Questions: Calculate the percent ionic character for the sodium chloride, NaCl, molecule. The measured dipole moment, μ for sodium chloride is 9.00 D and the Na-Cl bond length is 236 pm, 1 D=3.34 x 10^-20 Cm.

Solution Steps

We need to calculate the percent ionic character of the sodium chloride (NaCl) molecule using the given dipole moment and bond length. The percent ionic character is determined by comparing the actual dipole moment to the dipole moment if the bond were 100% ionic.

The theoretical dipole moment (\(\mu_{\text{theoretical}}\)) for a 100% ionic bond is calculated using the formula: \[ \mu_{\text{theoretical}} = q \times d \] where \(q\) is the charge of an electron (\(1.602 \times 10^{-19} \, \text{C}\)) and \(d\) is the bond length in meters. Given \(d = 236 \, \text{pm} = 236 \times 10^{-12} \, \text{m}\), we have: \[ \mu_{\text{theoretical}} = (1.602 \times 10^{-19} \, \text{C}) \times (236 \times 10^{-12} \, \text{m}) = 3.777 \times 10^{-29} \, \text{Cm} \]

The measured dipole moment is given as \(9.00 \, \text{D}\). Convert this to Coulomb-meters (Cm) using the conversion \(1 \, \text{D} = 3.34 \times 10^{-30} \, \text{Cm}\): \[ \mu_{\text{measured}} = 9.00 \, \text{D} \times 3.34 \times 10^{-30} \, \text{Cm/D} = 3.006 \times 10^{-29} \, \text{Cm} \]

The percent ionic character is calculated using the formula: \[ \text{Percent Ionic Character} = \left( \frac{\mu_{\text{measured}}}{\mu_{\text{theoretical}}} \right) \times 100\% \] Substituting the values: \[ \text{Percent Ionic Character} = \left( \frac{3.006 \times 10^{-29} \, \text{Cm}}{3.777 \times 10^{-29} \, \text{Cm}} \right) \times 100\% = 79.6\% \]

Final Answer