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Based on their positions in the periodic table, predict which atom of the following pairs will have the larger first ionization energy: (a) \(\mathrm{Cl}, \mathrm{Ar} ;\) (b) Be, Ca; (c) \(\mathrm{K}, \mathrm{Co}\); (d) S, Ge; (e) Sn, Te.

Short Answer

Expert verified
Based on the positions in the periodic table, the elements with larger first ionization energy in each pair are: (a) Ar, (b) Be, (c) Co, (d) S, and (e) Te.

Step by step solution

01

Compare Ionization Energies for Cl and Ar

Looking at the periodic table, we can see that Cl and Ar are in the same period with positions: Cl (period 3, group 17) and Ar (period 3, group 18). Since they are in the same period, we just need to check their positions within that period. As the ionization energy increases from left to right across a period, Ar should have a larger first ionization energy than Cl.
02

Compare Ionization Energies for Be and Ca

Be and Ca are in the same group (group 2), but in different periods with positions: Be (period 2, group 2) and Ca (period 4, group 2). Since they are in the same group, we just need to check their positions within that group. As the ionization energy decreases from top to bottom within the same group, Be should have a larger first ionization energy than Ca.
03

Compare Ionization Energies for K and Co

K and Co are in the same period (period 4) but in different groups with positions: K (period 4, group 1) and Co (period 4, group 9). As they are in the same period, we only need to check their positions within that period. As ionization energy increases from left to right across a period, Co should have a larger first ionization energy than K.
04

Compare Ionization Energies for S and Ge

Looking at the periodic table, we find S and Ge in the same period (period 4) with positions: S (period 4, group 16) and Ge (period 4, group 14). Since they are in the same period, we only need to check their positions within that period. As ionization energy increases from left to right across a period, S should have a larger first ionization energy than Ge.
05

Compare Ionization Energies for Sn and Te

Sn and Te are in the same period (period 5) with positions: Sn (period 5, group 14) and Te (period 5, group 16). Since they are in the same period, we just need to check their positions within that period. As ionization energy increases from left to right across a period, Te should have a larger first ionization energy than Sn.
06

Summary of Results

(a) Ar has a larger first ionization energy than Cl. (b) Be has a larger first ionization energy than Ca. (c) Co has a larger first ionization energy than K. (d) S has a larger first ionization energy than Ge. (e) Te has a larger first ionization energy than Sn.

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Key Concepts

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

Periodic Table Trends
Understanding how elements behave is key to solving problems in chemistry, and the periodic table is a chemist’s roadmap. Ionization energy, which is the energy required to remove an electron from the atom in its gaseous state, exhibits notable trends in the periodic table.

Generally, ionization energy increases from left to right within a period. This happens because as we move across a period, the atomic number increases, adding more protons to the nucleus and therefore more positively charged nuclear pull. This increased pull makes it more difficult to remove an electron.

Conversely, ionization energy decreases from top to bottom within a group. As we move down a group, new electron shells are added, which means that the outer electrons are farther from the nucleus and experience a lesser nuclear attraction. Additionally, the effect of electron shielding, where inner electrons block some of the nuclear charge from reaching the outermost electrons, also contributes to this decrease.

The interplay of these trends helps predict which element in a pair might have a larger ionization energy based on its position in the periodic table.
Atomic Structure
The atomic structure of an element underlies its position in the periodic table and governs its properties, including ionization energy. Each atom consists of a dense nucleus made up of protons and neutrons, surrounded by a cloud of electrons in orbitals at various energy levels.

The arrangement of electrons in these orbitals is crucial as it impacts how tightly bound they are to the nucleus. Electrons in orbitals closer to the nucleus are more strongly attracted, making them harder to remove. As the number of protons increases, the effective nuclear charge felt by the valence electrons also increases, generally leading to higher ionization energies for elements with more protons, provided they are in the same period.

Furthermore, the distribution of electrons among the subshells within a period can lead to slight variations in ionization energies due to electron-electron repulsions in the same orbital, also known as Hund's Rule effects, which can impact the ease with which an electron can be removed.
Comparing Elements
When comparing elements to determine which has a higher ionization energy, the atomic structure and periodic table trends provide the necessary context. To improve students' abilities to predict ionization energies, consider the following focal points:
  • Atomic number (amount of protons) – Higher atomic numbers within the same period usually correspond to higher ionization energies due to increased nuclear charge.
  • Energy levels (electron shells) – Elements with more electron shells have their valence electrons further from the nucleus, which typically results in lower ionization energies.
  • Position in the periodic table – Specifically, look at whether elements are in the same period or group to apply the correct trend.
  • Electron configurations – These can cause anomalies in the general trend, such as the stability provided by full or half-full orbitals, which may result in unexpectedly higher ionization energies.

By focusing on these aspects and considering the underpinning principles of electron shielding and nuclear charge, one can make informed comparisons on the relative ionization energies of different elements.

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Most popular questions from this chapter

Little is known about the properties of astatine, \(\mathrm{At}\), 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) What is the chemical formula of the compound it forms with Na?

(a) If the core electrons were totally effective at shielding the valence electrons and the valence electrons provided no shielding for each other, what would be the effective nuclear charge acting on the \(3 s\) and \(3 p\) valence electrons in \(\mathrm{P}\) ? (b) Repeat these calculations using Slater's rules. (c) Detailed calculations indicate that the effective nuclear charge is \(5.6+\) for the 3 s electrons and \(4.9+\) for the \(3 p\) electrons. Why are the values for the \(3 s\) and \(3 p\) electrons different? (d) If you remove a single electron from a \(\mathrm{P}\) atom, which orbital will it come from? Explain.

Consider the change in effective nuclear charge experienced by a \(2 p\) electron as we proceed from \(C\) to \(N\). (a) Based on a simple model in which core electrons screen the valence electrons completely and valence electrons do not screen other valence electrons, what do you predict for the change in \(Z_{\text {eff }}\) from \(C\) to \(\mathrm{N}\) ? (b) What change do you predict using Slater's rules? (c) The actual change in \(Z_{\text {eff }}\) from \(C\) to \(N\) is \(0.70+.\) Which approach to estimating \(Z_{\text {eff }}\) is more accurate? (d) The change in \(\mathrm{Z}_{\text {eff }}\) from \(\mathrm{N}\) to \(\mathrm{O}\) is smaller than that from \(\mathrm{C}\) to \(\mathrm{N}\). Can you provide an explanation for this observation?

Identify the element whose ions have the following electron configurations: (a) a 2+ ion with \([\operatorname{Ar}] 3 d^{9}\), (b) a 1+ ion with [Xe]4f \(^{14} 5 d^{10} 6 s^{2}\). How many unpaired electrons does each ion contain?

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like a nonmetal. 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.

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