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Chapter 3: 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 given 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

Know Electronegativity Values

We must find the electronegativity values for the elements involved in the given bonds: - Nitrogen (N): 3.04 - Oxygen (O): 3.44 - Calcium (Ca): 1.00 - Carbon (C): 2.55 - Fluorine (F): 3.98 - Bromine (Br): 2.96 - Potassium (K): 0.82
02

Calculate Electronegativity Difference

Now, we need to find the difference in electronegativity values for each bond: - N-O: \(|3.44-3.04|\) = 0.40 - Ca-O: \(|3.44-1.00|\) = 2.44 - C-F: \(|3.98-2.55|\) = 1.43 - Br-Br: \(|2.96-2.96|\) = 0 - K-F: \(|3.98-0.82|\) = 3.16
03

Rank Bonds

Rank the bonds in order of increasing ionic character based on the electronegativity differences calculated in Step 2: 1. Br-Br (0.00) 2. N-O (0.40) 3. C-F (1.43) 4. Ca-O (2.44) 5. K-F (3.16) So, the given 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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