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Chapter 5: Problem 184

Using only the periodic table, determine which bond in each pair is more ionic: (a) \(\mathrm{H}-\mathrm{F}\) or \(\mathrm{H}-\mathrm{Cl}\) (b) \(\mathrm{O}-\mathrm{F}\) or \(\mathrm{C}-\mathrm{F}\)

Short Answer

Expert verified
(a) The \(\mathrm{H}-\mathrm{F}\) bond is more ionic, with an electronegativity difference of 1.88 compared to 1.06 for \(\mathrm{H}-\mathrm{Cl}\). (b) The \(\mathrm{C}-\mathrm{F}\) bond is more ionic, with an electronegativity difference of 1.43 compared to 0.54 for \(\mathrm{O}-\mathrm{F}\).

Step by step solution

01

Determine electronegativity values from the periodic table

Using the periodic table, we look up the electronegativity values for H, F, Cl, O, and C: - H (Hydrogen) has an electronegativity value of 2.1 - F (Fluorine) has an electronegativity value of 3.98 - Cl (Chlorine) has an electronegativity value of 3.16 - O (Oxygen) has an electronegativity value of 3.44 - C (Carbon) has an electronegativity value of 2.55
02

Calculate the electronegativity differences for each bond pair

Now, we will compute the electronegativity difference for each bond pair: (a) H-F and H-Cl - H-F: 3.98 - 2.1 = 1.88 - H-Cl: 3.16 - 2.1 = 1.06 (b) O-F and C-F - O-F: 3.98 - 3.44 = 0.54 - C-F: 3.98 - 2.55 = 1.43
03

Determine which bond is more ionic in each pair

Now that we have the electronegativity differences calculated, we can identify the more ionic bond in each pair: (a) H-F or H-Cl As the electronegativity difference between H and F (1.88) is greater than the electronegativity difference between H and Cl (1.06), the H-F bond is more ionic. (b) O-F or C-F Since the electronegativity difference between O and F (0.54) is less than the electronegativity difference between C and F (1.43), the C-F bond is more ionic. So, the final answer is: (a) The \(\mathrm{H}-\mathrm{F}\) bond is more ionic. (b) The \(\mathrm{C}-\mathrm{F}\) bond is more ionic.

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