P-Block Elements Ques 616
Question: Excess of KI reacts with $ CuSO_4 $ solution and then $ Na_2S_2O_3 $ solution is added to it. Which of the statements is incorrect for this reaction?
Options:
A) $ Na_2S_2O_3 $ is oxidized
B) $ CuI_2 $ is formed
C) $ Cu_2I_2 $ is formed
D) Evolved $ I_2 $ is reduced
Show Answer
Answer:
Correct Answer: B
Solution:
$ 4\overset{-1}{\mathop{KI}}+2CuSO_4\to \overset{0}{\mathop{I_2}}+Cu_2I_2+2K_2SO_4 $ $ {{\overset{0}{\mathop{I}}}_2}+2\overset{2+}{\mathop{Na_2S_2O_3}}\to \overset{-2.5}{\mathop{Na_2S_4O_6}}+2\overset{-1}{\mathop{NaI}} $ In this $ CuI_2 $ is not formed.
