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

Question

Match the following: List I (n = un-paired e \(\mathbf{~}^{-}\)) \(\quad\) List II (Meg. moment) 1\. \(\mathrm{Sc}^{3+}=0\) (i) 5.92 B.M. 2\. \(\mathrm{V}^{2+}=3\) (ii) \(1.73\) B.M. 3\. \(\mathrm{Fe}^{3+}=5\) (iii) zero 4\. \(\mathrm{Cu}^{2+}=1\) (iv) \(3.87\) B.M. The correct matching is: \(\begin{array}{llll}1 & 2 & 3 & 4\end{array}\) (a) (ii) (i) (iii) (iv) (b) (iii) (iv) (ii) (i)

Step-by-Step Solution

Answer

The correct matching is option (b): 1-iii, 2-iv, 3-i, 4-ii.

Step 1: Understand Magnetic Moments

The magnetic moment formula is given by \( \mu = \sqrt{n(n+2)} \text{ B.M.} \), where \( n \) is the number of unpaired electrons. Thus, it's important to match the unpaired electrons from List I with their corresponding magnetic moments from List II.

Step 2: Calculate Magnetic Moments for List I

1. For \( \mathrm{Sc}^{3+} \), \( n = 0 \). \( \mu = \sqrt{0(0+2)} = 0 \text{ B.M.} \).2. For \( \mathrm{V}^{2+} \), \( n = 3 \). \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.} \).3. For \( \mathrm{Fe}^{3+} \), \( n = 5 \). \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ B.M.} \).4. For \( \mathrm{Cu}^{2+} \), \( n = 1 \). \( \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ B.M.} \).

Step 3: Match List I with List II Based on Calculations

- \( \mathrm{Sc}^{3+} = 0 \text{ B.M.} = \text{zero} = \text{(iii)} \)- \( \mathrm{V}^{2+} = 3.87 \text{ B.M.} = \text{(iv)} \)- \( \mathrm{Fe}^{3+} = 5.92 \text{ B.M.} = \text{(i)} \)- \( \mathrm{Cu}^{2+} = 1.73 \text{ B.M.} = \text{(ii)} \)

Step 4: Write the Correct Matching

The correct matching of List I with List II based on calculated magnetic moments is: 1. (iii) 2. (iv) 3. (i) 4. (ii)

Step 5: Compare with Given Options

The given option (b) matches our calculated correct matching of List I with List II.

Key Concepts

Unpaired Electrons

Understanding unpaired electrons is crucial in chemistry, especially when dealing with magnetic properties of atoms and ions. Electrons in an atom fill orbitals in a specific order, and each orbital can hold up to two electrons with opposite spins. When an orbital has only one electron, it is termed as having an unpaired electron.
These unpaired electrons contribute to the magnetic moment of an atom or ion. A greater number of unpaired electrons usually leads to a larger magnetic moment.

• Atoms or ions with unpaired electrons are often paramagnetic, meaning they are attracted by a magnetic field.
• In contrast, atoms or ions with all paired electrons are diamagnetic and are typically repelled by a magnetic field.

The presence of unpaired electrons in transition metal ions is the key factor that determines their magnetic properties.

Transition Metal Ions

Transition metal ions play a special role in determining magnetic properties because they often contain unpaired electrons. Transition metals are elements found in the d-block of the periodic table, and they typically form cations by losing electrons from their outermost shell.
For example:

• Scandium (Sc): As a 3+ ion, it loses its three outer electrons leading to no unpaired electrons. Thus, it has a magnetic moment of zero.
• Vanadium (V): The 2+ ion typically has three unpaired electrons.
• Iron (Fe): As Fe³⁺, it usually has five unpaired electrons.
• Copper (Cu): In Cu²⁺, there is typically one unpaired electron.

The behavior of these metal ions under magnetic fields can be predicted accurately by identifying the number of unpaired electrons they contain.

Magnetic Moment Formula

The magnetic moment is an essential concept in understanding the magnetism of ions and is given by the formula:\[ \mu = \sqrt{n(n+2)} \text{ B.M.} \]where \( n \) represents the number of unpaired electrons.
It is often expressed in Bohr magnetons (B.M.), and provides a numerical value that helps grasp the magnetic power of a material influenced by its electron configuration.
Here’s how this translates in practice:

• For an ion with zero unpaired electrons, like \( \text{Sc}^{3+} \), the magnetic moment is 0 B.M.
• If an ion such as \( \text{V}^{2+} \) has three unpaired electrons, its magnetic moment is calculated as:\[\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.}\]
• For \( \text{Fe}^{3+} \) with five unpaired electrons, its magnetic moment is near 5.92 B.M.
• \( \text{Cu}^{2+} \), with one unpaired electron gives:\[\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ B.M.}\]

This equation helps in predicting and matching magnetic characteristics with theoretical outcomes and is a powerful tool in both academic and applied chemistry.