Coordination Compounds · Lesson 4 of 12
Geometrical Isomerism
“Use donor positions and symmetry to identify cis/trans and fac/mer arrangements without double-counting rotations.”
• Distinguish geometrical isomers from rotated drawings. • Construct square-planar and octahedral isomers. • Explain why tetrahedral MA₂B₂ has no cis/trans pair. • Count geometrical arrangements with chelating ligands.
Same bonds, different positions
Cisplatin and its trans counterpart have the same platinum, ammonia and chloride composition, but their ligand arrangements differ. Geometrical isomerism changes the relative positions of ligands without changing which donor atoms attach to the metal. Before counting isomers, establish the geometry. Four donors drawn as a cross represent a square plane only if that is the actual molecular geometry; flattening a tetrahedron on paper does not create cis and trans forms.
A drawing rotated through space remains the same structure. Compare relationships that survive rotation: which ligand is opposite which, which pairs are adjacent, and which donor pairs belong to the same chelate. An octahedral drawing is a projection of six positions, not a flat six-membered ring. Its three opposite pairs can be labelled +x/−x, +y/−y and +z/−z; any pair from different axes is adjacent.
Square-planar complexes
For square-planar MA₂B₂, place the two A ligands adjacent at 90° to get cis, or opposite at 180° to get trans. Choosing a different adjacent pair merely rotates the cis drawing. Thus [PtCl₂(NH₃)₂] has two geometrical isomers. Neither ideal planar arrangement has a distinct non-superimposable mirror image when the ligands themselves are achiral; the plane of the molecule is a reflection plane.
Problem
Count geometrical isomers of square-planar [PtBrCl(NH₃)(py)], where py is pyridine.
- 1.Fix NH₃ at one corner. Its opposite ligand can be Br, Cl or py, giving three possibilities.
- 2.For each choice the remaining two ligands occupy the remaining opposite pair; swapping them is achieved by rotating the molecule in three-dimensional space, so it does not double the count.
- 3.The three distinct sets of trans partners are NH₃/Br with Cl/py; NH₃/Cl with Br/py; and NH₃/py with Br/Cl. All are achiral in the ideal square-planar model with achiral ligands.
Tetrahedral complexes
All six pairs of vertices in an ideal tetrahedron have the same angular relationship, about 109.5° from the centre. There is no opposite pair at 180°. Consequently tetrahedral MA₂B₂ has one geometrical arrangement: changing which two corners carry A is equivalent to a rotation. This does not mean all tetrahedral complexes are achiral. A tetrahedral centre with four different ligands may have optical isomers; absence of cis/trans is a narrower statement.
Octahedral MA₄B₂ and MA₃B₃
For [CoCl₂(NH₃)₄]⁺, put chlorido ligands on +z and −z to obtain trans. Put them on +z and +x to obtain cis. Both complexes have CN 6, Co +3 and entity charge +1. Neither a charge balance nor the empirical formula distinguishes them. The same method applies to [Fe(NH₃)₂(CN)₄]⁻: use the two NH₃ ligands as the marked pair, leaving four C-bound cyanido donors at the other vertices.
For MA₃B₃, three identical ligands may lie on one triangular face, such as +x,+y,+z. They are pairwise cis; this is facial, fac. Alternatively they lie on a meridional plane, such as +z,−z,+x, containing one trans pair and two cis pairs; this is meridional, mer. Both fac-[Co(NH₃)₃(NO₂)₃] and mer-[Co(NH₃)₃(NO₂)₃] can be described after fixing the same nitrite donor mode throughout. Changing the donor atom as well would introduce linkage isomerism, a different classification.
Problem
Three NH₃ ligands in octahedral [Co(NH₃)₃Cl₃] have pairwise angles 90°, 90° and 180°. Identify the isomer.
- 1.A facial set has all three pairwise angles 90°. One 180° pair rules it out.
- 2.The set lies in a plane through the metal with two NH₃ opposite and one adjacent to both. It is mer.
- 3.There are two geometrical isomers overall, fac and mer; ideal rotations do not generate additional ones.
Chelates restrict which vertices can be paired
In the en and oxalate examples, each short bidentate ligand spans adjacent octahedral positions. A drawing that joins opposite vertices with one ordinary en ligand does not represent the stated geometry. For [MCl₂(en)₂], the chlorido ligands may still be cis or trans. The four remaining vertices must be paired into two adjacent donor pairs, one pair per en. The cis form also has a mirror-image pair, introduced in the next lesson; that optical subdivision does not create a third geometrical class.
| Octahedral family | Geometrical classes | Spatial test |
|---|---|---|
| M(AA)₃ with three identical symmetric chelates | One | Every donor belongs to one of three equivalent adjacent pairs |
| MA₄B₂ | Two | B–M–B is 90° or 180° |
| MA₃B₃ | Two | All A pairwise cis, or one A trans pair |
| M(AA)₂B₂ | Two | B pair cis or trans |
| M(AA)A₂B₂ | Three | A trans; B trans; or both A and B cis |
For [Co(NH₃)₂Cl₂(en)]⁺, choose en on +x,+y. The four remaining vertices are −x,−y,+z,−z. A trans NH₃ pair must occupy +z,−z; a trans chlorido pair can instead occupy those positions; or neither identical pair is trans. These are three geometrical classes. In the both-cis class there is additionally an enantiomeric pair, so the full stereoisomer count differs from the geometrical count. Keeping these counts separate prevents many apparent contradictions.
Quick Quiz
Quiz
How many geometrical isomers does ideal square-planar MA₂B₂ have?
Adjacent and opposite identical ligand pairs give cis and trans.
Which relationship identifies a facial MA₃B₃ arrangement?
A triangular face contains no opposite vertices.
Why is tetrahedral MA₂B₂ not a cis/trans pair?
There are no opposite tetrahedral vertices corresponding to the square-planar trans arrangement.
How many distinct trans-partner patterns exist for square-planar MABCD?
Fix A and choose B, C or D as its trans partner.
Which is invalid for an ordinary octahedral en chelate?
The short en ligand spans cis positions in these complexes, not a 180° separation.
Practice Problems
- Describe both geometrical isomers of [Fe(NH₃)₂(CN)₄]⁻ using octahedral vertex labels.
- Construct fac and mer [Co(NH₃)₃Cl₃] and explain their counts.
- List the isomers of square-planar [PtBrCl(NH₃)(py)] by trans partners and discuss optical isomerism.
- Compare geometrical counts for [Cr(C₂O₄)₃]³⁻ and [CoCl₂(en)₂]⁺.
- Enumerate geometrical classes for [Co(NH₃)₂Cl₂(en)]⁺ without treating rotations as new isomers.
- Explain why different colours alone cannot establish which isomer is cis.
1. Use ±x, ±y, ±z for the six donor positions. 2. Put NH₃ at +z and −z for trans; cyanido occupies ±x and ±y. 3. Put NH₃ at +z and +x for cis; cyanido occupies the other four vertices. 4. In each case Fe is +3 since x − 4 = −1; CN remains 6.
Key Takeaways
• Geometrical isomers differ in ligand positions while retaining donor connectivity. • Rotating a molecule does not produce a new isomer. • Square-planar MA₂B₂ has cis and trans arrangements. • Tetrahedral MA₂B₂ has no cis/trans distinction. • Octahedral MA₃B₃ has fac and mer arrangements. • Short chelates in these examples join adjacent donor positions. • Geometrical classes and total stereoisomer counts must be distinguished.