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Chapter 7: Problem 55

Consider the first ionization energy of neon and the electron affinity of fluorine. (a) Write equations, including electron configurations, for each process. (b) These two quantities will have opposite signs. Which will be positive, and which will be negative? (c) Would you expect the magnitudes of these two quantities to be equal? If not, which one would you expect to be larger? Explain your answer.

Short Answer

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(a) Equations and electron configurations: - First ionization energy of Ne: Ne (g) \( \rightarrow \) Ne\(^+\) (g) + e\(^-\); Ne: \(1s^2 2s^2 2p^6\) - Electron affinity of F: F (g) + e\(^-\) \( \rightarrow \) F\(^-\) (g); F: \(1s^2 2s^2 2p^5\) (b) Neon's first ionization energy is positive and fluorine's electron affinity is negative. (c) The magnitudes are not equal; neon's first ionization energy is larger due to its complete and stable electron configuration, while fluorine's electron affinity is smaller due to its one-electron-short configuration.

Step by step solution

01

(a) Writing the equations and electron configurations

The first ionization energy refers to the energy required to remove an electron from a neutral atom, while the electron affinity refers to the energy change that occurs when an electron is added to the atom. Thus, for the first ionization energy of neon, we can write: Ne (g) \( \rightarrow \) Ne\(^+\) (g) + e\(^-\) Electron configuration of Ne: \(1s^2 2s^2 2p^6\) For the electron affinity of fluorine, we can write: F (g) + e\(^-\) \( \rightarrow \) F\(^-\) (g) Electron configuration of F: \(1s^2 2s^2 2p^5\)
02

(b) Determining the sign of the quantities

The first ionization energy is always a positive value because energy is required to remove an electron from an atom. Thus, the first ionization energy of neon will be positive. The electron affinity can be either positive or negative, depending on whether energy is released or absorbed when an electron is added. In the case of fluorine, energy is released when an electron is gained because it completes the stable electron configuration. Therefore, the electron affinity of fluorine will be negative.
03

(c) Comparing the magnitudes of the quantities and providing an explanation

We wouldn't expect the magnitudes of these two quantities to be equal. This is because they depend on different factors, such as the effective nuclear charge experienced by the electron and the electron shielding effect. The first ionization energy of neon would be expected to be larger than the electron affinity of fluorine. Neon has a complete and stable electron configuration, which means it takes a significant amount of energy to remove an electron and disrupt its stability. On the other hand, fluorine has an electron configuration that is one electron short of being complete, so when an electron is added, the energy released is not as large as the energy needed to ionize neon. In conclusion, the first ionization energy of neon is larger than the electron affinity of fluorine, and the first ionization energy is positive while the electron affinity is negative.

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Key Concepts

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

Electron Affinity
Electron affinity is a crucial concept in understanding chemical reactivity and trends across the periodic table. It measures the energy change when an electron is added to a neutral atom in the gaseous state. Typically, if energy is released during this process, the electron affinity is considered to be negative, suggesting that the atom has a tendency to gain an electron.

For instance, halogens like fluorine have high electron affinities because adding an electron fills their outermost p-orbital, leading to a more stable electron configuration. This release of energy when an atom like fluorine accepts an electron is indicative of the atom's high reactivity and its strong desire to attain a noble gas configuration. Subsequently, when solving problems involving electron affinity, it is essential to consider not only the energy change but also the resulting electron configuration.

This clarification helps students understand why, in some cases like with noble gases, the electron affinity can be positive since they already have a full valence shell and do not naturally attract additional electrons.
Electron Configuration
Electron configuration is the arrangement of electrons in an atom's orbitals and is pivotal for predicting an atom's chemical behavior. In the context of first ionization energy and electron affinity, electron configuration helps explain why certain atoms require more energy to remove an electron or why they release energy when gaining one. For example, the noble gas neon has a full outer shell with the electron configuration of 1s2 2s2 2p6, making it stable and less prone to losing an electron.

In contrast, atoms like fluorine have a high affinity for electrons due to their one-electron short of a full outer shell configuration of 1s2 2s2 2p5. Understanding electron configuration is fundamental when conceptualizing why elements react in certain ways, and how they either release or absorb energy during ionization processes.
Chemical Stability
Chemical stability refers to the likelihood of an atom or compound to maintain its original structure and resist changes or reactions under specified conditions. It is closely tied to an element's electron configuration. Atoms with a full valence shell, such as noble gases, are generally more chemically stable because their electron configuration does not lend itself easily to gaining or losing electrons. This explains why the ionization energy for neon is quite high; a stable atom like neon would require a substantial amount of energy to remove an electron.

Meanwhile, atoms that are one electron away from a full valence shell, such as fluorine, are more reactive but can achieve chemical stability by gaining that one electron, as reflected in their negative electron affinity. Understanding the direct relationship between electron configuration and chemical stability assists in comprehending why atoms undergo certain reactions and the energy implications of those reactions.

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Most popular questions from this chapter

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S,\) calculate \(Z_{\mathrm{eff}}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, S. (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

(a) If the core electrons were totally effective at screening the valence electrons and the valence electrons provided no screening 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.

Make a simple sketch of the shape of the main part of the periodic table, as shown. (a) Ignoring \(\mathrm{H}\) and He, write a single straight arrow from the element with the smallest bonding atomic radius to the element with the largest. Ignoring \(\mathrm{H}\) and He, write a single straight arrow from the element with the smallest first ionization energy to the element with the largest. (c) What significant observation can you make from the arrows you drew in parts (a) and (b)? [Sections 7.3 and 7.4]

It is possible to define metallic character as we do in this book and base it on the reactivity of the element and the ease with which it loses electrons. Alternatively, one could measure how well electricity is conducted by each of the elements to determine how "metallic" the elements are. On the basis of conductivity, there is not much of a trend in the periodic table: Silver is the most conductive metal, and manganese the least. Look up the first ionization energies of silver and manganese; which of these two elements would you call more metallic based on the way we define it in this book?

Some metal oxides, such as \(\mathrm{Sc}_{2} \mathrm{O}_{3},\) do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect \(\mathrm{Sc}_{2} \mathrm{O}_{3}\) to react when the solution becomes acidic or when it becomes basic? Write a balanced chemical equation to support your answer.
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