Question

Depict high spin and low spin configurations for each of the following complexes. Tell whether each is diamagnetic or paramagnetic. Give the number of unpaired electrons of the paramagnetic complexes:
\[{\left[ {Cr{F_6}} \right]^{4 - }}\]

Answer 1

Hint: We must have to know that in the case of a paramagnetic compound, it should contain the unpaired electrons. And it has small, positive sensitivity to the magnetic field and they are very weak to attract the external magnetic field. But in the case of diamagnetic molecules, it will not have any unpaired electrons. And the diamagnetic atom will repel the magnetic field.

Complete answer:
We need to know that here, the central atom is chromium which is attached with six fluorine ligands. And the fluorine is a very weak field ligand.
In\[{\left[ {CrF6} \right]^{4 - }}\], the oxidation state of chromium is, \[ + 2\].
And the electronic configuration of \[C{r^{2 + }}\]is \[4{s^0}3{d^3}\].
We can draw the electronic configuration of $C{r^{2 + }}$ as,

And fluorine is a weak ligand. According to crystal field theory, the number of unpaired electrons present in\[{\left[ {Cr{F_6}} \right]^{4 - }}\]is equal to four. And it is paramagnetic in nature.

Note:
We must have to know that the crystal field theory shows the model for the interaction of bonding that takes place between the ligands and the transition metals. And there is an attraction between, negative charge of the non- bonding electron present in the ligand and positive charge of the metal cation. Here, the fluorine acts as the ligand and chromium is the metal cation. We have to remember that when the pairing energy is less than crystal field splitting energy, the electrons will fill the orbital with lower energy and after that only it moves to the higher orbitals.