Choose the correct statement(s) among the following.

A solution of

P b C l_{2}

H C l

contains

P b^{2 +}

and

C l^{-}

ions.

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S n C l_{2} .2 H_{2} O

is a reducing agent.

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S n O_{2}

reacts with

K O H

to form

K_{2} [S n (O H)_{6}]

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The reaction of

P b_{3} O_{4}

with hot dilute nitric acid to give

P b O_{2}

is a redox reaction.

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Solution

The correct option is C $S n O_{2}$ reacts with $K O H$ to form $K_{2} [S n (O H)_{6}]$ .
Answer:A, B
a) $S n C l_{2} .2 H_{2} O$ is a reducing agent since $S n^{2 +}$ is lss stable and tends to convert into $S n^{4 +}$

(b) $S n O_{2}$ is amphoteric and it dissolves in $K O H$
$S n O_{2} + K O H \to K_{2} S n O_{3} + H_{2} O$ or amphoteric $K_{2} [S n (O H)_{6}]$

(c)First group cations( $P b^{2 +}$ ) form insoluble chloride with $H C l i . e . P b C l_{2}$ , however it is slightly soluble in water and therefore $P b^{2 +}$ ion is never completely precipitated on adding hydrochloric acid in test sample of Pb²⁺. Rest of the $P b^{2 +}$ ions are quantitatively precipitated with $H_{2} S$ in acidic medium.

So that we can say that filtrate of first group containing solution of $P b C l_{2}$ in $H C l$ contains $P b^{2 +}$ and $C l^{-}$ .

However, in the presence of conc. $H C l$ or excess $H C l$ , it can produce $H_{2} [P b C l_{4}]$ .

$P b C l_{2} + H C l \to H_{2} [P b C l_{4}]$

(d) $P b_{3} O_{4} o r 2 P b O . P b O_{2} + H N O_{3} \to P b O_{2} + 2 P b (N O_{3}) + 6 H_{2} O$
It is not a redox reaction because there is no change in oxidation state of $P b$ . The $P b$ in oxidation state of $+ 4$ remains unreactive and only $P b$ in oxidation state of $+ 2$ reacts but its oxidation number remains unchanged.
So, we can conclude A, B or A,B,C should be answers.

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