Which solid in each pair has the higher melting point and why? a. TiO 2 (s) or HOOH(s) b. CCl 4 (s) or SiCl 4 (s) c. Kr(s) or Xe(s)

hey everyone in this example, we need to determine which of these atoms has a lower melting point. And we need to explain our answer. So what we should recognize is that we have the atoms sodium and RB for rubidium which are located both in group one A of our periodic tables which we should recognize as our alkali metal group. So these are alkali metals. And when we recall our trend for atomic radius on our periodic table, we would recall that atomic radius is going to be increasing as we go towards the bottom left of our periodic table. So looking at first sodium on the periodic table again, we started seeing group but it's across period three vs rubidium we would see is also in group one a. But across period five. And so we would say that because rubidium is in period five, it's going to therefore be a larger adam because it has a larger atomic radius based on the trend that we've recalled and so therefore the ions or the ion. We should say that rubidium forms being in group one A. It should form a plus one ion. Well have a decreased ion density or we can say a decreased charge density, weakening the metallic bonds and decreasing its melting point. And so to overall explain our answer, we would say that rubidium has the lower melting point, sorry, lower melting point due to its weaker metallic bonds. And so this statement here is our final answer. As long or as well as our explanation as to why it has the lower melting point. I hope that everything I reviewed was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

Ch.12 - Solids and Modern Material

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Tro 4th Edition

Ch.12 - Solids and Modern Material

Problem 43

Chapter 12, Problem 43

Which solid in each pair has the higher melting point and why?
a. TiO₂(s) or HOOH(s)
b. CCl₄(s) or SiCl₄(s)
c. Kr(s) or Xe(s)

Verified step by step guidance

Identify the type of forces involved: Both krypton (Kr) and xenon (Xe) are noble gases, and in their solid forms, they are held together by London dispersion forces, which are a type of van der Waals force.

Understand the relationship between atomic size and dispersion forces: London dispersion forces increase with the size of the electron cloud, which is related to the atomic size.

Compare the atomic sizes: Xenon (Xe) has a larger atomic size compared to krypton (Kr) because it is located further down the group in the periodic table.

Relate atomic size to melting point: Since xenon has a larger atomic size, it has stronger London dispersion forces compared to krypton. Stronger intermolecular forces generally lead to a higher melting point.

Conclude which has the higher melting point: Therefore, Xe(s) is expected to have a higher melting point than Kr(s) due to its larger atomic size and stronger dispersion forces.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Intermolecular Forces

Intermolecular forces are the attractions between molecules that influence physical properties such as melting and boiling points. In solids, these forces can include van der Waals forces, dipole-dipole interactions, and hydrogen bonding. The strength of these forces determines how much energy is required to break them apart, thus affecting the melting point.

Atomic Size and Mass

Atomic size and mass play a significant role in determining the melting points of solids. Larger atoms generally have more electrons, which can lead to stronger van der Waals forces due to increased polarizability. In the case of noble gases like Kr and Xe, the larger atomic size of xenon contributes to a higher melting point compared to krypton.

Noble Gases and Their Properties

Noble gases are characterized by their complete valence electron shells, making them chemically inert. However, they still exhibit physical properties influenced by their atomic structure. Among noble gases, melting points increase with atomic number due to greater van der Waals forces, which is relevant when comparing the melting points of solid krypton and xenon.