Given below are two statements:
Statement I: [Ni(CN)4]2– is square planar and diamagnetic complex, with dsp2 hybridization for Ni but [Ni(CO)4] is tetrahedral, paramagnetic and with sp3-hybridization for Ni.
Statement II: [NiCl4]2– and [Ni(CO)4] both have same d-electron configuration, have same geometry and are paramagnetic.
In light of the above statements, choose the correct answer from the options given below :
To determine the accuracy of the given statements, let's analyze each statement based on the properties of the nickel complexes involved.
The complex [\text{Ni}(\text{CN})_4]^{2-} is known to be square planar and diamagnetic. This arrangement occurs because the cyanide ion (\text{CN}^-) is a strong field ligand, causing pairing of electrons in the nickel ion (Ni2+), and it uses \text{dsp}^2 hybridization.
For [\text{Ni}(\text{CO})_4], the nickel is in a zero oxidation state (Ni0). Carbon monoxide (\text{CO}) is a neutral and strong field ligand but does not lead to electron pairing here. This complex is tetrahedral and diamagnetic, with \text{sp}^3 hybridization. Thus, the part about [Ni(CO)4] being paramagnetic is incorrect, making this statement false.
The complex [NiCl4]2– has Ni2+, with a configuration of 3d^8. Chloride (\text{Cl}^-) is a weak field ligand, so it does not cause electron pairing in the d-orbitals, and the complex is tetrahedral and paramagnetic, with unpaired electrons.
On the other hand, as discussed above, [Ni(CO)4] with Ni0 is tetrahedral and diamagnetic. Since they do not have the same electron configuration or magnetic properties, the statement is false.
Both Statement I and Statement II have been found to be false based on the discussed coordination chemistry of the nickel complexes and their respective properties.
Hybridisation and geometry of [Ni(CN)$_4$]$^{2-}$ are