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a) Which of the elements \(\mathrm{Be}, \mathrm{N}, \mathrm{O}\), and \(\mathrm{F}\) has the most negative electron affinity? Explain. (b) Which of the ions \(\mathrm{Se}^{2-}, \mathrm{F}, \mathrm{O}^{2-}\), and \(\mathrm{Rb}^{+}\) has the largest radius? Explain.

Short Answer

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(a) F has the most negative electron affinity. (b) Se^{2-} has the largest radius.

Step by step solution

01

Understanding Electron Affinity

Electron affinity refers to the energy change when an electron is added to a gaseous atom. Elements with a more negative electron affinity more readily accept an electron.
02

Determine Electron Affinity of Given Elements

The elements given are Be, N, O, and F. Typically, electron affinity becomes more negative across a period. However, exceptions occur, like between N and O. Among these, F has the most negative electron affinity because it has a high tendency to accept an electron due to its high electronegativity and electron affinity trend in halogens.
03

Understanding Ionic Radii

Ionic radius refers to the size of an ion. When an atom gains electrons to become an anion, its size increases, whereas it decreases when it loses electrons to become a cation.
04

Comparing Ionic Radii of Given Ions

The ions given are Se^{2-}, F, O^{2-}, and Rb^{+}. Typically, anions (negative ions) are larger than cations (positive ions) because of additional electron-electron repulsion and electron addition to outer shells.
05

Determine the Largest Ion by Radius

Among the given ions, O^{2-} and Se^{2-} are anions, while Rb^{+} is a cation and F remains a neutral atom. Se^{2-} is likely the largest due to it having more electron shells and being a negative ion, increasing its electron cloud.

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Key Concepts

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

Ionic Radius
An ion's radius, or ionic radius, is essentially a measure of its size. The ionic radius can change significantly depending on whether an ion is an anion or a cation.
  • Anions, which are negatively charged ions, increase in size compared to their neutral atoms. This is because they have gained additional electrons, augmenting electron-electron repulsion which causes the electron cloud to expand.
  • Cations, on the other hand, are positively charged ions that have lost one or more electrons. As a result, they generally have a smaller radius because the loss of electrons reduces the electron-electron repulsion, enabling the positively charged nucleus to pull the remaining electrons closer.
For example, the ion Se\(^{2-}\), which has gained electrons, has a larger ionic radius than the neutral F atom or the Rb\(^{+}\) cation, illustrating how electron gain and outer shell occupation lead to size changes.
Electron Affinity Trend
Electron affinity refers to the energy released when an electron is added to a neutral gaseous atom, forming a negative ion. This concept plays a key role in understanding why certain elements more effectively form anions. Typically, electron affinity becomes more negative as we move across a period from left to right in the periodic table. This trend is due to increasing nuclear charge with the same number of electron shells, pulling electrons more strongly toward the nucleus. However, there are some exceptions. For instance, while one might expect nitrogen to have a higher electron affinity than oxygen, it's actually lower due to extra stability from its half-filled p orbitals. In the case of our given elements, fluorine (F) possesses the most negative electron affinity, making it eager to gain an electron, as it's located at the far right of its period under the halogens, all of which are highly electronegative.
Anions and Cations
Anions and cations are types of ions that occur through either the gain or loss of electrons. Understanding them is crucial in appreciating how ionic compounds are formed and their overall characteristics.
  • Anions are negative ions formed when atoms gain electrons. This addition causes the atom to have more electrons than protons, resulting in a negative charge. Common examples include O\(^{2-}\) and Se\(^{2-}\), which gain electrons to fulfill their electron shells, becoming more stable and increasing in radius.
  • Cations, in contrast, occur when atoms lose electrons, resulting in a positive charge. They have fewer electrons, lowering electron-electron repulsion and allowing the remaining electrons to be drawn closer to the nucleus. An example is Rb\(+\), which loses electrons and exhibits a smaller size.
The interplay of these oppositely charged ions leads to the formation of ionic compounds, held together by the electrostatic force between them.
Electron-Electron Repulsion
Electron-electron repulsion is an essential concept that greatly influences atomic and ionic sizes. When two electrons are added to an orbital or an electron cloud, their like charges repel each other, leading to a spread out or expanded electron cloud.This principle is especially relevant in the context of anions. When an atom becomes an anion by gaining electrons, these added electrons increase the repulsion among the electrons already present, leading to an enlargement of the ion’s radius. It explains why larger anionic radii, such as that of Se\(^{2-}\), expand more significantly compared to their parent atoms or to cations. Thus, electron-electron repulsion not only affects the size of anions but also explains how electron affinity varies, as it modifies how comfortably added electrons are accommodated within an atom's existing electron cloud.

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Most popular questions from this chapter

What is the electron configuration of \(\mathrm{Ca}^{2+}\) ? What is the electron configuration of \(\mathrm{Ti}^{2+}\) ?

Which atom or ion in each of the following pairs would you expect to be larger? (a) \(\mathrm{O}\) or \(\mathrm{O}^{2-}\) (b) Fe or \(\mathrm{Fe}^{3+}\) (c) \(\mathrm{H}\) or \(\mathrm{H}^{-}\)

Given the following information, construct a Born-Haber cycle to calculate the lattice energy of \(\mathrm{CrCl}_{2} \mathrm{I}(s)\) : Net energy change for the formation of \(\mathrm{CrCl}_{2} \mathrm{I}(s)=-420 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{Cl}_{2}(g)=+243 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{I}_{2}(s)=+151 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{I}_{2}(s)=+62 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Cr}(s)=+397 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{il}}\) for \(\mathrm{Cr}(g)=652 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{i} 2}\) for \(\mathrm{Cr}(g)=1588 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{i} 3}\) for \(\mathrm{Cr}(\mathrm{g})=2882 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{ca}}\) for \(\mathrm{Cl}(g)=-349 \mathrm{~kJ} / \mathrm{mol}\) \(E_{e a}\) for \(\mathrm{I}(g)=-295 \mathrm{~kJ} / \mathrm{mol}\)

What structural features do ionic liquids have that prevent them from forming solids easily?

The following ions all have the same number of electrons: \(\mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-} .\) Order them according to their expected sizes, and explain your answer. 6.88 Calculate overall energy changes in kilojoules per mole for the formation of \(\mathrm{MgF}(s)\) and \(\mathrm{MgF}_{2}(s)\) from their elements. In light of your answers, which compound is more likely to form in the reaction of magnesium with fluorine, \(\mathrm{MgF}\) or \(\mathrm{MgF}_{2}\) ? The following data are needed: \(E_{\mathrm{ea}}\) for \(\mathrm{F}(\mathrm{g})=-328 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\mathrm{Mg}(g)=+737.7 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Mg}(g)=+1450.7 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Mg}(s)=+147.7 \mathrm{~kJ} / \mathrm{mol}\)Bond dissociation energy for \(\mathrm{F}_{2}(\mathrm{~g})=+158 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{MgF}_{2}(s)=+2952 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{MgF}(s)=930 \mathrm{~kJ} / \mathrm{mol}\) (estimated)

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