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Chapter 8: Problem 29

Without using Fig. 8.3, predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{F}, \mathrm{Si}-\mathrm{F}, \mathrm{Ge}-\mathrm{F}\) b. \(\mathrm{P}-\mathrm{Cl}\) or \(\mathrm{S}-\mathrm{Cl}\) c. \(\mathrm{S}-\mathrm{F}, \mathrm{S}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}\) d. \(\mathrm{Ti}-\mathrm{Cl}, \mathrm{Si}-\mathrm{Cl}, \mathrm{Ge}-\mathrm{Cl}\)

Short Answer

Expert verified
The most polar bonds within each group are: a. \(C-F\) b. \(P-Cl\) c. \(S-F\) d. \(Ti-Cl\)

Step by step solution

01

Recall electronegativity trends in the periodic table

Electronegativity generally increases from left to right across a period and decreases down a group. Recall that Fluorine has the highest electronegativity value.
02

Compare electronegativity differences for group (a)

In group (a), all bonds are with Fluorine (F). As we move down the group from C to Si and Ge, electronegativity decreases. Thus, the difference in electronegativity will be greatest for C-F. Therefore, C-F will be the most polar bond in group (a).
03

Compare electronegativity differences for group (b)

In group (b), we have P-Cl and S-Cl bonds. Both P and S are in the same period, but P is to the left of S. This means P has a lower electronegativity value than S. The difference in electronegativity will be greater for P-Cl than S-Cl. Therefore, P-Cl will be the most polar bond in group (b).
04

Compare electronegativity differences for group (c)

In group (c), all bonds are with Sulfur (S). Fluorine (F) has higher electronegativity than Chlorine (Cl), and Chlorine has higher electronegativity than Bromine (Br). The difference in electronegativity will be greatest for S-F, followed by S-Cl, and least for S-Br. Therefore, S-F will be the most polar bond in group (c).
05

Compare electronegativity differences for group (d)

In group (d), all bonds are with Chlorine (Cl). As we move down the group from Ti to Si and Ge, electronegativity decreases. This means that the difference in electronegativity will be greatest for Ti-Cl. Therefore, Ti-Cl will be the most polar bond in group (d). In conclusion, the most polar bonds within each group are: a. C-F b. P-Cl c. S-F d. Ti-Cl

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Most popular questions from this chapter

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

Which compound in each of the following pairs of ionic substances has the most exothermic lattice energy? Justify your answers. a. \(\mathrm{NaCl}, \mathrm{KCl}\) b. \(\mathrm{LiF}, \mathrm{LiCl}\) c. \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{MgO}\) d. \(\mathrm{Fe}(\mathrm{OH})_{2}, \mathrm{Fe}(\mathrm{OH})_{3}\) e. \(\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{O}\) f. \(\mathrm{MgO}, \mathrm{BaS}\)

Two different compounds have the formula \(\mathrm{XeF}_{2} \mathrm{Cl}_{2} .\) Write Lewis structures for these two compounds, and describe how measurement of dipole moments might be used to distinguish between them.

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)

The second electron affinity values for both oxygen and sulfur are unfavorable (endothermic). Explain.
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