molecules - Identifying the number of bonding and lone pairs without a dot and cross diagram

I've just learned how to predict the shapes of molecules in class today using VSEPR theory. I would like to ask is there anyway to find the number of bond pairs and lone pairs without drawing a dot-and-cross diagram?

Answer

There is a simple, four-step calculation that you can perform:

1. 
   

   count the valence electrons the atoms you are bonding have.

   
   


2. 
   

   count the number of valence electrons the atoms would like to have for a noble gas’ valence shell. (i.e. eight for everything main group, two for hydrogen.)

   
   



3. 
   

   substract $2.-1.$, i.e. the first (existing) from the second (desired). This is your number of bonding electrons. Divide by two for bonding electron pairs.

   
   


4. 
   

   substract $1.-3.$, i.e. the third (bonding) from the first (existing). This is your number of free electrons. Divide by two for lone pairs.

   
   

Using a simple example such as sulphur dioxide:

1. 
   
   

   $6\ (\ce{S}) + 2 \cdot 6\ (\ce{2 O}) = 18$

   
   


2. 
   

   $8\ (\ce{S}) + 2 \cdot 8\ (\ce{2 O}) = 24$

   
   


3. 
   

   $24 - 18 = 6$, i.e. three bonding electron pairs.

   
   


4. 
   

   $18 - 6 = 12$ i.e. six lone pairs.

   
   
   

Unfortunately, that by itself does not allow you to write the structure; you need to know which element is in the centre and how they are connected. Here, we have a central sulphur bonded to two oxygens. The final result is:

$$\ce{O=S+-O-}$$

With two lone pairs on the left-hand oxygen, one on the sulphur and three on the right hand oxygen. The formal charges are important.