M^{n+}(aq.) + KI to Xdownarrow text{ (ppt.)} | vedclass

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(B) $1$. $Pb^{2+} + 2KI 	o PbI_2\downarrow$ (yellow ppt.). $PbI_2$ is soluble in excess $KI$ due to the formation of a soluble complex $[PbI_4]^{2-}$

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$Cu^{2+}(aq.)$

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(B) $1$. $Pb^{2+} + 2KI 	o PbI_2\downarrow$ (yellow ppt.). $PbI_2$ is soluble in excess $KI$ due to the formation of a soluble complex $[PbI_4]^{2-}$.$2$. $Cu^{2+} + 2KI 	o CuI_2 	o Cu_2I_2\downarrow + I_2$. $Cu_2I_2$ is insoluble in excess $KI$.$3$. $Bi^{3+} + 3KI 	o BiI_3\downarrow$ (black ppt.). $BiI_3$ is soluble in excess $KI$ due to the formation of $[BiI_4]^-$.$4$. $Hg^{2+} + 2KI 	o HgI_2\downarrow$ (red ppt.). $HgI_2$ is soluble in excess $KI$ due to the formation of $[HgI_4]^{2-}$.Therefore,$Cu^{2+}$ is the cation that forms a precipitate which remains insoluble in excess $KI$.

List-$I$ (Ion)	List-$II$ (Group Number in Cation Analysis)
$A. Co^{2+}$	$I. Group-I$
$B. Mg^{2+}$	$II. Group-III$
$C. Pb^{2+}$	$III. Group-IV$
$D. Al^{3+}$	$IV. Group-VI$

$M^{n+}(aq.) + KI \to X\downarrow \text{ (ppt.)}$ $\xrightarrow{Excess KI} \text{ppt. remains insoluble in excess } KI \text{ solution. Then cation } M^{n+}(aq.) \text{ can be}$

Similar Questions

Match List-$I$ with List-$II$:

List-$I$ (Ion) List-$II$ (Group Number in Cation Analysis)

$A. Co^{2+}$ $I. Group-I$

$B. Mg^{2+}$ $II. Group-III$

$C. Pb^{2+}$ $III. Group-IV$

$D. Al^{3+}$ $IV. Group-VI$

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