Question

Which of the following pairs is predicted to exhibit the same color in solution?
A. \[VOC{l_2}:FeC{l_2}\]
B. \[CuC{l_2}:VOC{l_2}\]
C. \[MnC{l_2}:FeC{l_2}\]
D. \[FeC{l_2}:CuC{l_2}\]

Answer 1

Hint: We have to remember that the rationale for color in any metal compounds/ion is presence of free unpaired electrons in metal atom/ion. Check the number of unpaired electrons in each of the given solutions and decide the color. We must remember that the colour of components in d-block elements due to the presence of d-d transition.

Complete step by step answer:
From the given,
In case of option A, we have to remember that the ${V^{4 + }}$ in $VOC{l_2}$ have an unpaired electron so which gives blue colour whereas $F{e^{2 + }}$ in $FeC{l_2}$ have four unpaired electron which give greenish white colour in solution. From the above data the colour of this pair is differ from each other. Therefore, the option A is incorrect.
In case of option B, as we know that the $C{u^{2 + }}$ in $CuC{l_2}$ have one unpaired electron and which gives blue colour in solution whereas the ${V^{4 + }}$ in $VOC{l_2}$ have an unpaired electron so which also gives blue colour. From the above we can see the colour in the solution of both compounds is the same. Therefore, the option B is correct.
In case of option C, we have to remember that the $M{n^{2 + }}$ in $MnC{l_2}$have five unpaired electron and which gives pink colour in solution whereas the $C{u^{2 + }}$ in $CuC{l_2}$ have one unpaired electron and which gives blue colour in solution. From the above data the colour of this pair is differ from each other. Therefore, the option C is incorrect.
In case of option D, $F{e^{2 + }}$ in $FeC{l_2}$ have four unpaired electron which give greenish white colour in solution whereas $C{u^{2 + }}$ in $CuC{l_2}$ have one unpaired electron and which gives blue colour in solution. From the above data the colour of this pair is differ from each other. Therefore, the option D is incorrect.

So, the correct answer is Option B.

Note: As we know that there are specific colours of specific compounds that they are going to impart. The color of the compound will always be complementary to the color of the number of unpaired electrons present within the given compound. Sometimes, the color of the compound are often decided through the wavelength absorbed, it is the cation which imparts the colour to that compound.