Chemistry 3730 Fall 2000 Assignment 5 Solutions

1.

The energy for a Morse oscillator is given by

> Evib := n -> w0*(n+1/2) - wa*(n+1/2)^2;

$$\mathit{Evib} := n\rightarrow \mathit{w0}\,(n + {\displaystyl... ...{2}} ) - \mathit{wa}\,(n + {\displaystyle \frac {1}{2}} )^{2}$$

where we are using the shorthand $\texttt{w0} = \hbar\omega_0$ and $\texttt{wa} = \hbar\omega_0\chi$. It is a relatively straightforward matter to fit the data to this expression. The data are

> nvals := [0,1,2,3];

$$\mathit{nvals} := [0, \,1, \,2, \,3]$$

> Evals := [2.8368,8.4883,14.110,19.702];

$$\mathit{Evals} := [2.8368, \,8.4883, \,14.110, \,19.702]$$

> with(stats):
> fitted_eq :=
  fit[leastsquare[[n,En],En=Evib(n),{w0,wa}]]([nvals,Evals]);

$$\mathit{fitted\_eq} := \mathit{En}=5.681132253\,n + 2.840566127 - .01485366631\,(n + {\displaystyle \frac {1}{2}} )^{2}$$

Thus we have

> hbar := 6.626069e-34/(2*Pi);

$$\mathit{hbar} := .3313034500\,10^{-33}\,{\displaystyle \frac {1}{ \pi }}$$

> omega0 := 5.681132253e-21/hbar;

$$\omega 0 := .1714782099\,10^{14}\,\pi$$

The anharmonicity constant is

> chi := .1485366631e-1/5.681132253;

$$\chi := .002614560909$$

The anharmonicity constant is dimensionless.

The isotopic mass of sodium is (in amu)

> mNa := 22.98976967;

$$\mathit{mNa} := 22.98976967$$

and for iodine, we have

> mI := 126.904468;

$$\mathit{mI} := 126.904468$$

The reduced mass (in kg) is therefore

> mu := 1/(1/mNa+1/mI)/1000/6.02214199e23;

$$\mu := .3232031645\,10^{-25}$$

The bond stiffness is therefore

> k := evalf(omega0^2*mu);

k := 93.79792525

Thus, $k=93.80\,\mathrm{N/m}$. The Morse parameter D is given by

> DM := 5.681132253e-21/(4*chi);

$$\mathit{DM} := .5432204919\,10^{-18}$$

To get the dissociation energy, we need to subtract off the zero-point energy:

> De := DM - 1/2*5.681132253e-21*(1-1/2*chi);

$$\mathit{De} := .5403836392\,10^{-18}$$

Our spectroscopic calculation therefore gives a dissociation energy of $5.404\times 10^{-19}\,\mathrm{J}$.

2.

I performed a UHF geometry optimization with the STO-3G^* basis set for NaI. The energy of the molecule was

$$E_\mathrm{NaI}^\mathrm{STO-3G^*} = -4\,402\,302.00\,\mathrm{kcal/mol}.$$

I saved this structure to disk, then converted the sodium atom to a ghost atom. Thus I computed the energy of an iodine atom avoiding basis set superposition error:

$$E_\mathrm{I}^\mathrm{STO-3G^*} = -4\,301\,886.09\,\mathrm{kcal/mol}.$$

I loaded the optimized NaI structure back into my workspace and converted the iodine atom to a ghost atom. The following single-point calculation gave me

$$E_\mathrm{Na}^\mathrm{STO-3G^*} = -100\,377.38\,\mathrm{kcal/mol}.$$

The energy of separated sodium and iodine atoms is therefore

$$E_\mathrm{Na}^\mathrm{STO-3G^*} + E_\mathrm{I}^\mathrm{STO-3G^*} = -4\,402\,263.47\,\mathrm{kcal/mol}.$$

The difference in energy between the molecule and separated atoms (the dissociation energy) is

$$E_{d,\mathrm{NaI}} = 38.53\,\mathrm{kcal/mol} \equiv 2.677\times 10^{-19}\,\mathrm{J}.$$

The answers we get from the spectroscopic calculation and from the computational method are wildly different from each other.

3.

The discussion below refers to the lines as numbered here:

Here is my identification of the modes:

1

This is a symmetric stretching mode involving mostly the C-Cl bonds.

2

This is the corresponding C-Cl antisymmetric stretch.

3

This normal mode is a ``wagging'' mode in which the two H atoms rock back-and-forth in concert.

4

The identification of this and the next mode are much less certain. However, it seems likely that this mode is a symmetric stretch involving mostly the C-H bonds.

5

If the prior identification is correct, this would be the corresponding antisymmetric stretch.

4.

Here are my results: