Excess of $KI$ reacts with $CuS{O_4}$ solution and then $N{a_2}{S_2}{O_3}$ solution is added to it . Which of the following is correct for this reaction .
A. $C{u_2}{I_2}$ is formed
B. $Cu{I_2}$ is formed
C. $N{a_2}{S_2}{O_3}$ is oxidised
D. Evolved ${I_2}$ is reduced

Hint: The given question is an example of iodometry in which first an unknown excess of $KI$ is added to the weakly acidic solution and then the${I_2}$ which is released is titrated with a standard sodium thiosulphate solution .

Complete step by step answer: When we add excess of $KI$ to$CuS{O_4}$ , copper sulphate reacts with potassium iodide to form cuprous iodide and iodine .
The reaction which takes place is as follow :
$2CuS{O_4} + 4KI \to C{u_2}{I_2} \downarrow + {I_2} + 2{K_2}S{O_4}$
As we can see in the reaction \[C{u_2}{I_2}\] is formed and ${I_2}$ is evolved , it is clear from the above reaction that $Cu{I_2}$ is not formed .
Hence , Option B is not correct and Option A is correct .
Now , further it is given that $N{a_2}{S_2}{O_3}$ is added to it .
So , the liberated iodine reacts with sodium thiosulphate to form sodium tetrathionate .
The reaction which takes place is
$2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI$
From the above reaction we infer that iodine is reduced ( its oxidation state has increased from 0 to 1 ) , also sodium thiosulphate is oxidised .
Hence option C and option D are also correct as $N{a_2}{S_2}{O_3}$ is oxidised and evolved ${I_2}$ is reduced .

Note: The above reaction has to be done in a weak acid environment as sodium thiosulphate needs a neutral or weak acid environment to oxidise with tetrathionate because in strong acid environment thiosulphate decomposes to and in acidic environment the iodide is oxidised to iodine .