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Isomerism in MABCDEF Octahedral Complexes

Isomerism in MABCDEF Octahedral Complexes | ChemCa.in
Inorganic Chemistry / Coordination Compounds

Geometrical Isomerism in $MABCDEF$ Complexes

Understanding the absolute limits of octahedral stereochemistry with 6 different ligands.

1. The Ultimate Asymmetry

Most octahedral complexes studied in undergraduate chemistry have high degrees of symmetry (like $MA_6$, $MA_4B_2$, or $MA_3B_3$). However, the theoretical limit of complexity is reached with the general formula $MABCDEF$.

In this complex, the central metal ion ($M$) is coordinated to six completely different monodentate ligands. Because every position is occupied by a unique group, the molecule lacks any plane of symmetry, center of inversion, or improper axis of rotation.

The $MABCDEF$ Octahedron

6 Unique Ligands

M A B C D E F

Real-World Example

Synthesizing a complex with 6 different ligands is incredibly difficult. One of the few verified examples synthesized by platinum chemists is:

$[Pt(py)(NH_3)(NO_2)(Cl)(Br)(I)]$

Where 'py' is pyridine. Isolating the distinct isomers of this complex is a monumental task in inorganic chemistry.

2. Calculating the Number of Geometrical Isomers

Trying to draw all isomers randomly will lead to duplicates and confusion. Instead, chemists use a systematic mathematical approach to find that there are exactly 15 geometrical isomers. Here is the logic:

Step 1: Fix one ligand and choose its "Trans" partner

Let's fix ligand A at the top axial position. We must place another ligand exactly opposite (trans) to it at the bottom axial position. We have 5 choices for this trans partner: B, C, D, E, or F.

This gives us 5 primary spatial arrangements.

Step 2: Arrange the remaining 4 ligands on the equator

Let's look at one of those primary arrangements (e.g., A is trans to B). The remaining 4 ligands (C, D, E, F) must be placed on the 4 equatorial positions (the square plane).

If we fix C at one of the equatorial positions, we must choose which ligand sits trans to it across the square. We have 3 choices left: D, E, or F.

Once this trans pair is chosen, the positions of the final two ligands are locked in geometrically. Thus, there are 3 distinct equatorial arrangements for every axial pair.

Total Calculation

$5 \text{ (axial pairs)} \times 3 \text{ (equatorial arrangements)} = \mathbf{15 \text{ Geometrical Isomers}}$

3. The Optical Isomerism Connection (Stereoisomers)

The story doesn't end with 15 isomers. Because all 6 ligands are different, none of these 15 geometrical isomers possess a plane of symmetry ($\sigma$) or a center of inversion ($i$).

Therefore, every single one of the 15 geometrical isomers is chiral (optically active). Each geometrical isomer exists as a pair of non-superimposable mirror images (enantiomers).

Total number of Stereoisomers for MABCDEF = $15 \text{ Geometrical} \times 2 \text{ (Enantiomers)} =$ 30 Stereoisomers

Test Your Knowledge

Practice MCQs on MABCDEF Complexes

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