Electron configuration in the 2p orbital

When drawing orbital diagrams I know electrons are placed in the 2p boxes singly first then paired because of they have parallel rotations and the Pauli exclusion principle states that an orbital can only house 2 electrons and those electrons must be of opposite spin. My question is why do electrons behave this way in the 2p orbital but not the 1s or 2s? And why does the p subshell contain three 2p orbitals?

As far as electron configurations and orbital diagrams are concerned I've gotten to Ne and the 2p configuration confused me so I though I'd ask before moving on.

• Cross posted to Chemistry SE. – Jon Custer Nov 14 '15 at 20:40
• @JonCuster: is cross posting different from migration? – Gert Nov 14 '15 at 20:48
• He posted the same question on both sites at the same time - pretty uncool. If it might be better somewhere else, it can be migrated. But only post it once. – Jon Custer Nov 14 '15 at 20:49
• I dont understand why that isnt cool. Someone on the other site might have an answer that makes more sense to me. – Chinasa Nov 14 '15 at 20:50
• @Chinasa read this – Danu Nov 14 '15 at 21:54

Orbitals fill from lowest energy to highest energy, so for example $\mathrm{1s,2s,2p,3s,3p,4s,}$ etc.

No two electrons can have the same set of four quantum numbers, in the same orbital. In practice this means atomic (or molecular) orbitals can only accomodate 2 electrons, one with spin quantum number $+\frac{1}{2}$, one with spin quantum number $-\frac{1}{2}$.

In practice also known as the 'bus rule': electrons will fill up $p,d\:\text{and}\:\ f$ orbitals like bus passengers fill the seats on a bus; by occupying seats singularly first and sharing only when all seats have been half-filled already, see these example for nitrogen and oxygen.

And why does the p subshell contain three 2p orbitals?

For $n=2$, $m_l=-1,0,+1$ which gives three $p$ orbitals.

For $n=3$, three $p$ but also $m_l=-2,-1,0,+1,+2$ which gives five $d$ orbitals.

For $n=4$, three $p$, five $d$ but also $m_l=-3,-2,-1,0,+1,+2,+3$ which gives seven $f$ orbitals.