Indicate the complex ion which does not show geometrical isomerism:

The correct answer is Option (3) → \([Pt(NH_3)_6]^{3+}\)

Geometrical (cis–trans) isomerism needs at least two different kinds of ligands, which can be placed next to each other (cis) or opposite each other (trans). It is shown by square planar MA₂B₂ and by octahedral MA₄B₂ complexes, among others.

• Option (1) \([Cr(H_2O)_4Cl_2]^+\) — Shows geometrical isomerism. An octahedral MA₄B₂ complex; the two \( \ce{Cl^-} \) can be cis or trans.

• Option (2) \([Pt(NH_3)_2Cl_2]\) — Shows geometrical isomerism. A square planar MA₂B₂ complex; it exists as cis- and trans-isomers.

• Option (3) \([Pt(NH_3)_6]^{3+}\) — Does not show geometrical isomerism. All six ligands are the same (type MA₆), so every arrangement is identical and no cis or trans form is possible.

• Option (4) \([Co(CN)_4(NC)_2]^{3–}\) — Shows geometrical isomerism. Four C-bound \( \ce{CN^-} \) and two N-bound \( \ce{NC^-} \) make it an octahedral MA₄B₂ type, so the two \( \ce{NC^-} \) can be cis or trans.

NCERT (2026-27 edition):
Geometric Isomerism – “This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour are found with coordination numbers 4 and 6. 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.”