CuSO 4 - KI - CuI 2- CuI 2 - Cu 2 I 2 - I 2- I 2 Na 2 S 2 O 3 - Na 2 S 4 O 6 - NaI In this reaction- Which of the following product does not exit-

Excess of KI reacts with $C u S O_{4}$ solution and then $N a_{2} S_{2} O_{3}$ solution is added to it. Which of the following statement is incorrect from this reaction

Q. In the reaction,

2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -}

, the eq. wt. of

{N a}_{2} S_{2} O_{3}

is equal to its:

Q. In the reaction,

2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{4}^{2 -} + 2 I^{-}

, the eq.wt of

{N a}_{2} S_{2} O_{3}

is equal to its:

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CuSO4 + KI → CuI2, CuI2 → Cu2I2 + I2, I2 Na2S2O3→ Na2S4O6 + NaI In this reaction, Which of the following product does not exit?

The correct option is A CuI2Iodine is a strong reducing agent.It reduces CuI2 which is highly unstable to CuI.Hence CuI2 does not exist.

${C u S O}_{4}$ + KI $\to$ ${C u I}_{2}$ , ${C u I}_{2}$ $\to$ ${C u}_{2} I_{2}$ + $I_{2}$ , $I_{2}$ ${N a}_{2} S_{2} O_{3} \to$ ${N a}_{2} S_{4} O_{6}$ + NaI
In this reaction, Which of the following product does not exit?

The correct option is A ${C u I}_{2}$
Iodine is a strong reducing agent.
It reduces $C u I_{2}$ which is highly unstable to $C u I$ .
Hence $C u I_{2}$ does not exist.