# Arrow pushing: why does the O become positive when its lone pair becomes a double bond?

In the following image: 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?

• 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. Dec 12, 2016 at 9:35
• Thank you for the answer @QuantumAMERICCINO . In the carbocation case, does the circled plus sign now refer to valence electrons and not formal charge? Dec 12, 2016 at 10:07
• 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. Dec 12, 2016 at 10:39
• 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. Dec 12, 2016 at 17:01
• @Quantum goldbook.iupac.org/L03513.html "on the formal basis that bonding electrons are shared equally between atoms they join" Dec 12, 2016 at 17:22