1. Vibrational frequency :
Hooks law shows relationship between the wavenumber or frequency with force constant (K) and reduce mass (μ) .
υ=1/λ , υ = C/ λ = C υ
υ = frequency
υ = wavenumber
C = velocity of light
μ = reduce mass
K = force constant of the bond
υ/ = υ / C

2. The frequency of the stretching vibration depends on two factors:
1- Masses of the Atoms : Heavier atoms vibrate more slowly than lighter ones which means frequency decreases with increasing atomic weight. e.g. The characteristic frequency of a C-D bond is lower than that of a C-H bond.
2- Stiffness of the Bond: Stronger bonds are generally stiffer, they are requiring more force to stretch or compress them. Thus, stronger bonds usually vibrate faster than weaker bonds (assuming the atoms have similar masses).
In a group of bonds having atoms of similar masses, K frequency increases with bond energy.
e.g.
1- O-H bonds are stronger than C-H bonds, and O-H bonds vibrate at higher frequencies.
2- Triple bonds are stronger than double bonds, so triple bonds vibrate at higher frequencies than double bonds. 6

4. No. of vibrations = 3N-5 linear molecules
Number of Fundamental Vibrations
The IR spectra of polyatomic molecules may exhibit more than one vibrational absorption bands. The number of these bands corresponds to the number of fundamental vibrations in the molecule which can be calculated from the degrees of freedom of the molecule.
No. of vibrations = 3N linear molecules
No. of vibrations = 3N non- linear molecules

5. H2O , N=3
No. of vibrations = 3N-6
No. of vibrations = 3x3-6 =9 – 6 = 3
CO2 , N=3
No. of vibrations = 3N-5
No. of vibrations = 3x3-5 =9 – 5 = 4
C2H6
N=8
No. of vibrations = 3N-6
No. of vibrations = 3x8-6 =24 – 6 = 18

6. The carbon dioxide molecule is linear and has four fundamental vibrations
Thus, four theoretical fundamental bands are expected but actually it shows only two because :
1- The symmetrical Stretching vibration in carbon dioxide is IR inactive because it produces no change in the dipole moment of the
molecule.
2- The two bending vibrations are equivalent and absorb at the same wavenumber (667.3 cm-1)

7. Solved Problems
1- Convert the following wavelengths into the corresponding wavenumbers in cm-1:

8. 2- Convert wavenumber 1755 cm-1 into the corresponding wavelength in µm.