Question

When KI (excess) is added to I : $CuS{{O}_{4}}$ II: $HgC{{l}_{2}}$ III: $Pb{{\left( N{{O}_{3}} \right)}_{2}}$
(A) a white ppt. of CuI in I, a orange ppt in $Hg{{I}_{2}}$in II and a yellow ppt. $Pb{{I}_{2}}$ in III
(B) a white ppt. of CuI in I, a orange ppt dissolves to $HgI_{4}^{2-}$ in II and a yellow ppt. $Pb{{I}_{2}}$ in III.
(C) a white ppt. of CuI $Hg{{I}_{2}}$ and $Pb{{I}_{2}}$ in each case
(D) None of the above is correct

Answer 1

When potassium iodide (KI) is added in excess to certain compounds, several iodide compounds are formed and each of this iodide gives a characteristic precipitate. By identifying this characteristic precipitate , we can identify the iodide formed.

Complete answer:
- Potassium iodide (KI) will react with copper sulphate($CuS{{O}_{4}}$) and will form a precipitate of CuI and the corresponding reaction can be represented as follows
$4KI+2CuS{{O}_{4}}\to C{{u}_{2}}{{I}_{2}}\downarrow +{{I}_{2}}+2{{K}_{2}}S{{O}_{4}}$
- The second given compound is $HgC{{l}_{2}}$.$HgC{{l}_{2}}$ will react with KI and will form an orange precipitate of $Hg{{I}_{2}}$ which can dissolve in excess of KI to give the Nessler’s solution and the corresponding reaction can be written as follows
$HgC{{l}_{2}}+2KCl\to \underset{orange}{\mathop{Hg{{I}_{2}}}}\,\downarrow +2KCl$
The excess of reaction with KI can be written as below
$Hg{{I}_{2}}+\underset{excess}{\mathop{KI}}\,\to \underset{nessler's\text{ }solution}{\mathop{{{K}_{2}}Hg{{I}_{4}}}}\,$
- The third compound is $Pb{{\left( N{{O}_{3}} \right)}_{2}}$. KI can react with $Pb{{\left( N{{O}_{3}} \right)}_{2}}$ to form an yellow precipitate of $Pb{{I}_{2}}$ and the reaction can be written as follows
$2KI+Pb{{\left( N{{O}_{3}} \right)}_{2}}\to \underset{yellow}{\mathop{Pb{{I}_{2}}}}\,+2KN{{O}_{3}}$
As we can see , the reactions of the three compounds with KI involves a white ppt. of CuI in I, a orange ppt dissolves to $HgI_{4}^{2-}$ in II and a yellow ppt. $Pb{{I}_{2}}$ in III.

Thus the answer is option (B).

Note: There might be a confusion with the reaction of KI with $HgC{{l}_{2}}$. In this reaction , only in the presence of excess potassium iodide, the Nessler's solution will be formed and otherwise an orange precipitate will be formed.