Question

HgCl₂ and I₂ both when dissolved in water containing I^– ions the pair of species formed is

• A. HgI₂, I₃^-
• B. HgI₂, I^-
• C. HgI₄^2-, I₃^-
• D. Hg₂I₂, I^-

Answer 1

The Correct Option is C

To determine the species formed when HgCl₂ and I₂ are dissolved in water containing I^– ions, let's examine the reactions step-by-step:

1. When Iodine (I₂) is added to a solution containing iodide ions (I^–), it forms the triiodide ion (I₃^–). This is given by the reversible reaction:
   I_2 + I^- \rightleftharpoons I_3^-
   In this reaction, I₂ acts as a Lewis acid by accepting an electron pair from I^–, forming the stable I₃^– complex.
2. When mercuric chloride (HgCl₂) is introduced to a solution with an excess of iodide ions, a series of reactions takes place:
   • First, mercury ions (Hg²⁺) react with two iodide ions to form mercurous iodide (HgI₂).
     Hg^{2+} + 2I^- \rightarrow HgI_2
   • However, in the presence of excess iodide ions, the HgI₂ further reacts to form a soluble complex ion:
     HgI_2 + 2I^- \rightarrow [HgI_4]^{2-}
     The [HgI₄]^2– is a stable, soluble complex.

Combining both results from the reactions above:

• The addition of I₂ to a solution containing I^– results in the formation of I₃^–.
• The addition of HgCl₂ to a solution with an excess of I^– results in the formation of the complex ion [HgI₄]^2–.

Therefore, when both HgCl₂ and I₂ are introduced to a solution containing I^– ions, the resulting species are [HgI₄]^2– and I₃^–.

Thus, the correct option is:

HgI₄^2-, I₃^-