Question

When $HgCl_2$ and ${I_2}$ both dissolved in water containing ${I^ - }$ ions. Which of following pair of species will formed:
A.$H{g_2}{I_2},{I^ - }$
B.$Hg{I_2},I_3^ -$
C.$Hg{I_2},{I^ - }$
D.$HgI_4^{2 - },I_3^ -$

Answer 1

$HgC{l_2}$ when dissolved in water it will dissociate into ions so first of all make its ions as Ions will react with ${I^ - }$. In this case multiple reactions will occur in the solution as the Product formed will also react with iodide ions.

Complete step by step answer:
When $HgC{l_2}$ mix with water it will dissociate into two ions namely $H{g^{ + 2}}$ and $C{l^ - }$.
In the solution ${I^ - }$ ions are present as given in the question. So $H{g^{ + 2}}$ will react with iodide ion.
And the precipitate of $Hg{I_2}$ will form and $C{l^ - }$ left behind unreacted. So the reaction can be
Shown as;
$HgC{l_2}$ (aq) + $2{I^ - }$(aq)$\to$ $Hg{I_2}$ (s) + $2C{l^ - }$(aq)
But as per given in the solution iodide ion is in excess amount so it will further react with $Hg{I_2}$ and form a solution of $HgI_4^{2 - }$and the reaction can be depicted as;
$Hg{I_2}$ (s) + ${I^ - } \rightleftharpoons HgI_4^{2 - }$(soluble Complex)
${I_2}$ (iodine) will also react with iodide and form a water soluble compound that is $I_3^ -$.
${I_2}$ + ${I^ - }$ $\to I_3^ -$ (water soluble)
So mainly three products will form but as iodide ion is present in excess amount it will completely consume $Hg{I_2}$. Hence in the solution $Hg{I_2}$ will not be observed in this condition only $HgI_4^{2 - }$ and $I_3^ -$ can be observed. The formation constant for the $HgI_4^{ - 2}$ is far greater than $I_3^{ - 2}$. Therefore, ${I^ - }$ will preferentially combine with $HgC{l_2}$.

Note:
When $H{g^{ + 2}}$ combine with or react with iodide ion it will form a red coloured precipitate of $Hg{I_2}$ but if iodide ion present in excess amount it will further react with the red precipitate and form a water soluble complex.
Iodine can be used as a broad spectrum of antimicrobial activity.