Complete Detailed Solutions · 2022-23 Semester 1 (BSCH101)
(I) Which spectroscopic technique is most useful to identify the presence of known impurity in a drug substance?
Answer: Infrared (IR) Spectroscopy or High-Performance Liquid Chromatography (HPLC) coupled with UV detection.
Explanation: IR spectroscopy (specifically the fingerprint region) is excellent for identifying specific functional groups and known impurities by matching spectra. However, in modern pharmaceutical analysis, HPLC is the gold standard.
(II) Van der Waal's forces are directly proportional to which factor?
Answer: Surface area and polarizability of the molecule.
Explanation: As the size (molecular weight) and surface area of a molecule increase, its electron cloud becomes more polarizable, leading to stronger London dispersion forces (the primary van der Waals force).
(III) Give one example of a state function.
Answer: Enthalpy (H), Internal Energy (U), Entropy (S), or Gibbs Free Energy (G).
(IV) Give an example of soft acid.
Answer: Ag⁺, Au⁺, Hg²⁺, Cd²⁺
Explanation: According to HSAB theory, soft acids are large, highly polarizable metal ions with low positive charge.
(V) Give one example of ionization isomerism.
Answer: [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br
Explanation: They yield different ions in solution (sulfate vs. bromide).
(VI) What is the reactivity order of alkyl halide in SN2 mechanism?
Answer: Methyl > 1° > 2° > 3°
Explanation: The S_N2 mechanism involves a backside attack. Less steric hindrance (smaller groups) leads to a faster reaction rate.
(VII) What is the fingerprint region range in IR spectra?
Answer: 1500 cm⁻¹ to 400 cm⁻¹
Explanation: This region contains a complex pattern of bending vibrations that are unique to a specific molecule.
(VIII) Write down the formula of critical volume for Van der Waal's gas.
Answer: V_c = 3b
(IX) State one application of the first law of thermodynamics.
Answer: Hess's Law of Constant Heat Summation.
Explanation: The first law (conservation of energy) allows us to calculate the enthalpy change of a reaction from the enthalpies of formation, regardless of the path taken.
(X) What is the shielding constant value for 1s orbital electron?
Answer: 0.30
Explanation: According to Slater's rules, an electron in the 1s orbital shields another electron in the same 1s orbital by 0.30 (not the usual 0.35 for other groups).
(XI) What will be the absorbance if % of Transmittance = 80?
Answer: 0.097
Explanation: A = 2 - log(%T) = 2 - log(80) = 2 - 1.903 = 0.097
(XII) Write down the relation of critical temperature for Van der Waal's gas.
Answer: T_c = 8a / (27Rb)
Germanium as n-type:
Pure germanium (Ge) is a Group 14 element and an intrinsic semiconductor. It does not naturally act as an n-type semiconductor. It becomes an n-type semiconductor only when it is doped with a Group 15 element (such as Phosphorus, Arsenic, or Antimony). Group 15 elements have 5 valence electrons. Four form covalent bonds with the Ge lattice, and the fifth extra electron is loosely bound and free to conduct electricity, making it n-type (negative type).
Difference between n-type and p-type:
| Feature | n-type Semiconductor | p-type Semiconductor |
|---|---|---|
| Dopant | Group 15 elements (P, As, Sb) - Pentavalent | Group 13 elements (B, Al, Ga) - Trivalent |
| Majority Charge Carriers | Electrons | Holes (positive vacancies) |
| Minority Charge Carriers | Holes | Electrons |
| Energy Level | Donor energy level is just below the conduction band. | Acceptor energy level is just above the valence band. |
Factors affecting the intensity of a spectral line:
1. Transition Probability: Governed by quantum mechanical selection rules. Allowed transitions produce intense lines, while forbidden transitions produce very weak lines.
2. Population of Energy Levels: According to the Boltzmann distribution, the intensity is proportional to the number of atoms/molecules present in the initial state from which the transition originates.
3. Concentration: Higher concentration of the absorbing/emitting species leads to higher intensity.
4. Path Length: The thickness of the sample the light passes through.
Lambert - Beer Law:
The Lambert-Beer Law states that when a beam of monochromatic light passes through an absorbing medium, the absorbance (A) is directly proportional to the concentration of the absorbing species (c) and the path length (l) of the medium.
Formula: A = ε · c · l
Where:
- A = Absorbance (log(I₀/I))
- ε = Molar absorptivity (molar extinction coefficient), a constant for a specific substance at a specific wavelength.
- c = Concentration of the solution (mol/L)
- l = Path length of the cuvette (usually 1 cm)
The total energy E of a quantum mechanical system is the sum of its Kinetic Energy (K) and Potential Energy (V):
E = K + V
In classical mechanics, Kinetic Energy (K) is given by p² / 2m, where p is momentum. Thus:
E = p²/2m + V
In quantum mechanics, the momentum operator is: p = -i(h/2π)∇
Squaring it gives: p² = - (h²/4π²)∇²
Substituting into the kinetic energy term:
K = p²/2m = (-h²/4π²)∇² / 2m = -(h² / 8π²m)∇²
The Hamiltonian operator H (total energy) is: H = -(h² / 8π²m)∇² + V
The Schrödinger equation is HΨ = EΨ. Substitute H into it:
(-(h² / 8π²m)∇² + V)Ψ = EΨ
Rearranging the terms on the left side gives the proof:
(V - (h² / 8π²m)∇²)Ψ = EΨ
Molecules showing IR spectra:
Only molecules that experience a change in their net dipole moment during a vibrational mode will absorb IR radiation and show an IR spectrum.
- IR Active: Heteronuclear diatomic molecules (e.g., HCl, CO) and polyatomic molecules (e.g., H₂O, CO₂) exhibit dipole moment changes during stretching/bending.
- IR Inactive: Homonuclear diatomic molecules (e.g., O₂, N₂, H₂) have zero dipole moment and no change occurs during vibration, so they do not absorb IR light.
Justification of "Molecular Fingerprint":
An IR spectrum is generally divided into two regions:
1. Functional Group Region (4000 - 1500 cm⁻¹): Identifies specific functional groups (e.g., -OH, C=O).
2. Fingerprint Region (1500 - 400 cm⁻¹): This region contains complex bending vibrations of the entire molecular skeleton. The pattern of peaks here is extraordinarily intricate and unique to every individual molecule, much like a human fingerprint. Even if two different molecules have the exact same functional groups (e.g., two structural isomers), their skeletal vibrations in the fingerprint region will be noticeably different. Thus, comparing the fingerprint region of an unknown sample to a reference spectrum provides definitive identification.
Ground State Energy (n=1):
Formula: E_n = n²h² / (8ma²)
Given: h = 6.627 × 10⁻³⁴ Js, m = 9.11 × 10⁻³¹ kg, a = 10⁻¹⁰ m.
E₁ = (1)²(6.627 × 10⁻³⁴)² / [8 × 9.11 × 10⁻³¹ × (10⁻¹⁰)²]
E₁ = (43.917 × 10⁻⁶⁸) / (72.88 × 10⁻⁵¹)
E₁ ≈ 6.026 × 10⁻¹⁸ J
Separation between levels n=2 and n=3:
ΔE = E₃ - E₂ = (3² - 2²)h² / (8ma²) = (9 - 4)E₁ = 5E₁
ΔE = 5 × 6.026 × 10⁻¹⁸ J = 30.13 × 10⁻¹⁸ J
(a) Why UV-Visible absorption bands are broad:
Electronic transitions in molecules (excited by UV-Vis light) do not occur in isolation. An electronic state consists of numerous vibrational energy levels, and each vibrational level contains numerous rotational energy levels. When a molecule absorbs UV-Vis light, transitions occur between various closely spaced rotational-vibrational levels of the ground electronic state to various rotational-vibrational levels of the excited state. Because these sublevels are very close in energy, the individual transition lines overlap and merge, producing a broad continuous absorption band rather than a sharp line.
(b) Why 1,3-butadiene possesses higher λmax than ethane:
- Ethane (CH₃-CH₃): Contains only σ bonds. The only possible transition is the high-energy σ → σ* transition, which occurs in the vacuum UV region (λmax < 150 nm).
- 1,3-Butadiene (CH₂=CH-CH=CH₂): Contains a conjugated system of alternating double and single bonds. Conjugation causes the π molecular orbitals to delocalize, which significantly lowers the energy gap (ΔE) between the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO). Since Energy and wavelength are inversely related (E = hc/λ), a smaller energy gap results in a π → π* transition at a longer wavelength (λmax ~ 217 nm). Increased conjugation shifts absorption to higher wavelengths (bathochromic shift).
(c) Why the colour of the sky is blue:
The sky is blue due to Rayleigh Scattering. As sunlight passes through the Earth's atmosphere, gases and particles scatter the light in all directions. According to Rayleigh's law, the intensity of scattered light is inversely proportional to the fourth power of the wavelength (I ∝ 1/λ⁴). Blue light has a shorter wavelength than red light, so it is scattered much more strongly. Our eyes are also more sensitive to blue light, so we perceive the scattered light in the sky as blue.
(d) Predict the proton NMR spectra of CH₄:
Methane (CH₄) consists of a central carbon atom bonded to four entirely equivalent hydrogen atoms. Because all four protons are in the exact same chemical and magnetic environment (chemically equivalent), they will all resonate at the exact same frequency. Thus, the ¹H NMR spectrum of CH₄ will show a single sharp peak (a singlet). Due to the lack of electronegative atoms, it will appear highly shielded, typically around δ = 0.23 ppm.
(e) Name any four surface characterization techniques:
1. Scanning Electron Microscopy (SEM)
2. Transmission Electron Microscopy (TEM)
3. X-ray Photoelectron Spectroscopy (XPS)
4. Atomic Force Microscopy (AFM)
(a) Types and conditions of hydrogen bonding:
Hydrogen bonding is a strong dipole-dipole interaction between a hydrogen atom covalently bonded to a highly electronegative atom (N, O, F) and a lone pair on another electronegative atom.
Types: 1. Intermolecular: Occurs between two different molecules (e.g., between H₂O molecules), causing high boiling points. 2. Intramolecular: Occurs within the same molecule (e.g., o-nitrophenol).
Conditions: High electronegativity of the atom bonded to H, small size of the electronegative atom, and a lone pair of electrons.
(b) Equation of state for real gas and significance of a and b:
Van der Waals equation: (P + an²/V²)(V - nb) = nRT
- 'a' corrects for intermolecular attractive forces. Significance: Higher 'a' means stronger intermolecular forces and easier liquefaction.
- 'b' corrects for the finite volume occupied by the gas molecules (excluded volume). Significance: Indicates the effective size of the molecules.
(c) Critical phenomenon of real gas:
The critical phenomenon is the continuous transition of a gas to a liquid at the critical temperature and critical pressure. At the critical point, the liquid and vapor phases have identical densities and become indistinguishable. Above the critical temperature (T_c), a gas cannot be liquefied no matter how much pressure is applied.
(d) Calculate Boyle Temperature:
Formula: T_B = a / (Rb)
Given: a = 7.18 L² atm/mol², b = 0.854 L/mol, R = 0.082 L atm/(K mol).
T_B = 7.18 / (0.082 × 0.854) = 7.18 / 0.070028 ≈ 102.5 K
(a) Nernst equation for Daniel cell:
Cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Nernst Eq: E_cell = E°_cell - (0.0591 / 2) log ([Zn²⁺] / [Cu²⁺]) at 298K.
Effect of increasing [Zn²⁺]: From the equation, increasing the concentration of Zn²⁺ increases the log term, which is subtracted from E°_cell. Therefore, E_cell will decrease.
(b) Hardness and Alkalinity of water:
Hardness is due to dissolved Ca²⁺ and Mg²⁺ salts, which react with soap to form insoluble scum instead of lather. Alkalinity is the water's capacity to neutralize acids, caused by bicarbonates (HCO₃⁻), carbonates (CO₃²⁻), and hydroxides (OH⁻).
(c) Corrosion and its types:
Corrosion is the gradual destruction of metals by chemical/electrochemical reaction with the environment. Types: Dry/chemical corrosion, Wet/electrochemical corrosion, Galvanic corrosion, Pitting corrosion, and Crevice corrosion.
(a) Shielding and deshielding in NMR:
- Shielding: Electron clouds around a nucleus circulate in a magnetic field, inducing an opposing local field. This reduces the effective field felt by the nucleus, shifting the NMR signal upfield (lower ppm).
- Deshielding: Electronegative atoms withdraw electron density, reducing the opposing local field. The nucleus feels a stronger effective field, shifting the signal downfield (higher ppm).
(b) Microwave Spectrum of ²⁷Al¹H:
Spacing between lines = 2B = 12.604 cm⁻¹. So, rotational constant B = 6.302 cm⁻¹ = 630.2 m⁻¹.
B = h / (8π²Ic) => I = h / (8π²Bc)
I = 6.626×10⁻³⁴ / (8 × π² × 630.2 × 3×10⁸) ≈ 4.44 × 10⁻⁴⁷ kg·m².
Reduced mass μ = (m₁m₂) / (m₁+m₂) = (26.981 × 1.008) / (27.989) amu ≈ 0.971 amu ≈ 1.61 × 10⁻²⁷ kg.
I = μr² => r = √(I/μ) = √(4.44×10⁻⁴⁷ / 1.61×10⁻²⁷) ≈ 1.66 × 10⁻¹⁰ m = 1.66 Å.
(c) ¹³C is NMR active while ¹²C is not:
NMR requires a nucleus with a non-zero spin (I ≠ 0). ¹²C has an even number of protons (6) and neutrons (6), so I=0. ¹³C has an odd number of neutrons (7), giving it a net spin of I=1/2, making it NMR active.
(d) Electronic transition in Cl₂ and Carbonyl group:
- Cl₂: Has non-bonding electrons and sigma bonds. Typical transition is n → σ* (weak intensity).
- Carbonyl (C=O): Has pi bonds and lone pairs. Transitions are π → π* (allowed, high intensity) and n → π* (forbidden, low intensity).
(a) Complexes of 1st vs 2nd/3rd transition series:
1st transition series (3d) metals often form high-spin complexes with weak field ligands because the crystal field splitting (Δo) is relatively small. In contrast, 4d and 5d orbitals are larger and interact more strongly with ligands, leading to a much larger Δo. This large splitting almost always overcomes pairing energy, resulting in low-spin complexes.
Ionization energies (5d > 3d/4d): 5d elements experience the lanthanide contraction due to poor shielding by 4f electrons. The increased effective nuclear charge tightly holds the valence electrons, making them harder to remove.
(b) Second IE of Cu and Cr:
Cu (3d¹⁰4s¹) and Cr (3d⁵4s¹) both achieve highly stable half-filled or fully-filled d-subshells after losing their first electron. The second electron must be removed from this stable d-core, which requires an enormous amount of energy.
Fe vs Na: Fe has an incompletely filled d-subshell (3d⁶), characterizing it as a transition metal. Na (3s¹) has a completely empty d-subshell and belongs to the s-block.
(c) First IE of Cu vs Alkali:
Cu has a higher first IE than alkali metals because its 3d electrons shield the nucleus poorly compared to the p-electrons in alkali metal cores, leading to a higher effective nuclear charge in Cu. However, its 2nd/3rd IEs are lower because the overall ionic charge density allows stabilization via hydration or lattice energies that compensate.
Lanthanides and Actinides: Placed at the bottom to maintain the structural integrity and periodicity of the main table, as they involve the filling of f-orbitals (f-block), which would make the table impractically wide.