Coordination number and geometry — prediction from formula

Question

How do you determine the coordination number and predict the geometry of a coordination compound from its formula? What geometries correspond to coordination numbers 2, 4, and 6?

(JEE Main, NEET, CBSE 12 — coordination number and geometry questions appear every year)

Solution — Step by Step

The coordination number (CN) is the total number of donor atoms directly bonded to the central metal ion — not the number of ligands.

Rules:

Monodentate ligands (NH3, Cl-, H2O) contribute 1 to CN
Bidentate ligands (ethylenediamine/en, oxalate) contribute 2 to CN
Polydentate ligands (EDTA) contribute up to 6 to CN

Example: In $[\text{Co(NH}_3\text{)}_4\text{Cl}_2]^+$ , CN = 4 (from NH3) + 2 (from Cl) = 6

Example: In $[\text{Ni(en)}_3]^{2+}$ , en is bidentate, so CN = 3 ligands x 2 donor atoms = 6

CN	Geometry	Examples
2	Linear	$[\text{Ag(NH}_3\text{)}_2]^+$ , $[\text{CuCl}_2]^-$
4	Tetrahedral	$[\text{NiCl}_4]^{2-}$ , $[\text{CoCl}_4]^{2-}$
4	Square planar	$[\text{Ni(CN)}_4]^{2-}$ , $[\text{Pt(NH}_3\text{)}_2\text{Cl}_2]$
6	Octahedral	$[\text{Co(NH}_3\text{)}_6]^{3+}$ , $[\text{Fe(CN)}_6]^{4-}$

For CN = 4, the geometry depends on the metal and ligand field strength. d8 metals with strong-field ligands (Ni2+, Pd2+, Pt2+ with CN-) prefer square planar. Most others prefer tetrahedral.

To find CN, we first need to identify the metal’s oxidation state.

For $[\text{Fe(CN)}_6]^{4-}$ :

CN- has charge -1, so 6 CN- gives -6
Complex charge is -4
Fe oxidation state: $x + (-6) = -4$ , so $x = +2$

Fe is Fe2+ with d6 configuration. With strong-field CN-, it forms a low-spin octahedral complex.

graph TD
    A["Coordination Number"] --> B["CN = 2"]
    A --> C["CN = 4"]
    A --> D["CN = 6"]
    B --> E["Linear"]
    C --> F{"d8 + strong field?"}
    F -->|"Yes"| G["Square Planar"]
    F -->|"No"| H["Tetrahedral"]
    D --> I["Octahedral"]
    J["Finding CN"] --> K["Count donor atoms, not ligands"]
    K --> L["Bidentate = 2 per ligand"]

Why This Works

The geometry of a complex is determined by the need to minimise repulsion between ligands around the central metal. With 6 ligands, octahedral geometry places them as far apart as possible. With 4 ligands, tetrahedral does the same — unless electronic factors (like the stability of a d8 square planar configuration) override the steric preference.

The crystal field theory explains why d8 metals prefer square planar: in a square planar field, the $d_{x^2-y^2}$ orbital is highest in energy and empty, while all 8 electrons fill the lower four orbitals. This gives extra stability (CFSE) that compensates for the steric cost of having ligands closer together.

Alternative Method

For JEE, a quick rule: Pt2+, Pd2+, and Ni2+ with strong-field ligands (CN-, CO) always form square planar complexes (CN = 4). Ni2+ with weak-field ligands (Cl-, H2O) forms tetrahedral complexes. This metal + ligand combination determines geometry faster than calculating CFSE.

Common Mistake

The biggest error: confusing the number of ligands with the coordination number. In $[\text{Ni(en)}_3]^{2+}$ , there are 3 ligands but CN = 6 (each en donates 2 atoms). In $[\text{Co(EDTA)}]^-$ , there is 1 ligand but CN = 6 (EDTA is hexadentate). Always count donor atoms, not ligand molecules.

Also, students assume CN = 4 always means tetrahedral. For d8 metals (Ni2+, Pd2+, Pt2+) with strong-field ligands, CN = 4 gives square planar, not tetrahedral. JEE tests this distinction with $[\text{Ni(CN)}_4]^{2-}$ (square planar) vs $[\text{NiCl}_4]^{2-}$ (tetrahedral).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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