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This is taken from the book Organic Chemistry as a Second Language from page 35. In the following image: enter image description here

It is seen that when the electrons from the double bond are pushed into the lone pair electrons, we get 2 electrons as a lone pair. And when a lone pair electrons are pushed to from a double bond we remove 2 electrons from the oxygen.

The part that confuses me is how the electrons are being counted (especially for octet rule). Let's take the example of oxygen that starts out double bonded (the molecule left of the double headed arrows) and then the double bond electrons get pushed into a lone pair (Oxygen on the right of the double headed arrow with a formal charge of 1). If I count the first oxygen (double bonded oxygen to the left of the double headed arrow), I get 6 electrons where the double bond counts as 1 electron for each bond, now if I go to the right of the arrow I see that same oxygen I count 7 electrons. How can a lone pair which is made up of 2 electrons result as being counted as 1 electron when it becomes part of a bond or vice versa?

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  • $\begingroup$ I'm glad you re-posted this question. I tried to post a comment suggesting that you not delete the question altogether, but you had already deleted it! This question is much better stated and easier to read than your original post; nice job. $\endgroup$ – airhuff Feb 20 '17 at 18:23
  • $\begingroup$ Yea, I did it because I became unsure as to what I was trying to ask, and I just wanted to have a fresh start. Deleting wasn't necessary but I did it anyways. Thanks for the advice! $\endgroup$ – ahat Feb 20 '17 at 18:58
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Count 2 electrons per bond and you'll be fine. The partners of a bond usually share two electrons in their mutual orbitals. While you are correct in thinking 2 divided by 2 is 1, thats not how the octet rule works - count them all.

Left w/ dbl bond: 4 lone pair electrons + 4 bond electrons = 8.

Left w/o dbl bond: 6 lone pairs electrons + 2 bond electrons = 8.

Octet rule satisfied.

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  • $\begingroup$ Then what about formal charge? If octet rule is being satisfied how do I figure out the formal charge for an atom? How can I know that the oxygen on the left with no double bond has a formal charge of 1? $\endgroup$ – ahat Feb 20 '17 at 20:39
  • $\begingroup$ Formal charge can be tricky sometimes, with ambiguous atoms. H, S, etc. Oxygen is usually easier. Count the lone pairs as 2, count bonds as 1. Then compare with the state of the atom. Oxygen has 6 electrons to start with. So, 4 lone pairs + 4 bond electrons (/2) gives 6 and net formal charge 0. It helps "knowing" that Oxygen needs 2 bonds to be 0 formal charge, "knowing" that N needs 3, C needs 4, all these things which you learn in a while. $\endgroup$ – Stian Yttervik Feb 21 '17 at 7:49
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this image is wrong, you are trying to draw acetic acid and that is why you are confused.

there is a hydrogen on the second oxygen you have drawn the 6 valence electrons of oxygen but have magically bonded it the carbon?

6 electrons on oxygen 1 goes to an hydrogen the other 1 goes to the carbon bond.

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