Classification of Molecular Rotors in Rotational Spectroscopy
Important for: CSIR-NET, GATE, SET, and other competitive exams.
The rotation of a three dimensional body may be quite complex and it is convenient to resolve it into rotational components about three mutually perpendicular direction through the centre of gravity - the principal axes of rotation. Thus a body has three principal moments of inertia, one about each axis, usually Ia, Ib, and Ic.
Molecules are classified as rigid rotors based on their three principal moments of inertia (Ia, Ib, Ic), conventionally ordered as Ia ≤ Ib ≤ Ic. Rotational constants are defined as A = h/(8π²c Ia), B = h/(8π²c Ib), C = h/(8π²c Ic) in cm⁻¹ (A ≥ B ≥ C).
Summary Table
| Type | Moments of Inertia | Rotational Constants | Energy Levels (Rigid Rotor) | Examples | Microwave Spectrum |
|---|---|---|---|---|---|
| Linear Molecules | Ia ≈ 0, Ib = Ic | A >> B = C | EJ = B J(J+1) (hc cm⁻¹) | CO2, HC≡CH, OCS, HCl, CO | Yes (if μ ≠ 0) |
| Symmetric Tops | Two equal | Two equal | EJ,K = B J(J+1) + (A - B) K² | See sub-types | Yes (if μ along symmetry axis) |
| • Prolate | Ia < Ib = Ic | A > B = C | E = B J(J+1) + (A - B) K² | CH3Cl, CH3CN, CH3F | |
| • Oblate | Ia = Ib < Ic | A = B > C | E = B J(J+1) + (C - B) K² | BF3, NH3, C6H6 | |
| Spherical Tops | Ia = Ib = Ic | A = B = C | EJ = B J(J+1) | CH4, SF6, CCl4 | No (μ = 0; weak due to distortion) |
| Asymmetric Tops | Ia ≠ Ib ≠ Ic | A > B > C | No closed form; numerical (Wang basis) | H2O, H2CO, NO2, CH3OH | Yes (complex spectrum) |
1. Linear Molecules
- Atoms arranged linearly; rotation about bond axis has negligible moment of inertia (Ia ≈ 0).
- Treated as rigid rotor with single rotational constant B.
- Energy levels: F(J) = BJ(J+1) (in cm⁻¹)
- Degeneracy: (2J+1)
- Selection rule: ΔJ = ±1 → lines at 2B(J+1)
- Non-polar (e.g., CO2, N2) show no pure rotational spectrum.
2. Symmetric Tops
- Two moments equal; possess C3 or higher symmetry axis.
- Quantum numbers: J (total angular momentum), K (projection along symmetry axis, |K| ≤ J).
- Energy levels: F(J,K) = BJ(J+1) + (A-B)K²
- Degeneracy: (2J+1) for K=0; 2(2J+1) for K≠0
- Selection rules: ΔJ = ±1, ΔK = 0
- Spectrum similar to linear molecules (spacing 2B).
Prolate Symmetric Tops
- Cigar-shaped; symmetry axis has smaller inertia.
- Ia < Ib = Ic → A > B = C
- Examples: CH3F, CH3Cl, CH3C≡CH
Oblate Symmetric Tops
- Disc-shaped; symmetry axis has larger inertia.
- Ia = Ib < Ic → A = B > C
- Examples: NH3, BF3, benzene (C6H6)
3. Spherical Tops
- All moments equal; high symmetry (tetrahedral/octahedral).
- Energy: F(J) = BJ(J+1)
- Degeneracy: (2J+1)2
- No permanent dipole → no pure rotational microwave spectrum (weak lines due to centrifugal distortion).
- Examples: CH4, SF6, SiH4, OsO4, CCl4
4. Asymmetric Tops
- All three moments different; most molecules belong here.
- No analytical energy expression; use basis of symmetric top functions and diagonalize Hamiltonian.
- Ray's asymmetry parameter κ = (2B - A - C)/(A - C) (-1 for prolate limit, +1 for oblate limit).
- Complex spectra; near-symmetric cases approximate symmetric top behavior.
- Examples: H2O (near prolate), H2CO, SO2, CH3OH, CH3=CH-Cl
Practice Exam Questions (MCQs)
Suitable for CSIR-NET, GATE, SET, JAM, etc.
Explanation: Spherical tops have Ia = Ib = Ic (e.g., tetrahedral/octahedral symmetry). CH4 is tetrahedral.
Explanation: In linear molecules, Ia ≈ 0 → A → ∞, making it an extreme prolate case (A >> B = C).
Explanation: CO2 is linear but non-polar (μ = 0). Pure rotational transitions require a permanent dipole moment.
Explanation: Planar hexagonal structure → I⊥ (out-of-plane) smaller than in-plane → oblate (A = B > C).
Explanation: κ = (2B - A - C)/(A - C); κ → -1 (prolate limit), κ → +1 (oblate limit).
Explanation: Standard expression for prolate (A > B = C).
Explanation: High symmetry (e.g., Td, Oh) results in μ = 0.
Explanation: Bent structure → all three moments of inertia different.
Explanation: ΔK = 0 → transitions depend only on J → spacing 2B, independent of K.
Explanation: C3v symmetry, elongated along C–Cl axis → Ia < Ib = Ic.