Q114P - Chemistry: Molecular Nature Of Matter And Change

Question

Hexafluorocobaltate(III) ion is a high-spin complex. Draw the orbital-energy splitting diagram for its d orbitals.

Step-by-Step Solution

Answer

In this complex, there are two higher energy  $\left({\mathrm{e}}_{\mathrm{g}}\right)$ orbitals and three lower energy $\left({\mathrm{t}}_{2\mathrm{g}}\right)$ orbitals formed, with a total of 6 electrons.

Step 1: Give data

Hexafluorocobaltate(III) ion is a high-spin complex

Step 2: Concept Introduction

The complex with the same number of unpaired electrons as the free metal ion is termed a high-spin complex and it has the maximum number of unpaired electrons. High-spin complexes occur when weak-field ligands bond with central metal ions because they cause small splitting energy.

 

Weak-field ligands give smaller splitting energy $(∆)$  because they don't cause large splitting of orbitals (F- is a weak-field ligand).

Step 3: Solution

${\left[{\mathrm{CoF}}_6\right]}^{3-}$ in this complex Co has an oxidation state 3+, so the number of electrons in its d subshell is 6. The electron configuration of Co is $3{\mathrm{d}}^{7}4{\mathrm{s}}^2$. In this complex, there are two higher energy ( ${\mathrm{e}}_{\mathrm{g}}$) orbitals and three lower energy ( ${\mathrm{t}}_{2\mathrm{g}}$ ) orbitals formed.