Answer
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Hint: Approach this question by relating the given properties of the unknown compounds in the question with the inorganic compounds which have either the same cation or anion from the known compounds. Also colours of the compound can help in the prediction of the reaction and compound formed.
Complete step by step solution: Let us start with the first reaction as follows:-
-As it is given that ‘P’ is a metal nitrate compound that reacts with sodium bromide solution to give a compound ‘Q’ which is used in photography. So it is sure that the product formed will also be a nitrate and a bromide compound because metal nitrate shows displacement reaction with other ionic compounds.
-Since no nitrate compound is used in photography, so ‘Q’ is not a nitrate compound. Normally silver halides like AgCl or AgBr are used in photography films, so compound ‘Q’ is definitely silver bromide (as chloride ion is not present in any of the reactants as suffix) which is a yellow colour compound as well. This also brings us to the conclusion that the metal of nitrate compound is silver.
So ‘P’ = $AgN{{O}_{3}}$ and ‘Q’ = AgBr
-The reaction of silver nitrate with sodium bromide is shown below:-
$AgN{{O}_{3}}+NaBr\to AgBr+NaN{{O}_{3}}$
On the completion of the above reaction, a yellow colour precipitate is formed in the solution. This reaction is a ‘double displacement reaction’ because the metal ions of both the ionic compounds replace each other during the process.
-When silver bromide is exposed to light, following reaction takes place:-
$2AgBr\to 2Ag+B{{r}_{2}}$
When silver bromide gets decomposed, it releases reddish brown fumes of bromine gas and silver metal is left alone. This is known as ‘decomposition reaction’.
Note:
-Remember that the another reason to not to choose silver chloride was that when it is a white colour compound and it was already mentioned that the precipitate formed was yellow in colour. Therefore it is important to learn the colours of various compounds along with their properties to solve these types of questions.
Complete step by step solution: Let us start with the first reaction as follows:-
-As it is given that ‘P’ is a metal nitrate compound that reacts with sodium bromide solution to give a compound ‘Q’ which is used in photography. So it is sure that the product formed will also be a nitrate and a bromide compound because metal nitrate shows displacement reaction with other ionic compounds.
-Since no nitrate compound is used in photography, so ‘Q’ is not a nitrate compound. Normally silver halides like AgCl or AgBr are used in photography films, so compound ‘Q’ is definitely silver bromide (as chloride ion is not present in any of the reactants as suffix) which is a yellow colour compound as well. This also brings us to the conclusion that the metal of nitrate compound is silver.
So ‘P’ = $AgN{{O}_{3}}$ and ‘Q’ = AgBr
-The reaction of silver nitrate with sodium bromide is shown below:-
$AgN{{O}_{3}}+NaBr\to AgBr+NaN{{O}_{3}}$
On the completion of the above reaction, a yellow colour precipitate is formed in the solution. This reaction is a ‘double displacement reaction’ because the metal ions of both the ionic compounds replace each other during the process.
-When silver bromide is exposed to light, following reaction takes place:-
$2AgBr\to 2Ag+B{{r}_{2}}$
When silver bromide gets decomposed, it releases reddish brown fumes of bromine gas and silver metal is left alone. This is known as ‘decomposition reaction’.
Note:
-Remember that the another reason to not to choose silver chloride was that when it is a white colour compound and it was already mentioned that the precipitate formed was yellow in colour. Therefore it is important to learn the colours of various compounds along with their properties to solve these types of questions.
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