What is the hybridization of [Ni(CN)6]4-?
The correct answer and explanation is :
The hybridization of the [Ni(CN)₆]⁴⁻ complex is d²sp³.
Explanation:
To determine the hybridization of the [Ni(CN)₆]⁴⁻ complex, we need to consider the following key factors: the oxidation state of nickel, the electron configuration, the coordination number, and the geometry of the complex.
- Oxidation State of Nickel:
The cyanide (CN⁻) ligand is a monodentate ligand and carries a -1 charge. Since the overall charge of the complex is -4, the nickel must have an oxidation state of +2. This means Ni²⁺ is the central metal ion. - Electron Configuration of Ni²⁺:
Nickel (Ni) has an atomic number of 28, so the electron configuration of neutral nickel is [Ar] 3d⁸ 4s². When it loses two electrons to become Ni²⁺, the configuration becomes [Ar] 3d⁸. In the case of Ni²⁺, the 4s orbitals are empty, and the 3d orbitals are partially filled with eight electrons. - Geometry of the Complex:
The cyanide ligands are linear and strongly bound to the metal. The six cyanide ions form an octahedral arrangement around the central nickel ion. This arrangement results in a coordination number of 6. - Hybridization:
In an octahedral geometry, six atomic orbitals participate in bonding with the ligands. These orbitals come from the d, s, and p orbitals of the metal atom. The specific hybridization for a metal ion in this configuration is d²sp³, where:
- Two electrons from the d orbitals (from the 3d orbitals) mix with one s orbital (from the 4s orbital) and three p orbitals (from the 3p orbitals) to form six hybrid orbitals that are directed toward the ligands in an octahedral arrangement.
Thus, the hybridization of the [Ni(CN)₆]⁴⁻ complex is d²sp³, which is consistent with the octahedral geometry and coordination number of 6.