List the values of all four quantum numbers for each of the valence electrons of radium (Ra). State how these two sets of quantum numbers satisfy Pauli exclusion principle. List all the possible values of magnetic quantum number for the electrons in praseodymium? How many magnetic quantum number values are possible for the electrons in the energy level that is closest to the nucleus of praseodymium? List the values.

Electronic configuration of Radium (Ra) : 88 : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 5d10 5f14 6s2 6p6 7s2 The valence electrons are the electrons present in the outermost orbital. The valence electrons for Ra are : 7s2 The possible 4 quantum numbers for the 2 valance electrons present are: Principal quantum number (n) = 7 Since it is an s-orbital Azimuthal quantum number (l) = 0 Since it is an s-orbital magnetic quantum number for both the electrons (ml) = 0 As both the electrons are present in same shell the have an opposite spin orientation So, spin quantum number for the electrons is (ms) = +1/2 for one electron and -1/2 for other electron. Therefore quantum numbers for 2 valance electrons is: n=7,l=0, ml=0, ms=+1/2, n=7,l=0, ml=0, ms=-1/2, According to Pauli's exclusion principle no two electrons can have all the 4 quantum numbers same. As there is a difference in the spin quantum number for the 2 electrons present it follows the Pauli's exclusion principle.Electronic configuration of Praseodymium: 59 : [Xe] 4f3 5d0 6s2 Possible magnetic quantum numbers for 2 electrons in 6s2 : s=0: So ml = 0 Possible magnetic quantum numbers for 3 electrons in 4f3 : f=3 : ml = +3, +2. +1, 0, -1, -2, -3 The energy level that is closest to the nucleus of Praseodymium is 1s orbital. And 1s orbital contains 2 electrons in it. The possible magnetic quantum numbers for electrons in s-orbital is: +1/2 and -1/2