Drawing lewis structures correctly (possibly without trial and error)..

[mahela007]

I've tried to learn the rules for drawing the lewis structures (otherwise knows as dot-cross structures) of molecules but there are a few problems..

 

First of all in some instances the octet simply cannot be completed and in others there are more than 8 electrons in the outer shell. If I encounter an unfamiliar molecule how should I set about drawing the lewis structure?

 

Secondly, I ran into a few problems with the SO4^ 2- molecule.. Why is it that the "correct" structure is a central S atom with single bonds to two O atoms and double bonds to the other two O atoms like this?

O

|

O=S=O

|

O

 

and why isn't it this..

O

||

O=S=O

||

O

Both the structures seem to be correct according to the octet rule.. I even tried using the "formal charge" but even that is the same for both molecules (correct me if I'm wrong.. .I'm not very comfortable with all this yet..)

 

Thanks in advance

Edited June 11, 2009 by mahela007

[hermanntrude]

bear in mind that the octet "rule" is more of a guideline.

 

There are three classes of atoms which can disobey the octet rule:

 

1) the atoms Beryllium and boron (and also technically lithium helium and hydrogen, but no-one uses lithium much in Lewis diagrams for molecules and He and H are usually dealt with by mentioning that they like tohave only two within the original statement of the octet rule). These atoms like to have 4 and 6 electrons respectively. The usual stated reason for this is simply that they are quite small. I suspect there's more to it than that, but don't freak about it. There are really only six molecules to look out for in a question: [ce]BeH2[/ce], [ce]BeF2[/ce], [ce]BeCl2[/ce], [ce]BH3[/ce], [ce]BF3[/ce], [ce]BCl3[/ce]. Others do occur but don't show up very often in exams. Basically, if you see B or Be, expect them to have incomplete octets

 

2) radicals. these have odd numbers of electrons and so they also fit into category 1. The most common example of a stable radical is NO (nitrogen monoxide). If your electron count comes to an odd number, first check to see if you forgot to add or deduct electrons for the charge on the ion, and if you didnt, then you're probably seeing an example of category 3. The unpaired electron usually goes on the central atom.

 

3) atoms which can have expanded valence shells. These are found in the 3rd period and lower. NEVER draw a Lewis diagram with more than 8 electrons on an atom in the second period. The reason these atoms can do this is something you may or may not have learned yet: They have access to their empty 3d subshells, which can participate in hybridization to form either sp3d or sp3d2 hybrid orbitals. the sulfate ion is a good example of this kind of species. Sulfur is in the 3rd period so it can safely break the octet rule. You can tell from the Lewis diagram in which all the atoms have complete octets that the formal charges aren't happy.

 

The structure with four single bonds has four negative formal charges, one on each oxygen and a 2+ formal charge on the central sulfur atom. For a good Lewis structure, we want the formal charges to be as small as possible. In an atom from the second period we would simply have to suck it up and obey the octet rule, but since sulfur is in the third period, it can disobey the octet rule and get some double bonds. Notice also that sulfur is on the list of atoms which tend to form double bonds (N, O, C, P and S... there are others but these are the most common).

 

What you didn't notice when you wrote about the two structures you drew was that neither of them obeys the octet rule. A double bond contains 4 electrons, not two.

 

the first structure you drew has complete octets for the oxygen atoms, but the sulfur has 12 electrons.

 

the second structure you drew also has complete octets for the oxygen atoms but the sulfur now has 16 electrons.

 

How to decide between the two? well let's look again at the formal charges. remember that the formal charge is calculated by taking the number of valence electrons on the atom itself, then subtracting one for each electron in a lone pair and one for each single bond. The first diagram you drew has two oxygens with zero formal charges, two with negative formal charges, and the sulfur has a zero formal charge too. The second one has all of the oxygen atoms with zero formal charges and a 2- formal charge on the sulfur. the reason this is no good is that if there has to be a negative formal charge (and there does have to be, since the total charge is -2), it prefers to be on the most electronegative atom, in this case, oxygen. It's also true that it's very rare for any atom to have more than 6 bonds on it. The only example I can think of are XeO4, which has 8 bonds (four doubles). So in this case the best compromise between formal charges and the octet rule is to have the first structure you drew, which has sulfur with 12 electrons, which fit nicely into the 6 sp3d2 hybrid orbitals.

 

In some cases (particularly the oxyanions of the halogens... things like [ce]ClO4-[/ce], [ce]BrO3-[/ce], etc), it becomes difficult to tell exactly how much the central atom ought to break the octet rule by, and there are some nit-picky rules to follow for that decision. However, most courses just skim over that and never ask you to draw a lewis diagram for those ions. However, a useful rule I like to follow is that for oxyanions (polyatomic ions which contain a central atom and several oxygen atoms), there are as many singly-bonded oxygens as the charge on the overall ion.

 

Here is a website with loads of examples of lewis diagrams, nicely presented to help you learn the methods. Note that it does include some of the nit-picky rules I mentioned so if you've learned another method, just use that.

 

I hope that helps you out :0)

[mahela007]

Thank you VERY much!

[hermanntrude]

Thank you VERY much!

 

Did that help, then?

 

I was bored that day... you were lucky... lol

[NATT]

Hey hey guyz.Imma new to this site and...whats upp???

[Mark_Schumacher]

I registered to this site just to thank Hermanntrude. Thank you so much.

 

I recieve a oxyanions on a worksheet today . To be specific the ClO4- . And I thought I had lewis structures mastered, I had done hours and hours of homework.

And my T.A said , this stuff is simple compared to the exam. Yah...... ? We didnt even over oxyanions or for that matter ions in lewis structure.

 

Honestly it took alot of time for you to write that , and it was crystal clear. Thanks so much. No other websites explain this

Edited October 19, 2010 by Mark_Schumacher