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Chapter 5: Problem 182

Arrange the following sets of bonds in order of increasing ionic character. Use the symbols \(\delta+\) and \(\delta-\) to indicate partial charges, if any, in the bonds: \(\mathrm{O}-\mathrm{Cl}, \mathrm{C}-\mathrm{F}, \mathrm{N}-\mathrm{Cl}, \mathrm{O}-\mathrm{H}, \mathrm{S}-\mathrm{O}\)

Short Answer

Expert verified
The bonds in order of increasing ionic character are: N-Cl, O-Cl, S-O, O-H, C-F With partial charges assigned as: 1. N\(\delta+\)-Cl\(\delta-\) 2. O\(\delta-\)-Cl\(\delta+\) 3. S\(\delta+\)-O\(\delta-\) 4. O\(\delta-\)-H\(\delta+\) 5. C\(\delta+\)-F\(\delta-\)

Step by step solution

01

Electronegativities of the elements involved

First, we need to know the electronegativities of the elements involved in the given bonds, which can be found in a periodic table or related resources. I will provide the approximate electronegativities below: 1. Oxygen (O): 3.44 2. Chlorine (Cl): 3.16 3. Carbon (C): 2.55 4. Fluorine (F): 3.98 5. Nitrogen (N): 3.04 6. Hydrogen (H): 2.20 7. Sulfur (S): 2.58 Now we will calculate the electronegativity differences for each bond.
02

Calculate electronegativity differences

1. O-Cl: \(\Delta EN = 3.44 - 3.16 = 0.28\) 2. C-F: \(\Delta EN = 3.98 - 2.55 = 1.43\) 3. N-Cl: \(\Delta EN = 3.16 - 3.04 = 0.12\) 4. O-H: \(\Delta EN = 3.44 - 2.20 = 1.24\) 5. S-O: \(\Delta EN = 3.44 - 2.58 = 0.86\)
03

Arrange bonds in increasing order of ionic character

Here, higher electronegativity differences correspond to higher ionic characters. Thus, we can arrange these differences in increasing order: 1. N-Cl \(\Delta EN = 0.12\) 2. O-Cl \(\Delta EN = 0.28\) 3. S-O \(\Delta EN = 0.86\) 4. O-H \(\Delta EN = 1.24\) 5. C-F \(\Delta EN = 1.43\) Based on these values, the bonds in order of increasing ionic character are: N-Cl, O-Cl, S-O, O-H, C-F
04

Assign partial charges to the bonds

Next, let's indicate the partial charges in the bonds using the symbols \(\delta+\) (partial positive charge) and \(\delta-\) (partial negative charge): 1. N-Cl: N\(\delta+\)-Cl\(\delta-\), as Cl is more electronegative than N 2. O-Cl: O\(\delta-\)-Cl\(\delta+\), as O is more electronegative than Cl 3. S-O: S\(\delta+\)-O\(\delta-\), as O is more electronegative than S 4. O-H: O\(\delta-\)-H\(\delta+\), as O is more electronegative than H 5. C-F: C\(\delta+\)-F\(\delta-\), as F is more electronegative than C

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