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When metal 'M' is treated with NaOH, a w...

When metal 'M' is treated with `NaOH`, a white gelatinous precipitate 'X' is obtained which is soluble in excess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography as an adsorbent. Then metal 'M'

A

Zn

B

Ca

C

`Al`

D

`Fe`

Text Solution

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To solve the problem, we need to identify the metal 'M' based on the given conditions. Let's break down the information step by step: ### Step 1: Identify the reaction with NaOH When metal 'M' is treated with sodium hydroxide (NaOH), it produces a white gelatinous precipitate 'X'. **Hint:** Think about metals that typically react with NaOH to form precipitates. ### Step 2: Characteristics of precipitate 'X' The precipitate 'X' is described as white and gelatinous, and it is soluble in excess NaOH. This suggests that 'X' is likely a hydroxide of a metal that can form complex ions in the presence of excess hydroxide ions. **Hint:** Consider common metal hydroxides that fit this description. ### Step 3: Determine the nature of 'X' Given that 'X' is soluble in excess NaOH, we can infer that it is likely aluminum hydroxide (Al(OH)₃). Aluminum hydroxide is known to be a white gelatinous precipitate and is soluble in excess NaOH to form sodium aluminate. **Hint:** Recall the solubility rules for metal hydroxides in alkaline solutions. ### Step 4: Heating 'X' to form an oxide When 'X' (Al(OH)₃) is heated strongly, it decomposes to form aluminum oxide (Al₂O₃) and water (H₂O). Aluminum oxide is known to be used as an adsorbent in chromatography. **Hint:** Think about the thermal decomposition of metal hydroxides. ### Step 5: Conclusion about metal 'M' Based on the above deductions, the metal 'M' that reacts with NaOH to form the white gelatinous precipitate 'X' (Al(OH)₃) is aluminum (Al). **Final Answer:** The metal 'M' is Aluminum (Al). ### Summary of Steps: 1. Identify the reaction with NaOH. 2. Determine the characteristics of precipitate 'X'. 3. Conclude that 'X' is aluminum hydroxide (Al(OH)₃). 4. Recognize that heating 'X' produces aluminum oxide (Al₂O₃). 5. Conclude that metal 'M' is aluminum.
To solve the problem, we need to identify the metal 'M' based on the given conditions. Let's break down the information step by step: ### Step 1: Identify the reaction with NaOH When metal 'M' is treated with sodium hydroxide (NaOH), it produces a white gelatinous precipitate 'X'. **Hint:** Think about metals that typically react with NaOH to form precipitates. ### Step 2: Characteristics of precipitate 'X' ...
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Knowledge Check

  • An alkaline metal oxide which is soluble in NaOH has

    A
    `NaCl` structure
    B
    wurzite structure
    C
    `CsCl` structure
    D
    rutile structure

  • The hydroxide of which metal ion(s) which is/are soluble in excess of NaOH solution

    A
    `A1^(3+)`
    B
    `zn^(2+)`
    C
    `Fe^(3+)`
    D
    `Cu^(2+)`

  • Which of the followig metal in solution forms a precipitate with NaOH which is not soluble in an excess of the base

    A
    Fe
    B
    Sn
    C
    Pb
    D
    Zn

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