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The element A burns in nitrogen to give an ionic compound B. The compound B reacts with water to give C and D. A solution of C becomes milky on bubbling carbon dioxide. What is the nature of compound (D)?
A. Acidic
B. Basic
C. Amphoteric
D. Neutral

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Last updated date: 14th Jun 2024
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Answer
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Hint: First synchronize the whole process using a flowchart, so that reactions and all elements are considered, without any confusion. The flowchart of the above mentioned reactions is: $\begin{align}
  & \text{A}\xrightarrow{{{\text{N}}_{2}}}\underset{\text{ionic}}{\mathop{\text{B}}}\,\xrightarrow{{{\text{H}}_{2}}\text{O}}\text{D}+\left( \text{C} \right)\xrightarrow{\text{C}{{\text{O}}_{2}}}\text{milkiness} \\
 & \text{ } \\
\end{align}$. We should follow this flowchart from the backside to solve the question easily. The milkiness appears because of formation of white-coloured precipitate of carbonate.

Complete answer:
Let us start solving this question from the back, as it would be easier to find the compounds from the back side.
- When C reacts with carbon dioxide, the solution turns milky. This milkiness is observed due to formation of calcium carbonate in the solution. The reaction would be $\text{Ca}{{\left( \text{OH} \right)}_{2}}\left( \text{aq}\text{.} \right)+\text{C}{{\text{O}}_{2}}\left( \text{g} \right)\to \text{CaC}{{\text{O}}_{3}}\downarrow \left( \text{s} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)$. So, it is clear that the compound $\left( \text{C} \right)$ is calcium hydroxide.
- If element ‘A’ reacts with nitrogen gas then, the ionic compound formed must have nitrogen in it. Further, on hydrolysis, the nitrogenous compound would have been formed along with calcium hydroxide. So, it is clear that element ‘A’ is calcium because calcium is not introduced in between the reaction.
- Reaction of element ‘A’ with nitrogen would be ionic in nature. The reaction would be $\text{3Ca}+{{\text{N}}_{2}}\to \text{C}{{\text{a}}_{3}}{{\text{N}}_{2}}$. The compound formed is calcium nitride. So, the ionic compound ‘B’ is calcium nitride.
- When calcium nitride undergoes hydrolysis, the compounds formed are $\text{C}{{\text{a}}_{3}}{{\text{N}}_{2}}+6{{\text{H}}_{2}}\text{O}\to 2\text{N}{{\text{H}}_{3}}+3\text{Ca}{{\left( \text{OH} \right)}_{2}}$. Where compound ‘D’ is ammonia gas and compound ‘C’ is calcium hydroxide.
The reactions and flowchart is completed; $\begin{align}
  & \text{Ca}\xrightarrow{{{\text{N}}_{2}}}\underset{\text{ionic}}{\mathop{\text{C}{{\text{a}}_{3}}{{\text{N}}_{2}}}}\,\xrightarrow{{{\text{H}}_{2}}\text{O}}\text{N}{{\text{H}}_{3}}+\left[ \text{Ca}{{\left( \text{OH} \right)}_{2}} \right]\xrightarrow{\text{C}{{\text{O}}_{2}}}\text{milkiness} \\
 & \text{ } \\
\end{align}$ . The nature of compound ‘D’ is basic in nature. We know ammonia has a lone pair, so it denotes its lone pair to electrophiles. The basic nature of ammonia can be explained by a reaction$\text{N}{{\text{H}}_{3}}+{{\text{H}}_{2}}\text{O}\to \text{N}{{\text{H}}_{4}}^{+}+\text{O}{{\text{H}}^{-}}$ by the formation of $\text{O}{{\text{H}}^{-}}$ ions.
So, the correct answer is “Option B”.

Note: In this question, the points to remember are the basic equations and the nature of the compounds formed. Such questions can be easily tackled if we make a proper flowchart, considering all the reactions and reagents along with their properties.