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.

Column-$I$	Column-$II$ (Shape)
$A$. $SF_4$	$1$. Tetrahedral
$B$. $BrF_3$	$2$. Pyramidal
$C$. $BrO_3^-$	$3$. See-Saw shaped
$D$. $NH_4^+$	$4$. Bent $T$-shaped

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

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Match the species given in Column-$I$ with the shape given in Column-$II$ and mark the correct option:-

Column-$I$ Column-$II$ (Shape)

$A$. $SF_4$ $1$. Tetrahedral

$B$. $BrF_3$ $2$. Pyramidal

$C$. $BrO_3^-$ $3$. See-Saw shaped

$D$. $NH_4^+$ $4$. Bent $T$-shaped

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