Q.Draw all the isomers (geometrical and optical) of:

Concept: Geometrical Isomerism in octahedral complexes with bidentate ligands (en = ethylenediamine). Optical activity arises when the complex lacks a plane of symmetry.

(i) $[CoCl_2(en)_2]^{+}$

• Two Cl ligands can be cis (90°) or trans (180°).
• cis form has no plane of symmetry → exists as a pair of non-superimposable mirror images (optical isomers). …

(i) $[CoCl_2(en)_2]^{+}$

Co(III), octahedral; two en chelates and two $Cl^-$.

• trans: the two $Cl^-$ are $180^\circ$ apart and the two en lie in the equatorial plane. A plane of symmetry is present -> optically inactive.
• cis: the two $Cl^-$ are $90^\circ$ apart. The ion has no plane of symmetry -> optically active, existing as a non-superimposable d/l enantiomeric pair.

Total = 3 isomers (1 trans + cis d and l).

(ii) $[Co(NH_3)Cl(en)_2]^{2+}$

Type $[M(en)_2bc]$ with $b = NH_3$, $c = Cl$.

• trans (NH$_3$ and Cl axial, $180^\circ$): a plane of symmetry is present -> optically inactive.
• cis (NH$_3$ and Cl at $90^\circ$): no plane of symmetry -> optically active, a d/l pair.

Total = 3 isomers (1 trans + cis d and l).

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

Type $[M(en)a_2b_2]$ with $a = NH_3$, $b = Cl$. en takes two cis sites; the remaining four sites hold two NH$_3$ and two Cl. Only one trans pair of sites is available, so exactly three geometrical arrangements exist: …

Method: Coordination Isomer Enumeration via Coordination Sphere Analysis

We use the Coordination Geometry + Ligand Arrangement method for octahedral complexes. The steps:

1. Identify coordination number → 6 (octahedral) for all.
2. List all ligands and their denticity (monodentate vs bidentate).
3. Determine possible geometric arrangements for bidentate ligands (en = ethylenediamine, bidentate).
4. Check for optical activity — look for absence of plane of symmetry / centre of inversion.
5. Draw all distinct isomers — geometric first, then optical.

(i) $[CoCl_2(en)_2]^{+}$

Step 1: Co(III), coordination number 6. Ligands: 2 × Cl⁻ (monodentate), 2 × en (bidentate).

Step 2: The two en ligands can occupy positions relative to each other:

• cis — both en rings adjacent (90° apart)
• trans — both en rings opposite (180° apart)

Step 3: Check optical activity:

• cis isomer: No plane of symmetry → optically active (exists as a pair of enantiomers)
• trans isomer: Has a plane of symmetry → optically inactive

Isomers:

• Geometrical: cis and trans
• Optical: cis has two enantiomers (d and l); trans has none

Total isomers: 3 (cis-d, cis-l, trans)

(ii) $[Co(NH_3)Cl(en)_2]^{2+}$

Step 1: Ligands: 1 × NH₃, 1 × Cl⁻ (both monodentate), 2 × en (bidentate).

Step 2: The two en ligands are bidentate, so (as in part (i)) they can only occupy cis positions relative to each other — a chelate ring cannot span trans sites. This leaves two remaining coordination positions for NH₃ and Cl⁻, and these two remaining positions can themselves be cis or trans to each other.

Step 3: So there are exactly two geometrical arrangements:

• cis: NH₃ and Cl⁻ occupy adjacent (cis) positions → 1 geometrical isomer
• trans: NH₃ and Cl⁻ occupy opposite (trans) positions → 1 geometrical isomer

Step 4: Check optical activity:

• cis (NH₃/Cl cis): No plane of symmetry → optically active (2 enantiomers)
• trans (NH₃/Cl trans): Has a plane of symmetry → optically inactive

Total isomers: 3 (cis: d and l; trans: 1)

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

Step 1: Ligands: 2 × NH₃, 2 × Cl⁻ (monodentate), 1 × en (bidentate).

Step 2: The en ligand is fixed in one position. The four monodentate ligands occupy the remaining four positions.

Step 3: Possible arrangements of the two NH₃ and two Cl: with en chelating two adjacent sites, only one trans axis survives among the four remaining positions — so the two NH₃ and the two Cl can never BOTH be trans at the same time. The distinct arrangements are: …

🧠 Core Concept Recap (Why This Matters)

Geometrical isomerism in coordination compounds arises when ligands can occupy different positions around the metal ion. Optical isomerism occurs when the complex is non-superimposable on its mirror image (chiral).

Key conditions for geometrical isomerism:

• Coordination number 4 or 6 (most common)
• Presence of at least two different types of ligands (or ambidentate ligands)
• Ma₂b₂, Ma₂bc, M(AA)₂, M(AA)bc type complexes (where AA = bidentate ligand)

✗ Common Mistake #1: Forgetting that en is a bidentate ligand

What students do wrong:

They treat en (ethylenediamine) as two separate monodentate ligands. This leads to wrong isomer counts and wrong geometries.

Example: In [CoCl₂(en)₂]⁺, some students draw en as two separate NH₂CH₂CH₂NH₂ groups — but it’s one chelating ligand that occupies two adjacent positions.

How to avoid:

• Always remember: en is bidentate — it occupies two coordination sites but acts as one ligand unit.
• In drawings, show en as a curved line connecting two coordination positions.

✗ Common Mistake #2: Confusing cis and trans for bidentate ligands

What students do wrong:

For [CoCl₂(en)₂]⁺, they think en can be placed trans to each other — but bidentate ligands cannot span trans positions (too far apart).

How to avoid:

• A single bidentate ligand can never span two trans positions — each en must chelate two adjacent (cis) sites; the chelate ring is too short to reach across the octahedron.
• The cis/trans labels of [CoCl₂(en)₂]⁺ therefore refer to the two Cl⁻ ligands, not to the en rings — and the cis form is optically active.

✗ Common Mistake #3: Missing optical isomerism in [CoCl₂(en)₂]⁺

What students do wrong:

They draw only cis and trans geometrical isomers, but forget that the cis isomer is chiral (non-superimposable mirror image).

How to avoid:

• Check for plane of symmetry in the cis isomer. In [CoCl₂(en)₂]⁺, the cis form has no plane of symmetry → it exists as a pair of enantiomers (d and l).
• Trans isomer has a plane of symmetry → no optical isomerism.

Correct answer for (i):

• Geometrical: cis and trans
• Optical: cis → 2 enantiomers; trans → none

✗ Common Mistake #4: Claiming [Co(NH₃)Cl(en)₂]²⁺ has only ONE arrangement because en "fixes" everything

What students do wrong:

They reason that since both en ligands must chelate cis positions, the NH₃ and Cl have no freedom left — and conclude only one geometrical isomer exists. Others try to draw a trans arrangement of a single en ligand itself.

How to avoid:

• Each en must chelate two adjacent sites, but the two REMAINING sites (holding NH₃ and Cl) can still be cis or trans to each other — that is exactly the cis/trans choice.
• cis (NH₃ and Cl at 90°): no plane of symmetry → chiral → 2 enantiomers
• trans (NH₃ and Cl at 180°): plane of symmetry → optically inactive

Correct answer for (ii):

• Geometrical: cis and trans
• Optical: the cis form exists as a d/l pair → 3 stereoisomers in total

✗ Common Mistake #5: Overcounting or undercounting isomers for [Co(NH₃)₂Cl₂(en)]⁺

What students do wrong:

They treat en as two separate ligands and try to draw many cis/trans combinations — but en is bidentate, so it occupies two adjacent sites. Some students also wrongly assume NH₃ and Cl can BOTH be trans to each other at once.

How to avoid:

• First, fix en as a cis unit (it must be adjacent) — this leaves only one genuine trans axis among the remaining four sites, so NH₃ and Cl can never both be trans simultaneously.
• Then, place the two NH₃ and two Cl in the remaining four positions.
• Possible arrangements:
  • Both NH₃ cis to each other and both Cl cis to each other → one isomer (chiral, gives a d/l pair) …