|  |  |  |  |  |  |  |  |  |  |  |  | 
   
    | • | 3. Subtract the
    no. of e-’s calculated in step 
 |  | 
   
    |  | 
   
    |  | 1 from the no.
    in step 2. This gives you the 
 |  | 
   
    |  | 
   
    |  | no. of e-’s that
    must be shared to get an 
 |  | 
   
    |  | 
   
    |  | octet around all
    atoms in the molecule. 
 |  | 
   
    |  | 
   
    | • | 4. no. of e-’s
    that must be shared /2 gives 
 |  | 
   
    |  | 
   
    |  | you the no. of
    bonds. 
 |  | 
   
    |  | 
   
    | • | 5. subtract the
    no. of e-’s that are shared 
 |  | 
   
    |  | 
   
    |  | (from step 3)
    from the total no. of valence 
 |  | 
   
    |  | 
   
    |  | e-’s. This gives
    you the no. of unshared e-’s. 
 | 
   
    |  | 
   
    | • | If you divide the
    no. of unshared e-’s by 2 
 |  | 
   
    |  | 
   
    |  | you get the no.
    of lone pairs. 
 |  |