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

##### 1 Answer

The wavenumber

#\mathbf(tildenu = 1/(2pic)sqrt(k/mu))# where:

#c# is thespeed of light,#2.998xx10^(10)# #"cm/s"# .#k# is theforce constantin#"kg/s"^2# of the bond between the two atoms in theharmonic oscillatormodel, which can alternatively be labeled theball-and-springmodel. The force constant value is generally in the hundreds.#mu# is thereduced mass;#mu = (m_1m_2)/(m_1 + m_2)# , where#m_i# is the molar mass of atom#i# .

Since for the same molecule, **higher** **higher**.

A ** higher** force constant

**"spring" (i.e. stronger bond).**

*stiffer*Therefore, a ** stronger** bond has a

**IR frequency when comparing the**

*higher***type of vibrational motion (e.g. symmetric stretch with symmetric stretch, asymmetric bend with asymmetric bend, etc).**

*same**CHALLENGE: Can you show me why the wavenumber frequency for* *for the symmetric stretch is* *whereas that of* *is* *? What does that tell you about the bond strength of* *?*