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An inorganic compound 'X' on treatment with concentrated H2SO4 produces brown fumes and gives dark brown ring with FeSO4 in presence of concentrated H2SO4. Also, compound 'X' gives precipitate 'Y', when its solution in dilute HCI is treated with H2S gas. The precipitate 'Y' on treatment with concentrated HNO3 followed by excess of NH4OH further gives deep blue coloured solution. Compound 'X' is:
- a)Cu(NO3)2
- b)Pb(NO3)2
- c)Pb(NO2)2
- d)Co(NO3)2
Correct answer is option 'A'. Can you explain this answer?
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Explanation:
The given information suggests that compound X is an inorganic compound that reacts with concentrated H2SO4, FeSO4, dilute HCl, H2S gas, concentrated HNO3, and NH4OH to produce specific color changes or precipitates. Let's analyze each step to determine the identity of compound X.
Reaction with concentrated H2SO4:
When compound X is treated with concentrated H2SO4, it produces brown fumes. This indicates the presence of a volatile compound that is being released. This observation is not sufficient to determine the identity of compound X.
Reaction with FeSO4 in the presence of concentrated H2SO4:
Compound X reacts with FeSO4 in the presence of concentrated H2SO4 to form a dark brown ring. This reaction is characteristic of the presence of nitrate ions (NO3-). The formation of a dark brown ring is due to the formation of Fe(NO3)3, which is brown in color. This suggests that compound X contains nitrate ions (NO3-).
Reaction with dilute HCl and H2S gas:
When the solution of compound X in dilute HCl is treated with H2S gas, it forms a precipitate Y. This precipitate is likely to be a sulfide compound. The formation of a sulfide precipitate indicates the presence of a metal cation in compound X.
Reaction with concentrated HNO3 followed by excess NH4OH:
When precipitate Y is treated with concentrated HNO3 followed by excess NH4OH, it gives a deep blue-colored solution. This color change suggests the presence of copper ions (Cu2+). The formation of a deep blue-colored solution is indicative of the formation of a complex ion, most likely [Cu(NH3)4]2+, which is responsible for the blue color.
Conclusion:
Based on the observations from the reactions, compound X is most likely Cu(NO3)2, copper nitrate. This compound contains nitrate ions (NO3-) as indicated by the reaction with FeSO4 and concentrated H2SO4, and it forms a deep blue-colored solution when treated with concentrated HNO3 followed by excess NH4OH, indicating the presence of copper ions (Cu2+). The other options (Pb(NO3)2 and Pb(NO2)2) do not exhibit these specific reactions and color changes, so they can be eliminated as possibilities. Co(NO3)2 also does not show the formation of a deep blue-colored solution, eliminating it as an option as well.
Therefore, the correct answer is option 'A', Cu(NO3)2.
The given information suggests that compound X is an inorganic compound that reacts with concentrated H2SO4, FeSO4, dilute HCl, H2S gas, concentrated HNO3, and NH4OH to produce specific color changes or precipitates. Let's analyze each step to determine the identity of compound X.
Reaction with concentrated H2SO4:
When compound X is treated with concentrated H2SO4, it produces brown fumes. This indicates the presence of a volatile compound that is being released. This observation is not sufficient to determine the identity of compound X.
Reaction with FeSO4 in the presence of concentrated H2SO4:
Compound X reacts with FeSO4 in the presence of concentrated H2SO4 to form a dark brown ring. This reaction is characteristic of the presence of nitrate ions (NO3-). The formation of a dark brown ring is due to the formation of Fe(NO3)3, which is brown in color. This suggests that compound X contains nitrate ions (NO3-).
Reaction with dilute HCl and H2S gas:
When the solution of compound X in dilute HCl is treated with H2S gas, it forms a precipitate Y. This precipitate is likely to be a sulfide compound. The formation of a sulfide precipitate indicates the presence of a metal cation in compound X.
Reaction with concentrated HNO3 followed by excess NH4OH:
When precipitate Y is treated with concentrated HNO3 followed by excess NH4OH, it gives a deep blue-colored solution. This color change suggests the presence of copper ions (Cu2+). The formation of a deep blue-colored solution is indicative of the formation of a complex ion, most likely [Cu(NH3)4]2+, which is responsible for the blue color.
Conclusion:
Based on the observations from the reactions, compound X is most likely Cu(NO3)2, copper nitrate. This compound contains nitrate ions (NO3-) as indicated by the reaction with FeSO4 and concentrated H2SO4, and it forms a deep blue-colored solution when treated with concentrated HNO3 followed by excess NH4OH, indicating the presence of copper ions (Cu2+). The other options (Pb(NO3)2 and Pb(NO2)2) do not exhibit these specific reactions and color changes, so they can be eliminated as possibilities. Co(NO3)2 also does not show the formation of a deep blue-colored solution, eliminating it as an option as well.
Therefore, the correct answer is option 'A', Cu(NO3)2.
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An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer?
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An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer?.
An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer?.
Solutions for An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice An inorganic compound X on treatment with concentrated H2SO4produces brown fumes and gives dark brown ring with FeSO4in presence of concentrated H2SO4. Also, compound X gives precipitate Y, when its solution in dilute HCI is treated with H2S gas. The precipitate Y on treatment with concentrated HNO3followed by excess of NH4OH further gives deep blue coloured solution. Compound X is:a)Cu(NO3)2b)Pb(NO3)2c)Pb(NO2)2d)Co(NO3)2Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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