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

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(D) To determine the molecular shapes and lone pairs,we calculate the number of lone pairs on the central atom for each molecule:$1$. For $SF_4$: The

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Different with $1, 0$ and $2$ lone pairs of electrons on the central atom,respectively

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(D) To determine the molecular shapes and lone pairs,we calculate the number of lone pairs on the central atom for each molecule:$1$. For $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $(6 - 4) / 2 = 1$ lone pair. The shape is see-saw.$2$. For $CF_4$: The central atom $C$ has $4$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $(4 - 4) / 2 = 0$ lone pairs. The shape is tetrahedral.$3$. For $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms. Lone pairs = $(8 - 4) / 2 = 2$ lone pairs. The shape is square planar.Since the shapes and the number of lone pairs are different for all three,option $D$ is correct.

List-$I$ (Compound / Species)	List-$II$ (Shape / Geometry)
$A$. $SF_4$	$I$. Tetrahedral
$B$. $BrF_3$	$II$. Pyramidal
$C$. $BrO_3^{-}$	$III$. See-saw
$D$. $NH_4^{+}$	$IV$. Bent $T$-shape

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

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What is the geometry of the $PCl_5$ molecule as per $VSEPR$ theory?

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

List-$I$ (Compound / Species) List-$II$ (Shape / Geometry)

$A$. $SF_4$ $I$. Tetrahedral

$B$. $BrF_3$ $II$. Pyramidal

$C$. $BrO_3^{-}$ $III$. See-saw

$D$. $NH_4^{+}$ $IV$. Bent $T$-shape

Choose the correct answer from the options given below :

Which of the following is the incorrect order of bond angle?

The $H-O-H$ bond angle in a water molecule is about .............. $^\circ$.

Explain: The shape of $H_2O$ is $V$ (angular).

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