3.4.4 · D5Coordination Chemistry
Question bank — Coordination number and geometry — 2 (linear), 4 (tetrahedral - square planar), 6 (octahedral)
First, the pictures you need in your head
Before the traps, lock in three pieces of vocabulary that the questions below keep re-using. Each is drawn so you never have to imagine the 3D shape.
The three core geometries — and the trick of building square planar by deleting the two axial ligands of an octahedron — are shown here. Refer back to the labelled corners whenever a trap mentions "axial" or "equatorial".
True or false — justify
A ligand and a donor atom are the same thing.
False. A ligand is the whole molecule/ion; a donor atom is the specific atom that shares its lone pair. One ligand can carry several donor atoms (see Chelation and Denticity).
has coordination number 3 because it has 3 ligands.
False. CN counts donor atoms, not molecules. Each en is bidentate (2 N donors), so CN and the geometry is octahedral.
Every 4-coordinate complex is tetrahedral.
False. CN = 4 splits into tetrahedral () and square planar (); ions with strong-field ligands choose square planar (see figure s04).
All complexes have the same geometry because the metal is fixed.
False. is tetrahedral (weak-field Cl⁻) while is square planar (strong-field CN⁻). The ligand flips the shape.
Octahedral is the most common geometry for transition-metal complexes.
True. Six ligands at balance good bonding/charge satisfaction against manageable repulsion, so most ions (Co, Fe, Cr) are octahedral.
A higher coordination number always means a bigger crystal-field stabilisation.
False. CFSE depends on d-electron count, geometry and field strength, not merely on how many ligands are attached; a ion gains none regardless of CN.
Linear geometry always implies exactly two ligand molecules.
False. Linear means CN = 2, i.e. two donor atoms (two domains, figure s01). It usually is two monodentate ligands, but it is the donor count of 2 that matters.
Square planar can be pictured as an octahedron with its two axial ligands removed.
True. Deleting the top and bottom ligands of an octahedron leaves the four equatorial ones in a plane — the square-planar arrangement (figure s02).
Only CN = 2, 4 and 6 exist for coordination complexes.
False. CN = 5 (trigonal bipyramidal or square pyramidal) is real and common; this page focuses on the three most frequent values, not an exhaustive list — see the edge-case note below.
Spot the error
" is bent because two electron pairs repel."
Error: with two domains (figure s01), maximum separation puts them at opposite poles → , i.e. linear, not bent. Bending needs extra lone pairs on the central atom, which Ag⁺ () doesn't force.
" is a strong-field ligand, so is square planar."
Error: Cl⁻ is a weak-field ligand. Weak field on gives tetrahedral and paramagnetic, not square planar.
"EDTA contributes 4 to the coordination number."
Error: EDTA is hexadentate (2 amine N + 4 carboxylate O = 6 donors), so one EDTA contributes 6 → octahedral.
" ions like Cu⁺ prefer octahedral because 6 ligands are most stable."
Error: a filled shell is spherical and gains no crystal-field energy, so it minimises ligand repulsion by holding just two ligands → linear.
" is square planar because Zn²⁺ is a ion."
Error: Zn²⁺ is , not . With no CFSE preference it takes the VSEPR optimum → tetrahedral.
"The bond angle in a tetrahedral complex is ."
Error: the ideal tetrahedral angle is ; belongs to octahedral and square-planar adjacent angles.
"Since is square planar, it must be paramagnetic like most Ni complexes."
Error: strong-field CN⁻ pairs the 8 electrons into the four lower orbitals, leaving none unpaired → diamagnetic (figure s03).
Why questions
Why do we count donor atoms rather than ligand molecules for CN?
Because the metal's bonds form to individual lone-pair-donating atoms; a chelate with several hands makes several bonds from one molecule, and only counting donors captures the true coordination.
Why does CN = 4 have two possible geometries when CN = 2 and 6 do not?
At CN = 4 the pure-VSEPR tetrahedron and the CFSE-favoured square plane give two separate energy minima (figure s04), so d-electron effects (especially + strong field) can tip the balance — see Crystal Field Theory.
Why does a ion with a strong field become square planar?
Square planar pushes very high (figure s03); 8 electrons fill the four lower orbitals and leave that orbital empty. A strong field makes this large splitting worth the pairing energy.
Why is octahedral described as "a great compromise"?
Six ligands satisfy bonding and charge well while spacing keeps ligand–ligand repulsion tolerable — more ligands would crowd, fewer would waste bonding capacity.
Why do bulky ligands push CN = 4 toward tetrahedral rather than square planar?
A tetrahedron spreads four ligands at , giving more room; the square-planar neighbours would clash sterically for large ligands.
Why doesn't ligand-field strength change the shape of an complex?
The shell gains no CFSE from field splitting, so the geometry is set purely by VSEPR repulsion (linear) and is insensitive to whether the ligand is strong or weak field.
Edge cases
Does a ion ever adopt square planar at CN = 4?
Typically no — with no d-electrons there's no CFSE gain from square planar, so (like ) takes the tetrahedral VSEPR optimum.
Why does this page skip CN = 5?
CN = 5 (trigonal bipyramidal or square pyramidal) does exist, but it is rarer and its two shapes are very close in energy so it interconverts easily; the parent topic concentrates on the far more common CN = 2, 4 and 6.
If a complex has one hexadentate ligand only, what is its geometry?
CN , so it is octahedral, with the single ligand wrapping around all six coordination sites (e.g. ).
Can the same metal ion show tetrahedral in one complex and square planar in another?
Yes — () is tetrahedral with weak-field Cl⁻ but square planar with strong-field CN⁻; the ligand, not the metal alone, decides.
Is coordination number the same as oxidation state?
No — CN counts donor-atom bonds (a geometric/steric count); oxidation state is the metal's formal charge. They can differ freely (e.g. Co³⁺ with CN = 6).
What happens to geometry if a bidentate ligand is replaced by two monodentate ones donating the same atoms?
The CN stays the same because donor-atom count is unchanged, so the geometry (e.g. octahedral) is preserved; chelation affects stability, not CN — see Chelation and Denticity.
Does square planar vs tetrahedral affect magnetism for a ion?
Yes — square planar (strong field) pairs all electrons → diamagnetic, while tetrahedral (weak field) leaves two unpaired → paramagnetic; this is a key diagnostic in Magnetic properties of complexes.
Recall One-line survival kit
Count donor atoms for CN; use VSEPR for the default shape (2→linear, 4→tetrahedral, 6→octahedral); only + strong field overrides 4 into square planar. When in doubt, ask what does the ligand field do to the d-electrons? (Crystal Field Theory).