predict the number of unpaired electrons in `[CoF_6]^(3-)` and `[Co(NH_3)_6]^(3+)`

Explanation

In both the complexes, Co is present as Co (Ill). It has `d^6` configuration i.e. four unpaired electrons. Now `NH_3` is a strong ligand so these electrons will pair up i.e. number of unpaired electrons = 0. On the other hand `F^(-)` is a weak ligand, electrons will not pair up i.e. number of unpaired electrons = 4.