9. Find all of the term symbols for the s^{1}p^{1} electron configuration, including J values. Energy order these states.

Answer

In this case there are no Pauli forbidden states because each electron is required to have a different *l* quantum number. Thus, L = 1 and S = 1. The total number of microstates is (6!/1!5!)(2!/1!1!) = 12.

M |
|||

1 |
_{s}^{+}, 1_{p}^{+}) |
_{s}^{+}, 1_{p}^{–}) (0_{s}^{–}, 1_{p}^{+}) |
_{s}^{–}, 1_{p}^{–}) |

0 |
_{s}^{+}, 0_{p}^{+}) |
_{s}^{+}, 0_{p}^{–}) (0_{s}^{–}, 0_{p}^{+}) |
_{s}^{–}, 0_{p}^{–}) |

–1 |
_{s}^{+}, –1_{p}^{+}) |
_{s}^{+}, –1_{p}^{–}) (0_{s}^{–}, –1_{p}^{+}) |
_{s}^{–}, –1_{p}^{–}) |

Find the states:

M

_{L}= 1, M_{S}= 1:^{3}P (9 states), J = 2, 1, 0M

_{L}= 1, M_{S}= 0:^{1}P (3 states), J = 1

Since the configuration is less than half–filled, the smallest J value is the lowest energy, so the energy order is ^{3}P_{0} < ^{3}P_{1} < ^{3}P_{2} < ^{1}P_{1}.