Chemistry-1 Previous Year Questions

(I) Which type of isomerism is observed in CH₃CH₂OH and CH₃OCH₃?

Answer: Functional Isomerism
Explanation: Both have the same molecular formula (C₂H₆O) but possess different functional groups. CH₃CH₂OH is an alcohol (ethanol), while CH₃OCH₃ is an ether (dimethyl ether).

(II) A nucleophile must possess _______.

Answer: an unshared electron pair (lone pair) or a negative charge.
Explanation: Nucleophiles are "nucleus-loving" species that donate an electron pair to an electrophile to form a chemical bond.

(III) Which type of semiconductor is formed when Germanium is doped with Aluminium?

Answer: p-type semiconductor
Explanation: Germanium is a Group 14 element. Aluminium is a Group 13 (trivalent) element. Doping Ge with Al creates a deficiency of one electron, creating a "hole" which acts as a positive charge carrier.

(IV) In UV spectroscopy, shift of λmax towards shorter wavelength is called _______.

Answer: Hypsochromic shift (or Blue shift)

(V) Write the expression of critical pressure.

Answer: P_c = a / (27b²)
Explanation: This relates critical pressure to the van der Waals constants 'a' and 'b'.

(VI) Give example of a reference electrode.

Answer: Standard Hydrogen Electrode (SHE) or Saturated Calomel Electrode (SCE).

(VII) Write two types of luminescence.

Answer: Fluorescence and Phosphorescence.

(VIII) When do real gases behave as ideal gases?

Answer: At high temperature and low pressure.
Explanation: Under these conditions, the kinetic energy of molecules is high (overcoming intermolecular forces) and the volume occupied by the gas molecules is negligible compared to the total volume.

(IX) Write 3 ions which cause alkalinity of water.

Answer: Hydroxide (OH⁻), Carbonate (CO₃²⁻), and Bicarbonate (HCO₃⁻).

(X) Arrange NaF, NaCl, NaBr, NaI in order of increasing melting point.

Answer: NaI < NaBr < NaCl < NaF
Explanation: Melting point generally correlates with lattice energy for ionic compounds. Lattice energy is inversely proportional to the sum of ionic radii. Since F⁻ is the smallest halide, NaF has the highest lattice energy and highest melting point.

(XI) Write the criteria for a compound to be aromatic.

Answer: It must be (1) cyclic, (2) planar, (3) completely conjugated (a continuous ring of overlapping p-orbitals), and (4) follow Hückel's Rule by containing exactly (4n + 2) π electrons, where n is an integer.

(XII) How many NMR signal is obtained for isopropanol [CH₃CH(OH)CH₃]?

Answer: 3 signals
Explanation: There are three sets of chemically non-equivalent protons: the 6 protons of the two equivalent -CH₃ groups, the 1 proton of the -CH- group, and the 1 proton of the -OH group.

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.

Chromophore:
A chromophore is a covalently bonded functional group with extended π-electron systems (conjugation) that absorbs electromagnetic radiation in the UV or visible region, imparting color to the molecule.
Example: Nitro group (-NO₂), Azo group (-N=N-), Carbonyl group (C=O), Conjugated dienes.

Auxochrome:
An auxochrome is a functional group containing non-bonding (lone pair) electrons that does not absorb UV-Vis radiation by itself, but when attached to a chromophore, it alters both the wavelength (shifts to longer wavelength - bathochromic shift) and the intensity of the absorption (hyperchromic effect). It is a "color enhancer."
Example: Hydroxyl group (-OH), Amino group (-NH₂), Halogens (-X).

Range of UV Spectra:
The standard Ultraviolet (UV) spectroscopic region used in general laboratory analysis is from 200 nm to 400 nm. (The vacuum UV region lies below 200 nm, typically 10-200 nm, but requires specialized equipment as air absorbs in this range).

Magnetic Resonance Imaging (MRI):
MRI is a non-invasive medical imaging technique that utilizes strong magnetic fields, magnetic field gradients, and radio waves to generate detailed images of the organs in the body. It is fundamentally based on the principles of Nuclear Magnetic Resonance (NMR) spectroscopy, specifically exploiting the NMR signal of hydrogen nuclei (protons) present abundantly in the water and fat of the human body.

Uses:
- Diagnostic imaging of the brain and spinal cord (e.g., detecting tumors, strokes, or multiple sclerosis).
- Evaluating joint abnormalities, torn ligaments, and cartilage damage.
- Examining soft tissue structures like the heart, liver, and abdominal organs for diseases without using ionizing radiation (like X-rays).

Observation 1: Boiling point of n-pentane > neo-pentane
Both are structural isomers with the formula C₅H₁₂. Boiling points of non-polar alkanes depend entirely on the strength of London dispersion forces, which are proportional to the surface area of the molecule.
- n-Pentane is a straight-chain alkane. It has a large, elongated surface area, allowing extensive intermolecular contact and stronger dispersion forces.
- neo-Pentane (2,2-dimethylpropane) is highly branched and nearly spherical. Its compact shape severely reduces its surface area, resulting in weaker intermolecular contact and much weaker dispersion forces, leading to a significantly lower boiling point.

Observation 2: H₂O is liquid while H₂S is gas at room temperature
The boiling point of a substance depends on the strength of its intermolecular forces.
- H₂O: Oxygen is highly electronegative and very small. This results in the formation of strong, extensive intermolecular Hydrogen Bonds between water molecules. Breaking these strong H-bonds requires significant thermal energy, making water a liquid at room temperature (boiling point 100°C).
- H₂S: Sulfur is much larger and less electronegative than oxygen. The S-H bond is less polar, and sulfur's large size prevents it from forming effective hydrogen bonds. Therefore, H₂S molecules are held together only by weak dipole-dipole interactions and London dispersion forces, causing it to be a gas at room temperature.

The definition of Gibbs free energy is:
G = H - TS --- (1)

Differentiating equation (1) completely, we get:
dG = dH - TdS - SdT --- (2)

We know that Enthalpy H = U + PV. Differentiating this gives:
dH = dU + PdV + VdP

From the First Law of Thermodynamics, dU = dq - PdV. If the process is reversible, dq = TdS. Thus:
dU = TdS - PdV

Substitute dU into the dH equation:
dH = (TdS - PdV) + PdV + VdP
dH = TdS + VdP

Now substitute this expression for dH back into equation (2):
dG = (TdS + VdP) - TdS - SdT
dG = VdP - SdT --- (3)

At constant pressure, dP = 0. Therefore, equation (3) becomes:
dG = -SdT
or, (∂G/∂T)_P = -S --- (4)

Now, substitute S from equation (4) back into the original definition G = H - TS:
G = H - T[-(∂G/∂T)_P]
G = H + T(∂G/∂T)_P

Rearranging this yields the Gibbs-Helmholtz Equation:
ΔG = ΔH + T[∂(ΔG)/∂T]_P

(Note: This is often expressed in terms of finite changes ΔG and ΔH between two states).

(a) Splitting of d-orbitals in a tetrahedral field:
In a tetrahedral complex, four ligands approach the central metal ion along the diagonals of a cube (between the axes), rather than along the Cartesian axes (x, y, z) as in an octahedral field.
- The t₂ orbitals (dxy, dyz, dzx) lie between the axes and point directly toward the approaching ligands. Therefore, they experience greater electrostatic repulsion and are raised in energy.
- The e orbitals (dx²-y², dz²) point along the axes, exactly bisecting the ligand approach paths. They experience less repulsion and are lowered in energy.
This results in a splitting pattern exactly opposite to that of an octahedral field. The lower energy level consists of two 'e' orbitals, and the higher energy level consists of three 't₂' orbitals. The energy gap is denoted by Δt.

(b) Low spin complexes are not obtained in tetrahedral crystal field - Give reason:
Because there are only 4 ligands (compared to 6 in octahedral) and none of them point directly at the d-orbitals, the crystal field splitting energy (Δt) in a tetrahedral complex is significantly smaller than in an octahedral complex (specifically, Δt ≈ 4/9 Δo).
Because Δt is always very small, it is virtually always smaller than the pairing energy (P). Thus, it is always energetically favorable for electrons to jump to the higher t₂ set rather than pair up in the lower e set. As a result, almost all tetrahedral complexes are high-spin.

(c) On the basis of band theory differentiate between conductors, semiconductors and insulators:
Band theory explains electrical conductivity based on the energy gap (Band Gap, Eg) between the Valence Band (highest filled band) and the Conduction Band (lowest empty band).
- Conductors (Metals): The valence band and conduction band overlap, meaning there is zero band gap (Eg = 0). Electrons can freely move into the conduction band at any temperature, providing excellent conductivity.
- Insulators: There is a very large band gap (Eg > 3 eV, typically 5-10 eV) between the valence and conduction bands. Thermal energy at room temperature is entirely insufficient to promote electrons across this gap, resulting in zero conductivity.
- Semiconductors: They have a small, finite band gap (Eg ≈ 1 eV, e.g., Si is 1.1 eV). At absolute zero (0 K), they act as insulators. But at room temperature, some thermal energy is sufficient to promote a small fraction of electrons into the conduction band, allowing moderate, tunable conductivity.

(d) What are anti-aromatic compounds? Give examples:
Anti-aromatic compounds are cyclic, planar, completely conjugated systems, but instead of following Hückel's rule (4n+2 π electrons), they possess exactly 4n π electrons (where n = 1, 2, 3...).
Because of the 4n π electron count, they suffer from extreme thermodynamic instability (paramagnetic ring current) and are highly reactive, often distorting their geometry to avoid being planar if possible.
Examples:
- Cyclobutadiene (4 π electrons, n=1)
- Cyclooctatetraene (if it were planar, it has 8 π electrons, n=2. However, it adopts a tub shape to avoid anti-aromaticity).

(a) Lambert-Beer's Law:
It states that the absorbance (A) of a solution is directly proportional to the concentration (c) of the absorbing species and the path length (l) of the light through the solution.
A = εcl (where ε is the molar absorptivity).
Proof of linearity to concentration: The rate of decrease of intensity of light (-dI) with thickness (dx) is proportional to the intensity (I) and concentration (c).
-dI/dx = k'Ic => -dI/I = k'c dx. Integrating gives -ln(I/I₀) = k'cl.
Absorbance A = log(I₀/I) = (k'/2.303)cl. Thus, A is linearly proportional to c.

(b) IR Inactive Molecules:
Molecules that do not experience a change in their net dipole moment during a vibrational mode are IR inactive. This includes perfectly symmetrical, homonuclear diatomic molecules.
Examples: N₂, O₂, H₂, Cl₂.

(c) Bathochromic and Hypsochromic shift:
- Bathochromic Shift (Red Shift): A shift of the absorption maximum (λmax) to a longer wavelength. This occurs due to the presence of an auxochrome or an increase in conjugation.
- Hypsochromic Shift (Blue Shift): A shift of the absorption maximum (λmax) to a shorter wavelength. This can occur due to the removal of conjugation or a change in solvent polarity.

(d) Shift observed if conjugation is increased:
Bathochromic (Red) Shift is observed.
Reason: Increasing conjugation delocalizes the π electrons over a larger area, which decreases the energy gap (ΔE) between the HOMO (Highest Occupied Molecular Orbital) and LUMO (Lowest Unoccupied Molecular Orbital). Since E = hc/λ, a smaller energy gap results in absorption at a longer wavelength.

(a) Fluorescence process and its uses:
Fluorescence occurs when a molecule absorbs a photon, exciting an electron from the singlet ground state (S₀) to a higher singlet excited state (S₁ or S₂). The electron rapidly relaxes to the lowest vibrational level of S₁ via non-radiative internal conversion. It then drops back to the ground state (S₀), emitting a photon of lower energy (longer wavelength) than the absorbed photon. This emission happens almost instantaneously (10⁻⁸ seconds).
Uses: Fluorescent lighting, biological imaging/tagging, forensics, and analytical sensors.

(b) Electronic transitions in UV for formaldehyde (H₂C=O):
Formaldehyde contains a π bond and lone pairs (n electrons) on the oxygen. Therefore, it exhibits two primary transitions:
1. π → π* transition: High energy (shorter wavelength, ~180 nm). It is a quantum mechanically allowed transition, hence it has a high intensity (large molar absorptivity, ε).
2. n → π* transition: Lower energy (longer wavelength, ~290 nm). It is a quantum mechanically forbidden transition, hence it has a low intensity (small ε).

(c) Which atoms are NMR inactive and why?
Atoms whose nuclei possess an even number of protons and an even number of neutrons have a net nuclear spin (I) of zero. Because they have no magnetic moment, they cannot interact with an external magnetic field and are NMR inactive.
Examples: ¹²C, ¹⁶O, ³²S.

(d) Chemical Shift of Proton:
Chemical shift (δ) is the resonant frequency of a nucleus relative to a standard reference (usually TMS, Tetramethylsilane), normalized by the operating frequency of the spectrometer. It is expressed in parts per million (ppm). It arises because the local electron cloud shields the nucleus from the external magnetic field to varying degrees depending on the chemical environment.

(a) Van der Waal equation:
(P + an²/V²)(V - nb) = nRT
where P = pressure, V = volume, T = temperature, n = moles, R = gas constant. 'a' corrects for intermolecular forces, 'b' corrects for molecular volume.
- At high pressure: The volume V is small, so the volume correction 'nb' cannot be ignored. However, P is very large compared to an²/V², so the pressure correction can be ignored. Equation becomes: P(V - nb) = nRT => PV = nRT + Pnb.
- At low pressure: V is very large. 'nb' is negligible compared to V. The pressure P is small, so an²/V² is significant. Equation becomes: (P + an²/V²)V = nRT => PV = nRT - an/V.

(b) Boyle Temperature:
Boyle temperature (T_B) is the temperature at which a real gas obeys ideal gas laws over an appreciable range of pressure.
Relation: T_B = a / (Rb)

(c) Compressibility factor (Z):
It is the ratio of the molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure. Z = PV / nRT.
Value for ideal gas: For an ideal gas, PV = nRT exactly, so Z = 1 under all conditions.

(d) Can we liquefy a gas by increasing pressure alone?
No, not if the gas is above its Critical Temperature (T_c).
Reason: Above the critical temperature, the kinetic energy of the gas molecules is so high that intermolecular attractive forces cannot pull them together into a liquid state, regardless of how much pressure is applied to force them closer.

(e) Van der Waal forces:
Weak, short-range electrostatic attractive forces between uncharged molecules, arising from the interaction of permanent or transient electric dipole moments. They include London dispersion forces, dipole-dipole forces, and dipole-induced dipole forces.

(a) Electronic transitions between molecular orbitals:
The possible transitions in increasing order of energy (from smallest energy gap to largest) are:
1. n → π* : Lowest energy gap, highest wavelength (occurs in carbonyls).
2. n → σ* : Intermediate energy gap (occurs in alcohols, amines, halides).
3. π → π* : Intermediate to high energy gap (occurs in alkenes, alkynes, aromatics).
4. σ → σ* : Highest energy gap, shortest wavelength (occurs in saturated alkanes).
Energy order: n→π* < n→σ* < π→π* < σ→σ*.

(b) Differentiate 1,3-pentadiene and 1,4-pentadiene by UV spectroscopy:
- 1,3-pentadiene (CH₂=CH-CH=CH-CH₃): This is a conjugated diene. The conjugation lowers the HOMO-LUMO energy gap, shifting the π → π* absorption to a longer wavelength (λmax ~ 220 nm) with a high intensity.
- 1,4-pentadiene (CH₂=CH-CH₂-CH=CH₂): This is an isolated diene. Since there is an sp³ carbon breaking the conjugation, the two double bonds act independently. It has a larger energy gap and absorbs at a much shorter wavelength (λmax ~ 175 nm) in the far-UV region. Thus, they are easily distinguishable.

(c) Force constant (k) for harmonic oscillator:
Formula for vibrational frequency (in wavenumbers, v̄):
v̄ = (1 / 2πc) * √(k / μ)
Squaring both sides:
v̄² = (1 / 4π²c²) * (k / μ)
Rearranging for k:
k = 4π²c²v̄²μ
Given: v̄ = 2880 cm⁻¹ = 288000 m⁻¹; μ = 1.63 × 10⁻²⁴ g = 1.63 × 10⁻²⁷ kg; c = 3 × 10⁸ m/s.
k = 4 × (3.14159)² × (3 × 10⁸)² × (288000)² × (1.63 × 10⁻²⁷)
k = 39.478 × (9 × 10¹⁶) × (8.294 × 10¹⁰) × (1.63 × 10⁻²⁷)
k = 39.478 × 9 × 8.294 × 1.63 × 10⁽¹⁶ ⁺ ¹⁰ ⁻ ²⁷⁾
k ≈ 4802.7 × 10⁻¹ = 480.27 N/m (or J/m²).