Hybridization with ligand electrons

The majority of RX and AnX compounds with X being a pnictide (N, P, As, Sb, Bi) or a chalcogenide (S, Se, Tfe) crystallize in the NaCl fee cubic crystal structure. The materials formed with the anomalous f-elements have played a central role in unravelling the electronic structure properties where the f-electron orbitals are partially delocalized from the R or An ion. As mentioned, hybridization with ligand electrons is the important ingredient for the NaCl compounds. Rather complex spin structures are often present despite the simple crystal structure. The latter does ease the theoretical interpretation. 

Discussion Since dv v exceeds the Hill limit in these materials the hybridization with ligand states dominates the mechanisms of the 5f electron delocalization. The d-states are situated well below EF in AnUX3 compounds and therefore they do not hybridize with the 5f states of the actinide atoms. If d-states with relatively high N(E) are present in the valence band (which is undoubtedly the case in UT3 compounds) the strength of the 5f-d hybridization becomes a very critical parameter. It depends mostly on details of the mutual position (overlap) of the d-states and the 5f-states which remain pinned at (or near) EF. 

The free valence electron pairs on the central atom seek high -character i.e., sp" hybridization. If the number of ligands is larger than 4 and one or more of them are free valence election pairs, then as many F ligands form linear semi-ionic 3 center-4 electron bonds as are required to allow the free electron pairs to form an sp" hybrid with the remaining F ligands. These semi-ionic 3c-4e bonds are considerably weaker and longer than the mainly covalent sp hybrid bonds. 

In open shell metals, these empty states can be d- or f-states somewhat hybridized with band states (see Chap. A). In a metal, these states may be pulled down into the conduction band (as a virtual state, see Chap. A) in a compound, presenting a ligand valence band (insulator or semiconductor), they may be pulled down to an energy position coinciding with or very near to this valence band (as a true impurity level). The two possible final states (Eqs. 22 a and 22 b) explain the occurrence of a split response one of the crystal band electrons occupies either the outer hole level P (Eq. 22 a) or the more bandlike hole B " (Eq. 22 b). 

In square planar complexes, the metal uses a set of four hybrid orbitals called dsp2 hybrids, which point toward the four corners of a square. By pairing up the two unpaired d electrons in one d orbital, we obtain a vacant 3d orbital that can be hybridized with the 4s orbital and two of the 4p orbitals to give the square planar dsp2 hybrids. These hybrids form bonds to the ligands by accepting a share in the four pairs of ligand electrons ... 

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