Why does Methyl Carbocation have p orbital For a carbocation to form, it needs to lose an electron. In the example below the leaving group is Br and the products are a methyl carbocation and a bromide ion. 



What I am unsure about is why the carbocation needs an empty p orbital. For a carbon atom with 6 valence electrons, the molecular orbital diagram looks like the one below. So when an electron is lost in heterolysis the $\ce{2p}$ orbital will be empty and I assume that is why the p orbital is empty. However, why do you include the empty p orbital in the shape. Does it serve a purpose?
You might want to read up on hyperconjugation. 

Basically, the R groups (if NOT a hydrogen) can donate some electron density into the carbocation’s empty p orbital, stabilising the positive charge and increasing the ion’s stability. This is the reason for Markovnikov’s Rule when it comes to Sn1 mechanisms; the tertiary carbocation is most stable, followed by a secondary one, while a primary/methyl carbocation hardly exists at all.