Problem 19 - A Complete Resource Book in Chemistry for JEE Main

Question

Ferrous ion changes to \(\mathrm{X}\) ion, on reacting with acidified hydrogen peroxide. The number of d-electrons present in \(\mathrm{X}\) and its magnetic moment (in B.M.) are respectively (a) 5 and \(4.9\) (b) 4 and \(5.92\) (c) 6 and \(6.95\) (d) 5 and \(5.92\)

Step-by-Step Solution

Answer

(d) 5 and 5.92 B.M.

Step 1: Determine the Oxidation States

Ferrous ion, which is Fe²⁺, loses electrons when oxidized. Hydrogen peroxide serves as the oxidizing agent. The process converts Fe²⁺ to Fe³⁺.

Step 2: Calculate d-electrons in Fe³⁺

Fe has an atomic number of 26. In the neutral state, it has a configuration of \[ ext{[Ar]} \, 3d^6 \, 4s^2 \]. In Fe³⁺, 3 electrons are removed, one from the 4s and two from the 3d, resulting in \[ 3d^5 \]. So, there are 5 d-electrons in Fe³⁺.

Step 3: Calculate Magnetic Moment

The formula for magnetic moment (in Bohr Magnetons, B.M.) is given by \[ \mu = \sqrt{n(n+2)} \], where \ n \ is the number of unpaired electrons. Since Fe³⁺ has 5 unpaired d-electrons (from \ 3d^5 \ configuration), the magnetic moment is \ \mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \text{ B.M.} \

Key Concepts

Magnetic Moment Calculation

The magnetic moment provides insight into an ion's magnetic properties. For an iron ion like Fe³⁺, this refers to the spinning of unpaired electrons, which influences how the ion interacts with magnetic fields. The calculation begins by identifying the number of unpaired electrons, denoted as \( n \). Then, we use the formula for magnetic moment, \( \mu = \sqrt{n(n+2)} \), where \( \mu \) is in Bohr Magnetons (B.M.). For Fe³⁺, which has a \( 3d^5 \) configuration, there are 5 unpaired electrons. Substitute \( n = 5 \) into the formula:

• \( \mu = \sqrt{5(5+2)} \)
• \( \mu = \sqrt{35} \)
• \( \mu \approx 5.92 \) B.M.

This calculation shows that Fe³⁺ has a significant magnetic moment due to the five unpaired d-electrons. This unpaired configuration allows it to exhibit paramagnetic properties.

d-electrons Configuration

An understanding of the d-electrons configuration is crucial when dealing with transition metals like iron. These electrons affect various properties, including color, magnetism, and chemical reactivity. The d-electron configuration of a ferric ion (Fe³⁺) can be determined based on its position in the periodic table. Iron in its neutral state has the atomic number 26, with the electron configuration [Ar] 3d\(^6\) 4s\(^2\), where

• The 4s electrons are filled first, followed by d-orbitals.

When iron is oxidized to Fe³⁺, three electrons are removed:

• 1 electron from the 4s orbital
• 2 electrons from the 3d orbital

This results in a new configuration, [Ar] 3d\(^5\), with five electrons in the d-orbital. All these electrons are unpaired, significantly impacting the ion's magnetic properties. The presence of these unpaired electrons is why Fe³⁺ shows magnetic behavior.

Ferrous Ion Oxidation

The ferrous ion, denoted by Fe²⁺, undergoes oxidation to form the ferric ion, Fe³⁺, in chemical reactions. This transformation typically occurs in the presence of oxidizing agents, such as acidified hydrogen peroxide. Understanding this process involves knowing the rules of oxidation states and electron transfer. Oxidation is defined as the loss of electrons. For Fe²⁺:

• The iron ion initially possesses an electron configuration of [Ar] 3d\(^6\).
• Upon oxidation, it loses one electron to transition to Fe³⁺.

The new configuration becomes [Ar] 3d\(^5\) after losing a total of three electrons (one from 4s and two 3d). This shift highlights the changes in electronic structure and magnetic properties. Oxidation impacts the chemical behavior and reactivity of the ion, not just its magnetic characteristics. Transformations like this are critical in various industrial and biological processes where the conversion of ferrous to ferric is essential, such as in mineralization and metabolic pathways.