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A number of salts containing the tetrahedral polyatomic anion, \(\mathrm{BF}_{4}^{-}\), are ionic liquids, whereas salts containing the somewhat larger tetrahedral ion \(\mathrm{SO}_{4}{ }^{2-}\) do not form ionic liquids. Explain this observation.

Short Answer

Expert verified
Salts containing the tetrahedral polyatomic anion \(\mathrm{BF}_{4}^{-}\) form ionic liquids due to their larger and more diffuse charges, which result in weaker electrostatic forces and reduced lattice energies. This allows for relatively easy separation of ions at lower temperatures. In contrast, salts with the tetrahedral ion \(\mathrm{SO}_{4}{ }^{2-}\) have higher charge density and stronger electrostatic forces, leading to higher lattice energies and melting points, thus preventing the formation of ionic liquids at lower temperatures.

Step by step solution

01

Understanding ionic liquids

Ionic liquids are salts that exist in the liquid state at room temperature or below (typically below 100°C). They have unique properties like high thermal stability, low vapor pressure, and good solvating ability. To become an ionic liquid, a compound must have ionic species with a suitable balance between electrostatic forces and van der Waals forces. Generally, large and asymmetric ions with lower charge density form ionic liquids due to weaker lattice energies and reduced melting points.
02

Comparing the structure of the two ions

Now let's look at the two ions mentioned, \(\mathrm{BF}_{4}^{-}\) and \(\mathrm{SO}_{4}{ }^{2-}\). Both ions have a tetrahedral structure. The \(\mathrm{BF}_{4}^{-}\) anion has one boron atom surrounded by four fluorine atoms, while the \(\mathrm{SO}_{4}{ }^{2-}\) anion has one sulfur atom surrounded by four oxygen atoms.
03

Comparing the charges and polarizability

The \(\mathrm{BF}_{4}^{-}\) anion has a -1 charge whereas the \(\mathrm{SO}_{4}{ }^{2-}\) has a -2 charge. In the case of \(\mathrm{BF}_{4}^{-}\), the charge is distributed over the fluorine atoms due to the electronegativity of the fluorine atom, which results in more diffuse charges and lower overall charge density. In contrast, the oxygen atoms in the \(\mathrm{SO}_{4}{ }^{2-}\) anion have more localized charges due to their lesser electronegativity compared to fluorine, which leads to a higher overall charge density.
04

Relating the properties of the ions to ionic liquids formation

The larger, more diffuse charges present in \(\mathrm{BF}_{4}^{-}\) results in weaker electrostatic forces and reduced lattice energies. The lower lattice energies enable relatively easy separation of the ions at lower temperatures, leading to the formation of ionic liquids. On the other hand, in \(\mathrm{SO}_{4}{ }^{2-}\), due to the higher charge density, the electrostatic forces are stronger, leading to higher lattice energies. This results in higher melting points and prevents the formation of ionic liquids at lower temperatures. Thus, it can be concluded that salts containing the tetrahedral polyatomic anion \(\mathrm{BF}_{4}^{-}\) form ionic liquids due to the more diffuse charges and lower lattice energies, whereas salts with the somewhat larger tetrahedral ion \(\mathrm{SO}_{4}{ }^{2-}\) do not form ionic liquids due to their higher charge density and lattice energies.

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Key Concepts

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

Tetrahedral Anions
Tetrahedral anions are ions where the central atom is surrounded by four other atoms or groups in a shape resembling a pyramid with a triangular base. This shape is called a tetrahedron. Common examples include \([BF_4]^−\) and \([SO_4]^{2−}\). Both have a similar structure, but the atoms involved and the charges are different.

The tetrahedral geometry contributes to the distribution of electronic charge in the ion, affecting their behavior in forming ionic liquids. This geometry facilitates even charge distribution across the ion. For example, in \([BF_4]^−\), the boron atom is smaller and surrounded by more electronegative fluorine atoms. This helps distribute the negative charge more evenly, resulting in a more diffuse charge.

In contrast, \([SO_4]^{2−}\) has a sulfur atom bonded to four oxygen atoms, carrying a higher -2 charge. The oxygen's lower electronegativity compared to fluorine results in localized charges, making the entire anion more tightly bound. This restricts the ion’s ability to flow freely at lower temperatures, thus preventing the formation of ionic liquids.
Charge Density
Charge density refers to how much electric charge is present per unit volume of an ion. In simple terms, it's like the concentration of charge in a given space.

For tetrahedral anions like \([BF_4]^−\) and \([SO_4]^{2−}\), the charge density significantly impacts their ability to form ionic liquids. The negative charge in \([BF_4]^−\) is spread over a larger volume due to the more electronegative fluorine atoms, resulting in a lower charge density. This means that ions do not cling tightly together, which reduces the strength of electrostatic forces, allowing the substance to exist as a liquid at room temperature.

On the other hand, in \([SO_4]^{2−}\), the charge is more concentrated due to the -2 charge spread across the sulfur-oxygen structure. This results in a higher charge density, causing the ions to attract each other more strongly. As a result, the \([SO_4]^{2−}\) ions form more stable, solid structures at room temperature, making it difficult for them to transform into ionic liquids.
Lattice Energy
Lattice energy is the amount of energy required to separate one mole of a solid ionic compound into its gaseous ions. It is an indicator of the strength of bonds between ions in a crystal lattice.

Ionic liquids benefit from having lower lattice energies, as these are easier to separate into individual ions which can then move freely and form a liquid. The \([BF_4]^−\) ion, with its low charge density, results in reduced lattice energy. As a result, \([BF_4]^−\) forms weaker bonds with cations, allowing the solid to melt at lower temperatures and become an ionic liquid.

Conversely, \([SO_4]^{2−}\) ions have high lattice energies due to their high charge density. Strong electrostatic attractions among the ions require more energy to overcome. This makes \([SO_4]^{2−}\) salts typically solid at room temperature, making them incapable of forming ionic liquids under these conditions. Ultimately, the ability of ions to separate easily is crucial for the formation of ionic liquids, and lattice energy plays a pivotal role in this process.

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Most popular questions from this chapter

The following data present the temperatures at which certain vapor pressures are achieved for dichloromethane \(\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) and methyl iodide \(\left(\mathrm{CH}_{3} \mathrm{I}\right):\) (a) Which of the two substances is expected to have the greater dipole-dipole forces? Which is expected to have the greater dispersion forces? Based on your answers, explain why it is difficult to predict which compound would be more volatile. (b) Which compound would you expect to have the higher boiling point? Check your answer in a reference book such as the CRC Handbook of Chemistry and Physics. (c) The order of volatility of these two substances changes as the temperature is increased. What quantity must be different for the two substances for this phenomenon to occur? (d) Substantiate your answer for part (c) by drawing an appropriate graph.

As a metal such as lead melts, what happens to (a) the average kinetic energy of the atoms and (b) the average distance between the atoms?

(a) Would you expect the viscosity of isopropanol, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH},\) to be larger or smaller than the viscosity of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) ? (b) Would you expect the viscosity of isopropanol to be smaller or larger than the viscosity of 1-propanol, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{OH}\) ?

Describe the intermolecular forces that must be overcome to convert these substances from a liquid to a gas: (a) \(\mathrm{CF}_{4}\), (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{BCl}_{3}\)

Rationalize the difference in boiling points in each pair: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\left(-23^{\circ} \mathrm{C}\right)\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \quad\left(78^{\circ} \mathrm{C}\right),\) (b) \(\mathrm{CO}_{2}\left(-78.5^{\circ} \mathrm{C}\right)\) and \(\mathrm{CS}_{2}\left(46.2^{\circ} \mathrm{C}\right),(\mathbf{c}) \mathrm{CH}_{3} \mathrm{COCH}_{3}\left(50.5^{\circ} \mathrm{C}\right)\) and \(\mathrm{CH}_{3} \mathrm{COOH}\left(101^{\circ} \mathrm{C}\right)\)

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