The hybridisation in BF_3 molecule is | vedclass

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(B) In the $BF_3$ molecule,the central Boron atom has $3$ valence electrons. It forms $3$ single bonds with $3$ Fluorine atoms. Using the formula for

Vedclass · Accepted Answer

$sp^2$

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(B) In the $BF_3$ molecule,the central Boron atom has $3$ valence electrons. It forms $3$ single bonds with $3$ Fluorine atoms. Using the formula for steric number: $	ext{Steric Number} = \frac{1}{2} 	imes (V + M - C + A)$,where $V = 3$ (valence electrons of $B$),$M = 3$ (monovalent atoms),$C = 0$,and $A = 0$. $	ext{Steric Number} = \frac{1}{2} 	imes (3 + 3) = 3$. $A$ steric number of $3$ corresponds to $sp^2$ hybridisation.

Column $I$ (Molecule/ion)	Column $II$ (Hybridisation)
$A$. $H_3O^{+}$	$P$. $sp$
$B$. $NH_2^{-}$	$Q$. $sp^2$
$C$. $NO_3^{-}$	$R$. $sp^3$
$D$. $ClF_3$	$S$. $sp^3d$

The hybridisation in $BF_3$ molecule is

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Column $I$ (Molecule/ion) Column $II$ (Hybridisation)

$A$. $H_3O^{+}$ $P$. $sp$

$B$. $NH_2^{-}$ $Q$. $sp^2$

$C$. $NO_3^{-}$ $R$. $sp^3$

$D$. $ClF_3$ $S$. $sp^3d$

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