Why is the groundstate electronic degeneracy of monoatomic O(g) 5? A problem in my Pchem textbook states that the electronic degeneracy of an oxygen atom is 5. My reasoning leads me to believe that the degeneracy is actually 6 (S = 1, L = 1, so (S+1)(L+1) = 6. This would indicate there are two unpaired electrons and a pair in the P orbitals... 

The only way I can reason a 5 degeneracy is if the ground state has a pair in the d-orbitals, which I do not believe is the case.
S = 1, L = 1, so (S+1)(L+1) = 6


You are using here not just alternative physics, but also alternative math. For, even if the formula were right, you should get $2 \cdot 2 = 4$ and not $6$.

Anyway, the right formula for the degeneracy is $(2S + 1)(2L + 1)$ which being used for the ground $^1D$ state of oxygen atom (i.e. for $S = 0, L=2$) indeed gives $1 \cdot 5 = 5$.