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The generic structural formula for a 1 -alkyl-3-methylimidazolium cation is where \(\mathrm{R}\) is \(\mathrm{a}-\mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{n} \mathrm{CH}_{3}\) alkyl group. The melt- ing points of the salts that form between the 1 -alkyl3-methylimidazolium cation and the \(\mathrm{PF}_{6}^{-}\) anion are as follows: \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} \cdot \mathrm{p} .=60^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(\left(\mathrm{m} \cdot \mathrm{p} .=40^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} \cdot \mathrm{p} .=10^{\circ} \mathrm{C}\right),\) and \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} \cdot \mathrm{p} .=-61^{\circ} \mathrm{C}\right) .\) Why does the melting point decrease as the length of alkyl group increases?

Short Answer

Expert verified
The melting point of salts formed between the 1-alkyl-3-methylimidazolium cation and the PF6^- anion decreases as the length of the alkyl group increases because the increase in van der Waals forces between the alkyl chains is not sufficient to compensate for the decrease in electrostatic forces between the cations and anions due to the larger separation of charges. This leads to weaker overall intermolecular forces, requiring a lower temperature to break these forces and melt the substance.

Step by step solution

01

Identify the given information

We are given the melting points of salts formed between the 1-alkyl-3-methylimidazolium cation and the PF6^- anion for different alkyl groups R: - R = CH2CH3 (m.p. = 60°C) - R = CH2CH2CH3 (m.p. = 40°C) - R = CH2CH2CH2CH3 (m.p. = 10°C) - R = CH2CH2CH2CH2CH2CH3 (m.p. = -61°C) Here, the melting point decreases as the length of the alkyl group increases.
02

Examine the structure of the cation and anion

The 1-alkyl-3-methylimidazolium cation has a positively charged nitrogen atom in the imidazolium ring and a long alkyl chain on one side, while the PF6^- anion contains a central phosphorus atom surrounded by six fluorine atoms with a negative charge, making it quite symmetric.
03

Analyze the forces affecting the melting point

The melting point of a substance is influenced by the strength of the intermolecular forces between its molecules. In this case, those forces come from two main sources: electrostatic forces between the cations and anions, and van der Waals forces between the alkyl chains.
04

Evaluate the effect of increasing alkyl chain length

As the length of the alkyl chain (R) increases, the van der Waals forces between the alkyl chains will also increase. These forces arise from momentary fluctuations in electron distribution within the molecules, leading to temporary dipoles that attract the neighboring molecules. On the other hand, the overall strength of the electrostatic forces between the cations and anions will decrease. This happens due to the larger separation of charges as the bulky alkyl chains push the charged nitrogen atom farther away from the anion.
05

Conclude why the melting point decreases

The melting point decreases as the length of the alkyl group increases because the increase in van der Waals forces between the alkyl chains is not sufficient to overcome the decrease in electrostatic forces between the cations and anions due to the larger separation of charges. The weaker overall intermolecular forces result in a lower temperature needed to break these forces and cause the substance to melt.

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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 a crucial physical property that indicates the temperature at which it changes from a solid to a liquid state. It is primarily determined by the intermolecular forces within the substance. The stronger these forces, the higher the melting point. In substances composed of ionic compounds, such as salts, melting points can be quite high due to the strong electrostatic forces between oppositely charged ions.

When analyzing the melting points of salts involving the 1-alkyl-3-methylimidazolium cation, as the length of the alkyl chain attached to this cation increases, the melting point decreases significantly. This effect can be understood by examining how different types of intermolecular forces behave as the molecular structure changes.
Alkyl Chain Length
Alkyl chains are hydrocarbon chains attached to a central molecule. In this exercise, we consider the 1-alkyl-3-methylimidazolium cation, where the alkyl group (denoted as R) varies in length. As the alkyl chain length increases, it affects several molecular properties, significantly influencing the compound's physical properties, such as its melting point.

Here's how alkyl chain length impacts molecular interactions:
  • Longer alkyl chains yield increased van der Waals forces due to a larger area of contact among molecules.
  • As chains become longer, they can also create more significant separation between the charged regions of the cation and its paired anion, like PF6^-, weakening electrostatic interactions.

Thus, increasing the alkyl chain length in these salts decreases their melting points by changing the balance of forces acting within the lattice structure.
Electrostatic Forces
Electrostatic forces play a pivotal role in determining the melting point of ionic compounds. These forces are the attractions between positively charged cations and negatively charged anions. Typically, such solid ionic compounds exhibit high melting points due to these strong attractions which require more energy to overcome in the form of heat.

However, with longer alkyl chains on the 1-alkyl-3-methylimidazolium cations, the electrostatic forces become less dominant. This is because the extended alkyl chains effectively increase the distance between the charged cation sites and the PF6^- anion. As this separation increases, the strength of the electrostatic forces decreases, leading to a lower energy requirement to melt the substance, resulting in a lower melting point.
Van der Waals Forces
Van der Waals forces are weak inter-molecular attractions that arise from temporary dipoles in molecules. They do not stem from any fixed charge or permanent dipole like those seen in electrostatic forces. Instead, these forces depend significantly on how close molecules can get to one another, often increasing with larger surface areas available for contact.

In the context of the increasing alkyl chain length, van der Waals forces become more noticeable. Longer alkyl chains have more surface area for these interactions, which can slightly increase the overall interactions between molecules. Although these forces do increase with alkyl chain length, they are weaker than electrostatic forces, which explains why their enhancement isn't enough to counterbalance the reduction in electrostatic forces caused by increased chain length.
  • Van der Waals forces contribute but do not dominate the melting behavior in ionic compounds.
  • They work collectively with other forces to determine total intermolecular interactions.
These effects make understanding the delicate balance between different types of intermolecular forces critical when predicting physical properties such as melting points.

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Most popular questions from this chapter

Ethyl chloride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) boils at \(12^{\circ} \mathrm{C}\). When liquid \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) under pressure is sprayed on a room-temperature \(\left(25^{\circ} \mathrm{C}\right)\) surface in air, the surface is cooled considerably. (a) What does this observation tell us about the specific heat of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g)\) as compared with that of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(l) ?(\mathbf{b})\) Assume that the heat lost by the surface is gained by ethyl chloride. What enthalpies must you consider if you were to calculate the final temperature of the surface?

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from 35 to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water? (The heat of vaporization of water in this temperature range is \(2.4 \mathrm{~kJ} / \mathrm{g} .\) The specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} .)\)

Name the phase transition in each of the following situations and indicate whether it is exothermic or endothermic: (a) Iodine solid turns to iodine gas when it is heated. (b) Snowflakes turn into water when they fall on an open palm. (c) Droplets of water appear on grass in a cold humid morning. (d) Dry ice gradually disappears when left at room temperature for some period of time.

True or false: \((\mathbf{a}) \mathrm{CBr}_{4}\) is more volatile than \(\mathrm{CCl}_{4}\). (b) \(\mathrm{CBr}_{4}\) has a higher boiling point than \(\mathrm{CCl}_{4}\). (c) \(\mathrm{CBr}_{4}\) has weaker intermolecular forces than \(\mathrm{CCl}_{4}\). (d) \(\mathrm{CBr}_{4}\) has a higher vapor pressure at the same temperature than \(\mathrm{CCl}_{4}\).

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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