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Chapter 3: Problem 33

Without using Fig. \(3-4,\) predict which bond in each of the following groups will be the most polar. a. \(C-F, S i-F, G e-F\) b. \(P-C\) or \(S-C\) \(\mathbf{c} . \mathbf{S}-\mathbf{F}, \mathbf{S}-\mathbf{C} \mathbf{l}, \mathbf{S}-\mathbf{B r}\) d. \(\mathrm{Ti}-\mathrm{Cl}, \mathrm{Si}-\mathrm{Cl}, \mathrm{Ge}-\mathrm{Cl}\)

Short Answer

Expert verified
The most polar bonds for each group are: a. Si-F b. P-C c. S-F d. Ti-Cl

Step by step solution

01

Group a: C-F, Si-F, Ge-F

To determine the most polar bond, compare the difference in electronegativity for each bond: 1. C-F: Electronegativity of C = 2.55, Electronegativity of F = 3.98, Difference = \(|3.98-2.55|\) = 1.43. 2. Si-F: Electronegativity of Si = 1.90, Electronegativity of F = 3.98, Difference = \(|3.98-1.90|\) = 2.08. 3. Ge-F: Electronegativity of Ge = 2.01, Electronegativity of F = 3.98, Difference = \(|3.98-2.01|\) = 1.97. Since Si-F has the greatest difference in electronegativity, the most polar bond is Si-F.
02

Group b: P-C, S-C

Compare the difference in electronegativity for each bond: 1. P-C: Electronegativity of P = 2.19, Electronegativity of C = 2.55, Difference = \(|2.55-2.19|\) = 0.36. 2. S-C: Electronegativity of S = 2.58, Electronegativity of C = 2.55, Difference = \(|2.58-2.55|\) = 0.03. Since P-C has the greater difference in electronegativity, the most polar bond is P-C.
03

Group c: S-F, S-Cl, S-Br

Compare the difference in electronegativity for each bond: 1. S-F: Electronegativity of S = 2.58, Electronegativity of F = 3.98, Difference = \(|3.98-2.58|\) = 1.40. 2. S-Cl: Electronegativity of S = 2.58, Electronegativity of Cl = 3.16, Difference = \(|3.16-2.58|\) = 0.58. 3. S-Br: Electronegativity of S = 2.58, Electronegativity of Br = 2.96, Difference = \(|2.96-2.58|\) = 0.38. Since S-F has the greatest difference in electronegativity, the most polar bond is S-F.
04

Group d: Ti-Cl, Si-Cl, Ge-Cl

Compare the difference in electronegativity for each bond: 1. Ti-Cl: Electronegativity of Ti = 1.54, Electronegativity of Cl = 3.16, Difference = \(|3.16-1.54|\) = 1.62. 2. Si-Cl: Electronegativity of Si = 1.90, Electronegativity of Cl = 3.16, Difference = \(|3.16-1.90|\) = 1.26. 3. Ge-Cl: Electronegativity of Ge = 2.01, Electronegativity of Cl = 3.16, Difference = \(|3.16-2.01|\) = 1.15. Since Ti-Cl has the greatest difference in electronegativity, the most polar bond is Ti-Cl.

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Most popular questions from this chapter

The designations \(1 \mathrm{A}\) through \(8 \mathrm{A}\) used for certain families of the periodic table are helpful for predicting the charges on ions in binary ionic compounds. In these compounds, the metals generally take on a positive charge equal to the family number, while the nonmetals take on a negative charge equal to the family number minus \(8 .\) Thus the compound between sodium and chlorine contains \(\mathrm{Na}^{+}\) ions and \(\mathrm{Cl}^{-}\) ions and has the formula NaCl. Predict the formula and the name of the binary compound formed from the following pairs of elements. a. Ca and N b. \(K\) and 0 c. \(\mathrm{Rb}\) and \(\mathrm{F}\) d. \(\mathrm{Mg}\) and \(\mathrm{S}\) e. Ba and I f. Al and Se g. Cs and \(P\) h. In and Br

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