(B) When $Cu^{2+}$ salts react with potassium iodide $(KI)$,the $Cu^{2+}$ ions are reduced to $Cu^{+}$ ions by the iodide ions $(I^{-})$.The chemical

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(B) When $Cu^{2+}$ salts react with potassium iodide $(KI)$,the $Cu^{2+}$ ions are reduced to $Cu^{+}$ ions by the iodide ions $(I^{-})$.The chemical reaction is as follows:$2Cu^{2+} + 4I^{-} \longrightarrow 2CuI(s) + I_{2}$Here,$CuI$ is formed as a white precipitate,and $I_{2}$ is liberated.Therefore,the correct product is $CuI$.