Q.Draw the structures of optical isomers of:

Key idea: Optical isomerism occurs when a complex lacks a plane of symmetry and is non-superimposable on its mirror image — typically due to chelating ligands creating a chiral arrangement.

(i) $[Cr(C_2O_4)_3]^{3-}$ …

Concept and Intuition

Optical isomerism in coordination compounds is a geometric property: a molecule is chiral if it cannot be superimposed on its mirror image. The key condition is the absence of any improper rotation axis ($S_n$), which includes planes of symmetry ($\sigma$) and centers of inversion ($i$). For octahedral complexes, chirality often arises when:

• The complex has bidentate ligands (like oxalate $C_2O_4^{2-}$ or ethylenediamine $en$) that create a helical twist.
• The arrangement of monodentate ligands around the metal is asymmetric enough to break all symmetry planes.

A common shortcut: if a complex has no plane of symmetry, it is almost certainly chiral. But be careful — some complexes with planes can still be chiral if the plane is not a symmetry plane of the whole molecule (e.g., a plane that cuts through a ligand but not the metal center in a symmetric way).

Let’s work through each complex step by step.

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

Step 1: Identify the coordination geometry.

Chromium(III) is $d^3$, octahedral. Each oxalate ion ($C_2O_4^{2-}$) is a bidentate ligand, so three oxalates occupy all six coordination sites. The complex is analogous to $[Co(ox)_3]^{3-}$.

Step 2: Visualize the structure.

Each oxalate forms a five-membered chelate ring with the metal. The three rings are arranged around the metal in a propeller-like shape. Imagine the three oxalate planes — they are not coplanar; they twist to minimize steric clash. This twist gives the complex a handedness: a left-handed propeller (Λ) and a right-handed propeller (Δ).

Step 3: Check for symmetry.

Does the complex have any plane of symmetry? No — if you try to reflect it, the chelate rings swap positions in a way that cannot be superimposed. The complex belongs to the $D_3$ point group (a threefold rotation axis through the metal, perpendicular to the plane of the three oxalates, with three perpendicular $C_2$ axes, but no mirror plane and no centre of inversion). Since $D_3$ contains no improper symmetry element ($\sigma$, $i$, or $S_n$) at all, the complex is chiral.

Step 4: Draw the enantiomers.

We represent the octahedron as a perspective drawing. For Λ, the three oxalate rings form a left-handed helix when viewed along the $C_3$ axis. For Δ, the helix is right-handed.

Λ-[Cr(C₂O₄)₃]³⁻          Δ-[Cr(C₂O₄)₃]³⁻
   (left-handed)            (right-handed)

(Imagine the oxalate ligands as curved arcs connecting adjacent vertices of the octahedron.)

(ii) $[PtCl_2(en)_2]^{2+}$

Step 1: Identify the metal and geometry.

Platinum(IV) is $d^6$, octahedral. Ethylenediamine ($en$) is a bidentate ligand, so two $en$ ligands occupy four coordination sites. The remaining two sites are occupied by chloride ions.

Step 2: Consider possible geometric isomers.

With two bidentate ligands and two monodentate ligands, there are two geometric arrangements:

• cis: the two chlorides are adjacent (90° apart).
• trans: the two chlorides are opposite (180° apart).

Step 3: Check chirality of each.

• trans isomer: The two $en$ ligands lie in a plane, and the two chlorides are opposite. This complex has a plane of symmetry that passes through the metal, the two chlorides, and bisects the $en$ ligands. It also has a center of inversion at the metal. So it is achiral — no optical isomers.
• cis isomer: The two chlorides are adjacent. The $en$ ligands are now forced into a non-planar arrangement. Does this complex have any plane of symmetry? If you try to find a plane, you’ll see that the two $en$ rings are oriented perpendicular to each other, and the chlorides break any symmetry. The complex is chiral — it exists as a pair of enantiomers.

Step 4: Draw the enantiomers for the cis form.

We show the two chlorides at adjacent positions. The two $en$ ligands are drawn as arcs connecting the remaining four vertices. The enantiomers are mirror images.

cis-[PtCl₂(en)₂]²⁺ (enantiomer 1)    cis-[PtCl₂(en)₂]²⁺ (enantiomer 2)
   Cl at positions 1,2                    Cl at positions 1,2
   en rings twist one way                en rings twist opposite way

(For the trans isomer, simply draw the two chlorides opposite, and the two $en$ ligands in a plane — no mirror image needed.)

(iii) $[Cr(NH_3)_2Cl_2(en)]^{+}$

Step 1: Identify the geometry and ligands. …

Method: Chirality Check via Symmetry & Non-Superimposable Mirror Images

This is the standard method for identifying optical isomers in coordination compounds. Optical isomers (enantiomers) exist when a complex is chiral — meaning it has no plane of symmetry, no centre of symmetry, and its mirror image is non-superimposable.

Steps:

1. Draw the complex in a clear 3D representation (usually octahedral).
2. Draw its mirror image.
3. Check for superimposability — if the mirror image cannot be rotated to match the original, the complex is chiral and shows optical isomerism.
4. Identify the pair of non-superimposable mirror images (enantiomers).

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

• Ligand: Oxalate ($C_2O_4^{2-}$) is a bidentate ligand — it occupies two adjacent coordination sites.
• Geometry: Octahedral, with three bidentate ligands.

Structure:

• The three oxalate rings create a propeller-like shape.
• This complex has no plane of symmetry — it is chiral.

Optical isomers (enantiomers):

The three oxalate ligands each span two adjacent sites with their two oxygen donor atoms, wrapping around the chromium centre in a three-bladed, propeller-like (helical) arrangement — analogous to $[Co(ox)_3]^{3-}$.

Mirror images are non-superimposable — like left and right hands.

Result: Two enantiomers exist (a Δ and Λ pair).

(ii) $[PtCl_2(en)_2]^{2+}$

• Ligand: en (ethylenediamine) is bidentate.
• Geometry: Octahedral.
• Key point: The two Cl ligands can be cis or trans.

Check for chirality:

• trans isomer: Cl ligands opposite — has a plane of symmetry → achiral (no optical isomers).
• cis isomer: Cl ligands adjacent — no plane of symmetry → chiral.

Optical isomers (only for cis form):

The two Cl⁻ ligands occupy adjacent sites, and the two en ligands occupy the remaining four sites in a puckered (non-coplanar) arrangement around the platinum centre.

Mirror image is non-superimposable.

Result: Only the cis isomer shows optical isomerism — two enantiomers.

(iii) $[Cr(NH_3)_2Cl_2(en)]^{+}$

• Ligands: Two NH₃ (monodentate), two Cl (monodentate), one en (bidentate).
• Geometry: Octahedral.

Step 1: Identify possible geometrical isomers

• The en ligand is fixed in a cis position (bidentate) — this leaves only one genuine trans axis among the remaining four sites, so NH₃ and Cl cannot both be trans to each other at the same time.
• The two NH₃ and two Cl can be arranged as:
  • cis-cis (both NH₃ adjacent, both Cl adjacent)
  • NH₃-trans, Cl-cis (NH₃ occupies the one surviving trans axis) …

Common Mistakes in Geometrical & Optical Isomerism (Coordination Compounds)

Students often confuse geometrical isomerism (cis/trans) with optical isomerism (d/l or non-superimposable mirror images). Here are the most frequent errors and how to avoid them.

Mistake 1: Assuming all octahedral complexes show optical isomerism

The error: Thinking that any complex with bidentate ligands will be optically active.

Why it's wrong: Optical activity requires no plane of symmetry and no centre of symmetry. Even with bidentate ligands, if the complex retains a plane of symmetry, it is achiral and optically inactive.

Example:

For $[Cr(C_2O_4)_3]^{3-}$ — this is tris(oxalato)chromate(III).

• Oxalate ($C_2O_4^{2-}$) is a bidentate ligand.
• The complex has a trigonal twist (like $[M(AA)_3]$ type).
• It lacks any plane of symmetry → exists as a pair of non-superimposable mirror images → optical isomers exist.

✓ How to avoid:

• Draw the complex in 3D (use wedge-dash).
• Check for any plane that cuts the complex into two mirror halves.
• If no such plane exists → optical isomers are possible.

Mistake 2: Mis-assigning the oxidation state and geometry of $[PtCl_2(en)_2]^{2+}$

The error: Treating the complex as square-planar Pt(II), then reasoning about its isomerism from the wrong geometry.

Why it's wrong: Count the donor atoms: two bidentate en (4 sites) + two Cl⁻ = coordination number 6, so this is octahedral Pt(IV) ($x + 2(-1) = +2 \Rightarrow x = +4$). A square-planar complex would in any case be optically inactive at this level — its molecular plane is itself a mirror plane; the optical activity here arises from the octahedral cis arrangement.

Example:

For $[PtCl_2(en)_2]^{2+}$ — this is dichlorobis(ethylenediamine)platinum(IV).

• Ethylenediamine (en) is bidentate.
• The complex is octahedral (Pt is in +4 oxidation state, coordination number 6).
• Two en ligands can be arranged cis or trans.
  • cis isomer: no plane of symmetry → optical isomers exist (d and l).
  • trans isomer: has a plane of symmetry → no optical isomers.

✓ How to avoid:

• First determine geometry (octahedral vs square planar).
• For $[PtCl_2(en)_2]^{2+}$, Pt(IV) is d⁶, always octahedral.
• Draw both cis and trans — only cis gives optical activity.

Mistake 3: Mixing up geometrical and optical isomerism in mixed-ligand complexes

The error: Thinking that geometrical isomers (cis/trans) automatically give optical isomers.

Why it's wrong: Geometrical isomers are diastereomers — they have different physical properties. Optical isomers are enantiomers — mirror images. A geometrical isomer may or may not be chiral.

Example:

For $[Cr(NH_3)_2Cl_2(en)]^{+}$ — this is diamminedichloro(ethylenediamine)chromium(III).

• en is bidentate, NH₃ and Cl are monodentate.
• Because en fixes one cis pair of sites, only ONE trans axis survives among the rest — so NH₃ and Cl can never both be trans simultaneously. Possible geometrical isomers:
  • cis-cis (both Cl adjacent, both NH₃ adjacent)
  • NH₃-trans, Cl-cis (NH₃ occupies the surviving trans axis)
  • Cl-trans, NH₃-cis (Cl occupies the surviving trans axis)
• cis-cis isomer: no plane of symmetry → optical isomers exist (d and l).
• The two trans-type isomers: each has a plane of symmetry → no optical isomers.

✓ How to avoid:

• Draw all possible geometrical isomers first.
• For each, check symmetry:
  • If the isomer has a plane of symmetry → optically inactive.
  • If no plane → optically active (pair of enantiomers).
• Common trap: Students draw only one cis and claim optical activity — but must check if the cis form is actually chiral.

Mistake 4: Incorrectly drawing mirror images as identical

The error: Drawing two structures that are actually the same (superimposable) and calling them optical isomers.

Why it's wrong: Optical isomers must be non-superimposable mirror images. If you can rotate one structure to match the other, they are identical.

Example: …