(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 $CuI$.

Class 12 Chemistry d-and f-Block Elements Chemical properties

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$Cu^{2+}$ salt reacts with potassium iodide to give :

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A
$Cu_{2}I_{3}$
B
$CuI$
C
$Cu_{2}I_{2}$
D
$Cu(I_{3})_{2}$

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Class 12

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d-and f-Block Elements

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Chemical properties

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The highest oxidation state of manganese in fluorides is $+4$ $(MnF_4)$, but the highest oxidation state in oxides is $+7$ $(Mn_2O_7)$, because

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The compound $AgF_2$ is an unstable compound. However,if formed,the compound acts as a very strong oxidizing agent. Why?

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Which of the following orders is not correct?

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The compound $AgF_2$ is an unstable compound. However,if formed,the compound acts as a very strong oxidizing agent. Why?

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For the four successive transition elements $(Cr, Mn, Fe, \text{ and } Co)$,the stability of $+2$ oxidation state will be in which of the following order?
(Atomic numbers: $Cr = 24, Mn = 25, Fe = 26, Co = 27$)

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$Cu^{2+}$ salt reacts with potassium iodide to give :

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The highest oxidation state of manganese in fluorides is $+4$ $(MnF_4)$, but the highest oxidation state in oxides is $+7$ $(Mn_2O_7)$, because

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For the four successive transition elements $(Cr, Mn, Fe, \text{ and } Co)$,the stability of $+2$ oxidation state will be in which of the following order?
(Atomic numbers: $Cr = 24, Mn = 25, Fe = 26, Co = 27$)