Problem 23 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In Wilkinson's catalyst, the hybridisation of central metal ion and its shape are respectively: (a) \(s p^{3} d\), trigonal bipyramidal (b) \(d^{2} s p^{3}\), octahedral (c) \(d s p^{2}\), square planar (d) \(s p^{3}\), tetrahedral

Step-by-Step Solution

Answer

Wilkinson's catalyst has a hybridization of ds{p^2} and a square planar geometry.

Step 1: Understanding the Problem

Wilkinson's catalyst is ext{RhCl(PPh}_3 ext{)}_3. We need to find the hybridization of the central metal ion Rhodium (Rh) and deduce the geometry of the compound.

Step 2: Identify the Ligands and Oxidation State

In ext{RhCl(PPh}_3 ext{)}_3, the Rhodium is bonded to three phosphine ligands (PPh) and one chloride ion. Chloride is a monodentate ligand and typically has a charge of -1. Phosphines are neutral ligands. Thus, the oxidation state of Rhodium (Rh) remains +1. Therefore, Rhodium exists as Rh(I).

Step 3: Determine the Hybridization

Rhodium is in the +1 oxidation state and has a low-spin electronic configuration. Rh(I) typically leads to ds{p^2} hybridization when forming square planar complexes, as the metal's d orbitals are involved.

Step 4: Determine the Geometry

The hybridization ext{ds}{p^2} indicates a square planar geometry. Thus, Wilkinson's catalyst in its given coordination forms a square planar structure instead of any alternative spatial arrangement.

Key Concepts

Hybridization

Hybridization is a concept that helps us understand how atomic orbitals mix to form new hybrid orbitals. This is critical in predicting the geometry and bonding properties of molecules. In the context of Wilkinson's catalyst, we focus on the central metal, Rhodium (Rh). Rhodium, in this catalyst, exists in a +1 oxidation state. In this specific catalyst, Rhodium undergoes hybridization to accommodate its ligands, which are the atoms or molecules bonded to the central metal. Since Rh exists as Rh(I) with one unpaired d-electron, it favors a dsp\(^2\) hybridization.

• "d" orbitals are involved, especially because we're dealing with a transition metal like Rhodium.
• "s" and "p" orbitals contribute to the formation of four hybrid orbitals that can readily bind with ligands.

In essence, the dsp\(^2\) hybridization allows for the formation of a stable and specific bonding environment, leading to the square planar geometry discussed in the next section.

Square Planar Geometry

Square planar geometry is a molecular shape that results when there are four ligands symmetrically arranged around a central atom. This arrangement forms a plane, resembling a square when viewed from above, hence the name. In the case of Wilkinson's catalyst, the dsp\(^2\) hybridization of Rhodium dictates this square planar geometry.

• The four ligands are positioned at the corners of a square.
• The central Rhodium atom sits at the center of this square plane.

This configuration leads to the minimization of electron repulsion, making it energetically favorable for transition metals with a +1 oxidation state, like Rhodium in Wilkinson's catalyst. This structure is common with low-spin complexes, where the energy conditions favor pairing of d-electrons in lower-energy orbitals.

Rhodium Oxidation State

The oxidation state of an element describes the degree of oxidation of an atom within a compound; essentially, it provides a gauge for the distribution of electrons around the atom. In Wilkinson's catalyst, the Rhodium atom is in the +1 oxidation state.

• Rhodium is bonded to three triphenylphosphine (PPh\(_3\)) ligands, which are neutral and do not affect the oxidation state.
• It is also bonded to a chloride ion, which carries a -1 charge.

By balancing the charges, we determine that Rhodium maintains its oxidation state at +1. This oxidation state is critical in determining the hybridization and geometry of the complex, as it ensures that Rhodium remains in a low-spin state, favoring the dsp\(^2\) hybridization and the resulting square planar geometry.