Assertion (A): CuI_2 cannot be prepared by th | vedclass

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(B) The reaction between $Cu^{2+}_{(aq)}$ and $I^{-}_{(aq)}$ leads to the reduction of $Cu^{2+}$ to $Cu^{+}$ and the oxidation of $I^{-}$ to $I_2$,for

Vedclass · Accepted Answer

Both $(A)$ and $(R)$ are correct but $(R)$ is not the correct explanation of $(A)$

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(B) The reaction between $Cu^{2+}_{(aq)}$ and $I^{-}_{(aq)}$ leads to the reduction of $Cu^{2+}$ to $Cu^{+}$ and the oxidation of $I^{-}$ to $I_2$,forming $CuI$ precipitate instead of $CuI_2$.The reaction is: $2Cu^{2+}_{(aq)} + 4I^{-}_{(aq)} \longrightarrow 2CuI_{(s)} + I_{2(s)}$.This occurs because $Cu^{2+}$ is a strong enough oxidizing agent to oxidize $I^{-}$ to $I_2$.Therefore,Assertion $(A)$ is correct.The reason $(R)$ stating that aqueous $Cu^{2+}$ is blue is a true statement regarding the color of the hydrated ion $[Cu(H_2O)_6]^{2+}$,but it does not explain why $CuI_2$ cannot be prepared.Thus,both $(A)$ and $(R)$ are correct,but $(R)$ is not the correct explanation of $(A)$.

Assertion $(A)$: $CuI_2$ cannot be prepared by the reaction of $Cu^{2+}_{(aq)}$ with $I^{-}_{(aq)}$.
Reason $(R)$: Aqueous $Cu^{2+}$ solution is blue in colour.
The correct answer is:

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Assertion $(A)$: $CuI_2$ cannot be prepared by the reaction of $Cu^{2+}_{(aq)}$ with $I^{-}_{(aq)}$.Reason $(R)$: Aqueous $Cu^{2+}$ solution is blue in colour.The correct answer is: