Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 10

Consider the ionic compounds \(\mathrm{NaF}\) and \(\mathrm{NaCl}\) : a) In which compound is the Coulombic force of attraction greater? b) \(\mathrm{NaCl}\) has a melting point of \(801{ }^{\circ} \mathrm{C}\). Which of these would you predict is the melting point of \(\mathrm{NaF}: 609^{\circ} \mathrm{C}, 800^{\circ} \mathrm{C}, 993{ }^{\circ} \mathrm{C}\) ? Explain your reasoning.

Short Answer

Expert verified
a) The Coulombic force of attraction is greater in NaF. b) The predicted melting point of NaF is 993 °C.

Step by step solution

01

Identify the ionic radii of Na, F and Cl

The ionic radius of sodium (Na) in nm is 0.102, for fluoride (F) is 0.133 and for chloride (Cl) is 0.181.
02

Compare the Coulombic force of attraction

The Coulombic force of attraction is inversely proportional to the squared distance between the ions. Given that the distances are Na-F and Na-Cl, and knowing that F has a smaller ionic radius than Cl, it is clear that the distance is shorter in NaF than NaCl. Therefore, NaF has a greater Coulombic force of attraction.
03

Compare the melting points based on Coulombic force

Since NaF has a greater force of attraction, it would require more heat to overcome this force, which means that NaF should have a higher melting point than NaCl (801 °C). The only option given higher than 801 °C is 993 °C.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 charge on the cation formed from \(\mathrm{Mg}\) ? b) What is the charge on the anion formed from \(\mathrm{O}\) (called the oxide ion)? c) What is the formula for the ionic compound containing magnesium ions and oxide ions?

In \(\mathrm{NaCl}\), what are the charges on the ions? Explain your reasoning.

Consider the ionic compounds \(\mathrm{MgO}\) and \(\mathrm{MgS}\) : a) In which compound is the Coulombic force of attraction greater? b) \(\mathrm{MgO}\) has a melting point of \(2852{ }^{\circ} \mathrm{C}\). Which of these would you predict is the melting point of \(\mathrm{MgS}\) : about \(2000^{\circ} \mathrm{C}\), about \(2850^{\circ} \mathrm{C}\), about 4000 \({ }^{\circ} \mathrm{C}\) ? Explain your reasoning.

The ions formed in molecules from Group 2 atoms (the alkaline earth metals, such as \(\mathrm{Mg}\) ) are almost exclusively \(\mathrm{M}^{2+}\) ions rather than \(\mathrm{M}^{3+}\) ions. Explain this result in a manner analogous to your analysis from CTQ \(2 .\)

a) Determine the core charge and valence shell for \(\mathrm{F}^{-}\), and \(\mathrm{F}^{2-}\). Drawing diagrams may be helpful. b) Based on your answer to part a, explain why an attempt to add an electron to \(\mathrm{F}^{-}\) does not result in the formation of \(\mathrm{F}^{2-}-\) that is, the "added" electron does not stay "attached" to the original \(\mathrm{F}^{-}\) ion. c) The ions formed in molecules from Group 17 atoms (the halogens, such as \(\mathrm{F}\) ) are almost exclusively \(\mathrm{X}^{-}\) ions rather than \(\mathrm{X}^{2-}\) ions. Explain this result based on your answers to parts a and b.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

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.

Sign-up for free