The geometry around boron in the product 'B' | vedclass

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(B) The given reactions are:$2BF_{3} + 6NaH \xrightarrow{450 \ K} B_{2}H_{6} + 6NaF$$B_{2}H_{6} + 2NMe_{3} \rightarrow 2[BH_{3} \leftarrow NMe_{3}]$He

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tetrahedral

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(B) The given reactions are:$2BF_{3} + 6NaH \xrightarrow{450 \ K} B_{2}H_{6} + 6NaF$$B_{2}H_{6} + 2NMe_{3} \rightarrow 2[BH_{3} \leftarrow NMe_{3}]$Here,$A$ is $B_{2}H_{6}$ (diborane) and $B$ is the adduct $[BH_{3} \leftarrow NMe_{3}]$.In the adduct $[BH_{3} \leftarrow NMe_{3}]$,the boron atom is bonded to three hydrogen atoms and one nitrogen atom.The boron atom is $sp^{3}$ hybridized and possesses a tetrahedral geometry.

The geometry around boron in the product '$B$' formed from the following reaction is
$BF_{3} + NaH \xrightarrow{450 \ K} A + NaF$
$A + NMe_{3} \rightarrow B$

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The geometry around boron in the product '$B$' formed from the following reaction is$BF_{3} + NaH \xrightarrow{450 \ K} A + NaF$$A + NMe_{3} \rightarrow B$