The equilibrium constant for the disproportionation of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ into ${\text{HgC}}{{\text{l}}^ + }$ and ${\text{HgCl}}_3^ - $ is ______
Given, ${\text{HgC}}{{\text{l}}^ + } + {\text{C}}{{\text{l}}^ - } \rightleftharpoons {\text{HgC}}{{\text{l}}_{\text{2}}};{\text{ }}{{\text{K}}_1} = 3 \times {10^6}$
${\text{HgC}}{{\text{l}}_{\text{2}}} + {\text{C}}{{\text{l}}^ - } \rightleftharpoons {\text{HgCl}}_3^ - ;{\text{ }}{{\text{K}}_2} = 9.0$
A.$27 \times {10^6}$
B.$3.3 \times {10^{ - 7}}$
C.$3.3 \times {10^{ - 6}}$
D.$3 \times {10^{ - 6}}$

Hint: We know that the reaction in which one reactant gets oxidized and the same reactant gets reduced is known as disproportionation reaction. The disproportionation reaction is also known as dismutation reaction. The ratio of the product of concentrations of products to the product of concentrations of reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation is known as the equilibrium constant. The equilibrium constant is denoted by ${\text{K}}$.

Complete step by step solution:
We know that the ratio of the product of concentrations of products to the product of concentrations of reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation is known as the equilibrium constant. The equilibrium constant is denoted by ${\text{K}}$.
We are given the first reaction as follows:
${\text{HgC}}{{\text{l}}^ + } + {\text{C}}{{\text{l}}^ - } \rightleftharpoons {\text{HgC}}{{\text{l}}_{\text{2}}}$
The expression for the equilibrium constant for the first reaction is as follows:
${{\text{K}}_{\text{1}}} = \dfrac{{\left[ {{\text{HgC}}{{\text{l}}_{\text{2}}}} \right]}}{{\left[ {{\text{HgC}}{{\text{l}}^ + }} \right]\left[ {{\text{C}}{{\text{l}}^ - }} \right]}}$ …… (1)
We are given the second reaction as follows:
${\text{HgC}}{{\text{l}}_{\text{2}}} + {\text{C}}{{\text{l}}^ - } \rightleftharpoons {\text{HgCl}}_3^ - $
The expression for the equilibrium constant for the second reaction is as follows:
${{\text{K}}_{\text{2}}} = \dfrac{{\left[ {{\text{HgCl}}_3^ - } \right]}}{{\left[ {{\text{HgC}}{{\text{l}}_{\text{2}}}} \right]\left[ {{\text{C}}{{\text{l}}^ - }} \right]}}$ …… (2)
We have to calculate the equilibrium constant for the disproportionation of ${\text{HgC}}{{\text{l}}_{\text{2}}}$. ${\text{HgC}}{{\text{l}}_{\text{2}}}$ disproportionates into ${\text{HgC}}{{\text{l}}^ + }$ and ${\text{HgCl}}_3^ - $. The disproportionation reaction of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ is as follows:
${\text{HgC}}{{\text{l}}_{\text{2}}} \rightleftharpoons {\text{HgC}}{{\text{l}}^ + } + {\text{HgCl}}_3^ - $
The expression for the equilibrium constant for the disproportionation reaction of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ is as follows:
${{\text{K}}_{\text{3}}} = \dfrac{{\left[ {{\text{HgC}}{{\text{l}}^ + }} \right]\left[ {{\text{HgCl}}_3^ - } \right]}}{{\left[ {{\text{HgC}}{{\text{l}}_{\text{2}}}} \right]}}$ …… (3)
Equation (3) is the ratio of equation (2) to equation (1). Thus,
${{\text{K}}_{\text{3}}} = \dfrac{{\dfrac{{\left[ {{\text{HgCl}}_3^ - } \right]}}{{\left[ {{\text{HgC}}{{\text{l}}_{\text{2}}}} \right]\left[ {{\text{C}}{{\text{l}}^ - }} \right]}}}}{{\dfrac{{\left[ {{\text{HgC}}{{\text{l}}_{\text{2}}}} \right]}}{{\left[ {{\text{HgC}}{{\text{l}}^ + }} \right]\left[ {{\text{C}}{{\text{l}}^ - }} \right]}}}} = \dfrac{{\left[ {{\text{HgC}}{{\text{l}}^ + }} \right]\left[ {{\text{HgCl}}_3^ - } \right]}}{{\left[ {{\text{HgC}}{{\text{l}}_{\text{2}}}} \right]}}$
Thus,
${{\text{K}}_{\text{3}}} = \dfrac{{{{\text{K}}_{\text{2}}}}}{{{{\text{K}}_{\text{1}}}}}$
Substitute ${{\text{K}}_2} = 9.0$ and ${{\text{K}}_1} = 3 \times {10^6}$. Thus,
${{\text{K}}_{\text{3}}} = \dfrac{{9.0}}{{3 \times {{10}^6}}}$
$\Rightarrow {{\text{K}}_{\text{3}}} = 3 \times {10^{ - 6}}$
Thus, the equilibrium constant for the disproportionation of ${\text{HgC}}{{\text{l}}_{\text{2}}}$ into ${\text{HgC}}{{\text{l}}^ + }$ and ${\text{HgCl}}_3^ - $ is $3 \times {10^{ - 6}}$.

Thus, the correct option is (D) $3 \times {10^{ - 6}}$.

Note: The reverse of disproportionation reaction is called comproportionation reaction. In a comproportionation reaction, a compound in an intermediate oxidation state is formed from the lower or higher oxidation state.