Metal Carbonyls — Synergic Bonding and the IR Evidence for It
Carbon monoxide is a poor Lewis base and, on its own, a rather unreactive molecule. Yet it binds transition metals so strongly that it forms neutral compounds such as Ni(CO)4 which distil below 50 °C. The explanation is synergic bonding, and the beauty of the topic for examiners is that the explanation is testable: a single infrared spectrum both proves the bonding model and reveals the geometry. This is standard CSIR-NET and GATE organometallic material, and every argument below can be checked against a number.
How CO binds — two flows of electrons
The bonding has two components acting in opposite directions:
- σ donation. The highest occupied orbital of CO is a lone pair localised mainly on carbon, and it is slightly antibonding with respect to C–O. Donating it to an empty metal orbital therefore strengthens C–O very slightly. This is also why CO binds through carbon, not oxygen.
- π back-donation. Filled metal d orbitals of the right symmetry overlap with the empty π* antibonding orbitals of CO and push electron density into them. Filling an antibonding orbital weakens the C–O bond and at the same time strengthens the M–C bond.
The two reinforce each other — donation makes the metal more electron rich in the σ sense, which encourages back-donation, which makes the metal a better acceptor. That mutual reinforcement is what "synergic" means. In practice back-donation dominates the observable effects, so the working rule is simple:
The reference number and what shifts it
Free CO absorbs at about 2143 cm−1. Almost every terminal metal carbonyl absorbs below that, and how far below is a direct measure of how much back-donation the metal is doing. The classic demonstration is an isoelectronic series where only the charge changes:
| Species | Charge on the complex | Approximate ν(CO) / cm−1 | Interpretation |
|---|---|---|---|
| Free CO | — | 2143 | No metal, no back-donation |
| [Mn(CO)6]+ | +1 | ≈ 2090 | Positive metal holds its d electrons tightly; least back-donation |
| Cr(CO)6 | 0 | ≈ 2000 | Intermediate |
| [V(CO)6]− | −1 | ≈ 1860 | Electron-rich metal; most back-donation, weakest C–O |
All three complexes have the same 18-electron d6 octahedral structure, so geometry is not the variable — only electron density is. A drop of roughly 230 cm−1 across one unit of charge in each direction is a large, unambiguous effect, and it is the single most quoted piece of evidence for back-donation. The same logic explains why replacing one CO by a strong σ-donor phosphine lowers the frequency of the CO ligands that remain: the metal now has more electron density to give away.
Bridging carbonyls give themselves away
A CO bridging two or three metals shares its π* system with more than one metal, so it receives more back-donation and drops further:
| Bonding mode | Typical ν(CO) range / cm−1 | Notes |
|---|---|---|
| Free CO | 2143 | Reference |
| Terminal M–CO | about 2120–1850 | Higher end for cationic complexes, lower end for anionic |
| Doubly bridging μ2-CO | about 1850–1750 | Two metals back-donating |
| Triply bridging μ3-CO | about 1730–1620 | Three metals; lowest of all |
These ranges overlap slightly and shift with charge, so treat them as strong evidence rather than proof. A band at 1800 cm−1 in a neutral cluster is a bridging carbonyl for all practical purposes; the same band in a heavily reduced anion could still be terminal.
Counting the electrons
Each terminal CO is a two-electron donor, so the 18-electron rule predicts the stable stoichiometries directly:
| Complex | Metal d electrons | From CO | M–M bond | Total |
|---|---|---|---|---|
| Ni(CO)4 | Ni(0) = 10 | 4 × 2 = 8 | — | 18 |
| Fe(CO)5 | Fe(0) = 8 | 5 × 2 = 10 | — | 18 |
| Cr(CO)6 | Cr(0) = 6 | 6 × 2 = 12 | — | 18 |
| Mn2(CO)10 (per Mn) | Mn(0) = 7 | 5 × 2 = 10 | 1 | 18 |
| V(CO)6 | V(0) = 5 | 6 × 2 = 12 | — | 17 |
V(CO)6 is the well-known exception — it is one electron short, and it is correspondingly reactive and easily reduced. Adding that missing electron gives [V(CO)6]−, which is 18-electron, stable, and sits at the bottom of the frequency table above. Two independent arguments, electron counting and infrared, point the same way.
Counting bands — the structural test
The number of CO stretching bands that appear in the infrared spectrum is fixed by the symmetry of the molecule, and it is often different for two isomers of identical formula. This makes IR a quick structural assignment, no crystals required:
| Species | Point group | CO stretching modes | IR-active bands |
|---|---|---|---|
| Ni(CO)4 | Td | A1 + T2 | 1 |
| Cr(CO)6 | Oh | A1g + Eg + T1u | 1 |
| Fe(CO)5 (trigonal bipyramid) | D3h | 2A1′ + A2″ + E′ | 2 |
| cis-M(CO)4L2 | C2v | 2A1 + B1 + B2 | 4 |
| trans-M(CO)4L2 | D4h | A1g + B1g + Eu | 1 |
| fac-M(CO)3L3 | C3v | A1 + E | 2 |
| mer-M(CO)3L3 | C2v | 2A1 + B1 | 3 |
Four bands versus one settles cis against trans; two bands versus three settles fac against mer. The underlying reason is that a vibration is IR-active only if it changes the dipole moment, and the highly symmetric modes of a centrosymmetric isomer do not. Note the related rule of mutual exclusion: in a centrosymmetric molecule such as Cr(CO)6 or the trans isomer, no vibration can be active in both IR and Raman, so the modes missing from the IR spectrum can be recovered from a Raman spectrum. That is how the "silent" A1g and Eg modes are actually observed.
Worked example — turning a frequency into a force constant
The stretching frequency of a diatomic oscillator is governed by the harmonic-oscillator relation:
with ν̄ in cm−1, c the speed of light in cm s−1, μ the reduced mass in kg, and k the force constant in N m−1.
Part A — free CO at 2143 cm−1.
Step 1 — reduced mass. Using 12C = 12.000 u and
16O = 15.995 u:
μ = (12.000 × 15.995) ÷ (12.000 + 15.995)
= 191.940 ÷ 27.995 = 6.8562 u
In kilograms: 6.8562 × 1.66054 × 10−27 = 1.1385 × 10−26 kg
Step 2 — frequency in s−1.
ν = c ν̄ = (2.998 × 1010 cm s−1) × 2143 cm−1
= 6.425 × 1013 s−1
Step 3 — force constant.
ν² = (6.425 × 1013)² = 4.128 × 1027 s−2
4π² = 39.478
k = 39.478 × 4.128 × 1027 × 1.1385 × 10−26
= (39.478 × 4.128 = 162.97) → 1.6297 × 1029 × 1.1385 × 10−26
= 1855 N m−1 (about 18.6 N cm−1)
That is a very stiff bond, as expected for a triple bond.
Part B — Cr(CO)6 at 2000 cm−1. The reduced mass is
unchanged, so k scales as ν̄²:
k = 1855 × (2000 ÷ 2143)²
2000 ÷ 2143 = 0.93327; squared = 0.87099
k = 1855 × 0.87099 = 1616 N m−1
Interpretation. The drop is 1855 − 1616 = 239 N m−1, which is 239 ÷ 1855 = 12.9% of the free-CO value. Coordination has genuinely weakened the C–O bond by about an eighth — and it did so by putting electron density into an antibonding orbital, exactly as the back-donation model predicts.
Common mistakes that cost marks
- Reading a low ν(CO) as a weak bond overall. A low CO frequency means a weak C–O bond and a strong M–C bond. The two move in opposite directions.
- Saying CO binds through oxygen because oxygen is more electronegative. It binds through carbon, because that is where the donor orbital is concentrated.
- Confusing the ν(CO) region with the ν(C=O) of organic carbonyls. Ketones and aldehydes absorb near 1700 cm−1. A band at 1750–1850 cm−1 in an organometallic sample is far more likely a bridging carbonyl.
- Using the number of CO ligands as the number of bands. Cr(CO)6 has six CO ligands and one IR band. Symmetry, not stoichiometry, sets the count.
- Forgetting the M–M bond when electron counting. Mn2(CO)10 reaches 18 only when the metal–metal bond contributes one electron to each metal.
- Using cm−1 directly as a frequency in the force-constant formula. You must multiply by c first, or the answer will be wrong by more than twenty orders of magnitude.
- Treating overlapping bands as one. Weak or coincident bands can hide; a "missing" band in a cis complex is usually resolution, not symmetry.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CSIR-NET | Ordering ν(CO) across a charge or ligand series; band counting for cis/trans and fac/mer |
| GATE (Chemistry) | 18-electron counting; number of IR-active CO bands; force-constant numericals |
| IIT-JAM | Synergic bonding described qualitatively; identifying terminal versus bridging CO |
| CUET-PG / M.Sc. entrance | Structures of the common binary carbonyls and their electron counts |
Check the current official syllabus for your paper — the organometallic section is one of the more frequently revised parts.
The constants this calculation needs, in one place. A force-constant problem uses the speed of light, the atomic mass unit in kilograms and π to enough figures to survive a squaring step. The Scientific Constants reference in the suite gives all of them with their units, so you are not copying values from memory into an exponent-heavy calculation.
Open the Scientific Constants Reference →Preparing for CSIR-NET, GATE, IIT-JAM or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches at the coaching centre and as live online classes for students across India — details at abcchemistry.in.