How to convert soluble colorless [HgI₄]²⁻ back to insoluble red HgI₂? [closed]








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When a solution contains $\ce{Hg^2+}$ and $\ce{I-}$, depend on the concentration ratio of $\ce{Hg^2+}$ and $\ce{I-}$, the following reactions can happen:


$$\ce{Hg^2+ + 2 I- ⟶ \underset{insoluble red color solid}{HgI2}}$$


If there is excess $\ce{I-}$, then this red solid will continue to react with excess $\ce{I-}$ to form an colorless soluble complex:


$$\ce{HgI2 + I- ⟶ [HgI3]^-}$$


$$\ce{[HgI3]^- + I- ⟶ [HgI4]^2-}$$


Is there a way to revert $\ce{[HgI4]^2-}$ back to $\ce{HgI2}$?




As noted in the comments, to attain mercuric iodide, you need to maintain a balance between the concentration of $\ce{Hg^2+}$ and iodide ion. If there is slight excess of iodide ion, it will form a complex (much like $\ce{KI3}$ from $\ce{KI}$). The complex is popularly known as Nessler's reagent and is prepared pretty much in this way.


Another solution is to decompose the complex to revert back into original products. But I won't recommend this solution. Though, it is expected to give the original reactants, occasionally, decomposition reactions give rise to undesired side products and [may release toxic fumes of mercury, iodine, iodide, potassium oxide($\ce{K2O)}$ etc.](https://www.chemicalbook.com/ChemicalProductProperty_EN_CB2481117.htm) **Do not try to heat the complex.** Instead opt for the 1st solution.


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