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

Question

Match the following: List I (n= un-paired e-) \(\quad\) 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: \(\begin{array}{lll}1 & 2 & 3\end{array}\) 4 (a) (ii) (i) (iii) (iv) (b) (iii) (iv) (ii) (i)

Step-by-Step Solution

Answer

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

Step 1: Understanding Unpaired Electrons and Magnetic Moments

The magnetic moment in Bohr Magnetons (B.M.) for an ion can be calculated using the formula: \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons. We will use this to match the unpaired electron count with the given magnetic moments.

Step 2: Calculate Magnetic Moment for \(\mathrm{Sc}^{3+}\)

\(\mathrm{Sc}^{3+}\) has \( n = 0 \) unpaired electrons. The magnetic moment is \( \mu = \sqrt{0(0+2)} = 0 \). This corresponds to zero, matching (iii).

Step 3: Calculate Magnetic Moment for \(\mathrm{V}^{2+}\)

\(\mathrm{V}^{2+}\) has \( n = 3 \) unpaired electrons. The magnetic moment is \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.} \). This corresponds to (iv).

Step 4: Calculate Magnetic Moment for \(\mathrm{Fe}^{3+}\)

\(\mathrm{Fe}^{3+}\) has \( n = 5 \) unpaired electrons. The magnetic moment is \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ B.M.} \). This corresponds to (i).

Step 5: Calculate Magnetic Moment for \(\mathrm{Cu}^{2+}\)

\(\mathrm{Cu}^{2+}\) has \( n = 1 \) unpaired electron. The magnetic moment is \( \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ B.M.} \). This corresponds to (ii).

Step 6: Match List I and List II

Now, we can match each element with its corresponding magnetic moment: \(\mathrm{Sc}^{3+} = (iii)\), \(\mathrm{V}^{2+} = (iv)\), \(\mathrm{Fe}^{3+} = (i)\), \(\mathrm{Cu}^{2+} = (ii)\).

Key Concepts

Unpaired Electrons

Understanding unpaired electrons is crucial for addressing many chemistry topics, including the calculation of magnetic moments. In each atom, electrons can be either paired or unpaired. Paired electrons occupy the same orbital and have opposite spins, cancelling each other's magnetic effects.
However, unpaired electrons exist alone in their orbitals, contributing to the magnetic properties of the atom. Certain ions, especially those in transition metals, often have unpaired electrons.

• When calculating magnetic moments, more unpaired electrons generally mean a stronger magnetic effect.
• The presence of unpaired electrons is key to forming coordination compounds and can impact the color and reactivity of a substance.

Knowing the number of unpaired electrons helps predict an ion's behavior in a magnetic field, providing insight into its chemical properties.

Transition Metals

Transition metals are fascinating elements that have unique chemical properties due to their electron configurations. These metals, found in groups 3 to 12 of the periodic table, typically have partially filled d orbitals.

• Their ability to have multiple oxidation states is due to the energy closeness between their s and d orbitals.
• Having unpaired electrons contributes to their magnetic properties and potential as catalysts in various reactions.

For example, ions like \( ext{V}^{2+}\), \( ext{Fe}^{3+}\), and \( ext{Cu}^{2+}\) have differing numbers of unpaired electrons, influencing their magnetic moments.
Transition metals also have considerable economic and industrial significance. They are used extensively in construction, manufacturing, and the creation of alloys due to their physical robustness and versatility.

Bohr Magneton

The Bohr Magneton (\( ext{B.M.}\)) is a physical constant that acts as the unit of the magnetic moment for an electron due to its orbit around the nucleus and its intrinsic spin.

• The magnetic moment is a vector quantity that indicates the magnetic strength and orientation.
• It is directly related to the number of unpaired electrons in an ion, calculated using the formula \(\mu = \sqrt{n(n+2)}\).
• Here, \(n\) represents the number of unpaired electrons in the atom or ion.

Calculating with Bohr Magnetons gives insights into the magnetic properties of transition metals and their ions. The more unpaired electrons an ion has, the larger its magnetic moment will be in \( ext{B.M.}\). Understanding this helps chemists and physicists in various applications, including magnetic resonance imaging and the study of magnetic materials.