Chemistry 2720 Fall 2000 Assignment 8 Solutions

1.

Rn

7s²5f¹⁴6d¹⁰7p²

(a)

There should be a Uuq²⁺ ion obtained by removing the p electrons with a ground-state configuration of [Rn]7s²5f¹⁴6d¹⁰. There should also be a Uuq⁴⁺ ion whose ground state is [Rn]5f¹⁴6d¹⁰.

(b)

The atom should be paramagnetic since its p electrons are unpaired. On the other hand, both of the ions have only paired electrons, so they should be diamagnetic.

2.

(a)

The chromium atom has six valence electrons, so the Cr₂ molecule has twelve valence electrons. The orbitals are filled as follows:

(b)

Since there are unpaired electrons, Cr₂ should be paramagnetic.

(c)

i.

It is a bonding orbital with an enhanced electron density between the two nuclei.

ii.

This orbital looks like it results from adding 3d₀ atomic orbitals on each chromium atom.

3.

(a)

Raman only

(b)

both absorption and Raman

(c)

both absorption and Raman

(d)

neither

4.

Straightforward method:

The lines of the rotational spectrum of a diatomic molecule are spaced by $\Delta_r=\hbar^2/(\mu R^2)$. The spacings (obtained by subtracting the frequencies of adjacent lines) are as follows: 624, 617, 617, 619, 615, 610, 611 and 607GHz. The average is $615\pm 5\,\mathrm{GHz}$.¹ We need to convert this quantity to energy units:

$$\begin{eqnarray*}\Delta_r & = & h\Delta\nu_r\\ & = & (6.626\,068\,8\times 10^... ...athrm{Hz}]\\ & = & (4.08\pm 0.03)\times 10^{-22}\,\mathrm{J}. \end{eqnarray*}$$

We also need the reduced mass:

$$\begin{eqnarray*}\mu & = & (m_\mathrm{H}^{-1}+m_\mathrm{Cl}^{-1})^{-1}\\ & = &... ...,amu)^{-1}\right]^{-1}\\ & = & 0.979\,592\,539\,\mathrm{amu}. \end{eqnarray*}$$

To convert amu to kg, since an amu is the same as a g/mol, we divide by Avogadro's number and by 1000. The result is $\mu = 1.626\,651\times 10^{-27}\,\mathrm{kg}$. We can now calculate R:

$$\begin{eqnarray*}R & = & \frac{\hbar}{\sqrt{\mu\Delta_r}}\\ & = & \frac{h}{2\p... ...rm{J}]}}\\ & = & (1.295\pm 0.005)\times 10^{-10}\,\mathrm{m}. \end{eqnarray*}$$

By fitting:

We don't know the transition (i.e. the value of J) to which the first line reported corresponds. However, we do know that adjacent lines differ by one unit of J. Suppose that the first line corresponds to J=J₀. Then the second line is J₀+1, the next J₀+2, and so on. We can therefore rewrite the equation for the transition energies in the form

$$E_\mathrm{photon} = \frac{\hbar^2}{\mu R^2}(J_0+1+j)$$

where $j=0,1,2,\ldots,7$. If we divide this equation by h, we get

$$\nu = \frac{h}{4\pi^2\mu R^2}(J_0+1+j).$$

If we plot the frequencies vs our (arbitrary) numbering (j), we should get a straight line of slope $h/(4\pi^2\mu R^2)$:

The slope of this line is $614.6\,\mathrm{GHz}$. Thus

$$\begin{eqnarray*}% latex2html id marker 150 R^2 & = & \frac{h}{4\pi^2\mu(\mathrm... ...m{m}^2\\ \therefore R & = & 1.296\times 10^{-10}\,\mathrm{m}. \end{eqnarray*}$$

Footnotes

...$615\pm 5\,\mathrm{GHz}$.¹

Note that it was not necessary for you to carry out an error analysis. I am performing such an analysis to show you the accuracy of this method for determining bond lengths.