What is the hybridization of [Cu(NH3)4]2+?
The correct answer and explanation is :
The hybridization of the central copper ion in the complex $[Cu(NH_3)_4]^{2+}$ is d^2sp^3.
Explanation:
The complex $[Cu(NH_3)_4]^{2+}$ involves a copper ion (Cu²⁺) coordinated to four ammonia (NH₃) ligands. To determine the hybridization of the central metal ion (Cu²⁺), we need to consider its electronic configuration, the nature of the ligands, and the coordination number of the complex.
- Oxidation State of Copper: In $[Cu(NH_3)_4]^{2+}$, the copper ion is in the +2 oxidation state. The electronic configuration of copper (Cu) in its ground state is [Ar] 3d¹⁰ 4s¹. Upon losing two electrons to form Cu²⁺, the electron configuration becomes [Ar] 3d⁹. This means that Cu²⁺ has a partially filled d-orbital, which is available for bonding with ligands.
- Ligands and Coordination Number: Ammonia (NH₃) is a neutral ligand that donates a lone pair of electrons to the metal ion. In this complex, there are four NH₃ ligands, which means the coordination number of copper is 4. A coordination number of 4 generally results in a square planar or tetrahedral geometry, depending on the metal’s electronic structure.
- Geometry and Hybridization: Copper(II) typically prefers a square planar geometry when coordinated by four ligands, especially for d⁸ metal ions, such as Cu²⁺. This is because the d-orbitals of Cu²⁺ will interact with the ligands in such a way that the hybridization leads to a square planar arrangement. In the square planar geometry, the copper ion undergoes d^2sp^3 hybridization, where two d-orbitals, one s-orbital, and one p-orbital mix to form six hybrid orbitals. Four of these hybrid orbitals are used to form bonds with the four NH₃ ligands, while the remaining two hybrid orbitals can accommodate lone pairs if necessary.
Therefore, the hybridization of Cu²⁺ in $[Cu(NH_3)_4]^{2+}$ is d^2sp^3, leading to a square planar geometry.