Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 69

Would you expect each of the following atoms to gain or lose electrons when forming ions? What ion is the most likely in each case? a. \(\mathrm{Ra}\) b. In c. \(\mathrm{P}\) d. Te e. Bi f. Rb

Short Answer

Expert verified
a. Ra loses 2 electrons to form Ra\(^{2+}\). b. In loses 3 electrons to form In\(^{3+}\). c. P gains 3 electrons to form P\(^{3-}\). d. Te gains 2 electrons to form Te\(^{2-}\). e. Bi most likely gains 3 electrons to form Bi\(^{3-}\), but it can also lose 5 electrons to form Bi\(^{3+}\). f. Rb loses 1 electron to form Rb\(^{+}\).

Step by step solution

01

a. Ra

Ra (Radium) has an atomic number of 88. Thus, it has 2 valence electrons in the 7s orbital. Its electron configuration is \([Rn]7s^2\), where Rn denotes the electron configuration of the preceding noble gas Radon. Considering its position in the periodic table as an alkaline earth metal (Group 2), Ra will tend to lose these 2 valence electrons to have a full 6s orbital and achieve a stable electronic configuration like a noble gas. Thus, the most likely ion for Ra is Ra\(^{2+}\).
02

b. In

In (Indium) has an atomic number of 49. It has 3 valence electrons in the 5s and 5p orbitals. Its electron configuration is \([Kr]5s^2 4d^{10} 5p^1\), where Kr represents the electron configuration of the preceding noble gas Krypton. As a member of Group 13 (post-transition metals), In will tend to lose these 3 valence electrons to have a full 4d orbital and achieve a stable electronic configuration like a noble gas. The most likely ion for In is In\(^{3+}\).
03

c. P

P (Phosphorus) has an atomic number of 15. It has 5 valence electrons in the 2s and 2p orbitals. Its electron configuration is \([Ne]3s^2 3p^3\), where Ne represents the electron configuration of the preceding noble gas Neon. Phosphorus belongs to Group 15 (pnictogens), and it will tend to gain 3 electrons to fill its 2p orbital and achieve a stable electronic configuration like a noble gas. Thus, the most likely ion for P is P\(^{3-}\).
04

d. Te

Te (Tellurium) has an atomic number of 52. It has 6 valence electrons in the 5s and 5p orbitals. Its electron configuration is \([Kr]5s^2 4d^{10} 5p^4\), where Kr represents the electron configuration of the preceding noble gas Krypton. Tellurium belongs to Group 16 (chalcogens), and it tends to gain 2 electrons to fill its 5p orbital and achieve a stable electronic configuration like a noble gas. The most likely ion for Te is Te\(^{2-}\).
05

e. Bi

Bi (Bismuth) has an atomic number of 83. It has 5 valence electrons in the 6s and 6p orbitals. Its electron configuration is \([Xe]6s^2 4f^{14} 5d^{10} 6p^3\), where Xe represents the electron configuration of the preceding noble gas Xenon. Bismuth belongs to Group 15 (pnictogens), and it will tend to gain 3 electrons to fill its 6p orbital and achieve a stable electronic configuration like a noble gas. The most likely ion for Bi is Bi\(^{3-}\). However, in some cases, Bi can also form Bi\(^{3+}\) ions by losing its 5 valence electrons.
06

f. Rb

Rb (Rubidium) has an atomic number of 37. It has 1 valence electron in the 5s orbital. Its electron configuration is \([Kr]5s^1\), where Kr represents the electron configuration of the preceding noble gas Krypton. As a member of Group 1 (alkali metals), Rb will tend to lose its single valence electron to achieve a stable electronic configuration like a noble gas. The most likely ion for Rb is Rb\(^{+}\).

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

Write the formula for each of the following compounds: a. chromium(VI) oxide b. disulfur dichloride c. nickel(II) fluoride d. potassium hydrogen phosphate e. aluminum nitride \(\mathbf{f}\). ammonia g. manganese(IV) sulfide h. sodium dichromate i. ammonium sulfite i. carbon tetraiodide

Identify the elements that correspond to the following atomic numbers. Label each as either a noble gas, a halogen, an alkali metal, an alkaline earth metal, a transition metal, a lanthanide metal, or an actinide metal. a. 17 b. 4 c. 63 d. 72 e. 2 f. 92 g. 55

When mixtures of gaseous \(\mathrm{H}_{2}\) and gaseous \(\mathrm{Cl}_{2}\) react, a product forms that has the same properties regardless of the relative amounts of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) used. a. How is this result interpreted in terms of the law of definite proportion? b. When a volume of \(\mathrm{H}_{2}\) reacts with an equal volume of \(\mathrm{Cl}_{2}\) at the same temperature and pressure, what volume of product having the formula \(\mathrm{HCl}\) is formed?

Which of the following is(are) correct? a. \(4^{40} \mathrm{Ca}^{2+}\) contains 20 protons and 18 electrons. b. Rutherford created the cathode-ray tube and was the founder of the charge- to-mass ratio of an electron. c. An electron is heavier than a proton. d. The nucleus contains protons, neutrons, and electrons.

What is the systematic name of \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) ? If the charge on the metal remained constant and then sulfur was substituted for oxygen, how would the formula change? What is the difference in the total number of protons between \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) and its sulfur analog?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear 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