Draw two HCl molecules and show how they would be attracted to each other. Show the partial charges and dipole moment vectors for both molecules, and orient the molecules properly with respect to each other.

Short Answer

Expert verified
Two HCl molecules can be represented as \(δ^+H-Cl\:δ^-\) and \(δ^+H-Cl\:δ^-\), where H has a partial positive charge (δ+) and Cl has a partial negative charge (δ-). The dipole moment vectors (arrows) point from hydrogen to chlorine atoms. To show the dipole-dipole interaction, orient the molecules so that the oppositely charged ends are close to each other, as shown: \(δ^+H \overset{\underset{\smile}{\longrightarrow}}{|\vec{d}|}Cl \: δ^-\) and \(δ^+H \overset{\underset{\smile}{\longrightarrow}}{|\vec{d}|}Cl \: δ^-\).

Step by step solution

01

Draw the HCl molecules with their respective atoms

Draw two HCl molecules with hydrogen (H) connected to chlorine (Cl) via a single bond: \(H - Cl\) and \(H - Cl\)
02

Indicate the partial charges on each atom

Since chlorine is more electronegative than hydrogen, it will have a partial negative charge (δ-) and hydrogen will have a partial positive charge (δ+). The molecules will now look like: \(δ^+ H - Cl \: δ^-\) and \(δ^+ H - Cl \: δ^-\)
03

Represent the dipole moment vectors

A dipole moment vector indicates the direction of the dipole moment, from the positive end to the negative end. Draw an arrow from the hydrogen to the chlorine atom in each molecule, representing the dipole moment. The molecules with the dipole moment vectors will look like: \(δ^+ H - \vec{d} \to Cl \: δ^-\) and \(δ^+ H - \vec{d} \to Cl \: δ^-\)
04

Orient the molecules with respect to each other to show their attraction

In order to show the attraction between the two HCl molecules, orient them so that the positively charged hydrogen in one molecule is close to the negatively charged chlorine in the other molecule, and vice versa. This interaction is known as a dipole-dipole interaction. The final arrangement of the two HCl molecules with their attractions will look like: \(δ^+ H \overset{\underset{\smile}{\longrightarrow}}{|\vec{d}|}Cl \: δ^-\) and \(δ^+ H \overset{\underset{\smile}{\longrightarrow}}{|\vec{d}|}Cl \: δ^-\)

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