Question

A metal ion in a high-spin octahedral complex has two more unpaired electrons than the same ion does in a low spin octahedral complex. Name some possible metal ions for which this arrangement would apply.

Short answer

Here, ${\text{CO}}^{2+},{\text{Ni}}^{3+}$has two more unpaired electrons in a high spin state compared to a low spin state.

Step-by-step solution

A concept:

The crystal field splitting is the difference in energy between the d orbitals of the ligands. The division number of the crystal field is denoted by the capital Greek letter $\mathbf{∆}$. Crystal field splitting explains the difference in color between two similar metal-ligand complexes.

Name some possible metal ions:

The possible metal ions are ${\text{CO}}^{2+},{\text{Ni}}^{3+}$etc.

The ${\text{CO}}^{2+},{\text{Ni}}^{3+}$ions have${\text{d}}^7$ configuration in a high spin octahedral complex has two more unpaired electrons (total three electrons) than the same ion does in low spin octahedral complex (one unpaired electron). The electron distribution diagram for high spin and low spin complex of ${\text{CO}}^{2+}$ion are as shown below.

The splitting diagram for high spin octahedral complex it possesses three unpaired electrons.

The splitting diagram for low spin octahedral complex it possesses one unpaired electron.