Problem 96 - A Complete Resource Book in Chemistry for JEE Main

Question

The number of unpaired electrons in the complex ion \(\left[\mathrm{CoF}_{6}\right]^{3}\) is (Atomic number of \(\mathrm{Co}=27\) ) (a) 4 (b) zero (c) 2 (d) 3

Step-by-Step Solution

Answer

The number of unpaired electrons is 4 (option a).

Step 1: Determine the Oxidation State

In the complex ion \(\left[\mathrm{CoF}_{6}\right]^{3-}\), let the oxidation state of \(\mathrm{Co}\) be \(x\). Since fluoride (\(F^-\)) has a charge of -1, we can set up the equation: \(x + 6(-1) = -3\). This simplifies to \(x = +3\). Thus, the oxidation state of \(\mathrm{Co}\) is +3.

Step 2: Identify the Electronic Configuration

The atomic number of \(\mathrm{Co}\) is 27, so the neutral Co atom has the electron configuration \([\mathrm{Ar}] 3d^7 4s^2\). With a +3 oxidation state, three electrons are removed, primarily from the 4s and 3d orbitals, resulting in \([\mathrm{Ar}] 3d^6\).

Step 3: Understand the Ligand Field

Fluoride ions \((F^-)\) are weak field ligands, which means they do not cause a significant splitting in the 3d orbitals. Due to this weak field effect, the 3d orbitals are expected to remain in a high spin state.

Step 4: Calculate Number of Unpaired Electrons

In a high spin state for \([\mathrm{CoF}_6]^{3-}\), the electron configuration remains as \(t_{2g}^4 e_g^2\). In this configuration, 4 electrons in the \(t_{2g}\) orbitals and 2 electrons in the \(e_g\) orbitals suggest the presence of unpaired electrons. By distributing the 6 d-electrons, we find 4 electrons remain unpaired.

Key Concepts

Oxidation State

To determine the oxidation state of an element in a complex ion, we need to consider the charges of all components. In the complex ion \([\mathrm{CoF}_6]^{3-}\), cobalt (Co) is the central metal atom, and fluoride (F\(^-\)) acts as the ligand. Each fluoride ion has a charge of -1. In the formula, there are 6 fluoride ions, contributing a total charge of -6. To solve for the oxidation state of Co, assigned as \(x\), we set up the equation: \[ x + 6(-1) = -3 \]This equation simplifies to \(x = +3\). Thus, the oxidation state of cobalt in this complex is +3, indicating it has lost three electrons compared to its elemental form. Understanding the oxidation state is crucial as it helps determine the electronic configuration after the ionization process.

Electron Configuration

The electron configuration describes the distribution of electrons in an atom or ion. Cobalt (Co) has an atomic number of 27, which means in its neutral state, it has 27 electrons with a configuration of \([\mathrm{Ar}] 3d^7 4s^2\). However, in the \([\mathrm{CoF}_6]^{3-}\) complex, Co is in a +3 oxidation state. Removing three electrons affects the 4s and 3d orbitals. Typically, electrons are removed from the 4s orbital before the 3d. Therefore, the electron configuration for Co\(^{3+}\) becomes \([\mathrm{Ar}] 3d^6\). This configuration will be affected by the type of ligands present and their ability to split the d orbitals, which we'll explore further in the ligand field theory.

Ligand Field Theory

Ligand field theory helps explain how ligands affect the energy and distribution of electrons in the d orbitals of a transition metal complex. In \([\mathrm{CoF}_6]^{3-}\), fluoride ions \((F^-)\) are considered weak field ligands. Weak field ligands result in a small splitting of the d orbitals. These orbitals are denoted as \(t_{2g}\) and \(e_g\) in an octahedral arrangement. Due to the minor splitting caused by fluoride, the electrons prefer to stay in separate orbitals to maintain higher spin states. This results in the electron configuration \(t_{2g}^4 e_g^2\), where the distribution of the six electrons across these orbitals leads to 4 unpaired electrons. Understanding ligand strength is key to predicting whether a complex will have a high or low spin state, essential in explaining the number of unpaired electrons.