(a) Show the splitting of 'd' orbitals in square planar field according to Crystal Field Theory (CFT). Comment on the following…

(a) Square-planar splitting and the dx²−y²/dxy statement

A square-planar complex can be derived from an octahedron by withdrawing the two ligands on the z-axis. As the axial ligands recede:

• orbitals with a z-component are stabilised: dz² strongly, and dxz and dyz slightly;
• the orbitals lying wholly in the xy plane (dx²−y² and dxy) rise, to keep the barycentre at zero.

In the square-planar limit (point-charge CFT, same ligands, same M–L distance), the energies in units of Δₒ are:

• dx²−y²: +1.228 (lobes point directly at the four ligands, so it is highest)
• dxy: +0.228
• dz²: −0.428
• dxz, dyz: −0.514

Comment on the statement. In the octahedron, dx²−y² (e_g) and dxy (t₂g) are separated by exactly Δₒ. In the square plane, E(dx²−y²) − E(dxy) = 1.228 − 0.228 = 1.000 Δₒ. The statement is therefore correct within the crystal-field model. Both orbitals lie in the xy plane, so their interaction with the four in-plane ligands is unchanged when the z-ligands are removed. Both are raised by the same amount, and their separation stays Δₒ.

Qualifications

• Only this pair keeps the octahedral separation. The overall pattern is quite different: a total spread of about 1.74 Δₒ, and dz² falls below dxy.
• The large gap above dxy is why d⁸ ions (Ni²⁺ with strong ligands, and Pd²⁺, Pt²⁺, Au³⁺) form diamagnetic square-planar complexes, such as [Ni(CN)₄]²⁻ and [PtCl₄]²⁻. Their eight electrons fill the four lower orbitals and leave dx²−y² empty.
• In real complexes the exact ordering of the lower levels (dz² relative to dxz/dyz) depends on s–dz² mixing and π-bonding. The equality is a feature of the simple CFT model, not an exact experimental law.

(b)(i) Bond orders in Ni(CO)₄, [Co(CO)₄]⁻ and [Fe(CO)₄]²⁻

The three species are isoelectronic: Ni(0), Co(−I) and Fe(−II) are all d¹⁰. All are tetrahedral 18-electron complexes. They differ only in the nuclear charge of the metal and the overall negative charge.

M–CO bonding is synergic:

• σ-donation from the carbon lone pair (5σ) of CO into an empty metal orbital;
• π back-donation from filled metal dπ orbitals into the empty π* orbitals of CO.

Back-donation strengthens the M–C bond (giving partial M=C character) and weakens the C–O bond, because the electrons enter a C–O antibonding orbital.

Going from Ni(CO)₄ to [Co(CO)₄]⁻ to [Fe(CO)₄]²⁻:

• The metal's nuclear charge falls (28, 27, 26) while the total charge becomes more negative (0, −1, −2).
• The metal d orbitals therefore become higher in energy and more diffuse, closer in energy to the CO π* orbitals and better able to overlap with them.
• Back-donation increases steadily. The negative charge is dispersed onto the CO ligands.

Hence:

• the M–C bond order rises: 1.33 → 1.89 → 2.16 (the bond acquires more double-bond character);
• the C–O bond order falls: 2.64 → 2.14 → 1.85, compared with 3 in free CO.

The sum of the two bond orders stays close to 4 (3.97, 4.03, 4.01). The M–C–O unit behaves like a continuum between the resonance forms M–C≡O and M=C=O, with electron density shifting from C–O to M–C.

Supporting IR evidence: ν(CO) falls from 2143 cm⁻¹ in free CO to about 2060 cm⁻¹ in Ni(CO)₄, about 1890 cm⁻¹ in [Co(CO)₄]⁻ and about 1790 cm⁻¹ in [Fe(CO)₄]²⁻.

(b)(ii) Identification of A, B and C

The scheme: [Co(NH₃)₅Cl]Cl₂ + NaNO₂ → A; A —hν→ B; B —stand/warm (HCl)→ A; [Co(NH₃)₅Cl]Cl₂ + dilute aqueous NH₃ → C; C + NaNO₂/HCl → B.

A = [Co(NH₃)₅(NO₂)]Cl₂, nitropentaamminecobalt(III) chloride (yellow-brown).

• Nitrite replaces chloride and binds through N (Co–NO₂).
• This is the thermodynamically stable isomer, which is why B reverts to A on standing or warming.

B = [Co(NH₃)₅(ONO)]Cl₂, nitritopentaamminecobalt(III) chloride (red).

• Nitrite binds through O (Co–ONO). This is the metastable isomer.
• It forms from A by UV irradiation (photo-isomerisation, N → O).
• It also forms from C by treatment with NaNO₂/HCl in the cold. Nitrosation (by N₂O₃/HNO₂) of the Co–OH group builds the O–N=O unit on the oxygen already bound to cobalt; ¹⁸O-labelling shows that the Co–O bond is never broken.

C = [Co(NH₃)₅(OH)]²⁺, as the chloride [Co(NH₃)₅(OH)]Cl₂ (pink). In acid it becomes the aqua complex [Co(NH₃)₅(H₂O)]³⁺.

• It forms by base hydrolysis of the Co–Cl bond in dilute aqueous ammonia (the conjugate-base mechanism).
• Dilute NH₃ does not substitute Co(III) to give [Co(NH₃)₆]³⁺, because Co(III) is kinetically inert.

Relationship between A and B: linkage isomers.

• They have the same composition, but the ambidentate NO₂⁻ ligand is attached through different donor atoms: N in A and O in B.
• IR distinguishes them. The nitro isomer shows ν(NO₂) at about 1430 and 1310 cm⁻¹. The nitrito isomer shows ν(N=O) at about 1470 cm⁻¹ and ν(N–O) at about 1065 cm⁻¹.
• This pair, described by Jørgensen (1894), was the first recognised case of linkage isomerism.

(c)(i) Why [CoF₆]³⁻ is paramagnetic but [Co(CN)₆]³⁻ is diamagnetic

In both complexes cobalt is Co(III), d⁶, in an octahedral field. Whether the six electrons pair up in t₂g depends on whether the crystal-field splitting Δₒ exceeds the mean pairing energy P (about 21 000 cm⁻¹ for Co³⁺).

[CoF₆]³⁻

• F⁻ is a weak-field π-donor ligand, low in the spectrochemical series. Δₒ ≈ 13 000 cm⁻¹, which is less than P.
• It is cheaper to promote electrons to e_g than to pair them, giving high spin, t₂g⁴e_g².
• Four unpaired electrons: paramagnetic. μ(spin-only) = √(4 × 6) = 4.90 BM.
• CFSE = 4(−0.4Δₒ) + 2(0.6Δₒ) = −0.4Δₒ.

[Co(CN)₆]³⁻

• CN⁻ is a strong-field ligand: a good σ-donor and π-acceptor, high in the series. Δₒ ≈ 33 500 cm⁻¹, which is greater than P.
• All six electrons pair in t₂g, giving low spin, t₂g⁶e_g⁰.
• No unpaired electrons: diamagnetic.
• CFSE = −2.4Δₒ, which more than repays the two extra pairings.

(Spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO. Almost all Co(III) complexes are low-spin; [CoF₆]³⁻ and [CoF₃(H₂O)₃] are the classic high-spin exceptions.)

(c)(ii) Structures and magnetism of Co₂(CO)₈

Co(0) is d⁹. An isolated Co(CO)₄ fragment would have 9 + 8 = 17 valence electrons and be a paramagnetic radical. It achieves 18 electrons by dimerising through a Co–Co bond, so the compound is Co₂(CO)₈ (orange crystals, m.p. 51 °C).

1. Solid state: bridged C₂ᵥ isomer, (CO)₃Co(μ-CO)₂Co(CO)₃.

• Each Co carries three terminal CO groups, and the two Co atoms are spanned by two bridging (μ₂) CO groups.
• The Co₂(μ-CO)₂ ring is folded along the Co···Co axis, a "butterfly" shape, with Co–Co ≈ 252 pm.
• Electron count per Co: 9 (Co) + 6 (three terminal CO) + 2 (half of two μ-CO) + 1 (Co–Co) = 18.
• IR: bridging ν(CO) at about 1860 cm⁻¹, plus terminal bands at about 2020–2070 cm⁻¹.
• Modern calculations suggest the Co–Co interaction is a bent bond largely mediated by the bridges, but the 18-electron formalism counts one Co–Co bond.

2. In solution: the bridged form is in rapid equilibrium with non-bridged isomers, (CO)₄Co–Co(CO)₄, of D₃d (staggered) and D₂d symmetry.

• These have eight terminal CO groups and an unsupported Co–Co σ bond.
• Electron count per Co: 9 + 8 + 1 = 18.
• Their IR spectra lack the bridging band.

Magnetic behaviour. In every isomer the odd electron of each d⁹ Co(CO)₄ unit is paired in the Co–Co bond, so there are no unpaired electrons: Co₂(CO)₈ is diamagnetic. This is consistent with the 18-electron rule. The easy homolysis of the Co–Co bond (to the ·Co(CO)₄ radical) and its reduction to [Co(CO)₄]⁻ underlie its use as a catalyst precursor in hydroformylation (the oxo process).