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NEET CHEMISTRYCoordination CompoundsMedium

Question

The hybridization involved in the complex [Ni(CN)4]2−[Ni(CN)_4]^{2-} is: (Atomic number of Ni = 28)

A

dsp²

B

sp³

C

d²sp²

D

d²sp³

Step-by-Step Solution

  1. Oxidation State: In the complex ion [Ni(CN)4]2−[Ni(CN)_4]^{2-}, the cyano group (CN−CN^-) is a monodentate negative ligand. Let xx be the oxidation state of Nickel. x+4(−1)=−2⇒x=+2x + 4(-1) = -2 \Rightarrow x = +2. Thus, Nickel is in the +2+2 oxidation state (Ni2+Ni^{2+}).
  2. Electronic Configuration: The ground state configuration of Ni (Z=28) is [Ar]3d84s2[Ar]3d^8 4s^2. For Ni2+Ni^{2+}, the configuration is [Ar]3d8[Ar]3d^8 (electrons are removed from 4s4s first).
  3. Ligand Field Effect: The cyanide ion (CN−CN^-) is a strong field ligand according to the spectrochemical series. It causes the pairing of the two unpaired electrons present in the 3d3d orbitals against Hund's rule.
  4. Hybridisation: After pairing, one 3d3d orbital becomes empty. This empty inner 3d3d orbital mixes with the empty 4s4s and two empty 4p4p orbitals to form four equivalent dsp2dsp^2 hybrid orbitals.
  5. Geometry: The dsp2dsp^2 hybridisation corresponds to a square planar geometry. The complex is diamagnetic (no unpaired electrons) .

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Coordination Compounds. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYCoordination Compoundshybridizationinvolvedcomplexatomicnumber

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