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Heteronuclear diatomics and isoelectronic series, carbon monoxide's donor and acceptor orbitals, and the C-O stretch as a measure of back-donation.
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By the end of this lesson you will be able to find the bond order of a heteronuclear diatomic, explain how carbon monoxide bonds to a metal, and read back-donation from a C-O stretching frequency.
Lesson 24 built the molecular orbitals of two identical atoms and read bond order and magnetism from them. Lesson 15 counted carbon monoxide as a two-electron donor and said it is a $\pi$ acceptor at the strong end of the spectrochemical series. This lesson explains both facts from carbon monoxide's own orbitals, and turns the infrared stretch of a bound carbonyl into a measurement of what the metal is doing.
| Term | What it means |
|---|---|
| Heteronuclear diatomic | A molecule of two different atoms, such as CO or NO. |
| Isoelectronic | Having the same number of electrons, and so the same orbital diagram: CO, $\mathrm{CN^-}$ and $\mathrm{NO^+}$. |
| HOMO | The highest occupied molecular orbital; for CO, a $\sigma$ lone pair on carbon. |
| LUMO | The lowest unoccupied molecular orbital; for CO, the $\pi^*$ pair, largest on carbon. |
| $\sigma$ donation | A ligand giving a lone pair into an empty metal orbital along the bond. |
| $\pi$ back-donation | A metal giving electron density from a filled d orbital into an empty ligand $\pi^*$ orbital. |
| Stretching frequency | The wavenumber at which a bond vibrates, which falls as the bond weakens. |
When the two atoms of a diatomic differ, their atomic orbitals start at different energies: oxygen's lie lower than carbon's, because oxygen holds its electrons more tightly. The molecular orbitals still form as bonding and antibonding pairs, but unevenly. A bonding orbital lies closer in energy to the more electronegative atom and has more of its density there; an antibonding orbital lies closer to the less electronegative atom and is larger on it.
Carbon monoxide has ten valence electrons, the same as dinitrogen, and fills the same way: bond order three. But its two key orbitals are lopsided. The HOMO is a $\sigma$ orbital that behaves as a lone pair on carbon, pointing away from the oxygen. The LUMO is the empty $\pi^*$ pair, also largest on carbon. So CO bonds to a metal through carbon, and it does two things at once:
The two reinforce each other: the more the carbon donates, the richer the metal, and the more it can give back. Back-donation is what makes CO a strong-field ligand and what stabilizes metals in low oxidation states, the metal carbonyls of lesson 15.
It also leaves a fingerprint. Electrons pushed into CO's $\pi^$ orbital sit in an antibonding orbital of the C-O bond, so they weaken it. Free carbon monoxide stretches at $2143$ cm$^{-1}$; in $\mathrm{Cr(CO)_6}$ the stretch falls to about $2000$, and in the anion $\mathrm{[V(CO)_6]^-}$ to $1860$. The drop below $2143$* measures how much the metal back-donates: the more negative and electron-rich the metal, the larger the drop.
Another way: picture
Picture a handshake that goes both ways. Carbon monoxide reaches out with its lone pair and the metal takes it; the metal reaches back with a filled d orbital into CO's empty $\pi^*$. The firmer the metal's grip, the more it tugs the C-O bond apart from the inside, and the lower the note the bond sings at in the infrared.
Another way: steps
Carbon monoxide, the cyanide ion and the nitrosonium ion, $\mathrm{NO^+}$, all have ten valence electrons and the orbital diagram of dinitrogen: bond order three, no unpaired electrons. They differ in charge and in how lopsided their orbitals are, and that decides how they behave as ligands. Cyanide, negatively charged, is a stronger $\sigma$ donor and a weaker $\pi$ acceptor than CO; $\mathrm{NO^+}$, positively charged, is the reverse.
Add an electron to $\mathrm{NO^+}$ and it becomes nitric oxide, $\mathrm{NO}$, with eleven: the extra electron goes into $\pi^*$, the bond order falls to two and a half, and one electron is unpaired. Nitric oxide is a radical, which is why it reacts readily and why, in the body, it serves as a short-lived signaling molecule that relaxes blood vessels. Add another and $\mathrm{NO^-}$, with twelve, has two unpaired electrons and bond order two, just like dioxygen, with which it is isoelectronic.
The C-O stretch sits in a part of the infrared spectrum, roughly $1{,}700$ to $2{,}150$ cm$^{-1}$, where almost nothing else absorbs, and it is intense, so chemists use it constantly. The position of the band reports the electron density at the metal. Across the isoelectronic series $\mathrm{[Ti(CO)_6]^{2-}}$, $\mathrm{[V(CO)_6]^-}$, $\mathrm{Cr(CO)_6}$ and $\mathrm{[Mn(CO)_6]^+}$, every metal has six d electrons, but the charge changes, and the stretch climbs from $1750$ through $1860$ and $2000$ to $2090$ cm$^{-1}$ as the metal gets less electron-rich.
Other ligands on the metal change the stretch too. A ligand that donates strongly makes the metal richer and lowers the carbonyl stretch; a ligand that accepts strongly competes with CO for the metal's electrons and raises it. That is the basis of a widely used scale, described in the applications, for ranking how strongly a ligand donates. The number of carbonyl bands, counted with the symmetry methods of lesson 23, gives the arrangement of the carbonyls as well.
Carbon monoxide is the clearest case of back-bonding, but not the only one. Phosphines, $\mathrm{PR_3}$, donate through the lone pair on phosphorus and accept back-donation into empty orbitals of the phosphorus-carbon bonds, more strongly when the groups on phosphorus are electronegative. That is why trifluorophosphine behaves much like carbon monoxide, and why chemists can tune a metal's electron density by choosing the groups on a phosphine.
Alkenes bind the same two ways. The filled $\pi$ orbital of the carbon-carbon double bond donates into the metal, and a filled metal d orbital back-donates into the alkene's empty $\pi^*$. This picture, named for Michael Dewar, Joseph Chatt and Leonard Duncanson, explains why a bound alkene's C=C bond lengthens and its stretch falls, just as for carbon monoxide, and it underlies every catalytic reaction in which a metal handles an alkene, from polymerizing ethylene to adding hydrogen across a double bond.
Even dinitrogen, the most unreactive of common gases, binds to electron-rich metals in the same way, weakly donating and accepting back-donation into its $\pi^$ orbitals. The more back-donation, the more the N-N bond is weakened, and chemists trying to turn atmospheric nitrogen into ammonia under mild conditions design metal complexes to push as much density into dinitrogen's $\pi^$ as possible. The nitrogenase enzymes that do this in soil bacteria bind dinitrogen at an iron-molybdenum cluster, and the first step of their chemistry is thought to be exactly this weakening of the triple bond.
Three checks catch most slips. First, the direction: back-donation always lowers the C-O stretch, never raises it, because it fills an antibonding orbital. A bound carbonyl stretching above $2143$ cm$^{-1}$ is unusual and means the metal is so electron-poor that $\sigma$ donation dominates, as in some cationic metal carbonyls of late metals.
Second, compare like with like. A tetracarbonyl and a hexacarbonyl, or complexes of different metals, differ in more than charge, so the trend with charge is cleanest within an isoelectronic series. Third, the bond order of CO itself stays near three in all of these: back-donation lowers it by a fraction, not by a whole bond, which is why even the lowest carbonyl stretches stay far above the $1{,}200$ or so of a C-O single bond.
A last check is the electron count of lesson 15. A carbonyl complex that obeys the eighteen-electron rule has its metal's $t_{2g}$ orbitals full, and those are exactly the orbitals that back-donate. A complex with fewer d electrons has less to give, so its carbonyls stretch higher, and a trend that runs the other way deserves a second look at the electron counts before it is believed, and at whether the complexes being compared really share a structure.
Hemoglobin carries oxygen on iron(II), and carbon monoxide binds to the same iron about two hundred times more tightly than oxygen does. The reason is back-donation: iron(II) in the heme has filled d orbitals that push electron density into carbon monoxide's empty $\pi^*$ orbitals, adding a strong $\pi$ component to the iron-carbon bond that oxygen cannot match as well.
Once carbon monoxide occupies a heme, that site cannot carry oxygen, and the remaining sites in the same hemoglobin hold their oxygen more tightly, so less is released to the tissues. Treatment is breathing pure oxygen, sometimes under pressure in a hyperbaric chamber, which shifts the competition back toward oxygen. Home carbon monoxide detectors, required in new U.S. homes by many state building codes, exist because the gas is colorless and odorless and its binding is so strong.
Chemists designing catalysts need to know how strongly each candidate ligand donates electron density to the metal, and they measure it with carbon monoxide. In 1977 Chadwick Tolman proposed a standard: make the complex $\mathrm{Ni(CO)_3L}$ with the ligand $L$ and record the frequency of its main C-O stretch. A strongly donating ligand makes the nickel richer, the nickel back-donates more to the three carbonyls, and the stretch falls.
Tri-tert-butylphosphine, a strong donor, gives $2056.1$ cm$^{-1}$; trifluorophosphine, a poor donor and good acceptor, gives $2110.8$. The Tolman electronic parameter is still quoted for new ligands today, and catalyst makers use it to choose ligands that make a metal richer, for oxidative addition, or poorer, for reductive elimination, steps met in unit 8.
Oxygen is more electronegative and carries more of the electron density, so it seems natural that it should be the atom that bonds to a metal. It is not. The orbital that donates, carbon monoxide's highest occupied orbital, is a lone pair on carbon, and the orbital that accepts back-donation, the $\pi^*$ pair, is largest on carbon too. Both point the carbon at the metal, and metal carbonyls are M-C-O, not M-O-C.
The companion error is thinking back-donation strengthens every bond involved. It strengthens the metal-carbon bond, which gains $\pi$ character, but weakens the C-O bond, whose antibonding orbital is being filled. The falling C-O stretch and the stronger metal-carbon bond are two sides of the same effect.
Count the valence electrons.
$4 + 6 = 10$
Carbon brings four, oxygen six.
Choose the ordering.
$\text{as for } \mathrm{N_2}$
Ten electrons, with the $\sigma$ orbital above the $\pi$ pair.
Fill the orbitals.
$(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4(\sigma_{2p})^2$
Every electron paired.
Count bonding and antibonding electrons.
$8 \text{ and } 2$
As in dinitrogen.
Find the bond order.
$\tfrac{1}{2}(8 - 2) = 3$
A triple bond, with the highest filled orbital a lone pair on carbon.
Count the valence electrons.
$5 + 6 = 11$
One more than carbon monoxide.
Place the eleventh electron.
$(\pi^*_{2p})^1$
The first antibonding $\pi$ electron.
Count bonding and antibonding electrons.
$8 \text{ and } 3$
Two in $\sigma^_{2s}$, one in $\pi^$.
Find the bond order.
$\tfrac{1}{2}(8 - 3) = 2.5$
Between a double and a triple bond.
Count the unpaired electrons.
$1$
A radical, reactive and paramagnetic.
Compare with the nitrosonium ion.
$\mathrm{NO^+}: \ 10 \text{ electrons, order } 3$
Removing the $\pi^*$ electron strengthens the bond.
List the stretches.
$\mathrm{[V(CO)_6]^-}: 1860, \quad \mathrm{Cr(CO)_6}: 2000, \quad \mathrm{[Mn(CO)_6]^+}: 2090$
All three metals have six d electrons.
Find each drop below free CO.
$2143 - 1860 = 283, \quad 2143 - 2000 = 143, \quad 2143 - 2090 = 53$
Subtract each from $2143$ cm$^{-1}$.
Relate the drops to the charges.
$-1 \to 283, \quad 0 \to 143, \quad +1 \to 53$
The more negative the complex, the larger the drop.
Explain the trend.
$\text{more negative metal} \Rightarrow \text{more back-donation}$
Extra electron density flows into CO's $\pi^*$.
Relate it to the C-O bond.
$\text{filled } \pi^* \Rightarrow \text{weaker bond}$
An antibonding orbital lowers the bond order a little.
Predict the dianion.
$\mathrm{[Ti(CO)_6]^{2-}}: \text{ lower still, } 1750$
A charge of $-2$ back-donates most of all.
Subtract from the free stretch.
$2143 - 2060 = 83$
The drop in wavenumbers.
Compare with the anion.
$\mathrm{[Co(CO)_4]^-}: 2143 - 1890 = 253$
Its isoelectronic anionic partner.
State the conclusion.
Match each species to the wavenumber of its C-O stretch.
| $1750$ cm⁻¹ | $1890$ cm⁻¹ | $2060$ cm⁻¹ | $2090$ cm⁻¹ | |
|---|---|---|---|---|
| $\mathrm{[Ti(CO)_6]^{2-}}$ | ||||
| $\mathrm{[Co(CO)_4]^-}$ | ||||
| $\mathrm{Ni(CO)_4}$ | ||||
| $\mathrm{[Mn(CO)_6]^+}$ |
Complete the worked solution: fill the valence orbitals of $\mathrm{NO^-}$ and find its bond order.
Count the electrons in the unstarred orbitals.
$\text{bonding electrons} =$ b
They hold the two atoms together.
Count the electrons in the starred orbitals.
$\text{antibonding electrons} =$ a
They cancel part of the bonding.
Halve the difference.
$\text{bond order} =$ o
Net bonding pairs.
Which of these carbonyls has the lowest C-O stretching frequency?
$\mathrm{NO^+}$ has $10$ valence electrons and $\mathrm{CN}$ has $9$. For each, in that order, fill in the electrons in antibonding orbitals, the bond order and the unpaired electrons.
| antibonding electrons | bond order | unpaired electrons | |
|---|---|---|---|
| the first species | |||
| the second species |
Free carbon monoxide stretches at $2143$ cm$^{-1}$ and $\mathrm{[Mn(CO)_6]^+}$ at $2090$ cm$^{-1}$. By how many wavenumbers has binding to the metal lowered the C-O stretch?
Answer: cm⁻¹ below free CO
$\mathrm{CO^+}$ has $9$ valence electrons. Fill its molecular orbitals and find its bond order.
Answer: bond order
A catalysis chemist choosing a phosphorus ligand ranks candidates by the C-O stretch of their $\mathrm{Ni(CO)_3L}$ complexes: $2064.1$ cm$^{-1}$ for $\mathrm{PMe_3}$, $2068.9$ for $\mathrm{PPh_3}$ and $2079.5$ for $\mathrm{P(OMe)_3}$. Free CO is at $2143$ and the trifluorophosphine complex at $2110.8$. For each ligand, in that order, fill in the drop below free CO and the drop below the trifluorophosphine complex.
| drop below free CO (cm⁻¹) | drop below the PF3 complex (cm⁻¹) | |
|---|---|---|
| the first ligand | ||
| the second ligand | ||
| the third ligand |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$\mathrm{CO}$ has $10$ valence electrons and $\mathrm{NO^-}$ has $12$. For each, in that order, fill in the electrons in antibonding orbitals, the bond order and the unpaired electrons.
| antibonding electrons | bond order | unpaired electrons | |
|---|---|---|---|
| the first species | |||
| the second species |
You can explain back-bonding. Say why the C-O stretch of $\mathrm{[V(CO)_6]^-}$ is lower than that of $\mathrm{Cr(CO)_6}$.
15. Your turn: $\mathrm{Ni(CO)_4}$ stretches at $2060$ cm$^{-1}$. How far below free CO is that, and what does it say?, step 3
$\text{neutral nickel back-donates less}$
Less negative charge, less density pushed into $\pi^*$.