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Hint: Separation of precipitates of $\text{AgCl}$ and $\text{PbC}{{\text{l}}_{2}}$ can take place, when one is soluble in one solvent but the other is compound is not soluble in that solvent.
Complete step by step answer:
$\text{PbC}{{\text{l}}_{2}}$ is an inorganic compound with lead chloride as its name. This compound is white solid. This compound is white in colour because $\text{P}{{\text{b}}^{+2}}$ has no unpaired electrons present in it, to this compound colour. The configuration of $\text{P}{{\text{b}}^{+2}}$ is $\left[ \text{Xe} \right]6{{\text{s}}^{2}}\text{4}{{\text{f}}^{14}}\text{5}{{\text{d}}^{10}}6{{\text{p}}^{0}}$.
$\text{AgCl}$ has its name as silver chloride. It is a crystalline solid and has poor solubility in water. $\text{AgCl}$ is white in colour because it too has no unpaired electrons present in it. The electronic configuration of $\text{A}{{\text{g}}^{+}}$ is $\left[ \text{Kr} \right]4{{\text{d}}^{10}}\text{5}{{\text{s}}^{0}}$.
Let us see the solubility of both the compounds with hot water and reaction with aqueous ammonia to differentiate between the two:
So, aqueous ammonia and hot water can be used to separate each other’s white precipitates. The correct answer is option ‘c’ both a and b.
Note: $\text{A}{{\text{g}}^{+}}$ and $\text{P}{{\text{b}}^{+2}}$ are the cations of Group (I). The $\text{PbC}{{\text{l}}_{2}}$ compound is not soluble in cold water as lattice energy is greater than interaction energy of molecules to dissolve in the solvent. It will dissolve but very slightly. On increasing the temperature, lattice energy is made to be overcome and show dissolution in water.
Complete step by step answer:
$\text{PbC}{{\text{l}}_{2}}$ is an inorganic compound with lead chloride as its name. This compound is white solid. This compound is white in colour because $\text{P}{{\text{b}}^{+2}}$ has no unpaired electrons present in it, to this compound colour. The configuration of $\text{P}{{\text{b}}^{+2}}$ is $\left[ \text{Xe} \right]6{{\text{s}}^{2}}\text{4}{{\text{f}}^{14}}\text{5}{{\text{d}}^{10}}6{{\text{p}}^{0}}$.
$\text{AgCl}$ has its name as silver chloride. It is a crystalline solid and has poor solubility in water. $\text{AgCl}$ is white in colour because it too has no unpaired electrons present in it. The electronic configuration of $\text{A}{{\text{g}}^{+}}$ is $\left[ \text{Kr} \right]4{{\text{d}}^{10}}\text{5}{{\text{s}}^{0}}$.
Let us see the solubility of both the compounds with hot water and reaction with aqueous ammonia to differentiate between the two:
S. No. | Name of compound | Solubility in hot water | Reaction with aqueous ammonia or $\text{N}{{\text{H}}_{4}}\text{OH}$ | Reason for ‘no’ reaction |
1. | Lead chloride or $\text{PbC}{{\text{l}}_{2}}$ | $\text{PbC}{{\text{l}}_{2}}$ is seems to be made up of ions. As, it readily dissolves in hot water to give a colourless solution. | Shows no reaction with aqueous ammonia. | $\text{PbC}{{\text{l}}_{2}}$ does not react as basic salt is insoluble in excess ammonia. |
2. | Silver chloride or $\text{AgCl}$ | $\text{AgCl}$ does not react with hot water. | Two molecules of ammonia form coordination compound with silver chloride to form diamine complex,$\text{AgCl}+2\text{N}{{\text{H}}_{3}}\to \left[ \text{Ag}{{\left( \text{N}{{\text{H}}_{3}} \right)}_{2}} \right]\text{Cl}$ | Solubility of $\text{AgCl}$ increases very little with temperature range. Hence, insoluble. |
So, aqueous ammonia and hot water can be used to separate each other’s white precipitates. The correct answer is option ‘c’ both a and b.
Note: $\text{A}{{\text{g}}^{+}}$ and $\text{P}{{\text{b}}^{+2}}$ are the cations of Group (I). The $\text{PbC}{{\text{l}}_{2}}$ compound is not soluble in cold water as lattice energy is greater than interaction energy of molecules to dissolve in the solvent. It will dissolve but very slightly. On increasing the temperature, lattice energy is made to be overcome and show dissolution in water.
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