Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 1

Why does \(\mathrm{CaF}_{2}\) have a higher melting point than \(\mathrm{NH}_{3} ?\) (A) \(\mathrm{CaF}_{2}\) is more massive and thus has stronger London dispersion forces. (B) CaF_2 exhibits network covalent bonding, which is the strongest type of bonding. (C) CaF_2 is smaller and exhibits greater Coulombic attractive forces. (D) \(\mathrm{CaF}_{2}\) is an ionic substance and it requires a lot of energy to break up an ionic lattice.

Short Answer

Expert verified
The correct answer is (D). \(\mathrm{CaF}_{2}\) has a higher melting point than \(\mathrm{NH}_{3}\) because it is an ionic substance and it requires a lot of energy to break up an ionic lattice.

Step by step solution

01

Understand the nature of bonding in \(\mathrm{CaF}_{2}\) and \(\mathrm{NH}_{3}\)

\(\mathrm{CaF}_{2}\) is an ionic compound as it is formed through the transfer of electrons between calcium (a metal) and fluorine (a non-metal). On the other hand, \(\mathrm{NH}_{3}\) is a covalent molecular compound formed through the sharing of electrons between nitrogen and hydrogen, both of which are non-metals.
02

Understand the effect of bonding on physical properties

Ionic compounds typically have high melting and boiling points due to the strong electrostatic forces of attraction between the oppositely charged ions. This force requires a considerable amount of energy to overcome, hence the high melting points. Covalent molecular compounds, like \(\mathrm{NH}_{3}\), typically have lower melting and boiling points because they depend on weak intermolecular forces (like London dispersion forces or dipole-dipole interactions), which require less energy to overcome.
03

Evaluate given options

Looking at the options given, (A) does not apply as London dispersion forces primarily apply to covalent molecular compounds, not ionic compounds. (B) is incorrect as network covalent bonding is not observed in ionic compounds like \(\mathrm{CaF}_{2}\). (C) is not applicable as the size relation does not explain the difference in melting points. (D) correctly states that \(\mathrm{CaF}_{2}\) is an ionic substance and requires a lot of energy to break up the ionic lattice, which is consistent with our understanding of the strong forces in ionic compounds causing higher melting points.

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 laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer? (A) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad K_{a}=5.9 \times 10^{-2}\) (B) \(\mathrm{H}_{3} \mathrm{AsO}_{4} \quad K_{a}=5.6 \times 10^{-3}\) (C) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \quad K_{a}=1.8 \times 10^{-5}\) (D) \(\mathrm{HOCl}\) \(\quad K_{a}=3.0 \times 10^{-8}\)

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . What are the concentrations of \(\mathrm{Sr}^{2+}\) and \(\mathrm{F}^{-}\) in the beaker? (A) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (B) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\) (C) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (D) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} \mathrm{M}\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\)

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following occurs at the cathode? (A) \(\mathrm{Cr}^{3+}\) is reduced to \(\mathrm{Cr}(\mathrm{s})\) (B) \(\mathrm{Pb}^{2+}\) is reduced to \(\mathrm{Pb}(\mathrm{s})\) (C) \(\mathrm{Cr}(s)\) is oxidized to \(\mathrm{Cr}^{3+}\) (D) \(\quad \mathrm{Pb}(s)\) is oxidized to \(\mathrm{Pb}^{2+}\)

Use the following information to answer questions 14-16 The radius of atoms and ions is typically measured in Angstroms \((A),\) which is equivalent to \(1 * 10^{-10} \mathrm{m} .\) Below is a table of information for three different elements. TABLE NOT AVAILABLE Neon has a smaller atomic radius than phosphorus because: (A) Unlike neon, phosphorus has electrons present in its third energy level. (B) Phosphorus has more protons than neon, which increases the repulsive forces in the atom. (C) The electrons in a neon atom are all found in a single energy level. (D) Phosphorus can form anions, while neon is unable to form any ions.

The value of \(K_{a}\) for \(\mathrm{HSO}_{4}^{-}\) is \(1 \times 10^{-2} .\) What is the value of \(K_{b}\) for \(\mathrm{SO}_{4}^{2-} ?\) (A) \(K_{b}=1 \times 10^{-12}\) (B) \(K_{b}=1 \times 10^{-8}\) (C) \(K_{b}=1 \times 10^{-2}\) (D) \(K_{b}=1 \times 10^{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear 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