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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 "Organotransition metal chemistry" 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 $\ce{e^-}$ "X-ligands", 2 $\ce{e^-}$ "L-ligands", 3 $\ce{e^-}$ "LX-ligands" 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 "anionic, 4-e ligand"), and 3 in the covalent model.

Table 1.1. Electron counts and changes of common ligands.
Table 1.1. Electron counts and changes of common ligands.

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 "donated" electrons are actually from the metals? Or the H-M composite ligand which we can view as then binding to the third metal?

Reference

  1. Hartwig, J. F. Organotransition Metal Chemistry: From Bonding to Catalysis; University Science Books: Sausalito, California, 2010. ISBN 978-1-891389-53-5
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  • $\begingroup$ You might want to have a look at the original Green's paper (doi.org/10.1016/0022-328X(95)00508-N, PDF) where the covalent bond classification (CBC) method is described in greater detail. Also, see this answer by Tyberius. $\endgroup$ – andselisk Jan 25 at 5:53
  • $\begingroup$ “Hydride” and “halide” are quite different. Which are you asking about? $\endgroup$ – Andrew Jan 26 at 2:18
  • $\begingroup$ @Andrew. Good looking out. I must have been typing fast. Definitely meant hydride. Fixed it. $\endgroup$ – Yajibromine Jan 26 at 13:05
  • $\begingroup$ But in the pasted table the 4and 3 values are for bridging halide. There is no entry for bridging hydride. $\endgroup$ – Andrew Jan 26 at 13:15
  • $\begingroup$ BTW, I addressed the discrepancy in the edit below. It's basically the same except the halide has a lone pair it can use. $\endgroup$ – Zhe Jan 26 at 14:00
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bridging hydride

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.

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  • $\begingroup$ For clarification, when we treat the M-H as an incoming L-type ligand to our second metal, and have it donate 2 $e^-$'s for a total of 3, are we not double-counting that first $e^-$? In other words: If the hydride H$^-$ acts as an X ligand upon binding to M$_1$, and then that system M-H acts as a L ligand upon binding to M$_2$, haven't we only really tracked 2 electrons? The first being donated by the hydride in an X fashion, and then the second being donated along with that first in a 2 electron L fashion to M$_2$? $\endgroup$ – Yajibromine Jan 25 at 22:23
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    $\begingroup$ We are double counting but we're supposed to double count because those electrons are contributing both to the first metal center and to the second metal center separately. $\endgroup$ – Zhe Jan 26 at 4:08

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