Excess $\text{KI}$ reacts with ${\text{CuSO}}_4$ solution and then ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ solution is added to it. Which of the following statements is incorrect for this reaction?

When ${\text{CuSO}}_4$ reacts with excess $\text{KI}$, it first forms ${\text{CuI}}_2$, which is unstable and immediately decomposes to give ${\text{Cu}}_2{\text{I}}_2$ (white precipitate) and ${\text{I}}_2$.
$2{\text{CuSO}}_4+4\text{KI}\to 2{\text{K}}_2{\text{SO}}_4+{\text{Cu}}_2{\text{I}}_2+{\text{I}}_2\uparrow$.
Then, ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ reduces the evolved ${\text{I}}_2$:
${\text{I}}_2+2{\text{Na}}_2{\text{S}}_2{\text{O}}_3\to 2\text{NaI}+{\text{Na}}_2{\text{S}}_4{\text{O}}_6$.
Statement D is incorrect because ${\text{CuI}}_2$ is not the final product; it is an unstable intermediate that converts to ${\text{Cu}}_2{\text{I}}_2$.