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I am trying to work out the ground state symmetry of electronic states. The paper states the GS symmetry is $^6A_1$ for the right hand side electron configuration and $^4B_1$ for the LHS in the D$_{2\mathrm{d}}$ point group. I understand that the symmetry is determined from direct products of the molecular orbital symmetries of the singly occupied orbitals. However I cannot get to the two symmetries listed. For the left electronic configuration it is $b_1 \times a_1 \times b_2 \times e \times e = [a_1+a_2+b_1+b_2]$ so I'm unsure of how to get $A_1$. Likewise the product e x e for the right electronic configuration produces the same results and again I'm unsure of how to arrive at $B_1$. enter image description here

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Each of the terms you generate has a different energy, and may have different spin multiplicity. The spin multiplicity $(2S+1)$ is given by the left superscript (6 and 4) . The quartet (right) it seems has the lowest energy with spin of a triplet (sym e) x doublet ($b_2) \to$ triplet + quartet. To do this add spin vectors triplet $S_T=1/2+1/2$ , doublet $S_D= 1/2$ and make series $S_T+S_D \to |S_T-S_D|$ in unit steps, (Clebsh-Gordon series) this gives total spin 3/2 & 1/2 or quartet($\uparrow\uparrow\uparrow$) + doublet($\uparrow\uparrow\downarrow$). The symmetry has next to be associated with the spin multiplicity and this is complicated. It is necessary to find the character ($\chi$) under, say, quartet multiplicity for each symmetry operation $R$ for each symmetry species $A_1,A_2 ..$ etc. $R^2$ means square the operation. In this case this is

$$\chi(quartet) = (1/6)\left([\chi(R)]^3-3\chi(R)\chi(R^2) +2\chi(R^3)\right)$$

The result has to be reduced into symmetric $\chi^+$ and antisymmetric components $\chi^-$ as $\chi^+=\left( [\chi(R)]^2+\chi(R^2)\right)/2 $ and as $\chi^-=\left( [\chi(R)]^2-\chi(R^2)\right)/2 $ and the set of characters produced reduced to find the symmetry species.

Chapter 5 of 'Symmetry & Spectroscopy' by Harris & Bertolucci gives a detailed account of how to do this.

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  • $\begingroup$ Thanks for the reply. I must add that the two electron configurations are for different metals. I'm reading Symmetry and Spectroscopy however is all this analysis necessary to derive the term symbol 4B1 for the given electron configuration on the left? $\endgroup$ – Hanros94 Sep 11 '18 at 15:49
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    $\begingroup$ unfortunately it seems so , the direct product gives you many possibilities, finding which ones have a given spin is the difficult part as you realise. $\endgroup$ – porphyrin Sep 11 '18 at 16:34
  • $\begingroup$ Ok thanks lot for the recommending "Symmetry & spectroscopy"..it's an absolute gold mine $\endgroup$ – Hanros94 Sep 11 '18 at 16:42

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