/*! 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 55 Consider the first ionization en... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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

Expert verified
The ionization energy of neon is positive and has a larger magnitude than the electron affinity of fluorine, which is negative. This is because neon has a very stable electron configuration, making it more difficult to remove an electron, while fluorine has a high electron affinity as it is one electron short of obtaining a stable electron configuration.

Step by step solution

01

Write the electron configuration for neon and fluorine

First, we need to write the electron configurations for neon and fluorine. Check the periodic table and find out the atomic numbers of neon and fluorine. The atomic number of neon (Ne) is 10, so its electron configuration is: \(1s^2 2s^2 2p^6\) The atomic number of fluorine (F) is 9, so its electron configuration is: \(1s^2 2s^2 2p^5\)
02

Write the equations for Ionization Energy and Electron Affinity processes

To find the first ionization energy of neon, we are removing one electron from the neutral atom. The equation for this process is: Ne(\(1s^2 2s^2 2p^6\)) 鈫� Ne\(^+\)(\(1s^2 2s^2 2p^5\)) + e\(^-\) For the electron affinity of fluorine, we are adding one electron to the neutral atom. The equation for this process is: F(\(1s^2 2s^2 2p^5\)) + e\(^-\) 鈫� F\(^-\)(\(1s^2 2s^2 2p^6\))
03

Determine the sign for Ionization Energy and Electron Affinity

When energy is required to remove an electron, the Ionization Energy is positive. When adding an electron the energy is released; hence, the Electron Affinity is negative. Therefore, the ionization energy of neon is positive, while the electron affinity of fluorine is negative.
04

Compare the magnitudes of Ionization Energy and Electron Affinity

We can expect that the magnitude of the ionization energy of neon and the electron affinity of fluorine won't be equal. This is because Neon is a noble gas with a completely filled electron shell (2p subshell is full), which makes it very stable, and it requires more energy to remove one electron from this stable configuration. Conversely, fluorine is one electron short of obtaining a stable electron configuration (2p subshell needs one more electron to become full). So, fluorine has a high electron affinity due to its high effective nuclear charge, which attracts and stabilizes the added electron. However, it will not be the same as the ionization energy of neon. So, the ionization energy of neon has a larger magnitude than the electron affinity of fluorine. In conclusion, the ionization energy of neon is positive and has a larger magnitude than the electron affinity of fluorine, which is negative. The ionization energy is larger because neon's electron configuration is very stable, making it more challenging to remove an electron.

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 91影视!

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

(a) What is the trend in first ionization energies as one proceeds down the group 7 A elements? Explain how this trend relates to the variation in atomic radii. (b) What is the trend in first ionization energies as one moves across the fourth period from \(\mathrm{K}\) to \(\mathrm{Kr}\) ? How does this trend compare with the trend in atomic radii?

An element \(X\) reacts with \(\mathrm{F}_{2}(g)\) to form the molecular product shown below. (a) Write a balanced equation for this reaction (do not worry about the phases for \(X\) and the product). (b) Do you think that \(X\) is a metal or nonmetal? Explain. [Section 7.6]

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Potassium metal burns in an atmosphere of chlorine gas, (b) Strontium oxide is added to water. (c) A fresh surface of lithium metal is exposed to oxygen gas. (d) Sodium metal is reacted with molten sulfur.

(a) Among the nonmetallic elements, the change in atomic radius in moving one place left or right in a row is smaller than the change in moving one row up or down. Explain these observations. (b) Arrange the following atoms in order of increasing atomic radius: \(S i\). \(\mathrm{A} 1\), Ge, Ga.

An element \(X\) reacts with oxygen to form \(\mathrm{XO}_{2}\) and with chlorine to form \(\mathrm{XCl}_{4} \mathrm{XO}_{2}\) is a white solid that melts at high temperatures (above \(\left.1000^{\circ} \mathrm{C}\right)\). Under usual conditions, \(\mathrm{XCl}_{4}\) is a colorless liquid with a boiling point of \(58^{\circ} \mathrm{C}\). (a) \(\mathrm{XCl}_{4}\) reacts with water to form \(\mathrm{XO}_{2}\) and another product. What is the likely identity of the other product? (b) Do you think that element \(X\) is a metal, nonmetal, or metalloid? Explain. (c) By using a sourcebook such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element \(\mathrm{X}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation
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.