
A solid (A) which has photographic effect reacts with the solution of a sodium salt (B) to give a pale yellow ppt (C). Sodium salt on heating gives brown vapour. Identify A, B and C.
(A) $AgN{O_3},NaBr,AgBr$
(B) $AgN{O_3},NaCl,AgC{l_2}$
(C) $AgN{O_3},NaBr,AgC{l_2}$
(D) $AgCl,NaBr,AgB{r_2}$
Answer
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Hint:Photo means light. Hence, photographic effect is the effect shown by coloured compounds that can absorb light. Solve the question based on Identification of compounds on the basis of given characteristics of the compound and option elimination.
Complete Step by Step Solution:
Step\[1\]: Prepare equation on the basis of given equation
$A$ $ + $ $B$ $ \to $ C
(Solid having photographic effect) (Sodium Salt) (Pale yellow ppt)
$B$$\xrightarrow{{heating}}$brown vapours
Step $2$: option elimination
A shows photographic effect thus, it must be a coloured compound. $AgN{O_3}$ is a back coloured compound hence, it shows photographic effect. But $AgCl$is a white coloured compound not showing photographic effect. Thus, rule out option D. thus, A is $AgN{O_3}$
Step $3$: identification of compound on basis of characteristics of the compound.
$AgN{O_3}$reacts with sodium salt to give yellow precipitate. Let the sodium salt is of the form $NaX$.
$AgN{O_3} + NaX \to AgX + NaN{O_3}$
$NaN{O_3}$is not yellow in colour. Thus, $NaX$ must be of yellow colour.
The silver salt which is of yellow coloured precipitate is $AgBr$. Thus, C is $AgBr$.
So the sodium salt B is $NaBr$.
Thus, the correct option is A.
Our answers are correct or not can be proved by the second reaction
That is, $NaBr\xrightarrow{{heating}}B{r_2}$
Sodium bromide on heating releases bromine gas which is brown in colour. Hence the brown vapours observed are of bromine gas.
Note: $AgI$is also a deep yellow coloured precipitate but it is not the compound C . this is because if compound C is$AgI$then B must be $NaI$which on heating gives iodine vapours which are violet in colour. This is the importance of the second reaction which clears the doubt of the compound C.
Complete Step by Step Solution:
Step\[1\]: Prepare equation on the basis of given equation
$A$ $ + $ $B$ $ \to $ C
(Solid having photographic effect) (Sodium Salt) (Pale yellow ppt)
$B$$\xrightarrow{{heating}}$brown vapours
Step $2$: option elimination
A shows photographic effect thus, it must be a coloured compound. $AgN{O_3}$ is a back coloured compound hence, it shows photographic effect. But $AgCl$is a white coloured compound not showing photographic effect. Thus, rule out option D. thus, A is $AgN{O_3}$
Step $3$: identification of compound on basis of characteristics of the compound.
$AgN{O_3}$reacts with sodium salt to give yellow precipitate. Let the sodium salt is of the form $NaX$.
$AgN{O_3} + NaX \to AgX + NaN{O_3}$
$NaN{O_3}$is not yellow in colour. Thus, $NaX$ must be of yellow colour.
The silver salt which is of yellow coloured precipitate is $AgBr$. Thus, C is $AgBr$.
So the sodium salt B is $NaBr$.
Thus, the correct option is A.
Our answers are correct or not can be proved by the second reaction
That is, $NaBr\xrightarrow{{heating}}B{r_2}$
Sodium bromide on heating releases bromine gas which is brown in colour. Hence the brown vapours observed are of bromine gas.
Note: $AgI$is also a deep yellow coloured precipitate but it is not the compound C . this is because if compound C is$AgI$then B must be $NaI$which on heating gives iodine vapours which are violet in colour. This is the importance of the second reaction which clears the doubt of the compound C.
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