
Silver, mercury and lead are grouped together in a scheme of qualitative analysis because they form:
A) Nitrates
B) Carbonates which dissolve in dil. $HN{O_3}$
C) Insoluble chlorides
D) Colourless compound
Answer
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Hint: As we know that silver, mercury and lead are placed in the sixth period of the periodic table and the qualitative analysis of is a common method in analytical chemistry to find the detection of ions present in aqueous solution.
Complete solution:
As we know that all the given metals are grouped together in qualitative analysis schemes because they are the metal cations which have the ability to form insoluble precipitates with chlorides.
As we all know that silver reacts with chlorine to produce silver chloride where silver is present in $ + 1$ oxidation state and the product so formed is insoluble in water as well as other solvents like alcohols and dilute acids. We can show this reaction with the help of a chemical equation as shown below:
$2Ag + C{l_2} \to 2AgCl$
Similarly, when chlorine reacts with mercury it forms mercurous chloride which is again insoluble in various solvents including water. We can show this reaction with the help of a chemical equation as follows:
$2Hg + C{l_2} \to H{g_2}C{l_2}$
Lastly we have lead and like silver and mercury, lead also forms an insoluble complex on reaction with chlorine. We can show this reaction using the chemical equation given below:
$Pb + C{l_2} \to PbC{l_2}$
Therefore, from the above explanation we can say that the correct answer is (C).
Note:Remember that the forces which hold the solid silver chloride lattice are very strong and thus it does not easily dissociate into its ions once this compound is formed. Similar is the case with mercury and lead, all of these metals are insoluble in water. Mercury compounds are slightly soluble in water and they are also toxic in nature.
Complete solution:
As we know that all the given metals are grouped together in qualitative analysis schemes because they are the metal cations which have the ability to form insoluble precipitates with chlorides.
As we all know that silver reacts with chlorine to produce silver chloride where silver is present in $ + 1$ oxidation state and the product so formed is insoluble in water as well as other solvents like alcohols and dilute acids. We can show this reaction with the help of a chemical equation as shown below:
$2Ag + C{l_2} \to 2AgCl$
Similarly, when chlorine reacts with mercury it forms mercurous chloride which is again insoluble in various solvents including water. We can show this reaction with the help of a chemical equation as follows:
$2Hg + C{l_2} \to H{g_2}C{l_2}$
Lastly we have lead and like silver and mercury, lead also forms an insoluble complex on reaction with chlorine. We can show this reaction using the chemical equation given below:
$Pb + C{l_2} \to PbC{l_2}$
Therefore, from the above explanation we can say that the correct answer is (C).
Note:Remember that the forces which hold the solid silver chloride lattice are very strong and thus it does not easily dissociate into its ions once this compound is formed. Similar is the case with mercury and lead, all of these metals are insoluble in water. Mercury compounds are slightly soluble in water and they are also toxic in nature.
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