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Coordination Compounds
Valence Bond Theory: Hybridisation
1 question from 1 paper, each with its answer.
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Concepts tested
- Hybridisation in nickel complexes
All Questions
2004 AIPMT 1 question
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Among $[Ni(CO)_4]$, $[Ni(CN)_4]^{2-}$, $[NiCl_4]^{2-}$ species, the hybridization states at the Ni atom are, respectively :- (At. No. of Ni = 28)Ni (Z = 28): $[Ar]\,3d^8\,4s^2$
$[Ni(CO)_4]$: Ni is in the 0 state. CO is a strong field ligand, so the two 4s electrons are paired into 3d, giving $3d^{10}$. The empty 4s and three 4p orbitals give $sp^3$ hybridisation; tetrahedral and diamagnetic.
$[Ni(CN)_4]^{2-}$: $Ni^{2+} = 3d^8$. $CN^-$ is a strong field ligand, so the eight d electrons pair up into four d orbitals, leaving one 3d orbital empty. One 3d, one 4s and two 4p orbitals give $dsp^2$ hybridisation; square planar and diamagnetic.
$[NiCl_4]^{2-}$: $Ni^{2+} = 3d^8$. $Cl^-$ is a weak field ligand, so no pairing occurs and two electrons stay unpaired. The 4s and three 4p orbitals give $sp^3$ hybridisation; tetrahedral and paramagnetic.
So the hybridisation states are $sp^3$, $dsp^2$ and $sp^3$ respectively.