Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.1P

(a)

Interpretation Introduction

Interpretation:

The way by which the ionization energy of the main group elements influences their metallic character is to be determined.

Concept introduction:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.

(a)

Expert Solution
Check Mark

Answer to Problem 9.1P

The metallic character of the main group elements decreases with an increase in ionization energy.

Explanation of Solution

On moving across the period in the periodic table, the size of the atoms decreases. Thus the outermost electrons in the atom are closer to the nucleus and are thus harder to be pulled out of the atom. Metals have the specific property of losing electrons. The easier it is for an element to lose an electron, the more is the metallic character of the element. Since a large value of the ionization energy implies more difficulty in extracting an electron from an atom, it thus also indicates a low metallic character as well. Hence, with an increase in the ionization energy of an element, the metallic character decreases.

Conclusion

The metallic character of the main group elements decreases with an increase in ionization energy.

(b)

Interpretation Introduction

Interpretation:

The way by which the atomic radius of the main group elements influences their metallic character is to be determined.

Concept introduction:

The atomic radius of an element is defined as the distance of the outermost electron in the atom from its nucleus.

The types of atomic radii are as follows:

1) Covalent radius – Covalent radius is calculated as one half of the distance of the two atoms of the same element that are covalently bonded to each other.

2) Van der Waals radius – Van der Waals radius is calculated as one half the distance between two nuclei of two atoms of the same element that are not bonded to each other.

3) Metallic radius – Metallic radius is calculated as one half the distance between the nuclei of two metallic atoms or ions in the metallic lattice.

(b)

Expert Solution
Check Mark

Answer to Problem 9.1P

The metallic character of the main group elements increases with an increase in the atomic radius.

Explanation of Solution

In the periodic table, on moving across the period, the radius of the elements decreases. As the radius decreases, the distance of the outermost electrons from the nucleus of the atom decreases. At a smaller distance from the nucleus, the outermost electrons experience greater forces of attraction from the nucleus and hence are harder to be knocked out of the atom. The atoms of an element have a greater metallic character if they can lose their outermost electrons easily. Hence with an increase in the atomic radius of an element, the metallic character increases.

Conclusion

The metallic character of the main group elements increases with an increase in the atomic radius.

(c)

Interpretation Introduction

Interpretation:

The way by which the number of outer electrons of the main group elements influences their metallic character is to be determined.

Concept introduction:

The atomic radius of an element is defined as the distance of the outermost electron in the atom from its nucleus.

The types of atomic radii are as follows:

1) Covalent radius – Covalent radius is calculated as one half of the distance of the two atoms of the same element that are covalently bonded to each other.

2) Van der Waals radius – Van der Waals radius is calculated as one half the distance between two nuclei of two atoms of the same element that are not bonded to each other.

3) Metallic radius – Metallic radius is calculated as one half the distance between the nuclei of two metallic atoms or ions in the metallic lattice.

(c)

Expert Solution
Check Mark

Answer to Problem 9.1P

The metallic character decreases with an increase in the number of outermost electrons on moving across a period in the periodic table.

Explanation of Solution

While moving across a period from left to right in the periodic table, the radius of the elements decreases. This happens because the increase in the number of electrons and the protons is the same, whereas on moving down a group in the periodic table, the outermost electrons due to electron shielding experience much lesser nuclear charge and hence are easily knocked out.

Thus while moving across a period, with the increase in the number of outermost electrons, the metallic character decreases due to a decrease in the atomic radius and hence an increase in the ionization potential.

Conclusion

The metallic character decreases with an increase in the number of outermost electrons on moving across a period in the periodic table.

(d)

Interpretation Introduction

Interpretation:

The way by which the effective nuclear charge of the main group elements influences their metallic character is to be determined.

Concept introduction:

The effective nuclear charge is the net nuclear charge an electron in an atom experiences. The electrons at the outermost orbitals experience lesser nuclear charge compared to the electrons in the inner orbitals. Thus the inner electrons shield the outer electrons from the attractive forces of the atomic nucleus.

The effective nuclear charge is calculated as follows:

Zeff=ZS

Here,

Zeff is the effective nuclear charge experienced by an electron.

Z is the atomic number.

S is the shielding constant.

(d)

Expert Solution
Check Mark

Answer to Problem 9.1P

The metallic character of an element decreases with an increase in the effective nuclear charge.

Explanation of Solution

In an atom, as the effective nuclear charge experienced by the outermost electrons increases, the electrons experience more attraction from the nucleus. The electrons experiencing greater nuclear charge are more firmly held in the atom and are thus harder to be knocked out. Elements, in which the outermost electrons are difficult to be knocked out, have decreased metallic character. Therefore, an increase in the effective nuclear charge decreases the metallic character.

Conclusion

The metallic character of an element decreases with an increase in the effective nuclear charge.

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Chapter 9 Solutions

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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