Identify the kinds of intermolecular forces that might arise between molecules of each of the following substances: (a) \(\mathrm{NH}_{2} \mathrm{OH}\); (b) \(\mathrm{CBr}_{4}\); (c) \(\mathrm{H}_{2} \mathrm{SeO}_{4}\); (d) \(\mathrm{SO}_{2}\).

Short Answer

Expert verified
For (a) \(\mathrm{NH}_{2} \mathrm{OH}\), hydrogen bonding, dipole-dipole interactions, and London dispersion forces are present. For (b) \(\mathrm{CBr}_{4}\), the primary intermolecular force is London dispersion, with weak dipole-dipole interactions. For (c) \(\mathrm{H}_{2} \mathrm{SeO}_{4}\), hydrogen bonding, dipole-dipole interactions, and London dispersion forces occur. For (d) \(\mathrm{SO}_{2}\), dipole-dipole interactions and London dispersion forces are observed.

Step by step solution

01

Understanding Intermolecular Forces

Intermolecular forces are interactions that occur between molecules. They include dipole-dipole interactions, hydrogen bonding, and London dispersion forces. The presence of certain elements and the molecular structure determines which forces are possible.
02

Analyzing \(\mathrm{NH}_{2} \mathrm{OH}\)

For \(\mathrm{NH}_{2} \mathrm{OH}\), hydrogen atoms are bonded to nitrogen and oxygen, both of which are highly electronegative. This means that hydrogen bonding can occur. Additionally, dipole-dipole interactions and London dispersion forces will also be present.
03

Analyzing \(\mathrm{CBr}_{4}\)

In \(\mathrm{CBr}_{4}\), carbon tetrabromide, there are no hydrogen atoms bonded to highly electronegative atoms, hence no hydrogen bonding. However, due to the differences in electronegativity between carbon and bromine, dipole-dipole interactions may occur but are weak due to the tetrahedral symmetry. Primarily London dispersion forces will be present, which increase with the large mass and size of the bromine atoms.
04

Analyzing \(\mathrm{H}_{2} \mathrm{SeO}_{4}\)

For \(\mathrm{H}_{2} \mathrm{SeO}_{4}\), selenic acid, there are hydrogen atoms bonded to oxygen, a highly electronegative atom, suggesting the presence of hydrogen bonding. There will also be dipole-dipole interactions since the molecule is polar, as well as London dispersion forces.
05

Analyzing \(\mathrm{SO}_{2}\)

In \(\mathrm{SO}_{2}\), sulfur dioxide, there is no hydrogen bonding as there are no hydrogen atoms. However, the molecule is polar as there is a difference in electronegativity between sulfur and oxygen, which gives rise to dipole-dipole interactions. London dispersion forces are also present.

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Key Concepts

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

Hydrogen Bonding
To fully understand why substances like NH2OH and H2SeO4 exhibit hydrogen bonding, it's essential to dive deeper into the concept. Hydrogen bonding is a specific type of intermolecular force that occurs when a hydrogen atom, bonded to a highly electronegative element such as nitrogen, oxygen, or fluorine, is attracted to another electronegative atom in a nearby molecule. The result is a strong interaction relative to other types of intermolecular forces.

Hydrogen bonds are responsible for many of the unique properties of water, such as its high boiling point and surface tension. In our specific cases, NH2OH exhibits this behavior due to nitrogen and oxygen having lone pairs of electrons that offer a site for hydrogen atoms of adjacent molecules to bond. H2SeO4, despite selenium being less electronegative than oxygen, can still participate in hydrogen bonding due to the presence of oxygen.
Dipole-Dipole Interactions
Dipole-dipole interactions are another significant type of force, arising between polar molecules where there is a partial positive charge on one end and a partial negative charge on the other. These are directional interactions, as positive ends of one molecule align with the negative ends of another. Molecules like SO2 exemplify dipole-dipole interactions due to the substantial difference in electronegativity between sulfur and oxygen.

This polarity creates a dipole moment, where you can imagine a molecule having a 'north' and 'south' pole similar to a magnet. In a substance such as CBr4, the symmetrical shape of the molecule makes the dipole moments cancel out, thus considerably weakening any dipole-dipole interactions.
London Dispersion Forces
The London dispersion forces are often thought of as the 'default' interaction because they can be found between all molecules, regardless of polarity. They are particularly significant in nonpolar compounds like CBr4. The size and shape of molecules influence the magnitude of these forces; larger atoms or molecules with more electrons often have stronger London dispersion forces.

This happens because electrons in larger atoms can become more easily polarizable, creating temporary dipoles that induce corresponding dipoles in adjacent molecules, thus attracting them to each other. While these are the weakest of the intermolecular forces, they are omnipresent and can become quite strong in large, heavy atoms or molecules.
Molecular Polarity
The term 'molecular polarity' refers to the distribution of electric charge over the atoms in a molecule. Compounds like SO2 and H2SeO4 are polar due to an unequal sharing of electrons resulting from bonds between atoms with different electronegativities. This disparity leads to a net dipole moment where one end of the molecule is relatively positive, and the other is relatively negative.

Molecular polarity is crucial because it influences a substance's physical and chemical properties, including its solubility, melting point, and how it interacts with other molecules. It's also a determining factor in the kinds of intermolecular forces that can occur within a sample of the substance.
Electronegativity
Electronegativity is a measure of how strongly an atom can attract and hold onto electrons in a chemical bond. Atoms like oxygen, nitrogen, and fluorine are among the most electronegative and can create polarized bonds leading to hydrogen bonding or dipole-dipole interactions. Understanding electronegativity helps us predict the kind of intermolecular forces that may arise in a substance.

For instance, in NH2OH, the significant difference in electronegativity between hydrogen and the atoms it's bound to (nitrogen and oxygen) allows for robust hydrogen bonding. However, in symmetrical molecules like CBr4, even with the difference in electronegativity between carbon and bromine, the symmetry results in a nonpolar molecule, minimizing the impact of electronegativity on intermolecular forces.

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Most popular questions from this chapter

In each pair, indicate which substance has the stronger intermolecular forces and explain your reasoning: (a) Ne, Ar; (b) \(\mathrm{NF}_{3}, \mathrm{BF}_{3} ;\) (c) \(\mathrm{SiH}_{4}, \mathrm{GeH}_{4}\); (d) \(\mathrm{NaF}, \mathrm{HF}\).

Draw the Lewis structure of (a) \(\mathrm{CF}_{4}\), (b) \(\mathrm{SF}_{4}\), name the molecular shape, and indicate whether each can participate in dipole- dipole interactions.

Rank the following molecules in order of increasing viscosity at \(50^{\circ} \mathrm{C}: \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SH}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}, \mathrm{C}_{6} \mathrm{H}_{6}\).

Indium arsenide crystallizes in the sphalerite structure (Fig. \(5.45\) ). (a) What are the coordination numbers of the indium and arsenide ions? (b) What is the formula of indium arsenide?

The following boiling points correspond to the substances listed. Match the boiling points to the substances by considering the relative strengths of their intermolecular forces. b.p. \(\left({ }^{\circ} \mathrm{C}\right)\) : $$ -162,-88.5,28,36,64.5,78.3,82.5,140,205,290 \text {; substance: } $$ \(\mathrm{CH}_{4}, \mathrm{CH}_{3} \mathrm{CHOHCH}_{3}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) (has a benzene ring), \(\mathrm{HOCH}_{2} \mathrm{CHOHCH}_{2} \mathrm{OH}, \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\). Hint: The boiling point of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}{ }_{2} \mathrm{CH}_{3}\) is \(28^{\circ} \mathrm{C}\) and that of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(64.5^{\circ} \mathrm{C}\).

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