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Jump to one of the following Lewis structures' parts:   Choose One Introduction to Lewis Structures Part 1- Basics Part 2- Multiple Bonding and Polyatomic Ions Part 3- Resonance Part 4- Hypervalency Part 5- Formal Charge I Part 6- Formal Charge II Part 7- Electron Deficiency Part 8- Oxoacids Part 9- Quizzes

Pull up the Periodic Table if you need one.

The first problem is to determine the total number of valence electrons in the molecule. Thankfully, you've already learned how to determine the number of valence electrons in an atom. All you need to to do is take the number due to each atom and sum them up. Remember to keep an eye on the big picture. You are doing this so you can find out how to distribute the valence (bonding) electrons in the molecule.

Easily enough, the number of valence electrons is the same as the atom's group number. Well, as long as you're using the US column/group number designations.

The next hurdle is to determine how the atoms should be connected. In other words, what's the skeleton structure of the molecule? You'll generally be writing Lewis structures for molecules that have a central atom and surrounded by one or more atoms.

Just how do you choose the central atom? Many times it's readily apparent and you don't need to use any chemical knowledge. You can probably intuit what the central atom is in SiBr₄ and in CH₄ without knowing the explicit rules.

Molecules are many times comprised of atoms of lesser electronegativity surrounded by atoms of greater electronegativity. That's fine for SiBr₄, but what about CH₄ when going by electronegativity? Common sense dictates that the carbon atom is surrounded by the hydrogen atoms. But, the electronegativity of hydrogen is actually less than that of carbon (2.1 versus 2.4). So, what gives?

Hydrogen will never be the central atom. While I'll give an explanation in a moment, simply trust me for right now. All that needs to be done now is to arrange the rest of the atoms symmetrically about the central atom. As in the following.

SiBr₄: CH₄:

OK, the atoms are now arranged but what about the electrons? Remember, the counting you did initially to determine the total number of valence electrons in the molecule... There are two things that can be done with them. Atoms are connected by bonds and these bonds are made up of pairs of electrons. It's the sharing of electrons that creates a bond between two atoms. I haven't forgotten about the second thing that can be done with electrons but let's set that thought aside for the moment. Now that you've got the skeleton structure, you can now put the electrons in to make bonds. A pair of electrons for each bond.

SiBr₄: or CH₄: or

What's the difference between the left and right structures? Nothing except for how I've represented the bonding electrons. You can either display them as on the left with the two dots for electrons or as on the right by using a line to represent the two electrons in the bond. Whichever one you choose to use, make sure that you are comfortable with each version. Also, you are allowed to use a line only to represent a bond.

If you've been counting, you may have noticed something about the SiBr₄ structure. You counted 32 valence electrons but you've only used 8 electrons (2 for each bond). What the heck do you do with the remaining 24? Two paragraphs ago I mentioned that there were two things you could do with the valence electrons. The first was to make bonds which you've done. Now for the second thing. You can also distribute electrons about atoms. But how to do it?

If you remember when you learned how to draw Lewis diagrams for monatomic ions, you discovered that the they tend to have the same number of valence electrons as their nearest Noble Gas. In other words, 8 valence electrons in the case of monatomic anions. This is better known as a full octet of electrons or the Octet Rule. In the structures we are now studying, atoms can fulfill the Octet Rule by sharing enough valence electrons to equal a total of 8 about the atom. After making bonds between the central atom and the surrounding atom(s), you distribute the remaining electrons in pairs first about the surrounding atom(s) and then distribute any remaining electrons about the central atom. Just make sure you keep an eye on the Octet Rule when you are passing out the electrons. Note that since you are distributing them in pairs, you have to begin with an even number of valence electrons. It's not that we don't ever have an odd number of valence electrons but that it's beyond the scope of this tutorial.

Back to SiBr₄. If you remember where you were with it, you had drawn the skeleton structure and put in bonds between silicon and each of the bromine atoms for a total of 8 electrons used and 24 electrons remaining. Well, let's look at it once it's finished so you can see how the Octet Rule is satisfied for each atom:

Move the cursor over each atom to see which valence electrons are included in its octet. Remember, each bond is a pair of electrons and is shared by both atoms in the bond. Therefore, the bond electrons count towards satisfying each atom's octet.

What about the Lewis structure for CH₄? The valence electron count is 8 and once the 4 C-H bonds are made there aren't any electrons left to distribute. Didn't I just make a big deal about the Octet Rule?

Look at the Octet Rule again. The name "Octet Rule" is derived from the notion that most atoms "want" to have a valence configuration similar to that of their nearest Noble Gas and thus the magic number "8". But, what's the nearest Noble Gas to hydrogen? It's helium. Helium is in the first period on the Periodic Table and only has a 1s orbital available. A 1s orbital can only hold 2 electrons. The rest of the Noble Gases have a valence electron configuration of nsnp and can hold 8 electrons. So, for hydrogen we have to modify our rules a little and make reference to not the Octet Rule but the Duet Rule or the Modified Octet Rule (take your pick as to which name you like).

So, what needs else needs to be done to the Lewis structure for CH₄ after we've dropped the valence electrons in to make the bonds? Nothing. You're finished! Taking a closer look at it:

Move the cursor over each atom to see which valence electrons are included in its octet/duet. Remember, each bond is a pair of electrons and is shared by both atoms in the bond. Therefore, the bond electrons count towards satisfying each atom's octet/duet.

How about a list of the rules to this point?

What about a molecule like PF₃? Here's a chance to do it step by step:

PF3
Your turn- enter your answer in the first box and hit "Verify" to see how you did.
1.	Total number of valence electrons.
2. Draw the skeleton structure. 3. Connect the atoms by making bonds. 4. Satisfy the octet rule for the surrounding atoms.

If you haven't yet done so, count how many valence electrons you've used and see how many you have left. You should have 2 electrons left after fulfilling the Octet Rule for all of the surrounding atoms. The last thing to do is to is to satisfy the Octet Rule for the central atom- P. Luckily enough, P is already sharing 6 six electrons via the 3 P-F bonds it makes. Take those shared electrons, add the remaining pair of electrons and you get a full octet around P. Examining the PF₃ in the same way as SiBr₄ and CH₄:

Move the cursor over each atom to see which valence electrons are included in its octet. Remember, each bond is a pair of electrons and is shared by both atoms in the bond. Therefore, the bond electrons count towards satisfying each atom's octet.

Let's look at the final structure and see how the fluorine atoms were arranged about the central phosphorus atom. In SiBr₄ and CH₄ there wasn't a choice since it was 4 atoms surrounding one central atom. In this case, there are only 3 surrounding atoms. Did I have to start on the right and work my way around? No. It's simply the way I've chosen to draw Lewis structures. Remember, used by themselves, Lewis structures tell nothing about the shape of the molecule. I could have done any of the following:

I urge you to pick a method and stick with it. I'll generally start by putting a surrounding atom on the right and place the others clockwise out of habit.