What is the hybridization for [Ni(CN)4] 2-?
The correct answer and explanation is :
The hybridization of the central metal atom in the complex $[Ni(CN)_4]^{2-}$ is sp³.
Explanation:
- Oxidation State of Nickel:
In the complex $[Ni(CN)_4]^{2-}$, the cyanide ions (CN⁻) each have a charge of -1. Since there are four CN⁻ ions, the total negative charge contributed by them is -4. The overall charge on the complex is -2, so to balance the charge, the oxidation state of nickel must be +2. Thus, the central metal ion is $Ni^{2+}$. - Coordination Number:
In the complex $[Ni(CN)_4]^{2-}$, nickel is surrounded by four cyanide ligands. This indicates that the coordination number of nickel is 4, which is typical for a tetrahedral geometry. - Electron Configuration:
The electron configuration of $Ni^{2+}$ is [Ar] 3d⁸, as the 4s electrons are lost first when forming the $Ni^{2+}$ ion. In the presence of the four cyanide ligands, these electrons are involved in bonding. - Type of Ligands and Bonding:
Cyanide (CN⁻) is a strong field ligand and can cause nickel to undergo a low-spin configuration. The cyanide ions form sigma bonds with the nickel atom by donating electron density through their lone pair of electrons. This bonding arrangement typically leads to a hybridization of the nickel’s d orbitals. - Hybridization:
The $Ni^{2+}$ ion in this complex undergoes sp³ hybridization to form four equivalent hybrid orbitals. These orbitals overlap with the lone pairs from the cyanide ligands to form sigma bonds, resulting in a tetrahedral geometry. In this case, the hybrid orbitals are made by mixing one s orbital and three p orbitals from the nickel atom. - Geometry:
The complex adopts a tetrahedral geometry, as predicted by the sp³ hybridization, which is a common geometry for complexes with a coordination number of 4.
Thus, the hybridization of the central nickel ion in $[Ni(CN)_4]^{2-}$ is sp³ with a tetrahedral geometry.