GATE Organometallics — The 18-Electron Rule Applied
Electron counting is the cheapest set of marks in inorganic chemistry: it is pure arithmetic once you know the donor number of each ligand. The rule is simple — a transition metal is most stable when its nine valence orbitals (one s, three p, five d) are filled, giving 18 electrons. What separates a confident answer from a guess is knowing which counting method you are in and staying there. This article counts real complexes both ways, then shows where 18 is the wrong target.
Two methods, one answer
Ionic (donor-pair) method: d-electron count of the metal in its assigned oxidation state + 2 electrons for each ligand pair donated (anionic ligands counted as anions)
Both must give the same total. Mixing them — taking the neutral metal but the anionic ligand — is the single most common error in this topic. Note the sign convention: in the neutral method you subtract a positive charge and add a negative one.
| Ligand | Neutral method | Ionic method |
|---|---|---|
| CO, PR₃, NH₃, R₂S, N₂ | 2 | 2 |
| H, Cl, Br, R (alkyl), OR, NR₂ | 1 | 2 (as H⁻, Cl⁻, R⁻ …) |
| η²-alkene, η²-alkyne (2e mode) | 2 | 2 |
| η³-allyl | 3 | 4 (as allyl anion) |
| η⁴-diene | 4 | 4 |
| η⁵-cyclopentadienyl | 5 | 6 (as Cp⁻) |
| η⁶-arene | 6 | 6 |
| NO (linear, bent M–N–O 180°) | 3 | 2 (as NO⁺) |
| =CR₂ (Fischer carbene) | 2 | 2 |
| μ₂-bridging CO | 1 to each metal | 1 to each metal |
| M–M single bond | 1 to each metal | 1 to each metal |
Group numbers you need for the neutral method: Ti 4, V 5, Cr/Mo/W 6, Mn/Re 7, Fe/Ru/Os 8, Co/Rh/Ir 9, Ni/Pd/Pt 10.
Counting real complexes
Worked example 1 — the straightforward carbonyls.
Ni(CO)₄: Ni = 10; 4 CO × 2 = 8. Total = 10 + 8 = 18 ✓
Fe(CO)₅: Fe = 8; 5 × 2 = 10. Total = 18 ✓
Cr(CO)₆: Cr = 6; 6 × 2 = 12. Total = 18 ✓
This is why the stable binary carbonyls of the first row are Ni(CO)₄, Fe(CO)₅ and Cr(CO)₆ and not, say, Fe(CO)₄ — the even-electron metals need exactly enough CO to reach 18.
Worked example 2 — charged and mixed-ligand complexes, both methods.
[Mn(CO)₅]⁻
Neutral: Mn = 7; 5 CO × 2 = 10; the −1 charge adds 1 → 7 + 10 + 1 = 18
Ionic: the charge sits on the metal, so Mn(−I) is d⁸ → 8; 5 CO × 2 = 10 → 18 ✓
Both agree.
(η⁵-C₅H₅)Fe(CO)₂Cl
Neutral: Fe 8 + Cp 5 + (2 × 2) + Cl 1 = 18
Ionic: Fe(II) is d⁶ → 6; Cp⁻ 6; 2 CO 4; Cl⁻ 2 = 18 ✓
Ferrocene, Fe(η⁵-C₅H₅)₂
Neutral: 8 + 5 + 5 = 18. Ionic: Fe(II) d⁶ + 6 + 6 = 18 ✓
(η⁴-C₄H₆)Fe(CO)₃: 8 + 4 + 6 = 18 ✓
Cr(η⁶-C₆H₆)₂: 6 + 6 + 6 = 18 ✓
(η⁵-C₅H₅)Mn(CO)₃: 7 + 5 + 6 = 18 ✓
[Fe(CO)₄]²⁻: 8 + 8 + 2 = 18 ✓
HMn(CO)₅: 7 + 1 + 10 = 18 ✓
Co(CO)₃(NO): 9 + 6 + 3 (linear NO) = 18 ✓
Worked example 3 — metal–metal bonds and clusters.
Mn₂(CO)₁₀. Per Mn: 7 + (5 CO × 2) = 17, one short. The Mn–Mn bond supplies 1 more → 18 each. That single M–M bond is what the electron count predicts, and it is what the structure shows.
Co₂(CO)₈, bridged isomer. Per Co: 9 + (3 terminal CO × 2 = 6) + (2 bridging CO × 1 = 2) + (M–M bond = 1) = 18 ✓
Counting M–M bonds in a cluster. Total valence electrons (TVE) are compared with 18 per metal; each shared pair supplies 2 electrons to the deficit:
M–M bonds = (18n − TVE) / 2
Os₃(CO)₁₂: TVE = 3 × 8 + 12 × 2 = 24 + 24 = 48. (18 × 3 − 48)/2 = (54 − 48)/2 =
3 bonds → a triangle.
Ir₄(CO)₁₂: TVE = 4 × 9 + 12 × 2 = 36 + 24 = 60. (72 − 60)/2 =
6 bonds → a tetrahedron, which has exactly 6 edges.
Know the limit: localised two-centre bond counting works for small clusters only. For larger cages the count stops matching the number of edges, and Wade's polyhedral skeletal electron pair rules are used instead.
When 18 is the wrong answer
GATE tests the exceptions as often as the rule, so learn the three families.
Worked example 4 — genuine 16-electron complexes.
Zeise's salt, [PtCl₃(η²-C₂H₄)]⁻
Ionic: Pt(II) is d⁸ → 8; 3 Cl⁻ × 2 = 6; ethene 2. Total = 16
Neutral: Pt 10 + (3 Cl × 1) + 2 + 1 (charge) = 16 ✓
Wilkinson's catalyst, RhCl(PPh₃)₃: Rh(I) d⁸ = 8; Cl⁻ 2; 3 PPh₃ × 2 = 6 →
16
Vaska's complex, trans-IrCl(CO)(PPh₃)₂: Ir(I) d⁸ = 8; Cl⁻ 2; CO 2;
2 PPh₃ 4 → 16
These are not failures of the theory. For late, heavy d⁸ metals in a square-planar field one metal orbital (pz) is left high and empty, so 16 is the closed configuration — and that vacancy is exactly what makes them catalysts, cycling 16 → 18 by oxidative addition and back by reductive elimination.
- Early transition metals frequently stop below 18 for steric reasons — they simply cannot fit enough ligands. Cp₂TiCl₂ counts 4 + 5 + 5 + 1 + 1 = 16, and WMe₆ counts 6 + 6 = 12. Both are perfectly stable.
- Odd-electron species exist. V(CO)₆ counts 5 + 12 = 17 and is an isolable radical; it is easily reduced to the 18-electron [V(CO)₆]⁻.
- Classical Werner complexes are outside the rule's scope. [Co(NH₃)₆]³⁺ counts Co(III) d⁶ + 6 × 2 = 18, but [Cu(NH₃)₄]²⁺ counts Cu(II) d⁹ + 4 × 2 = 17 and [Ni(H₂O)₆]²⁺ counts d⁸ + 12 = 20 — both perfectly ordinary compounds. The rule was framed for low-spin, strong-field, π-accepting environments and should not be forced onto weak-field Werner complexes.
Two related items GATE pairs with counting
Oxidation state. Remove every ligand with its own electron pair — neutral donors leave neutral, X-type ligands (H, halide, alkyl, Cp) leave as anions. The charge left on the metal is its oxidation state.
ν(CO) as a probe of back-bonding. CO donates a σ pair from carbon and accepts density from a filled metal d orbital into its π* orbital, so more back-donation means a weaker C–O bond and a lower stretching frequency. Across the isoelectronic series [Mn(CO)₆]⁺, Cr(CO)₆, [V(CO)₆]⁻ the frequency falls steadily — roughly 100 cm⁻¹ per unit of added negative charge. Bridging CO absorbs lower than terminal CO, and μ₃-CO lower still, which is how the bonding mode is assigned.
- Mixing the two methods. Pick one, write it at the top of your working and stay in it.
- Getting the charge sign backwards. In the neutral method a negative overall charge adds electrons.
- Assuming maximum hapticity. A Cp ring can be η⁵, η³ or η¹; the question tells you which, and ring slippage from η⁵ to η³ is a standard way a complex makes room for an incoming ligand.
- Forgetting the M–M contribution in dinuclear carbonyls — you will land on 17 and think the compound is a radical.
- Counting a bridging CO as 2 electrons to each metal. It gives 1 to each, 2 in total.
- Applying the rule to square-planar d⁸ or to early metals and marking a perfectly stable compound as impossible.
Exam relevance
| Question style | What to do |
|---|---|
| "Total valence electron count of X is ___" | Direct count; one method, shown in a line |
| "Value of n in [M(CO)ₙ]⁻" | Solve 18 = metal + charge + 2n for n |
| "Number of M–M bonds" | (18n − TVE)/2, for small clusters |
| "Which is a 16-electron complex" | Look for square-planar d⁸: Rh(I), Ir(I), Pd(II), Pt(II) |
| "Order the ν(CO) values" | More negative charge or more donating co-ligands → lower ν(CO) |
| "Oxidation state of the metal" | Strip X-type ligands as anions, L-type as neutral |
Quick self-test: how many CO ligands make [Co(CO)ₙ]⁻ an 18-electron species? Co = 9, the charge adds 1, so 2n = 8 and n = 4. Take the syllabus for your session from the current official notification.
Counting is arithmetic — check it like arithmetic. When a question chains a count into a percentage, a molar mass or a spectroscopic energy, run those steps through the calculator suite rather than doing them in the margin. The scientific calculator, molar mass and unit-conversion tools all sit on the same page.
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