The 18-Electron Rule
Questions 1–4The 18-electron rule states that thermodynamically stable organometallic complexes tend to possess a specific total valence electron count. What is that count, and what is its relationship to noble gas configuration?
The 18-electron rule is the transition-metal analogue of the octet rule. A transition metal has nine valence orbitals: one (n)s, three (n)p, and five (n−1)d. Filling all nine with electron pairs gives 18 electrons, which matches the electron configuration of the next noble gas (e.g., Kr for a 4d metal). Complexes obeying this rule are often kinetically inert and thermodynamically stable.
Count the total valence electrons in [Fe(CO)5]. Does this complex obey the 18-electron rule?
Iron is in group 8, so Fe(0) has 8 valence electrons. Each CO is a 2-electron donor (it donates its lone pair on carbon to the metal). Five CO ligands contribute 5 × 2 = 10 electrons. The total is 8 + 10 = 18 electrons. [Fe(CO)5] obeys the 18-electron rule, which explains its remarkable stability as a volatile liquid at room temperature.
Both [Cr(CO)6] and [Mn(CO)5]− are 18-electron species. Which electron count breakdown is correct for both complexes?
For [Cr(CO)6]: Cr is in group 6, so Cr(0) has 6 valence electrons. Six CO ligands donate 6 × 2 = 12 electrons. Total = 6 + 12 = 18e. For [Mn(CO)5]−: Mn is in group 7, so Mn(0) has 7 valence electrons. Five CO ligands donate 5 × 2 = 10 electrons. The negative charge adds 1 electron. Total = 7 + 10 + 1 = 18e. These species are isoelectronic — they have the same electron count and analogous structures.
Ferrocene, [Fe(Cp)2], is one of the most important organometallic compounds. Using the ionic model (where each Cp ring is Cp−), count the valence electrons. Does ferrocene obey the 18-electron rule?
In the ionic model, each cyclopentadienyl ring is treated as Cp−, a 6-electron donor (5 carbon atoms each contributing one electron, plus the extra negative charge). Iron is assigned the +2 oxidation state: Fe2+ is d6 with 6 valence electrons. Two Cp− rings donate 2 × 6 = 12 electrons. Total = 6 + 12 = 18 electrons. Ferrocene obeys the 18-electron rule, consistent with its exceptional thermal stability.
Metal Carbonyls & Bonding
Questions 5–8
When CO binds to a transition metal, the CO stretching frequency decreases from the free-ligand value of 2143 cm−1. What is the bonding explanation for this decrease?
Metal–CO bonding is synergic: CO donates its lone pair on carbon to an empty metal orbital (σ donation), while the metal simultaneously donates d-electron density back into the empty π* antibonding orbitals of CO (π back-bonding). Populating the π* orbitals weakens the C≡O bond, reducing its bond order from 3 toward 2 and lowering the IR stretching frequency. This is a direct, measurable signature of back-bonding.
Rank the following species in order of decreasing CO stretching frequency: free CO, [Ni(CO)4], [Co(CO)4]−, [Fe(CO)4]2−.
Free CO has the highest stretching frequency (2143 cm−1) because there is no back-bonding. As the negative charge on the complex increases, the metal center becomes more electron-rich and donates more electron density into CO π* orbitals. Ni(0) in [Ni(CO)4] back-bonds less than Co(−1) in [Co(CO)4]−, which in turn back-bonds less than Fe(−2) in [Fe(CO)4]2−. Greater back-bonding means a weaker C≡O bond and a lower stretching frequency.
The isolobal analogy is a powerful concept in organometallic chemistry. Which of the following best describes this analogy?
The isolobal analogy, developed by Roald Hoffmann, relates molecular fragments from organic and organometallic chemistry. Two fragments are isolobal (indicated by the double-headed arrow with a lobe, ⟷) if their frontier orbitals match in symmetry, energy, and electron occupation. For example, Mn(CO)5 has one half-filled frontier orbital pointing outward, just like CH3·. This allows prediction of bonding: since CH3–CH3 exists (ethane), (CO)5Mn–Mn(CO)5 should also form.
[Mn2(CO)10] contains a direct Mn–Mn bond with no bridging carbonyls. Why does this dimer form rather than a mononuclear [Mn(CO)5] species?
Mn is in group 7, so Mn(0) has 7 valence electrons. Five CO ligands donate 10 electrons, giving a total of 17 electrons for the Mn(CO)5 fragment — an odd, unstable count. By forming a direct Mn–Mn bond, each Mn shares one electron with the other, bringing each metal center to 18 electrons. The structure is staggered (D4d symmetry) with no bridging carbonyls, unlike [Co2(CO)8] which has bridging CO isomers.
Organometallic Reactions
Questions 9–12Oxidative addition is one of the most fundamental organometallic reactions. Which of the following correctly describes what happens to the metal center during oxidative addition of H2 to Vaska's complex, trans-[IrCl(CO)(PPh3)2]?
In oxidative addition, a substrate X–Y adds across the metal center. The metal formally inserts into the X–Y bond, breaking it and forming two new M–X and M–Y bonds. For Vaska's complex: Ir(I) (d8, 16e, 4-coordinate, square planar) adds H2 to become [IrClH2(CO)(PPh3)2] — Ir(III) (d6, 18e, 6-coordinate, octahedral). The metal is "oxidized" because it formally donates two electrons to form the new bonds.
Reductive elimination is the microscopic reverse of oxidative addition. Which statement best describes this process?
In reductive elimination, two ligands that are cis to each other on the metal couple together to form a new X–Y bond and depart as a single molecule. The metal center is formally reduced: its oxidation state decreases by 2, coordination number decreases by 2, and electron count decreases by 2. This step is critical for product-forming steps in catalytic cycles such as cross-coupling reactions (Suzuki, Heck, etc.).
Migratory insertion of CO into an M–CH3 bond is a key step in many catalytic processes. Which description of the mechanism is correct?
Despite being called "CO insertion," the actual mechanism involves the methyl group migrating to a cis-coordinated CO ligand to form an acyl group (M–C(=O)CH3). This is a 1,2-shift: the alkyl migrates with its bonding electrons onto the carbon of CO. The process creates a vacant coordination site on the metal and decreases the electron count by 2, which is typically filled by an incoming ligand (often another CO from solution).
β-Hydride elimination is a common decomposition pathway for metal alkyl complexes. Which of the following correctly states the structural requirement and outcome?
β-Hydride elimination requires: (1) a hydrogen on the β-carbon of the alkyl ligand, (2) a vacant coordination site on the metal, and (3) a syn-coplanar arrangement of the M–Cα–Cβ–H unit. The hydrogen transfers to the metal as a hydride while the alkyl becomes an olefin that remains coordinated. This reaction is both a common decomposition pathway for metal alkyls and a key step in catalytic cycles such as the Ziegler–Natta polymerization.
Catalytic Cycles
Questions 13–15The Monsanto process is an industrial Rh-catalyzed carbonylation of methanol to produce acetic acid. Which sequence correctly describes the key catalytic steps?
The Monsanto process uses [Rh(CO)2I2]− as the active catalyst. Methanol is first converted to CH3I (by HI). The cycle then proceeds: (1) oxidative addition of CH3I to Rh(I), giving a Rh(III)–CH3 species; (2) migratory insertion of a cis CO into the Rh–CH3 bond to form an acyl intermediate; (3) reductive elimination of acetyl iodide (CH3COI), regenerating the Rh(I) catalyst. Hydrolysis of CH3COI gives acetic acid and regenerates HI.
In olefin hydrogenation using Wilkinson's catalyst, [RhCl(PPh3)3], which step corresponds to oxidative addition and which to reductive elimination?
The catalytic cycle of Wilkinson's catalyst involves: (1) dissociation of one PPh3 to open a coordination site; (2) oxidative addition of H2 to the 14e Rh(I) center, forming a Rh(III) dihydride (16e); (3) coordination of the olefin; (4) migratory insertion of the olefin into the Rh–H bond to give a Rh–alkyl; (5) reductive elimination in which the remaining hydride and the alkyl couple to release the saturated alkane product, regenerating the Rh(I) catalyst.
Olefin metathesis is a reaction in which C=C double bonds are broken and reformed. Which of the following correctly describes this reaction and its development?
Olefin metathesis proceeds through a metal carbene (alkylidene) intermediate, typically on Mo, W, or Ru centers. The mechanism involves [2+2] cycloaddition of the olefin with the M=C bond to form a metallacyclobutane, followed by [2+2] cycloreversion to produce a new olefin and a new metal carbene. Robert Grubbs (along with Chauvin and Schrock) received the 2005 Nobel Prize in Chemistry for this work. Grubbs' Ru-based catalysts are air-tolerant and functional-group tolerant, making them widely used in organic synthesis and polymer chemistry.