Problem 101 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following: List I ( \(\mathbf{n}=\) un-paired \(\mathrm{e}^{-}\)) 1\. \(\mathrm{Sc}^{3+}=0\) 2\. \(\mathrm{V}^{2+}=3\) 3\. \(\mathrm{Fe}^{3+}=5\) 4\. \(\mathrm{Cu}^{2+}=1\) List II (Meg. moment) (i) \(5.92\) B.M. (ii) \(1.73\) B.M. (iii) zero (iv) 3.87 B.M.

Step-by-Step Solution

Answer

1. (iii), 2. (iv), 3. (i), 4. (ii)

Step 1: Understanding Unpaired Electrons

The number of unpaired electrons in an ion determines its magnetic moment. The magnetic moment (m) is given by the formula: \[ \mu = \sqrt{n(n+2)} \] B.M., where \(n\) is the number of unpaired electrons.

Step 2: Calculate Magnetic Moments for List I

For each ion in List I, evaluate the magnetic moment using \( \mu = \sqrt{n(n+2)} \):1. \(\mathrm{Sc}^{3+}\) with \(n = 0\): \( \mu = \sqrt{0(0+2)} = 0\) B.M.2. \(\mathrm{V}^{2+}\) with \(n = 3\): \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\) B.M.3. \(\mathrm{Fe}^{3+}\) with \(n = 5\): \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\) B.M.4. \(\mathrm{Cu}^{2+}\) with \(n = 1\): \( \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\) B.M.

Step 3: Match List I to List II

Using the calculated magnetic moments, match each ion from List I to the corresponding magnetic moment from List II:1. \(\mathrm{Sc}^{3+}\) has a magnetic moment of 0 B.M. b matched with (iii) zero.2. \(\mathrm{V}^{2+}\) has a magnetic moment of 3.87 B.M. b matched with (iv) 3.87 B.M.3. \(\mathrm{Fe}^{3+}\) has a magnetic moment of 5.92 B.M. b matched with (i) 5.92 B.M.4. \(\mathrm{Cu}^{2+}\) has a magnetic moment of 1.73 B.M. b matched with (ii) 1.73 B.M.

Key Concepts

Unpaired Electrons

Unpaired electrons are electrons in an atom or ion that do not have a fellow electron to partner with in an orbital. In chemistry, electrons are usually paired as they occupy orbitals; however, there are instances where one electron is found alone in an orbital, leading to what we call unpaired electrons. The presence of unpaired electrons influences an atom's or ion's magnetic properties. These unpaired electrons generate a magnetic field as they have a spin and a magnetic moment associated with their spin property.

This means the more unpaired electrons an ion possesses, the stronger its magnetic properties, which are quantified by its magnetic moment. To determine the number of unpaired electrons:

• Examine the electron configuration of the ion.
• Count the electrons in incomplete orbitals to find unpaired electrons.

Unpaired electrons are crucial in understanding why certain materials are attracted to magnets while others are not. They provide a foundational explanation for the magnetic behavior observed in transition metals and their ions.

Magnetic Properties of Ions

The magnetic properties of an ion are largely determined by its unpaired electrons. If an ion has no unpaired electrons, it is considered diamagnetic, which means it is not attracted to a magnetic field. On the other hand, ions with unpaired electrons are termed paramagnetic, and they exhibit attraction towards a magnetic field. This paramagnetic behavior can be quantified by the magnetic moment, often measured in Bohr Magnetons (B.M.).

The magnetic moment - is calculated using the formula: \[ \mu = \sqrt{n(n+2)} \] - correlates directly with the number of unpaired electrons \(n\). - the greater the number of unpaired electrons, the larger the magnetic moment and the greater the attraction to a magnetic field. This formula helps to predict the nature of ions in transition metals and their respective magnetic properties. Understanding the magnetic properties of ions is essential for many practical applications, from MRI technology to the creation of magnetic storage materials.

Transition Metal Ions

Transition metals occupy the d-block of the periodic table and are known for their ability to form ions with different charges. These metals are characterized by having partially filled d-orbitals, which often lead to the presence of unpaired electrons. This results in rich magnetic behavior, which serves as an identifying property of transition metal ions.

These ions typically show a variability in oxidation states, where the number of electrons removed from or added to the neutral atom can affect the configuration of d-orbitals significantly.

• This versatility leads to many transition metal ions having multiple unpaired electrons, which increases their magnetic properties.
• They are widely studied for their industrial and technological applications, particularly those that rely on magnetic characteristics.

The transition metals, like Sc, V, Fe, and Cu, showcase a diverse range of magnetic moments due to their varying numbers of unpaired electrons. For instance, a high oxidation state often correlates with a higher number of unpaired electrons and a stronger magnetic moment. Hence, transition metal ions are crucial to understanding and developing new materials with specific magnetic properties.