Calculate the minimum and maximum number of electrons which may have magnetic quantum number, m = +1 and spin quantum number, s = -$\frac{1}{2}$, in chromium (Cr):

Out of 6 electrons in 2p and 3p must have one electron with m = + 1 and s = -$\frac{1}{2}$ but in 3d-subshell an orbital having m = +1 may have spin quantum no.
Therefore, minimum an maximum possible values are 2 and 3 respectively.