# How does a d4 high-spin complex have a spin-allowed transition without violating the selection rule for total angular momentum?

1) I am confused by the fact that the spin-allowed transition for a high-spin d4 complex is 5E→T2

Doesn't this violate the spin selection rule for total angular momentum?

2) Also, for a d6 complex, some sources I've consulted with mention a one spin-allowed transition for a high-spin complex, and no spin-allowed transitions for a low-spin complex.

From my understanding of the spin selection rules and the use of the Tanabe-Sugano diagram, I found no spin-allowed transitions for a high-spin complex and 4 spin-allowed transitions for a low-spin complex (i.e. 1A1→1T1; 1A1→1T2; 1A1→1A2; 1A1→1A2)

• Can you please describe the selection rules you are using to determine whether a transition is allowed/forbidden? From your description I can’t figure out whether you are applying these correctly and I suspect you aren’t, because there is no spin selection rule forbidding E to T2 transitions (the spin needs to stay the same, which is satisfied as long as you are going from quintet state to quintet state) and as far as I know there is no selection rule for total ang mom either. The only other rule deals with the transition dipole moment but that is always violated for d-d transitions anyway. – orthocresol Apr 13 '18 at 0:56
• Thank you for the reply. I am referring to the following rule: "The selection Rule for total angular momentum, J, is Δ J = 0 or ± 1. The transitions such as 2P1/2 → 2D3/2 and 2P3/2 → 2D3/2 are allowed, but transition such as 2P1/2 → 2D5/2 is forbidden since Δ J= 2." So for the d4 and d6 high-spin complexes, isn't Δ J = 2? – Lola Apr 13 '18 at 9:10
• These rules refer to electronic transitions in isolated atoms, not complexes. Observe that the term symbols are entirely different. S/P/D/F term symbols only apply to spherically symmetric atoms and A1/A2/E/T1/T2 only apply to tetrahedral complexes (or octahedral if you also specify the g/u parity). – orthocresol Apr 13 '18 at 11:23

The photon has one unit of angular momentum and as angular momentum is conserved this must be taken up by the molecule; in terms of symmetry species this means that the product of the symmetry species of the two states involved and of the transition dipole must be worked out. The electron spin in not affected by the photon angular momentum so the spin selection rule applies $\Delta S=0$ unless there is also some spin - orbit coupling then otherwise spin-forbidden transitions can occur but with generally low probability compared to a spin allowed one.

To work out the symmetry species involved in a transition the value of the integral

$$\int\psi_i\mu\psi_fd\tau$$

between initial $i$ and final state $f$ has to be evaluated.; $\mu$ is the transition dipole. This integral is very hard to evaluate but fortunately symmetry can help.

Using symmetry the integral can be determined to be zero (transition forbidden) or not zero. The equation to use is

$$\Gamma\psi_i \Gamma\mu \Gamma\psi_f = \;'A_1\,'$$

where $\Gamma$ means the symmetry species, $A,\; E$ etc and $'A_1\,'$ means the product must contain the totally symmetric representation in the point group used. This is the label given to the top row of characters in the table ($A_1$ or $A_{1g}$).

The transition dipole symmetry is found in the column of the point group just to the right of the characters and are the symmetry species with x, y, and z. In the $T_d$ and $O$ types of point groups the transition dipoles are either the $T_2$ or $T_1$ symmetry species respectively.

Thus in your $A_1\to T_1$ example the symmetry product is

$$A_1 \otimes T_1 \otimes T_1 = A_1$$

in an $O$ point group. The symbol $\otimes$ means multiply characters in each symmetry species column by column. Hence you can work out the product by multiplying the characters together or more easily by looking up the values in direct product tables.

In this case this is easy to do as anything multiplied by $A_1$ is unchanged (as its totally symmetric) and any symmetry species multiplied by itself always contains the totally symmetric representation $A_1$. So the transition $A_1\to T_1$ is allowed in octadedral point groups.

In the tetrahedral point group $T_d$ where the transition dipoles each transform as $T_2$ the product is

$$A_1 \otimes T_2 \otimes T_1 = T_2 \otimes T_1$$

and now the characters have to be multiplied and the the irreducible representation found (see link to answer below). Instead looking up the product in direct product tables gives

$$T_2 \otimes T_1 = A_2+E+T_1+T_2$$

and in this case as the totally symmetric representation $A_1$ is absent the transition would not be allowed. This means that any transition will be weak, i.e. has a very low but finite extinction coefficient. There are effects in molecules that 'break' the symmetry, such as vibrations distorting the molecules 'ideal' symmetry and these can make forbidden transitions slightly allowed.

This answer gives some details on point groups. Understanding group theory easily and quickly

• What happens for geometries / point groups where the transition dipole moment is not represented by one symmetry element? In $D_3h$ for example z is represented by $A^"_2$ while x and y have $E'$. Is a transition allowed once it transforms by one of the two? – Justanotherchemist Jul 21 '18 at 14:31
• Molecules always have three dipoles in different directions and so belonging to different symmetry species. Transitions involving these will lead to different excited states. – porphyrin Jul 21 '18 at 16:50