Metal Hydrides

Introduction

Hydrogen is the smallest ligand, yet metal–hydrogen bonds sit at the centre of hydrogenation, hydroformylation, alkene isomerisation and many reductions in industry and in enzymes. A hydride ligand has no lone pairs left over after bonding, no π orbitals and almost no steric bulk, so it provides the cleanest example of an X-type σ ligand. This page explains how hydrides are counted, how they are detected, how they differ from bound H₂ and why the same M–H bond can deliver H⁻, H⁺ or H• depending on the metal.

Core explanation

Classification and counting. Hydride is an X ligand. In the neutral (covalent) method it contributes one electron; in the ionic method it is H⁻, contributing two electrons and raising the metal's oxidation state by one. For example, HMn(CO)₅ contains Mn(I), d⁶, and 18 valence electrons by either method. The name "hydride" reflects this formal counting; it does not mean the hydrogen actually carries a large negative charge.

Bonding. The M–H bond is a two-centre, two-electron σ bond formed mainly from hydrogen 1s and a metal s/d hybrid. Typical M–H distances are about 1.5–1.7 Å for first- and second-row metals. Because hydrogen scatters X-rays so weakly, its position in crystal structures is often located more reliably by neutron diffraction.

Spectroscopic signatures. Hydrides of d-block metals with partially filled d shells usually give ¹H NMR signals at unusually low chemical shift, commonly between 0 and −25 ppm, well away from organic signals. Coupling to ³¹P reveals geometry: a large ²J(P–H) of roughly 100–200 Hz indicates a trans phosphine, while cis couplings are typically 10–40 Hz. Infrared ν(M–H) bands appear around 1700–2300 cm⁻¹ and shift to lower wavenumber by about a factor of √2 on deuteration.

Dihydrogen versus dihydride. H₂ can bind side-on as an intact σ-complex, donating from its σ bonding orbital and accepting back-donation into its σ orbital. Kubas's tungsten complex W(CO)₃(PiPr₃)₂(η²-H₂) was the first recognised example. Strong back-donation stretches and eventually breaks the H–H bond, giving a dihydride by oxidative addition. Dihydrogen complexes show short H–H distances (about 0.8–1.0 Å), short NMR T₁ relaxation times and large ¹J(H–D) couplings in HD isotopologues, while dihydrides do not.

How hydrides form. Common routes include oxidative addition of H₂, β-hydride elimination from alkyl ligands, protonation of electron-rich metals, heterolytic splitting of H₂ across a metal and a basic ligand, and hydride transfer from main-group reagents such as borohydrides.

Three modes of reactivity. An M–H bond may transfer H⁻ to an electrophile (high hydricity, favoured by electron-rich metals and donor ligands), release H⁺ to a base (acidic hydrides such as HCo(CO)₄, which is a strong acid in water), or transfer a hydrogen atom to a radical (weak M–H bonds). The same metal can switch mode when ligands or charge change.

Step-by-step reasoning

1. Count each hydride as X: one electron (neutral method) or two electrons as H⁻ (ionic method). 2. Assign the oxidation state, adding +1 for each hydride. 3. Check the ¹H NMR shift and J(P–H) couplings to confirm the M–H bond and its position. 4. Decide whether the complex is likely to act as a hydride donor, proton donor or hydrogen-atom donor from charge and ligand environment.

Visual explanation

Draw three boxes. The first shows H–H far from the metal. The second shows H₂ side-on with a donation arrow from σ(H–H) to the metal and a back-donation arrow into σ (H–H). The third shows two separate M–H bonds with the hydrogens cis . The progression pictures bond weakening, then bond breaking.

Real-world analogy

A metal hydride is like a courier who can deliver a parcel in three different ways — handing it over complete, handing over only the box, or leaving just the contents — depending on who opens the door. The recipient (substrate) and the courier's employer (ligand set) decide which delivery happens.

Real-world example

Industrial hydroformylation catalysts such as HRh(CO)(PPh₃)₃ and HCo(CO)₄ operate through metal hydrides. The hydride adds across an alkene to create a metal alkyl, which then carries the carbon chain forward to the aldehyde product; H₂ later regenerates the hydride to close the cycle.

Why?

Why do hydride signals appear at such low chemical shift? Partially filled metal d orbitals generate local magnetic fields that strongly shield the adjacent hydrogen nucleus. The effect is related to the metal's electronic structure, not to a large negative charge on hydrogen, which is why d⁰ and d¹⁰ hydrides often appear at more normal shifts.

Common misconception

"Metal hydrides always behave like the hydride ion in NaH." Many do not: HCo(CO)₄ is acidic and releases H⁺, and some hydrides transfer hydrogen atoms. "Hydride" is a formal counting label; real reactivity depends on the metal and ligands.

Worked example

Question: Give the oxidation state, d-electron count and total valence electron count of RhH(CO)(PPh₃)₃.

Reasoning: Ionic method: H⁻ gives Rh(I), so rhodium is d⁸ (group 9 minus 1). Electrons: 8 (Rh⁺) + 2 (H⁻) + 2 (CO) + 3 × 2 (PPh₃) = 18.

Answer: Rh(I), d⁸, 18 electrons, a five-coordinate saturated complex that must lose a phosphine to bind an alkene.

Quick check

1. How does the ¹H NMR spectrum help show that H₂ binds as an intact dihydrogen ligand rather than as two hydrides? Answer: HD complexes show a large ¹J(H–D) coupling and short T₁ relaxation times, indicating an H–H bond is still present.

Exam focus

Count hydride as X and remember each one raises the oxidation state by one. Quote the characteristic negative ¹H chemical shifts and trans -phosphine couplings. Distinguish dihydrogen complexes from dihydrides and name the three reactivity modes.

Advanced insight

Hydricity and acidity can be placed on common thermodynamic scales, often measured in acetonitrile, which lets chemists design catalysts for CO₂ reduction or H₂ production by matching the hydride donor strength of a metal hydride to the hydride affinity of a substrate. Hydrogenase enzymes achieve heterolytic H₂ splitting using a pendant amine base next to iron or nickel sites.

Summary

Hydride is an X ligand: one electron by the neutral method or H⁻ by the ionic method, raising oxidation state by one. M–H bonds give low-shift ¹H NMR signals and characteristic IR bands. H₂ may bind intact as a σ-complex or split into a dihydride. Depending on the metal, M–H bonds donate H⁻, H⁺ or H•, making them versatile agents in catalytic hydrogen transfer.

Practice questions

1. What is the oxidation state of iron in FeH₂(CO)₄? Answer: Fe(II), since each hydride is counted as H⁻ and CO is neutral. 2. A hydride signal shows a doublet with J(P–H) = 150 Hz. What does this suggest? Answer: The hydride is trans to a single phosphorus ligand, since large couplings indicate a trans arrangement. 3. Why is HCo(CO)₄ acidic rather than hydridic? Answer: The strongly π-accepting carbonyls stabilise the conjugate base [Co(CO)₄]⁻, so loss of H⁺ is favourable. 4. Name two ways a metal hydride can be formed in a catalytic cycle. Answer: Oxidative addition of H₂ to the metal and β-hydride elimination from a metal alkyl.