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For each of the following pairs, indicate which element has the smaller first ionization energy: (a) Ti, Ba; (b) \(\mathrm{Ag}, \mathrm{Cu}\); (c) \(\mathrm{Ge}, \mathrm{Cl}\); (d) \(\mathrm{Pb}, \mathrm{Sb}\).

Short Answer

Expert verified
In the given pairs, the elements with the smaller first ionization energy are: (a) Ba (Barium) (b) Ag (Silver) (c) Ge (Germanium) (d) Pb (Lead)

Step by step solution

01

(a) Comparing Ti and Ba

Titanium (Ti) has an atomic number of 22 and is in Period 4 and Group 4 of the periodic table, whereas Barium (Ba) has an atomic number of 56 and is in Period 6 and Group 2. Since Ba is located towards the left and down from Ti in the periodic table, we can conclude that Ba has a larger atomic radius. The larger atomic radius of Ba results in weaker electrostatic attraction between its nucleus and outer electrons, which means it will have a smaller first ionization energy than Ti. Smaller first ionization energy: Ba
02

(b) Comparing Ag and Cu

Silver (Ag) has an atomic number of 47 and is in Period 5 and Group 11, whereas Copper (Cu) has an atomic number 29 and is in Period 4 and Group 11. They are both in the same group, meaning they have the same number of valence electrons and similar electronic configurations. However, Ag is in a higher period, which means it has a larger atomic radius. The larger atomic radius of Ag results in weaker electrostatic attraction between its nucleus and outer electrons, which means it will have a smaller first ionization energy than Cu. Smaller first ionization energy: Ag
03

(c) Comparing Ge and Cl

Germanium (Ge) has an atomic number of 32 and is in Period 4 and Group 14, whereas Chlorine (Cl) has an atomic number of 17 and is in Period 3 and Group 17. Since Cl is to the right and up from Ge in the periodic table, we can conclude that Cl has a smaller atomic radius. The smaller atomic radius of Cl results in stronger electrostatic attraction between its nucleus and outer electrons, which means it will have a larger first ionization energy than Ge. Smaller first ionization energy: Ge
04

(d) Comparing Pb and Sb

Lead (Pb) has an atomic number of 82 and is in Period 6 and Group 14, whereas Antimony (Sb) has an atomic number of 51 and is in Period 5 and Group 15. Pb is located beneath Sb in the periodic table, which means Pb has a larger atomic radius. The larger atomic radius of Pb results in weaker electrostatic attraction between its nucleus and outer electrons, which means it will have a smaller first ionization energy than Sb. Smaller first ionization energy: Pb

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atomic Radius
The atomic radius of an element is the distance from the nucleus of an atom to the outermost electrons. It is a key factor in understanding various chemical behaviors, such as ionization energy.
Atomic radius generally increases as you move down a group in the periodic table. This is because new electron shells are added, making the atoms larger. As you move across a period from left to right, the atomic radius decreases.
This decrease is due to protons being added to the nucleus, which increases the positive charge and pulls the electrons closer to the nucleus, thereby reducing the size.
  • Larger atomic radii are found in elements to the left and at the bottom of the periodic table.
  • Elements with larger atomic radii tend to have lower ionization energies because their outer electrons are farther from the nucleus.
Understanding the atomic radius helps in predicting how strongly an atom holds onto its outermost electrons, thus influencing its first ionization energy. For example, Barium (Ba) has a larger atomic radius than Titanium (Ti) and therefore a smaller ionization energy.
Periodic Table
The periodic table is a powerful tool that arranges all the known elements in an informative grid. It not only organizes the elements by increasing atomic number but also groups them by similar chemical properties.
The vertical columns are called groups, and they contain elements with similar valence electron configurations. This means elements within the same group display similar behavior chemically.
Horizontal rows are called periods. As you move across a period from left to right, the number of protons in the nucleus increases, resulting in a stronger attraction between the nucleus and the electrons.
  • Moving down a group, each element has an additional electron shell, making the atomic radius larger.
  • Elements in the same period have increasing atomic numbers and generally decreasing atomic radii.
The periodic table helps in predicting trends such as atomic size, ionization energies, and more. For example, comparing Germanium (Ge) and Chlorine (Cl), Cl has a smaller atomic radius because it is positioned to the right and above Ge.
Electrostatic Attraction
Electrostatic attraction is the force that holds the electrons close to the nucleus in an atom. This attraction is due to the opposite charges of the positively charged nucleus and the negatively charged electrons.
The strength of this attraction impacts how much energy is needed to remove an electron from an atom, known as ionization energy.
When the atomic radius is large, electrons are farther from the nucleus and experience weaker electrostatic attraction. Conversely, smaller atomic radii lead to stronger attraction, as electrons are closer to the nucleus.
  • Weaker electrostatic attraction means lower ionization energy, as less energy is needed to remove an outer electron.
  • Stronger attraction requires more energy for electron removal, resulting in higher ionization energy.
Understanding electrostatic attraction is essential for explaining why elements like Lead (Pb) have a smaller first ionization energy compared to Antimony (Sb), because Pb has a larger atomic radius resulting in weaker electrostatic attraction.

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Most popular questions from this chapter

Little is known about the properties of astatine, At, because of its rarity and high radioactivity. Nevertheless, it is possible for us to make many predictions about its properties. (a) Do you expect the element to be a gas, liquid, or solid at room temperature? Explain. (b) Would you expect At to be a metal, nonmetal, or metalloid? Explain. (c) What is the chemical formula of the compound it forms with \(\mathrm{Na}\) ?

Consider \(\mathrm{S}, \mathrm{Cl}\), and \(\mathrm{K}\) and their most common ions. (a) List the atoms in order of increasing size. (b) List the ions in order of increasing size. (c) Explain any differences in the orders of the atomic and ionic sizes.

Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Ru}^{3+}\), (b) \(\mathrm{As}^{3-}\), (c) \(\mathrm{Y}^{3+}\), (d) \(\mathrm{Pd}^{2+}\), (e) \(\mathrm{Pb}^{2+}\), (f) \(\mathrm{Au}^{3+}\).

The As - As bond length in elemental arsenic is \(2.48 \AA\). The \(\mathrm{Cl}-\mathrm{Cl}\) bond length in \(\mathrm{Cl}_{2}\) is \(1.99 \AA\). (a) Based on these data, what is the predicted As - Cl bond length in arsenic trichloride, \(\mathrm{AsCl}_{3}\), in which each of the three \(\mathrm{Cl}\) atoms is bonded to the As atom? (b) What bond length is predicted for \(\mathrm{AsCl}_{3}\), using the atomic radii in Figure 7.7?

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds. If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}\). Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

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