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Question: When a brown compound of Mn(A) is treated with HCl, it gives a gas B. The gas B taken in excess reac...

When a brown compound of Mn(A) is treated with HCl, it gives a gas B. The gas B taken in excess reacts with NH3N{{H}_{3}} to give an explosive compound C. The compounds A, B and C are:
[A] A = MnO2Mn{{O}_{2}}, B = Cl2C{{l}_{2}}, C = NCl3NC{{l}_{3}}
[B] A = MnO, B = Cl2C{{l}_{2}}, C = NH4ClN{{H}_{4}}Cl
[C] A = Mn3O4M{{n}_{3}}{{O}_{4}}, B = Cl2C{{l}_{2}}, C = NCl3NC{{l}_{3}}
[D] A = MnO3Mn{{O}_{3}} , B = Cl2C{{l}_{2}}, C = NCl2NC{{l}_{2}}

Explanation

Solution

HINT: Oxides of manganese upon reaction with hydrochloric acid gives us manganese chloride and water and a gas is evolved. This gas is yellow green in colour and has a pungent smell and on reaction with ammonia it gives us a yellow explosive liquid.

COMPLETE STEP BY STEP SOLUTION: In the question t is given to us that a brown compound of manganese is treated with hydrochloric acid and it gives a gas B. The gas B in excess will form an explosive compound C. So, let us discuss each option to find out the correct answer.
In the first option we have the compound A as MnO2Mn{{O}_{2}} which will react with HCl. We can write the reaction as-
MnO2+HClMnCl2+H2O+Cl2Mn{{O}_{2}}+HCl\to MnC{{l}_{2}}+{{H}_{2}}O+C{{l}_{2}}
Manganese dioxide is a blackish (might appear brown) coloured complex and it does evolve chlorine gas upon reaction with HCl.
Now let us see the reaction of excess of chlorine gas with ammonia which will give us an explosive compound-
NH3+3Cl2(excess)3HCl+NCl3N{{H}_{3}}+3C{{l}_{2}}(excess)\to 3HCl+NC{{l}_{3}}
When excess chlorine reacts with ammonia, it gives a yellow coloured explosive compound nitrogen chloride.
We can see here the compound A is MnO2Mn{{O}_{2}}, B is chlorine gas and C is nitrogen trichloride therefore, this option can be correct.
In the next option we have A as manganese oxide. When it reacts with HCl it gives us water and MnCl2MnC{{l}_{2}} and no other gas. Therefore, this option is incorrect.
In the third option we have Mn3O4M{{n}_{3}}{{O}_{4}} which can react with HCl and give us the same products as manganese dioxide did. We can write the reaction as-
Mn3O4+HClMnCl2+H2O+Cl2M{{n}_{3}}{{O}_{4}}+HCl\to MnC{{l}_{2}}+{{H}_{2}}O+C{{l}_{2}}
Again when gas B i.e. chlorine reacts with ammonia it will give us the same explosive nitrogen trichloride. Therefore, this option can be correct too.
And lastly we have MnO3Mn{{O}_{3}} which will evolve chlorine gas but C here is given as NCl2NC{{l}_{2}}, which is not the correct answer. Therefore, this is incorrect.
From the above discussion we can understand that both options [A] and [B] can form the required gases and explosives therefore both can be the correct answer here.
But, in the question it is mentioned that compound A is brown in colour. MnO2Mn{{O}_{2}} is blackish brown in colour but Mn3O4M{{n}_{3}}{{O}_{4}} is brown in colour. Therefore here [B] will be a more appropriate answer even though both are correct.

Therefore, the correct answer is option [C] A = Mn3O4M{{n}_{3}}{{O}_{4}}, B = Cl2C{{l}_{2}}, C = NCl3NC{{l}_{3}}

NOTE: Here if we added excess ammonia instead of excess chlorine we would get ammonium chloride, NH4ClN{{H}_{4}}Cl which is not an explosive.
The electronegativity difference between nitrogen and chlorine in nitrogen trichloride is not as high as that in nitrogen trifluoride. Thus, the N-Cl bond is weaker than N-F and therefore nitrogen trichloride is an explosive whereas nitrogen trifluoride is not.