How does a carbocation have a positive charge?

When an alkene bonds with an H+ ion, the electron pair from the pi bond goes towards a new dative covalent bond with the hydrogen ion, leaving, on one side of the old double bond, a carbon bonded to an extra hydrogen, and on the other side, a carbocation with a positive charge. If this gave, for example, a tertiary carbocation, I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge? I understand that it has lost electrons that were previously in the pi bond, but I don't see how it has a positive charge given that it still has more electrons than protons.

• Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the help center. Do you want to know more about this site, please take the tour.|| As for your question: A $\ce{H-}$ is not a free radical, but an anion with no unpaired electrons. If this adds to an alkene it will also result in an anion. Maybe you want to edit your question to change that. – Martin - マーチン Dec 24 '14 at 3:56
• @InquisitiveCheese: try counting up the total number of protons and electrons in your hypothetical ion. – Dissenter Dec 24 '14 at 13:24

I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge?

You need to count ALL the protons and electrons in the ion. What about the atoms other than the central carbon?

Well I get what your problem is.

The thing is that earlier C had 4 electrons in its valence shell. Now it is bonded to 3 other atoms giving it a total of 6 electrons in its shell. The octet rule says that it should have 8. Since the other carbon took away 2 electrons this means that our carbon should have a +2 charge.

Well this is where the confusion lies .

The trick is whenever we calculate formal charge we use

Formal Charge= (No.of valence electrons in unbonded state - no of lone pair electrons ) - (no. of bond pair electrons/2)

In this case the charge comes out to be (4-0) - (6/2) =+1 We usually associate half of the bond pair electrons to each atom while calculating charge. On the other hand When counting the no of electrons in valence shell we count both the electrons in the bond pair.

Good Luck :)

• Even at a lower level, when a positive species is added to a neutral species, math must apply: 0 + 1 = 1 !!! In addition, the atoms must balance. – user55119 Mar 5 '18 at 16:12