Complete Detailed Solutions · 2023-24 (BS-CH101)
(I) Write name of a molecule which have infrared active vibrations.
Answer: HCl, CO₂, or H₂O.
Explanation: Molecules that experience a change in their dipole moment during vibration are IR active.
(II) The strength of van der Waals forces depends upon which factor?
Answer: Polarizability and the surface area of the molecule.
(III) Write one process where entropy decreases.
Answer: Freezing of liquid water to solid ice (or condensation of a gas).
Explanation: The system becomes more ordered, hence entropy (ΔS) is negative.
(IV) What is the shape of XeF₄ molecule?
Answer: Square planar.
Explanation: Xenon is sp³d² hybridized (4 bonding pairs, 2 lone pairs). The lone pairs occupy axial positions to minimize repulsion, resulting in a square planar geometry.
(V) For n-butane which type of conformation is the least stable?
Answer: Fully eclipsed conformation.
Explanation: In the fully eclipsed conformation, the two bulky methyl groups are at a dihedral angle of 0°, causing maximum torsional and steric strain.
(VI) In SN1 type reaction which type of solvent is used?
Answer: Polar protic solvent.
Explanation: Solvents like water or alcohols stabilize the intermediate carbocation through solvation via hydrogen bonding.
(VII) If uncertainty in position and momentum are equal then what will be the uncertainty in velocity?
Answer: Δv = (1/2m)√(h/π)
Explanation: From Heisenberg's principle: Δx · Δp ≥ h/4π. Since Δx = Δp, we have (Δp)² = h/4π.
Since Δp = mΔv, then m²(Δv)² = h/4π, which gives Δv = (1/m)√(h/4π) = (1/2m)√(h/π).
(VIII) Which is detected by IR spectra?
Answer: Functional groups.
Explanation: Characteristic vibrational frequencies allow the identification of specific bonds and functional groups (like C=O, O-H).
(IX) Which interaction is the strongest interaction?
Answer: Ion-Ion interaction (among non-covalent forces) or Covalent bond (among all bonds).
(X) What is the internal energy change for a cyclic process?
Answer: Zero (ΔU = 0).
Explanation: Internal energy is a state function. In a cyclic process, the initial and final states are the same, so the change is zero.
(XI) Write the increasing order of effective nuclear charge in Na, Al, Mg and Si?
Answer: Na < Mg < Al < Si
Explanation: Effective nuclear charge (Z_eff) increases steadily across a period from left to right as protons are added to the nucleus without a proportional increase in shielding.
(XII) Give one example of ambidentate ligand.
Answer: Thiocyanate ion (SCN⁻).
Explanation: It can bind to a metal center through either the Sulfur atom or the Nitrogen atom. Another example is the Nitrite ion (NO₂⁻).
Van der Waals forces:
Van der Waals forces are relatively weak, short-range, distance-dependent intermolecular (or interatomic) attractions that exist between all molecules and atoms, including neutral and non-polar ones. They do not involve the sharing or transferring of electrons (unlike covalent or ionic bonds).
Nature of Van der Waals forces:
They are fundamentally electrostatic in nature, arising from the interactions of permanent or transient electric dipole moments. They are purely physical forces. They are divided into three types:
1. Keesom forces (Dipole-Dipole): Between two permanent dipoles.
2. Debye forces (Dipole-Induced Dipole): Between a permanent dipole and an induced dipole in a neighboring polarizable molecule.
3. London Dispersion forces: Present in all molecules (even non-polar). Caused by temporary fluctuations in electron distribution creating a transient dipole, which induces another transient dipole in a neighboring molecule. They are the weakest but often the most dominant because they occur everywhere.
(a) Chemical Potential (μ_i):
Chemical potential is defined as the partial molar Gibbs free energy of a component in a mixture. It represents the change in the total free energy of a system when one mole of component 'i' is added to an infinitely large amount of the system at constant temperature, pressure, and keeping the amounts of all other components constant. Mathematically: μ_i = (∂G / ∂n_i)_(T, P, n_j).
(b) Relation of EMF with ΔG and ΔH:
The maximum electrical work done by a galvanic cell is equal to the decrease in its Gibbs free energy.
ΔG = -nFE (where n = moles of electrons, F = Faraday's constant, E = EMF of the cell).
From the Gibbs-Helmholtz equation: ΔG = ΔH + T[∂(ΔG)/∂T]_P
Substituting ΔG = -nFE into the equation:
-nFE = ΔH + T[∂(-nFE)/∂T]_P
-nFE = ΔH - nFT(∂E/∂T)_P
Dividing by -nF, we get:
E = -ΔH/(nF) + T(∂E/∂T)_P
or ΔH = -nFE + nFT(∂E/∂T)_P. This is the required relation, where (∂E/∂T)_P is the temperature coefficient of the cell EMF.
(a) Chromium 4s Electron:
The expected electron configuration of Chromium (Z=24) is [Ar] 4s² 3d⁴. However, the actual configuration is [Ar] 4s¹ 3d⁵.
Reason: A exactly half-filled d-subshell (3d⁵) provides extra stability due to completely symmetrical electron distribution and maximum exchange energy. The energy difference between the 4s and 3d orbitals is very small, so one electron shifts from 4s to 3d to achieve this highly stable, half-filled state.
(b) Larger size between Mg²⁺ and N³⁻:
N³⁻ has a larger size.
Reason: Both Mg²⁺ and N³⁻ are isoelectronic species (both have 10 electrons; 1s² 2s² 2p⁶). However, Nitrogen has only 7 protons in its nucleus, while Magnesium has 12 protons.
In Mg²⁺, 12 protons pull the 10 electrons strongly inward, shrinking the electron cloud. In N³⁻, only 7 protons are pulling the same 10 electrons. The weaker effective nuclear charge in N³⁻ allows the electron cloud to expand due to inter-electronic repulsion, resulting in a significantly larger ionic radius.
(a) Constitutional vs Stereo isomers:
- Constitutional (Structural) Isomers: Molecules with the same molecular formula but different bonding connectivity (the atoms are linked in a different order). Example: Ethanol (CH₃CH₂OH) vs Dimethyl ether (CH₃OCH₃).
- Stereoisomers: Molecules with the same molecular formula and identical bonding connectivity, but a different spatial arrangement of atoms in 3D space. Example: cis-2-butene vs trans-2-butene.
(b) Chirality:
Chirality is the geometric property of a rigid object (or molecule) of being non-superimposable on its mirror image. Such molecules lack a plane of symmetry and a center of inversion. They are optically active (rotate plane-polarized light).
(c) Two chiral carbons and optical activity:
No, the presence of two chiral carbon atoms does NOT always make the molecule optically active.
Explanation: If a molecule has two (or any even number of) chiral centers but also possesses an internal plane of symmetry or a center of inversion, the optical activity of one half of the molecule cancels out the optical activity of the other half. This results in an optically inactive, achiral molecule called a Meso compound.
Example: meso-Tartaric acid has two chiral carbons (2R, 3S) but is optically inactive due to a plane of symmetry.
Comment: The statement is partially incorrect. It should read: "All adiabatic reversible expansions lead to a fall in temperature, while adiabatic reversible compressions lead to a rise in temperature."
Justification:
In an adiabatic process, there is no heat exchange with the surroundings (dq = 0).
From the First Law of Thermodynamics: dU = dq + dw. Since dq = 0, we have dU = dw.
- During Expansion: The gas does work on the surroundings (dw is negative). Therefore, dU is negative. Since the internal energy (U) of an ideal gas depends only on its temperature (dU = nCv dT), a decrease in internal energy results in a decrease in temperature (cooling).
- During Compression: Work is done on the gas by the surroundings (dw is positive). Therefore, dU is positive, which leads to an increase in temperature (heating).
(a) Bromination of Phenol:
- In non-polar solvent (CS₂) at low temperature: The -OH group is highly activating (ortho/para directing). However, in a non-polar solvent, phenol is not ionized to the more reactive phenoxide ion, so bromination stops at monosubstitution. It yields a mixture of o-bromophenol (X) and p-bromophenol (Y) (p-isomer is the major product).
Reaction: C₆H₅OH + Br₂ (in CS₂, 0°C) → o-Br-C₆H₄-OH + p-Br-C₆H₄-OH + HBr
- In polar solvent (Bromine Water): Water ionizes phenol into the highly reactive phenoxide ion. The ring becomes so activated that tri-substitution occurs instantaneously, yielding a white precipitate of 2,4,6-tribromophenol (Z).
Reaction: C₆H₅OH + 3Br₂ (in H₂O) → 2,4,6-tribromophenol↓ (White ppt) + 3HBr
(b) Enantiomer vs Diastereomer:
- Enantiomers: Stereoisomers that are non-superimposable mirror images of each other. They must occur in pairs. They have identical physical properties (boiling point, solubility) except for the direction they rotate plane-polarized light.
Example: (R)-lactic acid and (S)-lactic acid.
- Diastereomers: Stereoisomers that are NOT mirror images of each other. They occur when a molecule has multiple chiral centers and the configuration differs at one or more (but not all) centers. They have different physical properties.
Example: cis-2-butene and trans-2-butene, or D-Glucose and D-Galactose.
(c) Meso isomer vs Racemic mixture:
Both a meso isomer and a racemic mixture are optically inactive (net rotation = 0°), but for fundamentally different reasons:
- Meso isomer: A single, pure substance. It contains chiral centers, but the molecule as a whole is achiral due to an internal plane of symmetry. The optical inactivity is due to internal compensation (one half of the molecule cancels the other half).
- Racemic mixture (Racemate): A 50:50 mixture of two enantiomers. Each individual molecule is chiral and optically active, but because there are exactly equal amounts of the left-rotating and right-rotating enantiomers, the net rotation is zero. The optical inactivity is due to external compensation.
(a) Benzene MO Diagram & Aromaticity:
Benzene has 6 pi electrons. Using a Frost circle, the 6 p-orbitals combine to form 6 Molecular Orbitals: 3 bonding (lowest energy, filled with all 6 electrons) and 3 anti-bonding (higher energy, empty). Because all bonding MOs are completely filled and follow Hückel's 4n+2 rule (n=1), it is highly stable.
- (i) Furan: Has a 5-membered ring with 4 pi electrons from two double bonds and 2 from the oxygen lone pair (total 6 pi electrons). It is planar and fully conjugated, making it Aromatic.
- (ii) Cyclopentadienyl cation: Has a 5-membered ring with 4 pi electrons (from two double bonds). The positive charge means an empty p-orbital, completing conjugation. With 4 pi electrons (4n, n=1), it is Anti-aromatic.
(b) Synthesis of Paracetamol:
Paracetamol (p-hydroxyacetanilide) is synthesized from phenol:
1. Nitration: Phenol reacts with dilute HNO₃ to give a mixture of o- and p-nitrophenol. The para-isomer is separated.
2. Reduction: p-Nitrophenol is reduced to p-aminophenol using NaBH₄ or H₂/Pd.
3. Acetylation: p-Aminophenol reacts with acetic anhydride to selectively acetylate the amino group, forming Paracetamol.
(c) Mechanism of Nitration of Benzene:
The nitrating mixture is conc. HNO₃ + conc. H₂SO₄. H₂SO₄ acts as a strong acid to protonate HNO₃, facilitating the loss of water to generate the powerful electrophile, the nitronium ion (NO₂⁺).
Mechanism (Electrophilic Aromatic Substitution):
1. Generation of electrophile: HNO₃ + 2H₂SO₄ ⇌ NO₂⁺ + 2HSO₄⁻ + H₃O⁺
2. Attack on benzene: The pi-electrons of benzene attack NO₂⁺, destroying aromaticity and forming a resonance-stabilized arenium ion (sigma complex).
3. Loss of proton: The base (HSO₄⁻) removes the proton from the sp³ carbon, restoring the aromatic ring to yield nitrobenzene.
(a) Schrodinger equation for a particle in a 1D box:
Consider a particle of mass m in a 1D box of length L. Potential energy V(x) = 0 inside the box (0 < x < L) and V(x) = ∞ outside.
The time-independent Schrodinger equation inside the box is:
(-h² / 8π²m) (d²Ψ/dx²) = EΨ
d²Ψ/dx² + (8π²mE/h²)Ψ = 0
Let k² = 8π²mE/h². The solution is Ψ(x) = A sin(kx) + B cos(kx). Applying boundary conditions yields the quantized energy levels E_n = n²h² / (8mL²).
(b) Application to calculate energy spectra of polyene:
A conjugated polyene (like 1,3-butadiene) can be modeled using the "particle in a box" model (Free Electron Model). The delocalized π-electrons are treated as free particles moving in a 1D box whose length L is approximately the length of the carbon chain plus one bond length on each end. The energy levels E_n are calculated. The absorption wavelength corresponds to the transition of an electron from the HOMO (Highest Occupied MO) to the LUMO (Lowest Unoccupied MO). ΔE = E_LUMO - E_HOMO = hc/λ.
(c) Zero point energy:
The zero point energy is the lowest possible energy state (ground state, n=1). For a particle in a 1D box, E₁ = h² / (8mL²).
Why it cannot be zero: If energy were exactly zero, the momentum (p) would be exactly zero (since E = p²/2m). If p=0, then uncertainty in momentum Δp = 0. According to Heisenberg's Uncertainty Principle (Δx·Δp ≥ h/4π), if Δp=0, the uncertainty in position Δx must be infinite. But the particle is confined to a box of finite length L, so Δx cannot be infinite. Thus, the minimum energy must be greater than zero.
(a) Ionization Energy (IE) vs Electron Affinity (EA):
- IE: The energy required to remove an electron from an isolated gaseous atom to form a cation. It is always endothermic.
- EA: The energy released when an electron is added to an isolated gaseous atom to form an anion. It is usually exothermic.
IE of C vs B: Carbon has a higher 1st IE than Boron because of a higher effective nuclear charge holding the 2p electrons more tightly. However, for the 2nd IE, Boron requires removing an electron from the highly stable, fully-filled 2s subshell (2s²), which takes much more energy than removing a 2p electron from Carbon (2s² 2p¹). Thus, 2nd IE of B > C.
(b) Mn(II) electron configuration and Zn classification:
Manganese (Z=25) is [Ar] 4s² 3d⁵. When forming Mn(II), it loses the outermost 4s electrons first, resulting in [Ar] 3d⁵. This half-filled d-subshell is highly stable.
Zn as a transition element: Zinc (Z=30) has the configuration [Ar] 4s² 3d¹⁰. A transition element must have incompletely filled d-orbitals in its ground state or any of its common oxidation states. Zn and Zn²⁺ both have completely filled d¹⁰ subshells. Therefore, Zinc is NOT strictly classified as a transition element.
(c) IE of Neon and complex formation by transition elements:
- Neon's high IE: Neon has a highly stable, completely filled octet (1s² 2s² 2p⁶). Furthermore, it has the highest effective nuclear charge in the 2nd period, making its electrons incredibly tightly bound, resulting in the highest IE.
- Why transition elements form complexes readily: 1. They have small sizes and high nuclear charges (high charge density). 2. They have vacant d-orbitals of appropriate energy to accept lone pairs of electrons donated by ligands.
(a) Inelastic molecular collisions:
Kinetic molecular theory assumes perfectly elastic collisions (no loss of kinetic energy). If collisions were inelastic, the molecules would lose kinetic energy with every collision. They would eventually slow down and settle at the bottom of the container due to gravity, and the gas pressure would drop to zero. The gas phase would collapse.
(b) Why CO₂ doesn't form the lower layer of the atmosphere:
Although CO₂ (44 g/mol) is heavier than O₂ (32) and N₂ (28), it does not settle at the bottom because of continuous mixing caused by thermal convection currents and wind in the troposphere. Also, gases diffuse into each other (Graham's Law) regardless of their density to maximize entropy, keeping the atmosphere homogeneously mixed.
(c) Compressibility factor (Z) of Hydrogen and Helium:
For H₂ and He, the intermolecular attractive forces ('a' in van der Waals eq) are exceptionally weak due to their very small size and low polarizability. Therefore, the repulsive forces (excluded volume 'b') dominate at all pressures. The van der Waals equation becomes P(V-nb) = nRT, which yields Z = PV/nRT = 1 + Pb/RT. Thus, Z is always greater than 1 and increases linearly with increasing pressure.
(d) Liquefying gases by cooling:
Cooling a gas decreases the kinetic energy of its molecules. When the kinetic energy drops sufficiently low, the molecules move slowly enough that the inherent intermolecular attractive forces (van der Waals forces) can overcome the thermal motion. The molecules clump together to form a liquid.
(e) Boiling points of Polar vs Non-polar liquids:
Polar liquids generally have higher boiling points.
Reason: Non-polar liquids are held together only by weak London dispersion forces. Polar liquids are held together by stronger dipole-dipole interactions (and possibly hydrogen bonding). Stronger intermolecular forces require more thermal energy to break apart the molecules into the gas phase, resulting in a higher boiling point.