Solution Manual for Chemistry Atoms First 2nd Edition
Burdge Professor ISBN 0073511188 9780073511184
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Chapter 6
Representing Molecules
Practice Problems C
6.1 HCl (right), LiH (left)
6.2 Yellow. These two elements would form an ionic compound.
6.3 black < blue < green
6.4 (i) is not correct; there are only eight valence electrons (should have ten); neither N nor C has an octet.
(ii) is not correct; C has too many valence electrons (has twelve); O has too few (each has six).
(iii) is not correct; S has too few valence electrons (needs two more). 6.5
6.6 (i) 4A, (ii) 6A, (iii) 5A, (iv) 4A
6.7 (i) is not a resonance structure of the original because it is a different arrangement of atoms.
(iii) is not a resonance structure because it has too many electrons.
(ii) and (iv) are resonance structures of the original.
6.8 AB, AB2, and AB3
6.9 P can violate the octet rule and form more than the three bonds in PCl3 by accepting additional electrons in empty outer d orbitals; N does not have any d orbitals to violate the octet rule and so cannot form more than the three bonds in NCl3.
6.10 In the order of decreasing formal charge, the list of structures is ii > iii > i.
Key Skills
6.1
Questions and Problems
6.1 Lewis first suggested that a chemical bond involves atoms that share electrons. He summarized much of his theory of chemical bonding with the octet rule.
6.2 A Lewis symbol consists of an element’s symbol surrounded by dots representing the valence electrons. A Lewis structure is a representation of the covalent bonds in a molecule.
6.3 Lone pairs are nonbonding electrons. Example:
A Lewis structure is a representation of covalent bonding. Example:
The octet rule says that atoms will gain, lose, or share electrons to achieve noble gas configuration.
Example:
Bond length is defined as the distance between the nuclei of two covalently bonded atoms in a molecule.
Example:
6.4 The valence subshells of first-period elements can hold only two electrons, but in second-period elements the subshells can hold a total of eight electrons. So, it is mainly second-period elements that lose, gain, or share electrons attain the noble gas configuration. The octet rule can be violated by higher-period elements expanding their valence shells into empty d orbitals.
6.5 According to the octet rule, atoms without d orbitals will lose, gain, or share electrons to achieve a noble gas electron configuration For hydrogen, the octet rule dictates that it has two electrons, giving it the electronic configuration of the nearest noble gas, helium.
6.6 If the bond is formed between two atoms and consists of one shared pair of electrons, it is called a single bond. A double bond is a multiple bond in which the atoms share two pairs of electrons; in a triple bond, atoms share three pairs of electrons.
Single bond
e.g., C–C in ethane (C2H6)
Double bond
e.g., C=C in ethylene (C2H4)
Triple bond
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e.g., C≡C in acetylene (C2H2)
Single bonds are longer than double bonds, and double bonds are longer than triple bonds.
6.7 Bond length is defined as the distance between the nuclei of two covalently bonded atoms in a molecule. For a given pair of atoms (identical atoms), triple bonds are shorter than double bonds and double bonds are shorter than single bonds. The shorter multiple bonds are also shorter than single bonds. The bond strength can be quantified by measuring the quantity of energy required to break off. Because multiple bonds involve the sharing of multiple pairs of electrons, the bond energies of double and triple bonds between two atoms are higher than that of a single bond between the same atoms. In short, when bond length decreases, the bond strength increases.
6.8 a. I and Cl should form a covalent compound; both the elements are nonmetals. The only possibility would be ICl, iodine chloride.
b. Mg and F will form an ionic compound; Mg is a metal, whereas F is a nonmetal. The substance will be MgF2, magnesium fluoride.
6.9 a. covalent; BF3; boron trifluoride
b. ionic; KBr; potassium bromide
6.10 Electronegativity is the ability of an atom in a compound to draw electrons to itself that it shares in a chemical bond with another atom, whereas electron affinity refers to an isolated atom’s ability to attract an additional electron in the gas phase. Electron affinity is an experimentally measurable quantity, whereas electronegativity is an estimated number that cannot be measured directly.
In general, electronegativity increases from left to right across a period in the periodic table. Within a group, electronegativity decreases with increasing atomic number.
6.11 A polar covalent bond is a bond in which electrons are shared but are not shared equally by the two atoms.
e.g., water molecule: ; hydrogen chloride molecule:
6.12 Strategy: Rearrange Equation 6.1 to solve for Q. Convert the resulting charge in coulombs to charge in units of electronic charge.
Setup: The dipole moments must be converted from debye to C m and the distance between the ions must be converted to meters.
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6.13 Strategy:
Setup:
In units of electronic charge:
Therefore, the partial charges on Cl and O in ClO are +0.164 and –0.164, respectively.
Use Equation 6.1 to solve for the dipole moment, µ, of AlF.
The partial charges on Al and F are +0.019 and –0.019, respectively. In units of electronic charge, this is 0.019 e –The charge units of electronic charge must be converted to coulombs and the distance between the partial charges converted to meters. The calculated dipole moments should be expressed as debyes (1 D = 3.336 × 10–30 C · m).
6.14 Strategy: Use Equation 6.1 to calculate the dipole moment of BrF assuming that the charges on Br and F are +1 and –1, respectively; and use Equation 6.2 to calculate percent ionic character. The magnitude of the charges must be expressed in coulombs (1 e – = 1.6022 × 10–19 C); the bond length (r) must be expressed in meters (1 Å = 1 × 10–10 m); and the calculated dipole moment should be expressed in debyes (1 D = 3.336 × 10–30 C m).
Setup: The bond length is 1.76 Å (1.76 × 10–10 m) and the measured dipole moment of BrF is 1.42 D.
Solution: The expected dipole moment if the magnitude of charges were 1.6022 × 10–19 C is
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Converting to debyes gives:
The percent ionic character of the Br
F bond is:
6.15 Strategy: Use Equation 6.1 to calculate the dipole moment of BaO assuming that the charges on Ba and O are +2 and –2, respectively. Then, rearrange Equation 6.2 to solve for the measured (observed) dipole moment. The magnitude of the charges must be expressed in coulombs (1 e – = 1.6022 × 10–19 C; therefore 2 e – = 2 × 1.6022 × 10–19 C, or 3.2044 × 10–19 C). The bond length (r) must be expressed in meters (1 pm = 1 × 10–12 m); and the measured dipole moments should be expressed in debyes (1 D = 3.336 × 10–30 C·m).
Setup: The bond length is 194 pm (1.94 × 10–10 m) and the percent ionic character of the Ba–O bond is 85.3% or 85.3/100 = 0.853.
Solution: The expected dipole moment if the magnitude of charges were 3.2044 × 10
19 C is:
6.16 We calculate the electronegativity differences for each pair of atoms:
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Covalent bond character increases as the difference in electronegativity decreases. The order of increasing covalent bond character is: BG < GY < BY = RY.
6.17 Strategy: We can look up electronegativity values in Figure 6.4 of the text. The amount of ionic character is based on the electronegativity difference between the two atoms. The larger the electronegativity difference, the greater the ionic character.
Solution: Let EN electronegativity difference. The bonds arranged in order of increasing ionic character are: C
6.18 The degree of ionic character in a bond is a function of the difference in electronegativity between the two bonded atoms. Figure 6.4 of the text lists electronegativity values of the elements. The bonds in order of increasing ionic character are: N N (zero difference in electronegativity) < S O (difference 1.0) = Cl F (difference 1.0) < K O (difference 2.7) < Li F (difference 3.0).
6.19 a. The bond is covalent The two silicon atoms are the same.
b. The bond is polar covalent. The electronegativity difference between Cl and Si is 3.0 1.8 = 1.2
c. The bond is ionic. The electronegativity difference between F and Ca is 4.0 1.0 = 3.0.
d. The bond is polar covalent. The electronegativity difference between N and H is 3.0 2.1 = 0.9
6.20 a. The two carbon atoms are the same. The bond is covalent
b. The electronegativity difference between K and I is 2.5 0.8 = 1.7. The bond is polar covalent
c. The electronegativity difference between N and B is 3.0 2.0 -= 1.0. The bond is polar covalent.
d. The electronegativity difference between C and F is 4.0 2.5 = 1.5. The bond is polar covalent.
6.21 The order of increasing ionic character is:
Cl Cl (zero difference in electronegativity) < Br Cl (difference 0.2) < Si C (difference 0.7) < Cs F (difference 3.3)
6.22 The Lewis structures are:
a. b.
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6.26 Formal charge is a way of keeping track of the valence electrons in a species. Formal charge on an atom in a Lewis structure is equal to the number of valence electrons (group number) minus half of the electrons it shares with other atoms in the structure. Formal charge is determined by comparing the number of electrons associated with an atom in a Lewis structure with the number of electrons that would be associated with the isolated atom.
6.27 No, formal charges do not represent an actual separation of charges because the bonding electrons are not shared equally among the resonance forms. Formal charges merely help us keep track of the electrons involved in bonding in the molecule.
6.29 Strategy: We follow the procedure for drawing Lewis structures outlined in Section 6.3 of the text.
Solution: a. Step 1: The atoms in the Lewis structure are: O N O
Step 2: The outer-shell electron configuration of O is 2s22p4. Each O has six valence electrons. The outer-shell electron configuration of N is 2s22p3. N has five valence electrons. Also, we must subtract an electron to account for the positive charge. Thus, there are:
(2 6) + 5 1 = 16 valence electrons
Step 3: We draw single covalent bonds between the atoms and then attempt to complete the octets first for the O atoms.
Because this uses all the available electrons, Step 5 outlined in the text is not required, but Step 6 is required to move electron pairs to satisfy the octet for nitrogen.
We determine the formal charges on the atoms using the method outlined in Section 6.4 and Equation 6.2. For each oxygen atom, formal charge = 6 6 = 0. For the nitrogen atom, formal charge = 5 4 = +1.
Check: As a final check, we verify that there are 16 valence electrons in the Lewis structure of 2NO
Follow the same procedure as part (a) for parts (b), (c), and (d).
b. C N Sor
c. -S S or
d. Cl F F + or 6.30 a. Too many electrons. The correct structure is:
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6.31
b. Hydrogen atoms do not form double bonds. The correct structure is:
c. Too few electrons.
d. Too many electrons. The correct structure is:
e. Fluorine has more than an octet. The correct structure is:
f. Oxygen does not have an octet. The correct structure is:
g. Too few electrons. The correct structure is:
a. Neither oxygen atom has a complete octet, and the leftmost hydrogen atom shows two bonds (four electrons). Hydrogen can hold only two electrons in its valence shell.
b. The correct structure is:
6.32 A resonance structure is one of two or more Lewis structures for a single molecule or a polyatomic ion that cannot be represented accurately by only one Lewis structure. None of the resonance structures actually exists the actual molecule has its own unique, stable structure so no resonance structure can be isolated.
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6.33 Resonance structures must follow the rules for drawing Lewis structures and only have differences in the arrangement of electrons, not the location of atoms.
6.34 Three reasonable resonance structures with the formal charges indicated are:
6.35 Strategy: We follow the procedure for drawing Lewis structures outlined in Section 6.3 of the text. After we complete the Lewis structure, we draw the resonance structures.
Solution: Following the procedure in Sections 6.3 and 6.5 of the text, we come up with the following resonance structures: a. b.
6.36 The structures of the most important resonance forms are:
6.37 The structures of the most important resonance forms are:
6.38 Three reasonable resonance structures with the formal charges indicated are
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The structure on the left has formal charges of 1 on N and 0 on O, which does not make sense because O is more electronegative than N. The middle structure has formal charges that make sense with regard to the relative electronegativities of O and N. The structure on the right has a large negative formal charge on N and a positive formal charge on O. This is the least important of the resonance structures.
6.39 Three reasonable resonance structures for OCN–are:
6.40 Resonance structures differ only in the positions of their electrons, not in the positions of their atoms. Structures (a), (b), and (d) are resonance structures of BeCl2. The structures differ only in the positions of their electrons. Structures (c) and (e) are not resonance structures of BeCl2 because the positions of the atoms are different
6.41 Resonance structures differ only in the positions of their electrons, not in the positions of their atoms. Structures (b) and (e) are resonance structures of COCl2. The structures differ only in the positions of their electrons. Structures (a) and (c) are not resonance structures because they violate the octet rule for C (carbon cannot have more than four bonds). Structure (d) is not a resonance structure because C and O do not obey the octet rule in this structure.
6.42 Strategy: Draw two Lewis structures for the adenine molecule in which the atoms are arranged in the same way, but the electrons are arranged differently.
Setup: A correct structural formula for adenine is
Solution:
But the double bonds can be put in different positions in the molecule.The resulting Lewis structures are shown below.
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6.43 Strategy: Draw a Lewis structure for the guanine molecule in which the atoms are arranged in the same way but the electrons are arranged differently.
Setup: A correct structural formula for guanine is
Solution:
But the double bonds can be put in different positions in the molecule.
In order for all the atoms in the second structure to obey the octet rule, there are two atoms with nonzero formal charges. When there are nonzero formal charges, they should be consistent with the electronegativities of the atoms in the molecule. A positive formal charge on oxygen, for example, is inconsistent with oxygen’s high electronegativity.
6.44 Elements in the third period and beyond have d orbitals that can be used to hold additional electrons beyond an octet.
6.45 BeH2: Be has only four electrons.
NO2: There is an odd number of electrons. N has only seven electrons.
XeF4: Xe has an expanded octet with 12 electrons.
6.46 Fluorine is the most electronegative element, so it would not share the seven electrons needed to form these compounds.
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6.47 A coordinate covalent bond is a covalent bond in which one of the atoms donates both electrons. The properties of coordinate covalent bonds do not differ from those of a normal covalent bond. Sharing of electrons takes place in both bonds.
6.48 The resonance structures are:
6.49 Strategy: We follow the procedure outlined in Section 6.3 of the text for drawing Lewis structures. We assign formal charges as discussed in Section 6.4 of the text.
Solution: Drawing the structure with single bonds between Be and each of the Cl atoms, we see that Be is “electron deficient.” No, the octet rule is not followed for Be. The Lewis structure is:
An octet of electrons on Be can only be formed by making two double bonds as shown below:
It is not a plausible Lewis structure because this places a high negative formal charge on Be and positive formal charges on the Cl atoms. This structure distributes the formal charges counter to the electronegativities of the elements.
6.50 For simplicity, the three nonbonding pairs of electrons around the fluorine atoms are omitted.
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6.51 The reaction can be represented as:
The new bond formed is called a coordinate covalent bond
6.52 For simplicity, the three nonbonding pairs of electrons around the fluorine are omitted.
The octet rule is not satisfied for Se in both compounds.
6.53 The outer electron configuration of antimony is 5s25p3. The Lewis structure is shown below. All five valence electrons are shared in the five covalent bonds. No, the octet rule is not obeyed
6.54 Strategy: For ClF2, the structural formula is:
Cl F F
This puts the unique atom, Cl, in the center and puts the more electronegative F atoms in terminal positions.
For BrO2, the structural formula is:
Br O O
This puts the unique atom, Br, in the center and puts the more electronegative O atoms in terminal positions.
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Setup: ClF2: There are a total of 21 valence electrons (seven from the Cl and seven from each of the two F atoms). We subtract four electrons to account for the two bonds in the skeleton, leaving us with 17 electrons to distribute as follows: three lone pairs on each F atom, two lone pairs on the Cl atom, and the last remaining electron also on the Cl atom.
BrO2: There are a total of 19 valence electrons (seven from the Br and six from each of the two O atoms). We subtract four electrons to account for the two bonds in the skeleton, leaving us with 15 electrons to distribute at follows: three lone pairs on each O atom, one lone pair on the Br atom, and the last remaining electron also on the Br atom.
Solution: For ClF2, the Lewis structure is:
For BrO2, the Lewis structure is:
6.55 Strategy: For HO2, the structural formula is:
Because hydrogen has only one electron, it can form only one covalent bond.
For ClO, the structural formula is:
Setup: HO2: There are a total of 13 valence electrons (one from the H atom and six from each of the two O atoms). We subtract four electrons to account for the two bonds in the skeleton, leaving us with nine electrons to distribute as follows: three lone pairs on the terminal O atom, one lone pair on the central O atom, and the last remaining electron also on the central O atom.
ClO: There are a total of 13 valence electrons (seven from the Cl and six from the O atom). We subtract two electrons to account for the bond in the skeleton, leaving us with 11 electrons to distribute at follows: three lone pairs on the O atom, two lone pairs on the Cl atom, and the last remaining electron also on the Cl atom.
Solution: For HO2, the Lewis structure is:
For ClO, the Lewis structure is
6.56 Completed octet on Br: ; zero formal charge on Br:
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6.57 Completed octet on S: ; zero formal charge on S:
6.58 The three pairs of nonbonding electrons around each fluorine have been omitted for simplicity.
The octet rule is not obeyed in any of these compounds.
6.59 The resonance structures are:
6.60 A resonance structure is:
6.61 a. An example of an aluminum species that satisfies the octet rule is the anion 4AlCl . The Lewis dot structure is shown below.
b. An example of an aluminum species containing an expanded octet is anion 3 6AlF The Lewis dot structure is shown below.
c. An aluminum species that has an incomplete octet is the compound AlCl3. The dot structure is shown below.
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6.62 Four resonance structures together with the formal charges are:
6.63 CF2 would be very unstable because carbon does not have an octet
LiO2 would not be stable because the lattice energy between Li+ and superoxide 2O would be too low to stabilize the solid. A charge of +2 would be impossible on Li anyway, because it has only one valence electron.
CsCl2 requires a Cs2+ cation. The second ionization energy is too large to be compensated by the increase in lattice energy
PI5 appears to be a reasonable species. However, the iodine atoms are too large to have five of them “fit” around a single P atom.
6.64 Reasonable resonance structures are:
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6.65
a. False
b. False
c. False
6.66 If the central atom were more electronegative, there would be a concentration of negative charges at the central atom. This would lead to instability.
6.67 Only N2 has a triple bond. Therefore, it has the shortest bond length.
6.68
6.69 To be isoelectronic, molecules must have the same number and arrangement of valence electrons. CH4 and + 4NH are isoelectronic (8 valence electrons), as are N2 and CO (10 valence electrons), as are C6H6 and B3N3H6 (30 valence electrons). Draw Lewis structures to convince yourself that the electron arrangements are the same in each isoelectronic pair.
6.70 The complete Lewis structure is:
6.71 The complete Lewis structure is:
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6.73 Strategy: For 2SO , the structural formula is:
This puts the unique atom, S, in the center and puts the more electronegative O atoms in terminal positions.
For 3SO , the structural formula is:
This puts the unique atom, S, in the center and puts the more electronegative O atoms in terminal positions.
Setup: 2SO : There are a total of 18 valence electrons (six from the S and six from each of the two O atoms). We add one electron to account for the –1 charge, which gives us a total of 19 electrons. We subtract four electrons to account for the two bonds in the skeleton, leaving us with 15 electrons. Distribute the electrons as three lone pairs on each O atom and one lone pair as well as the last remaining electron on the S atom. Since the S atom has less than an octet, use one lone pair from one of the O atoms to make a double bond to the S atom.
3SO : There are a total of 24 valence electrons (six from the S and six from each of the three O atoms). We add one electron to account for the –1 charge which gives us a total of 25 electrons. We subtract six electrons to account for the three bonds in the skeleton, leaving us with 18 electrons. Distribute the electrons as three lone pairs on each O atom and the last remaining electron on the S atom. Since the S atom has less than an octet, use one lone pair from two of the O atoms to make two double bonds to the S atom.
Solution: Lewis structure of 2SO
6.75 The central iodine atom in 3I has ten electrons surrounding it: two bonding pairs and three lone pairs. The central iodine has an expanded octet. Elements in the second period such as fluorine cannot have an expanded octet, as would be required for 3F .
6.76 The Lewis structures of C2H3F, C3H6, and C4H8 are:
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6.77 The atoms in methyl isocyanate are The number of valence electrons is:
We can draw two resonance structures for methyl isocyanate.
6.78 A reasonable Lewis structure is:
6.79 Form (a) is the most important resonance structure with no formal charges and all satisfied octets. Form (b) is likely not as important as (a) because of the positive formal charge on O. Forms (c) and (d) do not satisfy the octet rule for all atoms and are likely not important.
6.80 The structures for C4H10 and C5H12 given below show only the isomers in which each C atom is bonded to no more than two other C atoms
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6.84
6.85 The noble gases violate the octet rule. This is because each noble gas atom already has completely filled ns and np subshells.
6.86 The resonance structures are:
a.
b.
In both cases, the most likely structure is on the left and the least likely structure is on the right.
6.87 a.
The first structure is the most important. Both N and O have formal charges of zero. In the second structure, the more electronegative oxygen atom has a formal charge of 1. Having a positive formal charge on a highly electronegative atom is not favorable. In addition, both structures leave one atom with an incomplete octet. This cannot be avoided due to the odd number of electrons.
b. No, it is not possible to draw a structure with a triple bond between N and O
Any structure drawn with a triple bond will lead to an expanded octet. Elements in the second row of the periodic table cannot exceed the octet rule.
6.88 a. Strategy: Rearrange Equation 6.1 to solve for Q. Convert the resulting charge in coulombs to charge in units of electronic charge.
Setup: The dipole moment must be converted from debye to C·m and the distance between the ions must be converted to meters.
Solution: In coulombs:
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6.89
Therefore, the partial charges on S and N in SN are +0.253 and –0.253, respectively.
b. Strategy: Use Equation 6.1 to solve for the dipole moment, µ, of SO.
Setup: The partial charges on S and O are +0.022 and –0.022. In units of electronic charge, this is 0.022 e
The charge in units of electronic charge must be converted to coulombs and the distance between the partial charges converted to meters.
The arrows indicate coordinate covalent bonds. This dimer does not possess a dipole moment.
6.90 One resonance form showing formal charges is:
6.91 Strategy: Use Equation 6.1 to calculate the dipole moments in RbF and RbI assuming that the charges on Rb, F, and I are +1, –1, and –1, respectively. Use Equation 6.2 to calculate percent ionic character. The magnitude of the charges must be expressed in coulombs (1 e – = 1.6022 × 10–19 C); the bond length (r) must be expressed in meters (1 pm = 1 × 10–12 m); and the calculated dipole moments should be expressed in debyes (1 D = 3.336 × 10–30 C m).
Setup: The dipole moments of RbF and RbI are 8.55 D and 11.48 D, respectively. Using Figure 4.13, the bond distance of RbF is 285 pm (152 pm (Rb+) + 133 pm (F ) = 285 pm) and the bond distance of RbI is 372 pm (152 pm (Rb+) + 220 pm (I ) = 372 pm). In meters, this gives 2.85 × 10–10 m and 3.72 × 10–10 m, respectively.
Solution: RbF:
For a magnitude of charges of 1.6022 × 10–19 C is, the expected dipole moment is:
µ = Q × r
Converting to debyes gives:
The percent ionic character of the Rb–F bond is:
For a magnitude of charges of 1.6022 × 10–19 C, the expected dipole moment is:
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The percent ionic character of the Rb
I bond is
We would expect the percent ionic character to be higher for RbF than it is for RbI, given that fluorine is more electronegative.
6.92 The unpaired electron on each N atom will be shared; each N attains an octet of electrons to form a covalent bond.
6.93 Strategy: Use Equation 6.1 to solve for the dipole moments, µ. The electrostatic potential map shows regions where electrons spend a lot of time in red and regions where electrons spend very little time in blue. (Regions where electrons spend a moderate amount of time are in green.) The electrons in covalent bonds, with low dipole moments, spend roughly the same amount of time in the vicinity of each atom. The electrostatic potential map will be symmetrical, with the electrons spending time in the vicinity of each atom. The higher the dipole moment, the more time an electron will spend in the vicinity of one atom over the other. In this case, the electrostatic potential map will show the electrons spending most of their time near one atom (red) and not much time near the other (blue).
Setup: First calculate the dipole moments:
Partial charges of +0.19 and –0.19:
In units of electronic charge, this is 0.19 e –
The charge units of electronic charge must be converted to coulombs and the distance between the partial charges converted to meters.
6.95 Pyridine has a structure similar to benzene (see Section 6.5), but one of the carbon atoms is replaced by a nitrogen atom. The two resonance structures of pyridine are:
6.96
In each structure, there is one single C–N bond and one double C–N bond. Therefore, the C–N bond length will be shorter than that of a single bond (143 pm) and longer than that of a double bond (138 pm), approximately 141 pm.
6.97
6.98 Strategy: Determine the total number of electrons in each of the structures and then determine the number of valence electrons on the unknown atoms.
Setup: Structure containing Z: There are 32 total electrons (5 bonds give 10 electrons and 11 lone pairs give 22 electrons) in this structure. There are four O atoms, each of which has 6 valence electrons. The hydrogen atom has 1 valence electron. The –1 charge contributes 1 valence electron. These account for a total of 26 valence electrons. This leaves 6 valence electrons for Z. The third-period element with 6 valence electrons is S.
Structure containing X: There are 28 total electrons (3 bonds give 6 electrons and 11 lone pairs give 22 electrons) in this structure. There are three F atoms, each of which has 7 valence electrons. These account for 21 valence electrons. This leaves 7 valence electrons for X. The third-period element with 7 valence electrons is Cl.
Solution: The structural formula for compound ZX3 or SCl3 is:
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The structure has 27 valence electrons (6 from the S and 7 from each of the three Cl atoms). We subtract 6 electrons to account for the three bonds in the skeleton, leaving us with 21 electrons to distribute as follows: three lone pairs on each Cl atom, one lone pair on the S atom, and the last remaining electron also on the S atom.
6.99 Using Mulliken’s definition,
When converted to the Pauling scale,
These values compare to the Pauling values for oxygen of 3.5, fluorine of 4.0, and chlorine of 3.0
6.100 Strategy: Use Equation 6.1 to solve for the dipole moment, µ, partial charge, Q, or distance between charges, r. The electrostatic potential map shows regions where electrons spend a lot of time in red and regions where electrons spend very little time in blue. (Regions where electrons spend a moderate amount of time are in green.) The electrons in covalent bonds, with low dipole moments, spend roughly the same amount of time in the vicinity of each atom. The electrostatic potential map will be symmetrical, with the electrons spending time in the vicinity of each atom. The higher the dipole moment, the more time an electron will spend in the vicinity of one atom over the other. In this case, the electrostatic potential map will show the electrons spending most of their time near one atom (red) and not much time near the other (blue).
Setup: Calculate the dipole moment:
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Partial charges of +0.16 and –0.16: In units of electronic charge, this is 0.16
The charge units of electronic charge must be converted to coulombs and the distance between the partial charges converted to meters.