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Boiling point of chloride, bromide and iodide is higher due to (A) High dipole and van der waal forces (B) Higher molecular masses (C) Greater development of polarity (D) All of these

Short Answer

Expert verified
The higher boiling points of chloride, bromide, and iodide can be attributed to a combination of factors including strong intermolecular forces (dipole-dipole and van der Waals), higher molecular masses, and greater development of polarity. Therefore, the correct answer is (D) All of these.

Step by step solution

01

Intermolecular Forces

There are three main types of intermolecular forces: dipole-dipole, hydrogen bonding, and van der Waals (London dispersion) forces. Compounds that can form dipoles have a difference in electronegativity between their atoms, while hydrogen bonding occurs between a highly electronegative atom (such as oxygen, nitrogen, or fluorine) and hydrogen. Van der Waals forces are temporary dipoles created by the uneven distribution of electrons in molecules, which is always present regardless of polarity. In the case of chloride, bromide, and iodide, they do not have any hydrogen bonding but can form dipoles and van der Waals forces. Therefore, there is a possibility that these intermolecular forces contribute to higher boiling points
02

Molecular Masses

Higher molecular masses usually lead to higher boiling points due to increased van der Waals forces between molecules. This is because larger molecules have more electrons and greater surface area, making it more likely for temporary dipoles (van der Waals forces) to form and hold the molecules together. Since chloride, bromide, and iodide generally have higher molecular masses than other compounds within the same group, there is a likelihood that higher molecular masses contribute to the high boiling points.
03

Polarity

Polarity arises when there is a difference in electronegativity between the atoms in a molecule, leading to an uneven distribution of electron density. Greater polarity means stronger dipole-dipole forces, which can result in higher boiling points for polar molecules. Chloride, bromide, and iodide form polar molecules with higher electronegativity differences. Thus, it can be inferred that the greater development of polarity might be another reason for higher boiling points.
04

Concluding the Answer

Based on our understanding of intermolecular forces, molecular masses, and polarity, we can see that all these factors contribute to the higher boiling points of chloride, bromide, and iodide. Answer: (D) All of these

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Key Concepts

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

Boiling Point
The boiling point is the temperature at which a substance transitions from a liquid to a vapor. It is crucial to understand that the boiling point is directly influenced by intermolecular forces acting between molecules. Intermolecular forces need to be overcome to convert a liquid into a gas, and stronger forces typically result in a higher boiling point. This is why compounds like chloride, bromide, and iodide exhibit higher boiling points.

Several factors contribute to the boiling point:
  • Type and strength of intermolecular forces: Molecules with strong intermolecular forces have higher boiling points.
  • Molecular mass: Generally, larger molecules with more mass have higher boiling points due to increased surface area and the strength of van der Waals forces.
  • Polarity: Molecules with greater polarity usually exhibit higher boiling points due to stronger dipole-dipole interactions.
In the case of chloride, bromide, and iodide, their higher boiling points can be attributed to a combination of these factors.
Dipole-Dipole Forces
Dipole-dipole forces are a type of intermolecular force that occur between polar molecules. These forces result from the attraction between the positive end of one polar molecule and the negative end of another. The presence of a permanent dipole moment in a molecule is what gives rise to these forces.

Here's how dipole-dipole forces work in simple terms:
  • A molecule with a positive charge on one end and a negative charge on the other creates a dipole.
  • When such molecules are in close proximity, the positive side of one molecule is attracted to the negative side of another, resulting in an intermolecular force.
Chloride, bromide, and iodide form polar compounds with notable electronegativity differences. This means that dipole-dipole forces play a significant role in explaining their relatively high boiling points, as these forces necessitate additional energy to be overcome during a phase change.
Van der Waals Forces
Van der Waals forces, also known as London dispersion forces, are a type of weak intermolecular force that exist among all molecules, whether they are polar or nonpolar. They occur due to the momentary distribution of electrons in molecules. These temporary dipoles arise because the electron cloud of a molecule is constantly moving, sometimes creating an uneven distribution of charge.

Essence of van der Waals forces:
  • They are the weakest among intermolecular forces but are universal.
  • Greater molecular size and increased number of electrons enhance these forces.
  • Larger surface area enables stronger van der Waals forces by increasing the possible area for contact between molecules.
In larger molecules like those of chloride, bromide, and iodide, van der Waals forces significantly contribute to the boiling point. These molecules have a greater number of electrons, leading to stronger van der Waals interactions, and consequently, a higher boiling point.

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Most popular questions from this chapter

The major product R obtained in the following reaction is/are : $$ \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \frac{\mathrm{O}_{3}}{\underset{\left(\mathrm{l}, \mathrm{H}_{2}\right)}{\mathrm{Z}} \mathrm{O}} \mathrm{P}+\mathrm{Q} \frac{\mathrm{H}_{2} / \mathrm{Pd}}{\mathrm{B}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{R} $$ (A) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{OH}\) (B) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{O}\) (C) \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{OH}\) (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\)

The incorrect option for the following reaction sequences is \(\mathrm{P} \frac{\mathrm{H}_{2} / \mathrm{Pd}}{\mathrm{leq}} \mathrm{Q} \frac{\mathrm{O}_{3}}{\mathrm{Zn}, \mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\) (Only product) (A) \(\mathrm{P}\) may be \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (B) \(\mathrm{P}\) may be \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) (C) \(Q\) is cis 2-butene (D) \(\mathrm{Q}\) is trans 2 -butene

Identify set of compound in which first compound react by higher rate with HBr than second compound. (A) \(\mathrm{HC} \equiv \mathrm{CH}, \mathrm{CH}_{2}=\mathrm{CH}_{2}\) (B) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{O}-\mathrm{C}_{6} \mathrm{H}_{5}, \mathrm{C}_{6} \mathrm{H}_{\mathrm{i1}}-\mathrm{O}-\mathrm{C}_{6} \mathrm{H}_{11}\) (C) (D)

Geometry of transition state during chlorination and bromination in Ist step of propagation respectively is: (A) Tetrahedral, planar (B) Planar, tetrahedral (C) Planar, Planar (D) Tetrahedral, Tetrahedral

Driving force for Finkelstein reaction is (A) Nal is a very good nucleophile (B) NaI is a very good leaving group (C) Formation of ppt. facilitates forward reaction (D) None of these

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