The 18-electron rule · printable packet

Counting valence electrons by the ionic method, the 18-electron rule for carbonyls and sandwich compounds, predicting carbonyl formulas, and the 16-electron complexes that make catalysts.

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1. What you will learn

By the end of this lesson you will be able to count the valence electrons of a complex, predict the formula of a metal carbonyl, and find the open sites of a catalyst.

2. What you already have

From lesson 2 you can find a metal's oxidation state and its d electrons from the group number. From lesson 3 you know that every donor atom gives the metal a lone pair. And from general chemistry you know the octet rule: main-group atoms are most stable with eight electrons, filling one s and three p orbitals. This lesson extends that rule to a transition metal, which has five d orbitals as well.

3. Words for this lesson

4. Nine orbitals, eighteen electrons

Term	What it means
18-electron rule	The observation that many stable complexes have eighteen electrons around the metal, filling its nine valence orbitals.
Ionic counting	Counting the metal's d electrons in its oxidation state and adding two for every donor pair.
Metal carbonyl	A complex of carbon monoxide, bonded through carbon, usually with the metal at a low oxidation state.
Back-bonding	Electron density flowing from filled metal d orbitals into empty orbitals of a ligand such as carbon monoxide.
Sandwich compound	A metal held between two flat rings, such as ferrocene, $\mathrm{Fe(C_5H_5)_2}$.
Coordinatively unsaturated	Having fewer than eighteen electrons, so a site is open for another ligand.
Organometallic compound	A compound with at least one bond between a metal and carbon.

A first-row transition metal has nine valence orbitals: five 3d, one 4s and three 4p. Just as a main-group atom is most stable with eight electrons filling its s and p orbitals, many transition-metal complexes are most stable with eighteen electrons filling all nine. This is the 18-electron rule.

Chromium hexacarbonyl, $\mathrm{Cr(CO)_6}$, has chromium(0), with six d electrons, and six carbon monoxides donating twelve: eighteen. Ferrocene, $\mathrm{Fe(C_5H_5)_2}$, has iron(II), with six d electrons, and two rings donating six each: eighteen again.

The rule also predicts formulas. A neutral carbonyl of a metal in group $g$ has $g$ d electrons, so it needs $(18 - g)/2$ carbon monoxides: six for chromium, five for iron, four for nickel, exactly the compounds $\mathrm{Cr(CO)_6}$, $\mathrm{Fe(CO)_5}$ and $\mathrm{Ni(CO)_4}$. A metal in an odd group cannot reach eighteen this way. Manganese, group 7, forms $\mathrm{Mn_2(CO)_{10}}$, in which each manganese has seventeen electrons from its own ligands and gains the eighteenth by sharing a metal-metal bond with the other. Vanadium stays at seventeen in $\mathrm{V(CO)_6}$ and readily takes an electron to become the 18-electron anion $\mathrm{[V(CO)_6]^-}$.

The rule holds best for ligands that bond strongly and accept electron density back from the metal, carbon monoxide above all, and for metals in low oxidation states: the world of organometallic chemistry. Many Werner complexes ignore it: $\mathrm{[Cr(H_2O)_6]^{3+}}$ has fifteen electrons and $\mathrm{[Ni(H_2O)_6]^{2+}}$ twenty, and both are perfectly stable.

Another way: picture

Think of the metal's nine valence orbitals as nine seats. In a carbonyl, each carbon monoxide brings two electrons and takes one seat's worth of bonding; the metal fills the rest with its own d electrons. When every seat is taken, eighteen electrons, no more ligands can sit down. When one is free, sixteen, the complex is ready to take on a new partner, and that readiness is what a catalyst needs.

Another way: steps

Find the oxidation state from the charges (neutral ligands add nothing).
$d$ = group number minus oxidation state.
Add two electrons per donor pair; six per $\mathrm{C_5H_5^-}$ or benzene ring.
Total $= d +$ ligand electrons; compare with eighteen.
To predict a carbonyl: $n = (18 - d)/2$.

5. Why eighteen, and why carbon monoxide

Carbon monoxide bonds to a metal in two ways at once. Its carbon lone pair donates into an empty metal orbital, a $\sigma$ bond. And its empty $\pi^$ orbitals accept electron density from the metal's filled $t_{2g}$ orbitals, back-bonding*. That second interaction lowers the energy of $t_{2g}$ and makes $\Delta_o$ very large, which is why carbon monoxide sits at the strong end of the spectrochemical series.

With six such ligands, the metal's orbitals sort into nine low-energy ones, six used for the bonds to the ligands and three $t_{2g}$ now stabilized by back-bonding, and a set of high-energy antibonding orbitals far above them. Eighteen electrons fill the nine low orbitals exactly. A nineteenth would have to go into an antibonding orbital, and a seventeenth leaves a low orbital half-empty, so both are less stable. For weak-field ligands such as water, the $t_{2g}$ set is barely bonding and the $e_g$ set barely antibonding, so the count matters much less, which is why Werner complexes so often break the rule.

6. Ferrocene and the sandwich compounds

In 1951 two research groups, trying to make quite different things, obtained an orange iron compound of formula $\mathrm{Fe(C_5H_5)_2}$ that was remarkably stable: it melts without decomposing and resists air, water and acids. Geoffrey Wilkinson and Ernst Otto Fischer showed independently that its iron sits between two parallel five-membered rings, bonded equally to all ten carbon atoms, a sandwich. They shared the Nobel Prize in Chemistry in 1973.

The 18-electron rule explains the stability. Each cyclopentadienide ring donates six $\pi$ electrons, and iron(II) brings six d electrons, for eighteen. The neighbors confirm it: cobaltocene, $\mathrm{Co(C_5H_5)_2}$, has nineteen electrons and gives one up very easily to become the stable 18-electron cobaltocenium ion; nickelocene, with twenty, is reactive and paramagnetic. The same count explains dibenzenechromium, $\mathrm{Cr(C_6H_6)_2}$: chromium(0), six d electrons, and two benzene rings donating six each.

Ferrocene's stability has made it useful far beyond the textbook. It loses one electron cleanly and reversibly to become the blue ferrocenium ion, and that couple is the standard reference that electrochemists quote potentials against when they work in organic solvents, where the usual hydrogen electrode cannot be used. Medicinal chemists have attached ferrocene to known drugs to change how they behave in the body; ferroquine, a ferrocene version of the antimalarial chloroquine, reached clinical trials because malaria parasites that resist chloroquine do not resist it as easily.

7. Sixteen electrons and catalysis

Square-planar $d^8$ complexes, such as those of rhodium(I), iridium(I), palladium(II) and platinum(II), are stable at sixteen electrons. In a square plane the $d_{x^2-y^2}$ orbital, pointing at the ligands, is so strongly antibonding that it stays empty, leaving eight orbitals and sixteen electrons.

A sixteen-electron complex has room for one more ligand, and that is what makes these metals such good catalysts. Wilkinson's catalyst, $\mathrm{RhCl(PPh_3)_3}$, adds hydrogen to alkenes. In its cycle it loses a phosphine to become a fourteen-electron species, binds a hydrogen molecule and an alkene in the open sites, transfers the hydrogen atoms to the alkene, and returns to where it started. Counting electrons at each step, sixteen, fourteen, eighteen and back, is how chemists check that a proposed mechanism is reasonable, and how they decide which ligand to change to make a catalyst faster.

8. Checking a count

Three checks catch most slips. First, use one method throughout: in the ionic count, chloride and cyclopentadienide are anions donating two and six, and the metal's d electrons are those of its oxidation state. Starting from the neutral metal's group number while treating the ligands as anions counts electrons twice, and inflates the count by the oxidation state.

Second, the oxidation state must make the charges balance: add the charges of the anionic ligands to the metal's oxidation state and you must get the charge of the whole complex. Third, the total should be even for any stable, diamagnetic compound, since electrons come in pairs; an odd count, such as seventeen, means a radical that will dimerize or be reduced, as vanadium hexacarbonyl is. A count above eighteen, as in nickelocene, predicts a compound eager to lose electrons or ligands. Finally, compare with a neighbor you know: moving one group to the right adds a d electron, so the same ligand set on the next metal needs one more positive charge, or one fewer pair, to stay at eighteen.

9. In the world: purifying nickel with carbon monoxide

In 1890 Ludwig Mond and his assistants found that finely divided nickel reacts with carbon monoxide at about $50\ ^{\circ}\text{C}$ to form a volatile liquid, nickel tetracarbonyl, $\mathrm{Ni(CO)_4}$, the 18-electron carbonyl of nickel(0). No other metal in crude nickel does the same under those conditions, so the carbonyl can be carried away as a gas, leaving iron, cobalt and copper behind. Heated to about $200\ ^{\circ}\text{C}$, it falls apart again and deposits nickel more than $99.9$ percent pure.

The Mond process is still used, in refineries in Wales and Canada, to make high-purity nickel pellets and powders. It has a serious hazard: nickel tetracarbonyl is one of the most toxic substances handled in industry, because it passes easily into the body and delivers both nickel and carbon monoxide to the lungs. The same 18-electron stability that makes it volatile and easy to separate makes it survive long enough to be absorbed.

10. In the world: the oxo process

Hydroformylation, also called the oxo process, adds hydrogen and carbon monoxide across the double bond of an alkene to make an aldehyde with one more carbon. The aldehydes are turned into alcohols for plasticizers, detergents and solvents, and industry makes more than ten million tonnes of them a year.

The first catalyst, discovered by Otto Roelen in 1938, was cobalt: $\mathrm{HCo(CO)_4}$, an 18-electron hydride. It must lose one carbon monoxide to become the 16-electron $\mathrm{HCo(CO)_3}$ before an alkene can bind. Modern plants mostly use rhodium with phosphine ligands, $\mathrm{HRh(CO)(PPh_3)_3}$, which works at lower temperature and pressure and gives more of the straight-chain product manufacturers want. Its cycle too begins by losing a phosphine to open a site. Counting electrons at each step tells the process chemist which step needs a ligand to leave, and so which conditions will speed it up.

11. Count one way, not two at once

There are two common ways to count electrons. The ionic method used here gives the metal its oxidation state and treats chloride, hydride and cyclopentadienide as anions donating two, two and six. The neutral method counts the metal as neutral and every ligand as a neutral fragment donating one, one and five. Both give the same total, eighteen for ferrocene. Mixing them does not: take iron's full eight electrons and count each ring as an anion donating six, and ferrocene appears to have twenty.

A second misunderstanding is to treat the rule as a law. It is a guide to stability for strong-field, back-bonding ligands and low oxidation states. Many ordinary coordination compounds, especially aqua and ammine complexes of metals in higher oxidation states, have counts from fifteen to twenty and are entirely stable.

12. Chromium hexacarbonyl

13. Counting ferrocene

14. Predicting the carbonyls of iron and manganese

15. Your turn: count the valence electrons of $\mathrm{[Mn(CO)_6]^+}$.

16. Guided practice

Match each metal to the number of carbon monoxide ligands in its stable, neutral, 18-electron carbonyl.

17. Guided practice

Complete the worked solution: count the valence electrons of $\mathrm{[Ni(CN)_4]^{2-}}$.

18. Guided practice

The metal in $\mathrm{[Ni(CN)_4]^{2-}}$ is in group $10$. How many valence electrons does it have by the ionic count?

19. Practice

For $\mathrm{Cr(CO)_6}$ (metal in group $6$, oxidation state $0$), $\mathrm{[Mn(CO)_6]^+}$ (group $7$, $1$) and $\mathrm{[Fe(CN)_6]^{4-}}$ (group $8$, $2$), in that order, fill in the metal's d electrons, the electrons the ligands donate, and the total.

20. Practice

An 18-electron carbonyl of $\mathrm{Re}$, a group $7$ metal, carries an overall charge of $1$. How many carbon monoxide ligands does it have?

21. Practice

Count the valence electrons of $\mathrm{[Co(NH_3)_6]^{3+}}$ by the ionic method. The metal is in group $9$ and its oxidation state is $3$.

22. Somewhere new

A process chemist following a catalyst through its cycle counts each species to see where a new molecule can bind. For $\mathrm{Fe(CO)_4}$ ($8$ d electrons), $\mathrm{Ni(CO)_3}$ ($10$) and $\mathrm{HRh(CO)(PPh_3)_2}$ ($8$), in that order, fill in the total valence electrons and the number of open sites, one for every two electrons short of eighteen.

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

For $\mathrm{Fe(C_5H_5)_2}$ (metal in group $8$, oxidation state $2$), $\mathrm{Cr(C_6H_6)_2}$ (group $6$, $0$) and $\mathrm{[Fe(CN)_6]^{3-}}$ (group $8$, $3$), in that order, fill in the metal's d electrons, the electrons the ligands donate, and the total.

25. What you can do now

You can count electrons at a metal. Explain why iron forms $\mathrm{Fe(CO)_5}$ and manganese forms $\mathrm{Mn_2(CO)_{10}}$ rather than a monomer.

Working for the steps left to you

15. Your turn: count the valence electrons of $\mathrm{[Mn(CO)_6]^+}$., step 3

	$6$ CO	$5$ CO	$4$ CO
$\mathrm{Mo}$
$\mathrm{Fe}$
$\mathrm{Ni}$

	valence electrons	open sites
the first species
the second species
the third species