Question

a) Schematically plot IR spectra of a N2, 14^NO and 15^NO. List all differences between these spectra. How many translational, rotational and vibrational degrees of freedom are in these molecules?

b) Sketch the potential that more accurately represents vibrations in diatomic molecules and compare it to the harmonic potential. Compare the energies in both cases. List all differences.

Answer 1

There is a quantum mechanical condition that in order for a molecular vibration to be observed in the IR there must be a dipole moment change accompanying the vibration. It is called a selection rule. For homonuclear diatomic molecules, A2, the stretch, ↔A-A↔ , is silent in the IR because the molecule does not have a dipole. H2, N2, O2 fit into this category.

The IR spectrum peak for 14NO is at 1875 cm^-1

This can be calculated using the formula

= 1/2k/

where k the force constant for NO is 1530 N/m

and can be calculated using the formula (m1xm2)/(m1 + m2) where m1 =14 and m2 =16 for 14 NO and M1=15 and m2 =16 for 15 NO

so for 14 NO is 7.4666 and for 15 NO is 7.742

for for 14 NO is 5.592 x 10¹³ sec^-1 this corresponds to 1864 cm^-1 using the formula 1/ = /c

and for 15 NO is 5.491 x 10¹³ sec^-1 this corresponds to 1830 cm^-1

3n degrees of freedom describe the motion of a molecule in relation to the coordinates (x,y,z). The 3n degrees of freedom also describe the translational, rotational, and vibrational motions of the molecule.

A diatomic gas molecule thus has 6 degrees of freedom. This set may be decomposed in terms of translations, rotations, and vibrations of the molecule. The center of mass motion of the entire molecule accounts for 3 degrees of freedom. In addition, the molecule has two rotational degrees of motion and one vibrational mode. The rotations occur around the two axes perpendicular to the line between the two atoms. The rotation around the atom–atom bond is not a physical rotation. This yields, for a diatomic molecule, a decomposition of: