Question

Give three examples of ions that have an electron configuration of \(n d^{8}(n=3,4,5, \ldots).\)

Short answer

The three examples of ions with the given electron configurations are: 1. Ni^2+ ion with a 3d^8 electron configuration 2. Pd^4+ ion with a 4d^8 electron configuration 3. Pt^2+ ion with a 5d^8 electron configuration

Step-by-step solution

Identify the neutral element with the desired electron configuration

We are looking for elements with \(nd^8\) electron configuration. As the given values of n are 3, 4, and 5, let's find elements having 3d^8, 4d^8, and 5d^8 electron configurations. 3d^8: This corresponds to an element with 3d subshell having 2 less electrons than the completely filled 3d subshell, which has 10 electrons. The 3d^10 electron configuration belongs to Zinc (Zn) with atomic number 30. So, the element with 3d^8 electron configuration would have 2 less electrons than Zinc, making it an element with atomic number 28, which is Nickel (Ni). 4d^8: This corresponds to an element with 4d subshell having 2 less electrons than the completely filled 4d subshell. The 4d^10 electron configuration belongs to Cadmium (Cd) with atomic number 48. So, the element with 4d^8 electron configuration would have 2 less electrons than Cadmium, making it an element with atomic number 46, which is Palladium (Pd). 5d^8: This corresponds to an element with 5d subshell having 2 less electrons than the completely filled 5d subshell. The 5d^10 electron configuration belongs to Hafnium (Hf) with atomic number 72. So, the element with 5d^8 electron configuration would have 2 less electrons than Hafnium, making it an element with atomic number 70, which is Ytterbium (Yb).

Determine ions that have the identified electron configurations

Now, we have identified the neutral elements with the desired electron configurations. Let's find three examples of ions from these elements or other elements by considering the charge of the ion and the difference in the number of electrons. 1. Nickel (Ni): Nickel, a transition metal, generally forms Ni^2+ ions by losing two 4s electrons. As the electron configuration of neutral Nickel is [Ar] 3d^8 4s^2, when it loses 2 electrons to form Ni^2+ ion, it will have an electron configuration of [Ar] 3d^8, which is one of the desired electron configurations. 2. Palladium (Pd): Palladium mostly forms Pd^4+ ions by losing four electrons from its 4d and 5s orbitals. As the electron configuration of neutral Palladium is [Kr] 4d^8 5s^2, when it loses 4 electrons to form Pd^4+ ion, it will have an electron configuration of [Kr] 4d^8, which fulfills our requirement. 3. Platinum (Pt): Platinum has an electron configuration of [Xe] 4f^14 5d^9 6s^1. When it forms Pt^2+ ion by losing both the 6s electron and one 5d electron, the electron configuration becomes [Xe] 4f^14 5d^8, which matches our required electron configuration.

Summarizing the results

The three examples of ions with the given electron configurations are: 1. Ni^2+ ion with a 3d^8 electron configuration 2. Pd^4+ ion with a 4d^8 electron configuration 3. Pt^2+ ion with a 5d^8 electron configuration

Key concepts

Understanding Transition Metals

Transition metals are a group of elements found in the middle of the periodic table. They are known for their unique ability to form various oxidation states and complex ions.
This flexibility comes from their partially filled d subshells.
Here's what you need to know about them:

• They often have metallic properties like conductivity and malleability.
• Transition metals can form colored compounds due to d-d electron transitions.
• They exhibit magnetic properties.
• Common examples include iron, copper, and gold.

These characteristics are crucial when studying electron configurations and the behavior of ions.

Nickel Ion (Ni²⁺) Configuration

Nickel, with atomic number 28, is a well-known transition metal. In its neutral state, nickel's electron configuration is [Ar] 3d^8 4s^2.
To form the nickel ion (Ni²⁺), nickel loses two electrons from the 4s orbital, resulting in the configuration [Ar] 3d^8.
Here's how it works:

• Nickel loses 2 electrons = Ni²⁺.
• Email configuration: [Ar] 3d^8.
• Ni²⁺ ion is stable with its 3d subshell holding 8 electrons.

This configuration is particularly stable, making Ni²⁺ a common ion in chemistry.

Palladium Ion (Pd⁴⁺) Configuration

Palladium, a transition metal with atomic number 46, has an interesting electron configuration. In its neutral state, palladium is [Kr] 4d^10.
To form the palladium ion (Pd⁴⁺), palladium loses four electrons from its 5s and 4d orbitals.
This gives it the configuration [Kr] 4d^8.

• Pd loses 4 electrons = Pd⁴⁺.
• Email configuration: [Kr] 4d^8.
• Pd⁴⁺ ion fits our target configuration of 4d^8.

This ion is less common in nature but essential for specific reactions in chemistry.

Platinum Ion (Pt²⁺) Configuration

Platinum, with atomic number 78, is another transition metal renowned for its catalytic properties. The electron configuration of neutral platinum is [Xe] 4f^14 5d^9 6s^1.
To generate the platinum ion (Pt²⁺), platinum loses electrons from its 5d and 6s orbitals.
The result is [Xe] 4f^14 5d^8.

• Pt loses 2 electrons = Pt²⁺.
• Email configuration: [Xe] 4f^14 5d^8.
• Pt²⁺ ion is stable and commonly found in catalytic processes.

This configuration highlights platinum's ability to stabilize and participate in various chemical reactions, especially in industrial applications.