Which one of the following compounds does not show geometrical isomer…

The correct answer is Option (1) → MnO₄⁻

Geometrical isomerism appears in heteroleptic complexes, where two ligands (or two donor atoms) of the same kind can sit either next to each other (cis) or opposite each other (trans). It is common in:

• square planar complexes of the type MX₂L₂,

• octahedral complexes of the types MX₂L₄ and M(AA)₂X₂ (AA = a didentate ligand such as en or ox).

It is not possible in a tetrahedral species, because every position in a tetrahedron is adjacent to every other one, so no “opposite” arrangement exists. It is also impossible when all the groups around the central atom are the same.

• Option (1) MnO₄⁻ — Does not show geometrical isomerism. \( \ce{MnO4^{-}} \) is a tetrahedral ion with four identical oxygen atoms around manganese, so no cis or trans arrangement is possible.

• Option (2) [Fe(NH₃)₂(CN)₄]⁻ — Shows geometrical isomerism. It is an octahedral MX₂L₄ complex; the two \( \ce{NH3} \) ligands can be cis (90°) or trans (180°).

• Option (3) [CrCl₂(ox)₂]³⁻ — Shows geometrical isomerism. It is an octahedral M(AA)₂X₂ complex; the two \( \ce{Cl^{-}} \) ions can be cis or trans.

• Option (4) [CoCl₂(en)₂] — Shows geometrical isomerism. It is also of the M(AA)₂X₂ type, so it has cis and trans forms.

NCERT (2026-27 edition):
Geometric Isomerism – …