Question

An aqueous solution containing one mole of $HgI_2$ and two moles of $NaI$ is orange in colour. On addition of excess $NaI$ the solution becomes colourless. The orange colour reappears on subsequent addition of $NaOCl$. Explain with equations.

Answer 1

Nal on reaction with $Hgl _{2}$ gives a complex salt :
$2 NaI + Hgl _{2} \rightleftharpoons Na _{2}\left[ Hgl _{4}\right]$
The orange colour is due to residual $HgI _{2}$. On addition of excess $NaI$, whole $Hgl _{2}$ is converted to complex $Na _{2}\left[ HgI _{4}\right]$ as colour disappear. The orange colour of $HgI _{2}$ reappear duc to conversion of $Na _{2}\left[ HgI _{4}\right]$ into $HgI _{2}$ on treatment with $NaOCl$.
$3 Na _{2}\left[ Hgl _{4}\right]+2 NaOCl +2 H _{2} O \longrightarrow 3 HgI _{2} +2 NaCl $
$+4 NaOH +2 NaI$