Question

Which is a high spin complex?
A) ${{\left[ Co{{F}_{6}} \right]}^{3-}}$
B) ${{\left[ Fe{{(CN)}_{6}} \right]}^{3-}}$
C) ${{\left[ Fe{{(CN)}_{6}} \right]}^{4-}}$
D) None of these

Answer 1

Hint:A high spin complex refers to the complex in which the d-orbitals of the central metal atom is having greater number of unpaired electrons and it is called so, as each electrons possess a spin and each electron can be considered as a spin.

Complete answer:
 In the question we are provided with four complexes and we have to say which complex is having the highest spin or comes under the classification of high spin complexes.
Before solving the question, we should have an idea about what a high spin complex and low spin complex refers to.
High spin-complex- is the complex which possesses a greater number of unpaired electrons in the d-orbitals of the central metal atom and it will be paramagnetic in nature.
Low-spin complexes-is the complex which does not possess any unpaired electron in the d-orbitals of the central metal atom of the complex and will be diamagnetic in nature.
Now let’s compare the options given,
Option (A)- ${{\left[ Co{{F}_{6}} \right]}^{3-}}$
 In this complex cobalt is the central metal atom and Co possess 27 electrons ,since its atomic number is 27 and Co is in +3 oxidation state ,so the electronic configuration will be,
\[C{{o}^{+3}}=\left[ Ar \right]3{{d}^{6}}4{{s}^{0}}\]
When the d-orbitals splits into ${{t}_{2g}}$ and ${{e}_{g}}$ in the presence of ligand, the electrons are filled according to the Hund's rule and as the ligand approaching here is Fluorine, which is a weak ligand.
As the ligand is a weak ligand the splitting of d-orbitals will be less .So the splitting energy is less than the pairing energy of the orbitals and the electrons will be filled as, there will be 4 electrons in ${{t}_{2g}}$ and 2 electrons in ${{e}_{g}}$.So there will be in total four unpaired electrons.
So the complex is a high spin complex and the hybridization of ${{\left[ Co{{F}_{6}} \right]}^{3-}}$ is $s{{p}^{3}}{{d}^{2}}$.
If we compare the options ${{\left[ Fe{{(CN)}_{6}} \right]}^{3-}}$ and${{\left[ Fe{{(CN)}_{6}} \right]}^{4-}}$, both are having the same ligand $C{{N}^{-}}$.
$C{{N}^{-}}$ Is one of the strongest ligands in the spectrochemical series and in these complexes the splitting of the d-orbitals is more. Hence splitting energy will be more than the pairing energy and the electrons will pair up and form low spin complexes which are diamagnetic in nature. Both the complexes are having the hybridization ${{d}^{2}}s{{p}^{3}}$.

So the correct option for this question is option (A).

Note:
The high spin complex is also called the outer orbital complex, and low spin complex is called the inner orbital complex. It is called so, since in the hybridization of outer orbital complexes the d-orbital involved is the d-orbitals of the outermost shell whereas in the inner orbital complex, the inner d-orbitals are involved in the hybridization.