Q7(a)(i). Metal carbonyls; Mond's process; tetrahedral Ni(CO)₄
What are metal carbonyls? How Ni(CO)₄ is obtained by Mond's process? Show that the molecule Ni(CO)₄, has a tetrahedral structure.
Answer in brief
Metal carbonyls are compounds of metals (usually in low oxidation states) with carbon monoxide bonded through carbon by σ donation and π back-donation. Mond's process: impure Ni + 4CO at about 50°C, 1 atm → volatile Ni(CO)₄, decomposed at about 220°C to pure Ni. Ni(CO)₄ is tetrahedral: Ni(0) d¹⁰, sp³, diamagnetic, zero dipole, one IR C–O band.
Key points
- Complexes in which CO is a ligand, bonded to the metal through carbon; binary carbonyls contain CO only: Ni(CO)₄, Fe(CO)₅, Cr(CO)₆ (mononuclear), Mn₂(CO)₁₀, Co₂(CO)₈, Fe₂(CO)₉ (polynuclear, with M–M bonds and sometimes…
- CO donates its lone pair (on C) into an empty metal orbital (σ), and the filled metal d orbitals give electrons back into the empty π* orbitals of CO (π back-donation). Each strengthens the other. Back-donation lowers the C–O…
- Metals are in low oxidation states (often 0), and most stable carbonyls obey the 18-electron rule: Ni 10+4×2=18; Fe 8+5×2=18; Cr 6+6×2=18.
- Impure nickel (made by reducing nickel oxide ore with H₂ and CO) is treated with CO at about 50°C and 1 atm: Ni(s) + 4CO(g) → Ni(CO)₄(g)
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Metal carbonyls
- Complexes in which CO is a ligand, bonded to the metal through carbon; binary carbonyls contain CO only: Ni(CO)₄, Fe(CO)₅, Cr(CO)₆ (mononuclear), Mn₂(CO)₁₀, Co₂(CO)₈, Fe₂(CO)₉ (polynuclear, with M–M bonds and sometimes bridging CO).
- Synergic bonding: CO donates its lone pair (on C) into an empty metal orbital (σ), and the filled metal d orbitals give electrons back into the empty π* orbitals of CO (π back-donation). Each strengthens the other. Back-donation lowers the C–O bond order: ν(CO) falls from 2143 cm⁻¹ (uncombined CO) to 1850–2120 cm⁻¹ (terminal).
- Metals are in low oxidation states (often 0), and most stable carbonyls obey the 18-electron rule: Ni 10+4×2=18; Fe 8+5×2=18; Cr 6+6×2=18.
Mond's process
- Impure nickel (made by reducing nickel oxide ore with H₂ and CO) is treated with CO at about 50°C and 1 atm: Ni(s) + 4CO(g) → Ni(CO)₄(g)
- Ni(CO)₄ (b.p. 43°C) is a gas at this temperature and is carried away from the impurities, which do not react under these mild conditions (Fe and Co form carbonyls only slowly, at high pressure).
- The gas is passed over pellets heated to about 220°C; the reaction reverses: Ni(CO)₄ → Ni + 4CO. Very pure nickel deposits, and CO is recycled. The process exploits the reversibility of a single equilibrium at two temperatures.
Why Ni(CO)₄ is tetrahedral
- Valence bond: Ni(0) is 3d⁸4s². CO is a strong-field ligand; the 4s electrons are pushed into 3d, giving 3d¹⁰. The empty 4s and three 4p orbitals form sp³ hybrids, which point to the corners of a tetrahedron, and accept four CO lone pairs.
- Crystal field: a d¹⁰ ion has no CFSE in any geometry, so the shape is fixed by ligand–ligand repulsion alone: four ligands are farthest apart at 109.5°, a tetrahedron.
- Magnetism and dipole: all electrons are paired (diamagnetic), and the molecule has zero dipole moment; a square-planar or distorted shape with d¹⁰ Ni gives no gain.
- Vibrational spectrum: the four C–O stretches of a T_d molecule transform as A₁+T₂; only T₂ is IR active, so the IR spectrum shows one C–O band (Raman: two, A₁ and T₂). A square-planar D₄ₕ structure would give a different pattern (A_(1g)+B_(1g)+Eᵤ).
- Electron diffraction confirms four equal Ni–C bonds and linear Ni–C–O at tetrahedral angles.
Application. Mond's process brought nickel ore from Canada to Wales for purification; the same reversibility makes Ni(CO)₄ extremely toxic (it releases CO and Ni inside the body), so it is handled only in closed systems.
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All parts of Question 7
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