When $Cu^{2+}$ ion is treated with $KI$,a white precipitate is formed. Explain the reaction with the help of a chemical equation.

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(N/A) When $Cu^{2+}$ ions are treated with $KI$,the $Cu^{2+}$ ions are reduced to $Cu^{+}$ ions by iodide ions $(I^-)$.
The resulting $Cu^{+}$ ions react with the remaining iodide ions to form a white precipitate of copper$(I)$ iodide $(Cu_2I_2)$.
The balanced chemical equation for this reaction is:
$2Cu^{2+}_{(aq)} + 4I^{-}_{(aq)} \rightarrow Cu_2I_{2(s)} + I_{2(s)}$

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$C$. Aqueous solution of $Cr^{2+}$ can liberate hydrogen from dilute acid.
$D$. Magnetic moment of $V^{2+}$ is observed between $4.4-5.2 \ BM$.
Choose the correct answer from the options given below:

Which of the following ions can reduce dilute acid to give hydrogen gas?

$E^{\circ}_{M^{3+} \mid M^{2+}} \text{ (in } V\text{) is highest for}$

Which of these statements is not true?

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