Question

Consider the following reactions and find the correct option(s).

• A. B is $CuI$ and the colour is due to the presence of impurity
• B. A is $C{u}_2{I}_2$
• C. Central atom of anion part of C has linear geometry due to $sp$ hybridization
• D. Compound B and compound D are same

Answer 1

Answer: (A), (D)

The correct options are
A B is $CuI$ and the colour is due to the presence of impurity
D Compound B and compound D are same

This is not a simple reaction but a mixture of processes taking place in the same reaction vessel.
First step:
$CuS{O}_4+2KI\to \underset{A}{Cu{I}_2}+K_{2}S{O}_4$
Second step:
$\underset{A}{2Cu{I}_2}\to \underset{B}{C{u}_2{I}_2}+I_2$
The precipitate appears to be a brown sludge because of the iodine. Iodine is sparingly soluble in water but it dissolves in the presence of $I^{-}$ ions to form a dark yellow/brown solution due to the formation of a complex ion. There is almost certainly some $KI$ left to achieve this.

Third step:
$I_2+KI\to K{I}_3$
If thiosulfate solution is used, then the Iodine is re-reduced back to Iodide and the brown discolouration disappears leaving the white precipitate of $CuI$.
$K{I}_3+2N{a}_2{S}_2{O}_3\to KI+2NaI+N{a}_2{S}_4{O}_6$
All of the processes 1, 2 and 3 are taking place simultaneously. The thiosulfate is just an added part to help illustrate the chemistry and, if necessary, help to obtain a clean sample of CuI.