The geometry around boron in the product 'B' formed from the following reaction is

$B F_{3} + N a H \overset{450 K}{\to} A + N a F$

A + NMe₃ $\overset{}{\to} B$

Option 1 - Trigonal planar

Option 2 - Tetrahedral

Option 3 - Pyramidal

Option 4 - Square planar

2 Views|Posted 6 months ago

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Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 2

Detailed Solution:

$2 B F_{3} + 6 N a H \to B_{2} H_{6} + 6 N a F$

$B_{2} H_{6} + 2 \ddot{N} {(M e)}_{3} \to 2 H_{3} B \leftarrow : N {(M e)}_{3}$

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Chemistry Ncert Solutions Class 11th 2023

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The geometry around boron in the product 'B' formed from the following reaction is B F 3 + N a H → 4 5 0 K A + N a F A + NMe3 → B

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