Question

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. \(\mathrm{H}-\mathrm{Cl}\)

Short answer

The bond polarity for each bond is as follows: a. C-O: polar, C(δ+) - O(δ-) b. P-H: nonpolar c. H-Cl: polar, H(δ+) - Cl(δ-) d. Br-Te: polar, Br(δ-) - Te(δ+) e. Se-S: nonpolar

Step-by-step solution

Determine electronegativities for Carbon and Oxygen

Using a periodic table or a chemistry textbook, find that the electronegativity of Carbon (C) is 2.5 and that of Oxygen (O) is 3.5.

Calculate the difference in electronegativities

For the C-O bond: subtract the electronegativity of Carbon from Oxygen: \[\text{Electronegativity difference} = 3.5 - 2.5 = 1.0\]

Determine bond polarity

Since there is a significant difference in electronegativities, the bond between Carbon and Oxygen is polar. The more electronegative atom, Oxygen, will have a partial negative charge (marked as δ-) and the less electronegative atom, Carbon, will have a partial positive charge (marked as δ+): \[\mathrm{C \,(\delta+)} - \mathrm{O\,(\delta-)}\] b. \(\mathrm{P}-\mathrm{H}\):

Determine electronegativities for Phosphorus and Hydrogen

Using a periodic table or a chemistry textbook, find that the electronegativity of Phosphorus (P) is 2.1 and that of Hydrogen (H) is 2.2.

Calculate the difference in electronegativities

For the P-H bond: subtract the electronegativity of Phosphorus from Hydrogen: \[\text{Electronegativity difference} = 2.2 - 2.1 = 0.1\]

Determine bond polarity

Since the difference in electronegativities is low, the bond between Phosphorus and Hydrogen is considered nonpolar. c. \(\mathrm{H}-\mathrm{Cl}\):

Determine electronegativities for Hydrogen and Chlorine

Using a periodic table or a chemistry textbook, find that the electronegativity of Hydrogen (H) is 2.2 and that of Chlorine (Cl) is 3.2.

Calculate the difference in electronegativities

For the H-Cl bond: subtract the electronegativity of Hydrogen from Chlorine: \[\text{Electronegativity difference} = 3.2 - 2.2 = 1.0\]

Determine bond polarity

Since there is a significant difference in electronegativities, the bond between Hydrogen and Chlorine is polar. The more electronegative atom, Chlorine, will have a partial negative charge (marked as δ-) and the less electronegative atom, Hydrogen, will have a partial positive charge (marked as δ+): \[\mathrm{H \,(\delta+)} - \mathrm{Cl\,(\delta-)}\] d. \(\mathrm{Br}-\mathrm{Te}\):

Determine electronegativities for Bromine and Tellurium

Using a periodic table or a chemistry textbook, find that the electronegativity of Bromine (Br) is 2.96 and that of Tellurium (Te) is 2.1.

Calculate the difference in electronegativities

For the Br-Te bond: subtract the electronegativity of Tellurium from Bromine: \[\text{Electronegativity difference} = 2.96 - 2.1 = 0.86\]

Determine bond polarity

Since there is a significant difference in electronegativities, the bond between Bromine and Tellurium is polar. The more electronegative atom, Bromine, will have a partial negative charge (marked as δ-) and the less electronegative atom, Tellurium, will have a partial positive charge (marked as δ+): \[\mathrm{Br \,(\delta-)} - \mathrm{Te\,(\delta+)}\] e. \(\mathrm{Se}-\mathrm{S}\):

Determine electronegativities for Selenium and Sulfur

Using a periodic table or a chemistry textbook, find that the electronegativity of Selenium (Se) is 2.55 and that of Sulfur (S) is 2.58.

Calculate the difference in electronegativities

For the Se-S bond: subtract the electronegativity of Selenium from Sulfur: \[\text{Electronegativity difference} = 2.58 - 2.55 = 0.03\]

Determine bond polarity

Since there is a very small difference in electronegativities, the bond between Selenium and Sulfur is considered nonpolar.