1 Answer | Add Yours
First we have to identify the charge of the cobalt ion in an octahedral complex `[CoF_6]^(3-)` .
`x + (-1 *6) = -3`
`x = -3 + 6`
`x = + 3 = Co^(3+)`
Looking at the periodic table we can see that the element cobalt has a configuration of `4s^2 3d^7` and the cobalt (III) ion would be `3d^6` .
For octahedral complexes, the crystal field splitting diagram will have its `e_g` orbitals above the `t_(2g)` orbitals and is affected by the ligand that is attached to the cobalt ion.
F- is a weak ligand and therefore will have a high spin state. Consequently, the six electrons are distributed to each of the orbitals.
__ __ `e_g`
__ __ __ `t_(2g)`
--- --- `e_g`
`uarr darr` `uarr` `uarr`
---- --- --- `t_(2g)`
4 unpaired electrons
We’ve answered 330,809 questions. We can answer yours, too.Ask a question