GATE IR and Raman Spectroscopy — Selection Rules and Calculations
Vibrational spectroscopy gives GATE two very different kinds of question. One kind is purely logical — count the modes, decide whether a band is IR active, Raman active or both. The other is numerical — get a force constant from a wavenumber, or predict how a band shifts when hydrogen is replaced by deuterium. Both are learnable in an evening because the number of underlying ideas is small. This guide covers the selection rules, the mode-counting formulas, the harmonic-oscillator equation, and three worked calculations with the arithmetic written out in full.
Check the official GATE notification for your year for the exact syllabus wording and the current paper pattern; never rely on a website for that.
The two selection rules, side by side
Everything in this topic follows from one sentence per technique.
Raman active: the vibration must change the molecular polarisability.
Polarisability is how easily the electron cloud is distorted by an external field. A vibration that makes the molecule alternately bigger and smaller changes it strongly, which is why symmetric stretches are usually strong in Raman even when they are silent in the IR.
The two rules combine into the rule of mutual exclusion: in a molecule that possesses a centre of symmetry, no vibration can be both IR active and Raman active. This is a powerful structural tool. If a band appears in both spectra at the same wavenumber, the molecule cannot be centrosymmetric — that alone decides between a cis and a trans isomer, or between a linear symmetric and a bent structure.
Counting the vibrational modes
Linear molecule: 3N − 5 vibrational modes
(N = number of atoms)
A molecule of N atoms has 3N degrees of freedom in total. Three are used by translation. Rotation takes three for a non-linear molecule but only two for a linear one, because rotation about the molecular axis moves no atom. That missing rotation is exactly why a linear molecule gets one extra vibration.
Worked example 1 — modes and activity for CO₂ and H₂O
CO₂ (linear, N = 3): 3(3) − 5 = 4 vibrational modes — a symmetric stretch, an asymmetric stretch, and two bends that are degenerate (they are the same motion in two perpendicular planes, so they appear as one band).
CO₂ has a centre of symmetry, so mutual exclusion applies:
- Symmetric stretch — both oxygens move out together, the dipole stays zero, so it is IR inactive; the size of the electron cloud changes, so it is Raman active.
- Asymmetric stretch and the bends — the dipole changes, so they are IR active and therefore Raman inactive.
H₂O (bent, N = 3): 3(3) − 6 = 3 vibrational modes — symmetric stretch, asymmetric stretch and bend. Water has no centre of symmetry, so mutual exclusion does not apply and all three modes are active in both IR and Raman.
Extending it: benzene, C₆H₆, has N = 12 and is non-linear, so 3(12) − 6 = 30 vibrational modes. Benzene is centrosymmetric, so no band appears in both spectra — a standard one-line question.
The harmonic oscillator — where the numbers come from
- ν̄ — wavenumber in cm⁻¹ (this is what an IR spectrum's x-axis shows).
- c — speed of light, 2.998 × 10¹⁰ cm s⁻¹ when ν̄ is in cm⁻¹.
- k — force constant of the bond in N m⁻¹, a direct measure of bond stiffness.
- µ — reduced mass in kg. Convert from atomic mass units with 1 u = 1.66054 × 10⁻²⁷ kg.
Two consequences you should be able to state without calculating: a stiffer bond (larger k) absorbs at higher wavenumber, which is why C≡C > C=C > C−C; and a heavier atom (larger µ) absorbs at lower wavenumber, which is why C−H is near 3000 cm⁻¹ but C−Cl is near 700 cm⁻¹.
Worked example 2 — force constant of H−Cl from its IR band
The fundamental vibrational band of H³⁵Cl appears at 2886 cm⁻¹. Find the force constant. (Masses: H = 1.008 u, ³⁵Cl = 34.969 u.)
Step 1 — reduced mass in u.
µ = (1.008 × 34.969) ÷ (1.008 + 34.969)
numerator = 35.2488
denominator = 35.977
µ = 35.2488 ÷ 35.977 = 0.97976 u
Step 2 — convert to kg.
µ = 0.97976 × 1.66054 × 10⁻²⁷ = 1.6269 × 10⁻²⁷ kg
Step 3 — rearrange the formula. From ν̄ = (1/2πc)√(k/µ),
k = µ (2πcν̄)².
cν̄ = 2.998 × 10¹⁰ × 2886 = 8.6522 × 10¹³ s⁻¹
2πcν̄ = 6.28319 × 8.6522 × 10¹³ = 5.4363 × 10¹⁴ rad s⁻¹
(2πcν̄)² = (5.4363 × 10¹⁴)² = 2.9553 × 10²⁹ s⁻²
Step 4 — multiply.
k = 1.6269 × 10⁻²⁷ × 2.9553 × 10²⁹ = 481 N m⁻¹
Is that sensible? Single bonds between light atoms typically come out in the range of roughly 400–600 N m⁻¹, double bonds around 1000 N m⁻¹ and triple bonds around 1500–1900 N m⁻¹. A value of 481 N m⁻¹ for a single H−Cl bond fits, so the arithmetic has not gone astray by a power of ten — which is the usual failure mode in this calculation.
Worked example 3 — the deuterium isotope shift
Predict the fundamental wavenumber of D³⁵Cl, assuming the force constant is unchanged. (D = 2.014 u.)
Step 1 — reduced mass of DCl.
µ(DCl) = (2.014 × 34.969) ÷ (2.014 + 34.969) = 70.4276 ÷ 36.983 = 1.9043 u
Step 2 — use the ratio. Because k is the same, ν̄ ∝ 1/√µ:
ν̄(DCl) ÷ ν̄(HCl) = √(µ(HCl) ÷ µ(DCl)) = √(0.97976 ÷ 1.9043) = √0.51449 = 0.71728
Step 3 — multiply.
ν̄(DCl) = 2886 × 0.71728 = 2070 cm⁻¹
Honest note: the observed fundamental of DCl is near 2090 cm⁻¹, a little above this prediction. The difference is not an arithmetic error — it is anharmonicity. The harmonic-oscillator model assumes a perfect parabola for the potential energy, and a real bond deviates from it. GATE questions of this type expect the harmonic answer, but you should know why the real number differs.
The shortcut worth memorising: replacing H by D roughly divides a stretching wavenumber by √2 ≈ 1.41, because the reduced mass of an X−H bond roughly doubles. This is exactly how a chemist assigns an O−H or N−H stretch — run the spectrum again in D₂O and see which band moves.
Group frequencies to have at your fingertips
These approximate ranges are the working vocabulary of IR interpretation. Learn the order and the reasoning, not just the numbers.
| Bond / group | Approximate region (cm⁻¹) | What identifies it |
|---|---|---|
| O−H (alcohol, H-bonded) | 3200–3600 | Broad; sharpens on dilution as H-bonding breaks |
| O−H (carboxylic acid) | 2500–3300 | Very broad, overlapping the C−H region |
| N−H | 3300–3500 | Sharper than O−H; two bands for a primary amine |
| C−H (sp³ / sp² / sp) | ≈2850–2960 / ≈3020–3100 / ≈3300 | Rises with s-character of the carbon |
| C≡N | ≈2250 | Sharp, medium intensity |
| C≡C | ≈2100–2260 | Weak or absent in a symmetric alkyne (Raman instead) |
| C=O | ≈1650–1820 | Strong; position shifts with conjugation and ring strain |
| C=C | ≈1620–1680 | Weak; strong in Raman |
| Terminal M−CO (metal carbonyl) | ≈1850–2125 | Falls below free CO (2143) as back-bonding increases |
The metal-carbonyl row is a recurring GATE favourite. Free CO absorbs at about 2143 cm⁻¹. In a complex, the metal donates electron density into the π* orbital of CO, which weakens the C−O bond and lowers ν(CO). More electron density on the metal means more back-bonding and a lower wavenumber, giving the order [Mn(CO)₆]⁺ > Cr(CO)₆ > [V(CO)₆]⁻ in ν(CO) — around 2090, 2000 and 1860 cm⁻¹ respectively. Bridging carbonyls appear lower still, roughly 1700–1860 cm⁻¹, because two metals push electron density into the same π* orbital. You can therefore read the charge on a complex and the bonding mode of CO straight off an IR spectrum.
Raman-specific points
Raman scattering is inelastic scattering of monochromatic light. A photon that loses energy to the molecule gives a Stokes line; one that gains energy from an already-excited molecule gives an anti-Stokes line. Because the excited vibrational level is thermally populated according to the Boltzmann distribution, anti-Stokes lines are weaker at ordinary temperatures.
Quick estimate. For a mode at 500 cm⁻¹ at 300 K, the population ratio
of the first excited level to the ground level is e−hcν̄/kT. Using
hc/k = 1.4388 cm K:
hcν̄/kT = 1.4388 × 500 ÷ 300 = 719.4 ÷ 300 = 2.398
e−2.398 = 0.091
So only about 9% of molecules are in the excited state, and the anti-Stokes line is correspondingly weak. Raising the temperature increases the anti-Stokes intensity — which is the basis of Raman thermometry. (A full treatment also includes a ν⁴ scattering-frequency factor, which slightly favours anti-Stokes; the Boltzmann term dominates.)
Two more practical points: water is a weak Raman scatterer but a strong IR absorber, which is why Raman suits aqueous and biological samples; and resonance Raman, where the laser wavelength matches an electronic transition, can amplify the bands of a chromophore enormously — the reason it is used to study metalloprotein active sites.
Common mistakes that cost marks
- Using 3N − 6 for a linear molecule. CO₂ and HCN are linear: 3N − 5.
- Applying mutual exclusion to a molecule with no centre of symmetry. The rule only holds for centrosymmetric molecules. Water, HCl and CH₄ are not centrosymmetric, so bands may be active in both.
- Saying homonuclear diatomics are spectroscopically silent. N₂ and O₂ are IR inactive because the dipole never changes, but they are Raman active.
- Leaving the reduced mass in atomic mass units. The formula needs kg; forgetting the 1.66054 × 10⁻²⁷ conversion changes the answer by 27 orders of magnitude.
- Confusing ν̄ (cm⁻¹) with ν (s⁻¹). They differ by a factor of c. Using 3 × 10⁸ instead of 3 × 10¹⁰ cm s⁻¹ silently introduces a factor of 100.
- Counting degenerate bends twice. CO₂ has four modes but only three distinct fundamental wavenumbers, because the two bends are degenerate.
- Assuming every band you see is a fundamental. Overtones, combination bands and Fermi resonance all appear in real spectra.
How to prepare this topic
| Sub-topic | Question style | What to practise |
|---|---|---|
| Mode counting | "How many vibrational modes does X have?" | Decide linear vs non-linear first, every time |
| Selection rules and mutual exclusion | Structure deduced from IR/Raman coincidence | cis/trans and linear/bent discrimination |
| Harmonic oscillator | k from ν̄, or ν̄ from k | Unit conversions; order-of-magnitude checking |
| Isotope effects | H → D shift of a stretching band | The √µ ratio method |
| Group frequencies | Identify a functional group from bands | C=O, O−H, N−H, C≡N first |
| Metal carbonyls | Order ν(CO) across a series | Back-bonding argument with charge |
Get comfortable moving between wavenumber, frequency and wavelength. Almost every vibrational-spectroscopy numerical begins by converting a wavenumber in cm⁻¹ into a frequency, and a slip of a factor of 100 in c is the commonest cause of a wrong force constant. The free Wave Equation calculator handles the v = fλ relationship between speed, frequency and wavelength, so you can check that conversion before you square anything.
Open the Wave Equation Calculator (v = fλ) →Preparing for GATE, IIT-JAM, CSIR-NET or CUET-PG chemistry? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and online for students anywhere in India — details at abcchemistry.in.