Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 21

\(\mathrm{VF}_{5}\) is a liquid that boils at \(48^{\circ} \mathrm{C},\) whereas \(\mathrm{VF}_{3}\) is a solid that melts above \(800^{\circ} \mathrm{C}\). Explain this difference in properties.

Short Answer

Expert verified
\(\text{VF}_3\) has stronger intermolecular forces than \(\text{VF}_5\), explaining its higher melting point.

Step by step solution

01

- Identify Physical States and Phase Transitions

Determine the physical states and corresponding temperatures at which \(\text{VF}_5\) and \(\text{VF}_3\) change phases. \(\text{VF}_5\) is a liquid that boils at \(48^{\circ} \text{C}\), and \(\text{VF}_3\) is a solid that melts above \(800^{\circ} \text{C}\).
02

- Analyze Molecular Structure and Intermolecular Forces

Evaluate the molecular structures of \(\text{VF}_5\) and \(\text{VF}_3\). \(\text{VF}_5\) is a molecular compound with relatively weaker intermolecular forces, such as Van der Waals forces. In contrast, \(\text{VF}_3\) likely has a more complex structure with stronger bonding, possibly ionic or covalent network bonds.
03

- Correlate Phase Transition Temperatures with Intermolecular Forces

Stronger intermolecular forces in \(\text{VF}_3\) mean that more energy (higher temperature) is needed to change its phase from solid to liquid compared to \(\text{VF}_5\), which has weaker intermolecular forces and thus transitions from liquid to gas at a lower temperature.
04

- Conclusion

Summarize the findings: \(\text{VF}_3\) has stronger intermolecular forces, requiring higher temperatures to melt, which explains its solid state until \(800^{\circ} \text{C}\). In contrast, \(\text{VF}_5\) has weaker intermolecular forces, resulting in its liquid state at lower temperatures and boiling at \(48^{\circ} \text{C}\).

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!

Key Concepts

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

Intermolecular Forces
Intermolecular forces are the forces that hold molecules together. They are crucial in determining the physical properties such as boiling and melting points of compounds. Here are a few types of intermolecular forces:
  • Van der Waals forces: These are weak forces arising from temporary dipoles that occur when molecules are close to each other.
  • Dipole-dipole interactions: These occur between molecules with permanent dipoles.
  • Hydrogen bonds: A particularly strong type of dipole-dipole interaction occurring between hydrogen and highly electronegative atoms like oxygen, nitrogen, or fluorine.
mainly exhibits Van der Waals forces, which are relatively weak. This is why it can transition from liquid to gas at a lower temperature (48°C). On the other hand, likely has stronger intermolecular forces, possibly ionic or covalent network bonds. That explains why a high temperature (above 800°C) is needed to change it from solid to liquid.
Boiling Point
The boiling point of a substance is the temperature at which its liquid phase changes to the gas phase. It signifies the point where the vapor pressure of the liquid equals the atmospheric pressure. For example:
  • boils at 48°C because the weaker Van der Waals forces holding its molecules together are easily overcome at this lower temperature.
  • Stronger intermolecular forces in a compound will result in a higher boiling point because more energy (in the form of heat) is required to overcome these forces and transition to the gas phase.
This means substances with stronger bonds, like ionic or network covalent bonds, have significantly higher boiling points than those with weaker bonds.
Melting Point
The melting point is the temperature at which a solid changes to a liquid. It equates the energy needed to overcome the forces holding the solid structure together.
  • has a melting point above 800°C, indicating it has very strong intermolecular forces, likely of ionic or covalent nature. More energy is required to break these strong bonds, leading to a high melting point.
  • In contrast, compounds with weaker bonds will have lower melting points. For example, substances held together by Van der Waals forces will melt at lower temperatures compared to ionic or covalent network solids.
Hence, the much higher melting point of compared to highlights the significant difference in the strength of their intermolecular forces.

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

According to valence bond theory, what set of orbitals is used by a Period 4 metal ion in forming (a) a square planar complex; (b) a tetrahedral complex?

Draw orbital-energy splitting diagrams and use the spectrochemical series to show the orbital occupancy for each of the following (assuming that \(\mathrm{H}_{2} \mathrm{O}\) is a weak-field ligand): (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) (c) \(\left[\mathrm{FeF}_{6}\right]^{3-}\)

You know the following about a coordination compound: (1) The partial empirical formula is \(\mathrm{KM}\left(\mathrm{CrO}_{4}\right) \mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\) (2) It has A (red) and B (blue) crystal forms. (3) When \(1.0 \mathrm{~mol}\) of \(\mathrm{A}\) or \(\mathrm{B}\) reacts with \(1.0 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}, 0.50 \mathrm{~mol}\) of a red precipitate forms immediately. (4) After the reaction in ( 3 ), \(1.0 \mathrm{~mol}\) of A reacts very slowly with \(1.0 \mathrm{~mol}\) of silver oxalate \(\left(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) to form \(2.0 \mathrm{~mol}\) of a white precipitate. (Oxalate can displace other ligands.) (5) After the reaction in ( 3 ), \(1.0 \mathrm{~mol}\) of \(\mathrm{B}\) does not react further with \(1.0 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}\). From this information, determine the following: (a) The coordination number of \(\mathrm{M}\) (b) The group(s) bonded to \(\mathrm{M}\) ionically and covalently (c) The stereochemistry of the red and blue forms

Using the periodic table to locate each element, write the electron configuration of (a) Os; (b) \(\mathrm{Co} ;\) (c) Ag.

A shortcut to finding optical isomers is to see if the complex has a plane of symmetry-a plane passing through the metal atom such that every atom on one side of the plane is matched by an identical one at the same distance from the plane on the other side. Any planar complex has a plane of symmetry, since all atoms lie in one plane. Use this approach to determine whether these exist as optical isomers: (a) \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) (tetrahedral); (b) \(\left[\mathrm{Pt}(\mathrm{en})_{2}\right]^{2+} ;\) (c) trans-[PtBr \(\left.\left._{4} \mathrm{Cl}_{2}\right]^{2-} ;\) (d) trans-[Co(en) \(\left._{2} \mathrm{~F}_{2}\right]^{+}\) (e) \(c i s-\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{~F}_{2}\right]^{+}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

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