CHAPTER 7. Lewis Formulas - General Chemistrymcquarriegeneralchemistry.com/pracprobs_final_7-11.pdf · atom. The Lewis formula is – H NH 7-5. See the text. 7-6. The Lewis formula

Practice-Problem-6660020 book December 22, 2010 8:7

CHAPTER 7. Lewis Formulas

7-1. See the text. Because silicon is below carbon in the periodic table, we can determine the Lewisformula of SiCl4 in the same manner as for CCl4.

7-2. Because phosphorus is the unique atom in this molecule, we shall assume that it is central andthat the four chlorine atoms are attached to it.

Cl P

Cl

Cl

Cl

The total number of valence electrons is (1 × 5) + (4 × 7) − 1 = 32. We use eight of the valenceelectrons to form phosphorus–chlorine bonds. We now place valence electrons as lone pairs onthe four chlorine atoms, accounting for the remaining 24 valence electrons. The completedLewis formula is

Cl P

Cl

Cl

Cl

7-3. We first arrange the atoms as

H P

H

H

The total number of valence electrons is (1 × 5) + (3 × 1) = 8. We use six valence electrons toform the P–H bonds. We place the remaining two valence electrons as a lone pair on the P atom.The Lewis formula is

H P

H

H


H N H

13


14 GENERAL CHEMISTRY, FOURTH EDITION | McQuarrie, Rock, and Gallogly

The total number of valence electrons is (1 × 5) + (2 × 1) + 1 = 8. We use four valence electronsto form the N–H bonds and place the remaining four valence electrons as lone pairs on the Natom. The Lewis formula is

–H N H

7-5. See the text.

7-6. The Lewis formula for the tetrafluoroborate ion is

–

F B

F

F

F

The formal charges on the boron atom and fluorine atoms are calculated using Equation 7.1.

formal charge on the B atom = 3 − 0 − 12

(4) = −1

formal charge on each F atom = 7 − 6 − 12

(2) = 0

Thus, the Lewis formula for a BF−4 ion with the formal charges indicated is

–F B

F

F

F

7-7. The formal charges on the oxygen and hydrogen atoms in Structure I are assigned according toEquation 7.1.

formal charge on end O atom = 6 − 6 − 12

(2) = −1

formal charge on bonded O atom = 6 − 2 − 12

(6) = +1

formal charge on each H atom = 1 − 0 − 12

(2) = 0

The formal charges in Structure II are

formal charge on each O atom = 6 − 6 − 12

(4) = 0

formal charge on each H atom = 1 − 0 − 12

(2) = 0

Structure II has zero formal charges everywhere and is the preferred Lewis formula for hydrogenperoxide.

7-8. (a) We first arrange the atoms as

OCH

H

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Chapter 7: Lewis Formulas 15

The total number of valence electrons is (1 × 4) + (1 × 6) + (2 × 1) = 12. As in Example 7-8, wecannot satisfy the octet rule with just single bonds. Therefore, we form a double bond betweenthe carbon atom and the oxygen atom, form two H–C bonds, and place the remaining fourvalence electrons as lone pairs on the oxygen atom to write

OCH

H

(b) The Lewis formula for the methyl group is

H C

H

H

Thus, the Lewis formula of ethanal is

OCH C

H

H H

7-9. As in Example 7-9, we cannot satisfy the octet rule with all single bonds or a double bond.Therefore, we form a triple bond between the two carbon atoms and two C–H bonds to write

H HC C


OC

O

O

The total number of valence electrons is (1 × 4) + (3 × 6) + (2 × 1) = 24. The resonanceformulas are

O

O

CO

– – – –

––

O

O

CO O

O

CO

The formal charges indicated are calculated by using Equation 7.1.

formal charge on C atom = 4 − 0 − 12

(4) = 0

formal charge on each single bonded O atom = 6 − 6 − 12

(2) = −1

formal charge on double bonded O atom = 6 − 4 − 12

(4) = 0



The superposition of these resonance forms gives the resonance hybrid

OC

O

O 2–

The three carbon–oxygen bonds in a CO2−3 ion are equivalent; they have the same bond length

and the same bond energy.

7-11. A nitrogen dioxide molecule has (1 × 5) + (2 × 6) = 17 valence electrons and so is a free radical.The two resonance forms of a NO2 molecule are

NOO

–N

O–

+ +

O


Cl P

Cl

Cl

O

The total number of valence electrons is (1 × 5) + (1 × 6) + (3 × 7) = 32. We use eight valenceelectrons to form three P–Cl bonds and one P–O bond. The remaining 24 valence electrons arelone pairs on the oxygen atom and three chlorine atoms. The Lewis formula is

Cl P

O

Cl

Cl

–

Both the oxygen atom and phosphorus atom have a formal charge. We can eliminate the formalcharges by expanding the the octet on the phosphorus atom to form a double bond between thephosphorus and oxygen atoms to obtain the Lewis formula

Cl P

Cl

Cl

O

Notice that all the atoms now have a zero formal charge.

7-13. A PCl−6 ion has (1 × 5) + (6 × 7) + 1 = 48 valence electrons. Twelve are used to form the six P–Clbonds. The remaining 36 valence electrons are placed as lone electron pairs on the six chlorineatoms. The Lewis formula is

P

Cl

ClCl Cl

Cl Cl

–



Chapter 7: Lewis Formulas 17

7-14. A PO3−4 ion has (1 × 5) + (4 × 6) + 3 = 32 valence electrons. The various resonance forms are

P

O

O

O O– –

–

P

O

O

four of theseone of these

O O– –

–

–

+

The superpositions of these resonance forms give the resonance hybrid

P

O

O

O O

3–

The four phosphorus–oxygen bonds in the PO3−4 ion are equivalent; they have the same bond

length and bond energy.

7-15. An oxygen atom is more electronegative than a hydrogen atom, and so we have

δ+

2δ–

δ+O

HH

7-16. The difference in the electronegativities of a hydrogen atom and a silicon atom is 2.1 − 1.90 =0.2. Thus, the bonding in a SiH4 molecule is covalent with a small degree of ionic character. Thedifference in the electronegativities of a beryllium atom and a chlorine atom is 3.16 − 1.57 =1.59. Thus, the bonding in a SiH4 molecule is polar covalent with a fair degree of ionic character.


CHAPTER 8. Prediction of Molecular Geometries

8-1. The Lewis formula of a CHCl3 molecule is

H C

Cl

Cl

Cl

There are four valence-shell electron pairs about the central carbon atom, so it is tetrahedral.The bond angles are 109.5◦.

8-2. The Lewis formula of a BF−4 molecule is

F B

F

F

F–

There are four valence-shell electron pairs about the central boron atom, and so the ion istetrahedral.

8-3. The Lewis formula of an AlF3−6 molecule is

Al

F

FF

F

F

F

3–

An AlF3−6 ion contains six covalent bonds, and so it is an octahedral ion with 90◦ bond angles.

8-4. The Lewis formula of a SF2 molecule is F S F . It is an AX2E2 molecule, and so it is bentwith bond angles less than the tetrahedral bond angle of 109◦.

8-5. The Lewis formula of a CS2 molecule is S C S . It is an AX2 molecule, and so it is linear.

18


Chapter 8: Prediction of Molecular Geometries 19

8-6. The Lewis formula of a ClO−2 ion is O Cl O

––+

. It is an AX2E2 ion, and so it is bent.The resonance formulas of a ClO−

3 ion are

––

–

O Cl

O

O O Cl

O

OO Cl

O

O

It is an AX3E ion, and so it is trigonal pyramidal.

8-7. The Lewis formulas of a SO2 molecule are

O S O– –

+

O S O

+

It is an AX2E molecule, and so is bent with a bond angle of less than 120◦ because of the two lonepairs.

8-8. The Lewis formula of a XeF2 molecule is F Xe F . It is an AX2E3 molecule, and so it is linear.

8-9. The Lewis formulas of a XeO3 molecule are

––

–

O Xe

O

O O (or)Xe

O

OO Xe

O

O

O

O Xe O

+ + +

Thus, it is an AX3E molecule, and so is trigonal pyramidal.

8-10. (a)δ+

δ–C

H

HH

Cltetrahedral with one chlorine atom; polar

(b) Br

F

F

Fδ+

δ–

δ–

δ–

T-shaped; polar

(c) BrF

F

F

F

F δ+

δ–δ–δ–

δ–δ–square pyramidal; polar

(d) As

F

FF

FF

δ+

δ–

δ–

δ–δ–

δ–

(symmetric) trigonal bipyramidal; nonpolar

8-11. See the text.


CHAPTER 9. Covalent Bonding

9-1. See the text.

9-2. Bond order = 9 − 92

= 0. See the text.

9-3. Using Figure 9.13, we have

CN+: 12 electrons (σ1s )2(σ ∗1s )

2(σ2s )2(σ ∗2s )

2(π2p)4 bond order = 8 − 42

= 2

CN: 13 electrons (σ1s )2(σ ∗1s )

2(σ2s )2(σ ∗2s )

2(π2p)4(σ2p)1 bond order = 9 − 42

= 212

CN−: 14 electrons (σ1s )2(σ ∗1s )

2(σ2s )2(σ ∗2s )

2(π2p)4(σ2p)2 bond order = 10 − 42

= 3

The species with the largest bond order and thus the largest bond energy is CN−.

9-4. See the text.

9-5. See the text.

9-6. See the text.

9-7. See the text.

9-8. See the text.

9-9. See the text.

20


CHAPTER 10. Chemical Reactivity

10-1. See the text.

10-2. See the text.

10-3. See the text.

10-4. See the text.

10-5. The oxalate ion has a charge of −2 (Table 10.1), and so we predict that it is a diprotic acid.

10-6. Using Tables 10.2 and 10.4 as a guide, we get the answers in the text.

10-7. The unreactive gas is N2(g) and the metallic mirrorlike deposit is Na(s). Thus, we predict thatthe decomposition reaction is described by

2 NaN3(s) −→ 2 Na(s) + 3 N2(g)

10-8. See the text.

10-9. Aluminum replaces the hydrogen atoms in H2SO4(aq ), giving us

2 Al(s) + 3 H2SO4(aq ) −→ Al2(SO4)3(aq ) + 3 H2(g)

10-10. See the text.

10-11. MgO(s) combines with SO2(g) to yield MgSO3(s) as described in Section 10-1.

10-12. Chlorine is more reactive than bromine and so liberates Br2(l) from solutions of bromide salts,and so

2 Br−(aq ) + Cl2(g) −→ 2 Cl−(aq ) + Br2(l)

10-13. (a) Soluble by Rule 1(b) Insoluble by Rule 6(c) Soluble by Rule 2(d) Soluble by Rule 1

10-14. (a) Hg2Br2 is insoluble by Rule 3.

Hg2(NO3)2(aq ) + 2 NaBr(aq ) −→ Hg2Br2(s) + 2 NaNO3(aq )

(b) No reaction. All possible products are soluble by Rule 1.

21



(c) CdS is insoluble by Rule 5.

Na2S(aq ) + Cd(NO3)2(aq ) −→ 2 NaNO3(aq ) + CdS(s)

10-15. Using the solubility rules, we find that the resulting double-replacement reaction produces asilver chromate precipitate. The complete equation for the reaction is

2 AgClO4(aq ) + Na2CrO4(aq ) −→ 2 NaClO4(aq ) + Ag2CrO4(s)

The net ionic equation is

2 Ag+(aq ) + CrO2−4 (aq ) −→ Ag2CrO4(s)

10-16. This is essentially an acid-base reaction, where Li2O(s) dissolves in water to yield a base (seeSection 10.3), according to

Li2O(s) + H2O(l) −→ 2 LiOH(aq )

Thus,

H2SO4(aq ) + 2 LiOH(aq ) −→ Li2SO4(aq ) + 2 H2O(l)

or

H2SO4(aq ) + Li2O(s) −→ Li2SO4(aq ) + H2O(l)

The net ionic equation is

2 H+(aq ) + Li2O(s) −→ 2 Li+(aq ) + H2O(l)

10-17. The chemical formula for sodium hydrogen carbonate is NaHCO3 (Table 10.1) and that ofacetic acid is CH3COOH (Table 10.3). Thus, the equation for the reaction is

NaHCO3(s) + CH3COOH(aq ) −→ NaCH3COO(aq ) + CO2(g) + H2O(l)

10-18. (a) The ionic charges on iron in iron metal, Fe(s), and chlorine in chlorine gas, Cl2(g), arezero. However, the ionic charge on a chloride ion, Cl−, is −1, and so that on the iron atomin a FeCl3 formula unit must be +3. Because the ionic charges of the iron and chlorineatoms are changing, this is an oxidation-reduction reaction.

(b) Because the nitrate ion, NO−3 , has an ionic charge of −1, the silver atom in a AgNO3

formula unit must have an ionic charge of +1. The ionic charges of the Na and S atoms ina Na2S formula unit are +1 and −2, respectively. Similarly, we find for the products thatthe ionic charges of the Ag and S atoms in a Ag2S formula unit are +1 and −2, respectively,and the ionic charge of the Na atom in a NaNO3 formula unit is +1. Thus, we concludethat this is not an oxidation-reduction reaction because the ionic charges of each atom inthe reactants are identical to their ionic charges in the products.

(c) The ionic charge on zinc in zinc metal, Zn(s), is zero, and that of the zinc atom in aZnCl2 formula unit is +2. Similarly, the ionic charge on the mercury atom in a HgCl2formula unit is +2, and that in mercury metal is 0. Thus, we conclude that this is anoxidation-reduction reaction.

10-19. The ionic charge of the aluminum atom changes from 0 in aluminum metal to +3 in aluminumoxide. Similarly, the ionic charge of the manganese atom changes from +3 in manganese oxideto 0 in manganese metal. Thus, in the chemical equation, two aluminum atoms are oxidized(0 → +3) and two manganese atoms are reduced (+3 → 0), and so Al(s) is the reducing agentand Mn2O3(s) is the oxidizing agent.



CHAPTER 11. Chemical Calculations

11-1. H2SO4(l)

number of moles = (37 515 × 103 metric ton)(

1000 kg1 metric ton

) (1000 g1 kg

) (1 mol H2SO4

98.078 g H2SO4

)

= 3.83 × 1011 mol H2SO4

N2(g)


1000 kg1 metric ton

) (1000 g1 kg

) (1 mol N2

28.013 g N2

)

= 9.52 × 1011 mol N2

C2H4(g)


1000 kg1 metric ton

) (1000 g1 kg

) (1 mol C2H4

28.053 g C2H4

)

= 9.15 × 1011 mol C2H4

O2(g)


1000 kg1 metric ton

) (1000 g1 kg

) (1 mol O2

31.9988 g O2

)

= 6.11 × 1011 mol O2

H2(g)


1000 kg1 metric ton

) (1000 g1 kg

) (1 mol H2

2.0158 g H2

)

= 8.78 × 1012 mol H2

On a mole basis, H2(g) is produced in the greatest quantity annually in the United States.

23



11-2. The mass of a CO2 molecule is

mass of one molecule =(

44.0095 g CO2

1 mol CO2

) (1 kg

1000 g

) (1 mol

6.022 × 1023

)

= 7.308 × 10−26 kg

The mass of a SF6 molecule is

mass of one molecule =(

146.055 g SF6

1 mol SF6

) (1 kg

1000 g

) (1 mol

6.022 × 1023

)

= 2.425 × 10−25 kg

11-3. The number of each is given by

number = (1.00 × 10−12 L)(

1000 mL1 L

) (1.00 g1 mL

) (1 mol H2O

18.015 g H2O

)

×(

6.022 × 1023 molecule1 mol

)

= 3.34 × 1013 H2O molecules

= 3.34 × 1013 O atoms

= 6.68 × 1013 H atoms

11-4. As usual, take a 100-gram sample and write

52.1 g C � 13.2 g H � 34.7 g O

Divide each value by the corresponding atomic mass

52.1 C12.01

� 13.2 H1.0079

� 34.7 O16.00

to get

4.34 mol C � 13.10 mol H � 2.17 mol O

Dividing by 2.17, the smallest number, we get

2.00 mol C � 6.04 mol H � 1.00 mol O

or C2H6O.

11-5. The amount of oxygen is given by

mass of oxygen = 3.37 g Sc O − 2.18 g Sc = 1.16 g

Thus, we have

2.18 g Sc � 1.16 g O

Dividing each value by the corresponding atomic mass gives

0.0485 mol Sc � 0.0725 mol O



Chapter 11: Chemical Calculations 25

Dividing by 0.0485, we get

1 mol Sc � 1.49 mol O

or, multiplying by 2, Sc2O3.

11-6. The amount of oxygen is given by (4.111 − 3.058) grams = 1.053 grams. Thus, we have

3.058 g M � 1.053 g O

The number of moles of oxygen is

moles of O = (1.053 g O)(

1 mol O16.00 g O

)= 0.06581 mol

The empirical formula of the oxide is M2O3, so we have

moles of M = (0.06581 mol O)(

2 mol M3 mol O

)= 0.04388 mol

and so

3.058 g M � 0.04388 mol M

or

69.69 g M � 1 mol M

The metal is gallium.

11-7. As usual, take a 100-gram sample and write

24.47 g C � 4.075 g H � 71.65 g Cl

2.037 mol C � 4.043 mol H � 2.021 mol Cl

1 mol C � 2 mol H � 1 mol Cl

or CH2Cl is the empirical formula. Given that its molecular mass is 98.95 and the empiricalformula mass is 49.48, the molecular formula must be 98.95/49.48 = 2 times the empiricalformula. Thus, the molecular formula is C2H4Cl2.

11-8. We have that

mass of C = (3.710 g CO2)

(12.01 g C

44.01 g CO2

)= 1.012 g C

mass of H = (1.013 g H2O)

(2 × 1.008 g H18.02 g H2O

)= 0.1133 g H

mass % C = 1.012 g C2.475 g sample

× 100 = 40.89%



mass % H = 0.1133 g H2.475 g sample

× 100 = 4.578%

mass % O = (100.00 − 40.89 − 4.578)% = 54.53%

As usual, take a 100-gram sample and write

40.89 g C � 4.578 g H � 54.53 g O

3.405 mol C � 4.542 mol H � 3.408 mol O

1.00 mol C � 1.33 mol H = 43

mol H � 1.00 mol O

3 mol C � 4 mol H � 3 mol O

The empirical formula is C3H4O3. Given that the empirical formula is 88.06 and that itsmolecular mass is 176, the molecular formula of vitamin C is C6H8O6.

11-9. (a) 2 KClO3(s) −→ 2 KCl(s) + 3 O2(g)

(b) moles of O2 = (0.50 mol KClO3)

(3 mol O2

2 mol KClO3

)= 0.75 mol

(c) mass of O2 = (30.6 g KClO3)

(1 mol KClO3

122.55 g KClO3

) (3 mol O2

2 mol KClO3

) (32.00 g O2

1 mol O2

)= 12.0 g

11-10. We have that

moles of ND3 = (7.15 mg ND3)(

1 g1000 mg

) (1 mol ND3

20.049 g ND3

)

= 3.566 × 10−4 mol

mass of D2O = (3.566 × 10−4 mol D2O)

(3 mol D2O1 mol ND3

) (20.027 g D2O

1 mol D2O

) (1000 mg

1 g

)

= 21.4 mg D2O

The number of millilters of D2O(l) required is given by

volume = (21.4 mg)

(1 × 10−3 g

1 mg

) (1 mL

1.106 g

)= 0.0194 mL

11-11. We have that

mass of SF6 = (5.00 g S)(

1 mol S32.065 g S

) (1 mol SF6

1 mol S

) (146.05 g SF6

1 mol SF6

)= 22.8 g

and

mass of F2 = (5.00 g S)(

1 mol S32.065 g S

) (3 mol F2

1 mol S

) (37.997 g F2

1 mol F2

)= 17.8 g

11-12. (a) The number of grams of PI3(s) prepared is given by

mass of PI3 = (1.25 g P)(

1 mol P30.974 g P

) (2 mol PI3

2 mol P

) (411.69 g PI3

1 mol PI3

)= 16.6 g



Chapter 11: Chemical Calculations 27

(b) The number of grams of I2(s) required is given by

mass of I2 = (1.25 g P)(

1 mol P30.974 g P

) (3 mol I2

2 mol P

) (253.81 g I2

1 mol I2

)= 15.4 g

11-13. All the copper in CuFeS2(s) ends up as copper metal. Thus,

mass of Cu = (1000 kg ore)(

30% CuFeS2

100% ore

) (1000 g1 kg

) (1 mol CuFeS2

183.52 g CuFeS2

)

×(

1 mol Cu1 mol CuFeS2

) (63.546 g Cu

1 mol Cu

)

= 1.04 × 105 g = 0.104 metric ton

11-14. Let x be the mass percentage of NaCl(s) in the mixture. The mass of NaCl(s) in the mixture isgiven by

mass of NaCl = (2.86 g)(

x100

)

and the mass of BaCl2 is given by

mass of BaCl2 = (2.86 g)(

1 − x100

)

The number of moles of each compound is

moles NaCl = (2.86 g NaCl)(

x100

) (1 mol NaCl

58.44 g NaCl

)= 4.89 × 10−4x mol

moles BaCl2 =[(2.86 g)

(1 − x

100

)BaCl2

] (1 mol BaCl2

208.23 g BaCl2

)= (0.0137 − 1.37 × 10−4x) mol

The number of moles of AgCl(s) is given by

moles AgCl = (4.81 g AgCl)(

1 mol AgCl143.32 g AgCl

)= 0.0336 mol AgCl

Because all the chloride in the AgCl(s) comes from the NaCl(s) and BaCl2(s), from thereaction stoichiometry, we have

moles AgCl = moles NaCl + 2 moles BaCl2

Therefore, we have

0.0336 mol = (4.89 × 10−4x) mol + (2)(0.0137 − 1.37 × 10−4x) mol

6.2 × 10−3 = 2.15 × 10−4x

or x = 29%. Therefore, the mixture contains 29% NaCl(s) and 71% BaCl2(s).

11-15. The balanced chemical equation is

CaSO4(s) + 4 C(s) −→ CaS(s) + 4 CO(g)



We must check to see if there is a limiting reactant.

moles of CaSO4 = (125 g CaSO4)

(1 mol CaSO4

136.14 g CaSO4

)= 0.918 mol

moles of C = (125 g C)(

1 mol C12.0107 g C

)= 10.4 mol

Dividing each by its respective stoichiometric coefficient in the balanced equation, we find thatthe carbon is in great excess. The number of grams of CaS(s) produced is given by

mass of CaS = (0.918 mol CaSO4)

(1 mol CaS

1 mol CaSO4

) (72.14 g CaS1 mol CaS

)= 66.2 g

11-16. For a theoretical yield, the number of grams of tin required is given by

mass of Sn = (0.106 g SnCl4)

(1 mol SnCl4

260.52 g SnCl4

) (1 mol Sn

1 mol SnCl4

) (118.71 g Sn

1 mol Sn

)= 0.0483 g

For a percentage yield of 64.3%, the masss of tin required is

mass of Sn = (0.0483 g Sn)(

100%64.3%

)= 0.0751 g


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CHAPTER 7. Lewis Formulas - General Chemistrymcquarriegeneralchemistry.com/pracprobs_final_7-11.pdf · atom. The Lewis formula is – H NH 7-5. See the text. 7-6. The Lewis formula