# 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 "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.

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
• 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. – andselisk Jan 25 at 5:53
• “Hydride” and “halide” are quite different. Which are you asking about? – Andrew Jan 26 at 2:18
• @Andrew. Good looking out. I must have been typing fast. Definitely meant hydride. Fixed it. – Yajibromine Jan 26 at 13:05
• But in the pasted table the 4and 3 values are for bridging halide. There is no entry for bridging hydride. – Andrew Jan 26 at 13:15
• BTW, I addressed the discrepancy in the edit below. It's basically the same except the halide has a lone pair it can use. – Zhe Jan 26 at 14:00

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.

• 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$_2$? – Yajibromine Jan 25 at 22:23
• 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. – Zhe Jan 26 at 4:08