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-$II$
$A$. $BF_3$	$II$. Trigonal planar
$B$. $ClF_3$	$III$. $T$-shape
$C$. $NH_3$	$IV$. Trigonal pyramidal
$D$. $NH_4^+$	$I$. Tetrahedral

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

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The number of molecules having a lone pair of electrons on the central atom in the following is: $BF_3, SF_4, SiCl_4, XeF_4, NCl_3, XeF_6, PCl_5, HgCl_2, SnCl_2$

Match the following:
List-$I$ List-$II$
$A$. $BF_3$ $II$. Trigonal planar
$B$. $ClF_3$ $III$. $T$-shape
$C$. $NH_3$ $IV$. Trigonal pyramidal
$D$. $NH_4^+$ $I$. Tetrahedral

The correct match is:

$SbF_5$ reacts with $XeF_4$ and $XeF_6$ to form ionic compounds $[XeF_3]^+ [SbF_6]^-$ and $[XeF_5]^+ [SbF_6]^-$ respectively. What are the molecular shapes of the $[XeF_3]^+$ and $[XeF_5]^+$ ions,respectively?

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