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Question: Both \([Ni{{(CO)}_{4}}]\) and \({{[Ni{{(CN)}_{4}}]}^{2-}}\) are diamagnetic. The hybridization of ni...

Both [Ni(CO)4][Ni{{(CO)}_{4}}] and [Ni(CN)4]2−{{[Ni{{(CN)}_{4}}]}^{2-}} are diamagnetic. The hybridization of nickel in these complexes, respectively are:
(a)- sp3,sp3s{{p}^{3}},s{{p}^{3}}
(b)- sp3,dsp2s{{p}^{3}},ds{{p}^{2}}
(c)- dsp2,sp3ds{{p}^{2}},s{{p}^{3}}
(d)- dsp2,dsp2ds{{p}^{2}},ds{{p}^{2}}

Explanation

Solution

First find the oxidation state of the central metal atom. Then according to the oxidation state, remove the electrons. If the ligand is a strong field then unpaired electrons will pair up and the number of electrons will occupy the next orbitals. If the ligand is a weak field ligand then, pairing will not occur and ligands will occupy the next orbitals. Based on the orbitals occupied by the hybridization is calculated.

Complete step by step answer:
In [Ni(CO)4][Ni{{(CO)}_{4}}], the oxidation number of nickel is:
The oxidation number of COCO is zero because it is a neutral ligand,
x+4(0)=0x+4(0)=0
x=0x=0
So, the oxidation state of nickel in [Ni(CO)4][Ni{{(CO)}_{4}}] is zero, the electronic configuration of the atomic orbital of nickel and in zero oxidation is given in the diagram.
The COCOis a strong field ligand, so the unpaired electrons will pair up, this will fill the 3d orbital. Now the orbitals next to 3d are 4s and 4p. So, there are 4 COCO molecules in [Ni(CO)4][Ni{{(CO)}_{4}}], hence they will occupy the 1 s-orbital and 3 p-orbitals.

So, the hybridization of nickel in [Ni(CO)4][Ni{{(CO)}_{4}}] is sp3s{{p}^{3}}.

In, [Ni(CN)4]2−{{[Ni{{(CN)}_{4}}]}^{2-}}the oxidation number of nickel is:
The oxidation number of CN−C{{N}^{-}} is -1 because it is a negative ligand. This overall charge on the molecule is -2.
x+4(−1)=−2x+4(-1)=-2
x=+2x=+2
So, the oxidation state of nickel in[Ni(CN)4]2−{{[Ni{{(CN)}_{4}}]}^{2-}} is +2, the electronic configuration of atomic orbital of nickel and in +2 oxidation is given in the diagram.
The CN−C{{N}^{-}} is a strong field ligand, so the unpaired electrons will pair up, this will fill the 3d orbital. Now the orbitals next to 3d are 4s and 4p. So, there are 4 COCO molecules in [Ni(CO)4][Ni{{(CO)}_{4}}], hence they will occupy the 1 d-orbital, 1 s-orbital, and 2 p-orbitals.

So, the hybridization of nickel in [Ni(CN)4]2−{{[Ni{{(CN)}_{4}}]}^{2-}} is dsp2ds{{p}^{2}}.
So, the correct answer is “Option B”.

Note: The pairing of electrons is based on strong field ligand and weak field ligand. The examples of strong-field ligands are NO2−,CN−,CON{{O}_{2}}^{-}, C{{N}^{-}}, CO, etc and examples of weak field ligands are all halogens, water molecule, etc.