Quantum Numbers

Ok so it sounds like some of you are confused about quantum numbers so here we go:

Each set of quantum numbers, which will be a set of four numbers separated by commas, represents one electron and said electrons "position" in the atom. According to the Pauli Exclusion Principle, which you should know, no two electrons can have the same 4 quantum numbers. The quantum numbers each have rules discussing which numbers they are "allowed" to be, which are as follows:

n = 1, 2, 3, 4,...,infinity

l = 0, 1, 2, 3, ... , n - 1

m_l = -l , ... ,+l

m_s = -0.5, +0.5

Therefore if I have a certain n number like n = 3 I can have electrons with angular momentum numbers, l numbers, up to n - 1; which for us is 3 - 1 = 2. Thus, the L-numbers possible are l = 0, 1, 2 and thats it. IN EACH OF THOSE I can have a magnetic quantum number, m_l-number, going from -l to +l. Then in each of those we can have an electron spin of -1/2 or +1/2. So for n = 3 we have these possible electron positions:

n = 3, l = 0, m_l = 0, m_s = +1/2

n = 3, l = 0, m_l = 0, m_s = -1/2

notice that these two electrons have all the same quantum numbers except for the m_s number, or the electron spin number. Continuing on:

n = 3, l = 1,

now since l = 1 my m_l numbers can either be -1, 0, +1

n = 3, l = 1, m_l = -1,

now there are two more possibilities for m_s:

n = 3, l = 1, m_l = -1, m_s = -1/2

n = 3, l = 1, m_l = -1, m_s = +1/2

and when m_l = 0:

n = 3, l = 1, m_l = 0, m_s = -1/2

n = 3, l = 1, m_l = 0, m_s = +1/2

and finally when m_l = +1:

n = 3, l = 1, m_l = +1, m_s = -1/2

n = 3, l = 1, m_l = +1, m_s = +1/2

Now try and write all of the configurations for when n = 3, l = 2