JEE Main 2020 · 2 September · evening · Single correct · +4 / −1
The one that is not expected to show isomerism is :
- (a) [Pt(NH₃)₂Cl₂]
- (b) [Ni(NH₃)₄(H₂O)₂]²⁺
- (c) [Ni(en)₃]²⁺
- (d) [Ni(NH₃)₂Cl₂]
Answer and solution
Answer (d) [Ni(NH₃)₂Cl₂]
Solution
Chemistry ma’am · 4 steps
Decide the shape first: a tetrahedron has every corner next to every other, so no cis or trans.
- Shape decides isomerism. Geometrical isomers need two different relative positions for a pair of ligands, either next to each other (cis) or opposite (trans). Square planar and octahedral shapes allow both. A tetrahedron does not.
- Pt versus Ni. Pt²⁺ is a 5d⁸ ion and is always square planar. Ni²⁺ is 3d⁸, and with weak-field Cl⁻ ligands its electrons do not pair, so it uses sp³ hybrid orbitals and becomes tetrahedral.
Complex Shape Isomerism [Pt(NH₃)₂Cl₂] square planar cis and trans [Ni(NH₃)₄(H₂O)₂]²⁺ octahedral cis and trans (two H₂O adjacent or opposite) [Ni(en)₃]²⁺ octahedral optical (non-superimposable mirror images) [Ni(NH₃)₂Cl₂] tetrahedral none - Why none. In a tetrahedron the two Cl⁻ ligands are always next to each other, so only one arrangement exists. The molecule also has a mirror plane, so it cannot be optically active.
- So. The complex not expected to show isomerism is [Ni(NH₃)₂Cl₂], option (d).
The idea to keep. Before hunting for isomers, fix the geometry. A tetrahedral MA₂B₂ complex shows neither geometrical nor optical isomerism. Square planar MA₂B₂ and octahedral MA₄B₂ both show cis and trans.
Answer from the NTA final answer key. Solution by PrepWiser.