Question

According to VBT, the shape of ${[Co{F_6}]^{ - 3}}$ is:

Answer 1

Hint: Many theories have been proposed to clarify the existence of coordination compound bonding. The Valence Bond (VB) Theory is one of them. The Valence Bond Theory was created in order to use quantum mechanics to describe chemical bonding. During the formation of a molecule, this principle focuses on the formation of individual bonds from the atomic orbitals of the involved molecules.

Complete answer:
The valence bond principle states that:
Rather than molecular orbitals, electrons in a molecule occupy atomic orbitals. On the bond forming, the atomic orbitals overlap, and the greater the overlap, the stronger the bond.
For the complex ${[Co{F_6}]^{ - 3}}$, We will find the hybridisation using VBT ,
We know that ${F^ - }$ ions are weak ligands and the central metal ion is $C{o^{3 + }}$.
Cobalt is a member of the periodic table's section $VIII$. Its physical properties are identical to those of iron and nickel. Cobalt has the electronic structure $[Ar]4{s^2}3{d^7}$ .
Now, the oxidation number of $Co$ in the above complex is $ + 3$, that means 3 electrons are removed from the compound. Then the electronic configuration of this compound will be $[Ar]3{d^6}4{s^0}$ .
In, $3d$ orbital, there are $6$ electrons. $4s,4p,4d$ orbitals are vacant. Hence, $2 - 4s,3 - 4p,2 - 4d$ orbitals will pair with $6{F^ - }$ compounds.
Hence, the hybridisation is $s{p^3}{d^2}$.
As $C{o^{3 + }}$ is connected with $6{F^ - }$ ions from $6$ different sides , it is an octahedral complex.

Note:
In the complex, $3d$ contains $4$ unpaired electrons, hence, it is paramagnetic (The poles of a magnet attract very weakly, so it does not hold any permanent magnetism.). Other four unpaired electrons are a high spin complex. Since, $4d$ is involved in hybridisation it is an outer orbital complex.