
Alkaline solution of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ and ${\text{KI}}$ is called ________.
Answer
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Hint: To solve this we must first write the reaction of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ and ${\text{KI}}$. In the reaction, a halide complex is formed. Name the complex using the rules for IUPAC nomenclature of coordination complexes.
Complete step by step answer:
The reaction of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ and ${\text{KI}}$ is as follows:
${\text{HgC}}{{\text{l}}_{\text{2}}} + {\text{4KI}} \to {{\text{K}}_{\text{2}}}\left[ {{\text{Hg}}{{\text{I}}_{\text{4}}}} \right]$
Thus, the reaction of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ and ${\text{KI}}$ produces a coordination complex ${{\text{K}}_{\text{2}}}\left[ {{\text{Hg}}{{\text{I}}_{\text{4}}}} \right]$.
Now we will name the coordination complex as per the rules of IUPAC nomenclature of coordination complexes:
The rules for writing the IUPAC nomenclature of complex are as follows:
> Identify the ligand and name the ligand with the prefix mono-, di-, tri-, tetra-, etc. depending on the number of ligands.
There are two ligands: The potassium ligand is outside the first coordination sphere and hence it is written outside the square parenthesis. Thus, we begin the name with potassium.
The second ligand is iodine and there are four iodine ligands. Thus, tetraiodo.
> Identify the central metal atom and write the name of the central metal atom ending with suffix –ate.
The central metal atom is mercury. Thus, mercurate.
> Determine the oxidation number of the central metal atom and write the charge in roman numeral in a bracket ( ) after writing the complete name.
Each iodine ligand has -1 charge and each potassium ligand has +1 charge. Thus, the charge on the central metal atom i.e. the mercury atom is +2 (II).
Thus, the name of the coordination complex ${{\text{K}}_{\text{2}}}\left[ {{\text{Hg}}{{\text{I}}_{\text{4}}}} \right]$ is potassium tetraiodomercurate (II).
Thus, the alkaline solution of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ and ${\text{KI}}$ is called potassium tetraiodomercurate (II) $\left( {{{\text{K}}_{\text{2}}}\left[ {{\text{Hg}}{{\text{I}}_{\text{4}}}} \right]} \right)$. It is also known as Nessler’s reagent.
Note: The Nessler’s reagent is mainly used for the determination of presence of ammonium compounds. In the test, the pale yellow solution of Nessler’s reagent becomes deep yellow in the presence of ammonia.
Complete step by step answer:
The reaction of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ and ${\text{KI}}$ is as follows:
${\text{HgC}}{{\text{l}}_{\text{2}}} + {\text{4KI}} \to {{\text{K}}_{\text{2}}}\left[ {{\text{Hg}}{{\text{I}}_{\text{4}}}} \right]$
Thus, the reaction of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ and ${\text{KI}}$ produces a coordination complex ${{\text{K}}_{\text{2}}}\left[ {{\text{Hg}}{{\text{I}}_{\text{4}}}} \right]$.
Now we will name the coordination complex as per the rules of IUPAC nomenclature of coordination complexes:
The rules for writing the IUPAC nomenclature of complex are as follows:
> Identify the ligand and name the ligand with the prefix mono-, di-, tri-, tetra-, etc. depending on the number of ligands.
There are two ligands: The potassium ligand is outside the first coordination sphere and hence it is written outside the square parenthesis. Thus, we begin the name with potassium.
The second ligand is iodine and there are four iodine ligands. Thus, tetraiodo.
> Identify the central metal atom and write the name of the central metal atom ending with suffix –ate.
The central metal atom is mercury. Thus, mercurate.
> Determine the oxidation number of the central metal atom and write the charge in roman numeral in a bracket ( ) after writing the complete name.
Each iodine ligand has -1 charge and each potassium ligand has +1 charge. Thus, the charge on the central metal atom i.e. the mercury atom is +2 (II).
Thus, the name of the coordination complex ${{\text{K}}_{\text{2}}}\left[ {{\text{Hg}}{{\text{I}}_{\text{4}}}} \right]$ is potassium tetraiodomercurate (II).
Thus, the alkaline solution of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ and ${\text{KI}}$ is called potassium tetraiodomercurate (II) $\left( {{{\text{K}}_{\text{2}}}\left[ {{\text{Hg}}{{\text{I}}_{\text{4}}}} \right]} \right)$. It is also known as Nessler’s reagent.
Note: The Nessler’s reagent is mainly used for the determination of presence of ammonium compounds. In the test, the pale yellow solution of Nessler’s reagent becomes deep yellow in the presence of ammonia.
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