Question

The oxygen atom has 4 electrons in the 2p shell. It's not easy to keep track of the orbital and spin angular momenta of 4 electrons. Fortunately, there is a trick. You can think of oxygen as having a filled 2p shell with two "holes" in it. The holes act just like electrons: s = 1/2, l=1. So the holes in oxygen have the same possible configuration as the electrons in carbon.

a) Neglecting spin-orbit coupling, make a table showing the possible electronic configurations for this shell. For each configuration, show the values of L and S, and then write it using the archaic spectroscope notation. What is the total degeneracy of each of the conficurations? (Put that in the table too.) Make sure that each of your configurations obeys the Spin-Statistics Theorem, i.e. the state is antisymmetric under exchange.

b) Which of those configurations has the lowest energy according to Hund's First Rule? c)Now include spin-orbit coupling, and determaine the value of J that minimizes the energy. Is your answer for oxygen the same as the answer you would have given for carbon? Why or why now?

Answer 1

	2	0	-2	1	1	1	1	0	0	0	0	-1	-1	-1	-1
	0	0	0	1	0	0	-1	1	0	0	-1	1	0	0	-1

the 2p⁴ electronic configuration is equivalent to 2p² configuration. the number of ways two electrons can be arranged in the p orbital is shown above in the table.( left arrow meaning spin up and right arrow meaning spin down). there are total 15 ways of doing so as shown above.

By noting the value of and for each configuration we can assign the term value for each configuration. one thing to note here is that p² electrons here being equivalent electrons we have to take paulis exclusion principle into account. M_l and M_s are projections of vectors L and S along a fixed laboratory axis. for L=2 the projections will be along 2,1,0,-1,-2. hence the pairs of (M_l , M_s) corresponding to (2,0),(1,0),(0,0), (-1,0),(-2,0) generate D term. now for L=1, the max S value along with it is S=1. the pairs corresponding to it generate the terms. simlarly for L=0 and S=0 we get the term. hence for equivalent p2 electrons the term symbols are , , ..

b) for finding the ground state configuration, we follow Hund's rule. the term with highest multiplicity lies lowest. also, among the terms with the same multiplicity, for more than half filled subshells, the term with the highest J value lies lowest. hence for p⁴ configuration is the ground state.

in the second part of the question, the ground state of carbon will be different. the term symbols will be the same, as the p² and p⁴ configurations have the same term symbols. however, carbon having less than half filled subshell, has its ground state term as