Question:

From the magnetic behaviour of \([NiCl_4]^2-\) (paramagnetic) and [Ni\((CO)_4\)] (diamagnetic), choose the correct geometry and oxidation state.

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Remember the spectrochemical series and how to determine the hybridization and geometry of coordination complexes. Also, paramagnetic complexes have unpaired electrons, while diamagnetic complexes do not.
Updated On: Mar 17, 2025
  • \([NiCl_4]^2-\) : Ni²⁺, square planar [Ni\((CO)_4\)] : Ni(0), square planar 
     

  • \([NiCl_4]^2-\) : Ni²⁺, tetrahedral [Ni\((CO)_4\)] : Ni(0), tetrahedral 
     

  • \([NiCl_4]^2-\): Ni²⁺, tetrahedral [Ni\((CO)_4\)] : Ni^2+, square planar 
     

  • \([NiCl_4]^2-\) : Ni(0), tetrahedral [Ni\((CO)_4\)] : Ni(0), square planar 
     

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The Correct Option is B

Solution and Explanation

\([NiCl_4]^2-\)
Ni²⁺ - [Ar] 3d⁸ 4s⁰ → sp³, Tetrahedral
Number of unpaired electron = 2 paramagnetic
[Ni\((CO)_4\)]
Ni(0) → [Ar] 3d¹⁰ 4s⁰ (After rearrangement)
No unpaired electron
sp^3³, Tetrahedral, Diamagnetic
 

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