Question

The correct order of number of unpaired electrons in the ion \(\mathrm{Cu}^{2+} \mathrm{Ni}^{2+}, \mathrm{Fe}^{3+}\) and \(\mathrm{Cr}^{3+}\) is: (a) \(\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Fe}^{3+}\) (b) \(\mathrm{Ni}^{2+}>\mathrm{Cu}^{2+}>\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}\) (c) \(\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}>\mathrm{Ni}^{2+}>\mathrm{Cu}^{2+}\) (d) \(\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}>\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}\)

Short answer

The correct order is (c) Fe³⁺ > Cr³⁺ > Ni²⁺ > Cu²⁺.

Step-by-step solution

Determine the electron configuration of each ion

To find the number of unpaired electrons, first determine the electron configuration of the neutral atoms and then the ions. - **Cu**: [Ar] 3d¹⁰ 4s¹; **Cu²⁺**: remove two electrons, resulting in [Ar] 3d⁹. - **Ni**: [Ar] 3d⁸ 4s²; **Ni²⁺**: remove two electrons, resulting in [Ar] 3d⁸. - **Fe**: [Ar] 3d⁶ 4s²; **Fe³⁺**: remove three electrons, resulting in [Ar] 3d⁵. - **Cr**: [Ar] 3d⁵ 4s¹; **Cr³⁺**: remove three electrons, resulting in [Ar] 3d³.

Calculate the number of unpaired electrons

Using the electron configurations determined in the previous step, calculate the unpaired electrons: - **Cu²⁺**: [Ar] 3d⁹ has 1 unpaired electron. - **Ni²⁺**: [Ar] 3d⁸ has 2 unpaired electrons. - **Fe³⁺**: [Ar] 3d⁵ has 5 unpaired electrons. - **Cr³⁺**: [Ar] 3d³ has 3 unpaired electrons.

Compare the number of unpaired electrons

Compare the number of unpaired electrons: 1. **Fe³⁺**: 5 unpaired electrons 2. **Cr³⁺**: 3 unpaired electrons 3. **Ni²⁺**: 2 unpaired electrons 4. **Cu²⁺**: 1 unpaired electron The decreasing order of unpaired electrons is Fe³⁺ > Cr³⁺ > Ni²⁺ > Cu²⁺.

Match with given options

The correct order based on comparison is: Fe³⁺ > Cr³⁺ > Ni²⁺ > Cu²⁺. Look at the provided options: - (a) Cu²⁺>Ni²⁺>Cr³⁺>Fe³⁺ - (b) Ni²⁺>Cu²⁺>Fe³⁺>Cr³⁺ - (c) Fe³⁺>Cr³⁺>Ni²⁺>Cu²⁺ - (d) Fe³⁺>Cr³⁺>Cu²⁺>Ni²⁺ The correct answer matches option (c).

Key concepts

Unpaired Electrons

Unpaired electrons refer to the electrons in an atom that do not have a corresponding electron with an opposite spin in the same orbital. These electrons are found singly in an orbital instead of in pairs, which occurs when an orbital has its full complement of two electrons. Unpaired electrons are important because they can affect the magnetic properties of an atom or ion.

For instance, the ion \( ext{Fe}^{3+}\) has 5 unpaired electrons in its 3d orbital. This high number of unpaired electrons makes \( ext{Fe}^{3+}\) strongly paramagnetic, meaning it is attracted to a magnetic field.

• The presence of unpaired electrons is what contributes to various properties of transition metals, such as their ability to form colored compounds, catalytic behavior, and magnetic properties.

Understanding the electron configurations helps in knowing exactly how many unpaired electrons are present in a particular ion.

Transition Metal Ions

Transition metal ions are formed when transition metals lose electrons to achieve stable electronic states. Transition metals are elements found in the d-block of the periodic table and are characterized by their ability to form different ions by losing electrons from the s and d orbitals.

Common examples include nickel forming the \( ext{Ni}^{2+}\) ion and iron forming the \( ext{Fe}^{3+}\) ion. The versatility of these ions

• Allows them to exhibit multiple oxidation states.
• Enables them to participate in complex reactions as they can change between these states easily.
• Gives them unique properties that are useful in various applications such as pigments, catalysts, and materials science.

Transition metal ions have unique electron configurations due to the involvement of d-orbitals, which can lead to various chemical and physical properties.

Ion Electron Removal

Ion electron removal is the process where electrons are taken away from neutral atoms to form ions. This happens particularly in transition metals, where s and then d orbital electrons are removed.

For instance, when copper forms \( ext{Cu}^{2+}\), it loses two electrons from its electron configuration, resulting in a change from \([ ext{Ar}] 3d^{10} 4s^1\) to \([ ext{Ar}] 3d^{9}\). Similarly, this change can be observed in \( ext{Cr}^{3+}\) and \( ext{Fe}^{3+}\) as well.

• This removal of electrons from specific orbitals does not just impact charge but also affects the stability and reactivity of the ions.
• It allows these ions to achieve electron configurations reminiscent of noble gases, albeit not always fully matching their stable forms.

The choice of which electrons to remove during the process is crucial as it determines the resulting ion's properties.

D-Orbital Electron Configuration

The d-orbital electron configuration is crucial for understanding the behavior of transition metals. Unlike other elements, transition metals fill their s-orbitals before their d-orbitals. However, upon ionization, the s-orbital electrons are typically removed first.

Consider the \( ext{Cu}^{2+}\) ion: the original configuration of copper is \([ ext{Ar}] 3d^{10} 4s^1\). When forming the ion, the two electrons are removed from these orbitals to form \([ ext{Ar}] 3d^{9}\).

• This means the d-orbitals become more important in determining ion characteristics.
• They dictate properties like magnetic attributes, color, and reactivity, essential for many transition metals.

Recognizing how these electrons are arranged lets us predict the number of unpaired electrons, aiding in predictions about magnetism and chemical behavior.