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Hydrogen chloride and hydrogen fluoride are both polar molecules…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Hydrogen chloride and hydrogen fluoride are both polar molecules. In each one the halogen end carries a partial negative charge and the hydrogen end an equal partial positive charge. A student wants to compare how ionic the two bonds are.

The data are as follows. For $\text{HCl}$ the dipole moment is $1.07\ \text{D}$ and the bond length is $127\ \text{pm}$ . For $\text{HF}$ the dipole moment is $1.82\ \text{D}$ and the bond length is $92\ \text{pm}$ . Take $1\ \text{D} = 3.33564 \times 10^{-30}\ \text{C m}$ and $e = 1.602 \times 10^{-19}\ \text{C}$ .

Suppose a bond were fully ionic. Then the hydrogen end would carry $+e$ and the halogen end $-e$ , at the same bond length. The per cent ionic character is the observed dipole moment as a percentage of this fully ionic dipole moment.

What is the dipole moment of $\text{HCl}$ in C m?
What would the dipole moment of $\text{HCl}$ be, in debye, if its bond were fully ionic?
What is the per cent ionic character of the H-Cl bond?
What is the per cent ionic character of the H-F bond?
What is the size of the partial charge on the hydrogen atom of $\text{HF}$ , in coulombs?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $\mu$ and $d$ for both molecules, and the size of $e$ .

Plan. Find the fully ionic dipole moment of each bond, $\mu_{\text{ion}} = e \times d$ . Then the per cent ionic character is $\dfrac{\mu}{\mu_{\text{ion}}} \times 100$ .

Step 1. (a) Observed dipole of HCl in SI units. $\mu = 1.07 \times 3.33564 \times 10^{-30}\ \text{C m}$ , which is $3.57 \times 10^{-30}\ \text{C m}$ .

Step 2. (b) Fully ionic HCl. $\mu_{\text{ion}} = 1.602 \times 10^{-19}\ \text{C} \times 127 \times 10^{-12}\ \text{m}$ . This is $2.035 \times 10^{-29}\ \text{C m}$ . Divide by $3.33564 \times 10^{-30}\ \text{C m}$ per debye: $6.10\ \text{D}$ .

Step 3. (c) Per cent ionic character of H-Cl. $\dfrac{1.07\ \text{D}}{6.10\ \text{D}} \times 100 = 17.5\ \%$ .

Step 4. (d) H-F. The fully ionic dipole is $1.602 \times 10^{-19}\ \text{C} \times 92 \times 10^{-12}\ \text{m}$ . In debye this is $4.42\ \text{D}$ . So the per cent ionic character is $\dfrac{1.82}{4.42} \times 100 = 41.2\ \%$ .

Step 5. (e) The partial charge. Each end of $\text{HF}$ carries $41.2\ \%$ of $e$ : $q = \dfrac{41.2}{100} \times 1.602 \times 10^{-19}\ \text{C}$ . This is $6.60 \times 10^{-20}\ \text{C}$ .

Answer: (a) $3.57 \times 10^{-30}\ \text{C m}$ . (b) $6.10\ \text{D}$ . (c) $17.5\ \%$ . (d) $41.2\ \%$ . (e) $6.60 \times 10^{-20}\ \text{C}$ .

Check. Go the other way for $\text{HF}$ : $q = \dfrac{\mu}{d}$ . That gives $\dfrac{1.82 \times 3.33564 \times 10^{-30}}{92 \times 10^{-12}}\ \text{C}$ , which is $6.60 \times 10^{-20}\ \text{C}$ again. The H-F bond is more ionic, as the larger electronegativity difference predicts.

Watch out. Do not reuse the fully ionic value of $\text{HCl}$ for $\text{HF}$ . That gives $29.8\ \%$ , which is wrong because the H-F bond is shorter. Also change picometres to metres before you multiply.

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