130_notes.dvi

26.6.4 Oxygen Ground State

Oxygen, withZ= 8 has the 1S and 2S levels filled givingj= 0 as a base. It has four valence 2P
electrons which we will treat as two valence 2P holes. Hund’s first rule, maximum totals, tells
us to couple the two hole spins tos= 1. This is the symmetric spin state so we’ll need to make
the space state antisymmetric. Hund’s second rule, maximumℓ, doesn’t play a role because only
theℓ= 1 state is antisymmetric. Since the shell is more than half full we couple to the the highest
j=ℓ+s= 2. So the ground state is^3 P 2.

mℓ e

1 ↑↓

0 ↑

-1 ↑

s=

∑

ms= 1

ℓ=

∑

mℓ= 1

26.7 Homework Problems

1. List the possible spectroscopic states that can arise in the following electronic configurations:
   (1s)^2 , (2p)^2 , (2p)^3 , (2p)^4 , and (3d)^4. Take the exclusion principle into account. Which should
   be the ground state?


2. Use Hund’s rules to find the spectroscopic description of the ground states of the following
   atoms: N(Z=7), K(Z=19), Sc(Z=21), Co(Z=27). Also determine the electronic configuration.


3. Use Hund’s rules to check the (S,L,J) quantum numbers of the elements withZ=14, 15, 24,
   30, 34.

26.8 Sample Test Problems

1. Write down the electron configuration and ground state for theelements fromZ= 1 toZ= 10.
   Use the standard^2 s+1Ljnotation.


2. Write down the ground state (in spectroscopic notation) for the element Oxygen (Z= 8).