Identifying the number of bonding and lone pairs without a dot and cross diagram
There is a simple, four-step calculation that you can perform:


count the valence electrons the atoms you are bonding have.
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.)
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.
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:


$6\ (\ce{S}) + 2 \cdot 6\ (\ce{2 O}) = 18$
$8\ (\ce{S}) + 2 \cdot 8\ (\ce{2 O}) = 24$
$24 - 18 = 6$, i.e. three bonding electron pairs.
$18 - 6 = 12$ 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.