/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Give three examples of ions that... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 47

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

Short Answer

Expert verified
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

01

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).
02

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.
03

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (Ni虏鈦�) 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 (Ni虏鈦�), nickel loses two electrons from the 4s orbital, resulting in the configuration [Ar] 3d^8.
Here's how it works:
  • Nickel loses 2 electrons = Ni虏鈦�.
  • Email configuration: [Ar] 3d^8.
  • Ni虏鈦� ion is stable with its 3d subshell holding 8 electrons.
This configuration is particularly stable, making Ni虏鈦� a common ion in chemistry.
Palladium Ion (Pd鈦粹伜) 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 (Pd鈦粹伜), palladium loses four electrons from its 5s and 4d orbitals.
This gives it the configuration [Kr] 4d^8.
  • Pd loses 4 electrons = Pd鈦粹伜.
  • Email configuration: [Kr] 4d^8.
  • Pd鈦粹伜 ion fits our target configuration of 4d^8.
This ion is less common in nature but essential for specific reactions in chemistry.
Platinum Ion (Pt虏鈦�) 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 (Pt虏鈦�), platinum loses electrons from its 5d and 6s orbitals.
The result is [Xe] 4f^14 5d^8.
  • Pt loses 2 electrons = Pt虏鈦�.
  • Email configuration: [Xe] 4f^14 5d^8.
  • Pt虏鈦� 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the first ionization energy of neon and the electron affinity of fluorine. (a) Write equations, including electron configurations, for each process. (b) These two quantities have opposite signs. Which will be positive, and which will be negative? (c) Would you expect the magnitudes of these two quantities to be equal? If not, which one would you expect to be larger?

Until the early 1960s, the group 8A elements were called the inert gases. (a) Why was the term inert gases dropped? (b) What discovery triggered this change in name? (c) What name is applied to the group now?

(a) As described in Section 7.7 , the alkali metals react with hydrogen to form hydrides and react with halogens to form halides. Compare the roles of hydrogen and halogens in these reactions. Write balanced equations for the reaction of fluorine with calcium and for the reaction of hydrogen with calcium. (b) What is the oxidation number and electron configuration of calcium in each product?

Discussing this chapter, a classmate says, "Since elements that form cations are metals and elements that form anions are nonmetals, elements that do not form ions are metalloids." Do you agree or disagree?

The first ionization energy of the oxygen molecule is the energy required for the following process: $$\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e}^{-}$$ The energy needed for this process is 1175 \(\mathrm{kJ} / \mathrm{mol}\) , very similar to the first ionization energy of \(\mathrm{Xe} .\) Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2} ?\) If so, suggest a product or products of this reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.