Excess of KI reacts with CuSO₄ solution and then Na₂S₂O₃ solution is added to it. Which of the statements is incorrect for this reaction?

$2\stackrel{+2}{\mathrm{Cu}}{\mathrm{SO}}_4+4\mathrm{K}\stackrel{-1}{\mathrm{I}}\to {\mathrm{Cu}}_2{\mathrm{I}}_2+\stackrel{0}{{\mathrm{I}}_2}+2{\mathrm{K}}_2{\mathrm{SO}}_4$

Thus, CuI₂​ is not formed in this reaction.

The liberated iodine is titrated with sodium thiosulphate to form sodium tetrathionate.

$2{\mathrm{Na}}_2{\stackrel{+2}{\mathrm{S}}}_2{\mathrm{O}}_3+{\mathrm{I}}_2\to {\mathrm{Na}}_2{\stackrel{2.5}{\mathrm{S}}}_4{\mathrm{O}}_6+2\mathrm{NaI}$

Iodine is reduced and sodium thiosulphate  is oxidized.