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Chapter 8: Problem 42

Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F}\) ; (b) \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P} ;(\mathbf{c}) \mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\)

Short Answer

Expert verified
The bonds arranged in order of increasing polarity are: (a) O-F, C-F, Be-F (b) S-Br, C-P, O-Cl (c) C-S, N-O, B-F

Step by step solution

01

Know the electronegativity trend in the periodic table.

Electronegativity generally increases across a period from left to right and decreases down a group. The most electronegative element is Fluorine (F), while the least electronegative element is Francium (Fr). Memorize or refer to the electronegativity values of the relevant elements to compare the electronegativity differences.
02

Calculate the electronegativity difference for each bond in group (a).

- For C-F, the electronegativity difference is \(|2.5 - 4.0| = 1.5\) - For O-F, the difference is \(|3.5 - 4.0| = 0.5\) - For Be-F, the difference is \(|1.5 - 4.0| = 2.5\)
03

Arrange group (a) bonds in order of increasing polarity.

Since the order of electronegativity differences is O-F < C-F < Be-F, the order of increasing polarity is: O-F, C-F, Be-F.
04

Calculate the electronegativity difference for each bond in group (b).

- For O-Cl, the difference is \(|3.5 - 3.0| = 0.5\) - For S-Br, the difference is \(|2.5 - 2.8| = 0.3\) - For C-P, the difference is \(|2.5 - 2.1| = 0.4\)
05

Arrange group (b) bonds in order of increasing polarity.

Since the order of electronegativity differences is S-Br < C-P < O-Cl, the order of increasing polarity is: S-Br, C-P, O-Cl.
06

Calculate the electronegativity difference for each bond in group (c).

- For C-S, the difference is \(|2.5 - 2.5| = 0.0\) - For B-F, the difference is \(|2.0 - 4.0| = 2.0\) - For N-O, the difference is \(|3.0 - 3.5| = 0.5\)
07

Arrange group (c) bonds in order of increasing polarity.

Since the order of electronegativity differences is C-S < N-O < B-F, the order of increasing polarity is: C-S, N-O, B-F. So, the final arrangement of all bonds in order of increasing polarity are: (a) O-F, C-F, Be-F (b) S-Br, C-P, O-Cl (c) C-S, N-O, B-F

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Most popular questions from this chapter

Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+},\) and energy is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm{O}^{2-} .\) Yet \(\mathrm{CaO}\) is stable relative to the free elements. Which statement is the best explanation? (a) The lattice energy of CaO is large enough to overcome these processes. (b) CaO is a covalent compound, and these processes are irrelevant. (c) CaO has a higher molar mass than either Ca or O. (d) The enthalpy of formation of CaO is small. (e) CaO is stable to atmospheric conditions.

Although \(\mathrm{I}_{3}\) is a known ion, \(\mathrm{F}_{3}^{-}\) is not. (a) Draw the Lewis structure for \(\mathrm{I}_{3}^{-}\) (it is linear, not a triangle). (b) One of your classmates says that \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{Fis}\) too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong. (c) Another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because it would violate the octet rule. Is this classmate possibly correct? (d) Yet another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{F}\) is too small to make bonds to more than one atom. Is this classmate possibly correct?

A common form of elemental phosphorus is the tetrahedral \(\mathrm{P}_{4}\) molecule, where all four phosphorus atoms are equivalent: At room temperature phosphorus is a solid. (a) Are there any lone pairs of electrons in the \(\mathrm{P}_{4}\) molecule? (b) How many \(\mathrm{p}-\mathrm{p}\) bonds are there in the molecule? (c) Draw a Lewis structure for a linear \(P_{4}\) molecule that satisfies the octet rule. Does this molecule have resonance structures? (d) On the basis of formal charges, which is more stable, the linear molecule or the tetrahedral molecule?

Fill in the blank with the appropriate numbers for both electrons and bonds (considering that single bonds are counted as one, double bonds as two, and triple bonds as three). (a) Fluorine has _____ valence electrons and makes______ bond(s) in compounds. (b) Oxygen has _____ valence electrons and makes ______ bond(s) in compounds. (c) Nitrogen has _____ valence electrons and makes ______ bond(s) in compounds. (d) Carbon has _____ valence electrons and makes ______ bond(s) in compounds.

(a) Construct a Lewis structure for \(\mathrm{O}_{2}\) in which each atom achieves an octet of electrons. (b) How many bonding electrons are in the structure? (c) Would you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) to be shorter or longer than the \(\mathrm{O}-\mathrm{O}\) bond in compounds that contain an \(\mathrm{O}-\mathrm{O}\) single bond? Explain.
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