Exercises
Exercise 1. Use periodic trends to order the following elements from smallest to largest electronegativity.
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Exercise 2. Classify each of the following bonds as nonpolar covalent, polar covalent, or ionic.
Al-Cl EN difference = 3.0 – 1.5 = 1.5 polar covalent
P-O EN difference = 3.5 – 2.1 = 1.6 polar covalent
H-S EN difference = 2.5 – 2.1 = 0.4 nonpolar covalent
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Exercise 3. Classify the following bonds as nonpolar, polar, or ionic.
Se-O EN difference = 3.5 – 2.4 = 1.1 polar covalent
P-I EN difference = 2.5 – 2.1 = 0.4 nonpolar covalent
Rb-S EN difference = 2.5 – 0.8 = 1.7 polar covalent
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Exercise 4. Calculate the percent ionic character for NaF. The dipole moment is 8.123 D and the distance between Na+ and F– is 231 pm.
μ = Q x r = 1.60 x 10-19 C x (2.31 x 10-10 m) = 3.696 x 10-29 C⋅m
\(\displaystyle 3.696\times\;10^{-29}\;C⋅m\times\frac{1\;D}{3.34\times\;10^{-30}C⋅m}\;=\;11.07\;D\)
\(\displaystyle \%\;ionic\;character\;=\;\frac{8.123\;D}{11.07\;D}\times\;100\;=\;\mathbf{73.38\%}\)
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Exercise 5. Calculate the percent ionic character for KBr if the dipole moment is 10.603 D and the distance between K+ and Br– is 282 pm.
μ = Q x r = 1.60 x 10-19 C x (2.82 x 10-10 m) = 4.512 x 10-29 C⋅m
\(\displaystyle 4.152\times\;10^{-29}\;C⋅m\times\frac{1\;D}{3.34\times\;10^{-30}C⋅m}\;=\;13.51\;D\)
\(\displaystyle \%\;ionic\;character\;=\;\frac{10.603\;D}{13.51\;D}\times\;100\;=\;\mathbf{78.48\%}\)
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Exercise 6. Use electronegativity values to determine which bond, Na-H or H-F, has the greater ionic character. You can then use the following data to check your answer:
Na-H: μ = 6.400 D, distance between ions = 190 pm
H-F: μ = 1.82 D, distance between ions = 92 pm
EN difference NaH: 2.1 – 0.9 = 1.2, the bond is polar covalent
EN difference H-F: 4.0 – 2.1 = 1.9 polar covalent but significant ionic character
Na-H:
μ = Q x r = 1.60 x 10-19 C x (1.90 x 10-10 m) = 3.04 x 10-29 C⋅m
\(\displaystyle 3.04\times\;10^{-29}\;C⋅m\times\frac{1\;D}{3.34\times\;10^{-30}C⋅m}\;=\;9.102\;D\)
\(\displaystyle \%\;ionic\;character\;=\;\frac{6.400\;D}{9.102\;D}\times\;100\;=\;\mathbf{70.31\%}\)
H-F:
μ = Q x r = 1.60 x 10-19 C x (0.92 x 10-10 m) = 1.47 x 10-29 C⋅m
\(\displaystyle 1.47\times\;10^{-29}\;C⋅m\times\frac{1\;D}{3.34\times\;10^{-30}C⋅m}\;=\;4.407\;D\)
\(\displaystyle \%\;ionic\;character\;=\;\frac{1.82\;D}{4.407\;D}\times\;100\;=\;\mathbf{41.3\%}\)
