Question

How can one decide whether the high spin or The Low spin configuration of a metal Complex will be more stable? ​

Answer 1

Tips For Determining High Spin or Low Spin Configurations

Determine the shape of the complex (i.e. octahedral, tetrahedral, square planar)

Determine the oxidation state of the metal center

Determine the d electron configuration of the metal center

Draw the crystal field diagram of the complex with regards to its geometry

Determine whether the splitting energy is greater than the pairing energy

Determine the strength of the of the ligand

example

What is the number of electrons of the metal in this complex: [CoF6]3- ?

SOLUTION

This coordination compound has Cobalt as the central transition metal and 6 Fluoro monodentate ligands. However, in this example as well as most other examples, we will focus on the central transition metal. We must determine the oxidation state of Cobalt in this example. Fluorine has a charge of -1 and the overall molecule has a charge of -3. Since there are six fluorines, the overall charge of fluorine is -6. The charge of Cobalt will add to this -6, so that the charge of the overall molecule is -3. Thus, we know that Cobalt must have a charge of +3 (see below). The electron configuration of Cobalt is [Ar]4s23d7. When observing Cobalt 3+, we know that Cobalt must lose three electrons. The first two to go are from the 4s orbital and Cobalt becomes:[Ar]4s03d7. Then, the next electron leaves the 3d orbital and the configuration becomes: [Ar]4s03d6. Thus, we can see that there are six electrons that need to be apportioned to Crystal Field Diagrams. The pairing of these electrons depends on the ligand. Since Fluorine is a weak field, it will be a high spin complex.

Cobalt charge        Fluorine charge      Overall charge

x              +              -1(6) =                       -3

x              +              -6 =                           -3   

 

x=-3+6

x=+3