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

Expert verified
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

01

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.
02

Calculate the difference in electronegativities

For the C-O bond: subtract the electronegativity of Carbon from Oxygen: \[\text{Electronegativity difference} = 3.5 - 2.5 = 1.0\]
03

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. \(\mathrm{P}-\mathrm{H}\):
04

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.
05

Calculate the difference in electronegativities

For the P-H bond: subtract the electronegativity of Phosphorus from Hydrogen: \[\text{Electronegativity difference} = 2.2 - 2.1 = 0.1\]
06

Determine bond polarity

Since the difference in electronegativities is low, the bond between Phosphorus and Hydrogen is considered nonpolar. c. \(\mathrm{H}-\mathrm{Cl}\):
07

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.
08

Calculate the difference in electronegativities

For the H-Cl bond: subtract the electronegativity of Hydrogen from Chlorine: \[\text{Electronegativity difference} = 3.2 - 2.2 = 1.0\]
09

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. \(\mathrm{Br}-\mathrm{Te}\):
10

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.
11

Calculate the difference in electronegativities

For the Br-Te bond: subtract the electronegativity of Tellurium from Bromine: \[\text{Electronegativity difference} = 2.96 - 2.1 = 0.86\]
12

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. \(\mathrm{Se}-\mathrm{S}\):
13

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.
14

Calculate the difference in electronegativities

For the Se-S bond: subtract the electronegativity of Selenium from Sulfur: \[\text{Electronegativity difference} = 2.58 - 2.55 = 0.03\]
15

Determine bond polarity

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

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Most popular questions from this chapter

Which of the following statements is(are) true? Correct the false statements. a. It is impossible to satisfy the octet rule for all atoms in \(\mathrm{XeF}_{2}\). b. Because \(\mathrm{SF}_{4}\) exists, \(\mathrm{OF}_{4}\) should also exist because oxygen is in the same family as sulfur. c. The bond in \(\mathrm{NO}^{+}\) should be stronger than the bond in \(\mathrm{NO}^{-}\). d. As predicted from the two Lewis structures for ozone, one oxygen-oxygen bond is stronger than the other oxygenoxygen bond.

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