Question

Among the following, the colored compound is:
A. $CuCl$
B. ${{\rm{K}}_{\rm{3}}}\left[ {{\rm{Cu}}{{\left( {{\rm{CN}}} \right)}_4}} \right]$
C. ${\rm{Cu}}{{\rm{F}}_{\rm{2}}}$
D. $\left[ {{\rm{Cu}}{{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CN}}} \right)}_4}} \right]{\rm{B}}{{\rm{F}}_{\rm{4}}}$

Answer 1

The crystal field theory explains that the colour of the coordination compounds is due to the transition of electrons from ${t_{2g}}$ level to ${e_g}$ level accompanying light absorption in the visible region.

Complete step by step answer:
We know that transition of electrons occurs when there are half filled electrons. If the electrons are completely filled in an orbital, no electronic transition takes place.
Here, all the compounds are compounds of copper. So, we have to find the oxidation states of copper in all the compounds. We know that the atomic number of copper is 29 and its electronic configuration is $\left[ {{\rm{Ar}}} \right]3{d^{10}}4{s^1}$. And copper possesses two oxidation states that is +1 and +2.
So, the configuration of ${\rm{C}}{{\rm{u}}^{ + 1}}$ is $\left[ {{\rm{Ar}}} \right]3{d^{10}}$. And this is a stable configuration and hence no electronic transition takes place. So, the compounds possessing ${\rm{C}}{{\rm{u}}^{ + 1}}$ show no colour.
Similarly the configuration of ${\rm{C}}{{\rm{u}}^{ + 2}}$ is $\left[ {{\rm{Ar}}} \right]3{d^9}$. This configuration is partially filled and hence undergoes electronic configuration. So compounds possessing ${\rm{C}}{{\rm{u}}^{ + 2}}$ are coloured compounds.
Now, we have to find the oxidation states of copper of all the compounds.
Option A is CuCl. We know that the oxidation state of chlorine is -1. We take the oxidation state of copper to be x. So,
$x - 1 = 0$
$\Rightarrow x = 1$
So, the oxidation state of copper in CuCl is +1. Therefore, it is not a colored compound.
Option B is ${{\rm{K}}_{\rm{3}}}\left[ {{\rm{Cu}}{{\left( {{\rm{CN}}} \right)}_4}} \right]$. We know oxidation states of K and CN are +1 and -1 respectively. We take the oxidation state of copper to be x. So,
$x - 4 = - 3$
$\Rightarrow x = 1$
So, oxidation state of Cu in ${{\rm{K}}_{\rm{3}}}\left[ {{\rm{Cu}}{{\left( {{\rm{CN}}} \right)}_4}} \right]$ is +1. Therefore, it is also not a colored compound.
Option C is ${\rm{Cu}}{{\rm{F}}_{\rm{2}}}$. We know the oxidation state of fluorine is -1. So,
$x - 2 = 0$
$\Rightarrow x = 2$
So, the oxidation state of copper in ${\rm{Cu}}{{\rm{F}}_{\rm{2}}}$ is +2. So, it is a colored compound.
Option D is $\left[ {{\rm{Cu}}{{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CN}}} \right)}_4}} \right]{\rm{B}}{{\rm{F}}_{\rm{4}}}$. We know that${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CN}}$ is a neutral ligand (oxidation state is zero) and oxidation state of ${\rm{B}}{{\rm{F}}_{\rm{4}}}$ is -1. So,
$x + 0 + \left( { - 1} \right) = 0$
$\Rightarrow x = 1$
So, the oxidation state of copper in $\left[ {{\rm{Cu}}{{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CN}}} \right)}_4}} \right]{\rm{B}}{{\rm{F}}_{\rm{4}}}$ is +1. Therefore, it is also not a colored compound.

So, the correct answer is Option C.

Note: Always remember that crystal field theory is a model that explains the bonding of transition metal complexes. This theory focuses on the energies of d-orbitals. Some assumptions of this theory are ligands are considered as negative charges and metal ligand bonding is entirely ionic.