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

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(D) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of

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The same,with $1, 0$,and $2$ lone pair of electrons on the central atom,respectively.

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(D) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (monovalent atoms).$1$. For $SF_4$: Central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (6 - 4) = 1$.$2$. For $CF_4$: Central atom $C$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (4 - 4) = 0$.$3$. For $XeF_4$: Central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (8 - 4) = 2$.Thus,the number of lone pairs on the central atoms of $SF_4$,$CF_4$,and $XeF_4$ are $1, 0$,and $2$ respectively.

List-$I$	List-$II$
$A. BrF_5$	$I. AB_4E$,see-saw
$B. SF_4$	$II. AB_4E_2$,square planar
$C. XeF_4$	$III. AB_5E$,square pyramidal
$D. ClF_3$	$IV. AB_3E_2$,$T$-shape

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

Similar Questions

What is the bond angle $O-S-O$ in the $SO_2$ molecule (in $^{\circ}$)?

$SbF_5$ reacts with $XeF_4$ and $XeF_6$ to form ionic compounds $[XeF_3^+][SbF_6^-]$ and $[XeF_5^+][SbF_6^-]$. Determine the molecular shape of $[XeF_3^+]$ ion and $[XeF_5^+]$ ion respectively.

In which molecule are all bonds not equal?

Match the following:

List-$I$ List-$II$

$A. BrF_5$ $I. AB_4E$,see-saw

$B. SF_4$ $II. AB_4E_2$,square planar

$C. XeF_4$ $III. AB_5E$,square pyramidal

$D. ClF_3$ $IV. AB_3E_2$,$T$-shape

The correct answer is:

Which one has a pyramidal structure?

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