Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

The d-orbitals participating in hybridization of central metal atom may be from the outermost shell or the penultimate shell. This depends on the nature of metal and the nature of ligand. The complexes involving the inner d level (inner orbital complexes) result when the ligand is a powerful or strong ligand resulting in diamagnetic or low spin complexes. A weak ligand usually results in the formation of outer orbital complex or high spin complex. The hybridization of \(\mathrm{Cu}\) in \(\left[\mathrm{CuCl}_{4}\right]^{2-}\) and \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) are respectively. (a) \(\mathrm{sp}^{3}, \mathrm{dsp}^{2}\) (b) \(\mathrm{dsp}^{2}, \mathrm{dsp}^{2}\) (c) \(\mathrm{dsp}^{2}, \mathrm{sp}^{3}\) (d) \(s p^{3}, s p^{3}\)

Step-by-Step Solution

Answer

(a) \(sp^3, dsp^2\)

Step 1: Determine the nature of ligands

First, assess the strength of the ligands. In the complex \([\mathrm{CuCl}_{4}]^{2-}\), the ligand is chloride (Cl), which is a weak ligand. In \([\mathrm{Cu(NH}_3)_4]^{2+}\), the ligand is ammonia (NH3), which is a strong ligand.

Step 2: Analyze ligand influence on d-orbitals

Since chloride is a weak ligand, it tends to form high-spin complexes, which typically use the outer d-orbitals (4d or 5d) depending on the metal. For \([\mathrm{CuCl}_{4}]^{2-}\), the hybridization involves 4p and the outer 4s and 3d, leading to \(sp^3\) hybridization. Ammonia, being a strong ligand, forms a low-spin complex using the inner d-orbitals (3d), leading to \(dsp^2\) hybridization in \([\mathrm{Cu(NH}_3)_4]^{2+}\).

Step 3: Determine the hybridization

For \([\mathrm{CuCl}_{4}]^{2-}\), the hybridization is \(sp^3\) due to weak ligand chloride. For \([\mathrm{Cu(NH}_3)_4]^{2+}\), the hybridization is \(dsp^2\) due to strong ligand ammonia.

Key Concepts

Hybridization

In coordination chemistry, hybridization is crucial in determining the geometric arrangement of orbitals in a complex. When forming a coordination complex, the central metal atom reorganizes its orbitals to accommodate the bonds made with ligands.
For instance, in the complex \([\mathrm{CuCl}_{4}]^{2-}\), the hybridization involving copper entails the mixing of 4s, 4p, and outer 3d orbitals. Since \([\mathrm{CuCl}_{4}]^{2-}\) features chloride, a weak ligand, the hybridization adopts an \(sp^3\) configuration. This configuration points towards a tetrahedral geometry.
On the other hand, for the \([\mathrm{Cu(NH}_3)_4]^{2+}\) complex, the presence of strong ligands like ammonia results in \(dsp^2\) hybridization of the central copper atom. Here, the inner d-orbitals (3d) participate, aligning the complex with a square planar geometry. Understanding these arrangements is key to predicting the physical and magnetic properties of the complex.

Ligand Strength

Ligands can significantly impact the properties of coordination complexes, especially in terms of electronic configuration and spin states. Ligand field theory explains these effects, highlighting the variance in electron donation and back-donation between ligands and metal centers.
Weak vs Strong Ligands:

• Weak Ligands: Electrostatic interaction dominates, allowing unpaired electrons and generally forming high-spin complexes. Chloride (Cl) is an example, leading to high-spin configurations like \(sp^3\) hybridization in \([\mathrm{CuCl}_{4}]^{2-}\).
• Strong Ligands: They possess greater  ability to pair up electrons within the d-orbitals, causing low-spin configurations. Strong ligands such as ammonia (NH3) in \([\mathrm{Cu(NH}_3)_4]^{2+}\) actively engage in electron pairing, resulting in \(dsp^2\) hybridization.

Understanding ligand strength is vital for predicting complex configuration, and consequently, its reactivity and color.

High-spin and Low-spin Complexes

High-spin and low-spin terminology refers to the electron configuration of the d-orbitals in a coordination complex, determined by the field strength of its ligands.
High-spin Complex: - When the ligand field strength is weak, as in the case of chloride (Cl), there isn't enough electrostatic force to pair up the electrons in the lower energy d-orbitals. This outcome maintains unpaired electrons, leading the \([\mathrm{CuCl}_{4}]^{2-}\) complex to be high-spin with an \(sp^3\) hybridization.
Low-spin Complex: - Conversely, strong ligands like ammonia (NH3) exert a significant pull to pair up electrons in the lower energy d-orbitals, minimizing unpaired electrons. This results in a low-spin configuration with \(dsp^2\) hybridization, as seen in the \([\mathrm{Cu(NH}_3)_4]^{2+}\) complex.
In essence, the spin state impacts the magnetic properties: high-spin complexes are often paramagnetic, while low-spin complexes are diamagnetic. This understanding is crucial when considering applications in materials science and catalysis.