The four quantum numbers (n, L, mL, ms) are well defined in the literature. Where; ms is magnetic moment associated with spin; ms = ±½ (spin up, spin down) It is reasonable to assume that the rotation of a bound electron may set up a magnetic dipole. If a magnetic dipole does exist then the magnetic moment “associated with orbit” (mn) may have two possible values; mn = ±½ Where; mn = -½ represents “dipole north” mn = +½ represents “dipole south” Does a bound electron have orbital magnetic moment? Reference; http://newstuff77.weebly.com page 01 The Pyramid Periodic Table
Yes, if it has orbital angular momentum. Any form of angular momentum of a charged body creates a magnetic dipole. You can see this in the splitting of spectral lines under the influence of an external magnetic field, in the Zeeman Effect (pronounced "zayman" by the way: he was Dutch). Here is a link: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html This is a bit complicated but essentially the various possible orientations of the magnetic dipoles due to both orbital and spin angular momentum with the external field lead to states that have different energy levels, thus causing more spectral lines to appear when a field is applied. Bear in mind though that in s orbitals the electron has zero orbital angular momentum.
The short answer is "yes", as exchemist said. But your post seems to be mixing up the orbital motion of an electron in an atom with the electron's spin. In fact, both of those motions (if you want to call spin a motion) contribute to the atomic magnetic moment.
