Q.How many geometrical isomers are possible in the following coordination entities?

Concept: Geometrical isomerism in coordination compounds arises when ligands occupy different relative positions around the central metal ion, and the complex has no plane of symmetry.

(i) $[Cr(C_2O_4)_3]^{3-}$

The oxalate ion ($C_2O_4^{2-}$) is a bidentate ligand. Three such ligands form an octahedral complex. This complex is chiral (like a three-bladed propeller) and exists as a pair of optical isomers, but it has no geometrical isomers — all positions are equivalent.

(ii) $[Co(NH_3)_3Cl_3]$

This is an octahedral complex with three identical ligands of one type and three of another. The possible arrangements are: …

Let’s understand why. Geometrical isomerism in coordination compounds arises when ligands can occupy different spatial positions around the metal ion, leading to distinct arrangements that are not superimposable by rotation. The key conditions are:

• The complex must have a coordination number of 4 (square planar or tetrahedral) or 6 (octahedral).
• There must be at least two different types of ligands (or polydentate ligands that create ring constraints).
• The arrangement must allow for non-identical spatial distributions — for example, cis/trans in square planar, or fac/mer in octahedral.

For octahedral complexes, the most common cases involve:

• Ma₃b₃ type (like your second complex): three of one ligand and three of another. Here, the isomers are facial (fac) — where the three identical ligands occupy one face of the octahedron — and meridional (mer) — where they lie in a plane that goes through the metal.
• Chelating ligands (like oxalate, C₂O₄²⁻): these are bidentate, meaning each ligand occupies two adjacent coordination sites. This restricts the geometry because the ligand’s two donor atoms must be cis to each other (they can’t span trans positions due to the small ring size). So, for a tris(bidentate) complex like $[Cr(C_2O_4)_3]^{3-}$, all three ligands are identical and each forces a cis arrangement. The only possible geometry is the one where all three chelate rings are arranged around the metal — this is actually a chiral structure (like a propeller), but it has no geometrical isomers because swapping any two ligands gives the same arrangement. There is no cis/trans variation possible.

Now, let’s work through each part step by step.

(i) $[Cr(C_2O_4)_3]^{3-}$

1. Identify the coordination number and geometry.

   Chromium(III) is in the +3 oxidation state, and oxalate ($C_2O_4^{2-}$) is a bidentate ligand. Three bidentate ligands give a coordination number of $3 \times 2 = 6$, so the geometry is octahedral.

2. Consider the ligand constraints.

   Each oxalate ligand must occupy two adjacent (cis) positions because the ligand’s two donor oxygen atoms are connected by a short carbon chain — they cannot span opposite (trans) positions. This means every ligand forces a cis arrangement at its binding site.

3. Check for possible different arrangements.

   Since all three ligands are identical, the only question is whether the three chelate rings can be arranged in more than one way. In an octahedron, three bidentate ligands can be placed such that the rings are all in one plane (like a belt) or twisted. But because each ligand is identical and each forces cis binding, the only distinct arrangement is the one where the three rings are mutually perpendicular — this is the familiar tris(chelate) structure. There is no alternative: you cannot have a “trans” version because that would require a ligand to span opposite sites, which is impossible. So, only one geometrical isomer exists.

   Watch out

   A common mistake is to think that tris(bidentate) complexes like this have cis/trans isomers. They don’t — because all ligands are identical and each is bidentate, the only variation is optical isomerism (left- and right-handed forms), not geometrical. Geometrical isomerism requires different spatial arrangements of different ligands or of the same ligand in non-equivalent positions.

4. Conclusion for (i).

   The complex has 0 geometrical isomers. Since all three oxalate ligands are identical and each is forced into a cis (chelating) arrangement, only one spatial arrangement of the ligands is possible — there is no cis/trans-type variation to distinguish. (The complex does exist as a pair of optical isomers, but that is a separate property from geometrical isomerism.)

(ii) $[Co(NH_3)_3Cl_3]$

1. Identify the coordination number and geometry. …

Method: Symmetry & Plane of Symmetry Approach

This method works by checking the number of unique spatial arrangements that cannot be superimposed by rotation — the key condition for geometrical isomerism.

Step 1 — Identify coordination number and geometry

• (i) $[Cr(C_2O_4)_3]^{3-}$

  • Coordination number = 6 (each oxalate is bidentate, 3 × 2 = 6)
  • Geometry = Octahedral

• (ii) $[Co(NH_3)_3Cl_3]$

  • Coordination number = 6
  • Geometry = Octahedral

Step 2 — Check for geometrical isomerism condition

Condition: The complex must have no plane of symmetry that makes all arrangements identical.

For (i) $[Cr(C_2O_4)_3]^{3-}$

• All three ligands are identical bidentate oxalate ions.
• In an octahedral complex with three identical bidentate ligands, the arrangement is always the same — it is a chiral pair (optical isomers), but no geometrical isomers exist.
• Reason: The three rings occupy equivalent positions; any attempt to change arrangement gives the same structure.

Result: $\boxed{0}$ geometrical isomers

For (ii) $[Co(NH_3)_3Cl_3]$

• This is an $MA_3B_3$ type octahedral complex. …

Common Mistakes in Geometrical Isomerism for Coordination Compounds

Mistake 1: Forgetting that $[Cr(C_2O_4)_3]^{3-}$ has no geometrical isomers

Why students get it wrong:

Students see three bidentate ligands and assume multiple arrangements exist — like in $[Ma_2b_2]$ or $[Ma_3b_3]$ complexes.

The correct reasoning:

• $C_2O_4^{2-}$ (oxalate) is a bidentate ligand — it occupies two adjacent coordination sites.
• In $[Cr(C_2O_4)_3]^{3-}$, all three ligands are identical and bidentate.
• The complex has an octahedral geometry with no possibility of different spatial arrangements — all positions are equivalent.
• This complex is chiral (exists as $d$ and $l$ optical isomers), but not geometrical isomers.

✓ How to avoid:

• For complexes with only one type of bidentate ligand, geometrical isomerism is impossible.
• Geometrical isomerism requires at least two different types of ligands (or ambidentate ligands) that can occupy different relative positions.

Mistake 2: Thinking $[Co(NH_3)_3Cl_3]$ has only 2 geometrical isomers

Why students get it wrong:

Many recall that $[Ma_3b_3]$ type octahedral complexes have two isomers — fac and mer. But they sometimes miss that both are geometrical isomers.

The correct reasoning:

• $[Co(NH_3)_3Cl_3]$ is an octahedral complex with three identical ligands of one type (NH$_3$) and three of another (Cl$^-$).
• Two distinct geometrical arrangements exist:

✓ How to avoid:

• For $[Ma_3b_3]$ octahedral complexes, always check for both fac and mer isomers.
• Draw the octahedron and place three identical ligands on one triangular face (fac) vs. in a plane through the centre (mer).
• Both are geometrical isomers (different spatial arrangements, same connectivity).

Mistake 3: Confusing geometrical isomerism with optical isomerism

Why students get it wrong: …