Question

Check whether the following statement is true or false.
${[CoC{l_4}]^{2 - }}$ (blue) and ${[Co{({H_2}O)_6}]^{2 + }}$ (pink) have different colors because ${\Delta _{tet}} < {\Delta _{oct}}$
1) True
2) False

Answer 1

Hint:This statement can be explained using the Crystal Field Theory or CFT. Using this theory, find the electron arrangement in the orbitals of the given compounds. Accordingly, we can relate the wavelength of the light absorbed to the energy required for the formation of these complexes.

Complete answer:According to the Crystal field Theory, the d-orbital degenerates into ${e_g}$ and ${t_{2g}}$ sets depending on the energy levels of that particular oribatal. The energy difference between these two sets is known as Crystal Field Splitting Energy or CFSE and is given as $\Delta $ . This value differs depending on the shape.
Firstly, we have ${[CoC{l_4}]^{2 - }}$ and its coordination number is $4$ so its shape could be either Tetrahedral or Square planar.
When it comes to ${[Co{({H_2}O)_6}]^{2 + }}$ , its coordination number is six so its shape would be Octahedral.
Now, we already know that the Crystal Field Splitting Energy is less for tetrahedral shaped compounds when compared to octahedral i.e. ${\Delta _{tet}} < {\Delta _{oct}}$
We can relate this energy directly to the wavelength absorbed as $\Delta = \dfrac{{hc}}{\lambda }$
Now, it is clear to us that the splitting energy is inversely proportional to the wavelength. We can substitute this equation in the given relation.
${\Delta _{tet}} < {\Delta _{oct}} \Rightarrow \dfrac{1}{{{\lambda _{tet}}}} < \dfrac{1}{{{\lambda _{oct}}}}$
By removing the inverse, we get ${\lambda _{oct}} < {\lambda _{tet}}$
Since ${[CoC{l_4}]^{2 - }}$ is tetrahedral and ${[Co{({H_2}O)_6}]^{2 + }}$ is octahedral, they absorb different wavelengths. Therefore, they emit different wavelengths and have different colors.

Hence the correct option is (1).

Note:Tetrahedral complexes usually take us less energy to form when compared to Octahedral. In this case, ${[CoC{l_4}]^{2 - }}$absorbs red so emits the complimentary color blue and therefore appears blue. Whereas ${[Co{({H_2}O)_6}]^{2 + }}$absorbs blue color and therefore appears to be pink.