# What are the microstates of the triplet oxygen molecule?

I am trying to understand molecular term symbols of oxygen molecule. Everywhere I looked, it is claimed that the molecule has 6 microstates, three of them are singlet, and three of them are triplet. Now I can draw the diagrams for the singlet states with no problem, but I am struggling to find the tree triplet configurations. According to my microstate matrix there should be a triplet state with a total spin multiplicity of zero - how is this possible?

• Are you able to find the term symbols? What are the degeneracies of the terms? Note that the triplet state has $S = 1$ which allows $M_S = 1, 0, -1$. May 21, 2017 at 13:05
• I can find the term symbols, I have uploaded a diagram here: imgur.com/a/quLxo . The singlet state and the triplet state are both triply degenerate, and the triplet state can have a total ang. mom. of 1, 0, and -1, exactly as you have pointed out. The question is: how is the triplet state having zero total ang. mom. possible? On the table I've linked, why is the triplet state in the middle of the matrix considered triplet? Why isn't it singlet?
– user23638
May 21, 2017 at 13:11
• $M_S$ doesn't refer to a total spin angular momentum, but rather to a projection of the spin angular momentum onto the z-axis. The triplet state still possesses nonzero angular momentum ($= \sqrt{S(S+1)}\hbar$) but the projection onto the axis ($= M_S\hbar$) can be zero. In fact, there has to be one component of the triplet state ($S = 1$) with a zero projection $(M_S = 0)$. May 21, 2017 at 13:43
• (By the way the symbol $M_L$ is inappropriate for a diatomic molecule; it should be $\Lambda$ as the projection of orbital angular momentum taken is not onto the z-axis but rather the internuclear axis. But that is a relatively minor point.) May 21, 2017 at 13:47

In a $(\pi2p)^2$ configuration the orbital $\lambda$ and spin s quantum numbers have to be tabulated and combinations removed which break the Pauli principle. The total angular momentum quantum number is $\Lambda$ and spin $\Sigma$, for example the combination
$\lambda_1 = 1, \lambda_2 =1 , s_1=1/2, s_2 =-1/2$
generates total angular momentum $\Lambda = 2, \Sigma =0$ and this is part of the $^1\Delta$ state.
There are only 6 valid combinations leading to $^1\Delta, ^3\Sigma^-,^1\Sigma^+$ states of which $^3\Sigma^-$ is the ground state.