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Chemistry LibreTexts

Molecular rotations and vibrations

Sample Problem 1. Calculate the probability of being outside the classical turning points for the v=1 wave function.

\[\Psi=e^{-\alpha x^2/2}where\,\alpha=\biggr(\frac{\mu k}{\hbar^2}\biggr)^{1/2}\]


First we need the normalized wave function. Given that the harmonic oscillator solutions have the form:

These can be normalized as follows.

\[N^2\int_{-\infty}^{\infty}e^{-\alpha x^2/2}e^{-\alpha x^2/2}dx=N^2\int_{-\infty}^{\infty}\,e^{-\alpha x^2}dx=N^2\sqrt\frac{\pi }{\alpha }=1N=\biggr(\frac{\alpha }{\pi }\biggr)^{1/4}\]

 The classical turning points occur at



So the value of x is:

k/\mu}}{k}=\biggr(\frac{\hbar^2}{\mu k}\biggr)^{1/4}=\frac{1}{\sqrt{\alpha }}\]

The trick is to calculate the probability of being inside the turning points and then for a normalized function the probability of being outside will be 1 – P(inside).

\[P(inside)=\sqrt\frac{\alpha }{\pi }\int_{-1/\sqrt{\alpha }}^{1/\sqrt\alpha }e^{-\alpha x^2/2}e^{-\alpha x^2/2}dx=\sqrt\frac{\alpha }{\pi }\int_{-1/\sqrt{\alpha }}^{1/\sqrt\alpha }e^{-\alpha x^2}dx\]

This can be integrated to give erf(1). Note that the definition of error function erf(x) is

\[erf(x)=\frac{2}{\pi }\int_{0}^{x}e^{-t^2}dt\]

Which is a tabulated function, so P(inside) = erf(1) = 0.843 and P(outside) = 1 - 0.843 = 0.157.


Sample problem 2.

Calculate the location of the classical turning point in the lowest vibrational state a for diatomic oxygen molecule 16O16O given that the wave number is 1556 cm-1.



This gives k = 1154 N/m.

\[x=\biggr(\frac{\hbar^2}{\mu k}\biggr)^{1/4}=\biggr(\frac{(1.054\times10^{-34}Js)^2}{\frac{16}{2}\times(1.672\times10^{-27}kg)1154\,N/m}\biggr)^{1/4}\]



Sample Problem 3.

The allowed rotational transitions in a diatomic molecule follow

the selection rule J ® J ± 1. The J ® J + 1 transition, derive an expression for DE = EJ+1 – EJ in terms of the quantum number J and the parameters of the molecule.



This energy is in joules. The rotational constant B is in cm-1. Recall that hc is the conversion between cm-1 and J. Therefore rotational lines will be observed at 2B, 4B, 6B etc. and the spacing between rotational lines is 2B.

b. Given that the rotational spacing is for diatomic oxygen is 2.89 cm-1 calculate the O-O bond length.


The rotation constant B = 1.445 cm-1 and therefore we can calculate the moment of inertia I.

\[\widetilde{B}=\frac{\hbar}{4\pi\,cI}so\, I=\frac{\hbar}{4\pi\,c\widetilde{B}}=\frac{1.054\times10^{-34}J}{4(3.141)(3\times10^{10}\,cm/s)(1.445\,cm^{-1})}\]


I = 1.936 x 10-46 kgm2 and since I = mr2 where m is the reduced mass we have


Sample Problem 4. For the following questions, assume that the molecular orbital energies are in the order 1p < 3s < 2p* < 4s*; in other words, 1p has the lowest energy and 4s* has the highest energy. Atomic numbers C = 6, F = 9.


Give the electron configuration (which molecular orbitals are occupied, and by how many electrons) for the C2 molecule.

There are four valence electrons for each carbon atom. Therefore we have eight total valence electrons and the configuration is

[He] 1s2 2s*2 1p4 3s 2p* 4s*

Calculate the bond order in the F2 molecule.


There are seven valence electrons for each F atom. Therefore we have 14 total valence electrons and the configuration is

\[[He]= 1\sigma^2  2{\sigma^*}^4   1\pi^4  3\sigma^2   2{\pi^*}^4  4\sigma^*\]

Bond Order = ½(bonding – anti-bonding) = ½(8-6) = 1


Calculate the change in the bond order of the F2 molecule if we remove an electron. In other words, upon removing an electron to make F2+, the bond order changes from _1__ to _1.5__.


If we remove one electron then we have 13 electrons remaining. The configuration is

\[[He]= 1\sigma^2  2{\sigma^*}^4   1\pi^4  3\sigma^2   2{\pi^*}^3  4\sigma^*\]






Since we removed on electron from an anti-bonding orbital we have

Bond Order = ½(bonding – anti-bonding) = ½(8-5) = 1.5