Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.20P

(a)

Interpretation Introduction

Interpretation:

The condensed electron configurations and Lewis symbols to depict the formation of ions formed from atoms Ba and Cl 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 9.20P

The condensed electronic configuration of Ba2+ is [Xe].

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

The Lewis orbital diagram is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.20P , additional homework tip  1

The formula of the compound formed is BaCl2.

Explanation of Solution

The condensed electronic configuration of a barium atom (Ba) is [Xe]6s2.

The condensed electronic configuration of the chlorine atom (Cl) is [Ne]3s23p5.

Barium atom loses two electrons to form Ba2+ and to attain noble gas configuration. The two electrons are gained by two chlorine atoms to form two Cl ions and attain noble gas configuration.

The condensed electronic configuration of Ba2+ is [Xe].

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

Barium atom loses two electrons to form Ba2+ and then react with two Cl ions formed from Cl to form barium chloride (BaCl2).

The Lewis orbital diagram is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.20P , additional homework tip  2

Conclusion

Ba belongs to Group 2A(2) and it loses two electrons to form Ba2+. Cl belongs to Group 7A(17) and it gains one electron to form Cl.

(b)

Interpretation Introduction

Interpretation:

The condensed electron configurations, partial orbital diagrams, and Lewis symbols to depict the formation of ions formed from atoms Sr and O 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 9.20P

The condensed electronic configuration of Sr2+ is [Kr].

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

The Lewis orbital diagram is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.20P , additional homework tip  3

The formula of the compound formed is SrO.

Explanation of Solution

The condensed electronic configuration of a strontium atom (Sr) is [Kr]5s2.

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

Strontium atom loses two electrons to form Sr2+ and to attain noble gas configuration. The two electrons are gained by the oxygen atoms to form O2 ions and to attain noble gas configuration.

The condensed electronic configuration of Sr2+ is [Kr].

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

Strontium atom loses two electrons to form Sr2+ and then react with O2 ions formed from O to form strontium oxide (SrO).

The Lewis orbital diagram is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.20P , additional homework tip  4

Conclusion

Sr belongs to Group 2A(2) and it loses two electrons to form Sr2+. 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 Al and F 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 9.20P

The condensed electronic configuration of Al3+ is [Ne].

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

The Lewis orbital diagram is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.20P , additional homework tip  5

The formula of the compound formed is AlF3.

Explanation of Solution

The condensed electronic configuration of an aluminium atom (Al) is [Ne]3s23p1.

The condensed electronic configuration of the fluorine atom (F) is [He]2s22p5.

Aluminium atom loses three electrons to form Al3+ and to attain noble gas configuration. The three electrons are gained by the three fluorine atoms to form F ions and to attain noble gas configuration.

The condensed electronic configuration of Al3+ is [Ne].

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

Aluminium atom loses three electrons to form Al3+ and then react with three F ions formed from F to form aluminium fluoride (AlF3).

The Lewis orbital diagram is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.20P , additional homework tip  6

Conclusion

Al belongs to Group 3A(13) and it loses three electrons to form Al3+. F belongs to Group 7A(17) and it gains one electron to form F.

(d)

Interpretation Introduction

Interpretation:

The condensed electron configurations, partial orbital diagrams, and Lewis symbols to depict the formation of ions formed from atoms Rb and O 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 9.20P

The condensed electronic configuration of Rb+ is [Kr].

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

The Lewis orbital diagram is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.20P , additional homework tip  7

The formula of the compound formed is Rb2O.

Explanation of Solution

The condensed electronic configuration of a rubidium atom (Rb) is [Kr]5s1.

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

Two rubidium atoms lose one electron respectively to form Rb+ and to attain noble gas configuration. The two electrons lost by Rb atoms are gained by oxygen atoms to form O2 ions and to attain noble gas configuration.

The condensed electronic configuration of Rb+ is [Kr].

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

Two rubidium atoms lose one electron respectively to form Rb+ and then react with O2 ions formed from O to form rubidium oxide (Rb2O).

The Lewis orbital diagram is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 9, Problem 9.20P , additional homework tip  8

Conclusion

Rb belongs to Group 1A(1) and it loses one electron to form Rb+. O belongs to Group 6A(16) and it gains two electrons to form O2.

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Chapter 9 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 9.5 - Prob. 9.5AFPCh. 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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