Which of the following complexes show linkage isomerism?

The correct answer is Option (3) → \( [Co(NH_3)_5(NO_2)]^{2+} \)

Linkage isomerism arises when a complex contains an ambidentate ligand, one that has two different donor atoms but binds through only one of them at a time. The two isomers have the same formula but differ in which atom is bonded to the metal. Common ambidentate ligands:

• NO₂⁻: through N (nitrito-N) or through O (nitrito-O, –ONO)

• SCN⁻: through S (thiocyanato-S) or through N (thiocyanato-N)

• CN⁻: through C (cyanido) or through N (isocyanido)

\( \ce{[Co(NH3)5(NO2)]^{2+}} \) contains the ambidentate NO₂⁻ ligand, so it exists as two linkage isomers:

    \( \ce{[Co(NH3)5(NO2)]^{2+}} \): bonded through N
    \( \ce{[Co(NH3)5(ONO)]^{2+}} \): bonded through O

Why the other options are incorrect

• Option (1) \( [Co(H_2O)_5CO]^{3+} \) — Incorrect. Water has only one donor atom (O), and carbon monoxide binds a metal through its carbon atom, as in metal carbonyls. Neither ligand gives a linkage isomer here.

• Option (2) \( [Co(NH_3)_3Cl_3] \) — Incorrect. \( \ce{NH3} \) and \( \ce{Cl^{-}} \) each have a single donor atom. This octahedral \( \ce{[MA3B3]} \) complex shows geometrical (fac-mer) isomerism, not linkage isomerism.

• Option (4) \( [Co(H_2O)_6]^{3+} \) — Incorrect. All six ligands are water molecules, which bind only through oxygen, so no linkage isomer is possible.

NCERT (2026-27 edition):
Linkage Isomerism

Note: The same/similar question has been asked in different years by NTA as per the details given below: