Indicate the bond polarity (show the partial positive and partial negative ends) in the following bonds. a. \(\mathrm{C}-\mathrm{O}\) d. \(\mathrm{Br}-\mathrm{Te}\) b. \(\mathrm{P}-\mathrm{H}\) e. \(\mathrm{Se}-\mathrm{S}\) c. \(\mathrm{H}-\mathrm{Cl}\)

Consult a periodic table or table of electronegativity values to find the electronegativity of each atom in each bond. a. C-O: Electronegativity of C: 2.55 Electronegativity of O: 3.44 d. Br-Te: Electronegativity of Br: 2.96 Electronegativity of Te: 2.1 b. P-H: Electronegativity of P: 2.19 Electronegativity of H: 2.20 e. Se-S: Electronegativity of Se: 2.55 Electronegativity of S: 2.58 c. H-Cl: Electronegativity of H: 2.20 Electronegativity of Cl: 3.16

Calculate the difference in electronegativity for each bond. The atom with the greater electronegativity will have a partial negative charge, while the atom with lower electronegativity will have a partial positive charge. a. C-O: Electronegativity difference: 3.44 - 2.55 = 0.89 Bond Polarity: \(\mathrm{C}^{δ+}-\mathrm{O}^{δ-}\) d. Br-Te: Electronegativity difference: 2.96 - 2.1 = 0.86 Bond Polarity: \(\mathrm{Te}^{δ+}-\mathrm{Br}^{δ-}\) b. P-H: Electronegativity difference: 2.20 - 2.19 = 0.01 Since the difference is very small, the bond between P and H is relatively non-polar. e. Se-S: Electronegativity difference: 2.58 - 2.55 = 0.03 Since the difference is very small, the bond between Se and S is relatively non-polar. c. H-Cl: Electronegativity difference: 3.16 - 2.20 = 0.96 Bond Polarity: \(\mathrm{H}^{δ+}-\mathrm{Cl}^{δ-}\) In summary: - C-O bond: \(\mathrm{C}^{δ+}-\mathrm{O}^{δ-}\) - Br-Te bond: \(\mathrm{Te}^{δ+}-\mathrm{Br}^{δ-}\) - P-H bond: Relatively non-polar - Se-S bond: Relatively non-polar - H-Cl bond: \(\mathrm{H}^{δ+}-\mathrm{Cl}^{δ-}\)