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Chapter 8: Problem 41

Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}\) , $\mathrm{O}-\mathrm{H}, \mathrm{N}-\mathrm{H}, \mathrm{F}-\mathrm{H}, \mathrm{C}-\mathrm{H} .$

Short Answer

Expert verified
The bonds are ranked in order of decreasing polarity as follows: \(\mathrm{F-H > O-H > N-H > C-H > P-H}\)

Step by step solution

01

Remember the electronegativity values and difference for each bond

To rank the bonds based on polarity, we need to know the electronegativity values for each of the elements involved. Here are the electronegativity values for the elements given: - Fluorine (F): 3.98 - Oxygen (O): 3.44 - Nitrogen (N): 3.04 - Carbon (C): 2.55 - Phosphorus (P): 2.19 - Boron (B): 2.04 - Hydrogen (H): (between B and C, identical to P, so approximately 2.19) Now, we will calculate the electronegativity difference for each bond: 1. P-H: 2.19 - 2.19 = 0 2. O-H: 3.44 - 2.19 = 1.25 3. N-H: 3.04 - 2.19 = 0.85 4. F-H: 3.98 - 2.19 = 1.79 5. C-H: 2.55 - 2.19 = 0.36
02

Rank the bonds based on decreasing electronegativity difference

Now that we have the electronegativity difference for each bond, we can rank them in order of decreasing polarity. A higher electronegativity difference corresponds to a higher polarity. So, we arrange the bonds in the decreasing order of their electronegativity difference: 1. F-H: 1.79 2. O-H: 1.25 3. N-H: 0.85 4. C-H: 0.36 5. P-H: 0
03

Write the final answer

The bonds are ranked in order of decreasing polarity as follows: \(\mathrm{F-H > O-H > N-H > C-H > P-H}\)

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Most popular questions from this chapter

Place the species below in order of the shortest to the longest nitrogen–oxygen bond. $$\mathrm{H}_{2} \mathrm{NOH}, \quad \mathrm{N}_{2} \mathrm{O}, \quad \mathrm{NO}^{+}, \quad \mathrm{NO}_{2}^{-}, \quad \mathrm{NO}_{3}^{-}$$ $\left(\mathrm{H}_{2} \mathrm{NOH} \text { exists as } \mathrm{H}_{2} \mathrm{N}-\mathrm{OH} .\right)$

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