Question

Classify each bond as nonpolar covalent or polar covalent or state that ions are formed. (a) \(\mathrm{S}-\mathrm{H}\) (b) \(\mathrm{P}-\mathrm{H}\) (c) \(\mathrm{C}-\mathrm{F}\) (d) \(\mathrm{C}-\mathrm{Cl}\)

Short answer

Question: Classify the following bonds as nonpolar covalent, polar covalent, or ionic: (a) S-H, (b) P-H, (c) C-F, (d) C-Cl. Answer: (a) S-H: Nonpolar covalent, (b) P-H: Nonpolar covalent, (c) C-F: Polar covalent, (d) C-Cl: Polar covalent.

Step-by-step solution

Find the electronegativity difference

For each bond, we will calculate the electronegativity difference between the bonded elements using the following electronegativity values: \(\mathrm{S}: 2.58\\ \mathrm{H}: 2.20\\ \mathrm{P}: 2.19\\ \mathrm{C}: 2.55\\ \mathrm{F}: 3.98\\ \mathrm{Cl}: 3.16\) (a) \(\mathrm{S} - \mathrm{H}\): \(2.58 - 2.20 = 0.38\) (b) \(\mathrm{P} - \mathrm{H}\): \(2.19 - 2.20 = -0.01\) (c) \(\mathrm{C} - \mathrm{F}\): \(2.55 - 3.98 = -1.43\) (d) \(\mathrm{C} - \mathrm{Cl}\): \(2.55 - 3.16 = -0.61\)

Classify the bond type

Based on electronegativity differences calculated in step 1 and the guidelines provided in the analysis, we will classify each bond as nonpolar covalent, polar covalent, or ionic: (a) \(\mathrm{S} - \mathrm{H}\) (0.38): Nonpolar covalent (b) \(\mathrm{P} - \mathrm{H}\) (-0.01): Nonpolar covalent (c) \(\mathrm{C} - \mathrm{F}\) (-1.43): Polar covalent (d) \(\mathrm{C} - \mathrm{Cl}\) (-0.61): Polar covalent