(a)
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)

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

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

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.
The metallic character decreases with an increase in the number of outermost electrons on moving across a period in the periodic table.
(d)
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:
Here,
(d)

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.
The metallic character of an element decreases with an increase in the effective nuclear charge.
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Chapter 9 Solutions
CHEM 212:CHEMISTSRY V 2
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