VBT vs CFT — compare approaches to bonding in coordination compounds

Question

Compare Valence Bond Theory (VBT) and Crystal Field Theory (CFT) as applied to coordination compounds. Using $[\text{Co(NH}_3)_6]^{3+}$ as an example, explain how each theory accounts for the geometry, magnetic behaviour, and bonding in this complex.

(JEE Advanced 2022, similar pattern)

Solution — Step by Step

Co³⁺ has configuration $[\text{Ar}]3d^6$ (lost 3 electrons from Co). Six NH₃ ligands coordinate to it. The complex is octahedral and diamagnetic (zero unpaired electrons). Both VBT and CFT must explain these observations.

In VBT, the central metal ion provides empty hybrid orbitals to accept lone pairs from ligands.

For $[\text{Co(NH}_3)_6]^{3+}$ :

NH₃ is a strong field ligand → forces pairing of 3d electrons
6 electrons pair up into 3 orbitals: $3d^6$ becomes $\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \_ \ \_$
Two empty 3d + one 4s + three 4p orbitals hybridise: $d^2sp^3$ hybridization
6 hybrid orbitals accept 6 lone pairs from 6 NH₃ → octahedral geometry

Since all d electrons are paired → diamagnetic. VBT calls this an inner orbital complex (uses 3d orbitals for bonding).

In CFT, ligands are treated as point charges that create an electric field, splitting the d orbitals.

In octahedral field:

d orbitals split into $t_{2g}$ (lower, 3 orbitals) and $e_g$ (higher, 2 orbitals)
NH₃ is a strong-field ligand → large $\Delta_o$ (crystal field splitting energy)
6 electrons fill $t_{2g}$ completely: $t_{2g}^6 e_g^0$ → all electrons paired
CFSE = $-6 \times 0.4\Delta_o = -2.4\Delta_o$ (maximum stabilisation for $d^6$ low spin)

Since $t_{2g}$ is fully occupied with all paired electrons → diamagnetic. The large $\Delta_o$ of NH₃ forces pairing rather than occupying the higher $e_g$ orbitals.

Feature	VBT	CFT
Nature of bonding	Covalent (orbital overlap)	Electrostatic (ion-dipole)
d-orbital splitting	Not considered	Central concept ( $\Delta_o$ , $\Delta_t$ )
Predicts geometry	Yes (from hybridization)	Yes (from ligand arrangement)
Predicts magnetism	Yes (counts unpaired e⁻)	Yes (from filling split d orbitals)
Explains colour	No	Yes (d-d transitions across $\Delta_o$ )
Spectrochemical series	Cannot explain	Naturally explains
Quantitative	No	Semi-quantitative (CFSE values)

Why This Works

VBT focuses on how orbitals overlap to form bonds — it uses hybridization as the main tool. It correctly predicts geometry and magnetic properties but struggles to explain why some ligands cause pairing (strong field) and others don’t. The classification into “inner” vs “outer” orbital complexes is somewhat arbitrary.

CFT treats the metal-ligand interaction as purely electrostatic. Ligands are point charges that repel d electrons differently depending on their orientation, creating the $t_{2g}$ - $e_g$ split. The magnitude of splitting ( $\Delta_o$ ) depends on the ligand’s field strength. This elegantly explains the spectrochemical series, colour, and magnetic behaviour in a unified framework.

Neither theory is complete. The more accurate Ligand Field Theory (LFT) combines both — it uses MO theory to account for both covalent and electrostatic aspects of metal-ligand bonding.

Alternative Method

A quick way to determine if a complex is inner or outer orbital (VBT) or high/low spin (CFT): check if the ligand is strong or weak field using the spectrochemical series:

\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO}

NH₃ and everything to its right → strong field → low spin (CFT) / inner orbital (VBT).

For JEE Advanced, CFT is far more important than VBT. Know how to calculate CFSE, determine number of unpaired electrons for any $d^n$ configuration in both strong and weak fields, and predict whether a complex will be coloured. VBT questions are rare in JEE Advanced but common in CBSE/NEET.

Common Mistake

Students assume that all $d^6$ complexes are diamagnetic. This is true only for strong-field (low-spin) $d^6$ complexes like $[\text{Co(NH}_3)_6]^{3+}$ . With weak-field ligands (like F⁻), $[\text{CoF}_6]^{3-}$ is paramagnetic with 4 unpaired electrons ( $t_{2g}^4 e_g^2$ , high spin). The ligand determines the spin state, not just the $d^n$ configuration.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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