Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Identify the most important types of interparticle forces present in the solids of each of the following substances. a. \(\mathrm{BaSO}_{4}\) b. \(\mathrm{H}_{2} \mathrm{S}\) c. \(\mathrm{Xe}\) d. \(\mathrm{C}_{2} \mathrm{H}_{6}\) e. \(\mathrm{CsI}\) f. \(P_{4}\) g. \(\mathrm{NH}_{3}\)

Short Answer

Expert verified
a. BaSO4: Ionic Bonds b. H2S: Dipole-Dipole Interactions and London Dispersion Forces c. Xe: London Dispersion Forces d. C2H6: London Dispersion Forces e. CsI: Ionic Bonds f. P4: Covalent Bonds and London Dispersion Forces g. NH3: Hydrogen Bonds, Dipole-Dipole Interactions, and London Dispersion Forces

Step by step solution

01

a. BaSO4: Ionic Bonds

Barium sulfate (BaSO4) is an ionic compound composed of a metal (Barium) and a polyatomic ion (sulfate). In the solid state, the particles are held together by the strong electrostatic attraction between the positive and negative ions (Barium and sulfate), forming an ionic crystal lattice. Thus, the most important type of interparticle force present in this solid is the ionic bond.
02

b. H2S: Dipole-Dipole Interactions and London Dispersion Forces

Hydrogen sulfide (H2S) is a polar covalent molecule. The electrostatic attraction between the oppositely charged ends of adjacent polar molecules results in dipole-dipole interactions, which significantly contribute to holding the molecules together in the solid state. Additionally, as a nonpolar molecule, H2S also exhibits London dispersion forces (temporary induced dipoles), although these forces are relatively weaker compared to dipole-dipole interactions in this case.
03

c. Xe: London Dispersion Forces

Xenon (Xe) is a noble gas and therefore exists as independent atoms in the solid state. It exhibits only London dispersion forces, which are temporary induced dipoles arising from the fluctuations of electron distribution in atoms. These weak van der Waals forces are responsible for holding the xenon atoms together in the solid state.
04

d. C2H6: London Dispersion Forces

Ethane (C2H6) is a nonpolar covalent molecule. Due to its nonpolarity, the primary interparticle forces are London dispersion forces, which are temporary induced dipoles resulting from the fluctuation of electron distribution around the atoms.
05

e. CsI: Ionic Bonds

Cesium iodide (CsI) is an ionic compound composed of a metal (Cesium) and a nonmetal (Iodine). In the solid state, they form an ionic crystal lattice, where the particles are held together by the strong electrostatic attraction between the positively charged cesium ions and the negatively charged iodide ions. Consequently, the most important interparticle force in this compound is the ionic bond.
06

f. P4: Covalent Bonds and London Dispersion Forces

Phosphorus (P4) is a nonmetal that forms a tetrahedral molecular structure, with covalent bonds between the phosphorus atoms. These covalent bonds hold the atoms together within the molecule. Additionally, as a nonpolar molecule, P4 also exhibits London dispersion forces (temporary induced dipoles) between the P4 molecules in the solid state.
07

g. NH3: Hydrogen Bonds, Dipole-Dipole Interactions, and London Dispersion Forces

Ammonia (NH3) is a polar covalent molecule consisting of nitrogen and hydrogen atoms. The primary interparticle force responsible for holding ammonia molecules together in the solid state is the hydrogen bond, which is a particularly strong dipole-dipole interaction between an electronegative atom (nitrogen in this case) and a hydrogen atom bonded to another highly electronegative atom. Additionally, there are other weaker forces such as general dipole-dipole interactions and London dispersion forces between the molecules.

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 Vaia!

Key Concepts

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

Ionic Bonds
Ionic bonds are one of the strongest types of intermolecular forces and are crucial in the formation of ionic compounds. These bonds occur when a metal transfers electrons to a nonmetal, resulting in the formation of positively and negatively charged ions.
This creates an electrostatic attraction between the ions, leading to the formation of a crystal lattice structure.
  • For example, in barium sulfate ( BaSO₄ ), barium donates electrons to the sulfate group, resulting in a compound held together by ionic bonds.
  • Similarly, in cesium iodide ( CsI ), cesium (a metal) and iodine (a nonmetal) form ions that create strong ionic bonds in the solid state.
These bonds yield high melting and boiling points due to the strong interactions between ions.
Dipole-Dipole Interactions
Dipole-dipole interactions occur between polar molecules, where there is an uneven distribution of electrons, resulting in a permanent dipole.
The positive end of one molecule is attracted to the negative end of another, leading to an intermolecular force that is significant in holding the molecules together in a substance.
  • In hydrogen sulfide ( H₂S ), which is a polar molecule, dipole-dipole interactions play a key role. The polar nature of H₂S creates regions of partial positive and negative charges that interact with neighboring molecules.
  • Ammonia ( NH₃ ) also exhibits dipole-dipole interactions due to its molecular shape and polar nature. These types of forces, while strong, are not as robust as ionic bonds.
London Dispersion Forces
London dispersion forces, or van der Waals forces, are the weakest type of intermolecular forces, present in all molecules, whether polar or nonpolar.
They occur as a result of temporary dipoles formed due to the movement of electrons in an atom or molecule.
  • Noble gases like xenon ( Xe ) only exhibit London dispersion forces. These forces allow xenon atoms to aggregate in the solid state, albeit weakly.
  • Similarly, nonpolar molecules like ethane ( C₂H₆ ) rely on dispersion forces as their primary intermolecular interaction.
  • Even in polar molecules like hydrogen sulfide ( H₂S ) and phosphorus ( P₄ ), these dispersion forces, although often overshadowed by stronger forces, contribute to the overall molecular cohesion.
Covalent Bonds
Covalent bonds involve the sharing of electrons between atoms, most commonly between nonmetals. These bonds are critical in the structure of molecular compounds and form when atoms achieve greater stability by sharing electron pairs.
  • Phosphorus ( P₄ ) forms a classic example, where each phosphorus atom shares electrons covalently, creating a stable tetrahedral structure.
  • While covalent bonds are primarily responsible for holding the atoms within a molecule together, the molecules themselves may still interact through weaker intermolecular forces like London dispersion forces.
Covalent bonds contribute significantly to the chemical properties of substances but are distinct from the forces between molecules in a substance.
Hydrogen Bonds
Hydrogen bonds are a special type of dipole-dipole interaction that occur when hydrogen is covalently bonded to a highly electronegative atom such as oxygen, nitrogen, or fluorine.
It results in a strong attraction between molecules, much stronger than regular dipole-dipole interactions.
  • In ammonia ( NH₃ ), the presence of nitrogen, a highly electronegative atom, allows the formation of hydrogen bonds with other NH₃ molecules, significantly contributing to its physical properties, such as a relatively high boiling point for a molecule of its size.
Hydrogen bonds are crucial in many biological and chemical processes, greatly affecting the structure and behavior of molecules in different environments.

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

Explain why water forms into beads on a waxed car finish.

Cobalt fluoride crystallizes in a closest packed array of fluoride ions with the cobalt ions filling one-half of the octahedral holes. What is the formula of this compound?

The molar heat of fusion of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 9.92 \(\mathrm{kJ} / \mathrm{mol}\) . Its molar heat of vaporization is 30.7 \(\mathrm{kJ} / \mathrm{mol}\) . Calculate the heat required to melt 8.25 \(\mathrm{g}\) benzene at its normal melting point. Calculate the heat required to vaporize 8.25 g benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?

The memory metal, nitinol, is an alloy of nickel and titanium. It is called a memory metal because after being deformed, a piece of nitinol wire will return to its original shape. The structure of nitinol consists of a simple cubic array of Ni atoms and an inner penetrating simple cubic array of Ti atoms. In the extended lattice, a Ti atom is found at the center of a cube of Ni atoms; the reverse is also true. a. Describe the unit cell for nitinol. b. What is the empirical formula of nitinol? c. What are the coordination numbers (number of nearest neighbors) of Ni and Ti in nitinol?

Which of the following statements is (are) true? a. LiF will have a higher vapor pressure at \(25^{\circ} \mathrm{C}\) than \(\mathrm{H}_{2} \mathrm{S}\) . b. HF will have a lower vapor pressure at \(-50^{\circ} \mathrm{C}\) than \(\mathrm{HBr}\) . c. \(\mathrm{Cl}_{2}\) will have a higher boiling point than Ar. d. HCl is more soluble in water than in \(\mathrm{CCl}_{4}\) e. \(\mathrm{MgO}\) will have a higher vapor pressure at \(25^{\circ} \mathrm{C}\) than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) .

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.

Sign-up for free