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 equation.

Explanation

Sodium iodide on reaction with `Hgl_2` gives complex salt, `Na[HgI_4]`

Orange colour is due to presence of residual `Hgl_2`.
But on addition of excess NaI, it becomes colourless due to change of `Hgl_2` into `Na[HgI_4]`
`HgI_2+2NaI` (excess) `toNa_2[HgI_4]` (colourless)
The orange colour of `HgI_2` reappears due to conversion of `Na_2[HgI_4]` into `HgI_2` by means of `NaOCl`.
`3Na_2[HgI_4]+2NaOCl+2H_2Oto`
`3HgI_2+2NaCl+4NaOH+2NaI_3`