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Hint: We realize that intermolecular forces are electrostatic in nature and incorporate van der Waals forces and hydrogen bonds. Particles in fluids are clung to different atoms by intermolecular forces that are more vulnerable than the intramolecular forces that hold the atoms together inside atoms and polyatomic particles. The three significant sorts of intermolecular forces are,
Dipole-dipole interactions
London dispersion interactions
Hydrogen bonds
Complete answer:
We are given that the aluminum trifluoride is not soluble in anhydrous hydrogen fluoride due to the covalent nature of the molecule of hydrogen fluoride. We have to know that strong hydrogen bonds are found in molecules of hydrogen fluoride. So, no ions of free fluoride are produced.
We know that sodium fluoride $\left( {NaF} \right)$ is an ionic compound and it contains ionic bonds. Free ions of fluoride are produced. We have to know that sodium hexafluoroaluminate (III), a soluble compound is formed when ions of fluoride are combined with $Al{F_3}$. We can write the chemical equation is,
$3NaF + Al{F_3}\xrightarrow{{}}N{a_3}\left[ {Al{F_6}} \right]$
We have to know that aluminum trifluoride is formed as precipitate when gaseous $B{F_3}$ is bubbled through. This is because $B{F_3}$ is a stronger Lewis acid when compared to $Al{F_3}$. This is because of the smaller size and more electronegativity of boron when compared to aluminum. So, boron has the ability to produce more complexes than aluminum. Thus, $B{F_3}$ gives complex compounds and $Al{F_3}$ gets precipitated. We can write the chemical equation is,
$N{a_3}\left[ {Al{F_6}} \right] + 3B{F_3}\xrightarrow{{}}3Na\left[ {B{F_4}} \right] + Al{F_3}$
Note:
We also need to know that in solids, the intermolecular forces are stronger since the particles are firmly loaded with one another. In fluids, the intermolecular forces are not sufficiently able to hold the atoms back from moving particles, henceforth fluids hold the state of the container. If there should be an occurrence of gases the intermolecular forces are extremely frail, and the feeble intermolecular forces allow gases to extend.
Dipole-dipole interactions
London dispersion interactions
Hydrogen bonds
Complete answer:
We are given that the aluminum trifluoride is not soluble in anhydrous hydrogen fluoride due to the covalent nature of the molecule of hydrogen fluoride. We have to know that strong hydrogen bonds are found in molecules of hydrogen fluoride. So, no ions of free fluoride are produced.
We know that sodium fluoride $\left( {NaF} \right)$ is an ionic compound and it contains ionic bonds. Free ions of fluoride are produced. We have to know that sodium hexafluoroaluminate (III), a soluble compound is formed when ions of fluoride are combined with $Al{F_3}$. We can write the chemical equation is,
$3NaF + Al{F_3}\xrightarrow{{}}N{a_3}\left[ {Al{F_6}} \right]$
We have to know that aluminum trifluoride is formed as precipitate when gaseous $B{F_3}$ is bubbled through. This is because $B{F_3}$ is a stronger Lewis acid when compared to $Al{F_3}$. This is because of the smaller size and more electronegativity of boron when compared to aluminum. So, boron has the ability to produce more complexes than aluminum. Thus, $B{F_3}$ gives complex compounds and $Al{F_3}$ gets precipitated. We can write the chemical equation is,
$N{a_3}\left[ {Al{F_6}} \right] + 3B{F_3}\xrightarrow{{}}3Na\left[ {B{F_4}} \right] + Al{F_3}$
Note:
We also need to know that in solids, the intermolecular forces are stronger since the particles are firmly loaded with one another. In fluids, the intermolecular forces are not sufficiently able to hold the atoms back from moving particles, henceforth fluids hold the state of the container. If there should be an occurrence of gases the intermolecular forces are extremely frail, and the feeble intermolecular forces allow gases to extend.
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