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Question:
Reaction of \(BeCl_2\) with \(LiAlH_4\) gives :- \(AlCl_3\)
- \(BeH_2\)
- \(LiH\)
- \(LiCl\)
- \(BeAlH_4\)
Choose the correct answer from the options given below :
Reaction of \(BeCl_2\) with \(LiAlH_4\) gives :
- \(AlCl_3\)
- \(BeH_2\)
- \(LiH\)
- \(LiCl\)
- \(BeAlH_4\)
Updated On: Sep 24, 2024
- (A), (D) and (E)
- (A), (B) and (D)
- (D) and (E)
- (B), (C) and (D)
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The Correct Option is B
Solution and Explanation
\(2BeCl_2 + LiAlH_4→ 2BeH_2 + LiCl + AlCl_3\)
So, the correct option is (B): (A), (B) and (D).
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Concepts Used:
Group 2 Elements
The group two or alkaline earth metals are s-block elements with two electrons in their s-orbital. They are alkaline earth metals. They are named so because of the alkaline nature of the hydroxides and oxides.
Alkaline earth metals are characterized by two s-electrons. This group of elements includes:
- Beryllium (Be)
- Magnesium (Mg)
- Calcium (Ca)
- Strontium (Sr)
- Barium (Ba)
- Radium (Ra)
Elements whose atoms have their s-subshell filled with their two valence electrons are called alkaline earth metals. Their general electronic configuration is [Noble gas] ns2. They occupy the second column of the periodic table and so-called as group two metals also.
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