JEE Main · 2020mediumCORD-210

The number of isomers possible for [Pt(en)(NO2)2] is:

Coordination Compounds · Class 12 · JEE Main Previous Year Question

Question

The number of isomers possible for [Pt(en)(NO2)2][\mathrm{Pt(en)(NO_2)_2}] is:

Options
  1. a

    2

  2. b

    4

  3. c

    1

  4. d

    3

Correct Answerd

3

Detailed Solution

🧠 Pt(en)(NO₂)₂ — Geometry First

[Pt(en)(NO2)2][\mathrm{Pt(en)(NO_2)_2}]: Pt(II) is d⁸ → square planar. Ligands: 1 en (bidentate, occupies adjacent positions) + 2 NO₂ (ambidentate).

In square planar with bidentate en, the two NO₂ are forced cis to each other (the only available positions).

🗺️ Linkage Isomerism of NO₂

NO₂⁻ is ambidentate: bonds via N (nitro, NO2\mathrm{-NO_2}) or O (nitrito, ONO\mathrm{-ONO}).

For 2 NO₂ ligands, the linkage combinations are:

  1. Both N-bonded: NO2,NO2-\mathrm{NO_2}, -\mathrm{NO_2}
  2. Both O-bonded: ONO,ONO-\mathrm{ONO}, -\mathrm{ONO}
  3. Mixed: NO2,ONO-\mathrm{NO_2}, -\mathrm{ONO} (only one mixed combination — the two NO₂ positions are equivalent by the en C₂ axis)

3 isomers.

🗺️ Geometric Isomerism Check

In sq pl with en: the two NO₂ must be cis. No trans option. So no geometric isomers. Only linkage isomers.

Counting Linkage Isomers

For nn ambidentate ligands, naive count = 2n2^n, but symmetry-equivalent positions reduce this. With en's C2C_2 axis swapping the two NO₂ positions, NN ≡ NN, OO ≡ OO, NO ≡ ON → 3 distinct.

⚠️ Pt(II) Always Square Planar

Pt(II) d⁸ is rigorously sq pl regardless of ligand strength — strong CFSE preference. No tetrahedral Pt(II).

Answer: (4) 3\boxed{\text{Answer: (4) 3}}

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