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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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-$I$ (Molecule)	List-$II$ (Shape)
$A$. $NH_3$	$I$. Square pyramid
$B$. $BrF_5$	$II$. Tetrahedral
$C$. $PCl_5$	$III$. Trigonal pyramidal
$D$. $CH_4$	$IV$. Trigonal bipyramidal

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

Similar Questions

In which of the following options,molecules are correctly arranged with respect to their bond angles?

Which of the following has the smallest bond angle?

Which of the following molecules have a distorted geometry?

How many lone pairs of electrons are present on the chlorine atom in chlorous acid $(HClO_2)$?

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

List-$I$ (Molecule) List-$II$ (Shape)

$A$. $NH_3$ $I$. Square pyramid

$B$. $BrF_5$ $II$. Tetrahedral

$C$. $PCl_5$ $III$. Trigonal pyramidal

$D$. $CH_4$ $IV$. Trigonal bipyramidal

Choose the correct answer from the options below:

Vedclass Test Series

Exam Paper Generator

Online Exam Module