone pairs on the central atom. The molecule adopts a perfectly symmetrical trigonal planar geometry. Although individual B-F bonds are highly polar, the three bond dipole vectors (separated by exactly 120°) completely cancel each other out (vector sum = 0). Thus, net dipole moment = 0.
- NF₃ (Nitrogen trifluoride): Nitrogen is sp³ hybridized. It forms 3 N-F bonds and has one lone pair. The molecule adopts an asymmetrical trigonal pyramidal shape. The three polar N-F bonds point downwards, and their dipoles do not cancel out. Additionally, the lone pair contributes to the net dipole. Therefore, NF₃ has a net, non-zero dipole moment (0.24D).
(a) Derive Schrodinger Equation:
Consider a particle of mass m moving with velocity v. The classical wave equation for a 1D standing wave is:
(∂²Ψ/∂x²) = (1/u²) * (∂²Ψ/∂t²), where u is wave velocity. For a stationary state, Ψ(x,t) = ψ(x)sin(2πνt).
Differentiating twice wrt x and t, and substituting gives the time-independent form:
d²ψ/dx² + (4π²/λ²)ψ = 0
From de Broglie relation, λ = h/mv, so 1/λ² = m²v²/h².
Substitute this: d²ψ/dx² + (4π²m²v²/h²)ψ = 0
Total Energy (E) = Kinetic (K) + Potential (V). So, K = E - V. Since K = 1/2 mv², we have m²v² = 2m(E - V).
Substituting this into the equation yields the Time-Independent Schrodinger Equation (1D):
d²ψ/dx² + (8π²m/h²)(E - V)ψ = 0
(b) Conditions of acceptable wave function Ψ:
To be physically meaningful (representing a real particle), the wave function Ψ must satisfy:
1. Continuous: Ψ must be continuous everywhere.
2. Single-valued: For any given position (x, y, z), Ψ must have only one unique value.
3. Finite: Ψ must be finite everywhere (must not go to infinity).
4. Smooth: Its first derivatives (dΨ/dx) must also be continuous (except at points where potential is infinite).
5. Normalized: The total probability of finding the particle somewhere in space must be 1 (integral of |Ψ|² dτ = 1).
(c) Extrinsic semiconductor and its types:
An extrinsic semiconductor is a pure semiconductor (intrinsic, like Si or Ge) that has been intentionally doped with a small, specific amount of impurity atoms to drastically increase its electrical conductivity.
- n-type (negative type): Formed by doping with Group 15 (pentavalent) elements like Phosphorus (P) or Arsenic (As). Four valence electrons form bonds, and the fifth extra electron is free to conduct electricity. The majority charge carriers are electrons.
- p-type (positive type): Formed by doping with Group 13 (trivalent) elements like Boron (B) or Gallium (Ga). The three valence electrons form bonds, leaving an electron deficiency or "hole" in the lattice. These holes act as positive charge carriers and can move, conducting electricity.
(a) Reaction of CH₃CH=CH₂ with HBr:
Propene reacts with HBr via electrophilic addition following Markovnikov's rule (the H atom adds to the carbon with more hydrogens, and the Br adds to the carbon with fewer hydrogens).
Product: CH₃-CHBr-CH₃ (2-Bromopropane).
Steps:
Step 1: The double bond attacks H⁺ from HBr. A secondary carbocation is formed because it is more stable than a primary one: CH₃-CH=CH₂ + H⁺ → [CH₃-C⁺H-CH₃] (2° carbocation).
Step 2: The bromide ion (Br⁻) attacks the carbocation to form the final product: [CH₃-C⁺H-CH₃] + Br⁻ → CH₃-CHBr-CH₃.
(b) Change if peroxide is added:
If an organic peroxide is added, the reaction follows the Anti-Markovnikov rule (Kharasch effect) via a free-radical mechanism.
New Product: CH₃-CH₂-CH₂Br (1-Bromopropane).
(c) Suitable substrate for E1 reaction (tert-butyl chloride vs methyl chloride):
tert-butyl chloride is the suitable substrate.
Reason: The E1 (Elimination Unimolecular) mechanism proceeds via the formation of a carbocation intermediate. The rate of the reaction depends entirely on the stability of this carbocation. tert-Butyl chloride forms a highly stable tertiary (3°) carbocation due to the inductive (+I) effect and hyperconjugation from three surrounding methyl groups. Methyl chloride would form a primary (1°) methyl carbocation, which is extremely unstable and practically impossible to form under normal conditions.
(d) Stereochemical aspects of SN1 and SN2 reactions:
- SN2 Reaction: Involves a single-step concerted mechanism where the nucleophile attacks the chiral carbon from the side directly opposite to the leaving group (backside attack). This leads to complete inversion of configuration (Walden inversion).
Example: Reaction of (S)-2-bromobutane with OH⁻ yields exclusively (R)-2-butanol.
- SN1 Reaction: Involves a two-step mechanism forming a planar carbocation intermediate. The nucleophile can attack this planar intermediate from either the front or the back face with roughly equal probability. This leads to a mixture of retention and inversion, resulting in racemization (or partial racemization).
Example: Reaction of (S)-3-bromo-3-methylhexane with H₂O yields a racemic mixture of (R)- and (S)-3-methyl-3-hexanol.
(a) Calculate CFSE and Magnetic moment of K₃[FeF₆]:
The complex ion is [FeF₆]³⁻. Iron is in the +3 oxidation state (Fe³⁺).
Fe is [Ar] 4s² 3d⁶, so Fe³⁺ is 3d⁵.
Fluoride (F⁻) is a weak field ligand, so the complex is high spin. The 5 electrons will fill the orbitals singly before pairing: t₂g³ eg².
CFSE: = [(-0.4 × number of t₂g e⁻) + (+0.6 × number of eg e⁻)] Δo
CFSE = [(-0.4 × 3) + (0.6 × 2)] Δo = (-1.2 + 1.2) Δo = 0 Δo.
Magnetic Moment (μ): Number of unpaired electrons (n) = 5.
μ = √(n(n+2)) = √(5(5+2)) = √35 ≈ 5.92 Bohr Magnetons (B.M.)
(b) CFSE of high spin and low spin complexes of d⁷ and d⁴:
- For d⁷ configuration:
High Spin: t₂g⁵ eg². CFSE = (-0.4×5 + 0.6×2)Δo = -2.0 + 1.2 = -0.8 Δo.
Low Spin: t₂g⁶ eg¹. CFSE = (-0.4×6 + 0.6×1)Δo = -2.4 + 0.6 = -1.8 Δo + P (where P is pairing energy).
- For d⁴ configuration:
High Spin: t₂g³ eg¹. CFSE = (-0.4×3 + 0.6×1)Δo = -1.2 + 0.6 = -0.6 Δo.
Low Spin: t₂g⁴ eg⁰. CFSE = (-0.4×4 + 0.6×0)Δo = -1.6 + 0 = -1.6 Δo + P.
(c) Molecular energy level diagram of CO:
CO has 14 valence electrons (4 from C + 6 from O). Because oxygen is much more electronegative than carbon, its atomic orbitals are lower in energy. This creates an asymmetric MO diagram, but the basic ordering is similar to N₂.
MO Configuration: σ(1s)² σ*(1s)² σ(2s)² σ*(2s)² π(2px)² = π(2py)² σ(2pz)²
- Bond Order: (Number of bonding e⁻ - Number of anti-bonding e⁻) / 2 = (10 - 4) / 2 = 6 / 2 = 3. (It implies a strong triple bond, C≡O).
- Magnetic Behavior: All electrons in the MO configuration are paired in the lowest energy orbitals. Since there are zero unpaired electrons, carbon monoxide (CO) is diamagnetic.