The molecular shapes of SF_4, CF_4,and XeF_4 | vedclass

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Question

(C) $1$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The geometry is see-saw.$2$.

Vedclass · Accepted Answer

different,with $1, 0$,and $2$ lone pairs of electrons respectively

Vedclass · Answer

(C) $1$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The geometry is see-saw.$2$. $CF_4$: The central atom $C$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms and has $0$ lone pairs. The geometry is tetrahedral.$3$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The geometry is square planar.Therefore,the shapes are different,and the number of lone pairs are $1, 0$,and $2$ respectively.

List-$I$ (species)	List-$II$ ($O-N-O$ angle)
$A$. $NO_2^+$	$1$. $180^{\circ}$
$B$. $NO_2$	$2$. $132^{\circ}$
$C$. $NO_2^-$	$3$. $115^{\circ}$
$D$. $NO_3^-$	$4$. $120^{\circ}$

The molecular shapes of $SF_4, CF_4$,and $XeF_4$ are:

Similar Questions

The bond angle in $PH_3$ is closest to which of the following $^o$?

$A$ pair of molecules with see-saw shape and linear shape,respectively,is

Match List-$I$ with List-$II$ and select the correct answer:

List-$I$ (species) List-$II$ ($O-N-O$ angle)

$A$. $NO_2^+$ $1$. $180^{\circ}$

$B$. $NO_2$ $2$. $132^{\circ}$

$C$. $NO_2^-$ $3$. $115^{\circ}$

$D$. $NO_3^-$ $4$. $120^{\circ}$

The molecule exhibiting the maximum number of non-bonding electron pairs $(l.p.)$ around the central atom is

The molecule having the smallest bond angle is:

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