An aqueous solution containing one mole of $Hg{I_2}$ and two moles of $NaI$ is orange in colour. In addition to excess $NaI$, the solution becomes colourless. The orange colour reappears on the subsequent addition of $NaOCl$.
If the statement is true enter 1 else enter 0.
Answer
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Hint:Sodium Iodide is an ionic compound formed by the chemical reaction of sodium metal and iodine comprising a 1:1 mix of sodium cations $\left( {N{a^ + }} \right)$ and iodide ions $\left( {{I^ - }} \right)$ in a crystal lattice. $NaI$ is a white odourless crystal or powder with a melting point of ${651^\circ }C$ and has a boiling point of ${1304^\circ }C$. $NaI$ is a deliquescent solid which absorbs moisture and turns into a solution. It is also a weak reducing agent. Sodium iodide is very soluble in water and some organic solvents.
Complete step by step answer:
As we know the colour of sodium iodide is white and the colour $Hg{I_2}$ is orange. If we prepare an aqueous solution containing one mole of $Hg{I_2}$ and two moles of $NaI$, then the orange colour compound is formed. If we add $NaI$ in excess, $Hg{I_2}$ is consumed completely, i.e. it is a limiting reagent. Hence the resulting solution becomes colourless due to formation of colourless $NaHg{I_4}$. The chemical reaction of $Hg{I_2}$ with excess of $NaI$ is given as:
$2NaI + Hg{I_2} \to NaHg{I_4}$
When we add sodium hypochlorite $NaOCl$ subsequently in the above solution, due to regeneration of $Hg{I_2}$, orange colour reappears. The chemical reaction for above is given below:
$N{a_2}Hg{I_4} + NaOCl + {H_2}O \to 2NaOH + NaCl + Hg{I_2} + {I_2}$
Hence the statement is true so press 1.
Note:
Sodium hypochlorite$\left( {NaOCl} \right)$ is a salt formed by the combination of negatively charged hypochlorite ion $(Cl{O^ - })$ and a positively charged sodium ion $\left( {N{a^ + }} \right)$. Hypochlorite decomposes to yield a variety of by-products in a solution. Hypochlorous acid which is a by-product of Hypochlorite acts as a powerful oxidizing agent and due to this hypochlorite exhibits the excellent ability of bleaching and disinfecting.
Complete step by step answer:
As we know the colour of sodium iodide is white and the colour $Hg{I_2}$ is orange. If we prepare an aqueous solution containing one mole of $Hg{I_2}$ and two moles of $NaI$, then the orange colour compound is formed. If we add $NaI$ in excess, $Hg{I_2}$ is consumed completely, i.e. it is a limiting reagent. Hence the resulting solution becomes colourless due to formation of colourless $NaHg{I_4}$. The chemical reaction of $Hg{I_2}$ with excess of $NaI$ is given as:
$2NaI + Hg{I_2} \to NaHg{I_4}$
When we add sodium hypochlorite $NaOCl$ subsequently in the above solution, due to regeneration of $Hg{I_2}$, orange colour reappears. The chemical reaction for above is given below:
$N{a_2}Hg{I_4} + NaOCl + {H_2}O \to 2NaOH + NaCl + Hg{I_2} + {I_2}$
Hence the statement is true so press 1.
Note:
Sodium hypochlorite$\left( {NaOCl} \right)$ is a salt formed by the combination of negatively charged hypochlorite ion $(Cl{O^ - })$ and a positively charged sodium ion $\left( {N{a^ + }} \right)$. Hypochlorite decomposes to yield a variety of by-products in a solution. Hypochlorous acid which is a by-product of Hypochlorite acts as a powerful oxidizing agent and due to this hypochlorite exhibits the excellent ability of bleaching and disinfecting.
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