Question

When aqueous solution of potassium fluoride is added to the blue coloured aqueous $CuS{O_4}$ solution, a green precipitate is formed. This observation can be explained as follows:
(A) On adding KF, ${H_2}O$ being weak field ligand is replaced by ${F^ - }$ ions forming ${\left[ {Cu{F_4}} \right]^{2 - }}$ which is green in colour.
(B) Potassium is coordinated to ${\left[ {Cu{{({H_2}O)}_4}} \right]^{ + 2}}$ ion present in $CuS{O_4}$ and gives green colour.
(C) On adding KF, $C{u^{ + 2}}$ are replaced by ${K^ + }$ forming a green complex.
(D) Blue colour of $CuS{O_4}$ and yellow colour of KI form green colour on mixing.

Answer 1

First write down the reactions involved and identify what whether ${H_2}O$ is replaced or not and if it is replaced then does it act like a weak field ligand. Then check for the product and its ionic form because ions are the ones which are responsible for colour of any solution.

Complete step by step answer:
-First of all, let us talk about aqueous copper sulphate solution and the way to represent it.
Aqueous copper sulphate solution: ${\left[ {Cu{{({H_2}O)}_4}} \right]^{ + 2}} + SO_4^{ - 2}$
Or we can write it as a complex: $\left[ {Cu{{({H_2}O)}_4}} \right]S{O_4}$

-And, as it is given in the above reaction that the aqueous copper sulphate solution is reacting with potassium fluoride. So, the reaction can be written as follows:
${\left[ {Cu{{({H_2}O)}_4}} \right]^{ + 2}} + KF \to {K_2}\left[ {Cu{F_4}} \right]$
So, in the above reaction we can see that on reacting potassium fluoride with copper sulphate solution there is formation of potassium tetra fluoro cuprate.
From the above reaction we can say that ${H_2}O$ is a weak field ligand and addition of KF to it replaces the weak field ligand ${H_2}O$.

-Aqueous copper can also be present as $CuS{O_4}.5{H_2}O$. And so on addition of KF and the replacement of ${H_2}O$ by the ${F^ - }$ ions can also be written in the form of following reaction:
$CuS{O_4}.5{H_2}O + 4KF \to 4{K^ + } + {\left[ {Cu{F_4}} \right]^{ - 2}} + SO_4^{ - 2} + 5{H_2}O$
From this reaction we can see that potassium tetra fluoro cuprate will be present in dissociated form in aqueous medium. So, It can be written as follows:
${K_2}\left[ {Cu{F_4}} \right]\xrightarrow{{Aq.medium}}{K^{ + 2}} + CuF_4^{ - 2}$

From the above reaction we can say that there is formation of green precipitate due to the formation of the ${\left[ {Cu{F_4}} \right]^{ - 2}}$. This given complex is green in colour.
-From the above discussion and observing the reaction we can say that on adding KF, ${H_2}O$ being weak field ligand is replaced by ${F^ - }$ ions forming ${\left[ {Cu{F_4}} \right]^{2 - }}$ which is green in colour. So, the correct answer is “Option A”.

Note: Let us know about the reason for coloured complexes. We should note that the colours that are obtained by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a d–d transition. Or we can say that when an electron jumps from lower energy level to higher energy level some amount of energy is absorbed. This energy corresponds to the frequency of light observed, which mostly lies in the visible region.