Question

What is the formula for hexafluorocobaltate(III)? Is it high spin or low spin?

Answer 1

Notice the "-ate" at the end. That indicates a negative overall charge to the complex. From there, we can deduce the oxidation state of cobalt.

"Hexafluoro" indicates six fluoro (`"F"^(-)`) ligands, and the `"III"` indicates the oxidation state of cobalt has a magnitude of `3`.

Since the overall charge is negative, and `9^-` would be an absurd charge, we therefore have, for hexafluorocobaltate(III):

`["CoF"_6]^(3-)`

where the oxidation state of cobalt here must be `color(blue)(+3)`, since...

`overbrace((+3))^(Co) + overbrace(6(-1))^(6xxF^(-)) = overbrace(-3)^("overall")` `color(blue)(sqrt"")`

Thus, with `"Co"^(3+)`, we have a `d^6` metal here. Since `"F"^(-)` is a weak-field ligand (a `pi` donor), the crystal field splitting energy is small, and we anticipate an (octahedral) high spin complex.

• Is this compound colored? Would you expect any charge-transfer bands in the UV-VIS spectrum?
  (There should be ligand-to-metal charge transfer, LMCT.)

• Is there Jahn-Teller distortion in this complex? Is it weak or strong?
  (should be weak)

• Is this compound paramagnetic, i.e. does it have any unpaired electrons?
  (yes. Should be 4.)

• What is the spin-only magnetic moment?
  (`mu_S = 2.0023sqrt(S(S+1))`, where `S` is the total electron spin.)

These are all questions that are totally valid on third-year or fourth-year undergraduate exams. The first and third bullet point questions, probably first or second year.