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CSIR-NET Organometallics and Homogeneous Catalysis

By Aniket Bhardwaj Β· 6 September 2026 Β· CSIR-NET Chemistry

Organometallic questions in CSIR-NET Chemical Sciences almost always reduce to two skills: counting electrons correctly, and recognising which of about five elementary steps has just happened. Every catalytic cycle in the syllabus is built from those steps. Get the counting clean and the cycles stop being things to memorise and become things you can derive.

Two counting methods β€” both correct, never mixed

Neutral (covalent) method: metal contributes its group number; every ligand is counted as a neutral fragment; then subtract the overall charge.

Ionic (donor-pair) method: assign an oxidation state, count the metal's d-electrons; every ligand is counted as an ion donating electron pairs.

Both give the same total. Choose one per question and stay in it β€” mixing them is the single most common source of a wrong answer.

LigandNeutral methodIonic method
CO, PR3, NH3, Ξ·2-alkene22
H, CH3, Cl, Br, I, acyl1 (as a radical)2 (as Xβˆ’)
Ξ·3-allyl34 (as allylβˆ’)
Ξ·5-Cp56 (as Cpβˆ’)
Ξ·6-benzene66
NO (linear, bent M–N–O 180Β°)32 (as NO+)
M–M single bond1 to each metal1 to each metal
Overall chargesubtract italready inside the oxidation state

Worked electron counts

Ferrocene, Fe(Ξ·5-C5H5)2
Neutral: Fe = 8, two Cp radicals = 2 Γ— 5 = 10 β†’ 8 + 10 = 18.
Ionic cross-check: Fe(II) is d6 = 6, two Cpβˆ’ = 2 Γ— 6 = 12 β†’ 6 + 12 = 18. The two routes agree, as they must.

Zeise's anion, [PtCl3(Ξ·2-C2H4)]βˆ’
Neutral: Pt = 10, three Cl = 3, ethene = 2, charge βˆ’1 adds 1 β†’ 10 + 3 + 2 + 1 = 16. Pt(II), d8, square planar β€” a genuinely stable 16-electron complex.

Wilkinson's catalyst, RhCl(PPh3)3
Neutral: Rh = 9, Cl = 1, three PPh3 = 6 β†’ 16. Rh(I), d8, square planar.

Dimanganese decacarbonyl, Mn2(CO)10
Per metal: Mn = 7, five CO = 10, one Mn–Mn bond = 1 β†’ 18. Without invoking the metal–metal bond you would get 17 and conclude, wrongly, that it is a radical.

[Mn(CO)5]βˆ’
Neutral: 7 + 10 + 1 (for the negative charge) = 18.
Ionic cross-check: Mn(βˆ’I) is d8 = 8, five CO = 10 β†’ 18. Agreed.

Cp2TiCl2
Neutral: Ti = 4, two Cp = 10, two Cl = 2 β†’ 16. Ionic: Ti(IV) is d0 = 0, two Cpβˆ’ = 12, two Clβˆ’ = 4 β†’ 16. An early transition metal that simply has no room for more ligands.

When the 18-electron rule genuinely fails

The rule works because filling the nine valence orbitals (one s, three p, five d) takes eighteen electrons. It fails in three predictable situations, and CSIR-NET tests exactly these.

The five elementary steps β€” and what each does to the count

StepOxidation stateCoordination numberElectron count
Oxidative addition+2+2+2
Reductive eliminationβˆ’2βˆ’2βˆ’2
Ligand association0+1+2
Ligand dissociation0βˆ’1βˆ’2
Migratory insertion0βˆ’1βˆ’2
Ξ²-hydride elimination0+1+2

Two conditions are worth remembering because questions are built on them. Oxidative addition needs a metal with a vacant site and an accessible higher oxidation state β€” which is why 16-electron d8 centres do it so readily. Ξ²-Hydride elimination needs a vacant coordination site and a Ξ²-hydrogen that can reach a syn-coplanar arrangement with the metal; remove either and the alkyl becomes stable, which is exactly how stable metal alkyls are designed.

Cycle 1 β€” Wilkinson hydrogenation

Overall: alkene + H2 β†’ alkane, with both hydrogens delivered to the same face (syn addition).

1. RhCl(PPh3)3 (16e) loses one PPh3 β†’ RhCl(PPh3)2 (14e).
2. Oxidative addition of H2 β†’ Rh(III) dihydride (16e). Rh(I) β†’ Rh(III).
3. The alkene coordinates β†’ 18e.
4. Migratory insertion: the alkene inserts into one Rh–H bond β†’ alkyl hydride (16e).
5. Reductive elimination releases the alkane and regenerates the 14-electron Rh(I) species.

The catalyst hydrogenates less hindered alkenes far faster than crowded ones, which is why a terminal double bond can be reduced while a tetrasubstituted one in the same molecule survives.

Cycle 2 β€” hydroformylation (the oxo process)

Overall: alkene + CO + H2 β†’ aldehyde, adding one carbon.

1. HCo(CO)4 (18e) loses CO β†’ HCo(CO)3 (16e).
2. Alkene coordinates β†’ 18e.
3. 1,2-insertion into the Co–H bond gives the alkyl, RCo(CO)3 (16e). Whether the metal ends up on the terminal or the internal carbon here is what decides linear versus branched product.
4. CO adds β†’ RCo(CO)4 (18e).
5. Migratory insertion of CO gives the acyl, RCOCo(CO)3 (16e).
6. Oxidative addition of H2 β†’ 18e, then reductive elimination releases RCHO and returns HCo(CO)3.

Rhodium catalysts with bulky phosphines give a much higher proportion of the linear aldehyde, because the bulk disfavours the branched insertion in step 3. That is a selectivity argument, not a rate argument β€” a distinction examiners like.

Cycle 3 β€” the Monsanto acetic acid process

Overall: CH3OH + CO β†’ CH3COOH. Methanol is first converted to CH3I by HI, and the iodide is regenerated at the end, so HI is a co-catalyst.

1. [Rh(CO)2I2]βˆ’ (16e, Rh(I)).
2. Oxidative addition of CH3I β†’ [CH3Rh(CO)2I3]βˆ’ (18e, Rh(III)). This step is rate-determining, which is why the rate depends on the rhodium and methyl iodide concentrations and is essentially independent of CO pressure.
3. Migratory insertion of CO β†’ the acyl complex (16e).
4. CO adds back β†’ 18e.
5. Reductive elimination of CH3COI regenerates [Rh(CO)2I2]βˆ’; hydrolysis of CH3COI gives acetic acid and returns HI.

The iridium-based Cativa process runs the same cycle with a different rate-determining step, which is the usual way this is examined β€” the mechanism is the same shape, the kinetics are not.

Mistakes that cost marks

  • Mixing the two counting methods. Counting Cp as 5 (neutral) while also taking the metal's d-electron count from an oxidation state (ionic) double-corrects and gives 17 or 19.
  • Forgetting the overall charge. In the neutral method a negative charge adds electrons and a positive charge removes them.
  • Ignoring metal–metal bonds in dimers, which costs exactly one electron per metal per bond.
  • Calling migratory insertion an oxidation. The oxidation state does not change β€” the alkyl or acyl group migrates onto an adjacent ligand, opening a site.
  • Expecting Ξ²-hydride elimination without a vacant site. An 18-electron complex must lose a ligand first.
  • Assuming 18 is compulsory. For Rh(I), Ir(I), Pd(II) and Pt(II), 16 electrons is the normal, stable state.

Where this appears in the exam

ExamTypical demand
CSIR-NET Chemical SciencesElectron count and oxidation state of a drawn complex; naming the elementary step; identifying the rate-determining step of a named cycle
GATE Chemistry18-electron counting, hapticity, metal carbonyl bonding and CO stretching frequencies
IIT-JAM / CUET-PGSimple carbonyls, ferrocene, EAN rule and oxidation states
MSc courseworkFull mechanism of an industrial homogeneous catalytic process

A last practical point: when a cycle is drawn in a question, label every intermediate with its electron count before you answer anything. The counts will move 16 β†’ 18 β†’ 16 β†’ 18 around the loop, and any step that breaks that rhythm is either a step you have named wrongly or the answer the examiner is looking for.

Keep the numbers beside the mechanism. Organometallic problems end in arithmetic β€” electron counts, CO stretching frequencies, turnover numbers, rate expressions. The ABC Chemistry Calculator Suite keeps the scientific constants, unit converter and general calculation tools in one page while you work through a cycle.

Open the ABC Chemistry Calculator Suite β†’

Preparing for CSIR-NET, GATE, IIT-JAM or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches at the coaching centre and online across India β€” details at abcchemistry.in.

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