# Lewis Structures

1. Oct 30, 2005

### jrd007

Lewis Structures....

Is there any place on the web that I can check to make sure I am doing my Lewis structures correctly?

2. Oct 30, 2005

### Cesium

3. Oct 31, 2005

### chem_tr

Hello,

In case you cannot find your Lewis structures in the site, let me instruct you some basic knowledge.

a) Find the neutral valence electron counts of the atoms, and find the total number of valence electrons. In case of negatively charged ions, this value will probably is an odd number, just add the extra electrons to yield an even number.

b) Find the number of electrons needed to reach the "octet state". This is a bit confusing sometimes, since some atoms may need more than 8 electrons. Hydrogen will need just two. If this is the case, just skip the atom and proceed the next atom. You will find it from step c.

c) Since you know the formula, you probably know how many bonds are there. If there is a confusion, turn back to step b and try some octet numbers starting from 8. Do not forget that you'll use even numbers such as 10 and 12. It is very unlikely that the octet is greater than 12, so you'll just try 3 numbers: 8, 10, and 12.

d) Now that you found valence shell electron count and the ideal octet count, just subtract to learn how many electrons are involved in the covalent bond. Divide this number by 2 to learn the number of bonds.

e) Subtract the valence shell electron number you found at step a from the electrons used in covalent bond (step d). This will give you the number of non-bonding electrons.

f) If you know about the electronegativity of the atoms involved, place the electropositive one at the center. Put the others around it. Start with single bonds to reach the total number of bonds you found at step d. If required, use double bonds when proceeding.

g) Put non-bonding electrons to the atoms involved. First, start with the atoms around, and if you get extra electrons after locating them, put these ones into the central atom. Note that this will yield a distorted (non-ideal) geometry as required by the VSEPR theory.

Now I'll give you an example about using this technique.

We'll try to find the Lewis dot structure of sulphuric acid (H2SO4). You'll use 1 for hydrogen, 6 for sulfur, and 6 for oxygen.

a) The total number of valence electrons is (2*1)+(1*6)+(4*6)=32.

b) The octet electron count is (2*2)+(1*?)+(4*8)=36+Soctet. If you use 8 for S, you'll get 44, for 10, you'll get 46, and for 12, you'll get 48 (Note these numbers and proceed with the next step).

c) The number of electrons involved in covalent bonding is 12, 14, and 16, respectively (for 8, 10, and 12 octet electrons for S). This means that there are 6, 7 or 8 bonds.

d) The number of n electrons is 20, 18 or 16.

The most electropositive element is sulfur here; you can find this from their respective pseudo-ionic charges. In most cases, hydrogen is 1+ and oxygen is 2-. Thus you'll find that sulfur is 6+. Place oxygens around sulfur and draw four single bonds. Since hydrogen cannot make more than one bond, you'll spare 2 for two hydrogens and you'll have the remaining: 0 for 6 bonds, 1 for 7 bonds, and 2 for 8 bonds.

Attach two hydrogens to two of the oxygens available. Now put the non bonding electrons around the surrounding atoms. All atoms must receive a total of 8 electrons, either by covalent bonding (1 for each atom) or n-electrons. Thus, two oxygens with hydrogens attached will have 4 n electrons, the others will have 6 each (if single bonded; for double-bonded oxygens, use 4 n electrons). You'll find the following structures:

For 6 bonded structure: H-O-S-(O-)2-O-H

For 7 bonded structure: H-O-S=O-(O-)-O-H

For 8 bonded structure: O=S-(O-H)2=O

The problem here is to deduce which one is correct. My quick "pseudo-ionic" calculation will guide you and you'll find the 8 bonded structure as the correct one.

Well, I tried to be as simple as possible, but I am using this technique almost all of the time and encountered little difficulty. I recommend (and hope) that you will find this technique useful.