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VKC Notes · Metal carbonyls · Civil Services P-I · IFoS P-I · Geo-Scientist P-I

13.5 Metal Carbonyls, Nitrosyls, 18-Electron Rule and Metal Clusters

★★★ 2021–2025: 26 parts, 244 marks (Civil Services 5 parts / 65 m · IFoS 5 / 54 m · Geo-Scientist 16 / 125 m). The most asked inorganic topic of the Geo-Scientist paper: electron counts, Co₂(CO)₈, IR bands and clusters come almost every year.

Syllabus lines. Civil Services and IFoS: “metal carbonyls … compounds with metal-metal bonds and metal atom clusters”. Geo-Scientist: “18-electron rule and its applications to carbonyls and nature of bonding involved therein. Simple examples of metal-metal bonded compounds and metal clusters”.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

13.5.1 CO as a ligand: synergic bonding ★★★

  • CO binds through carbon. Its HOMO (5σ) is a lone pair concentrated on C; its LUMO (2π*, a degenerate pair) also has its larger lobes on C. So one end of the molecule both gives and takes electrons.SN
  • σ donation: C lone pair → empty metal orbital (s, p, or eg-type d). π back-donation: filled metal t2g-type d → empty CO π*. Each step helps the other (synergic bonding): σ donation makes the metal richer and so a better π donor; back-donation removes the charge σ donation brought.SN
  • Because π* is bonding between M and C but antibonding between C and O, more back-donation gives a shorter, stronger M–C bond, a longer, weaker C–O bond and a lower ν(CO) CSE 2024 Q7(b)(i) 10m Geo 2021 Q1(h) 5m.SN
σ donation: C lone pair (5σ) → empty MMCO5σ is weakly C–O antibondingσ donation alone raises ν(CO) a littleπ back-donation: filled M dπ → CO π*MCOπ* is M–C bonding, C–O antibondingM–C bond order ↑, C–O ↓, ν(CO) ↓
Fig. 13.5.1 Synergic bonding in M–CO.
Draw it in 60 s: left: a metal lobe and the C lone-pair lobe facing it, arrow C → M; right: four-leaf d orbital, two π* lobes on C, arrow from the filled d lobe to the C π* lobe.

Solved PYQ IFoS 2024 Q7(a) 15m

N₂ and CO are isoelectronic. Why is CO a far better ligand?

  1. Both have 10 valence electrons and the configuration (1σ)²(2σ*)²(1π)⁴(3σ)². In N₂ the orbitals are symmetric (σg, πu); in CO the more electronegative O pulls the bonding orbitals towards itself, so the HOMO and LUMO are polarised towards C.
  2. HOMO energy: CO −14.0 eV, N₂ −15.6 eV. The CO lone pair is higher and points out from C: a better σ donor.
  3. LUMO: CO π* is lower and has big lobes on C (good overlap with metal d); N₂ π* is higher and shared equally: CO is the better π acceptor.
  4. So N₂ complexes are few and weak ([Ru(NH₃)₅(N₂)]²⁺, ν(N≡N) ≈ 2100 cm⁻¹ against 2331 cm⁻¹ free), while CO forms carbonyls with almost every d-metal in low oxidation states.SN
  • No ordinary lanthanoid carbonyls: 4f orbitals are contracted and shielded by 5s²5p⁶, so they cannot back-donate; Ln³⁺ is a hard acid that prefers O donors. Without back-donation the M–CO bond is too weak to isolate Geo 2024 Q1(j) 5m.SN

13.5.2 The 18-electron rule and the EAN rule ★★★

  • 18-electron rule: a stable low-valent complex has 18 valence electrons, filling the nine valence orbitals (one ns, three np, five (n−1)d). Best obeyed by carbonyls, because CO is small and a strong π acceptor that keeps all bonding orbitals filled and antibonding ones empty Geo 2023 Q9(a) 15m Geo 2025 Q4(b) 10m.SN
  • EAN rule (Sidgwick): the total electron count of the metal, core included, equals the next noble gas (36 Kr, 54 Xe, 86 Rn). It is the same idea with core electrons added Geo 2023 Q1(c) 5m IFoS 2021 Q5(d) 8m.SN
(13.5.1)
(13.5.2)
  • Use one method only. Neutral method: metal group number + CO 2 + Cl 1 + Cp 5 + NO (linear) 3 + M–M bond 1 + bridging CO 1 to each metal + μ₃-CO ⅔ each. Ionic method: dn of the ion + 2 per ligand (Cl⁻ 2, Cp⁻ 6).SN
SpeciesCountTotalShape
Cr(CO)₆ · Fe(CO)₅ · Ni(CO)₄18Oh · D3h (tbp) · Td
V(CO)₆17paramagnetic; reduced easily to [V(CO)₆]⁻
Mn₂(CO)₁₀ (per Mn)18no bridges
Co₂(CO)₈ bridged (per Co)18C2v
Fe₂(CO)₉ (per Fe)183 μ-CO
[(η⁶-C₆H₆)Cr(CO)₃]18piano stool
[Rh(CO)₂(μ-Cl)]₂ (per Rh, ionic)16square planar, no Rh–Rh
[Co(NH₃)₆]³⁺ · [Pt(NH₃)₆]⁴⁺ (EAN)36 · 86both d⁶, 18 e
  • Find n: K₂[Fe(CO)n]: Fe(−II) d¹⁰, 10 + 2n = 18, n = 4 (tetrahedral). MnCl(CO)n: Mn(I) d⁶, 6 + 2 + 2n = 18, n = 5 Geo 2022 Q1(h) 5m. “Fe(CO)₉” in the paper means Fe₂(CO)₉ IFoS 2021 Q5(d) 8m.SN
  • Square-planar d⁸ (Rh(I), Ir(I), Pd(II), Pt(II)) is stable at 16: the dx²−y² orbital is strongly antibonding and pz is high, so both stay empty Geo 2025 Q4(b) 10m.SN
  • 17-e V(CO)₆ substitutes much faster than 18-e [V(CO)₆]⁻: the 17-e radical can take PPh₃ into its half-empty orbital (associative, 19-e state); the 18-e anion must first lose a CO (dissociative, slow) Geo 2022 Q9(c) 5m.SN

Solved PYQ CSE 2022 Q5(d) 10m

Name [Ni(CO)₄] and [Ni(CN)₄]⁴⁻; give their structures.

  1. Tetracarbonylnickel(0) and tetracyanidonickelate(0) ion: both Ni(0), 3d¹⁰ (the strong-field ligands push 4s² into 3d).
  2. 4s and 4p empty → sp³: tetrahedral, diamagnetic, 10 + 8 = 18 e.
  3. CN⁻ is a poorer π acceptor and carries negative charge, so it cannot remove Ni(0)'s charge as CO does: K₄[Ni(CN)₄] (from K₂[Ni(CN)₄] + K in liquid NH₃) is air-sensitive and far less stable.SN

13.5.3 IR spectra of carbonyls ★★★

(13.5.3)
  • Back-donation lowers the C–O force constant k, so ν(CO) is a direct read-out of how electron-rich the metal is. Free CO 2143 cm⁻¹; terminal M–CO ≈ 2120–1850; μ₂-CO ≈ 1850–1750; μ₃-CO ≈ 1730–1620 (a bridge takes back-donation from two or three metals).SN
TrendDataReason
Isoelectronic d⁶[Mn(CO)₆]⁺ 2090 > Cr(CO)₆ 2000 > [V(CO)₆]⁻ 1860 cm⁻¹; C–O strength V⁻ < Cr < Mn⁺ Geo 2021 Q1(h) 5mmore negative metal, higher d levels, more back-donation
Isoelectronic d¹⁰Ni(CO)₄ 2060 > [Co(CO)₄]⁻ 1890 > [Fe(CO)₄]²⁻ 1790; M–C order 1.33 → 1.89 → 2.16, C–O 2.64 → 2.14 → 1.85 (sum ≈ 4) CSE 2024 Q7(b)(i) 10msame: charge 0, −1, −2
Other ligandsMo(CO)₆ ≈ 2000; fac-Mo(dien)(CO)₃ ≈ 1760 Geo 2023 Q1(i) 5mdien is a σ donor with no π acidity: three CO now share all the back-donation
  • [CpFe(CO)₂]₂: 1800 cm⁻¹ = bridging CO; ≈ 1980 and 2020 cm⁻¹ = in-phase and out-of-phase stretches of the two terminal CO (two oscillators couple) Geo 2024 Q1(h) 5m.SN
2200210020001900180017001600ν / cm−1terminal M–COμ2-COμ3-COfree CO 2143[Mn(CO)6]+Cr(CO)6[V(CO)6]−more negative metal →more back-donation → lower ν
Fig. 13.5.2 Where carbonyl bands fall.
Draw it in 60 s: a ruler 2200 → 1600 cm−1; boxes terminal, μ2, μ3 going right; mark free CO at 2143 and the d6 series Mn+ 2090, Cr 2000, V− 1860.

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