/*! 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 18 Write electron configurations fo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 18

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Zn}^{2+}\) (b) \(\mathrm{Te}^{2-}\) (c) \(\mathrm{Sc}^{3+}\), (d) \(\mathrm{Rh}^{3+}\),(e) \(\mathrm{Tl}^{+}\), (f) \(\mathrm{Bi}^{3+}\).

Short Answer

Expert verified
The electron configurations for the given ions are as follows: (a) \(\mathrm{Zn}^{2+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2\) (Ar) (b) \(\mathrm{Te}^{2-}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [5s]^2 [5p]^6\) (Xe) (c) \(\mathrm{Sc}^{3+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6\) (Ar) (d) \(\mathrm{Rh}^{3+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^8\) (e) \(\mathrm{Tl}^{+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\) (Xe) (f) \(\mathrm{Bi}^{3+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\) (Xe) All ions except \(\mathrm{Rh}^{3+}\) have noble-gas configurations.

Step by step solution

01

Write the electron configurations for each ion

Based on their atomic numbers and charges, we can find the electron configurations for the given ions: (a) \(\mathrm{Zn}^{2+}\): Zn is atomic number 30, so it has 30 electrons in its neutral state. Losing 2 electrons to become \(\mathrm{Zn}^{2+}\), it now has 28 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2\) (b) \(\mathrm{Te}^{2-}\): Te is atomic number 52, so it has 52 electrons in its neutral state. Gaining 2 electrons to become \(\mathrm{Te}^{2-}\), it now has 54 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [5s]^2 [5p]^6\) (c) \(\mathrm{Sc}^{3+}\): Sc is atomic number 21, so it has 21 electrons in its neutral state. Losing 3 electrons to become \(\mathrm{Sc}^{3+}\), it now has 18 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6\) (d) \(\mathrm{Rh}^{3+}\): Rh is atomic number 45, so it has 45 electrons in its neutral state. Losing 3 electrons to become \(\mathrm{Rh}^{3+}\), it now has 42 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^8\) (e) \(\mathrm{Tl}^{+}\): Tl is atomic number 81, so it has 81 electrons in its neutral state. Losing 1 electron to become \(\mathrm{Tl}^{+}\), it now has 80 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\) (f) \(\mathrm{Bi}^{3+}\): Bi is atomic number 83, so it has 83 electrons in its neutral state. Losing 3 electrons to become \(\mathrm{Bi}^{3+}\), it now has 80 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\)
02

Determine which ions have noble-gas configurations

Now, we need to see if the electron configurations of these ions match the electron configurations of noble gases. Noble gases have full electron shells, which means they have the following electron configurations: - He: \([1s]^2\) - Ne: \([1s]^2 [2s]^2 [2p]^6\) - Ar: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6\) - Kr: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6\) - Xe: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [5s]^2 [5p]^6\) - Rn: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\) Comparing the configurations of the ions with the noble gases, we can see that: - \(\mathrm{Zn}^{2+}\) has a noble-gas configuration (Ar) - \(\mathrm{Te}^{2-}\) has a noble-gas configuration (Xe) - \(\mathrm{Sc}^{3+}\) has a noble-gas configuration (Ar) - \(\mathrm{Rh}^{3+}\) does not have a noble-gas configuration - \(\mathrm{Tl}^{+}\) has a noble-gas configuration (Xe) - \(\mathrm{Bi}^{3+}\) has a noble-gas configuration (Xe) In conclusion, all the ions except \(\mathrm{Rh}^{3+}\) have noble-gas configurations.

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.

Noble Gas Configuration
Noble gas configuration is a way to express electron configurations by using the electron structure of the nearest noble gas. Noble gases are the elements in the far right of the periodic table, like helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). These elements are known for having full electron shells, which makes them very stable. When an element or ion has the same electron configuration as a noble gas, it is said to be isoelectronic with that noble gas.

For example, an ion such as Zn^{2+} loses two electrons to achieve the electron configuration of argon (Ar), and Te^{2-} gains two electrons to achieve the electron configuration of xenon (Xe). These full outer electron shells make them especially stable and less likely to react chemically. This concept is an essential key in understanding the electron reactivity and stability of elements.
Ion Electron Configurations
When dealing with ion electron configurations, you need to first understand what ions are. Ions are atoms that have either gained or lost electrons, resulting in a positive or negative charge. The process of finding an ion's electron configuration involves adjusting the normal electron configuration of the element by the number of electrons gained or lost.

Here's how you can do it:
  • If the ion is positive, like Sc^{3+}, you subtract electrons from the highest energy level in the electron configuration of the neutral atom.
  • If the ion is negative, like Te^{2-}, you add electrons to the highest energy level of the neutral atom's electron configuration.
In the original exercise, you see different ions such as Sc^{3+}, Rh^{3+}, and others. By following these steps, their electron configurations are adjusted to reflect the change in electron number. A fully understood electron configuration is vital for predicting chemical behavior.
Chemical Elements
Chemical elements are the fundamental substances that consist of one type of atom and are organized in the periodic table. Each element has a unique atomic number, which indicates the number of protons in its nucleus. In a neutral state, the number of electrons equals the number of protons.

The periodic table not only arranges these elements in order according to their atomic numbers but also groups them into categories, like metals, non-metals, and metalloids. By understanding elements and their locations on the periodic table, you can predict their electron configurations and chemical properties.

For instance, knowing that zinc (Zn) belongs to the transition metals gives you insights into its usual electron configuration and common ion state, Zn^{2+}. Recognizing the identity and properties of chemical elements is foundational for chemistry studies and helps with everything from predicting reactions to balancing equations.

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

(a) What is the trend in electronegativity going from left to right in a row of the periodic table? (b) How do electronegativity values generally vary going down a column in the periodic table? (c) How do periodic trends in electronegativity relate to those for ionization energy and electron affinity?

A common form of elemental phosphorus is the tetrahedral \(\mathrm{P}_{4}\) molecule, where all four phosphorus atoms are equivalent: At room temperature phosphorus is a solid. (a) Do you think there are any unshared pairs of electrons in the \(\mathrm{P}_{4}\) molecule? (b) How many \(\mathrm{P}-\mathrm{P}\) bonds are there in the molecule? (c) Can you draw a Lewis structure for a linear \(\mathrm{P}_{4}\) molecule that satisfies the octet rule? (d) Using formal charges, what can you say about the stability of the linear molecule vs. that of the tetrahedral molecule?

Explain the following trends in lattice energy: (a) \(\mathrm{CaF}_{2}>\mathrm{BaF}_{2} ;\) (b) \(\mathrm{NaCl}>\mathrm{RbBr}>\mathrm{CsBr} ;\) (c) \(\mathrm{BaO}>\mathrm{KF}\).

In the vapor phase, \(\mathrm{BeCl}_{2}\) exists as a discrete molecule. (a) Draw the Lewis structure of this molecule, using only single bonds. Does this Lewis structure satisfy the octet rule? (b) What other resonance forms are possible that satisfy the octet rule? (c) Using formal charges, select the resonance form from among all the Lewis structures that is most important in describing \(\mathrm{BeCl}_{2}\) :

(a) Describe the molecule chlorine dioxide, \(\mathrm{ClO}_{2}\), using three possible resonance structures. (b) Do any of these resonance structures satisfy the octet rule for every atom in the molecule? Why or why not? (c) Using formal charges, select the resonance structure(s) that is (are) most important.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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.