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Find Geometrical Isomers for Square Planar Complex [MABCD] | CHEMCA

Find Geometrical Isomers for Square Planar Complex [MABCD] | CHEMCA

How to Find Geometrical Isomers for Square Planar Complex [MABCD]

Published by Abhishek Sengar | CHEMCA India

In the Coordination Compounds chapter, identifying the exact number of Geometrical Isomers is a guaranteed question pattern in JEE and NEET exams. The challenge isn't just drawing them; the real challenge is making sure you don't accidentally draw duplicates!

Let's look at a complex with the general formula [MABCD], where the central metal ion (M) is surrounded by four completely different ligands (A, B, C, and D) in a square planar geometry.

Video Tutorial: The "Fix and Rotate" Trick

Watch Abhishek Sengar sir from CHEMCA demonstrate a foolproof method to draw every possible isomer without repeating a single structure.

Step-by-Step Method

The Core Strategy: To prevent drawing duplicates, lock one ligand in place (e.g., Ligand A) at the top-left position. Then, simply rotate the remaining three ligands (B, C, and D) clockwise for each new structure.

The easiest way to prove two isomers are different is to look at the trans (opposite) pairs. If the trans pairs are different, the isomer is unique!

  1. Isomer 1 (The Starting Position):
    Draw M in the center. Place A at top-left, B at top-right, C at bottom-right, and D at bottom-left.
    Check the opposite pairs: (A is opposite C) and (B is opposite D).
  2. Isomer 2 (First Rotation):
    Keep A exactly where it is. Move D to B's position, move B to C's position, and move C to D's position.
    Check the opposite pairs: Now, (A is opposite B) and (C is opposite D). Since the opposite pairs are new, this is a unique isomer!
  3. Isomer 3 (Second Rotation):
    Keep A fixed again. Rotate the others clockwise one more time. C moves to the top-right, D to the bottom-right, and B to the bottom-left.
    Check the opposite pairs: Now, (A is opposite D) and (C is opposite B). This is another unique isomer!
M A B C D Isomer 1 Opposites: (A-C), (B-D) Rotate M A D B C Isomer 2 Opposites: (A-B), (D-C) Rotate M A C D B Isomer 3 Opposites: (A-D), (C-B)

If you perform the rotation a third time, B moves back to the top-right, taking us exactly back to Isomer 1.

Conclusion: A square planar complex of type [MABCD] yields exactly 3 Geometrical Isomers.

Practice Questions for JEE & NEET

Test your knowledge of Geometrical Isomerism in coordination complexes with these targeted questions.

Question 1: Consider the square planar complex [Pt(NH3)2Cl2], which is an [MA2B2] type complex. How many geometrical isomers does it exhibit?

Answer: Exactly 2 Isomers (Cis and Trans).

Reasoning:

  • Cis-isomer: The two identical ligands (e.g., the two Chlorines) are placed at adjacent positions (90° apart). Fun fact: The cis-isomer is a famous anti-cancer drug known as Cisplatin!
  • Trans-isomer: The two identical ligands are placed at opposite positions across the metal center (180° apart).

Question 2: Why do complexes with a Tetrahedral geometry (coordination number 4) completely fail to show geometrical isomerism, regardless of the ligands attached?

Answer: Because all positions are perfectly equivalent.

In a tetrahedral geometry, the bond angle between any two ligands is exactly 109.5°. Because every position is adjacent to every other position, there is no structural difference between "cis" (next to) and "trans" (across). No matter how you arrange the ligands, the relative distance between them remains exactly the same, making geometrical isomerism impossible.

Ready for more Tricks?

Stop memorizing and start understanding! Visit www.chemca.in today to access Abhishek Sir's complete coordination chemistry crash course and practice exams.

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