Cis-trans isomerism is found in square planar complexes of which of the following formulae? (A and B are unidentate ligands)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → MA2B2
In a square planar complex the four ligands sit at the corners of a square.
In MA2B2 the two B can be adjacent to each other (cis) or opposite to each other (trans), so two geometrical isomers exist.
This is the square planar [MX2L2] type, with B in place of X and A in place of L.
Why the other options are incorrect
Option (1) MA4 — Incorrect. All four ligands are the same, so there is only one arrangement.
Option (2) MA3B — Incorrect. With a single B, any corner it takes has two A next to it and one A opposite, so every placement is the same.
Option (4) MAB3 — Incorrect. Now the single A is the odd ligand; for the same reason there is only one form.
NCERT (2026-27 edition):
5.4.1 Geometric Isomerism – “… In a square planar complex of formula [MX2L2] (X and L are unidentate), the two ligands X may be arranged adjacent to each other in a cis isomer, or opposite to each other in a trans isomer as depicted in Fig. 5.2.”
Fig. 5.2: Geometrical isomers (cis and trans) of [Pt(NH3)2Cl2]