IR spectroscopy

Model the bond as two point masses joined by a spring obeying Hooke's law, restoring force $F = -kx$, where $k$ is the force constant (bond stiffness, $\text{N m}^{-1}$) and $x$ is the displacement from the equilibrium bond length. Quantum mechanically this simple harmonic oscillator has equally spaced energy levels indexed by the vibrational quantum number $v = 0, 1, 2, \ldots$:

where: $h$ is Planck's constant and $\nu$ is the oscillator's natural (classical) vibrational frequency, given by

where: $\mu$ is the reduced mass of the two vibrating atoms, $\mu = m_1m_2/(m_1+m_2)$ — the same combination that appears in rotational spectroscopy. Even at $v=0$ the molecule keeps zero-point energy $E_0 = \tfrac12 h\nu$; a bond never stops vibrating entirely, not even at absolute zero. The selection rule for this ideal oscillator is $\Delta v = \pm 1$, and — exactly as in rotational spectroscopy — a vibration is only IR-active if it changes the dipole moment, $d\mu/dr \neq 0$. At room temperature the Boltzmann distribution keeps almost every molecule in $v=0$, so the dominant absorption is the fundamental, $v=0 \to v=1$, observed as a band at wavenumber $\tilde\nu = \nu/c$. Real bonds are not perfectly harmonic. As a bond stretches toward dissociation the restoring force weakens and the energy levels converge — anharmonicity, better described by a Morse-type potential than the symmetric SHO parabola. This is why weak overtone bands ($\Delta v = \pm2, \pm3, \ldots$) show up at slightly less than integer multiples of the fundamental, and why the $\Delta v=\pm1$ rule is an idealization rather than an absolute law. Beyond one isolated diatomic bond, a polyatomic molecule's many bonds and angles vibrate at frequencies that fall into recognizable ranges depending on the atoms and bond order involved: O–H stretch about $3200$ to $3550~\text{cm}^{-1}$, C–H stretch about $2850$ to $3000~\text{cm}^{-1}$, C≡C or C≡N stretch about $2100$ to $2260~\text{cm}^{-1}$, and C=O stretch about $1650$ to $1750~\text{cm}^{-1}$. These characteristic group frequencies are the practical payoff: scan a spectrum's high-frequency region and read off which functional groups are present. Below about $1500~\text{cm}^{-1}$ lies the fingerprint region, where bands arise from complex, coupled skeletal motions unique to each molecule and are not assignable bond by bond.

Problem: The fundamental vibrational absorption of $^{1}\text{H}^{35}\text{Cl}$ occurs at $\tilde\nu = 2886~\text{cm}^{-1}$. Calculate the force constant $k$ of the H–Cl bond, using $\mu = 1.6267\times10^{-27}~\text{kg}$.

Step 1. Convert wavenumber to frequency. $\nu = c\tilde\nu = (3.00\times10^{10}~\text{cm s}^{-1})(2886~\text{cm}^{-1}) = 8.658\times10^{13}~\text{s}^{-1}$.

Step 2. Rearrange the frequency formula for $k$. From $\nu = \dfrac{1}{2\pi}\sqrt{k/\mu}$, squaring both sides and solving gives $k = \mu(2\pi\nu)^2$. First compute the angular frequency: $2\pi\nu = 2\pi(8.658\times10^{13}) = 5.440\times10^{14}~\text{rad s}^{-1}$.

Step 3. Substitute and evaluate. $k = (1.6267\times10^{-27}~\text{kg})(5.440\times10^{14}~\text{s}^{-1})^2 = (1.6267\times10^{-27})(2.959\times10^{29})$, so

This is the bond force constant. It is a typical single-bond value, consistent with tabulated H–Cl data ($\approx 480$ to $516~\text{N m}^{-1}$); double and triple bonds give substantially larger $k$ and hence higher vibrational wavenumbers.

Competitive stretch

Problem: Using the same force constant $k = 481.5~\text{N m}^{-1}$ (bond stiffness barely changes on isotopic substitution), estimate the fundamental wavenumber of $^{2}\text{H}^{35}\text{Cl}$ (deuterium chloride, D–Cl).

Step 1. Since $\tilde\nu \propto \sqrt{k/\mu}$ at fixed $k$, the ratio of wavenumbers is $\tilde\nu_{\text{DCl}}/\tilde\nu_{\text{HCl}} = \sqrt{\mu_{\text{HCl}}/\mu_{\text{DCl}}}$; only the reduced masses need to be compared.

Step 2. Compute the reduced masses in atomic mass units: $\mu_{\text{HCl}} = \dfrac{1.008\times34.97}{1.008+34.97} = 0.980~\text{amu}$, and $\mu_{\text{DCl}} = \dfrac{2.014\times34.97}{2.014+34.97} = 1.904~\text{amu}$.

Step 3. Take the ratio: $\sqrt{\mu_{\text{HCl}}/\mu_{\text{DCl}}} = \sqrt{0.980/1.904} = \sqrt{0.5147} = 0.7174$.

Step 4. Scale the HCl wavenumber. $\tilde\nu_{\text{DCl}} = 0.7174 \times 2886~\text{cm}^{-1} \approx$ $2070~\text{cm}^{-1}$, close to the experimentally observed D–Cl fundamental near $2090~\text{cm}^{-1}$ — the small remaining gap comes from anharmonicity differences the ideal SHO model ignores. Heavier isotopes shift a band to lower wavenumber even though the bond stiffness $k$ is essentially unchanged.

• Confusing force constant $k$ with bond order casually. Higher bond order generally raises $k$ (C≡C > C=C > C–C), but $k$ also depends on which atoms are bonded (through $\mu$), so wavenumbers of different atom pairs cannot be ranked by bond order alone.
• Forgetting the changing-dipole-moment rule. A vibration only absorbs IR light if it changes the dipole moment. The symmetric stretch of $\text{CO}_2$, for instance, leaves the dipole moment at zero throughout and is IR-inactive, even though other modes of the same molecule are IR-active.
• Mistaking overtone or combination bands for new functional groups. Weak bands near integer multiples of a strong fundamental are anharmonicity-driven overtones, not evidence of a separate group.
• Reading the fingerprint region like the group-frequency region. Below about $1500~\text{cm}^{-1}$, bands come from coupled skeletal vibrations unique to each molecule and are not assignable to single bonds the way higher-frequency stretches are.

Ask the tutor: if you want more force-constant or isotope-shift calculations, or help reading a full IR spectrum peak by peak, just ask.