In the following image:

enter image description here

The oxygen's top pair of electrons forms a double bond.

But in a covalent bonds, aren't the electrons shared? So won't the oxygen still have these $2$ electrons (that it turned into a double bond) in its outer shell? Leaving its charge as $0$?

So why would its charge change?

I think what is confusing me is what the negative and positive signs mean and how this relates to it's formal charge and its octet.

I know the left structure's oxygen has a formal charge of $0$, and the right hand structure's oxygen has a formal charge of $+1$.

So are plus and minus signs showing formal charge, and not the number of valence electrons?

If so, does the plus sign have nothing to do with the number of valence electrons?

  • 1
    $\begingroup$ It is formal charge. Formal charges are usually expressed with the charge enclosed by a circle. However, this notation is also used for denoting carbocations. This explains why there's a positive charge enclosed by a circle in the left side of your drawing. $\endgroup$ Commented Dec 12, 2016 at 9:35
  • $\begingroup$ Thank you for the answer @QuantumAMERICCINO . In the carbocation case, does the circled plus sign now refer to valence electrons and not formal charge? $\endgroup$
    – K-Feldspar
    Commented Dec 12, 2016 at 10:07
  • $\begingroup$ Yes for carbocations. Electrons are "missing" in carbocations. The carbon in a carbocation does not have a full octect. This is not true for carbanions, however. A carbanion has a full octet. The minus sign in carbanions denote formal charge. $\endgroup$ Commented Dec 12, 2016 at 10:39
  • $\begingroup$ I don't really get what you are trying to say, @Quantum. Those charges are precisely formal charges. In $\ce{CH3+}$ if you cleave all three C-H bonds homolytically (as required for the determination of formal charge) you end up with three electrons on carbon and hence a +1 formal charge. $\endgroup$ Commented Dec 12, 2016 at 17:01
  • 1
    $\begingroup$ @Quantum goldbook.iupac.org/L03513.html "on the formal basis that bonding electrons are shared equally between atoms they join" $\endgroup$ Commented Dec 12, 2016 at 17:22

1 Answer 1


Ah, the confusion that layman’s terms can cause. You are assuming sharing to mean ‘both have the same electrons as before, they just share a little bit of them’ — but that’s not really the nature of a chemical bond let alone the idea behind formal charges. Rather, once a molecule forms the electrons are delocalised across the entire molecule and the Lewis structures we draw are a mere approximation that makes things easier for our brains.

When determining things such as formal charges, you should stick exactly to the book and perform it strictly as directed. To determine a formal charge, all bonds must be formally cleaved homolytically, i.e. oxygen only gets two electrons back from the double bond (and another from the other single bond). Then count the resulting electrons and compare to the initial state. Since you will arrive at five electrons for oxygen (which is one less than six), it gets a single, formal positive charge. And indeed physically, there is less negative charge density on oxygen than there was before. (But note that not all formal charges correspond well with actual charge density.)

This is a distinct process from determining oxidation states, whereby you must cleave the bonds heterolytically with the electronegative atom taking it all.

In none of these formal processes you can say ‘but it was originally oxygen’s electrons …’ You cannot make electrons wear pink and yellow hats (not even during Winterbash) and thus you cannot distinguish them.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.