Problem 115 - 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 number of unpaired electrons present in \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are, respectively (a) 0,0 (b) 0,4 (c) 1,2 (d) 2,4

Step-by-Step Solution

Answer

Option (b) 0,4. [Fe(CN)6]4-: 0 unpaired electrons; [Fe(H2O)6]2+: 4 unpaired electrons.

Step 1: Determine the oxidation states

To solve the problem, we need to first determine the oxidation state of Fe in both complexes. In \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\), the oxidation state of Fe is calculated as: \[ \text{Fe} + 6 \times (-1) = -4 \Rightarrow \text{Fe} = +2 \] In \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\), the oxidation state of Fe is already \(+2\) as the complex charge is \(+2\) and water is a neutral ligand.

Step 2: Electron configuration of Fe

The electron configuration of a neutral Fe atom is \([\mathrm{Ar}] 3d^6 4s^2\). When Fe is in the +2 oxidation state, it loses two electrons: \([\mathrm{Ar}] 3d^6\).

Step 3: Configuration with strong ligand (CN-)

\(\mathrm{CN}^-\) is a strong field ligand and forms low spin complexes by pairing electrons in the 3d orbitals. This results in electron configuration: \(t_{2g}^{6}e_{g}^{0}\), leading to \(\text{no unpaired electrons}\).

Step 4: Configuration with weak ligand (H2O)

\(\mathrm{H}_2\mathrm{O}\) is a weak field ligand, resulting in high spin complexes. In this case, the electron configuration remains \(t_{2g}^{4}e_{g}^{2}\), leading to \(\text{4 unpaired electrons}\).

Step 5: Count unpaired electrons and choose the correct option

From the calculations, \([\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}\) has \(0\) unpaired electrons, and \([\mathrm{Fe}(\mathrm{H}_{2} \mathrm{O})_{6}]^{2+}\) has \(4\) unpaired electrons. Therefore, the correct answer is (b) 0,4.

Key Concepts

d-orbital hybridization

In coordination chemistry, understanding which d-orbitals participate in hybridization is crucial to predict the behavior of a complex. Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals. For the central metal atom, this can involve various d-orbitals, depending on whether the orbitals are in the outermost shell or the penultimate shell. The configuration of these d-orbitals depends largely on the type of ligands attached to the metal center. A powerful or strong field ligand tends to force the use of inner or lower d-orbitals (also known as inner orbital complexes), while a weak field ligand uses the outer or higher d-orbitals (outer orbital complexes). This shift affects whether the resulting complex is low spin or high spin, impacting other properties like magnetism and stability.

strong and weak field ligands

Ligands play a pivotal role in determining the electronic structure of metal complexes. They can be categorized as strong or weak field ligands based on their ability to split d-orbital energies. 8Creating low spin complexes, strong field ligands like \(\mathrm{CN}^-\) cause greater splitting of the d-orbitals. This high degree of splitting causes electrons to pair up in the lower energy orbitals, reducing the number of unpaired electrons and often leading to a diamagnetic nature.8On the other hand, weak field ligands such as \(\mathrm{H}_2\mathrm{O}\) result in less splitting. Electrons remain unpaired and occupy higher energy orbitals, resulting in high spin complexes that are usually paramagnetic.AStrong and weak field ligands are crucial in shaping the electronic properties of coordination compounds. This classification helps predict a complex's properties, such as color, magnetism, and reactivity.

unpaired electrons

The presence or absence of unpaired electrons in a complex gives insight into its potential magnetic properties. In low spin complexes, strong field ligands cause electron pairing, leading to fewer or no unpaired electrons. An example of this is the \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\) complex, which exhibits diamagnetism due to zero unpaired electrons.In contrast, high spin complexes formed with weak field ligands have more unpaired electrons thanks to reduced d-orbital splitting. The \([\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}\) complex is an example where the presence of four unpaired electrons results in paramagnetism.Understanding the electron configuration allows chemists to predict the magnetic behavior of compounds, which can be seen as a valuable tool in various applications, such as material science and catalysis.

oxidation state determination

The oxidation state of a metal in a coordination complex is key to understanding its electronic configuration and overall reactivity. Determining this state involves considering the overall charge of the complex and the charges of the ligands. For example, in the complex \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\), the oxidation state of iron is \(+2\) because the cyanide ligand carries a charge of \(-1\) for each of its six ligands, totaling \(-6\), which when subtracted from the overall charge gives \(+2\).\11In \([\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}\), the oxidation state is straightforwardly \(+2\) since the ligands are neutral water molecules, and the charge of the complex reflects the oxidation state directly.\111The oxidation state affects the electron count and configuration of the central metal atom, impacting its chemical characteristics and the nature of the complex it's involved in. Understanding how to determine the oxidation state allows chemists to predict and manipulate the behavior of metallic compounds.