Michael B. answered 01/25/13
Seasoned and experienced tutor with extensive science background
Although Roman has answered the question quite thoroughly with respect to the accounting of electronic configurations one thing that was omitted is that the designations s, p, d, and f are referring to specific geometries for the orbitals themselves. Those designated "s" are spherical, "p" are teardrop (my term) shaped with a node at the origin, "d" are teardrop and toroidal with a node at the origin, and "f" are similarly teardrop and toroidal. Each increase in orbital complexity introduces more space for the electrons to occupy so that the "p" orbital have 2 lobes, "d" have 4 lobes, and "f" have 6-8 lobes; the exceptions being the "d" and "f" toroidal orbitals.
In essence, they vary with respect to their orientation about the origin and the number of nodes (where there is zero probability of an electron to occupy that space). The more nodes, the higher the energy level that the electrons must have in order to occupy them.
Michael B.
1s orbitals do not and cannot hybridize with 3d orbitals of the same atom, although they do occupy differently "shaped" regions of space that partially overlap. i.e. Even though a hydrogen atom technically has "d" orbitals, the probability of finding an electron occupying one is zero. Also, hybridizing is a summation of the entirety of the orbitals themselves (LCAO) and not just select portions of them.
03/03/13
Stanton D.
To refine Michael's note "introduces more space for the electrons to occupy", it's not that the new orbitals go out into new sections of space around the atom, it's that the same space can be re-used, over and over again. Because the wavefunctions for any two electrons, when multiplied by each other, and averaged over space, sum to zero, the electrons can just ignore each other. Pretty neat thing for them to do, especially when they start mix-and-matching portions of their orbitals ("hybridizing").
02/11/13