JEE Mains · Chemistry · STD 11 -p-Block elements - I
The geometry around boron in the product ' \(B\) ' formed from the following reaction is \(BF _{3}+ NaH \stackrel{450 K }{\longrightarrow} A + NaF\) \(A + NMe _{3} \rightarrow B\)
- A trigonal, planar
- B tetrahedral
- C pyramidal
- D square, planar
Answer & Solution
Correct Answer
(B) tetrahedral
Step-by-step Solution
Detailed explanation
\(BH _{3}+ NaH \stackrel{450 K }{\longrightarrow} \underset{\text { (diborane) }}{ B _{2} H _{6}+ NaF }\) \(B _{2} H _{6}+ NMe _{3} \longrightarrow 2\left[ BH _{3} \leftarrow NMe _{3}\right]\)
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