Posted by Vic on .
How can you find out how many unpaired electrons are present in gaseous Co3+ ion in its ground state?
According to the NBS Circular 467 "Atomic Energy Leves", Co(3+) has 24 electrons with structure
1s(2), 2s(2), 2p(6),3s(2), 3p(6), 3d(6), and spectroscopic designation 5D4. (The 5 is a superscript and the 4 is a subscript). Than means the total electron spin angular momentum is S = 2 and the total orbital angular momentum is L = 2. Spinwise, you need 4 unpaired electrons to get S=2. I hope this answers your question. Just knowing that you are dealing with Co+++ would not tell you the answer, unless you use the atomic energy level data derived from spectroscopy.
I was surprised to learn that the ground state configurations of neutral Cr and Co+++. which both have 24 electrons, are NOT the same. Cr has a 4s electron but Co+++ does not.
Both neutral Cr AND neutral Cu have 1 more 3d electron and 1 less 4s electron as is "expected" by the usual filling rules. And while the spectroscopy data are needed to tell you exactly what you want, you CAN make some educated guesses. For example, neutral Co is
1s2 2s2 2p6 3s2 3p6 3d7 4s2 = 27 and the 3d7 electrons are arranged with the 3d levels (5 of them) as .. .. . . .
1s2 2s2 2p6 3s2 3p6 3d6 4s0 = 24
The 6 4d6 electrons can arrange themselves as in the neutral Co above so it will have .. .. . . with an empty d orbital (i.e., two unpaired electrons) OR one of the paired electrons can fill the "unfilled" d orbital to make it .. . . . .
Remembering Hund's rule that atoms tend to maximum multiplicity, then the
.. . . . . would be the expected ground sate and that is 4 unpaired electrons.
I hope this helps.