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

Jul 7, 2017

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 (${\text{F}}^{-}$) ligands, and the $\text{III}$ 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):

${\left[{\text{CoF}}_{6}\right]}^{3 -}$

where the oxidation state of cobalt here must be $\textcolor{b l u e}{+ 3}$, since...

${\overbrace{\left(+ 3\right)}}^{C o} + {\overbrace{6 \left(- 1\right)}}^{6 \times {F}^{-}} = {\overbrace{- 3}}^{\text{overall}}$ color(blue)(sqrt"")

Thus, with ${\text{Co}}^{3 +}$, we have a ${d}^{6}$ metal here. Since ${\text{F}}^{-}$ 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?
(${\mu}_{S} = 2.0023 \sqrt{S \left(S + 1\right)}$, 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.