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JEE Main2018ChemistryCoordination CompoundsActual

The correct combination is

Options

  1. ANiCl 4 2 - - Square planar; Ni ( CN ) 4 2 - - paramagnetic
  2. BNi ( CN ) 4 2 - - tetrahedral; Ni ( CO ) 4 2 - - paramagnetic
  3. CNiCl 4 2 - - paramagnetic; Ni ( CO ) 4 - tetrahedral
  4. DNiCl 4 2 - - diamagnetic; Ni ( CO ) 4 - square-planar

Correct answer

C. NiCl 4 2 - - paramagnetic; Ni ( CO ) 4 - tetrahedral

Step-by-step solution

[ NiCl 4 ] − 2 Tetrahedral Paramagnetic [ Ni ( CO ) 4 ] Tetrahedral Diamagnetic Weak field ligand (WFL) Strong field ligand(SFL) NiCl 4 2 - Ni CO 4 Paramagnetic ( 2 unpaired electrons) Sp 3 Ni 2 + → Ar   3 d 8 ,   4 s 0 ,   4 p 0 Cl - (WFL) (No pairing) Ni O → Ar 3 d 8 ,   4 s 2 ,   4 p 0 CO is SFL Ar 3 d 10 ,   4 s 0 ,   4 p 0 (tetrahedral)

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