Coordination number and geometry — prediction from formula

Question

How do we determine the coordination number of a central metal ion from the formula of a coordination compound, and how does coordination number relate to the geometry of the complex?

(JEE Main, NEET, CBSE 12 — coordination number and geometry prediction is the foundation of the entire coordination compounds chapter)

Solution — Step by Step

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

For monodentate ligands (one donor atom each, like $Cl^-$ , $NH_3$ , $H_2O$ , $CN^-$ ), CN = number of ligands.

For polydentate ligands: ethylenediamine (en) is bidentate (2 donor atoms), EDTA is hexadentate (6 donor atoms). So one EDTA occupies 6 coordination sites.

Look at the coordination sphere (inside the square brackets):

$[Co(NH_3)_6]^{3+}$ : 6 $NH_3$ molecules, each monodentate. CN = 6
$[CoCl_2(en)_2]^+$ : 2 $Cl^-$ (monodentate) + 2 en (bidentate, 2 donors each) = 2 + 4 = CN = 6
$[Ni(CO)_4]$ : 4 CO (monodentate). CN = 4
$[PtCl_4]^{2-}$ : 4 $Cl^-$ (monodentate). CN = 4
$[Ag(NH_3)_2]^+$ : 2 $NH_3$ (monodentate). CN = 2

CN	Geometry	Examples
2	Linear	$[Ag(NH_3)_2]^+$ , $[CuCl_2]^-$
4	Tetrahedral	$[NiCl_4]^{2-}$ , $[Zn(NH_3)_4]^{2+}$
4	Square planar	$[Ni(CN)_4]^{2-}$ , $[PtCl_4]^{2-}$ , $[Pt(NH_3)_2Cl_2]$
6	Octahedral	$[Co(NH_3)_6]^{3+}$ , $[Fe(CN)_6]^{4-}$ , $[Cr(H_2O)_6]^{3+}$

CN = 4 ambiguity: How do we know if it is tetrahedral or square planar?

$d^8$ metals ( $Ni^{2+}$ , $Pd^{2+}$ , $Pt^{2+}$ ) with strong field ligands ( $CN^-$ , $CO$ ) prefer square planar
Most other CN = 4 complexes are tetrahedral (especially with weak field ligands like $Cl^-$ )
$Pt^{2+}$ and $Pd^{2+}$ are ALWAYS square planar regardless of ligand

flowchart TD
    A["Count donor atoms in coordination sphere"] --> B["Coordination Number"]
    B --> C{"CN = ?"}
    C -->|"2"| D["Linear (180°)"]
    C -->|"4"| E{"d⁸ metal + strong field ligand?<br/>Or Pt²⁺/Pd²⁺?"}
    E -->|"Yes"| F["Square planar"]
    E -->|"No"| G["Tetrahedral"]
    C -->|"6"| H["Octahedral"]

Why This Works

The geometry minimises electron-pair repulsion between ligands (analogous to VSEPR theory for simple molecules). CN = 6 naturally arranges into an octahedron (90-degree angles between adjacent ligands). CN = 4 can be tetrahedral (109.5 degrees) or square planar (90 degrees) — the preferred geometry depends on the electronic configuration of the metal and the crystal field splitting energy.

For $d^8$ metals with strong field ligands, the crystal field stabilisation energy (CFSE) gain from a square planar arrangement exceeds the energy cost of pushing ligands closer together — making square planar the thermodynamically favoured geometry.

Common Mistake

Students count the number of ligands instead of donor atoms. In $[Co(en)_3]^{3+}$ , there are only 3 ligands, but each en is bidentate (donates through 2 N atoms), so CN = $3 \times 2 = 6$ , not 3. The geometry is octahedral, not trigonal planar. Always count donor atoms, not ligand molecules.

For JEE: $Pt^{2+}$ complexes are always square planar — no exceptions in the syllabus. If you see Pt(II) in the formula, write square planar immediately and move on. This saves 30 seconds per question.

Question

Solution — Step by Step

Why This Works

Common Mistake

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