Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 31

(a) What is an isoelectronic series? (b) Which neutral atom is isoelectronic with each of the following ions: \(\mathrm{Ga}^{3+}, \mathrm{Zr}^{4+}\) \(\mathrm{Mn}^{7+}, \Gamma, \mathrm{Pb}^{2+} ?\)

Short Answer

Expert verified
(a) An isoelectronic series is a group of ions or atoms that share the same number of electrons but may have different charges and atomic numbers. (b) The neutral atoms isoelectronic with the given ions are: \(\mathrm{Ga}^{3+}\) - Nickel (Ni), \(\mathrm{Zr}^{4+}\) - Krypton (Kr), \(\mathrm{Mn}^{7+}\) - Argon (Ar), and \(\mathrm{Pb}^{2+}\) - Mercury (Hg).

Step by step solution

01

Understand an isoelectronic series

An isoelectronic series is a group of ions or atoms that share the same number of electrons but may have different charges and atomic numbers.
02

Calculate the number of electrons in each ion

1. For \(\mathrm{Ga}^{3+}\): The atomic number of Gallium (Ga) is 31. Since it has a charge of +3, it has lost 3 electrons. Therefore, the number of electrons in this ion is \(31-3=28\). 2. For \(\mathrm{Zr}^{4+}\): The atomic number of Zirconium (Zr) is 40. Since it has a charge of +4, it has lost 4 electrons. Therefore, the number of electrons in this ion is \(40-4=36\). 3. For \(\mathrm{Mn}^{7+}\): The atomic number of Manganese (Mn) is 25. Since it has a charge of +7, it has lost 7 electrons. Therefore, the number of electrons in this ion is \(25-7=18\). 4. For \(\mathrm{Pb}^{2+}\): The atomic number of Lead (Pb) is 82. Since it has a charge of +2, it has lost 2 electrons. Therefore, the number of electrons in this ion is \(82-2=80\).
03

Find the neutral atoms with the calculated electrons

1. For 28 electrons, the neutral atom is Nickel (Ni), which has an atomic number of 28. 2. For 36 electrons, the neutral atom is Krypton (Kr), which has an atomic number of 36. 3. For 18 electrons, the neutral atom is Argon (Ar), which has an atomic number of 18. 4. For 80 electrons, the neutral atom is Mercury (Hg), which has an atomic number of 80.
04

List the isoelectronic neutral atoms

Isoelectronic neutral atoms for the given ions are as follows: - \(\mathrm{Ga}^{3+}\): Nickel (Ni) - \(\mathrm{Zr}^{4+}\): Krypton (Kr) - \(\mathrm{Mn}^{7+}\): Argon (Ar) - \(\mathrm{Pb}^{2+}\): Mercury (Hg)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Some metal oxides, such as \(\mathrm{Sc}_{2} \mathrm{O}_{3},\) do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect \(\mathrm{Sc}_{2} \mathrm{O}_{3}\) to react when the solution becomes acidic or when it becomes basic? Write a balanced chemical equation to support your answer.

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both Si and Cl by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }} ?\) (d) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text {eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text {eff }}\) for a valence electron in P, phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S,\) calculate \(Z_{\mathrm{eff}}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, S. (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

(a) As described in Section \(7.7,\) the alkali metals react with hydrogen to form hydrides and react with halogens- for example, fluorine to form halides, Compare the roles of hydrogen and the halogen in these reactions. How are the forms of hydrogen and halogen in the products alike? (b) Write balanced equations for the reaction of fluorine with calcium and for the reaction of hydrogen with calcium. What are the similarities among the products of these reactions?

Chlorine reacts with oxygen to form \(\mathrm{Cl}_{2} \mathrm{O}_{7} .\) (a) What is the name of this product (see Table 2.6 )? (b) Write a balanced equation for the formation of \(\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) from the elements. (c) Under usual conditions, \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is a colorless liquid with a boiling point of \(81^{\circ} \mathrm{C}\). Is this boiling point expected or surprising? (d) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q) ?\) Explain. (e) If the oxygen in \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is considered to have the -2 oxidation state, what is the oxidation state of the Cl? What is the electron configuration of \(\mathrm{Cl}\) in this oxidation state?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Physical 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.

Sign-up for free