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-$I$	List-$II$
$(a)$ $PCl_{5}$	$(i)$ Square pyramidal
$(b)$ $SF_{6}$	$(ii)$ Trigonal planar
$(c)$ $BrF_{5}$	$(iii)$ Octahedral
$(d)$ $BF_{3}$	$(iv)$ Trigonal bipyramidal

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

Similar Questions

Identify the isostructural pairs from the following species: $NF_3, NO_3^-, BF_3, H_3O^+, HN_3$

The geometry with respect to the central atom of the following molecules are: $N(SiH_3)_3;\, Me_3N;\, (SiH_3)_3P$

The shapes of $XeF_2, XeF_4$,and $XeO_3$ respectively are

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$(a)$ $PCl_{5}$ $(i)$ Square pyramidal

$(b)$ $SF_{6}$ $(ii)$ Trigonal planar

$(c)$ $BrF_{5}$ $(iii)$ Octahedral

$(d)$ $BF_{3}$ $(iv)$ Trigonal bipyramidal

Choose the correct answer from the options given below.

The number of bonding and non-bonding electron pairs on the $N$ atom in $NO_3^-$ are respectively ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module