JEE Advanced 2026 — Paper 1 — CHEM-III solution

JEE Advanced2026Paper 1CHEM-III

The total number of all possible isomers for the square planar complex with formula K[M(NCS)(NO $_{2}$ )(gly)] is _____ ( $M$ = metal ion and gly $= \overline{NH_{2} CH_{2}} COO^{-}$ ).

Your answer

Solution

Setup. The square-planar complex contains the bidentate glycinate ligand (donating through both N and O), monodentate NCS (which can be linkage isomers N-bonded NCS or S-bonded SCN), and monodentate NO $_{2}$ (linkage isomers N-bonded NO $_{2}$ or O-bonded ONO).

Step 1: Geometrical isomers from the bidentate gly. In a square planar complex, the bidentate gly occupies two cis positions. The remaining two cis positions are filled by the two monodentate ligands. Holding the gly fixed, the two monodentate ligands can be swapped between the two remaining positions, giving 2 geometrical arrangements (NCS adjacent to N of gly vs adjacent to O of gly).

Step 2: Linkage isomerism. The NCS group has 2 linkage forms (N-bonded vs S-bonded), and the NO $_{2}$ group has 2 linkage forms (N-bonded vs O-bonded). For each geometrical arrangement, these multiply to give $2 \times 2 = 4$ linkage variants.

Total isomers = $2 \times 4 = 8.00$ .

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