Problem 26 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following ions exhibit highest magnetic moment? (a) \(\mathrm{Cu}^{2+}\) (b) \(\mathrm{Ti}^{3+}\) (c) \(\mathrm{Ni}^{2+}\) (d) \(\mathrm{Mn}^{2+}\)

Step-by-Step Solution

Answer

\(\mathrm{Mn}^{2+}\) exhibits the highest magnetic moment with approximately \(5.92\) Bohr magnetons.

Step 1: Determine the Electron Configuration

To find the magnetic moment, we first need to determine the electron configuration of each ion.- \(\mathrm{Cu}^{2+}\): Electron configuration is \([\mathrm{Ar}] \ 3d^9\).- \(\mathrm{Ti}^{3+}\): Electron configuration is \([\mathrm{Ar}] \ 3d^1\).- \(\mathrm{Ni}^{2+}\): Electron configuration is \([\mathrm{Ar}] \ 3d^8\).- \(\mathrm{Mn}^{2+}\): Electron configuration is \([\mathrm{Ar}] \ 3d^5\).

Step 2: Count the Number of Unpaired Electrons

The magnetic moment is determined by the number of unpaired electrons. More unpaired electrons imply a higher magnetic moment.- \(\mathrm{Cu}^{2+}\): 1 unpaired electron.- \(\mathrm{Ti}^{3+}\): 1 unpaired electron.- \(\mathrm{Ni}^{2+}\): 2 unpaired electrons.- \(\mathrm{Mn}^{2+}\): 5 unpaired electrons.

Step 3: Calculate the Magnetic Moment

The magnetic moment (in Bohr magnetons, \(\mu_B\)) can be calculated using the formula: \(\mu = \sqrt{n(n+2)}\), where \(n\) is the number of unpaired electrons.- \(\mathrm{Cu}^{2+}\): \(\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\).- \(\mathrm{Ti}^{3+}\): \(\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\).- \(\mathrm{Ni}^{2+}\): \(\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\).- \(\mathrm{Mn}^{2+}\): \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\).

Step 4: Compare Magnetic Moments

Compare the calculated magnetic moments to identify the ion with the highest value.- \(\mathrm{Cu}^{2+}: \approx 1.73\)- \(\mathrm{Ti}^{3+}: \approx 1.73\)- \(\mathrm{Ni}^{2+}: \approx 2.83\)- \(\mathrm{Mn}^{2+}: \approx 5.92\)\(\mathrm{Mn}^{2+}\) has the highest magnetic moment.

Key Concepts

Electron Configuration

Electron configuration is how electrons are distributed in an atom's orbitals. It helps to understand the chemical behavior of an element. For transition metal ions, the electron configuration typically involves 3d orbitals. For example, the electron configuration of

• \(\mathrm{Cu}^{2+}\) is \([\mathrm{Ar}] \ 3d^9\), which means after removing two electrons from copper, one of the 4s and one 3d orbital hold those electrons.
• \(\mathrm{Ti}^{3+}\) is \([\mathrm{Ar}] \ 3d^1\), after losing three electrons, titanium’s highest energy level remains with the single electron in the 3d orbital.
• \(\mathrm{Ni}^{2+}\) is \([\mathrm{Ar}] \ 3d^8\), leaving nickel with eight electrons distributed among the 3d orbitals.
• and \(\mathrm{Mn}^{2+}\) is \([\mathrm{Ar}] \ 3d^5\) indicating manganese loses two electrons leaving half-filled 3d orbitals.

Understanding electron configuration is critical for determining magnetic properties based on unpaired electrons.

Unpaired Electrons

Unpaired electrons are lone electrons in an atom's orbitals that do not link up with another electron. These electrons play a significant role in the magnetic behavior of the ions. The more unpaired electrons an ion has, the stronger its magnetic properties. For example:

• The ion \(\mathrm{Cu}^{2+}\) contains 1 unpaired electron, making its magnetic contribution relatively low.
• \(\mathrm{Ti}^{3+}\) also has 1 unpaired electron, resulting in a comparable low magnetic moment.
• \(\mathrm{Ni}^{2+}\) has 2 unpaired electrons, increasing its magnetic strength.
• In contrast, \(\mathrm{Mn}^{2+}\) has 5 unpaired electrons, which provides a high magnetic moment.

More unpaired electrons mean a higher potential magnetic moment, calculated through physical quantities known as Bohr magnetons.

Bohr Magnetons

The Bohr magneton is a physical constant and a unit to measure the magnetic moment of an electron caused by its spin and orbital motion. It helps quantify the magnetic properties of ions. Since the number of unpaired electrons influences the magnetic moment, it is represented in Bohr magnetons, \(\mu_B\). The magnetic moment \(\mu\) of an ion can be calculated using the formula: \[ \mu = \sqrt{n(n+2)} \]where \(n\) is the number of unpaired electrons. Here's how it applies:

• For \(\mathrm{Cu}^{2+},\) \(\mu = \approx 1.73\ \mu_B\)
• For \(\mathrm{Ti}^{3+},\) \(\mu = \approx 1.73\ \mu_B\)
• For \(\mathrm{Ni}^{2+},\) \(\mu = \approx 2.83\ \mu_B\)
• And for \(\mathrm{Mn}^{2+},\) \(\mu = \approx 5.92\ \mu_B\)

\(\mathrm{Mn}^{2+}\) exhibits the highest magnetic moment due to its high number of unpaired electrons.

Magnetic Properties of Ions

Magnetic properties of ions are determined by the presence of unpaired electrons and are quantified in Bohr magnetons. These properties are important in understanding the behavior of materials in a magnetic field:

• \(\mathrm{Cu}^{2+}\) and \(\mathrm{Ti}^{3+}\), each with one unpaired electron, display weak paramagnetism.
• \(\mathrm{Ni}^{2+}\) with two unpaired electrons is more paramagnetic since its magnetic moment is larger.
• Manganese ion \(\mathrm{Mn}^{2+}\), which has five unpaired electrons, shows strong paramagnetism, due to its significantly higher magnetic moment.

Knowing these properties is essential for predicting how a material will react in the presence of a magnetic field, which is vital in various applications like electronic devices and MRI technology.