Bonding in Coordination Compounds

Two models, two jobs. VBT says the metal ion offers empty hybrid orbitals sized to match the ligand count and geometry, and ligand lone pairs are donated into them: $sp^3$ gives tetrahedral, $dsp^2$ gives square planar, and octahedral complexes use either $d^2sp^3$ (inner orbital, using the metal's own $(n-1)d$ set, low spin) or $sp^3d^2$ (outer orbital, using outer $nd$ orbitals, high spin). VBT is a good geometry-and-magnetism sketch, but it cannot predict which ligands force pairing, and it says nothing about colour — for that CFT is needed.

CFT builds an energy-level picture. In a free ion the five d-orbitals are degenerate (same energy). Bring six ligands up along the $x$, $y$, $z$ axes (an octahedral field) and the picture changes: the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point directly at the incoming ligand charges and are pushed up in energy (the $e_g$ set), while $d_{xy}$, $d_{yz}$, $d_{xz}$ point between the ligands and are pushed down (the $t_{2g}$ set). The energy gap between the two sets is the crystal field splitting energy:

where: $\Delta_o$ (also written $10Dq$) is the octahedral splitting energy, $e_g$ is the higher-energy pair of orbitals, and $t_{2g}$ is the lower-energy trio. In a tetrahedral field the splitting inverts and shrinks to $\Delta_t = \tfrac{4}{9}\Delta_o$ for the same metal and ligands — small enough that tetrahedral complexes are almost always high spin.

The spectrochemical series ranks ligands by how large a $\Delta_o$ they produce, from weak field to strong field: $\text{I}^- < \text{Br}^- < \text{S}^{2-} < \text{SCN}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO}$. Weak-field ligands (left) give a small $\Delta_o$ smaller than the electron-pairing energy $P$, so electrons obey Hund's rule and spread out singly across all five orbitals first — a high-spin complex. Strong-field ligands (right) give $\Delta_o > P$, so it costs less energy to pair electrons in the lower $t_{2g}$ set than to promote them to $e_g$ — a low-spin complex.

Colour is a direct consequence of this same splitting: an electron absorbs a photon of visible light to jump from $t_{2g}$ to $e_g$ (a d-d transition), and the complex shows the colour complementary to the wavelength absorbed. Ions with no d-electrons to promote ($d^0$) or a completely filled d-set with no vacancy to jump into ($d^{10}$) show no d-d transition and are typically colourless.

Problem: Compare $[\text{Co}(\text{NH}_3)_6]^{3+}$ and $[\text{CoF}_6]^{3-}$ — both octahedral $\text{Co}^{3+}$ ($d^6$) — by finding the spin state, unpaired electrons, CFSE, and VBT hybridisation for each.

Step 1 — d-count and field strength. Both complexes start from $\text{Co}^{3+}$: $[\text{Ar}]3d^6$. $\text{NH}_3$ sits on the strong-field side of the spectrochemical series (large $\Delta_o$), while $\text{F}^-$ sits on the weak-field side (small $\Delta_o$).

Step 2 — Fill the orbitals for $[\text{Co}(\text{NH}_3)_6]^{3+}$ (strong field, low spin). With $\Delta_o > P$, all six electrons pack into the lower set first: $t_{2g}^6\,e_g^0$. This gives zero unpaired electrons (diamagnetic) and $n_{t_{2g}} = 6$, $n_{e_g} = 0$.

Step 3 — Compute its CFSE. Using the Step 4 formula from the derivation: $\text{CFSE} = (-0.4 \times 6 + 0.6 \times 0)\,\Delta_o = -2.4\,\Delta_o$, before subtracting any pairing-energy penalty for the three forced electron pairs. Its VBT picture pairs all six d-electrons within the inner $(n-1)d$ set, freeing two orbitals for $d^2sp^3$ hybridisation — an inner orbital octahedral complex.

Step 4 — Fill the orbitals for $[\text{CoF}_6]^{3-}$ (weak field, high spin) and compare. With $\Delta_o < P$, electrons spread out by Hund's rule: one in each of the five orbitals first, then the sixth pairs up in $t_{2g}$, giving $t_{2g}^4\,e_g^2$ and 4 unpaired electrons. Its CFSE is $(-0.4 \times 4 + 0.6 \times 2)\,\Delta_o = (-1.6 + 1.2)\,\Delta_o = -0.4\,\Delta_o$ — far weaker stabilisation than the low-spin case, exactly as expected since fewer electrons occupy the favourable $t_{2g}$ set. Its magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90\ \text{BM}$ (strongly paramagnetic), and since no inner d-orbital is free, it uses outer orbital $sp^3d^2$ hybridisation. Same metal, same $d^6$ count, opposite magnetism, opposite hybridisation, and a six-fold difference in CFSE magnitude — purely because of ligand field strength.

• Treating CFT as a covalent-bonding model — it is fundamentally electrostatic (point-charge repulsion); VBT is the model that treats the bond as covalent orbital overlap.
• Forgetting the CFSE sign convention: $t_{2g}$ occupation is stabilising ($-0.4\,\Delta_o$ per electron), $e_g$ occupation is destabilising ($+0.6\,\Delta_o$ per electron) — mixing these up flips every answer.
• Assuming $\Delta_t$ can ever exceed $\Delta_o$ for comparable metal and ligands; it is always smaller ($\Delta_t = \tfrac{4}{9}\Delta_o$), which is why tetrahedral complexes are almost never low spin.
• Believing VBT can explain colour — it only fixes geometry and rough magnetism; colour is explained solely by CFT's d-d transitions.

Ask the tutor: if the CFSE sign convention or the high-spin/low-spin split still feels shaky, ask the AI tutor to walk through two more $d^n$ ions computing CFSE and hybridisation side by side.