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Reducing Nature (Group 2)

A solution co...

Question

A solution contains ${C l}^{-}, I^{-}$ and $S O_{4}^{2 -}$ ions in it. Which of the following ion is capable to precipitate all of above when added in this solution?

{P b}^{2 +}

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{B a}^{2 +}

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{H g}^{2 +}

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{C u}^{2 +}

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Solution

The correct option is A ${P b}^{2 +}$
$P r e c i p i t a t i o n o r f o r m a t i o n o f p r e c i p i t a t e (p p t .)$ is a process in which solute particles find each other and form a insoluble solid together. Precipitation and solubility are the opposite phenomenons. So, to know weather a compound will form a ppt. or not its solubility must be known. If the compound is insoluble in nature it will form ppt. and vice versa.

Generally, the chlorides, bromides, and iodides of all metals except lead, silver, and mercury(I) are soluble in water. $H g I_{2}$ is insoluble in water. $P b C l_{2}, P b B r_{2}, a n d P b I_{2}$ are soluble in hot water.
The sulfates of all metals except lead, mercury (I), barium, and calcium are soluble in water.

Reaction involved are:

$P b^{2 +} + 2 C l^{-} ⟶ P b C l_{2} ↓$
$P b^{2 +} + 2 I^{-} ⟶ P b I_{2} ↓$
$P b^{2 +} + S O_{4}^{2 -} ⟶ P b S O_{4} ↓$

Barium forms precipitate with sulphate ion only, $B a C l_{2}$ and $B a I_{2}$ are soluble, hence they won't form ppt.-
$B a^{2 +} + S O_{4}^{2 -} ⟶ B a S O_{4} ↓$

$H g C l_{2}$ is soluble in nature. However, $H g I_{2}$ is an insoluble compound.
$H g^{2 +} + 2 I^{-} ⟶ H g I_{2} ↓$

$C u I_{2}$ is not a stable compound. It immediately decomposes to insoluble copper (I) iodide (CuI) , releasing $I_{2}$ . However, $C u C l_{2} a n d C u S O_{4}$ are soluble in nature. Hence, they won't form precipitate.

_{Reaction involved:}
_{$C u^{2 +} + 2 I^{-} ⟶ C u I_{2}$}
_{$2 C u I_{2} ⟶ 2 C u I + I_{2}$}

$P b^{2 +}$ is the only ion to form ppt. with all the given ions.
Hence, the correct option is (A).

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