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Coordination Compounds question

2009 · Shift 1 · Q11
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Coordination Compounds question

2009 · Shift 1 · Q11

JEE AdvancedChemistryCoordination CompoundsMultiple correct+4 / −2
The compound(s) that exhibit(s) geometrical isomerism is(are)
  1. A
    [Pt(en)Cl2]\mathrm{\left[ {Pt(en)C{l_2}} \right]}[Pt(en)Cl2​]
  2. B
    [Pt(en)2]Cl2\mathrm{\left[ {Pt{{(en)}_2}} \right]C{l_2}}[Pt(en)2​]Cl2​
  3. C
    [Pt(en)2Cl2]Cl2\mathrm{\left[ {Pt{{(en)}_2}C{l_2}} \right]C{l_2}}[Pt(en)2​Cl2​]Cl2​
  4. D
    [Pt(NH3)2Cl2]\mathrm{\left[ {Pt{{(N{H_3})}_2}C{l_2}} \right]}[Pt(NH3​)2​Cl2​]
View written solutionFree

Correct answer: C, D

Introduction

Geometrical isomerism, also known as cis-trans isomerism, arises in coordination compounds when ligands occupy different positions around the central metal ion. We will analyze each given compound to determine its geometry and the possibility of different spatial arrangements for its ligands.

Step-by-step Analysis

1. Analyze option A: \[Pt(en)Cl_2\]

  • Central Metal Ion: Platinum (Pt). Since 'en' is neutral and each Cl is -1, and the complex is neutral, the oxidation state of Pt is +2.
  • Ligands: One ethylenediamine (en, a bidentate ligand) and two chloride (Cl⁻, monodentate ligands).
  • Coordination Number (CN): The bidentate en ligand donates two pairs of electrons, and the two Cl⁻ ligands each donate one pair. So, CN = 2 + 2 = 4.
  • Geometry: Pt(II) complexes with CN=4 are typically square planar.
  • Isomerism: The complex is of the type [M(AA)b₂], where AA is a symmetric bidentate ligand. In a square planar complex, the bidentate ligand en must occupy two adjacent coordination sites. This forces the two Cl⁻ ligands to also be in adjacent positions. There is no other possible arrangement. For example, the en ligand cannot span opposite (trans) positions. Therefore, only one isomer exists, and this compound does not exhibit geometrical isomerism.

2. Analyze option B: \[Pt(en)_2\]Cl_2

  • Complex Ion: The compound is ionic, with the complex ion being \[Pt(en)_2\]^{2+}.
  • Central Metal Ion: Platinum (Pt). Since en is neutral, the oxidation state of Pt is +2.
  • Ligands: Two ethylenediamine (en) ligands.
  • Coordination Number (CN): Each en is bidentate, so CN = 2 × 2 = 4.
  • Geometry: Pt(II) with CN=4 gives a square planar geometry.
  • Isomerism: The complex is of the type [M(AA)₂]. Since both bidentate ligands are identical, there is only one way to arrange them in a square planar geometry. Therefore, it does not exhibit geometrical isomerism.

3. Analyze option C: \[Pt(en)_2Cl_2\]Cl_2

  • Complex Ion: The complex ion is \[Pt(en)_2Cl_2\]^{2+}.
  • Central Metal Ion: Platinum (Pt). Let the oxidation state be x. x + 2(0) + 2(-1) = +2, which gives x = +4. So, this is a Pt(IV) complex.
  • Ligands: Two en ligands (bidentate) and two Cl⁻ ligands (monodentate).
  • Coordination Number (CN): CN = 2×2 + 2 = 6.
  • Geometry: For CN=6, the geometry is octahedral.
  • Isomerism: The complex is of the type [M(AA)₂b₂]. In an octahedral geometry, the two b ligands (Cl⁻) can be arranged in two different ways:
    • cis-isomer: The two Cl⁻ ligands are in adjacent positions (with a Cl-Pt-Cl angle of 90°).
    • trans-isomer: The two Cl⁻ ligands are in opposite positions (with a Cl-Pt-Cl angle of 180°). Since two distinct geometrical isomers (cis and trans) are possible, this compound exhibits geometrical isomerism.

4. Analyze option D: \[Pt(NH_3)_2Cl_2\]

  • Central Metal Ion: Platinum (Pt). Since NH₃ is neutral and Cl⁻ has a -1 charge, the oxidation state of Pt is +2.
  • Ligands: Two ammonia (NH₃) ligands and two chloride (Cl⁻) ligands (all are monodentate).
  • Coordination Number (CN): CN = 2 + 2 = 4.
  • Geometry: Pt(II) with CN=4 is square planar.
  • Isomerism: The complex is of the type [Ma₂b₂]. This is a classic case for geometrical isomerism in square planar complexes.
    • cis-isomer (Cisplatin): The two identical ligands (Cl⁻ or NH₃) are adjacent to each other.
    • trans-isomer (Transplatin): The two identical ligands are opposite to each other. Since both cis and trans isomers exist, this compound exhibits geometrical isomerism.

Conclusion

The compounds that exhibit geometrical isomerism are \[Pt(en)_2Cl_2\]Cl_2 (Option C) and \[Pt(NH_3)_2Cl_2\] (Option D).

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