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Multiple Choice
What is the percent ionic character of a 125 pm bond which has a bond dipole of 4.5 D?
A
6.0%
B
75%
C
133%
D
0.75%
E
1.33%
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Verified step by step guidance
1
Understand that the percent ionic character of a bond is a measure of the degree to which the bond between two atoms is ionic versus covalent.
Use the formula for percent ionic character: \( \text{Percent Ionic Character} = \left( \frac{\mu_{\text{observed}}}{\mu_{\text{ionic}}} \right) \times 100 \), where \( \mu_{\text{observed}} \) is the observed dipole moment and \( \mu_{\text{ionic}} \) is the dipole moment if the bond were 100% ionic.
Calculate \( \mu_{\text{ionic}} \) using the formula \( \mu_{\text{ionic}} = q \times d \), where \( q \) is the charge of an electron (approximately \( 1.602 \times 10^{-19} \) C) and \( d \) is the bond length in meters. Convert the bond length from picometers to meters: \( 125 \text{ pm} = 125 \times 10^{-12} \text{ m} \).
Substitute the values into the formula for \( \mu_{\text{ionic}} \) to find the theoretical dipole moment if the bond were 100% ionic.
Substitute the observed dipole moment (4.5 D) and the calculated \( \mu_{\text{ionic}} \) into the percent ionic character formula to find the percent ionic character of the bond.