(a) What atoms must a molecule contain to participate in hydrogen bonding with other molecules of the same kind? (b) Which of the following molecules can form hydrogen bonds with other molecules of the same kind: \(\mathrm{CH}_{3} \mathrm{~F}, \mathrm{CH}_{3} \mathrm{NH}_{2}\), \(\mathrm{CH}_{3} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{Br} ?\)

Short Answer

Expert verified
(a) A molecule must contain hydrogen (H) bonded to nitrogen (N), oxygen (O), or fluorine (F) to participate in hydrogen bonding with other molecules of the same kind. (b) The molecules that can form hydrogen bonds with other molecules of the same kind are \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) and \(\mathrm{CH}_{3}\mathrm{OH}\).

Step by step solution

01

Understanding hydrogen bonding

Hydrogen bonding is a type of intermolecular force that occurs between molecules containing a hydrogen atom bonded to a highly electronegative element, usually nitrogen (N), oxygen (O), or fluorine (F). These special interactions occur when the positive end of a polar molecule (the hydrogen) is attracted to the negative end (N, O, or F) of another polar molecule.
02

Identifying atoms involved in hydrogen bonding for a molecule

For a molecule to participate in hydrogen bonding with other molecules of the same kind, it must have a hydrogen atom bonded to nitrogen, oxygen, or fluorine (N-H, O-H, or F-H). (a)
03

Identifying the elements required for hydrogen bonding

From Step 2, we know that a molecule must contain hydrogen (H) bonded to nitrogen (N), oxygen (O), or fluorine (F) to participate in hydrogen bonding with other molecules of the same kind. (b)
04

Determining if the given molecules can form hydrogen bonds

We will examine each molecule to see if it contains the necessary elements for hydrogen bonding. 1. \(\mathrm{CH}_{3}\mathrm{F}\): This molecule has a hydrogen atom, but it is bonded to a carbon atom instead of fluorine. Therefore, it does not have the necessary elements for hydrogen bonding. 2. \(\mathrm{CH}_{3}\mathrm{NH}_{2}\): In this molecule, hydrogen is bonded to nitrogen, so this molecule can form hydrogen bonds with other molecules of the same kind. 3. \(\mathrm{CH}_{3}\mathrm{OH}\): In this molecule, hydrogen is bonded to oxygen, so this molecule can form hydrogen bonds with other molecules of the same kind. 4. \(\mathrm{CH}_{3}\mathrm{Br}\): This molecule has a hydrogen atom, but it is bonded to a carbon atom instead of an electronegative atom (N, O, or F). Therefore, it does not have the necessary elements for hydrogen bonding.
05

Summarizing the results

The molecules that can form hydrogen bonds with other molecules of the same kind are \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) and \(\mathrm{CH}_{3}\mathrm{OH}\).

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Key Concepts

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

Intermolecular Forces
Intermolecular forces are the attractive forces that exist between individual molecules. They are responsible for determining the physical properties of substances, such as boiling and melting points, viscosity, and solubility. Different types of intermolecular forces include hydrogen bonds, dipole-dipole interactions, and London dispersion forces.

Hydrogen bonds are particularly strong intermolecular forces that form when a hydrogen atom bonded to a highly electronegative atom is attracted to another electronegative atom in a neighboring molecule. The strength of these bonds has a significant effect on the properties of the substance. For example, water's high boiling point and unique properties are primarily due to hydrogen bonding between the water molecules.

Dipole-dipole interactions occur between polar molecules, where the positive end of one molecule is attracted to the negative end of another. London dispersion forces are weak intermolecular forces that arise from the momentary changes in electron density in nonpolar molecules, causing temporary dipoles. All these interactions influence a substance's state, structure, and behavior.
Electronegativity
Electronegativity refers to the ability of an atom in a molecule to attract shared electrons to itself. On the periodic table, electronegativity values typically increase from left to right and bottom to top. Fluorine is the most electronegative element, followed by oxygen and nitrogen. These differences in electronegativity are crucial for understanding the formation of polar bonds and the emergence of intermolecular forces.

When there is a large difference in electronegativity between two atoms bonded together, the shared electrons are more likely to be found closer to the more electronegative atom. This creates a polar bond, with a partial negative charge on the more electronegative atom and a partial positive charge on the less electronegative one. In the context of hydrogen bonding, the high electronegativity of nitrogen, oxygen, or fluorine in relation to hydrogen is what enables these special intermolecular forces to exist. Electronegativity also helps to explain why certain molecules, despite containing hydrogen, cannot participate in hydrogen bonding if they lack sufficiently electronegative atoms bonded to hydrogen.
Polar Molecules
Polar molecules are molecules with an uneven distribution of charge due to the presence of polar bonds that are arranged asymmetrically. The existence of polar molecules is a result of differences in electronegativity between the atoms that form bonds in the molecule. A classic example is water (H_2O), where the oxygen atom is more electronegative than the hydrogen atoms, leading to a bent shape with a partial negative charge on the oxygen and a partial positive charge on the hydrogens.

Polarity in molecules is crucial for understanding solubility, reactivity, and the strength of intermolecular forces, such as hydrogen bonds. Polar molecules are often capable of forming dipole-dipole interactions with each other, and in cases where hydrogen is bound to nitrogen, oxygen, or fluorine, they can engage in hydrogen bonding, which is much stronger than regular dipole-dipole interactions. When molecules like CH_3NH_2 and CH_3OH, which have H atoms bonded to N or O, are present, they can align themselves such that opposite charges attract, forming hydrogen bonds. On the other hand, nonpolar molecules like CH_3Br cannot do this due to the uniform distribution of charge, which is why polar and nonpolar substances usually do not mix well.

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Most popular questions from this chapter

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

The boiling points, surface tensions, and viscosities of water and several alchohols are as follows: $$ \begin{array}{lrcc} & \begin{array}{l} \text { Boiling } \\ \text { Point }\left({ }^{\circ} \mathbf{C}\right) \end{array} & \begin{array}{l} \text { Surface } \\ \text { Tension }\left(\mathbf{J} / \mathbf{m}^{2}\right) \end{array} & \begin{array}{l} \text { Viscosity } \\ (\mathbf{k g} / \mathbf{m}-\mathbf{s}) \end{array} \\ \hline \text { Water, } \mathrm{H}_{2} \mathrm{O} & 100 & 7.3 \times 10^{-2} & 0.9 \times 10^{-3} \\ \text {Ethanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & 78 & 2.3 \times 10^{-2} & 1.1 \times 10^{-3} \\ \text {Propanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 97 & 2.4 \times 10^{-2} & 2.2 \times 10^{-3} \\ n \text { -Butanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 117 & 2.6 \times 10^{-2} & 2.6 \times 10^{-3} \\\ \text {Ethylene glycol, } \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 197 & 4.8 \times 10^{-2} & 26 \times 10^{-3} \end{array} $$ (a) For ethanol, propanol, and \(n\) -butanol the boiling points, surface tensions, and viscosities all increase. What is the reason for this increase? (b) How do you explain the fact that propanol and ethylene glycol have similar molecular weights \((60\) versus \(62 \mathrm{amu}),\) yet the viscosity of ethylene glycol is more than 10 times larger than propanol? (c) How do you explain the fact that water has the highest surface tension but the lowest viscosity?

Suppose you have two colorless molecular liquids, one boiling at \(-84^{\circ} \mathrm{C}\), the other at \(34{ }^{\circ} \mathrm{C},\) and both at atmospheric pressure. Which of the following statements is correct? For each statement that is not correct, modify the statement so that it is correct. (a) The higher-boiling liquid has greater total intermolecular forces than the lower- boiling liquid. (b) The lower-boiling liquid must consist of nonpolar molecules. (c) The lower-boiling liquid has a lower molecular weight than the higher-boiling liquid. (d) The two liquids have identical vapor pressures at their normal boiling points. (e) \(\mathrm{At}-84{ }^{\circ} \mathrm{C}\) both liquids have vapor pressures of \(760 \mathrm{~mm} \mathrm{Hg}\).

For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?

The following quote about ammonia \(\left(\mathrm{NH}_{3}\right)\) is from a textbook of inorganic chemistry: "It is estimated that \(26 \%\) of the hydrogen bonding in \(\mathrm{NH}_{3}\) breaks down on melting, \(7 \%\) on warming from the melting to the boiling point, and the final \(67 \%\) on transfer to the gas phase at the boiling point." From the standpoint of the kinetic energy of the molecules, explain (a) why there is a decrease of hydrogen-bonding energy on melting and (b) why most of the loss in hydrogen bonding occurs in the transition from the liquid to the vapor state.

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