Question

Write the hybridization and number of unpaired electrons in the complex ${\left[ {{{Co}}{{{F}}_6}} \right]^{3 - }}$
(Atomic number of ${{Co}}$$ = 27$ )

Answer 1

Hint: This question can be solved using valence bond theory. Using this theory we assume that the electrons are occupied in atomic orbitals of atoms and bonding occurs by overlapping of hybrid orbitals of individual atoms. The number of empty orbitals is the same as that of the coordination number.

Complete step by step solution:
Werner’s theory of coordination compounds couldn’t explain about the hybridization and magnetic properties of complexes. Thus valence bond theory was introduced. It was introduced to explain covalent bond formation quantitatively. There is an assumption that all bonds are localized bonds which are formed between two atoms by electron donation.
It could explain the electronic structure of central metal ions, hybridization, shape of complex, bonding representation and magnetic properties of complex.
In the complex, we have to find the oxidation number of cobalt first. The complex has a net charge $ - 3$ and fluorine has charge $ - 1$. Let the oxidation number of cobalt be ${{x}}$.
So by calculating the oxidation number, we get
${{x}} + - 6 = - 3 \Leftrightarrow {{x = + 3}}$
So cobalt has $ + 3$ oxidation state.
Atomic orbital electronic configuration of ${}^{27}{{Co}}$ is $\left[ {{{Ar}}} \right]3{{{d}}^7}4{{{s}}^2}$
Electronic configuration of ${{C}}{{{o}}^{3 + }}$ ion is $\left[ {{{Ar}}} \right]3{{{d}}^6}$
It may be represented as:

$ \uparrow \downarrow $

$ \uparrow $

$ \uparrow $

$ \uparrow $

$ \uparrow $

${{3d}}$ Excited state

Thus the number of unpaired electrons is $4$. Thus it is said to be paramagnetic. If it has all the electrons paired, then it is diamagnetic.
${{C}}{{{o}}^{3 + }}$ has vacant one ${{s}}$ and three ${{p}}$ orbital. So it forms high spin complexes.

Hence it has ${{s}}{{{p}}^3}{{{d}}^2}$ hybridization.

Note:
Fluorine is a weak ligand. Thus pairing does not occur in ${{3d}}$ orbital. Thus the complex formation involves ${{d - }}$ orbitals from which a high spin complex is obtained. The complex has octahedral shape.