The Question is from neet-chemistry classification-of-elements , and it was created by the author Mr. S M HUSSAIN

Question

${\rm{HgC}}{{\rm{l}}_{\rm{2}}}{\rm{\;and \;}}{I_2}$ both when dissolved in water containing ${I^ - }$ ions the pair of species formed is

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
**Primary Concept**: Complex Ion Formation and Triiodide Ion Formation
**Related Topics**: Chemical Equilibrium, Coordination Chemistry
**Prerequisites**: Understanding of ionic compounds, complex ions, and equilibrium reactions
## SOLUTION PATHWAY
**Given Information**:
    -  ${m{HgCl}}_2$ and $I_2$ are dissolved in water containing $I^-$ ions.
    -  We need to determine the species formed.
**Method & Steps**:
    1.  Reaction of ${m{HgCl}}_2$ with excess $I^-$ ions:
        Reasoning:  Mercury(II) halides tend to form complex ions with halide ions in solution. 
        Formula/Rule:  ${m{Hg}}^{2+} + 4I^- ightleftharpoons {m{HgI}}_4^{2-}$ 
    2.  Reaction of $I_2$ with $I^-$ ions:
        Reasoning: Iodine molecules react with iodide ions to form the triiodide ion.
        Formula/Rule:  $I_2 + I^- ightleftharpoons I_3^-$
    3.  Combining the two reactions:
        Working:  Both reactions occur simultaneously in the solution.
        Key Point: The presence of excess $I^-$ drives the formation of ${m{HgI}}_4^{2-}$ and $I_3^-$.
    4.  Final Result:
        Result: The species formed are ${m{HgI}}_4^{2-}$ and $I_3^-$.
        Verification:  This result aligns with the known behavior of mercury(II) ions and iodine in the presence of excess iodide.
 ## OPTION ANALYSIS
[A] :x: ${m{Hg}}_2{m{I}}_2$ is incorrect since ${m{HgCl}}_2$ forms ${m{HgI}}_4^{2-}$ complex with excess iodide. $I^-$ alone cannot be the product of $I_2$ in presence of excess $I^-$
[B] :x: ${m{HgI}}_2$ is incorrect since ${m{HgCl}}_2$ forms ${m{HgI}}_4^{2-}$ complex with excess iodide.
[C] :x: ${m{HgI}}_2$ is incorrect since ${m{HgCl}}_2$ forms ${m{HgI}}_4^{2-}$ complex with excess iodide. $I^-$ alone cannot be the product of $I_2$ in presence of excess $I^-$
[D] ✓: ${m{HgI}}_4^{2-}$ is the complex ion formed by the reaction of ${m{HgCl}}_2$ with excess $I^-$, and $I_3^-$ is formed by the reaction of $I_2$ with $I^-$.
 ## LEARNING AIDS
**Common Mistakes**:
    1.  Error 1: Forgetting that mercury(II) forms complex ions with excess halides. How to avoid:  Remember the coordination chemistry of transition metals.
    2. Error 2: Not recognizing the triiodide ion formation. How to avoid:  Be familiar with the reaction of halogens with their respective halide ions.
**Quick Tips**:
    1.  Complex ion formation is favored by excess ligand concentration.
    2.  The triiodide ion ($I_3^-$) is a common polyatomic ion formed by iodine.

Mr. S M HUSSAIN · Answer

${\rm{H}}{{\rm{g}}_2}{{\rm{I}}_{\rm{2}}}{\rm{,}}\,{{\rm{I}}^ - }$

Mr. S M HUSSAIN · Answer

${\rm{Hg}}{{\rm{I}}_{\rm{2}}}{\rm{,}}\,{\rm{I}}_3^ - $

Mr. S M HUSSAIN · Answer

${\rm{Hg}}{{\rm{I}}_{\rm{2}}}{\rm{,}}\,{{\rm{I}}^ - }$

Mr. S M HUSSAIN · Answer

The correct Option is "${\rm{HgI}}_4^{2 - }{\rm{,}}\,{\rm{I}}_3^ - $"

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