Question

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

Answer 1

The wavenumber `tildenu` is related to the force constant:

`\mathbf(tildenu = 1/(2pic)sqrt(k/mu))`

where:

• `c` is the speed of light, `2.998xx10^(10)` `"cm/s"`.
• `k` is the force constant in `"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 = (m_1m_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 `2pi` remains constant, for any higher `k`, you can tell that `tildenu` 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` `cm^(-1)`, whereas that of `H_2` is `4401` `cm^(-1)`? What does that tell you about the bond strength of `H_2` relative to that of `D_2`?