Question

Hybridizations of Ni(CO)_4 and \([{Ni(CN)}_4]^{3-}\) are respectively.} 

Hint:

In coordination chemistry, strong field ligands like CN\( ^- \) favor low-spin pairing and square planar hybridization (\( dsp^2 \)), while weak field ligands like CO favor tetrahedral hybridization (\( sp^3 \)).

Answer 1

Step 1: Understanding Hybridization 
Hybridization in coordination complexes is determined based on valence bond theory (VBT) and ligand field theory (LFT). The electronic configuration of Nickel (Ni, Z = 28) is: \[ [\text{Ar}] 3d^8 4s^2 \] 
Step 2: Hybridization of \([ \text{Ni(CO)}_4 ]\) 
- CO is a strong field ligand and causes low-spin or square planar geometry.
- Ni is in zero oxidation state (0).
- The valence shell configuration of Ni⁰ is: \[ 3d^8 4s^2 \] - CO ligand donates electron pairs, leading to sp³ hybridization and tetrahedral geometry. 
Step 3: Hybridization of \([ \text{Ni(CN)}_4 ]^{3-}\) 
- CN^- is a strong field ligand, which causes low-spin pairing. 
- Ni is in ⁺² oxidation state, so the valence shell is: \[ 3d^8 \] - The CN^- ligands cause electron pairing in the d-orbitals, leading to dsp² hybridization and square planar geometry. 
Step 4: Conclusion 
- \([ \text{Ni(CO)}_4 ]\) has sp³ hybridization (tetrahedral). 
- \([ \text{Ni(CN)}_4 ]^{3-}\) has dsp² hybridization (square planar).