Subject: Hund's rule for Phosphorus(15)

I think I understand Hund's rule but am having trouble with only the ground 
state for Phosphorus. Basically Phosphorus has 15 electrons so the outer config. 
is 3s(2) 3p(3), using Hund's I see that the three lone electrons will maximize S 
= 3/2, so multiplicity becomes (2X1.5 + 1 =4), then the highest L is 1(so P) 
since I can place one electron in each p-shell(px,py,pz), and since the 
occupancy is half I use j = L-S = -1.5, so my final config becomes 4P(-1/2), but  
actually it is 4S(3/2). Can you help please?

Re: Hund's rule for Phosphorus(15)