To determine the identity of the metal (M) based on the given information, let's analyze the situation step by step. ### Step 1: Understanding the Reaction with Nitric Acid When a metal (M) is dissolved in nitric acid (HNO₃), it typically forms a metal nitrate. The reaction can be generalized as follows: \[ \text{M} + 2 \text{HNO}_3 \rightarrow \text{M(NO}_3\text{)}_2 + \text{H}_2 \] The specific products depend on the metal, but for metals that can react with nitric acid, we often see the formation of soluble metal nitrates. ### Step 2: Reaction with Sodium Hydroxide The resulting solution is then treated with sodium hydroxide (NaOH). The formation of a white precipitate (B) indicates that the metal ion from the nitrate reacts with hydroxide ions to form a metal hydroxide. The general reaction can be represented as: \[ \text{M}^{2+} + 2 \text{OH}^- \rightarrow \text{M(OH)}_2 \] The fact that this precipitate is white suggests that the metal hydroxide formed is likely to be a common one, such as zinc hydroxide (Zn(OH)₂), which is white and precipitates out of solution. ### Step 3: Redissolving the Precipitate The problem states that the white precipitate (B) redissolves upon the addition of excess alkali. This behavior is characteristic of amphoteric hydroxides, which can react with excess hydroxide ions to form soluble complexes. For example, zinc hydroxide can react with excess NaOH to form sodium zincate: \[ \text{Zn(OH)}_2 + 2 \text{OH}^- \rightarrow \text{[Zn(OH)}_4\text{]}^{2-} \] This property of redissolving in excess alkali is a key indicator that the metal in question is likely to be zinc (Zn). ### Step 4: Ignition to Form the Metal Oxide The final part of the question states that when the precipitate (B) is ignited in a crucible, it produces the oxide of the metal (m). For zinc hydroxide, upon heating, it decomposes to form zinc oxide (ZnO): \[ \text{Zn(OH)}_2 \xrightarrow{\text{heat}} \text{ZnO} + \text{H}_2\text{O} \] This is consistent with the behavior described in the question. ### Conclusion: Identifying Metal M Based on the above analysis, the metal (M) is identified as zinc (Zn). ### Evaluating Other Options - **B. Cu (Copper)**: Copper does not form a white precipitate with NaOH; it forms a blue precipitate of copper(II) hydroxide (Cu(OH)₂), which does not redissolve in excess NaOH. - **C. Al (Aluminum)**: Aluminum hydroxide (Al(OH)₃) is also a white precipitate, but it does not redissolve in excess NaOH to form a soluble complex as readily as zinc does. It can dissolve, but the behavior is less straightforward compared to zinc. - **D. Au (Gold)**: Gold does not react with nitric acid in a way that would produce a hydroxide precipitate. Gold is a noble metal and does not form hydroxides that would precipitate in this manner. ### Summary - The metal (M) is zinc (Zn), which forms a white precipitate of zinc hydroxide (Zn(OH)₂) with NaOH. - Zinc hydroxide is amphoteric and redissolves in excess NaOH. - Upon ignition, zinc hydroxide produces zinc oxide (ZnO). - Other options do not fit the described behavior of forming a white precipitate that redissolves in excess alkali. ### Revision Summary - Metal (M) is identified as zinc (Zn). - Zinc reacts with nitric acid to form zinc nitrate. - Zinc hydroxide precipitates as a white solid and redissolves in excess NaOH. - Ignition of zinc hydroxide yields zinc oxide (ZnO).