Derive and expression for the tetrahedral and cubic crystal-field potential.

Short Answer

Expert verified
The tetrahedral crystal field potential is given by: \( V_{tetrahedral} = -2/5 Dq[r^2 + 3z^2] \) after substituting r and z in spherical coordinates using the tetrahedron's geometry. The cubic crystal field potential is zero in all three directions showing a degenerate ground state due to the symmetry in a cubic field.

Step by step solution

01

Understand and visualise the geometry for tetrahedral crystal field

A tetrahedral crystal field refers to a situation where a central ion is surrounded by four ions located at the corners of a regular tetrahedron. In this configuration, the angles between any two ligands are 109.5 degrees. Visualise this geometry.
02

Derive potential for tetrahedral crystal field

Remember that the crystal field potential is given by \(V = -2/5 Dq[r^2 + 3z^2]\) where Dq is the crystal field splitting energy, r is the distance from the central ion to the ion in consideration, and z is the height along the z axis. Using the geometry from step 1, rewrite r and z in terms of spherical coordinates and calculate the potential.
03

Understand and visualise the geometry for cubic crystal field

A cubic crystal field refers to a situation where a central ion is surrounded by eight ions located at the corners of a cube. In this configuration, there are three mutually perpendicular directions along which the ligands are placed. Visualise this geometry.
04

Derive potential for cubic crystal field

Again, use the general formula for crystal field potential. As in step 2, substitute the appropriate expressions for r and z using the geometry from step 3. After calculating, you'll find that the potential is zero along all three directions, leading to a degenerate ground state.

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