Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.21P

(a)

Interpretation Introduction

Interpretation:

The condensed electron configurations and Lewis symbols to depict the formation of ions formed from atoms Cs and S is to be determined. Also, the formula of the compound formed is to be given.

Concept introduction:

The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

Lewis electron-dot symbol is a representation employed to donate the valence electron present in the atom. It includes atom symbol to represent inner electrons and nucleus and the dots represent the valence present in the atom.

(a)

Expert Solution
Check Mark

Answer to Problem 9.21P

The condensed electronic configuration of Cs+ is [Xe].

The condensed electronic configuration of S2 is [Ne]3s23p6.

The Lewis orbital diagram is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 9, Problem 9.21P , additional homework tip  1

The formula of the compound formed is Cs2S.

Explanation of Solution

The condensed electronic configuration of a cesium atom (Cs) is [Xe]6s1.

The condensed electronic configuration of the sulfur atom (S) is [Ne]3s23p4.

Two cesium atoms lose one electron respectively to form Cs+ and to attain noble gas configuration. The two electrons lost by Cs atoms are gained by sulfur atoms to form S2 ions and to attain noble gas configuration.

The condensed electronic configuration of Cs+ is [Xe].

The condensed electronic configuration of S2 is [Ne]3s23p6.

Two cesium atoms lose one electron respectively to form Cs+ and then react with S2 ions formed from S to form cesium sulfide (Cs2S).

The Lewis orbital diagram is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 9, Problem 9.21P , additional homework tip  2

Conclusion

Cs belongs to Group 1A(1) and it loses one electron to form Cs+. S belongs to Group 6A(16) and it gains two electrons to form S2.

(b)

Interpretation Introduction

Interpretation:

The condensed electron configurations, partial orbital diagrams, and Lewis symbols to depict the formation of ions formed from atoms O and Ga is to be determined. Also, the formula of the compound formed is to be given.

Concept introduction:

The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

The partial orbital diagram is a pictorial representation of the electrons present in an orbital. Each orbital can occupy only two electrons of opposite spin.

Lewis electron-dot symbol is a representation employed to donate the valence electron present in the atom. It includes atom symbol to represent inner electrons and nucleus and the dots represent the valence present in the atom.

(b)

Expert Solution
Check Mark

Answer to Problem 9.21P

The condensed electronic configuration of Ga3+ is [Ar]3d10.

The condensed electronic configuration of O2 is [He]2s22p6.

The Lewis orbital diagram is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 9, Problem 9.21P , additional homework tip  3

The formula of the compound formed is Ga2O3.

Explanation of Solution

The condensed electronic configuration of gallium atom (Ga) is [Ar]3d104s24p1.

The condensed electronic configuration of oxygen atom (O) is [He]2s22p4.

Two gallium atoms lose three electrons respectively to form Ga3+ and to attain noble gas configuration. The six electrons lost by two Ga atoms are gained by three oxygen atoms to form O2 ions and to attain noble gas configuration.

The condensed electronic configuration of Ga3+ is [Ar]3d10.

The condensed electronic configuration of O2 is [He]2s22p6.

Two gallium atoms lose three electrons respectively to form Ga3+ and then react with three O2 ions formed from three O to form gallium oxide (Ga2O3).

The Lewis orbital diagram is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 9, Problem 9.21P , additional homework tip  4

Conclusion

Ga belongs to Group 3A(13) and it loses three electrons to form Ga3+. O belongs to Group 6A(16) and it gains two electrons to form O2.

(c)

Interpretation Introduction

Interpretation:

The condensed electron configurations, partial orbital diagrams, and Lewis symbols to depict the formation of ions formed from atoms N and Mg is to be determined. Also, the formula of the compound formed is to be given.

Concept introduction:

The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

The partial orbital diagram is a pictorial representation of the electrons present in an orbital. Each orbital can occupy only two electrons of opposite spin.

Lewis electron-dot symbol is a representation employed to donate the valence electron present in the atom. It includes atom symbol to represent inner electrons and nucleus and the dots represent the valence present in the atom.

(c)

Expert Solution
Check Mark

Answer to Problem 9.21P

The condensed electronic configuration of Mg is [Ne].

The condensed electronic configuration of N3 is [He]2s22p6.

The Lewis orbital diagram is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 9, Problem 9.21P , additional homework tip  5

The formula of the compound formed is Mg3N2.

Explanation of Solution

The condensed electronic configuration of magnesium atom (Mg) is [Ne]3s2.

The condensed electronic configuration of nitrogen atom (N) is [He]2s22p3.

Three magnesium atoms lose two electrons respectively to form Mg2+ and to attain noble gas configuration. The six electrons lost by three Mg atoms are gained by two nitrogen atoms to form N3 ions and to attain noble gas configuration.

The condensed electronic configuration of Mg is [Ne].

The condensed electronic configuration of N3 is [He]2s22p6.

Three magnesium atoms lose two electrons respectively to form Mg2+ and then react with two N3 ions formed from three N to form magnesium nitride (Mg3N2).

The Lewis orbital diagram is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 9, Problem 9.21P , additional homework tip  6

Conclusion

Mg belongs to Group 2A(2) and it loses two electrons to form Mg2+. N belongs to Group 5A(15) and it gains three electrons to form N3.

(d)

Interpretation Introduction

Interpretation:

The condensed electron configurations, partial orbital diagrams, and Lewis symbols to depict the formation of ions formed from atoms Br and Li is to be determined. Also, the formula of the compound formed is to be given.

Concept introduction:

The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

The partial orbital diagram is a pictorial representation of the electrons present in an orbital. Each orbital can occupy only two electrons of opposite spin.

Lewis electron-dot symbol is a representation employed to donate the valence electron present in the atom. It includes atom symbol to represent inner electrons and nucleus and the dots represent the valence present in the atom.

(d)

Expert Solution
Check Mark

Answer to Problem 9.21P

The condensed electronic configuration of Li+ is [He].

The condensed electronic configuration of Br is [Ar]3d104s24p6.

The Lewis orbital diagram is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 9, Problem 9.21P , additional homework tip  7

The formula of the compound formed is LiBr.

Explanation of Solution

The condensed electronic configuration of a lithium atom (Li) is [He]2s1.

The condensed electronic configuration of the bromine atom (Br) is [Ar]3d104s24p5.

Lithium atom loses one electron to form Li+ and to attain noble gas configuration. The electron is gained by the bromine atom to form Br ions and to attain noble gas configuration.

The condensed electronic configuration of Li+ is [He].

The condensed electronic configuration of Br is [Ar]3d104s24p6.

Lithium atom loses one electron to form Li+ and then react with Br ion formed from Br to form lithium bromide (LiBr).

The Lewis orbital diagram is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 9, Problem 9.21P , additional homework tip  8

Conclusion

Li belongs to Group 1A(1) and it loses one electron to form Li+. Br belongs to Group 7A(17) and it gains one electron to form Br.

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Chapter 9 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 9.5 - Arrange each set of bonds in order of increasing...Ch. 9.5 - Prob. 9.5BFPCh. 9 - Prob. 9.1PCh. 9 - Prob. 9.2PCh. 9 - What is the relationship between the tendency of a...Ch. 9 - Prob. 9.4PCh. 9 - Prob. 9.5PCh. 9 - State the type of bonding—ionic, covalent, or...Ch. 9 - State the type of bonding—ionic, covalent, or...Ch. 9 - State the type of bonding—ionic, covalent, or...Ch. 9 - Prob. 9.9PCh. 9 - Prob. 9.10PCh. 9 - Prob. 9.11PCh. 9 - Prob. 9.12PCh. 9 - Prob. 9.13PCh. 9 - Give the group number and condensed electron...Ch. 9 - Give the group number and condensed electron...Ch. 9 - Prob. 9.16PCh. 9 - Prob. 9.17PCh. 9 - Prob. 9.18PCh. 9 - Prob. 9.19PCh. 9 - Prob. 9.20PCh. 9 - Prob. 9.21PCh. 9 - Prob. 9.22PCh. 9 - Prob. 9.23PCh. 9 - Prob. 9.24PCh. 9 - Prob. 9.25PCh. 9 - For each pair, choose the compound with the larger...Ch. 9 - Prob. 9.27PCh. 9 - For each pair, choose the compound with the...Ch. 9 - Prob. 9.29PCh. 9 - Use the following to calculate of NaCl: Compared...Ch. 9 - Use the following to calculate of MgF2: Compared...Ch. 9 - Prob. 9.32PCh. 9 - Born-Haber cycles were used to obtain the first...Ch. 9 - Prob. 9.34PCh. 9 - Prob. 9.35PCh. 9 - Prob. 9.36PCh. 9 - How does the energy of the bond between a given...Ch. 9 - When liquid benzene (C6H6) boils, does the gas...Ch. 9 - Prob. 9.39PCh. 9 - Prob. 9.40PCh. 9 - Prob. 9.41PCh. 9 - Prob. 9.42PCh. 9 - The text points out that, for similar types of...Ch. 9 - Why is there a discrepancy between an enthalpy of...Ch. 9 - Which of the following gases would you expect to...Ch. 9 - Which of the following gases would you expect to...Ch. 9 - Use bond energies to calculate the enthalpy of...Ch. 9 - Prob. 9.48PCh. 9 - Prob. 9.49PCh. 9 - Prob. 9.50PCh. 9 - Prob. 9.51PCh. 9 - What is the general relationship between IE1 and...Ch. 9 - Is the H—O bond in water nonpolar covalent, polar...Ch. 9 - Prob. 9.54PCh. 9 - How is the partial ionic character of a bond in a...Ch. 9 - Using the periodic table only, arrange the...Ch. 9 - Prob. 9.57PCh. 9 - Prob. 9.58PCh. 9 - Prob. 9.59PCh. 9 - Prob. 9.60PCh. 9 - Use Figure 9.21 to indicate the polarity of each...Ch. 9 - Prob. 9.62PCh. 9 - Prob. 9.63PCh. 9 - Prob. 9.64PCh. 9 - Prob. 9.65PCh. 9 - Prob. 9.66PCh. 9 - Prob. 9.67PCh. 9 - Prob. 9.68PCh. 9 - Prob. 9.69PCh. 9 - Prob. 9.70PCh. 9 - Prob. 9.71PCh. 9 - Geologists have a rule of thumb: when molten rock...Ch. 9 - Prob. 9.73PCh. 9 - Use Lewis electron-dot symbols to represent the...Ch. 9 - Prob. 9.75PCh. 9 - Prob. 9.76PCh. 9 - By using photons of specific wavelengths, chemists...Ch. 9 - Prob. 9.78PCh. 9 - Prob. 9.79PCh. 9 - Prob. 9.80PCh. 9 - Prob. 9.81PCh. 9 - Prob. 9.82PCh. 9 - Prob. 9.83PCh. 9 - Find the longest wavelengths of light that can...Ch. 9 - The work function (ϕ) of a metal is the minimum...Ch. 9 - Prob. 9.86PCh. 9 - Prob. 9.87PCh. 9 - Prob. 9.88PCh. 9 - In a future hydrogen-fuel economy, the cheapest...Ch. 9 - Prob. 9.90PCh. 9 - Prob. 9.91P
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