The shape of AsF_5 according to VSEPR theory | vedclass

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(B) To determine the shape of $AsF_5$ using $VSEPR$ theory,we first calculate the number of valence electrons on the central atom $As$.$As$ belongs to

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Trigonal bipyramidal

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(B) To determine the shape of $AsF_5$ using $VSEPR$ theory,we first calculate the number of valence electrons on the central atom $As$.$As$ belongs to group $15$,so it has $5$ valence electrons.It forms $5$ bonds with $5$ $F$ atoms.Number of electron pairs $= \frac{1}{2} 	imes (V + M - C + A) = \frac{1}{2} 	imes (5 + 5 - 0 + 0) = 5$.Since there are $5$ bond pairs and $0$ lone pairs,the hybridization is $sp^3d$.$A$ molecule with $5$ bond pairs and $0$ lone pairs adopts a trigonal bipyramidal geometry.

List-$I$	List-$II$
$A$. $H_3O^+$	$I$. Tetrahedral
$B$. Acetylide anion	$II$. Linear
$C$. $NH_4^+$	$III$. Pyramidal
$D$. $ClO_2^-$	$IV$. Bent

The shape of $AsF_5$ according to $VSEPR$ theory $:-$

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Match List-$I$ with List-$II$ :

List-$I$ List-$II$

$A$. $H_3O^+$ $I$. Tetrahedral

$B$. Acetylide anion $II$. Linear

$C$. $NH_4^+$ $III$. Pyramidal

$D$. $ClO_2^-$ $IV$. Bent

Choose the correct answer from the options given below :

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