Question

\[HgC{l_2}\]and ${I_2}$ both when dissolved in water containing \[{\rm{ }}{I^ - }\] ions the pair of species formed is:
A.\[H{g_2}{I_2},{\rm{ }}{I^ - }\]
B.\[Hg{I_2},{I^{3 - }}\]
C.\[Hg{I_2},{I^ - }\]
D.\[HgI_4^{2 - },{\rm{ }}{I_3}^ - \]

Answer 1

Hint: Mercuric chloride i.e. $HgC{l_2}$is a laboratory reagent found in white crystalline solid form. It is obtained by the reaction of chlorine and mercury/ mercury (I) chloride. Mercuric chloride is soluble in water.

Step by step answer: When $HgC{l_2}$ and ${I_2}$ both are dissolved in water containing \[{\rm{ }}{I^ - }\] ions the reaction will take place in the following manner;
-Mercuric chloride will react with the iodide ion. Here, the nucleophilicity of iodide ion is more than the chlorine, that is why the iodide ion will easily replace the chlorine ion and will lead to the formation of mercury (II) iodide i.e. $Hg{I_2}$.
\[{I_2} + {I^ - } \to I_3^ - \]
-Further, the covalent compound formed i.e. mercury (II) iodide will react with the iodide ion and will lead to the formation of a complex ion named tetraiodomercurate. This complex ion is a soluble complex.
\[Hg{I_2} + 2{I^ - } \to HgI_4^{2 - }\]
Now, we also have the iodine in this mixture. The iodine will react with the iodide ion and will form the polyhalide ion. This is a combination reaction.
\[{I_2} + {I^ - } \to I_3^ - \]

Hence, option (D) \[Hg{I_4}^{2 - },{\rm{ }}{I_3}^ - \], is the correct option.

Note: Thus, one should know that when mercuric chloride and iodine dissolved in water containing \[{\rm{ }}{I^ - }\] ions, then a pair of a soluble complex i.e. tetraiodide mercurate and a polyhalide ion i.e. triiodide ion will be obtained. The complex will be formed as a result of reaction between mercuric chloride and iodide ion, whereas the polyhalide ion will be formed as a result of a combination reaction between iodine and iodide ion.