# When using IR spectroscopy, why does a stronger bond have a higher wavenumber?

Feb 7, 2016

The wavenumber $t i l \mathrm{de} \nu$ is related to the force constant:

$\setminus m a t h b f \left(t i l \mathrm{de} \nu = \frac{1}{2 \pi c} \sqrt{\frac{k}{\mu}}\right)$

where:

• $c$ is the speed of light, $2.998 \times {10}^{10}$ $\text{cm/s}$.
• $k$ is the force constant in ${\text{kg/s}}^{2}$ of the bond between the two atoms in the harmonic oscillator model, which can alternatively be labeled the ball-and-spring model. The force constant value is generally in the hundreds.
• $\mu$ is the reduced mass; $\mu = \frac{{m}_{1} {m}_{2}}{{m}_{1} + {m}_{2}}$, where ${m}_{i}$ is the molar mass of atom $i$.

Since for the same molecule, $\mu$ remains constant, $c$ remains constant, and obviously $2 \pi$ remains constant, for any higher $k$, you can tell that $t i l \mathrm{de} \nu$ is higher.

A higher force constant $k$ means a stiffer "spring" (i.e. stronger bond).

Therefore, a stronger bond has a higher IR frequency when comparing the same type of vibrational motion (e.g. symmetric stretch with symmetric stretch, asymmetric bend with asymmetric bend, etc).

CHALLENGE: Can you show me why the wavenumber frequency for ${D}_{2}$ for the symmetric stretch is $2990$ $c {m}^{- 1}$, whereas that of ${H}_{2}$ is $4401$ $c {m}^{- 1}$? What does that tell you about the bond strength of ${H}_{2}$ relative to that of ${D}_{2}$?