Question

for the complex (CoF6)3- write the hybridization magnetic character and spin of the complex

Answer 1

A diamagnetic species has no unpaired electrons, while a paramagnetic species has one or more unpaired electrons.

Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it.

So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3−

complex ions contain the cobalt(III) cation, Co3+, which has the following electron configuration

Co3+:1s2 2s2 2p6 3s2 3p6 3d6

For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical serieswill determine how these d-orbtals will split.

More specifically, you can say that

a strong field ligand will produce a more significant splitting energy, Δ
a weak field ligand will produce a less significant splitting energy, Δ

Now, the spectrochemical series looks like 1st pic
it means flouride  is weak legend 


now see 2nd pic

In the case of the hexafluorocobaltate(III) ion, the splitting energy is smaller than the electron pairing energy, and so it is energetically favorable to promote two electrons from the t2g orbitals to the eg orbitals → a high spin complex will be formed.

This will ensure that the hexafluorocobaltate(III) ion will have 4 unpaired electrons, and thus be paramagnetic, octahedral hybridisation .

Answer 2

Hey dear,

● Answer -
1) Hybridisation => sp3d2
2) Magnetic character=> 3) paramagnetic
Spin => High spin

● Explanation -
- Co (Z=27) has electronic configuration [Ar] 3d7 4s2.
- In [(CoF6)3-] complex, Co donates 3 e- to form Co3+ ion to which 6 F- ligands surround.
- This complex has thus sp3d2 hybridization.
- As F- is strong ligand it forms high spin complexes.
- 4 free electrons are present in 3d orbitals of Co, thus it shows paramagnetic behavior.

Hope this helped you...