IChO 2002 · Theory problem 9

Red Ruby

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Ruby crystals have a deep red colour and are well known for their use in jewellery. Not many people know that the heart of the first laser, built in 1960 by Maiman, was a big ruby crystal. The red colour of ruby originates from the absorption of light by \(\ce{Cr^3+}\) ions that are incorporated in colourless aluminium oxide (\(\ce{Al2O3}\)) crystals. The \(\ce{Cr^3+}\) ion has 3 electrons in the \(3d\) shell and the absorption of light is due to electronic transitions between \(3d\) orbitals of lower and higher energy.

Figure 1

9.1 Indicate which of the four absorption spectra belongs to ruby.

Figure 2

Figure 1

The rod used in ruby lasers is a cylinder with a length of 15.2 cm and a diameter of 1.15 cm. The amount of \(\ce{Cr^3+}\) ions is 0.050 mass %. The density of \(\ce{Al2O3}\) is \(4.05\ \mathrm{g\,cm^{-3}}\). The atomic mass of \(\ce{Cr}\) = \(52\,\mathrm{u}\). (\(1\,\mathrm{u}=1.67\times10^{-27}\ \mathrm{kg}\)).

9.2 Calculate how many \(\ce{Cr^3+}\) ions are in this laser rod.

In rubies the \(\ce{Cr^3+}\) ions are coordinated by an octahedron of 6 oxygen ions. The shape of the five \(3d\) orbitals is shown below. The box below shows the splitting of the five \(3d\) orbitals into a group of three orbitals at lower energy (\(t_{2g}\)) and a group of two at higher energy (\(e_g\))..

9.3 Indicate in the boxes below which of the \(3d\) orbitals (\(d_{z^2}\), \(d_{xy}\), \(d_{yz}\), \(d_{x^2-y^2}\), \(d_{xz}\)) belong to the \(t_{2g}\) group and which belong to the \(e_g\) group.

Figure 3

9.4 Indicate with arrows the distribution and the direction of the magnetic spin moment of the three \(3d\) electrons of \(\ce{Cr^3+}\) over the five \(d\) orbitals in the lowest energy state of \(\ce{Cr^3+}\).

Figure 4

The ruby is placed on a (non-magnetic) scale. When the scale is in balance (Figure 2) a magnet is placed directly under the side with the ruby.

Figure 5

Figure 2

9.5 Indicate what will happen with the ruby (mark the correct answer)

  • ☐ The magnet attracts the ruby (the ruby moves down)
  • ☐ The magnet has no influence on the ruby (the ruby does not move)
  • ☐ The magnet repels the ruby (the ruby moves up)
  • ☐ The magnet has an oscillating effect on the ruby (the ruby moves up and down)

Official English text of the problem.

Answers

  • 9.1: Spectrum (4).
  • 9.2: \(\boxed{3.7\times10^{20}}\) chromium(III) ions, \(\ce{Cr^3+}\), SMILES [Cr+3].
  • 9.3: \(t_{2g}:d_{xy},d_{yz},d_{xz}\); \(e_g:d_{z^2},d_{x^2-y^2}\). Fill the boxes from left to right with \(e_g,\ t_{2g},\ e_g,\ t_{2g},\ t_{2g}\).
  • 9.4: \(t_{2g}^{3}e_g^{0}\): one electron in each \(t_{2g}\) orbital, all three with parallel spins; both \(e_g\) orbitals empty.
  • 9.5: The magnet attracts the ruby (the ruby moves down).

Solution

9.1

Ruby appears red because it absorbs other visible wavelengths more strongly than red wavelengths.

Spectrum (4) has strong absorption in the violet–blue and green regions, with very little absorption in the red region, especially at approximately \(650\)–\(750\ \mathrm{nm}\). Thus red light is preferentially transmitted, giving the crystal its red colour. Mark the box beneath spectrum (4).

9.2

The cylindrical rod has volume

\[ V=\pi\left(\frac{d}{2}\right)^2L =\pi\left(\frac{1.15\ \mathrm{cm}}{2}\right)^2 (15.2\ \mathrm{cm}) =15.79\ \mathrm{cm^3}. \]

Using the supplied density, its mass is

\[ m_{\mathrm{rod}}=\rho V =(4.05\ \mathrm{g\,cm^{-3}})(15.79\ \mathrm{cm^3}) =63.94\ \mathrm{g}. \]

The chromium mass fraction is

\[ w=\frac{0.050}{100}=5.0\times10^{-4}. \]

Therefore, retaining guard digits,

\[ m_{\mathrm{Cr}}=wm_{\mathrm{rod}} =(5.0\times10^{-4})(63.94\ \mathrm{g}) =0.03197\ \mathrm{g} =3.197\times10^{-5}\ \mathrm{kg}. \]

Using the given atomic mass for each chromium(III) ion,

\[ m_{\mathrm{ion}} =(52)(1.67\times10^{-27}\ \mathrm{kg}) =8.684\times10^{-26}\ \mathrm{kg}. \]

Hence

\[ N=\frac{m_{\mathrm{Cr}}}{m_{\mathrm{ion}}} =\frac{3.197\times10^{-5}\ \mathrm{kg}} {8.684\times10^{-26}\ \mathrm{kg}} =3.682\times10^{20}. \]

The chromium mass percentage is given to two significant figures, so

\[ \boxed{N=3.7\times10^{20}\ \text{chromium(III) ions}.} \]

9.3

Each chromium(III) ion, \(\ce{Cr^3+}\) ([Cr+3]), is surrounded by six oxide ions, \(\ce{O^2-}\) ([O-2]), in an octahedral arrangement. These individual-ion SMILES do not encode the coordination geometry of the crystal.

Taking the metal–oxygen directions as the Cartesian axes:

  • The \(d_{z^2}\) and \(d_{x^2-y^2}\) orbitals point directly toward the surrounding ions. Their electrons experience greater electrostatic repulsion, so these orbitals form the higher-energy \(e_g\) group.
  • The \(d_{xy}\), \(d_{yz}\), and \(d_{xz}\) orbitals point between those directions. They experience less repulsion and form the lower-energy \(t_{2g}\) group.

Thus,

\[ \boxed{t_{2g}=\{d_{xy},d_{yz},d_{xz}\}}, \qquad \boxed{e_g=\{d_{z^2},d_{x^2-y^2}\}}. \]

The drawings are ordered \(d_{x^2-y^2},d_{yz},d_{z^2},d_{xz},d_{xy}\), so write in the boxes, from left to right:

\[ \boxed{e_g,\quad t_{2g},\quad e_g,\quad t_{2g},\quad t_{2g}}. \]

9.4

The three \(3d\) electrons occupy the three lower-energy \(t_{2g}\) orbitals. Hund’s rule requires singly occupied degenerate orbitals with parallel spins before pairing occurs.

The ground-state occupation is therefore

\[ \boxed{t_{2g}^{3}e_g^{0}}. \]

On the answer-sheet diagram, put one upward arrow on each of the three lower \(t_{2g}\) lines, and leave both upper \(e_g\) lines empty. Three downward arrows would describe the equivalent opposite orientation.

There are three unpaired electrons, with total spin

\[ S=3\left(\frac12\right)=\frac32. \]

9.5

The three unpaired electrons give the chromium(III) centres a paramagnetic response. In the treatment intended by the problem, the ruby is consequently attracted toward the stronger magnetic field near the magnet.

Since the magnet is underneath the ruby, the magnetic force is downward. Select the first option:

The magnet attracts the ruby (the ruby moves down).

The solution was checked against the official solution.