Intuition The big pictureA central metal ion is like a tiny planet surrounded by moons (ligands) . The number of moons it holds is the coordination number (CN) , and how those moons arrange themselves in 3D space is the geometry . Both are decided by a tug-of-war between how many ligands physically fit and how the metal's d-electrons want to arrange .
Definition Coordination number (CN)The number of donor atoms (not ligands!) directly bonded to the central metal ion through coordinate bonds.
For monodentate ligands: CN = number of ligands.
For chelating ligands: count each donor atom. e.g. ethylenediamine (en) is bidentate → contributes 2 to CN.
Common mistake Steel-man: "CN = number of ligand molecules"Why it feels right: Each ligand looks like one "thing" attached, so you count things.
Why it's wrong: A ligand can grab the metal with multiple hands (denticity). [ Co(en) 3 ] 3 + [\text{Co(en)}_3]^{3+} [ Co(en) 3 ] 3 + 3 ligands but CN = 6 (each en donates 2 N atoms).
Fix: Always count donor atoms , not molecules.
Intuition Minimise electron-pair repulsionLigands are electron-pair donors → they repel each other. So they spread out to be as far apart as possible around the metal. This is the same VSEPR idea you used for CH 4 \text{CH}_4 CH 4 BF 3 \text{BF}_3 BF 3
2 points on a sphere as far apart → opposite poles → linear (180°)
4 points → corners of a tetrahedron (109.5°)
6 points → corners of an octahedron (90°)
But for transition metals there's a twist: d-orbital electrons and crystal field stabilisation can override pure geometry, especially for CN = 4 (tetrahedral vs square planar).
WHY: With only two electron domains, maximum separation is to put them at opposite poles.
Who does this? Mostly d 10 d^{10} d 10
[ Ag(NH 3 ) 2 ] + [\text{Ag(NH}_3)_2]^+ [ Ag(NH 3 ) 2 ] + [ CuCl 2 ] − [\text{CuCl}_2]^- [ CuCl 2 ] − [ Au(CN) 2 ] − [\text{Au(CN)}_2]^- [ Au(CN) 2 ] −
d 10 d^{10} d 10 d 10 d^{10} d 10
This is the most interesting CN because two geometries compete .
Four ligands at the corners of a tetrahedron, bond angle 109.5 ° 109.5° 109.5°
WHY chosen: Pure VSEPR optimum (maximum spread of 4 points), favoured when:
the metal is d 0 d^0 d 0 d 5 d^5 d 5 d 10 d^{10} d 10
ligands are bulky (more room in tetrahedron),
ligands are weak field .
Examples: [ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 − [ CoCl 4 ] 2 − [\text{CoCl}_4]^{2-} [ CoCl 4 ] 2 − [ MnO 4 ] − [\text{MnO}_4]^- [ MnO 4 ] − [ Zn(NH 3 ) 4 ] 2 + [\text{Zn(NH}_3)_4]^{2+} [ Zn(NH 3 ) 4 ] 2 +
Four ligands at the corners of a square, all in one plane, angle 90 ° 90° 90°
WHY chosen: Favoured for d 8 d^8 d 8
d 8 d^8 d 8 d x 2 − y 2 d_{x^2-y^2} d x 2 − y 2 very high in energy, while the other four d-orbitals stay lower. A d 8 d^8 d 8 8 electrons , which can fill the lower 4 orbitals (8 electrons in 4 orbitals) and leave the high d x 2 − y 2 d_{x^2-y^2} d x 2 − y 2 empty . This is hugely stabilising — but only if the splitting (i.e. field) is strong enough to force pairing.
So: d 8 d^8 d 8 square planar . d 8 d^8 d 8 tetrahedral .
Examples: [ Ni(CN) 4 ] 2 − [\text{Ni(CN)}_4]^{2-} [ Ni(CN) 4 ] 2 − [ PtCl 4 ] 2 − [\text{PtCl}_4]^{2-} [ PtCl 4 ] 2 − [ Pd(NH 3 ) 4 ] 2 + [\text{Pd(NH}_3)_4]^{2+} [ Pd(NH 3 ) 4 ] 2 + Pt(II) \text{Pt(II)} Pt(II) Pd(II) \text{Pd(II)} Pd(II)
Common mistake Steel-man: "AllNi 2 + \text{Ni}^{2+} Ni 2 + Why it feels right: Same metal, same charge → should be same shape.
Why it's wrong: [ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 − tetrahedral & paramagnetic (weak field Cl⁻), but [ Ni(CN) 4 ] 2 − [\text{Ni(CN)}_4]^{2-} [ Ni(CN) 4 ] 2 − square planar & diamagnetic (strong field CN⁻). The ligand field strength flips the geometry.
Fix: Geometry of d 8 d^8 d 8
Six ligands: four in a square plane + one above + one below the metal. All adjacent angles 90 ° 90° 90°
WHY most common: Six is a great compromise — enough ligands to satisfy bonding/charge, while 90 ° 90° 90° Co 3 + \text{Co}^{3+} Co 3 + Fe 3 + \text{Fe}^{3+} Fe 3 + Cr 3 + \text{Cr}^{3+} Cr 3 +
Examples: [ Co(NH 3 ) 6 ] 3 + [\text{Co(NH}_3)_6]^{3+} [ Co(NH 3 ) 6 ] 3 + [ Fe(CN 6 ) ] 3 − [\text{Fe(CN}_6)]^{3-} [ Fe(CN 6 ) ] 3 − [ Cr(H 2 O ) 6 ] 3 + [\text{Cr(H}_2\text{O})_6]^{3+} [ Cr(H 2 O ) 6 ] 3 + [ Co(en) 3 ] 3 + [\text{Co(en)}_3]^{3+} [ Co(en) 3 ] 3 +
CN
Geometry
When
Example
2
Linear
d 10 d^{10} d 10 [ Ag(NH 3 ) 2 ] + [\text{Ag(NH}_3)_2]^+ [ Ag(NH 3 ) 2 ] +
4
Tetrahedral
d 0 , d 5 , d 10 d^0, d^5, d^{10} d 0 , d 5 , d 10 [ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 −
4
Square planar
d 8 d^8 d 8 [ Ni(CN) 4 ] 2 − [\text{Ni(CN)}_4]^{2-} [ Ni(CN) 4 ] 2 −
6
Octahedral
most ions
[ Co(NH 3 ) 6 ] 3 + [\text{Co(NH}_3)_6]^{3+} [ Co(NH 3 ) 6 ] 3 +
Worked example (1) Geometry of[ Ag(NH 3 ) 2 ] + [\text{Ag(NH}_3)_2]^+ [ Ag(NH 3 ) 2 ] + Step 1 — CN: 2 NH₃ (monodentate) → CN = 2. Why? count donor atoms = 2 N.
Step 2 — d-count: Ag is +1, [ Kr ] 4 d 10 [\text{Kr}]4d^{10} [ Kr ] 4 d 10 Why? d 10 d^{10} d 10 Step 3 — geometry: CN 2 → linear, 180 ° 180° 180° . ✓
[ Ni(CN) 4 ] 2 − [\text{Ni(CN)}_4]^{2-} [ Ni(CN) 4 ] 2 − [ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 − Step 1 — d-count: Ni²⁺ = d 8 d^8 d 8 Why? Ni is group 10, 3 d 8 3d^8 3 d 8 Step 2 — ligand field: CN⁻ is strong field ; Cl⁻ is weak field . Why this step? field strength decides if d 8 d^8 d 8 Step 3 — geometry: strong field d 8 d^8 d 8 square planar, diamagnetic ([ Ni(CN) 4 ] 2 − [\text{Ni(CN)}_4]^{2-} [ Ni(CN) 4 ] 2 − d 8 d^8 d 8 tetrahedral, paramagnetic ([ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 −
[ Cr(EDTA) ] − [\text{Cr(EDTA)}]^- [ Cr(EDTA) ] − Step 1 — denticity: EDTA is hexadentate (d = 6 d=6 d = 6 Why? 2 amine N + 4 carboxylate O donors.
Step 2 — CN: 1 × 6 = 6 1 \times 6 = 6 1 × 6 = 6 Why? one ligand, six donor atoms.
Step 3 — geometry: CN 6 → octahedral , with EDTA wrapping all 6 sites. ✓
Recall Feynman: explain to a 12-year-oldImagine a magnet ball in the middle and some sticky balls that want to grab it. Coordination number is just how many sticky balls grabbed on. Because the sticky balls push each other away, they spread out: 2 balls sit on opposite sides (a straight line), 4 balls make a little pyramid-ish shape (tetrahedron) or sometimes a flat square, and 6 balls make a perfect "double pyramid" (octahedron). One twist: some ball-grabbers have two hands (like a clip), so one of them counts as two grabs!
Mnemonic Remember the shapes"2 Lines, 4 Tents or Tables, 6 Octopus."
2 → Line ar4 → Tent (tetrahedral) or Table (square planar)6 → Octo hedral (octopus has 8, but think "octa = 6 here" → an octahedron has 8 faces, 6 vertices : ligands sit on the 6 vertices)
For the d 8 d^8 d 8 "d 8 d^8 d 8
What is coordination number? The number of donor atoms directly bonded to the central metal via coordinate bonds (count donor atoms, not ligand molecules).
Geometry for CN = 2? Linear, bond angle 180°, typical of
d 10 d^{10} d 10 ions like Ag⁺, Au⁺, Cu⁺.
Two possible geometries for CN = 4? Tetrahedral (109.5°) and square planar (90°).
When is CN = 4 square planar rather than tetrahedral? For
d 8 d^8 d 8 ions with strong-field ligands (e.g.
[ Ni(CN) 4 ] 2 − [\text{Ni(CN)}_4]^{2-} [ Ni(CN) 4 ] 2 − , Pt(II), Pd(II)).
Geometry for CN = 6? Octahedral, 90° angles; the most common geometry for transition complexes.
CN of [ Co(en) 3 ] 3 + [\text{Co(en)}_3]^{3+} [ Co(en) 3 ] 3 + 6, because en is bidentate and there are 3 of them (3×2).
Why is [ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 − [ Ni(CN) 4 ] 2 − [\text{Ni(CN)}_4]^{2-} [ Ni(CN) 4 ] 2 − Cl⁻ is weak field (tetrahedral, paramagnetic); CN⁻ is strong field forcing
d 8 d^8 d 8 pairing into square planar (diamagnetic).
Formula linking CN to denticity? CN = ∑ i n i d i \text{CN} = \sum_i n_i d_i CN = ∑ i n i d i (number of ligands × donor atoms each).
Which geometry minimises ligand repulsion for 4 monodentate ligands by pure VSEPR? Tetrahedral (109.5°), the maximum-spread arrangement.
Why do d 10 d^{10} d 10 Filled d-shell gains no extra crystal-field stabilisation, so it minimises repulsion by holding just two ligands far apart.
Crystal Field Theory — explains square-planar vs tetrahedral for d 8 d^8 d 8 VSEPR Theory — origin of linear/tetrahedral/octahedral shapesChelation and Denticity — why CN ≠ number of ligandsMagnetic properties of complexes — diamagnetic square planar vs paramagnetic tetrahedralIsomerism in coordination compounds — geometry decides cis/trans, optical isomers
Intuition Hinglish mein samjhoDekho, coordination number (CN) ka matlab hai metal ke saath kitne donor atoms judey hain — ligand molecules nahi, atoms ginne hain. Jaise en (ethylenediamine) ke do hath hote hain (bidentate), isliye [ Co(en) 3 ] 3 + [\text{Co(en)}_3]^{3+} [ Co(en) 3 ] 3 + donor atoms count karo .
Ab geometry. Ligands ek doosre ko push karte hain (electron pairs hain na), isliye woh maximum door spread hote hain — bilkul VSEPR jaisa. CN 2 → ek doosre ke opposite → linear (180°) , mostly d 10 d^{10} d 10 octahedral (90°) , sabse common, jaise [ Co(NH 3 ) 6 ] 3 + [\text{Co(NH}_3)_6]^{3+} [ Co(NH 3 ) 6 ] 3 +
Sabse interesting CN = 4 hai, kyunki yahan do shape compete karte hain: tetrahedral (109.5°) aur square planar (90°) . General rule: d 8 d^8 d 8 strong field ligand → square planar. Isiliye [ Ni(CN) 4 ] 2 − [\text{Ni(CN)}_4]^{2-} [ Ni(CN) 4 ] 2 − [ NiCl 4 ] 2 − [\text{NiCl}_4]^{2-} [ NiCl 4 ] 2 −
Yeh topic important isliye hai kyunki geometry se hi isomerism (cis/trans, optical), magnetic property aur colour sab decide hote hain. Shape samjhe toh poori coordination chemistry asaan ho jaati hai.
