Bridging Hydride Covalent Model Electron Count - Hartwig How can a bridging hydride be considered a 3 electron donor in the covalent model for electron counting?
On page 3 of John Hartwig's &quot;Organotransition metal chemistry&quot; textbook [1], he lists the number of donated electrons for common ligands according to the two standard models of electron counting in organometallic chemistry:
The ionic model: ligands are characterized according to their formal charges after cleaving the M-L bond under the assumption that both electrons reside on the ligand.
The covalent model: all ligands are treated as neutral, and the number of electrons they donate (1 $\ce{e^-}$ &quot;X-ligands&quot;, 2 $\ce{e^-}$ &quot;L-ligands&quot;, 3 $\ce{e^-}$ &quot;LX-ligands&quot; and so on) are determined by distributing the electrons from the M-L bond such that you generate a neutral ligand (he phrases this as a neutral organic group).
In Table 1.1 [1, pp. 3–4] a bridging hydride is listed as donating 4 electrons in the ionic model (in which it would be described as an &quot;anionic, 4-e ligand&quot;), and 3 in the covalent model.




The latter makes no sense to me. When the bonds connecting the bridging hydride are severed, the only way to produce a neutral ligand would be to generate atomic hydrogen, meaning 3 electrons from the M-H-M system would remain with the metals. How then, do we arrive at the hydride donating 3 electrons in this scheme? Is this a typo? Or is it written to somehow indicate that the 3 &quot;donated&quot; electrons are actually from the metals? Or the H-M composite ligand which we can view as then binding to the third metal?
Reference

Hartwig, J. F. Organotransition Metal Chemistry: From Bonding to Catalysis; University Science Books: Sausalito, California, 2010. ISBN 978-1-891389-53-5
A bridging hydride is both an X- and an L-type ligand. Since hydrogen has a valency of 1, the hydride itself can only be an X-type ligand for one of the metal centers. Then we treat the M-H σ bond as an L-type ligand to the other metal center. In the ionic method, this is 4 electrons. In the covalent method, this is 3 electrons.
EDIT:
It was noticed that in the table you provided, the example is actually for a bridging halide, not a hydride. This case is similar, but not exactly like the case of hydride. The main difference is that the halide has lone pairs, so those act as the L-type ligand for the second metal center instead of the M-X σ bond.