Oxidative Addition: Electron and Oxidation Accounting

Introduction

Oxidative addition is the step that lets a metal break strong bonds such as H–H, C–halogen and even C–H. It begins most cross-coupling cycles, activates hydrogen in hydrogenation and adds methyl iodide in acetic-acid manufacture. Before looking at mechanisms, it is essential to be able to do the bookkeeping: what happens to the oxidation state, the d-electron count, the coordination number and the total valence electron count when a metal adds a molecule A–B. Getting this accounting right is the key to reading any catalytic cycle.

Core explanation

The definition. In oxidative addition a metal complex LₙM reacts with a molecule A–B, the A–B bond breaks, and two new bonds, M–A and M–B, form:

LₙM + A–B → LₙM(A)(B)

Both A and B are X-type ligands afterwards, so each is counted as an anion in the ionic method.

The four changes for a mononuclear addition.

Quantity Change --- --- Formal oxidation state +2 d-electron count −2 Coordination number +2 Total valence electron count +2

The oxidation state rises because two new anionic ligands are attached, and the d-electron count falls accordingly: two metal electrons are used to form the two new M–X bonds. The total electron count rises by two because the metal gains two new X ligands, each counted as a two-electron donor in the ionic method, while losing two d electrons.

The classic example. Vaska's complex, trans -IrCl(CO)(PPh₃)₂, is square planar Ir(I), d⁸ and 16 electrons. It adds H₂ to give IrCl(CO)(H)₂(PPh₃)₂: octahedral Ir(III), d⁶ and 18 electrons, with the two hydrides cis to each other. All four changes in the table are visible.

Requirements. A complex can undergo oxidative addition only if:

- it has, or can generate, two vacant coordination sites (often it is 16 electrons or fewer, or loses a ligand first); - the oxidation state two units higher is accessible , which requires at least two d electrons (so d⁰ complexes cannot do it); - the metal is sufficiently electron-rich , since it is being oxidised.

Low-valent late metals such as Pd(0), Pt(0), Rh(I), Ir(I) and Ni(0) are therefore the usual candidates.

Ligand dissociation first. Many 18-electron precursors must lose ligands before adding. For example, Pd(PPh₃)₄ (18 electrons) loses two phosphines to give the reactive 14-electron PdL₂, which adds an aryl halide to give 16-electron square-planar Pd(Ar)(X)L₂ — a Pd(0) to Pd(II) change. The 16-electron product is the normal stable count for d⁸ square-planar complexes.

Binuclear oxidative addition. When A–B adds across two metals, each metal forms one new bond. For example, 2 [Co(CN)₅]³⁻ + H₂ → 2 [CoH(CN)₅]³⁻. Here each cobalt increases its oxidation state by one (Co(II) → Co(III)) and its electron count by one.

Step-by-step reasoning

1. Assign the oxidation state, d count, coordination number and electron count of the starting complex. 2. Check that the metal has at least two d electrons and room for two new ligands (after any dissociation). 3. Add A and B as X ligands: raise the oxidation state by two and lower the d count by two. 4. Recount the total electrons and confirm the product count is sensible (typically 16 or 18).

Visual explanation

Draw a square-planar d⁸ complex with an A–B molecule approaching perpendicular to the plane. Show the A–B bond breaking and A and B taking the two axial positions or two cis positions, producing an octahedral d⁶ complex. Label "+2 OS, −2 d, +2 CN, +2 e" beside the arrow.

Real-world analogy

Oxidative addition resembles a shop owner opening two new accounts for two new customers at once. The shop's cash in hand (d electrons) falls because it is committed to the new accounts, while the number of accounts (ligands) and the paperwork (electron count) both increase.

Real-world example

In Suzuki–Miyaura and other palladium-catalysed cross-coupling reactions used to make drugs and electronic materials, the cycle starts with oxidative addition of an aryl halide to Pd(0). Aryl iodides and bromides add readily; aryl chlorides, which are cheaper, need electron-rich, bulky phosphine ligands to make the metal reactive enough.

Why?

Why must the metal be electron-rich? Breaking A–B and forming two M–X bonds formally transfers two electrons from the metal to the incoming fragments. A metal that holds its electrons loosely, with donor ligands and low oxidation state, gives them up more readily, lowering the barrier and making the addition more favourable.

Common misconception

"Oxidative addition increases the electron count because the metal gains electrons." The metal is oxidised and loses d electrons. The total count rises only because two new ligands are added and counted as electron-pair donors.

Worked example

Question: Pd(PPh₃)₂ reacts with PhBr. Give the oxidation state, d count and electron count before and after.

Reasoning: Before: Pd(0), d¹⁰, 10 + 2 × 2 = 14 electrons, two-coordinate. After: Pd(Ph)(Br)(PPh₃)₂ with Ph⁻ and Br⁻, so Pd(II), d⁸. Electrons: 8 + 4 × 2 = 16. Coordination number rises from 2 to 4.

Answer: Pd(0), d¹⁰, 14 e → Pd(II), d⁸, 16 e.

Quick check

1. Why can a d⁰ complex such as Cp₂ZrCl₂ not undergo oxidative addition of H₂? Answer: It has no d electrons, so the metal cannot be oxidised by two units to form two new M–H bonds.

Exam focus

Memorise the table of changes (+2, −2, +2, +2) and apply it quickly. Examiners expect you to check both electron and site availability, to treat binuclear additions as +1 per metal and to use Vaska's complex as the standard example.

Advanced insight

Oxidative addition is the microscopic reverse of reductive elimination, so the position of equilibrium depends on the relative strengths of the A–B bond and the two new M–A and M–B bonds. Third-row metals such as iridium form stronger bonds and favour the oxidised product more than their second-row congeners, which is why iridium adducts are often isolable while rhodium analogues are fleeting intermediates.

Summary

Oxidative addition converts LₙM + A–B into LₙM(A)(B). For a single metal, oxidation state and coordination number rise by two, the d count falls by two and the total electron count rises by two. The metal needs two accessible d electrons, two vacant sites and enough electron richness. In binuclear additions each metal changes by one unit.

Practice questions

1. What is the change in oxidation state and electron count when Vaska's complex adds CH₃I? Answer: Ir(I) to Ir(III), and the electron count rises from 16 to 18. 2. Why must Pd(PPh₃)₄ usually lose ligands before oxidative addition? Answer: It is an 18-electron complex with no vacant sites; loss of phosphines creates the unsaturated PdL₂ that can accept two new ligands. 3. In 2 [Co(CN)₅]³⁻ + H₂ → 2 [CoH(CN)₅]³⁻, what is the oxidation-state change for each cobalt? Answer: +1, from Co(II) to Co(III), because the H₂ is shared across two metal centres. 4. Which is more likely to undergo oxidative addition: Ni(PEt₃)₄ or [Ni(H₂O)₆]²⁺? Explain. Answer: Ni(PEt₃)₄, because Ni(0) with donor phosphines is electron-rich and can reach Ni(II) after losing ligands, whereas Ni(II) aqua ions are already oxidised.