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 in \([\mathrm{CoF}_{6}]^{3-}\) is 4.

Step 1: Determine the Oxidation State of Cobalt

The complex ion given is \([\mathrm{CoF}_{6}]^{3-}\). Let the oxidation state of Co be \(x\). Fluoride (F) is a monodentate ligand with a charge of -1. The overall charge of the complex is -3. Therefore, we can set up the equation: \[ x + 6(-1) = -3 \] Solving for \(x\) gives \(x = +3\). This means cobalt is in the +3 oxidation state.

Step 2: Determine the Electron Configuration of Cobalt(III)

The electron configuration for neutral cobalt (Co) is \( [\mathrm{Ar}]\, 3d^7\, 4s^2 \). In the +3 oxidation state, cobalt loses three electrons (first from the 4s orbital and then from the 3d orbital), resulting in \( [\mathrm{Ar}]\, 3d^6 \).

Step 3: Determine the Type of Ligand and Complex Geometry

Fluoride is a weak field ligand, which does not cause significant splitting of the d orbitals (according to ligand field theory) and does not pair electrons. Thus, \([\mathrm{CoF}_{6}]^{3-}\) forms a high-spin octahedral complex.

Step 4: Analyze the Electron Configuration for High-Spin Complex

In a high-spin complex, the 6 electrons in the \(3d\) orbitals remain unpaired for as long as possible. For \(3d^6\): \(\uparrow\downarrow, \uparrow, \uparrow, \uparrow, \uparrow\). This results in four unpaired electrons.

Key Concepts

Oxidation State

The oxidation state is a way to determine the charge of an atom within a compound, assuming that electrons in bonds are distributed based on rules of electron counting. In the given complex ion \([{\mathrm{CoF}}_{6}]^{3-}\), we identify that cobalt is bonded with fluoride ions, each having a charge of -1, a typical property of halogens. Since there are six fluoride ions, the sum of their charges is -6. However, the overall complex is known to have a -3 charge. By setting up the equation like in Step 1, we can assess that the oxidation state of cobalt must be +3 to balance the charges: \(x + 6(-1) = -3\). Solving gives \(x = +3\). This oxidation state tells us about the number of electrons cobalt has lost compared to its neutral atom form. This is fundamental for further analysis of electronic structure and coordination chemistry.

Ligand Field Theory

Ligand Field Theory (LFT) expands on Crystal Field Theory by considering the effects of ligand bonding. It helps us understand how ligands like fluoride can impact the electronic levels of transition metals in complexes. In coordination chemistry, ligands cause the splitting of metal d orbitals into higher and lower energy levels. The degree of splitting depends on the field strength of the ligands. Fluoride, being a weak field ligand, causes minimal splitting of the d orbitals. This results in a high-spin complex, as seen with \([{\mathrm{CoF}}_{6}]^{3-}\), allowing electrons to remain unpaired across orbitals. LFT provides key insights into predicting magnetic properties and color based on the d orbital's electronic transitions.

Electron Configuration

Understanding electron configuration is crucial for predicting the distribution of electrons among orbitals in an atom or complex. For cobalt, which has an atomic number of 27, the neutral atom's configuration is \( [\mathrm{Ar}]\, 3d^7\, 4s^2\). When cobalt is in the +3 oxidation state, it loses three electrons, resulting in the configuration \( [\mathrm{Ar}]\, 3d^6\). Electrons are first removed from the 4s orbital before the 3d. This reveals how the electron configuration of cobalt within the coordination complex \([{\mathrm{CoF}}_{6}]^{3-}\) impacts its properties, indicating a change in electron count that influences reactivity and magnetic behavior.

High-Spin Complex

A high-spin complex is characterized by the maximum number of unpaired electrons, due to weak field ligands causing minimal orbital splitting. In the \([{\mathrm{CoF}}_{6}]^{3-}\) complex, fluoride is a weak field ligand, resulting in an electronic arrangement where the \(3d\) orbitals are populated with unpaired electrons. The electrons prefer occupying each of the orbitals individually before pairing, following Hund's Rule. This configuration (\(\uparrow\downarrow, \uparrow, \uparrow, \uparrow, \uparrow\)) results in four unpaired electrons within the \(3d^6\) system. High-spin complexes exhibit distinct magnetic properties, often showing paramagnetism due to these unpaired electrons.

Unpaired Electrons

Unpaired electrons play a significant role in determining the magnetic properties of a complex. In the case of \([{\mathrm{CoF}}_{6}]^{3-}\), the ligand field environment allows electrons to remain largely unpaired, with four out of six \(3d\) electrons unpaired. Unlike paired electrons, which have opposing spins and cancel out magnetic moments, unpaired electrons contribute to a net magnetic moment. This is why high-spin complexes like \([{\mathrm{CoF}}_{6}]^{3-}\) are often paramagnetic. Understanding the number and arrangement of unpaired electrons helps predict the complex's response to magnetic fields and its electronic spectrum. Each unpaired electron adds to the total magnetic effect, making the study of unpaired electrons crucial for coordination chemistry and material science.