Problem 27 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following is a diamagnetic ion? (a) \(\mathrm{Cu}^{2+}\) (b) \(\mathrm{Mn}^{2+}\) (c) \(\mathrm{S} \mathrm{c}^{3+}\) (d) \(\mathrm{Co}^{2+}\)

Step-by-Step Solution

Answer

Sc^{3+} is the diamagnetic ion.

Step 1: Understand Diamagnetism

Diamagnetism occurs when all the electrons in an ion are paired. Diamagnetic species are not attracted to a magnetic field.

Step 2: Write Electron Configurations

Determine the electron configurations of each ion by removing the appropriate number of electrons from the neutral atom. - For \({\mathrm{Cu}^{2+}}\): [Ar] 3d^9 - For \({\mathrm{Mn}^{2+}}\): [Ar] 3d^5 - For \({\mathrm{Sc}^{3+}}\): [Ar] - For \({\mathrm{Co}^{2+}}\): [Ar] 3d^7

Step 3: Identify Unpaired Electrons

Look at each configuration to identify the presence of unpaired electrons. - \({\mathrm{Cu}^{2+}}\) has 1 unpaired electron in 3d - \({\mathrm{Mn}^{2+}}\) has 5 unpaired electrons in 3d - \({\mathrm{Sc}^{3+}}\) has no 3d electrons unpaired - \({\mathrm{Co}^{2+}}\) has 3 unpaired electrons in 3d

Step 4: Conclusion

Since \({\mathrm{Sc}^{3+}}\) has no unpaired electrons, it is the only diamagnetic ion among the options.

Key Concepts

Electron Configurations

Electron configurations are a simple way to understand the arrangement of electrons within an atom or ion. They are written by following the order of energy levels, as illustrated by the Aufbau principle.
When determining electron configurations for ions, electrons are removed or added based on the charge of the ion.

• In a neutral atom, electrons fill atomic orbitals starting with the lowest energy level.
• For positive ions ( extit{cations}), electrons are removed starting from the highest energy level.
• In negative ions ( extit{anions}), we add electrons to the available orbitals.

For example, the electron configuration for \( ext{Cu}^{2+}\), derived from neutral copper, involves removing two electrons from its normal configuration: \([ ext{Ar}] 3d^{10} 4s^1\) becomes \([ ext{Ar}] 3d^9\). Similarly, \( ext{Sc}^{3+}\) starts as \([ ext{Ar}] 3d^1 4s^2\) and ends up as \([ ext{Ar}]\) after removing three electrons.

Unpaired Electrons

Unpaired electrons are single electrons present in an atomic or ionic orbital, which are not accompanied by a second electron in the same orbital. The presence or absence of unpaired electrons determines whether a substance is diamagnetic or paramagnetic.
Diamagnetic substances have only paired electrons, meaning all electrons in their orbitals are paired with another having opposite spin. Paramagnetic substances have one or more unpaired electrons and are attracted to magnetic fields.
Let's consider the ions mentioned:

• For \( ext{Cu}^{2+}\), the configuration \([ ext{Ar}] 3d^9\) shows one electron in "3d" that is unpaired.
• \( ext{Mn}^{2+}\) with \([ ext{Ar}] 3d^5\) has five unpaired electrons. This configuration makes it strongly paramagnetic.
• \( ext{Sc}^{3+}\) possesses no unpaired "3d" electrons, leading to its diamagnetic behavior.
• \( ext{Co}^{2+}\) has three unpaired electrons in \([ ext{Ar}] 3d^7\).

Magnetic Properties of Ions

The magnetic properties of ions are determined by their electron configurations, specifically the number of unpaired electrons.

• Diamagnetic ions like \( ext{Sc}^{3+}\) have no unpaired electrons and are not attracted to magnetic fields. They often exhibit weak magnetic properties.
• Paramagnetic ions possess one or more unpaired electrons, as seen in \( ext{Mn}^{2+}\) and \( ext{Co}^{2+}\). These ions are attracted to magnetic fields due to their unpaired electrons.

For students solving such exercises, recognizing the electron configuration and determining the number of unpaired electrons helps in categorizing the ions as diamagnetic or paramagnetic.
This identification is crucial for fields such as chemistry and physics, where understanding magnetic properties can impact predicting behaviors of materials.