KCET · Chemistry · d and f Block Elements
When a brown compound of \( M n(A) \) is treated with \( H C l \), it gives a gas (B). The gas (B) taken in
excess reacts with \( \mathrm{NH}_{3} \) to give an explosive compound (C).
The compounds A, B and C are;
- A \( \mathrm{A}=\mathrm{MnO}_{2}, \mathrm{~B}=\mathrm{Cl}_{2}, \mathrm{C}=\mathrm{NCl}_{3} \)
- B \( A=M n O, B=C l_{2}, C=N H_{4} C l \)
- C \( \mathrm{A}=\mathrm{Mn}_{3} \mathrm{O}_{4}, \mathrm{~B}=\mathrm{Cl}_{2}, \mathrm{C}=\mathrm{NCl}_{3} \)
- D \( \mathrm{A}=\mathrm{MnO}_{3}, \mathrm{~B}=\mathrm{Cl}_{2}, \mathrm{C}=\mathrm{NCl}_{2} \)
Answer & Solution
Correct Answer
(A) \( \mathrm{A}=\mathrm{MnO}_{2}, \mathrm{~B}=\mathrm{Cl}_{2}, \mathrm{C}=\mathrm{NCl}_{3} \)
Step-by-step Solution
Detailed explanation
When a brown compound of \( M n(A) \) is treated with \( H C l \), it gives a gas (B) as shown below.
\( \mathrm{MnO}_{2}(\mathrm{~A})+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2}(\mathrm{~B}) \)
The gas (B) taken in excess reacts with \( \mathrm{NH}_{3} \) to give an explosive compound (C) as shown below.
\( 3 \mathrm{Cl}_{2}(B)+\mathrm{NH}_{3} \rightarrow \mathrm{NCl}_{3}(C)+3 \mathrm{HCl} \)
Therefore, the compounds \( A, B \) and \( C \) are
\( \mathrm{MnO}_{2}, \mathrm{Cl}_{2} \), and \( \mathrm{NCl}_{3} \).
\( \mathrm{MnO}_{2}(\mathrm{~A})+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2}(\mathrm{~B}) \)
The gas (B) taken in excess reacts with \( \mathrm{NH}_{3} \) to give an explosive compound (C) as shown below.
\( 3 \mathrm{Cl}_{2}(B)+\mathrm{NH}_{3} \rightarrow \mathrm{NCl}_{3}(C)+3 \mathrm{HCl} \)
Therefore, the compounds \( A, B \) and \( C \) are
\( \mathrm{MnO}_{2}, \mathrm{Cl}_{2} \), and \( \mathrm{NCl}_{3} \).
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