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Chapter 8: Problem 42

Rank the following bonds in order of increasing ionic character: $\mathrm{N}-\mathrm{O}, \mathrm{Ca}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Br}-\mathrm{Br}, \mathrm{K}-\mathrm{F}$ .

Short Answer

Expert verified
The bonds ranked in order of increasing ionic character are: \(\mathrm{Br}-\mathrm{Br}, \mathrm{N}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Ca}-\mathrm{O}, \mathrm{K}-\mathrm{F}\).

Step by step solution

01

Determine Electronegativity Values for Each Atom

Consult a periodic table or another reference to find the electronegativity values for each of the atoms involved in the bonds: N: 3.04, O: 3.44, Ca: 1.00, C: 2.55, F: 3.98, Br: 2.96, K: 0.82
02

Calculate Electronegativity Difference for Each Bond

For each bond, subtract the lower electronegativity value from the higher value to find the electronegativity difference: N-O: |3.04 - 3.44| = 0.4 Ca-O: |1.00 - 3.44| = 2.44 C-F: |2.55 - 3.98| = 1.43 Br-Br: |2.96 - 2.96| = 0 K-F: |0.82 - 3.98| = 3.16
03

Rank the Bonds in Order of Increasing Ionic Character

Rank the bonds from lowest to highest electronegativity difference: 1. Br-Br: 0 2. N-O: 0.4 3. C-F: 1.43 4. Ca-O: 2.44 5. K-F: 3.16 So, the bonds ranked in order of increasing ionic character are: \(\mathrm{Br}-\mathrm{Br}, \mathrm{N}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Ca}-\mathrm{O}, \mathrm{K}-\mathrm{F}\).

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