Discuss the shape of the following molecules using the VSEPR model.
`BeCl_(2), BCl_(2), SiCl_(4) , AsF_(5), H_(2)S, PH_(3)`

Explanation

Shape of `BeCl_(2)` :
Be (Z = 4) `[He]^(2) 2s^(2)` and
Cl (z = 17) `[Ne]^(10) 3s^(2) 3p^(5)`
`BeCl_(2)` is `AB_(2)` type molecule. In it two Be - Cl bond and two-two bonding electron pair.
So two bonding pair remain at maximum distance so repulsion between then is minimum and so the shape is linear.

Shape of `BCl_(3)` :
In `BCl_(3)` , B is central atom and three Cl in and . So, three B - Cl bond in `BCl_(3)`.
B (Z = 5 ) , Thus `[He]^(2) 2s^(2)2p^(1)` . Three valence cell electron of B form covalent bond with three Cl. Non bonding electron is not present on B.
three boding pair arrange at maximum distance and form planar trigonal arrangement .
Shape of `SICl_(4)` :

In `SiCl_(4)` Si is central atom. Four Cl atom join with Si and form four Si - Cl bond.
Si (Z = 14) , So , `[Ne]^(10) 3s^(2)3p^(2) `. In Si four valence cell electrons and bonding pair with four Cl. So, non bonding electron pair is not present on Si.
Si has all four bonding pair so arrange at maximum distance at `109^(@)28.` and give tetrahedral arrangement.
Shape of `AsF_(5)`:

As is central atom. In it five bonding electron pair with five AS - F bond.
As present in 17 (VB) group so, according to`ns^(2)np^(3)` in valence cell five electrons and no nonbonding electrons.
As per VSEPR five bonding pair arrange at maximum distance `ltF `- As - F = `120^(@)` remain planar and `ltF - ` As - F = `90^(@)` So remain perpendicular & give trigonal bipyramidal arragement.

Shape of `H_(2)`S :
In `H_(2)`S, S is central atom & two H in end so , two S - H bond present.
In periodic table S in 16 (VI B) group , so electron configuration `ns^(2)np^(4)`. Two electron participate in bond and two electron act as nonboding `e^(-)` pair on S.
Two bonding pair S - H arrange in such a way that there will be least repulsion between two lone pair.
As a result this is not linear but the is angular so `H_(2)S " is " AB_(2)E_(2)` type molecule.

Shape of `PH_(3)` :
In `PH_(3)` central atom P and negative H in end, so , three P - H bond in `PH_(3)`.
Position of P in 17 0/ B) group, so electron configuration `ns^(2)np^(3)` and total 5 electrons in outer cell of P.
In `PH_(3)` three bonding electron pair of three P - H and one nonboding electron present on P.
In `PH_(3)` three boding electron pair of three P - H and one nonbonding electron present on P.
These electron pair arrange at maximum distance so arragnement of electron pair is tetrahedral & shape is trigonal pyramidal.