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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.

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Key Concepts

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

Melting Point
The melting point of a substance is the temperature at which it changes from a solid to a liquid. This property is influenced by the type of bonding and structure within a compound.
The stronger the bonds or interactions between the particles in a substance, the higher the melting point will be.
For example:
  • Ionic compounds, like \(\mathrm{CaF}_{2}\), have high melting points because they are composed of a lattice of ions held together by strong electrostatic forces.
  • Covalent compounds, such as \(\mathrm{NH}_{3}\), have relatively lower melting points. They depend on weaker intermolecular forces, which are easier to overcome.
Thus, understanding the type of bonds involved can provide insight into the melting point of the compound.
Intermolecular Forces
Intermolecular forces are the forces of attraction or repulsion between molecules. These forces play a crucial role in determining the properties of compounds, including whether they are solid, liquid, or gas at a given temperature. Some of the key types of intermolecular forces include:
  • London Dispersion Forces: Present in all molecules, they are the weakest and arise due to temporary shifts in electron density.
  • Dipole-Dipole Interactions: Occur between molecules that have permanent dipoles, like \(\mathrm{NH}_{3}\), and are stronger than dispersion forces.
  • Hydrogen Bonds: A special type of dipole-dipole interaction particularly strong in molecules containing \(-\)OH, \(-\)NH, or \(\)-FH bonds.
While these intermolecular forces can influence properties like melting points, they are typically weaker than the ionic bonds found in ionic compounds.
Ionic Bonding
Ionic bonding occurs when electrons are transferred from one atom to another, resulting in the formation of oppositely charged ions. These ions are then held together by strong electrostatic forces, creating a stable compound.
- In \(\mathrm{CaF}_{2}\), calcium donates electrons to fluorine atoms, forming \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\) ions.- This transfer leads to a structure known as an ionic lattice, characterized by high melting and boiling points.- Breaking these ionic bonds requires significant energy, contributing to the high melting point seen in ionic compounds like \(\mathrm{CaF}_{2}\).
The strong bonds and lattice structure make ionic compounds robust and durable, needing much more energy to be distorted or changed compared to the weaker forces in covalent bonds.
Covalent Bonding
Covalent bonding involves the sharing of electron pairs between atoms. This typically occurs between non-metals, like nitrogen and hydrogen in \(\mathrm{NH}_{3}\).
- These shared electrons form what are known as covalent bonds, holding the atoms together in a molecule.- Covalent compounds usually manifest as discrete molecules, with lower melting and boiling points compared to ionic compounds.- The forces between these molecules (intermolecular forces) determine their physical properties and usually require less energy to overcome than ionic bonds.
- Hence, covalent compounds tend to have softer, more flexible structures compared to the rigid lattices found in ionic substances.
Understanding covalent bonding helps explain why substances like \(\mathrm{NH}_{3}\) have lower energy requirements for phase changes.

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$$\mathrm{Br}_{2}(g)+\mathrm{I}_{2}(g) \leftrightarrow 2 \mathrm{IBr}(g)$$ At \(150^{\circ} \mathrm{C},\) the equilibrium constant, \(K_{c},\) for the reaction shown above has a value of \(300 .\) This reaction was allowed to reach equilibrium in a sealed container and the partial pressure due to IBr(g) was found to be 3 atm. Which of the following could be the partial pressures due to \(\operatorname{Br}_{2}(g)\) and \(I_{2}(g)\) in the container? \(\begin{array}{lll}{} & {\operatorname{Br}_{2}(g)} & {\mathrm{I}_{2}(g)} \\\ {\text { (A) }} & {0.1 \mathrm{atm}} & {0.3 \mathrm{atm}} \\ {\text { (B) }} & {0.3 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (C) }} & {1 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (D) }} & {1 \mathrm{atm}} & {3 \mathrm{atm}}\end{array}\)

Which of the following expressions is equal to the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{2} \mathrm{CO}_{3} ?\) (A) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CO}_{3}^{2-}\right]\) (B) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CO}_{3}^{2-}\right]^{2}\) (C) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CO}_{3}^{2-}\right]\) (D) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CO}_{3}^{2-}\right]^{2}\)

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