Discuss the shape of the following molecules using VSEPR model

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chemicalbonding

#1

Discuss the shape of the following molecules using VSEPR model.


#2

According to VSEPR theory, the shape of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom. Pairs of electrons in the valence shell repel each other. The order of their repulsions is as follows
Ip — lp> Ip — bp > bp — bp
(i) ${ BeCl }{ 2 }$
The central atom Be has only 2 valence electrons which are bonded to Cl, so there are only 2 bond pairs and no lone pairs. It is of the type ${ AB }
{ 2 }$ and hence, the shape is linear.

(ii) ${ Bcl }{ 3 }$

The central atom B has only 3 valence electrons which are bonded with three Cl atoms, so it contains only 3 bond pairs and no lone pair. It is of the type ${ AB }
{ 3 }$ and hence, the shape is trigonal planar.
(iii)${ SiCl }{ 4 }$

Similarly, the central atom Si has only 4 bond pairs and no lone pair. It is of the type ${ AB }
{ 4 }$ and hence, the shape is tetrahedral.
(iv)${ AsF }{ 5 }$

The central atom As has only 5 bond pairs and no lone pair. It is of the type ${ AB }
{ 5 }$ and hence, the shape is trigonal bipyramidal.
(v)$H { 2 }S$

The central atom S has 6 valence electrons. Out of these only two are used in bond formation with two H-atoms while four (two pairs) remains as non-bonding electrons (i.e., lone pairs). So, it contains 2 bond pairs and 2 lone pairs. It is of the type $AB
{2}E_{2}$ and hence, the shape is bent or V- shaped.
(vi)${ PH }{ 3 }$

The central atom P has 5 valence electrons. Out of which three are utilised in bonding with H atoms and one pair remains as lone pair. So, it contains 3 bond pairs and one lone pair. It is of the type ${ AB }
{ 3 }E$ and hence the shape is pyramidal.