**1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p 8s**

Fortunately, there is a much simpler method to get this order.

First, write a column of 's' orbitals from 1 to 8.

Second, write a second column for the 'p' orbitals starting at *n*=2. (1p is not an orbital combination allowed by quantum mechanics)

Third, write a column for the 'd' orbitals starting at *n*=3.

Fourth, write a final column for 4f and 5f. There are no elements that will need a 6f or 7f shell to fill.

Finally, read the chart by running the diagonals starting from 1s.

The graphic shows this table and the arrows follow the path to follow.

Now that the order of orbitals are known to fill, all that remains is memorizing how large each orbital is.

- s orbitals have 1 possible value of
*m*to hold 2 electrons - p orbitals have 3 possible value of
*m*to hold 6 electrons - d orbitals have 5 possible value of
*m*to hold 10 electrons - f orbitals have 7 possible value of
*m*to hold 14 electrons

For an example, take the element nitrogen. Nitrogen has seven protons and therefore seven electrons. The first orbital to fill is the 1s orbital. An s orbital holds two electrons, so five electrons are left. The next orbital is the 2s orbital and holds the next two. The final three electrons will go to the 2p orbital which can hold up to six electrons.