Birge–Sponer Extrapolation: Bond Dissociation Energy
In exams like CSIR NET, GATE, and IIT-JAM, vibrational spectroscopy and bond energetics are key physical chemistry topics. The Birge–Sponer extrapolation determines diatomic bond dissociation energies from vibrational spectra. While harmonic oscillators predict equal spacing and infinite bond strength, real bonds behave as anharmonic oscillators. As vibrational excitation increases, adjacent energy levels draw closer together until the bond breaks. Mastering this method connects observed spectral transitions directly to bond dissociation energies.
Anharmonicity and Birge–Sponer Method
In harmonic oscillators, vibrational levels follow G(v) = ωe(v + 1/2) with constant spacing ΔG = ωe. Real bonds obey an anharmonic Morse potential. Expressed in cm−1, the term value G(v) and the spacing between adjacent levels are:
Vibrational Term Value and Spacing:
G(v) = ωe(v + 1/2) − ωexe(v + 1/2)2
ΔGv+1/2 = G(v + 1) − G(v) = ωe − 2ωexe(v + 1)
Where ωe is the harmonic frequency in cm−1, ωexe is the anharmonicity constant in cm−1, and v is the vibrational quantum number.
Plotting ΔGv+1/2 against (v + 1) gives a straight line with intercept ωe and slope −2ωexe.
Dissociation Energy from the Plot
At the dissociation limit, level spacing drops to zero: ΔGv+1/2 = 0. Solving ωe − 2ωexe(vmax + 1) = 0 gives (vmax + 1) = ωe / (2ωexe).
The spectroscopic dissociation energy De, measured from the potential minimum, equals the triangular area under the Birge–Sponer line:
Dissociation Energy Formulas:
De = (1/2) × base × height = ωe2 / (4ωexe)
Zero-Point Energy: G(0) = (1/2)ωe − (1/4)ωexe ≈ (1/2)ωe
D0 = De − G(0) = (ωe − ωexe)2 / (4ωexe)
Conversions: 1 cm−1 = 1.19627 × 10−2 kJ/mol; 1 eV = 8065.54 cm−1
Because higher-order anharmonicity terms make real potential curves sag downward near dissociation, linear extrapolation overestimates De by 10% to 20%.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CSIR NET Chemical Sciences, GATE Chemistry, IIT-JAM Chemistry, CUET-PG
- Topic: Birge–Sponer Extrapolation and Dissociation Energies
- Difficulty: Advanced postgraduate entrance level
Step-by-Step Worked Examples
Question: A diatomic molecule has ωe = 1600.0 cm−1 and ωexe = 10.0 cm−1. Using linear Birge–Sponer extrapolation, calculate: (a) the maximum vibrational quantum number vmax at dissociation, (b) De in cm−1 and kJ/mol, and (c) the zero-point corrected bond dissociation energy D0 in kJ/mol.
Solution:
Step 1: Set spacing to zero to find the dissociation quantum number index.
ωe − 2ωexe(vmax + 1) = 0 ⇒ vmax + 1 = 1600.0 / 20.0 = 80
Therefore, vmax = 80 − 1 = 79.
Step 2: Calculate De from the area under the plot.
De = (1/2) × 80 × 1600.0 = 64000.0 cm−1.
Formula verification: De = ωe2 / (4ωexe) = 1600.02 / 40.0 = 2560000 / 40.0 = 64000.0 cm−1.
Step 3: Convert De to kJ/mol.
De = 64000.0 × 1.19627 × 10−2 = 765.61 kJ/mol.
Step 4: Compute zero-point energy G(0) and find D0.
G(0) = (0.5 × 1600.0) − (0.25 × 10.0) = 797.5 cm−1.
D0 = De − G(0) = 64000.0 − 797.5 = 63202.5 cm−1.
Formula verification: D0 = (1600.0 − 10.0)2 / 40.0 = 1590.02 / 40.0 = 2528100 / 40.0 = 63202.5 cm−1.
In kJ/mol: D0 = 63202.5 × 1.19627 × 10−2 = 756.07 kJ/mol.
Answer: vmax = 79, De = 64000.0 cm−1 (765.61 kJ/mol), D0 = 756.07 kJ/mol
Question: For a diatomic radical, the fundamental transition (v = 0 → 1) is at 2143.0 cm−1 and the first overtone (v = 0 → 2) is at 4258.0 cm−1. Calculate ωe, ωexe, and the dissociation energy De in cm−1 and kJ/mol using Birge–Sponer extrapolation.
Solution:
Step 1: Relate band origins to term values.
ΔG1/2 = ωe − 2ωexe = 2143.0 cm−1
ν̃0→2 = 2ωe − 6ωexe = 4258.0 cm−1
Step 2: Find second spacing ΔG3/2.
ΔG3/2 = 4258.0 − 2143.0 = 2115.0 cm−1.
Step 3: Solve for ωexe and ωe.
ΔG1/2 − ΔG3/2 = 2ωexe = 2143.0 − 2115.0 = 28.0 cm−1 ⇒ ωexe = 14.0 cm−1.
ωe = 2143.0 + 28.0 = 2171.0 cm−1.
Overtone check: 2(2171.0) − 6(14.0) = 4342.0 − 84.0 = 4258.0 cm−1. Matches.
Step 4: Compute De.
Route 1: De = ωe2 / (4ωexe) = 2171.02 / 56.0 = 4713241.0 / 56.0 = 84165.02 cm−1.
Route 2: Base = 2171.0 / 28.0 = 77.5357; Area = 0.5 × 77.5357 × 2171.0 = 84165.02 cm−1.
Convert to kJ/mol: 84165.02 × 1.19627 × 10−2 = 1006.84 kJ/mol.
Answer: ωe = 2171.0 cm−1, ωexe = 14.0 cm−1, De = 84165.02 cm−1 (1006.84 kJ/mol)
Common Mistakes Students Make in Exams
• Taking slope as −ωexe instead of −2ωexe: The slope of ΔGv+1/2 against (v + 1) is −2ωexe. Omitting 2 doubles the calculated dissociation energy.
• Confusing De with D0: De is measured from the potential well minimum, whereas D0 is measured from the v = 0 ground level. D0 equals De minus G(0).
• Ignoring overestimation in linear extrapolation: Real molecules show higher-order anharmonicity near dissociation. A linear Birge–Sponer plot overestimates true De by 10% to 20%.
• Unit conversion errors: Multiply wavenumbers (cm−1) by 1.19627 × 10−2 to obtain kJ/mol, or divide by 8065.54 to obtain eV.
Exam Relevance and Question Patterns
Vibrational spectroscopy and dissociation energy questions appear regularly across national postgraduate chemistry examinations. The summary below outlines key question formats.
| Examination | Question Format | Key Focus Areas |
|---|---|---|
| CSIR NET Chemical Sciences | Multiple-choice questions | Evaluating ωe and ωexe from band origins, calculating De and D0, Morse potential analysis |
| GATE Chemistry | Numerical Answer Type (NAT) and MCQs | Birge–Sponer plot intercept and slope interpretation, maximum vibrational quantum number, energy conversions |
| IIT-JAM Chemistry | Multiple-choice and NAT questions | Anharmonic oscillator spacing, zero-point energy calculation, level separation trends |
| CUET-PG Chemistry | Objective questions | Definitions of De versus D0, formulas relating ΔGv+1/2 to quantum numbers |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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