First slide

Orthoboric acid

Question

BCl₃ on hydrolysis forms

Easy

A

Square planar [B(OH)₄]^–
B

Octahedral [B(H₂O)₆]⁺³
C

Tetrahedral [B(OH)₄]^–
D

Tetrahedral [BCl₄]^–

Solution

$BC{l_3} + 3{H_2}O \to \,B{(OH)_3} + 3HCl$
$OR$
$OR$
$B{(OH)_3} + {H_2}O \to {[B{(OH)_4}]^ - } + {H^ + }$

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