Question

Which of the following would you expect to show dispersion forces? dipole forces? (a) \(\mathrm{GeBr}_{4}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (c) \(\mathrm{HF}(g)\) (d) \(\mathrm{TeCl}_{2}\)

Short answer

Answer: - \(\mathrm{GeBr}_{4}\) exhibits dispersion forces. - \(\mathrm{C}_{2} \mathrm{H}_{2}\) exhibits dispersion forces. - \(\mathrm{HF}(g)\) exhibits both dispersion forces and dipole forces. - \(\mathrm{TeCl}_{2}\) exhibits both dispersion forces and dipole forces.

Step-by-step solution

Determine the Polarity of \(\mathrm{GeBr}_{4}\)

Given the compound \(\mathrm{GeBr}_{4}\), we first consider the electronegativity difference between germanium (Ge) and bromine (Br). Ge has an electronegativity of 2.01 while Br has an electronegativity of 2.96. The difference is quite small, implying that the bonds in this compound may be slightly polar. However, the molecule has a tetrahedral shape, resulting in a symmetric distribution of the bond dipoles. This cancels out the dipoles, and the overall molecule is considered nonpolar.

Identify Forces in \(\mathrm{GeBr}_{4}\)

Because \(\mathrm{GeBr}_{4}\) is a nonpolar molecule, it does not exhibit dipole-dipole interactions, but it does exhibit dispersion forces due to the temporary fluctuations in electron distribution.

Determine the Polarity of \(\mathrm{C}_{2} \mathrm{H}_{2}\)

For the compound \(\mathrm{C}_{2} \mathrm{H}_{2}\), we have a linear molecule with two carbon (C) atoms double-bonded to each other and each carbon atom single-bonded to a hydrogen (H) atom. The electronegativity difference between C and H is small, and the molecule's shape results in a cancellation of any bond dipoles. Therefore, the overall molecule is considered nonpolar.

Identify Forces in \(\mathrm{C}_{2} \mathrm{H}_{2}\)

As a nonpolar molecule, \(\mathrm{C}_{2} \mathrm{H}_{2}\) does not have dipole-dipole interactions. However, it will experience dispersion forces due to the temporary fluctuations in electron distribution.

Determine the Polarity of \(\mathrm{HF}(g)\)

In the case of \(\mathrm{HF}(g)\), we have a single bond between hydrogen (H) and fluorine (F). The electronegativity difference between H and F is quite significant, making the bond polar. The molecule is also linear in shape, so there is no cancellation of bond dipoles. Therefore, \(\mathrm{HF}(g)\) is a polar molecule.

Identify Forces in \(\mathrm{HF}(g)\)

As \(\mathrm{HF}(g)\) is a polar molecule, it will experience dipole-dipole interactions due to the attraction between the positive and negative ends of the molecules. In addition, it will also have dispersion forces, as all molecules have this type of force.

Determine the Polarity of \(\mathrm{TeCl}_{2}\)

With the compound \(\mathrm{TeCl}_{2}\), we first consider the electronegativity difference between tellurium (Te) and chlorine (Cl). Te has an electronegativity of 2.1 while Cl has an electronegativity of 3.16. This relatively large difference suggests that the bonds in this compound can be considered polar. Moreover, the molecule has a bent shape, which does not cancel out the bond dipoles. As a result, the overall molecule is polar.

Identify Forces in \(\mathrm{TeCl}_{2}\)

Since \(\mathrm{TeCl}_{2}\) is a polar molecule, it will have dipole-dipole interactions due to the attraction between the positive and negative ends of the molecules. Additionally, it will experience dispersion forces, as all molecules exhibit this type of force. To summarize: - \(\mathrm{GeBr}_{4}\): Dispersion forces - \(\mathrm{C}_{2} \mathrm{H}_{2}\): Dispersion forces - \(\mathrm{HF}(g)\): Dispersion forces and dipole forces - \(\mathrm{TeCl}_{2}\): Dispersion forces and dipole forces