/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Consider the first ionization en... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 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?

Short Answer

Expert verified
(a) Equations with electron configurations: - Ionization energy of Ne: Ne(\([He]2s^22p^6\)) -> Ne+(\([He]2s^22p^5\)) + e^- - Electron affinity of F: F(\([He]2s^22p^5\)) + e^- -> F-(\([He]2s^22p^6\)) (b) Signs of quantities: - First ionization energy of Ne is positive. - Electron affinity of F is negative. (c) Magnitude comparison: - The first ionization energy of Ne is expected to be larger than the electron affinity of F, due to the larger difficulty in removing an electron from Ne's completely filled and stable 2p orbital compared to adding an electron to F's almost-filled 2p orbital.

Step by step solution

01

(a) Writing Equations with Electron Configurations

For the ionization energy of neon (Ne), we have to remove one electron from a neutral Ne atom, creating a positive ion. The electron configuration of a neutral Ne atom is \([He]2s^22p^6\). Its first ionization energy equation and electron configurations will be: Ne(\([He]2s^22p^6\)) -> Ne+(\([He]2s^22p^5\)) + e^- For the electron affinity of fluorine (F), we need to add an electron to a neutral F atom, creating a negative ion. The electron configuration of a neutral F atom is \([He]2s^22p^5\). Its electron affinity equation and electron configurations will be: F(\([He]2s^22p^5\)) + e^- -> F-(\([He]2s^22p^6\))
02

(b) Determining the Signs of the Quantities

Ionization energy is the energy needed to remove an electron, and it's always a positive value since the process requires energy input. So, the first ionization energy of neon will be positive. Electron affinity is the energy released when an electron is added to an atom, and it's usually negative since energy is released during the process. The electron affinity value of fluorine will be negative.
03

(c) Comparison of Magnitudes and Expectations

In general, the magnitudes of ionization energy and electron affinity can be different due to atomic structure and electron-electron repulsion forces. In this case, we're comparing the first ionization energy of neon and electron affinity of fluorine. Neon has a completely filled 2p orbital, making it very stable, and its ionization energy is expected to be high. Fluorine, on the other hand, has almost a completely filled 2p orbital and tends to gain an electron to achieve a stable electron configuration (similar to Ne). Therefore, fluorine has a high electron affinity. However, the first ionization energy of neon is expected to be larger than the electron affinity of fluorine. This is because removing an electron from the completely filled and stable 2p orbital of neon is more difficult compared to adding an electron to the almost-filled 2p orbital of fluorine.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Electron Configuration
Atoms are surrounded by electrons, which are arranged in specific orbitals around the nucleus. These arrangements are known as the electron configuration of an atom. The electron configuration helps to understand an atom's chemical behavior, its reactivity, and its placement in the periodic table.

Neon's electron configuration is particularly interesting as it is considered a noble gas. The configuration for neon (Ne) is \([He]2s^22p^6\). This notation indicates that neon has two electrons in the 2s subshell and six in the 2p subshell. This results in a stable configuration because all its available orbitals in the second shell are completely filled.

In contrast, fluorine (F) has the configuration \([He]2s^22p^5\). Here, you can notice it has one less electron in the 2p subshell compared to neon, which is why it's eager to gain an electron to achieve a stable electron configuration similar to that of neon. By understanding electron configurations, we can predict how and why substances like fluorine are highly reactive while neon remains inert.
Electron Affinity
Electron affinity refers to the amount of energy released when an atom in the gaseous state gains an extra electron. This process often results in the formation of a negative ion. Unlike ionization energy, which requires energy to remove an electron, electron affinity involves the energy change associated with the addition of an electron.

Fluorine is a great example of an element with high electron affinity. Its electron configuration \([He]2s^22p^5\) suggests a strong tendency to gain an electron to fill its 2p subshell completely. When an electron is added to fluorine, creating F\(^-\), its configuration becomes \([He]2s^22p^6\), similar to neon's stable configuration. This process releases energy, and thus fluorine has a negative electron affinity value.

Understanding electron affinity is crucial as it explains why certain elements, like fluorine, are more reactive. Highly negative electron affinities typically mean the element readily accepts electrons, which is why fluorine is one of the most electronegative elements known.
Neon
Neon is a noble gas element, characterized by its complete valence shell, making it extremely stable and chemically inert. The electron configuration of neon \([He]2s^22p^6\) clearly shows that all electrons occupy a full set of orbitals, underlining its stability and lack of reactivity.

Neon's lack of reactivity is mainly due to its full outer shell, making it energetically unfavorable to gain or lose electrons. This is why neon and other noble gases do not easily form chemical compounds or reactions under standard conditions.

When it comes to ionization energy, neon requires a considerable amount of energy to remove one of its electrons. Its high ionization energy is reflective of the strong attraction between its nucleus and the outer electrons. Unlike fluorine, which is eager to gain electrons, neon's stability means it's content staying in its current form, which explains its widespread use in lighting and displays, where it's known for creating bright, colorful lighting when electrically excited.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

Elemental barium reacts more violently with water than does elemental calcium. Which of the following best explains this difference in reactivity? (i) Calcium has greater metallic character than does barium. (ii) The electron affinity of calcium is smaller than that of barium. (iii) The first and second ionization energies of barium are less than those of calcium. (iv) The atomic radius of barium is smaller than that of calcium. (v) The ionic radius of the barium ion is larger than that of the calcium ion.

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 Xe. Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

If the electron affinity for an element is a negative number, does it mean that the anion of the element is more stable than the neutral atom? Explain.

Which neutral atom is isoelectronic with each of the following ions? \(\mathrm{H}^{-}, \mathrm{Ca}^{2+}, \mathrm{In}^{3+}, \mathrm{Ge}^{2+}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.