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Chapter 8: Problem 34

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(未+) - O(未-) b. P-H: nonpolar c. H-Cl: polar, H(未+) - Cl(未-) d. Br-Te: polar, Br(未-) - Te(未+) 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 未-) and the less electronegative atom, Carbon, will have a partial positive charge (marked as 未+): \[\mathrm{C \,(\delta+)} - \mathrm{O\,(\delta-)}\] 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 未-) and the less electronegative atom, Hydrogen, will have a partial positive charge (marked as 未+): \[\mathrm{H \,(\delta+)} - \mathrm{Cl\,(\delta-)}\] 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 未-) and the less electronegative atom, Tellurium, will have a partial positive charge (marked as 未+): \[\mathrm{Br \,(\delta-)} - \mathrm{Te\,(\delta+)}\] 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

A polyatomic ion is composed of \(\mathrm{C}, \mathrm{N}\), and an unknown element \(\mathrm{X} .\) The skeletal Lewis structure of this polyatomic ion is \([\mathrm{X}-\mathrm{C}-\mathrm{N}]^{-} .\) The ion \(\mathrm{X}^{2-}\) has an electron configuration of \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{6} .\) What is element \(\mathrm{X} ?\) Knowing the identity of \(\mathrm{X}\), complete the Lewis structure of the polyatomic ion, including all important resonance structures.

Give the formula of a negative ion that would have the same number of electrons as each of the following positive ions. a. \(\mathrm{Na}^{+}\) c. \(\mathrm{Al}^{3+}\) b. \(\mathrm{Ca}^{2+}\) d. \(\mathrm{Rb}^{+}\)

Place the species below in order of the shortest to the longest nitrogen- oxygen bond. \(\begin{array}{lllll}\mathrm{H}_{2} \mathrm{NOH}, & \mathrm{N}_{2} \mathrm{O}, & \mathrm{NO}^{+}, & \mathrm{NO}_{2}^{-}, & \mathrm{NO}_{3}^{-}\end{array}\) \(\left(\mathrm{H}_{2} \mathrm{NOH}\right.\) exists as \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{OH} .\).

\(\operatorname{LiI}(s)\) has a heat of formation of \(-272 \mathrm{~kJ} / \mathrm{mol}\) and a lattice energy of \(-753 \mathrm{~kJ} / \mathrm{mol}\). The ionization energy of \(\mathrm{Li}(g)\) is 520 . \(\mathrm{kJ} / \mathrm{mol}\), the bond energy of \(\mathrm{I}_{2}(g)\) is \(151 \mathrm{~kJ} / \mathrm{mol}\), and the electron affinity of \(\mathrm{I}(g)\) is \(-295 \mathrm{~kJ} / \mathrm{mol}\). Use these data to determine the heat of sublimation of \(\operatorname{Li}(s)\).

Use the following standard enthalpies of formation to estimate the \(\mathrm{N}-\mathrm{H}\) bond energy in ammonia: \(\mathrm{N}(\mathrm{g}), 472.7 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}(\mathrm{g})\), \(216.0 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{NH}_{3}(g),-46.1 \mathrm{~kJ} / \mathrm{mol}\). Compare your value to the one in Table \(8.4\).
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