Question

Carbon tetrachloride, \(\mathrm{CCl}_{4},\) and chloroform, \(\mathrm{CHCl}_{3},\) are common organic liquids. Carbon tetrachloride's normal boiling point is \(77^{\circ} \mathrm{C} ;\) chloroform's normal boiling point is \(61^{\circ} \mathrm{C} .\) Which statement is the best explanation of these data? (a) Chloroform can hydrogen-bond, but carbon tetrachloride cannot. (b) Carbon tetrachloride has a larger dipole moment than chloroform. (c) Carbon tetrachloride is more polarizable than chloroform.

Short answer

The best explanation for the difference in boiling points between carbon tetrachloride (CCl4) and chloroform (CHCl3) is (c) Carbon tetrachloride is more polarizable than chloroform. This is because CCl4 has a larger electron cloud, leading to stronger van der Waals forces and a higher boiling point.

Step-by-step solution

Analyze hydrogen bonding capabilities of the molecules

Hydrogen bonding occurs when a hydrogen atom is bonded to a highly electronegative atom (like O, N, or F), resulting in a strong dipole. We need to investigate if CCl4 and CHCl3 can form hydrogen bonds. Here, we find that neither of the molecules form hydrogen bonds. Both molecules have a central carbon atom, surrounded by four chlorine atoms in the case of CCl4, and three chlorine atoms and one hydrogen atom in the case of CHCl3. There is no electronegative atom bonded to hydrogen, hence, no hydrogen bonding occurs.

Analyze the dipole moments of the molecules

A dipole moment is formed when there is a separation of positive and negative charges in a molecule. In the case of CCl4, the four highly electronegative chlorine atoms arrange themselves symmetrically around the central carbon atom. This means the opposing dipoles effectively cancel each other out, and there is no net dipole moment. For CHCl3, on the other hand, the symmetry is broken by the hydrogen atom. Although the dipole from the hydrogen atom is not very strong, it still creates a net dipole moment as it takes on one of the positions occupied by chlorine atoms in CCl4.

Analyze polarizability of the molecules

Polarizability refers to the ease with which the electron cloud around a molecule can be distorted, which affects how the molecule interacts with other molecules. The electron cloud can be distorted by an external charge, leading to a temporary dipole moment and van der Waals forces. Carbon tetrachloride has a larger electron cloud than chloroform, due to the presence of four chlorine atoms compared to three in chloroform. Therefore, CCl4 will be more polarizable than CHCl3.

Choose the best explanation from the given choices

We have found that: (a) Neither CCl4 nor CHCl3 can form hydrogen bonds, so this statement is incorrect. (b) CHCl3 has a nonzero dipole moment, while CCl4 doesn't, but this information is insufficient, as the boiling point of CCl4 is higher than that of CHCl3 despite not having a dipole moment. (c) CCl4 is more polarizable than CHCl3, which can explain the higher boiling point of CCl4. The larger electron cloud of CCl4 leads to stronger van der Waals forces, leading to a higher boiling point. So, the correct answer is (c) Carbon tetrachloride is more polarizable than chloroform.

Key concepts

Hydrogen Bonding

Hydrogen bonding is a special type of dipole-dipole interaction, which is stronger than typical dipole interactions. For hydrogen bonding to occur, the hydrogens need to be directly bonded to highly electronegative atoms such as oxygen (O), nitrogen (N), or fluorine (F). This strong electronegativity creates a significant difference in the charges, leading to a strong dipole or hydrogen bond.

In the case of carbon tetrachloride (\(\mathrm{CCl}_{4}\)) and chloroform (\(\mathrm{CHCl}_{3}\)), neither forms hydrogen bonds because there isn't an appropriate electronegative atom bonded to hydrogen in these molecules. Specifically, in \(\mathrm{CHCl}_{3}\), the hydrogen is bonded to carbon, which isn't highly electronegative. As a result, neither molecule can participate in hydrogen bonding, impacting their boiling points differently.

Polarizability

Polarizability refers to the ability of a molecule's electron cloud to be distorted by external electric fields. This distortion creates what we call induced dipoles, which are temporary shifts in electron density across a molecule. Induced dipoles contribute to van der Waals forces, which play a significant role in determining physical properties like boiling and melting points.

Carbon tetrachloride (\(\mathrm{CCl}_{4}\)) is more polarizable than chloroform (\(\mathrm{CHCl}_{3}\)) due to its larger electron cloud size. This larger cloud arises because \(\mathrm{CCl}_{4}\) has four chlorine atoms compared to \(\mathrm{CHCl}_{3}\)'s three. Chlorine is relatively large with substantial electron density, enhancing \(\mathrm{CCl}_{4}\)'s polarizability. Thus, \(\mathrm{CCl}_{4}\) experiences stronger van der Waals forces, contributing to its higher boiling point.

Van der Waals Forces

Van der Waals forces are the collective group of intermolecular forces that arise between molecules, which include dipole-dipole interactions, London dispersion forces, and sometimes hydrogen bonds. These forces are weaker than ionic or covalent bonds but are crucial for the physical state of substances.

In our example, van der Waals forces can help explain the boiling point differences between \(\mathrm{CCl}_{4}\) and \(\mathrm{CHCl}_{3}\). With \(\mathrm{CCl}_{4}\), the molecule's symmetry results in no permanent dipole moment. However, its four large chlorine atoms contribute to significant London dispersion forces due to their increased polarizability. These forces are enough to elevate its boiling point compared to \(\mathrm{CHCl}_{3}\), where the presence of the slightly polar C-H bond does not contribute as significantly to intermolecular forces.