Discuss the shape of the following molecules using the $VSEPR$ model: $BeCl_2, BCl_3, SiCl_4, AsF_5, H_2S, PH_3$.
(N/A) $BeCl_2$: The central atom has no lone pair and two bond pairs ($AB_2$ type). Hence,it has a linear shape.
$BCl_3$: The central atom has no lone pair and three bond pairs ($AB_3$ type). Hence,it is trigonal planar.
$SiCl_4$: The central atom has no lone pair and four bond pairs ($AB_4$ type). Hence,it is tetrahedral.
$AsF_5$: The central atom has no lone pair and five bond pairs ($AB_5$ type). Hence,it is trigonal bipyramidal.
$H_2S$: The central atom has two lone pairs and two bond pairs ($AB_2E_2$ type). Hence,it is bent or $V$-shaped.
$PH_3$: The central atom has one lone pair and three bond pairs ($AB_3E$ type). Hence,it is trigonal pyramidal.
$BCl_3$: The central atom has no lone pair and three bond pairs ($AB_3$ type). Hence,it is trigonal planar.
$SiCl_4$: The central atom has no lone pair and four bond pairs ($AB_4$ type). Hence,it is tetrahedral.
$AsF_5$: The central atom has no lone pair and five bond pairs ($AB_5$ type). Hence,it is trigonal bipyramidal.
$H_2S$: The central atom has two lone pairs and two bond pairs ($AB_2E_2$ type). Hence,it is bent or $V$-shaped.
$PH_3$: The central atom has one lone pair and three bond pairs ($AB_3E$ type). Hence,it is trigonal pyramidal.
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View SolutionMatch the molecules in List-$I$ with their respective shapes in List-$II$.
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| List-$I$ (Molecule) | List-$II$ (Shape) |
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| $A. SF_4$ | $I. T\text{-shaped}$ |
| $B. ClF_3$ | $II. \text{Square planar}$ |
| $C. BrF_5$ | $III. \text{See-saw}$ |
| $D. XeF_4$ | $IV. \text{Square pyramidal}$ |
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