I'm going to assume that you don't know much. Forgive me if i start off a little too basic.
To start, a periodic table will be helpful. You will use it see how many valence electrons the atom has. I'll write more about that in a second. First, almost all atoms want 8 electrons in their outer valence shell. A molecule that they are a part of that doesn't satisfy it will be very unstable, or most likely will not even exist. Your job is to draw a structure that (gives each atom eight, while not using more electrons than are in their combined valence. If that doesn't make sense, read it again. I'll walk you through it some more, so bear with me. Let's skip hydrogen for now and work with Cl2.
OK, take a look at the periodic table. How many groups(columns) over is Cl? Count them. 1,2,3,4.....7! So, that means there are 7 electrons in its valence. So if there are 2 Chlorine atoms, there are going to be a total of 14 electrons we are going to have to represent with bonds(lines) or lone pair electrons(dots). We must use up all 14, and we must have satisfied their desire for 8 electrons(the octet rule). We are going to need to keep track of the total number of valence electrons in order to get these right. So write down the combined valence for each molecule. Also, you should be drawing these as I explain them, since I can't draw them for you. Take a look at illustration #1. See how the atom is drawn? Electrons are written in pairs, evenly spread out around the atom. If there is a loner, just put one dot. Now, let's make Cl2. It's an easy start. Just put two chlorines next to each other. They are going to share each other's single, lonely unpaired electron between the two of them. That's what a bond is. So your finished product should have two chlorine molecules with a single bond between them, and six dots evenly spaced around each one. Double check the valence and the total. 6 lone pairs +2 in the bond=8. So the valence is stable. Did we use up all of the possible electrons? 6+2+6=14. So, perfect. Now lets try one more, a little harder, and then I'll let you finish the rest. I don't have all day, I'm not getting paid.
Count up the total.
4+6+6. Got that? 16 total electrons.
Oh shoot. Here is a rule you MUST know.
Oxygen only likes 2 bonds.And carbon must have 4 bonds. Carbon can't stand having lone pairs. It's just the way it is. That definitely affects the way the structure will look.
Alright, back to the problem at hand. You are going to simply need to try drawing this. Draw a carbon and 2 oxygens. Only draw the letter that represents the atom, and single bonds to connect them. You could draw C-O-O or O-C-O or O-O-C. The first and the last are the same. So you really have two ways to go. Next you are going to fill in the rest of the octet for each atom. If it has one bond, it gets 3 lone pairs, and if it has 2 bonds it is going to get 2 lone pairs. Because one bond= 2 electrons, so 2+3(2)=8. And 4+2(2)=8. Just trying to satisfy the octet. Now count up all the total electrons. 4 in the bonds, another 16 in lone pairs makes 20. We've drawn more than we have. Now is the time for more bonds. Every time you draw a bond, erase a lone pair from each atom you are connecting to. So if you were using O-C-O you would make it., O=C-O. Count up the valence of each atom. Do they each have 8? What's the total? 18. Blast. Still too much. Time for another double bond. Now we have O=C=O. Look. Each oxygen has 2 bonds, Carbon has 4, each has a satisfied octet, and the total number of electrons in the molecule is 16. Perfect. If you ever have a difficult one where more than one seem possible, and you want to find out which is most stable, you use what is called the Formal Charge. In short, you want your molecule to be as close to zero as possible.
You calculate the FC of each atom by taking its # of valence e-, subtracting the number of non-bonding electrons and subtracting the number of bonds from that. The formula looks like this.
(# of Val. elec.) -(NB elec) -(# of bonds)
The formal charge of CO2 using our lewis structure is,
This will help you be able to weed out structures that have an ok valence and total, but simply don't exist because they are not the best choice. And you can be certain that you will see this material again. So learn it.
This is really easy once you know the rules. You just need to make sure you learn them, and practice. Look up the structures online after you think you have drawn them right. If you have more question, send me an email.