To determine the hybridization of nickel in , we need to find the oxidation state of nickel, the number of ligands, the geometry of the complex, and the electronic configuration.
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Step 1: Oxidation state of Ni
The complex is . Cyanide ( ) has a charge of . With four ligands, their total charge is . The overall charge of the complex is . Let the oxidation state of Ni be : So, Ni is in the oxidation state ( ). -
Step 2: Electronic configuration of
Nickel’s atomic number is 28, with a ground state configuration of . For , the two electrons are removed from the orbital: This gives eight electrons in the orbitals. -
Step 3: Ligand and coordination number
The complex has four ligands, so the coordination number is 4. is a strong-field ligand due to its position in the spectrochemical series, which causes significant splitting of the orbitals.
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Step 4: Geometry and hybridization
For coordination number 4, possible geometries are tetrahedral ( ) or square planar ( ). To determine the geometry, we consider the magnetic properties and ligand strength: - is known to be diamagnetic (no unpaired electrons), as confirmed by experimental data. - For ( ), in a tetrahedral field (weak-field, ), the electrons occupy the and orbitals. For , the configuration is typically , which has two unpaired electrons, making it paramagnetic. Since the complex is diamagnetic, tetrahedral geometry ( ) is ruled out.
- In a square planar field (strong-field, ), the orbitals split into , , , and . For a strong-field ligand like , the electrons fill the lower-energy orbitals (e.g., , , , ) completely, leaving no unpaired electrons, consistent with diamagnetism. The square planar geometry uses one orbital (usually ), one , and two orbitals, giving hybridization.
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Step 5: Confirm hybridization
The options are:
- : Tetrahedral, paramagnetic for , incorrect.
- : Square planar, diamagnetic with strong-field ligands, correct.
- : Linear geometry, not possible for coordination number 4.
- : Not typical for coordination number 4; often for higher coordination numbers.
Thus, the hybridization is , corresponding to a square planar geometry.