Q.Identify the optically active compounds from the following:

Concept: Coordination Compound Nomenclature & Optical Activity

Optical activity requires the complex to lack a plane of symmetry (i.e., be chiral).

Reasoning:

1. (i) $[Co(en)_3]^{3+}$ — The three bidentate en ligands form a propeller-like geometry; the complex is chiral and exists as non-superimposable mirror images (Δ and Λ isomers).
2. (ii) trans-$[Co(en)_2Cl_2]^{+}$ — The two Cl ligands are opposite each other; the complex has a plane of symmetry through the metal and the two en rings — achiral. …

Why optical activity matters here

Optical activity is a property of chiral molecules — those that are non-superimposable on their mirror image. In coordination chemistry, chirality usually appears when the complex lacks any plane of symmetry. The classic test: if you can draw a plane that cuts the complex into two mirror-image halves, the complex is achiral and optically inactive. If no such plane exists, the complex is chiral and can exist as a pair of enantiomers.

The ligands here are important:

• en (ethylenediamine) is a bidentate chelating ligand — it forms a ring with the metal.
• Cl and NH$_3$ are monodentate.

Let’s examine each complex one by one.

1. $[Co(en)_3]^{3+}$

This is an octahedral complex with three bidentate en ligands. Each en ligand forms a five-membered chelate ring. The three rings are arranged around the cobalt ion.

Think of the geometry: the three en ligands wrap around the metal like a three-bladed propeller. This arrangement has no plane of symmetry — the complex is chiral. It exists as a pair of enantiomers (often called $\Delta$ and $\Lambda$ forms).

So (i) is optically active.

2. trans-$[Co(en)_2Cl_2]^{+}$

Here, two en ligands occupy four positions in the octahedron, and the two Cl ligands are opposite each other (trans).

Draw the structure: the two en ligands lie in a plane, and the two Cl atoms are on opposite sides. This complex has several planes of symmetry — for instance, the plane that contains the metal and both Cl atoms, cutting through the middle of both en rings.

Because a plane of symmetry exists, the molecule is superimposable on its mirror image.

So (ii) is optically inactive.

3. cis-$[Co(en)_2Cl_2]^{+}$ …

Method: Optical Activity via Symmetry & Chirality Check

Optical activity arises when a compound is non-superimposable on its mirror image (i.e., it is chiral). For coordination compounds, the key is to check for the presence of a plane of symmetry or a centre of symmetry — if either exists, the compound is optically inactive.

Steps

1. Draw or visualise the geometry of the complex (octahedral for all given).
2. Identify the ligand arrangement — especially for chelating ligands like en (ethylenediamine).
3. Check for symmetry elements:
   • A plane of symmetry → optically inactive.
   • A centre of symmetry → optically inactive.
   • No symmetry plane or centre → optically active (chiral).
4. Classify each option.

Applying the steps

(i) $[Co(en)_3]^{3+}$

• Three bidentate en ligands form a propeller-like structure.
• The complex has no plane of symmetry — it exists as a pair of non-superimposable mirror images (Δ and Λ isomers).
• ✓ Optically active

(ii) trans-$[Co(en)_2Cl_2]^{+}$

• Two en ligands are in the same plane; the two Cl ligands are opposite (trans). …

🧠 Core Concept Recap

Optical activity in coordination compounds arises when the complex is non-superimposable on its mirror image — i.e., it lacks a plane of symmetry. This is most common in octahedral complexes with chelating ligands like ethylenediamine (en).

✗ Mistake 1: Thinking all chelate complexes are optically active

Why it happens:

Students see en (a bidentate ligand) and assume the complex must be chiral.

Example from the list:

• (i) $[Co(en)_3]^{3+}$ — is optically active (no plane of symmetry).
• (ii) trans-$[Co(en)_2Cl_2]^{+}$ — is not optically active (has a plane of symmetry).

How to avoid:

Always check for symmetry — draw the structure mentally or on paper. A complex with chelating ligands can still be achiral if it has a plane of symmetry.

✓ Rule: Chelating ligands can create chirality, but not always. Symmetry is the final judge.

✗ Mistake 2: Confusing cis/trans with optical activity

Why it happens:

Students think cis always means chiral and trans always means achiral.

Example from the list:

• (iii) cis-$[Co(en)_2Cl_2]^{+}$ — is optically active (no plane of symmetry).
• (ii) trans-$[Co(en)_2Cl_2]^{+}$ — is not optically active.

How to avoid:

Memorise the pattern for $[M(AA)_2X_2]$ type complexes:

• cis → optically active (two enantiomers exist)
• trans → optically inactive (has a plane of symmetry)

✓ Rule: For $[M(AA)_2X_2]$, only the cis isomer can be optically active.

✗ Mistake 3: Forgetting that monodentate ligands rarely give optical activity

Why it happens:

Students see $[Cr(NH_3)_5Cl]$ and think it might be chiral because it’s asymmetric.

Example from the list:

• (iv) $[Cr(NH_3)_5Cl]$ — not optically active (has a plane of symmetry through Cr, Cl, and the opposite NH₃).

How to avoid:

For octahedral complexes with only monodentate ligands, optical activity is extremely rare. You need at least three different bidentate ligands or a very specific arrangement.

✓ Rule: Monodentate ligands in an octahedral complex almost never give optical activity — check for symmetry first.

✗ Mistake 4: Not drawing the structure before deciding

Why it happens:

Students try to reason purely from the formula without visualising.

How to avoid: …