Class 11 Exam > Class 11 Questions > We can get beryllium hydride by:a)Trea...
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We can get beryllium hydride by:
- a)Treating Be with LiAlH4
- b)Treating BeCl2 with H2
- c)Treating BeCl2 with LiAlH4
- d)Treating Be with H2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
We can get beryllium hydride by:a)Treating Be with LiAlH4b)Trea...
BeCl2 + LiAlH4 give 2BeH2 + LiCl and AlCl3
However, other elements of group 2 combine with hydrogen upon heating to form MH2
However, other elements of group 2 combine with hydrogen upon heating to form MH2
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We can get beryllium hydride by:a)Treating Be with LiAlH4b)Trea...
To obtain beryllium hydride (BeH2), we can treat beryllium chloride (BeCl2) with lithium aluminum hydride (LiAlH4). Let's break down the process step-by-step:
1. Formation of Lithium Aluminum Hydride (LiAlH4):
- Lithium aluminum hydride is formed by reacting aluminum chloride (AlCl3) with lithium hydride (LiH) in a 1:1 molar ratio.
- The reaction is as follows:
4LiH + AlCl3 → LiAlH4 + 3LiCl
2. Reaction between Beryllium Chloride (BeCl2) and Lithium Aluminum Hydride (LiAlH4):
- Beryllium chloride reacts with lithium aluminum hydride to form beryllium hydride and lithium chloride (LiCl).
- The reaction is as follows:
BeCl2 + LiAlH4 → BeH2 + LiCl + AlCl3
Explanation of the Correct Answer (Option C):
- Option C, which involves treating beryllium chloride with lithium aluminum hydride, is the correct answer.
- This is because the reaction between BeCl2 and LiAlH4 yields beryllium hydride (BeH2), as well as lithium chloride (LiCl) and aluminum chloride (AlCl3) as byproducts.
Summary:
- Beryllium hydride (BeH2) can be obtained by treating beryllium chloride (BeCl2) with lithium aluminum hydride (LiAlH4).
- The reaction between BeCl2 and LiAlH4 results in the formation of BeH2, LiCl, and AlCl3.
Note: Beryllium hydride is a highly reactive compound and should be handled with caution due to its potential health hazards.
1. Formation of Lithium Aluminum Hydride (LiAlH4):
- Lithium aluminum hydride is formed by reacting aluminum chloride (AlCl3) with lithium hydride (LiH) in a 1:1 molar ratio.
- The reaction is as follows:
4LiH + AlCl3 → LiAlH4 + 3LiCl
2. Reaction between Beryllium Chloride (BeCl2) and Lithium Aluminum Hydride (LiAlH4):
- Beryllium chloride reacts with lithium aluminum hydride to form beryllium hydride and lithium chloride (LiCl).
- The reaction is as follows:
BeCl2 + LiAlH4 → BeH2 + LiCl + AlCl3
Explanation of the Correct Answer (Option C):
- Option C, which involves treating beryllium chloride with lithium aluminum hydride, is the correct answer.
- This is because the reaction between BeCl2 and LiAlH4 yields beryllium hydride (BeH2), as well as lithium chloride (LiCl) and aluminum chloride (AlCl3) as byproducts.
Summary:
- Beryllium hydride (BeH2) can be obtained by treating beryllium chloride (BeCl2) with lithium aluminum hydride (LiAlH4).
- The reaction between BeCl2 and LiAlH4 results in the formation of BeH2, LiCl, and AlCl3.
Note: Beryllium hydride is a highly reactive compound and should be handled with caution due to its potential health hazards.
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