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Chapter 7: Problem 44

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

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Si}\) and \(\mathrm{Cl}\) by assuming core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff? }}\) ? (d) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text {eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text {eff }}\) for a valence electron in \(\mathrm{P}\), phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).

Mercury in the environment can exist in oxidation states \(0,+1\), and \(+2\). One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111), but instead of using ultraviolet light to eject valence electrons, \(\mathrm{X}\) rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's \(4 f\) orbitals at \(105 \mathrm{eV}\), from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy. The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV}\), corresponding to the \(1 s\) orbital of oxygen. Overall the researchers concluded that oxidation states were \(+2\) for \(\mathrm{Hg}\) and \(-2\) for \(\mathrm{O}\). (a) Calculate the wavelength of the \(\mathrm{X}\) rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the \(1 s\) electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground-state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case? (d) Use Slater's rules to estimate \(Z_{\text {eff }}\) for the \(4 f\) and valence electrons of \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); assume for this purpose that all the inner electrons with \((n-3)\) or less screen a full \(+1\).

Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Co}^{2+}\), (b) \(\mathrm{Sn}^{2+}\), (c) \(\mathrm{Zr}^{4+}\), (d) \(\mathrm{Ag}^{+}\), (e) \(\mathrm{S}^{2-}\).

(a) Which ion is smaller, \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}^{4+}\) ? (b) In a lithium-ion battery that is discharging to power a device, for every \(\mathrm{Li}^{+}\)that inserts into the lithium cobalt oxide electrode, \(\mathrm{a} \mathrm{Co}^{4+}\) ion must be reduced to \(\mathrm{Co}^{3+}\) ion to balance charge. Using the \(C R C\) Handbook of Chemistry and Physics or other standard reference, find the ionic radii of \(\mathrm{Li}^{+}, \mathrm{Co}^{3+}\), and \(\mathrm{Co}^{4+}\). Order these ions from smallest to largest. (c) Will the lithium cobalt electrode expand or contract as lithium ions are inserted? (d) Lithium is not nearly as abundant as sodium. If sodium ion batteries were developed that function as lithium ion ones, do you think "sodium cobalt oxide" would still work as the electrode material? Explain. (e) If you don't think cobalt would work as the redox-active partner ion in the sodium version of the electrode, suggest an alternative metal ion and explain your reasoning.

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of \(\mathrm{Xe}\). Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.
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