Petrucci Chap 21 Answers

CHEMISTRY OF THE MAIN-GROUP ELEMENTS I:

GROUPS 1, 2, 13, AND 14

PRACTICE EXAMPLES

1A (E) From Figure 21-2, the route from sodium chloride to sodium nitrate begins with

electrolysis of NaCl(aq) to form NaOH(aq).

2 NaCl aq

 

+ 2 H₂O l

 

_electrolysis_{ 2 NaOH aq}

 

_{+ H}

2

 

g + Cl2

 

g followed by addition of

NO₂

 

g to NaOH(aq). 2 NaOH aq + 3 NO g

 

2

 

2 NaNO aq + NO g + H O(l)3

 

 

2

1B (E) From Figure 21-2, we see that the route from sodium chloride to sodium thiosulfate begins with the electrolysis of NaCl(aq) to produce NaOH(aq),

2 NaCl aq

 

+ 2 H₂O l

 

_electrolysis_{ 2 NaOH aq}

 

_{+ H}

2

 

g + Cl2

 

g

and continues through the reaction of SO2

 

g with the NaOH(aq) in an acid-base reaction

[ SO2

 

g is an acid anhydride] to produce

 

 

 

 

2 3 2 2 3 2

Na SO aq : 2 NaOH aq + SO g  Na SO aq + H O(l) and (3) concludes with the addition of S to the boiling solution:

 

boil

 

2 3 2 2 3

Na SO aq + S(s)Na S O aq .

2A (M) The first reaction indicates that 0.1 mol of NaNO2 reacts with 0.3 mol of Na to yield

0.2 mol of compound X and 0.05 mol of N2. Since this reaction liberates nitrogen gas, it is

very likely that compound X is an oxide of sodium. The two possibilities include Na2O

and Na2O2. Na2O2 will not react with oxygen, but Na2O will. Therefore, compound X is

most likely Na2O and compound Y is Na2O2. The balanced chemical equations for two

processes are:

NaNO₂(s)+3Na(s) 2Na₂O(s)+1 2N2(g) Na₂O(s)+1

2O2(g) Na2O2(s)

2B (M) Since compound X is used in plaster of Paris it must contain calcium. Calcium reacts with carbon to form calcium carbide, according to the following chemical equation:

Ca(s)+2C(s) CaC₂(s)

Furthermore, compound X or calcium carbide reacts with nitrogen gas to form compound Y or calcium cyanamide:

CaC₂(s)+N₂(g) CaCN₂(s)+C(s)

CN22- anion is isoelectronic with CO2 and it therefore contains 16 electrons. The structure

of this anion is: N C N

3A (M) The first two reactions, are those from Example 21-3, used to produce B O . _{2 3}

 

 

2 4 7 2 2 4 ₃ 2 4 2 2 3 2 3 Na B O 10 H O(s) + H SO (l) 4 B OH (s) + Na SO (s) + 5H O(l) 2 B OH (s)  B O (s) + 3H O(g)   

The next reaction is conversion to BCl with heat, carbon, and chlorine. ₃

 

 

 

 

 

2 3 2 3 2

2 B O s + 3C s + 6Cl g __4 BCl g + 3CO g 4

LiAlH is used as a reducing agent to produce diborane.

 

 

3 4 2 6 3

4 BCl g + 3LiAlH (s)2 B H g + 3LiCl(s) + 3AlCl (s)

3B (M) Na2B4O7.10 H2O(s) + H2SO4(aq)  4 B(OH)3(s) + Na2SO4(aq) + 5 H2O(l)

2 B(OH)3(s)  B2O3(s) + 3 H2O(l)

B2O3(s)+ 3 CaF2(s) + 3 H2SO4(l)  2 BF3(g) + 3 CaSO4(s) + 3 H2O(g)

INTEGRATIVE EXAMPLE

A (M) NaCN and Al(NO3)3 dissociate in water according to the following equations:

NaCN(aq) Na+_(aq)+CN-_(aq) Al(NO₃)₃(aq) Al3+_(aq)+3NO 3 -_(aq)

NaCN is a salt of strong base (NaOH) and weak acid (HCN) and its solution is therefore basic.

We proceed by first determining the [OH]- concentration in a solution which is 1.0M in NaCN: CN-+H2O É HCN+OH -1.0 / / 1.0-x x x K_b  Kw K_a(HCN ) 1.0 1014 6.2 1010  1.6  10 5 K_b [HCN ][OH] [CN_]  x x 1.0 x  x2 1.0 x  1.6  10 5 Assume, x<<1 and 1.0-x1.0 x2 _{ 1.6  10}5_{ x  1.6  10}5 _{ 4.0  10}3 [OH] 4.0  103

On mixing, [OH-]=2.010-3 _{M and [Al}3+_{]=0.50 M. The precipitation will occur according}

to the following equation:

Al3+(aq)+3OH-(aq) É Al(OH)3(s)

The product [Al3+][OH-]3=0.50(2.010-3₎3_=4.010-9 _{is much greater than K}

SP for Al(OH)3

B (M) When BeCl 4H O₂ ₂ is heated, it decomposes to Be(OH)2(s), H2O(g), and HCl(g), as

discussed in Section 21-3. BeCl 4H O₂ ₂ comprises [Be(H2O)4]2+ and Cl− ions. Because of

the high polarizing power of Be2+_{, it is difficult to remove the coordinated water molecules}

by heating the solid and the acidity of the coordinated H2O molecules is enhanced. When

2 2

BeCl 4H O is dissolved in water, [Be(H2O)4]2+ ions react with water, and

[Be(H2O)3(OH)]+ and H3O+ ions are produced. Hence, a solution of BeCl 4H O₂ ₂ is

expected to be acidic.

When CaCl 6H O₂ ₂ is heated, CaCl2(s) and H2O(g) are produced. Because the charge

density and polarizing power of Ca2+ is much less than that of Be2+, it is much easier to drive off the coordinated water molecules by heating the solid. When CaCl 6H O₂ ₂ is dissolved in water, Ca2+(aq) and Cl−(aq) are produced. However, the charge density of the Ca2+ ion is too low to affect the acidity of water molecules in the hydration sphere of the Ca2+ ion. Hence, a solution of CaCl 6H O₂ ₂ has a neutral pH.

EXERCISES

Group 1: The Alkali Metals

1. (E) (a) 2 Cs(s) + Cl (g) ₂  2 CsCl(s) (b) 2 Na(s) + O (g) ₂  Na O (s)_{2 2} (c) Li CO s2 3

 

 Li O s + CO g2

 

2

 

(d) Na SO s + 4 C s_{2 4}

 

 

 Na S s + 4 CO g₂

 

 

(e) K s + O g

 

₂

 

 KO s₂

 

2. (a) 2 Rb s

 

+ 2 H₂O l



 2RbOH aq

 

+ H₂

 

g (b) 2 KHCO s ₃

 

 K CO aq + H O(l) + CO g_{2 3}

 

_{2 2}

 

(c) 2 Li s + O g

 

₂

 

 Li O s_{2 2}

 

(d) 2 KCl s + H SO aq

 

_{2 4}

 

 K SO aq + 2 HCl g_{2 4}

 

 

(e) LiH s + H O l

 

₂

 

 LiOH aq + H g

 

₂

 

3. (E) Both LiCl and KCl are soluble in water, but Li PO_{3 4} is not very soluble. Hence the addition of K PO (aq)_{3 4} to a solution of the white solid will produce a precipitate if the white solid is LiCl, but no precipitate if the white solid is KCl. The best method is a flame test; lithium gives a red color to a flame, while the potassium flame test is violet.

4. (E) When heated, Li CO_{2 3} decomposes to CO (g)₂ and Li O(l). K CO (s) simply melts when _{2 2 3} heated. The evolution of CO (g)₂ bubbles should be sufficient indication of the difference in behavior. The flame test affords a violet flame if K+ is present, and a red flame if Li+ is present.

5. (M) First we note that sodium carbonate ionizes virtually completely when dissolved in H2O and

thus is described as highly soluble in water. This is made evident by the fact that there is no Ksp

value for Na2CO3. By contrast, the existence of Ksp values for MgCO3 and Li2CO3 shows that

these compounds have lower solubilities in water than Na2CO3. To decide whether MgCO3 is

more or less soluble than Li2CO3, we must calculate the molar solubility for each salt and then

compare the two values. Clearly, the salt that has the larger molar solubility will be more soluble in water. The molar solubilities can be found using the respective Ksp expressions for the two

salts. 1. MgCO3 -8 sp = 3.5 10 K 





Mg2+(aq) + CO32-(aq) Let “s” = molar solubility of MgCO3

excess – s s s s2 = 3.5×10-8 s = 1.9×10-4 M 2. Li2CO3 -2 sp = 2.5 10 K 





2 Li+(aq) + CO32-(aq) Let “s” = molar solubility of Li2CO3

excess – s 2s s

(2s)2×(s) = 2.5×10-8 4s3 = 2.5×10-8 s = 3 2.5×10-8

4 = 0.18 M

We can conclude that Li2CO3 is more soluble than MgCO3. Thus, the expected order of

increasing solubility in water is: MgCO3 < Li2CO3 < Na2CO3.

6. (M) We know sodium metal was produced at the cathode from the reduction of sodium ion, Na+_{. Thus, hydroxide must have been involved in oxidation at the anode. The}

hydrogen in hydroxide ion already is in its highest oxidation state and thus cannot be oxidized. This leaves oxidation of the hydroxide ion to elemental oxygen as the remaining reaction.

Cathode, reduction : {Na+_e _ Na(l)} _4 Anode, oxidation : 4 OH_ O (g) + 2 H O(g) + 4e_{2 2} 

Net : 4Na+ 4OH 4Na(l) + O (g) + 2 H O(g)

2 2

  

7. (M) (a) H (g)₂ and Cl (g)₂ are produced during the electrolysis of NaCl(aq). The electrode reactions are:

Anode, oxidation: 2 Cl (aq)  Cl (g) + 2e₂ 

Cathode, reduction: 2 H O(l) + 2e₂ H (g) + 2OH (aq)₂  We can compute the amount of OH_{produced at the cathode.}

mol OH  2.50 min  60s 1min  0.810C 1s  1mole 96,500C 2 molOH 2 mole  1.26  10 3_molOH

Then we compute the [OH] and, from that, the pH of the solution.

[OH mol OH Lsoln M _] .    _{ }  . . 126 10 0 872 145 10 3 3 _{pOH = log(1.45 10 } 3_{)2 839.} pH=14.000 2.839 11.161 

(b) As long as NaCl is in excess and the volume of the solution is nearly constant, the solution pH only depends on the number of electrons transferred.

8. (M) (a) total energy = _{3.0 V 0.50 A h 5.4 10 J}3600s 1C/s 1J 3

1hr 1A 1V C

     

We obtained the first conversion factor for time as follows. 5 0.  6 1 5 0 10. 6  W 1 10 W 1 W J / s 1W J 1s        time = 3 9 6 1s 1hr 1day 1y 5.4 10 J 1.1 10 s 34 y 5 10 J 3600s 24 hr 365days         

(b) The capacity of the battery is determined by the mass of Li present. 1 C/s 3600 s 1 mol e 1 mol Li 6.941 g Li mass Li = 0.50 A h = 0.13 g Li 1 A 1 h 96485 C 1 mol e 1 mol Li       

9. (M) (a) We first compute the mass of NaHCO3 that should be produced from 1.00 ton

NaCl, assuming that all of the Na in the NaCl ends up in the NaHCO3. We use the

unit, ton-mole, to simplify the calculations.

3

3 3

3 3

1 ton-mol NaCl 1ton-mol Na mass NaHCO 1.00 ton NaCl

58.4 ton NaCl 1ton-mol NaCl 1tol-mol NaHCO 84.0 ton NaHCO

= 1.44 ton NaHCO 1ton-mol Na 1 ton mol NaHCO

  

 

% yield ton NaHCO produced

ton NaHCO expected yield = 1.03

1.44 100% = 71.5%

3 3



(b) NH₃ is used in the principal step of the Solvay process to produce a solution in which NaHCO₃ is formed and from which it will precipitate. The filtrate contains NH Cl,₄ from which NH₃ is recovered by treatment with Ca(OH)₂. Thus, NH₃ is simply used

during the Solvay process to produce the proper conditions for the desired reactions. Any net consumption of NH₃ is the result of unavoidable losses during production.

10. (M) (a) 2

2 4 4

Ca(OH) (s) SO (aq)_CaSO (s) 2OH (aq)  (b) We sum two solubility reactions and combine their values of K_sp

2 2

Ca(OH) (s) Ca (aq) 2OH (aq) _  _ 

 K_sp _{55 10.}  6

2 2

4 4

Ca (aq) SO  (aq)_CaSO (s) _{1/K 1 9 1 10/ .} 6

sp  

   

2

 

 

 

4 4 2 Ca OH s + SO  aq CaSO s + 2 OH aq K_eq =5.5 10 9.1 10 = 0.60 6 6    

Because Keq is close to 1.00, we conclude that the reaction lies neither very far to the

right (it does not go to completion) nor to the left.

(c) Reaction: 2

2 4 4

Ca(OH) (s) SO_ (aq) _ CaSO (s) 2OH (aq)_   Initial: — 1.00 M —  0 M Changes: — -x M — +2x M Equil: — (1.00–x) M — 2x M K x x x x x x = = 0.60 = 4 1.00 4 = 0.60 0.60 4 + 0.60 0.60 = 0 2 4 2 2 2 2 OH SO   _   x b b ac a = 4 2 = 0.60 0.36 + 9.60 8 = 0.32 2      M SO₄2 _{= 1.00 x= 0.68} M OH _{= 2 = 0.64x} M

11. (M) Use Go_{f values to calculate G}o _{for the reaction and then the equilibrium constant}

K at 298 K. Na2O2(s) É Na2O(s)+ 1 2O2(g) G_fo_(kJmol-1_{) -449.63 -379.09} Go _{ 379.09  (449.63)  70.54kJmol}-1 Go _{ RT ln K} 8.314JK-1_mol-1_{ 298K  ln K  70.54  1000Jmol}-1 ln K 70.54 1000Jmol-1 8.314JK-1_mol-1_{ 298K}  28.47  K  e 28.47 _{ 4.32  10}13

SinceK  p_O

2

1 2  p

O2  K

2 _{ 1.87  10}25_{. The equilibrium constant and partial pressure of}

oxygen are both very small at 298 K. Therefore, Na2O2(s) is thermodynamically stable

with respect to Na2O(s) and O2(g) at 298 K.

12. (M) Use Go_{f values to calculate G}o _{for the reaction and then the equilibrium constant}

K at 298 K. 2KO2(s) É K2O(s)+ 3 2O2(g) G_fo_(kJmol-1_{) -240.59 -322.09} Go _{ 322.09  (2  240.59)  159.1kJmol}-1 Go _{ RT ln K} 8.314JK-1_mol-1_{ 298K  ln K  159.1  1000Jmol}-1 ln K 159.1 1000Jmol-1 8.314JK-1_mol-1_{ 298K}  64  K  e 64 _{ 1.604  10}28 Since K  p_O 2 3 2  p O2  K 2

3  2.952  1019_{. The equilibrium constant and partial pressure}

of oxygen are both very small at 298 K. Therefore, KO2(s) is thermodynamically stable

with respect to K2O(s) and O2(g) at 298 K.

Group 2: The Alkaline Earth Metals

13. (M)

   

 

 

 

 

 

   

 

 

 

 

 

 

   

 

 

2 2 2 ,electrolysis 2 2 ₂ 2 3 2 3 2 ₂ 2 4 4 2

The reactions are as follows. Ca OH s + 2 HCl aq CaCl aq + 2H O(l)

CaCl l Ca l + Cl g Ca OH s + CO g CaCO s + H O g

CaCO s CaO s + CO g Ca OH s + H SO aq CaSO s + 2 H O(l)       

   

2 3 4

 

4

 

2

Ca OH s + H PO aq  CaHPO aq + 2H O(l). Actually CaO s

 

is the industrial starting material from which Ca(OH)₂ is made. CaO(s) + H O(l)₂ Ca(OH) (s)₂

14. (M)

Once we return to Mg(OH)2 from MgSO4, the other substances can be made by the indicated

pathways. The return reaction is: MgSO (aq) + 2 NaOH(aq) ₄ Mg(OH) (s) + Na SO (aq)._{2 2 4} Then the other reactions are

CaO CaCO₃ CO2 _Ca(OH)

2 HCl CaCl₂ Ca H₂SO₄ CaSO₄ H₃PO₄ CaHPO₄  electrolysis

MgO MgCO₃ CO2 _Mg(OH)

2 HCl MgCl₂ Mg H₂SO₄ MgSO₄ H₃PO₄ MgHPO₄  electrolysis N_{2 Mg} 3N2

2 2 2 Mg(OH) (s) + 2 HCl(aq) MgCl (aq) +2 H O(l)

, electrolysis

2 2 2 2 3 2

MgCl (l)  Mg(l) + Cl (g) ; Mg(OH) (s) + CO (g) MgCO (s) + H O(g)

3 2 2 2 4 4 2

MgCO (s)__ MgO(s) + CO (g); Mg(OH) (s) + H SO (aq) _MgSO (s) + 2H O(l)

2 3 4 4 2

Mg(OH) (s) + H PO (aq)  MgHPO (aq) + 2H O(l)

15. (M) The reactions involved are:

Mg2+_{(aq) + Ca}2+_{(aq) + 2OH}_{(aq)  Mg(OH)}

2(s) + Ca2+(aq)

Mg(OH)2(s) + 2 H+(aq) + 2 Cl(aq)  Mg2+(aq) + 2 H2O(l) + 2 Cl(aq)

Mg2+(aq) + 2 Cl(aq) MgCl2(s)

MgCl2(s)  (electrolysis) Mg(l) + Cl2(g)

Mg(l)  Mg(s)

Mg2+ + 2 Cl(aq)  Mg(s) + Cl2(g) (Overall reaction)

As can be seen, the process does not violate the principle of conservation of charge.

16. (M) (a) MgO vs. BaO: MgO would have the higher melting point because, although Mg2+ and Ba2+ have the same charge, Mg2+ is a smaller ion. Smaller ions have a larger electrostatic attraction to anions (here, in both cases, the anion is O2), which is due to the smaller charge separation (Coulomb's law).

(b) MgF2 vs. MgCl2 solubility in water. F is smaller than Cl, hence, electrostatic attraction

between Mg2+ and the halide is greater in F than Cl. If we assume that hydration of the ions is similar, we expect that MgF2 is less soluble than MgCl2. (Note: Ksp given for

MgF2, which is sparingly soluble, while no Ksp value is given for readily soluble MgCl2.)

17. (M) (a) BeF (s) + Mg(s)_{2 }_ Be(s) + MgF s₂

 

(b) Ba s

 

+ Br₂

 

l  BaBr₂

 

s

(c) UO₂

 

s + 2 Ca s

 

 U s

 

+ 2 CaO s

 

(d) MgCO CaCO s_{3 3}

 

__MgO s + CaO s + 2 CO g

 

 

₂

 

(e) 2 H₃PO₄

 

aq + 3 CaO s

 

 Ca₃

 

PO_{4 2}

 

s + 3 H₂O l

 

18. (M) (a)



  

heat

 

 

3 2 2 2

Mg HCO s MgO s + 2  CO g + H O(l) (b)

 

electrolysis

 

 

2 2 BaCl l Ba l + Cl g (c) Sr s

 

+ 2 HBr aq

 

 SrBr₂

 

aq + H₂

 

g (d) H SO aq + Ca OH_{2 4}

 

   

₂ aq CaSO s + 2H O l₄

 

₂

 

(e)

 

1

 

3

 

4 2 4 2 2 2 2 CaSO 2 H O s_{ }_CaSO _ H O s + H O g

19. (D) Let us compute the value of the equilibrium constant for each reaction by combining the two solubility product constants. Large values of equilibrium constants indicate that the reaction is displaced far to the right. Values of K that are much smaller than 1 indicate that the reaction is displaced far to the left.

(a)

 

2+

 

2

 

10

4 4 sp

 

 

 

2+ 2 9 3 3 sp Ba aq + CO  aq _BaCO s 1/K = 1/ (5.1 10 )_  

 

2

 

 

2

 

10 2 4 3 3 4 9 1.1 10 BaSO s + CO aq BaCO s + SO aq = = 2.2 10 5.1 10 K       _   

Thus, the equilibrium lies slightly to the left.

(b)



  

2+

 

3

 

25 3 4 2 4 sp Mg PO s _3 Mg aq + 2 PO  aq K = 2.1 10 

 

 

 

2+ 2 3 8 3 3 3 sp 3{Mg aq + CO  aq _MgCO s } 1/ (K ) = 1/ (3.5 10 )_  



  

 

 

 





2 3 3 4 2 3 3 4 25 3 3 8 Mg PO s + 3CO aq 3 MgCO s + 2 PO aq 2.1 10

= = 4.9 10 Thus, the equilibrium lies to the left. 3.5 10 K       _    (c)

   

2+

 

 

6 sp 2 Ca OH s _Ca aq + 2 OH aq K = 5.5 10_  

 

 

 

2+ 9 2 sp Ca aq + 2 F aq _CaF s 1/K 1/ (5.3 10 ) 

 

 

 

 

2 2 Ca(OH) s + 2 F aq _CaF s + 2 OH aq  6 3 9 5.5 10 = = 1.0 10 5.3 10 K    _  Thus, the equilibrium lies to the right.

20. (M) We expect the reaction to occur to a significant extent in the forward direction if its equilibrium constant is >> 1. (a)

 

2+

 

2

 

9 3 3 sp BaCO s _Ba aq + CO  aq K = 5.1 10_  

 

₊

 

 

₂



₅



2 2 3 2 2 3 2 a 2 HC H O aq _2H aq + 2C H O  aq K = 1.8 10 

 

 

 

2 + 2 11 3 3 a H aq + CO  aq _HCO  aq 1/K = 1/ (4.7 10 )_  

 

 

 

1 + 7 3 2 3 a H aq + HCO  aq _H CO aq 1/K = 1/ (4.2 10 )_  

 

 



  

 

3 2 3 2 2 3 2 2 2 3 BaCO s + 2 HC H O aq _Ba C H O aq + H CO aq K = 5.1 10 9_{ 1.8  10}



5



2 4.7 1011_{ 4.2  10}7 = 8.4 10 2

Thus, no significant reaction occurs (the equilibrium mixture contains appreciable amounts of all four species involved in the reaction.

(b)

   

2+

 

 

6 sp 2 Ca OH s _Ca aq + 2 OH aq K = 5.5 10_  

 

 

 





2 + 2 5 4 3 2 b 2 NH aq + 2 OH aq _2 NH aq + 2 H O(l) 1/K = 1/ 1.8 10_  

   

 

 

 





+ 2+ 4 3 2 2 6 4 2 5 Ca OH s + 2 NH aq Ca aq + 2 NH aq + 2 H O(l) 5.5 10

= = 1.7 10 Thus, this reaction would occur to a significant extent. 1.8 10 K    _   

(c)

 

2+

 

 

6 2 sp BaF s _Ba aq + 2 F aq K = 1.0 10 

 

 

 





2 + 2 4 3 2 a 2 H O aq + 2 F aq _2 HF aq + 2 H O(l) 1/K = 1/ 6.6 10_  

 

 

 

 





6 2+ 2 3 2 ₄ 2 1.0 10 BaF s + 2H O aq Ba aq + 2HF aq + 2H O(l); = = 2.3 6.6 10 K       

This reaction would occur to some extent, but certainly not to completion.

21. (M) The SO42− ion is a large polarizable ion. A cation with a high polarizing power

will polarize the SO42− ion and kinetically assist the decomposition to SO3.

Because Be2+ _{has the largest charge density of the group 2 cations, we expect Be}2+

to be the most polarizing and thus, BeSO4 will be the least stable with respect to

decomposition.

22. (M) The CO32− ion is a large polarizable ion. A cation with a high polarizing power

will polarize the CO32− ion and kinetically assist the decomposition to CO2. A

cation with a low polarizing power will not polarize the anion as much; thus, the decomposition of CO32− to CO2 will not occur as readily in the presence of such a

cation. Because Ba2+ has the smallest charge density of the group 2 cations, we expect Ba2+ to be the least polarizing; thus, BaSO4 will be the most stable with

respect to decomposition.

Group 13: The Boron Family

23. (M) (a) B H contains a total of _{4 10} 4 3 +10 1 = 22  valence electrons or 11 pairs. Ten of these pairs could be allocated to form 10 B—H bonds, leaving but one pair to bond the four B atoms together, which is clearly an electron deficient situation.

(b) In our analysis in part (a), we noted that the four B atoms had but one electron pair to bond them together. To bond these four atoms into a chain requires three electron pairs. Since each electron pair in a bridging bond replaces two “normal” bonds, there must be at least two bridging bonds in the B H molecules. By analogy with_{4 10} B H , _{2 6} we might write the structure below left. But this structure uses only a total of 20 electrons. (The bridge bonds are shown as dots, normal bonds—electron pairs—as dashes.) In the structure at right below, we have retained some of the form of B H , _{2 6} and produced a compound with the formula B H and 11 electron pairs. (The _{4 10} experimentally determined structure of B H consists of a four-membered ring of _{4 10} alternating B and H atoms, held together by bridging bonds. Two of the B atoms have two H atoms bonded to each of them by normal covalent bonds. The other two B atoms have one H atom covalently bonded to each. One final B—B bond joins these last two B atoms, across the diameter of the ring.). See the diagram that follows:

(c) C H contains a total of _{4 10} 4 4 +10 1 = 26  valence electrons or 13 pairs. A plausible Lewis structure follows. (Note that each atom possess an octet of electrons.)

24. (E) (a) (b)

25. (M) (a) 2BBr3(l) + 3H2(g)  2B(s) + 6HBr(g)

(b) i) B2O3(s) + 3 C(s)  3 CO(g) + 2 B(s)

ii) 2 B(s) + 3 F2(g)  2 BF3(g)

(c) 2 B(s) + 3 N2O(g)  3 N2(g) + B2O3(s)

26. (M) Each boron atom has an oxidation number of +3. The hydroxyl oxygens are each –2, while the bridging oxygens are each –1. Finally, the hydroxyl H atoms are all in the +1 oxidation state. The oxidation numbers for the all the constituent atoms add up to the charge on the perborate ion, namely, 2.

27. (M) (a) 2 Al s

 

+ 6 HCl aq

 

 2 AlCl₃

 

aq + 3 H₂

 

g (b)

 

 

 

+

 

   

 

2 ₄ 2 2 NaOH aq + 2 Al s + 6 H O l 2 Na aq + 2 Al OH_ _ aq + 3 H g (c) Oxidation: _{{Al s}

 

_Al3+

 

_{aq + 3 e }} _2 Reduction: 2

 

+

 

 

4 2 2 {SO  aq + 4 H aq + 2e_SO g + 2 H O(l)} _ 3 _________________________________________________________________________________________________________________________________ Net:

 

2

 

+

 

3+

 

 

4 2 2 2 Al s + 3 SO  aq +12 H aq 2 Al aq + 3 SO aq + 6 H O(l)

28. (M) (a) 2 Al(s) + 3 Br2(l)  2 AlBr3(s)

(b) 2 Al s

 

+ Cr₂O₃

 

s _{ }heat_{ 2Cr l}

 

_{+ Al} 2O3

 

s H C C C C H H H H H H H H H F B F F F  F B F F N H H C H H C H H H _ ₊ H H H H H H B H B H B H B H             H H H H B H B H B H         B H H H

(c) Fe O s + OH aq_{2 3}

 



 

_no reaction

 

 

   

2 3 2 ₄

Al O s + 2 OH aq + 3 H O(l) _{ }2 Al OH _ aq

29. (M) One method of analyzing this reaction is to envision the HCO₃ ion as a combination of CO₂ and OH. Then the OH reacts with Al3+ _{and forms Al OH}

 

3. This method of

envisioning HCO₃ does have its basis in reality. After all,





2 3 2 2 3 2 H CO = H O + CO + OH_HCO + H O Al3+

 

_{aq + 3 HCO} 3 

 

_{aq  Al OH}

 

3

 

s + 3 CO2

 

g

Another method is to consider the reaction as, first, the hydrolysis of hydrated aluminum ion to produce Al OH

   

₃ s and an acidic solution, followed by the reaction of the acid with bicarbonate ion.





3+

 

  

  

₊

 

2 ₆ 2 ₃ 2 ₃ 3 Al H O aq + 3H O(l) Al OH H O s + 3H O aq     

 

 

 

+ 3 3 2 2

3H O aq + 3HCO  aq 6 H O(l) + 3CO g This gives the same net reaction:





3+

 

 

  

  

 

2 6 3 3 2 3 2 2

Al H O aq + 3HCO  aq Al OH H O s + 3CO g + 3H O(l)

  

 

30. (E) The Al3+

 

_{aq ion hydrolyzes.} 3+

 

   

+

 

2 3

Al aq + 3 H O(l)Al OH s + 3 H aq

Subsequently, the hydrogen ion that is produced reacts with bicarbonate ion to liberate CO g₂

 

.

 

 

 

+

3 2 2

H aq + HCO  aq H O(l) + CO g

31. (M) Aluminum and its oxide are soluble in both acid and base.

 

 

 

 

 

 

 

+ 3+ 2 + 3+ 2 3 2 2 Al s + 6 H aq 2 Al aq + 3 H g Al O s + 6 H aq 2 Al aq + 3 H O(l)  

 

 

2

   

₄ 2

 

2 Al s + 2 OH aq + 6 H O(l) _ 2 Al OH_{ } aq + 3 H g   

 

 

   

2 3 2 ₄ Al O s + 2 OH aq + 3 H O(l) _ 2 Al OH_{ } aq   

Al(s) is resistant to corrosion only over the pH range 4.5 to 8.5. Thus, aluminum is inert only when the medium to which it is exposed is neither highly acidic nor highly basic.

32. (M) Both Al and Mg are attacked by acid and their ions are both precipitated by hydroxide ion.

 

+

 

3+

 

 

2 2 Al s + 6 H aq 2 Al aq +3 H g

 

+

 

2+

 

 

2 Mg s + 2 H aq Mg aq + H g

 

 

 

 

3+ 3 Al aq +3 OH aq Al OH s 2+

 

 

 

2 Mg aq + 2 OHMg OH s

But, of these two solid hydroxides, only Al OH

 

₃

 

s redissolves in excess OH

 

_{aq .} Al OH

 

₃

 

s + OH

 

aq  Al OH_

 

_4

 

aq

Thus, the analytical procedure consists of dissolving the sample in HCl(aq) and then treating the resulting solution with NaOH(aq) until a precipitate forms. If this precipitate dissolves completely in excess NaOH(aq), the sample is aluminum 2S. If at least some of the precipitate does not dissolve, the sample was magnesium.

33. (M) CO2

 

g is, of course, the anhydride of an acid. The reaction here is an acid-base

reaction.

Al OH

 

₄

 

 

aq + CO₂

 

aq  Al OH

 

₃

 

s + HCO₃

 

aq

HCl(aq), being a strong acid, can’t be used because it will dissolve the Al(OH)3(s).

34. (M) (a) Oersted: 2 Al₂O₃

 

s + 3 C s

 

+ 6 Cl₂

 

g   4 AlCl₃

 

s + 3 CO₂

 

g

(b) Wöhler: AlCl₃

 

s + 3 K s

 

  Al s

 

+ 3 KCl s

 

35. (M) 2 KOH(aq) + 2 Al(s) + 6 H2O(l)  2 K[Al(OH)4](aq) + 3 H2(g)

2 K[Al(OH)4](aq) + 4 H2SO4(aq)  K2SO4(aq) + Al2(SO4)3(aq) + 8 H2O(l)

 crystallize 2 KAl(SO4)2(s)

36. (E) HCO3 and CO32 solutions are basic (they produce an excess of OH). Al3+(aq) in the

presence of OH(aq) will precipitate as the hydroxide Al(OH)3(s).

37. (E) The structure of, and bonding in, digallane is similar to the structure of and bonding in diborane, B2H6. Ga H H Ga H H H H

38. (M) The structure of, and bonding in, GaH2Cl2 is similar to the structure of and bonding

in Al2Cl6, except that the terminal Cl’s in Al2Cl6 are replaced by H’s.

Ga Cl Cl Ga H H H H

Group 14: The Carbon Family

39. (E) In the sense that diamonds react imperceptibly slowly at room temperature (either with oxygen to form carbon dioxide, or in its transformation to the more stable graphite), it is essentially true that “diamonds last forever.” However, at elevated temperatures, diamond will burn to form CO2(g) and thus the statement is false. Also, the transformation

C(diamond) → C(graphite) might occur more rapidly under other conditions. Eventually, of course, the conversion to graphite occurs.

40. (E) The graphite in the pencil “lead” is a good dry lubricant that will make slippery the stickiness (or reluctance) in the lock and enable it to work smoothly. The key carries the graphite to the site that needs to be lubricated within the lock mechanism.

41. (M) (a) 3 SiO2(s) + 4 Al(s)  2 Al2O3(s) + 3 Si(s)

(b) K2CO3(s) + SiO2(s)  CO2(g) + K2SiO3(s)

(c) Al4C3(s) + 12 H2O(l)  3 CH4(g) + 4Al(OH)3(s)

42. (M) (a) 2 KCN(aq) + AgNO3(aq)  KAg(CN)2(aq) + KNO3(aq)

or KCN(aq) + AgNO3(aq)  Ag(CN)(s) + KNO3(aq)

(b) Si3H8(l) + 5 O2(g)  3 SiO2(s) + 4 H2O(l)

(c) N2(g) + CaC2(s)  CaNCN(s) + C(s)

43. (M) A silane is a silicon-hydrogen compound, with the general formula Si H_{n 2 +2n} . A silanol is a compound in which one or more of the hydrogens of silane is replaced by an —OH group. Then, the general formula becomes Si H_{n 2 +1n}

 

OH . In both of these classes of compounds, the number of silicon atoms, n, ranges from 1 to 6. Silicones are produced when silanols condense into chains, with the elimination of a water molecule between every two silanol molecules.

2 2 2 2 2

HOSi H_{n n}OH + HOSi H_{n n}OHHOSi H_{n n} O Si H_{n n}OH + H O

44. (M) Both the alkali metal carbonates and the alkali metal silicates are soluble in water. Hence, they also are soluble in acids. However, the carbonates will produce gaseous carbon dioxide—CO₂—on reaction with acid,



2 +



3 2 3 2 2

CO (aq) + 2 H (aq) “H CO ”H O(l) + CO (g) , while the silicates will produce silica— SiO₂— in an analogous reaction



4 +



4 4 4 2 2

SiO (aq) + 4H (aq)“H SiO ”2 H O(l) + SiO (s) . The silica is produced in many forms depending on reaction conditions: a colloidal dispersion, a gelatinous precipitate, or a semisolid gel. Silicates of cations other than those of alkali metals are insoluble in water, as are the analogous carbonates. However, the carbonates will dissolve in acids (witness the reaction of acid rain on limestone carvings and marble statues, both forms of CaCO ), while ₃ the silicates will not. (Silicate rocks are not significantly affected by acid rain.)

45. (M) (1) 2 CH g + S g₄

 

₈

 

2 CS g + 4 H S g₂

 

₂

 

(2) CS g + 3 Cl g₂

 

₂

 

CCl l + S Cl l₄

 

_{2 2}

 

46. (M) (a)



CH₃



₃SiCl(l) + H O(l)₂ 



CH₃



₃SiOH(aq) + HCl(aq)



3



3



3



3



3



3 2

2 CH SiOH(aq)  CH Si O Si CH (s) + H O(l)

(b) A silicone polymer does not form from



CH_{3 3}



SiCl. Only a dimer is produced.

(c) The product that results from the treatment of CH SiCl is two long Si—O—Si _{3 3} chains, with CH groups on the outside, linked by Si—O—Si bridges. Part of one of ₃ these chains and the beginnings of the bridges are shown below.

47. (D) Muscovite or white mica has the formula KAl2(OH)2(AlSi3O10). Since they are not

segregated into O2 units in the formula, all of the oxygen atoms in the mineral must be in

the 2 oxidation state. Potassium is obviously in the +1 oxidation state, as are the hydrogen atoms in the hydroxyl groups. Up to this point, we have -24 from the twelve oxygen atoms and +3 from the potassium and hydrogen atoms for a net number of -21 for the oxidation state. We still have three aluminum atoms and three silicon atoms to account for. In oxygen-rich salts such as mica, we would expect that the silicon and the aluminum atoms would be in their highest possible oxidation states, namely +4 and +3, respectively. Since the salt is neutral, the oxidation numbers for the silicon and aluminum atoms must add up to +21. This is precisely the total that is obtained if the silicon and aluminum atoms are in their highest possible oxidation states: (3  (+3) + 3  (+4) = +21). Consequently, the empirical formula for white muscovite is consistent with the expected oxidation state for each element present.

48. (D) Crysotile asbestos has the formula [Mg3Si2O5(OH)4]. Since they are not segregated to

O2 units in the formula, all of the oxygen atoms in the mineral must be in the 2 oxidation

state. The three magnesium atoms are obviously in the +2 oxidation state, while hydrogen atoms in the hydroxyl groups have an oxidation state of +1. Up to this point, then, the sum of the oxidation numbers equals 8 (18 from O-atoms, +6 from Mg-atoms, and + 4 from H-atoms). Since the mineral is neutral, the two silicon atoms must have oxidation states that sum to +8 (8 + 8 = neutral mineral), therefore, each silicon would need to be in the +4 oxidation state. In oxygenrich salts such as asbestos, we would expect that the silicon atoms would be in their highest possible oxidation state, namely +4. Thus, the oxidation states for the element, in this mineral are precisely consistent with expectations.

49. (M) (a) PbO s + 2 HNO aq

 

₃

 

Pb NO



₃

  

₂ s + H O(l)₂ (b) SnCO₃

 

s   SnO s

 

+ CO₂

 

g (c) PbO s

 

+ C s

 

  Pb l

 

+ CO g

 

O Si O CH₃ O Si O O CH₃ Si O CH₃ O Si O O CH₃ Si O CH₃ O Si O O CH₃

(d) 2 Fe3+

 

_{aq + Sn}2+

 

_{aq  2 Fe}2+

 

_{aq + Sn}4 +

 

_aq (e) 2 PbS s + 3 O g

 

₂

 

2 PbO s + 2 SO g

 

₂

 

 

 

 

2 2 3 2 SO g + O g 2 SO g SO₃

 

g + PbO s

 

 PbSO₄

 

s

Or perhaps simply: PbS s + 2O g

 

₂

 

PbSO s₄

 

Yet a third possibility:

 

2

 

3

 

PbO s + SO s PbSO s , followed by 2 PbSO s + O s₃

 

₂

 

2PbSO s₄

 

50. (M) (a) Treat tin(II) oxide with hydrochloric acid.

 

 

2

 

2

SnO s + 2 HCl aq SnCl aq + H O(l)

(b) Attack tin with chlorine. Sn s

 

+ 2 Cl₂

 

g  SnCl₄

 

s

(c) First we dissolve PbO₂

 

s in HNO₃

 

aq and then treat the resulting solution with K₂CrO₄

 

aq to precipitate PbCrO4(s)

 

 



  

 

2 3 3 2 2 2

2 PbO s + 4 HNO aq 2 Pb NO aq + O g + 2 H O(l)



3

  

₂ 2 4

 

4

 

3

 

Pb NO aq + K CrO aq PbCrO s + 2 KNO aq

51. (M) We start by using the Nernst equation to determine whether the cell voltage still is positive when the reaction has gone to completion.

(a) Oxidation: _Fe2+

 

_{aq Fe}3+

 

_{aq + e } 2} _{ E} _{= 0.771 V} Reduction:

 

+

 

2+

 

2 2 PbO s + 4 H aq + 2 e_Pb aq + 2 H O(l) _E _{= +1.455 V} Net: 2+

 

 

+ 3+

 

2+

 

2 2

2 Fe aq + PbO s + 4 H (aq)2 Fe aq + Pb aq + 2 H O(l) E_cell = 0.771 +1.455 V V= +0.684V

In this case, when the reaction has gone to completion, Fe2+ _{= 0.001} M Fe_, 3+ _{= 0.999 M, and Pb}2+ _{= 0.500 M.} Ecell = Eocell 3 2 2 2 2 0.0592 [Fe ] [Pb ] log 2 [Fe ]     = 0.684 V 2 2 0.0592 [0.999] [0.500] log 2 [0.001]  = 0.515 V.

Thus, this reaction will go to completion.

(b) Oxidation: 2 SO₄2

 

aq S O_{2 8}2

 

aq + 2 e E = 2.01V Reduction:

 

+

 

2+

 

2 2 PbO s + 4 H aq + 2 ePb aq + 2 H O(l) E = +1.455 V

 

 

 

 

2 + 2 2+ 4 2 2 8 2

E_cell _{= 2.01+1.455 = 0.56 } V This reaction is not even spontaneous initially_. (c) Oxidation:{Mn2+ 1 10-4 M





+4H₂O(l) MnO₄

 

_{aq +8H}+

 

_{aq +5 e}-_{} 2 ;E}o_{= 1.51 V} Reduction:

 

+

 

2+

 

2 2 {PbO s +4 H aq +2 e_Pb aq +2 H O(l)}×5 _Eo_{=+1.455 V} Net: 2+



-4



 

+

 

2+

 

2 4 2 1 10 M

2Mn  + 5PbO s + 4H 2MnO  aq + 5Pb aq + 2H O(l)

o

cell= 1.51+1.455 = 0.06 V.

E   The standard cell potential indicates that this reaction is not spontaneous when all concentrations are 1 M. Since the concentration of a reactant

Mn2+





is lower than 1.00 M, this reaction is even less spontaneous than the standard cell potential indicates.

52. (M) A positive value of E_cell _{indicates that a reaction should occur.}

(a) _{Oxidation : Sn}2+

 

_{aq Sn}4+

 

_{aq + 2 e} _E _{= 0.154 V}

 

 

2 Reduction : I s + 2 e _2 I aq _E _{= +0.535 V} _________________________________________

 

 

 

 

2+ 4+ 2 Net : Sn aq + I s Sn aq + 2 I aq E_cell = +0.381 V Yes, Sn2+

 

_{aq will reduce I}

2 to I _. (b) _{Oxidation : Sn}2+

 

_{aq Sn}4+

 

_{aq + 2 e} _E _{= 0.154 V}

 

 

2+ Reduction : Fe aq + 2 e_Fe s E = 0.440 V_ ________________________________________

 

 

 

 

2+ 2+ 4+ cell Net : Sn aq + Fe aq _Sn aq + Fe s E = 0.594 V_

No, Sn2+

 

_{aq will not reduce Fe}2+

 

_{aq to Fe(s).}

(c) _{Oxidation : Sn}2+

 

_{aq Sn}4+

 

_{aq + 2 e} _E _{= 0.154 V}

 

 

2+ Reduction : Cu aq + 2 e_Cu s E = +0.337 V ______________________________________________________

 

 

 

 

2+ 2+ 4+ cell Net : Sn aq + Cu aq _Sn aq + Cu s E = +0.183 V

Yes, Sn2+

 

_{aq will reduce Cu}2+

 

_{aq to Cu(s).}

(d) _{Oxidation : Sn}2+

 

_{aq Sn}4+

 

_{aq + 2 e} _E _{= 0.154 V}

 

 

3+ 2+ Reduction : {Fe aq + e Fe aq } 2 E = +0.771 V ___________________________________________________________

 

 

 

 

2+ 3+ 4+ 2+ cell Net : Sn aq + 2 Fe aq Sn aq + 2 Fe aq E = +0.617 V

Yes, Sn2+

 

_{aq will reduce Fe}3+

 

_{aq to Fe}2+

 

_{aq .}

53. (M) As we move down a group, the lower oxidation state is generally favored. Thus, we expect PbCl2, with lead in the +2 oxidation state, to be the product.

54. (M) As we move down a group, the lower oxidation state is generally favored. The higher oxidation state is favored for elements higher in the group. Thus, we expect GeF4, with

lead in the +4 oxidation state, to be the product.

INTEGRATIVE AND ADVANCED EXERCISES

55. (M) What most likely happened is that a deliquescent solid has absorbed water vapor from the air over time and formed a saturated solution. This saturated solution can be used in any circumstance where the solid would be used to prepared an aqueous solution. Of course, the concentration will have to be determined by some means other than weighing, such as a gravimetric analysis or using a standard to react with the substance in a titration. Alternatively, the saturated solution can be heated in a drying oven to drive off the water and the solid may be recovered. Often, unfortunately, a solid with an unknown percent of water results from this treatment.

56. (D) (a) A solution of CO2(aq) has [CO32-] = Ka2[H2CO3] = 5.6×10-11 M. Since the Ksp =

2.8×10-9 for CaCO3, the [Ca2+] needed to form a precipitate from this solution can be

computed. [Ca2+] = -9 sp 2+ -11 2.8 10 = = 50. M [Ca ] 5.6 10 K _ 

This is too high to reach by dissolving CaCl2 in solution. The reason a solid forms is that

the OH- produced by the Ca(OH)2 neutralized some of the HCO3- from the ionization of

CO2(aq), thereby increasing the [CO32-] above a value of 5.6×10-11 M.

(b) The equation for redissolving can be obtained by combining several equations. CaCO3(s) _ Ca2+(aq) + CO32-(aq) Ksp=2.8×10-9

CO32-(aq) + H3O+(aq) _ HCO3-(aq) + H2O(l) 1/Ka2 = 1/(5.6×10-11)

CO2(aq) + 2 H2O(l) _ HCO3-(aq) + H3O+(aq) Ka1 = 4.2×10-7

CaCO3(s) + CO2(aq) + H2O(l)  Ca2+(aq) + 2 HCO3-(aq) K =

K_sp  K_a1 K_a2 K = (2.8 10 -9_{) (4.2  10}-7₎ (5.6 10-11₎ = 2.1 10 -5 ₌ [Ca 2+_][HCO 3 -_]2 [CO₂]

If CaCO3(s) is precipitated from 0.005 M Ca2+(aq) and then redissolved, [Ca2+] =

0.005 M and [HCO3-] = 2×0.005 M = 0.010 M. We use these values in the above

expression to compute the [CO2].

[CO₂] = [Ca

2+_][HCO 3 -_]2

(2.1 10-5₎ = 0.002 M

We repeat the calculation for saturated Ca(OH)2, in which [OH-] = 2 × [OH-], after

Ksp = [Ca2+][OH-]2 = 4×[Ca2+] = 5.5 × 10-6 [Ca2+] = 6 3 5.5 10 = 0.011 M 4   [CO₂] = [Ca 2+_][HCO 3 -_]2 (2.1 10-5₎ = (0.011)(0.022)2 2.1 10-5 = 0.25 M

Thus, to redissolve the CaCO3 requires that the [CO2] = 0.25 M, assuming the solution

is initially saturated with Ca(OH)2(aq).

(c) follow the solution for (b) above.

57. (M) There are two principal reasons why the electrolysis of NaCl(l) is used to produce sodium commercially rather than the electrolysis of NaOH(l). First, NaCl is readily available whereas NaOH is produced from the electrolysis of NaCl(aq). Thus the raw material NaCl is much cheaper than is NaOH. Second, but less important, when sodium is produced from the electrolysis of NaCl(l), a by-product is Cl2(g), whereas O2(g) is a by-product of the production of NaOH(l).

Since O2(g) can be produced more cheaply by the fractional distillation of liquid air, Cl2(g) can be

sold for a higher price than O2(g). Thus, the two reasons for producing sodium from NaCl(l)

rather than NaOH(l) are a much cheaper raw material and a more profitable by-product.

58. (M) The triiodide ion is linear (AX2E3). The Li+ ion has a high charge density and

significant polarizing power. The Li+ ion will polarize the I3− to a significant extent and

presumably assists the decomposition of I3− to I2 and I−.

I I I

59. (M) Use o f G

 values to calculate _Go _{for the reaction and then K. Because} 1 2 2

O K P , the value of K can then be used to calculate

2 O P . Li2O2(s) É Li2O(s)+ 1 2O2(g) G_fo_{[kJ / mol] -419.02 -466.40} Go _{ 466.40  (419.02)  47.38kJmol}-1 Go _{ RT ln K} 8.314JK-1_mol-1_{ 1000K  ln K  47.38  1000Jmol}-1 ln K 47.38  1000Jmol-1 8.314JK-1_mol-1_{ 1000K}  5.70  K  e 5.70 _{ 298} K  P_O2 1 2  P O2  K 2 _{ 298}2 _{ 8.8  10}4 60. (M) Use o 1



o o



hydr. T hydr. hydr.

S H G

     to calculate o hydr. S

S_hydr.o _(Li_) 1 298



522  (481)



 0.14J/Kmol S_hydr.o _(Na_) 1 298



407  (375)



 0.11J/Kmol S_hydr.o _(K_) 1 298



324  (304)



 0.07J/Kmol S_hydr.o _(Rb_) 1 298



299  (281)



 0.06J/Kmol S_hydr.o _(Cs_) 1 298



274  (258)



 0.05J/Kmol All the o hydr. S

 values are negative because, in the process, a gas is being converted into a liquid solution. However, o

hydr. S

 is most negative for Li+. The o hydr. S

 values increase as we move down the group from Li+ to Cs+. As the charge density and polarizing power of the metal cation decreases, so too does the degree of order (or organization) in the hydration sphere. Thus, the entropy change should be most negative for Li+ and least negative for Cs+.

61. (M) (a) Use the Kapustinskii equation in the form U 120, 200z z 1 34.5

r r r r           _  _  _  _, with r+ and r− in picometers, to calculate U in kJ mol−1. For LiO2, z+ = +1, z− = −1, and = 2.

(b) When we apply the Born–Fajans–Haber cycle to the reaction Li(s) + O2(g) → LiO2(s),

we get o f H  = o f,Li(g) H

 + IE(1)Li − 43 + U. Values of H_f,Li(g)o and IE(1)Li are given in

Appendix D and in Table 21.2, respectively. (c) For the reaction 2 LiO2(s) → Li2O(s) +

3 2O2(g), o H  = o ₂ f,Li O(s) H  − 2 o ₂ f,LiO (s) H

 . (d) The decomposition reaction is exothermic (_Ho _{< 0) and, with the assumption that entropy effects are not important, the}

decomposition of LiO2(s) to Li2O(s) and O2(g) is thermodynamically favorable.

62. (M) (a) The mass of MgO(s) that would be produced can be determined with information from the balanced chemical equation for the oxidation of Mg(s).

MgO(s) 2 ) g ( O Mg(s) 2  ₂  MgO g 332 . 0 MgO mol 1 MgO g 30 . 40 Mg mol 2 MgO mol 2 Mg g 305 . 24 Mg mol 1 Mg g 0.200 MgO mass     

(b) If the mass of product formed from the starting Mg differs from the 0.332 g MgO predicted, then the product could be a mixture of magnesium nitride and magnesium oxide. This scenario is not implausible since molecular nitrogen is readily available in the atmosphere. 3 Mg(s)N₂(g)Mg₃N₂(s) mass Mg₃N₂=0.200 g Mg 1 mol Mg 24.305 g Mg 1 mol Mg₃N₂ 3 mol Mg  100.93 g Mg₃N₂ 1 mol Mg₃N₂ =0.277 g Mg3N2

We can use a technique similar to determining the percent abundance of an isotope. Let f = the mass fraction of MgO in the product. The (1.000 – f) is the mass fraction of

Mg3N2 in the product. product mass = mass MgO × f + [mass Mg3N2 × (1.000 – f)]

0.315 g product = 0.332 × f + [0.277 × (1.000 – f)] = 0.277 + f(0.332 – 0.277) 69 . 0 277 . 0 332 . 0 277 . 0 315 . 0 _   

f The product is 69% by mass MgO.

63. (M) Reaction (21.4) is KCl(l)_Na(l)_850_{ }C NaCl(l)_K(g) _{It is practical when the} reactant element is a liquid and the product element is a gas at the same temperature.

(a) Since Li has a higher boiling point (1347°C) than K (773.9°C), a reaction similar to (21.4) is not a feasible way of producing Li metal from LiCl.

(b) On one hand, Cs has a lower boiling point (678.5°C) than K (773.9°C), and thus a reaction similar to (21.4) is a feasible method of producing Cs metal from CsCl. However, the ionization energy of Na is considerably larger than that of Cs, making it difficult to transfer an electron from Na to Cs+.

64. (M) (a) 2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s) H = -852 kJ

(b) 4 Al(s) + 3 MnO2(s)  2 Al2O3(s) + 3 Mn(s) H = -1792 kJ

(c) 2 Al(s) + 3 MgO(s)  Al2O3(s) + 3 Mg(s) H = +129 kJ

65. (M) Li+_{(aq) + e}- _{ Li(s) G = -293.3 kJ = -nFE = -1 mol e}- _{(96,485 C/mol e}-_)E

E = -3.040 V (this value is the same as the one that appears in Table 21.2.)

66. (M) The partial pressure of H2 is 748 mmHg – 21 mmHg = 727 mmHg

amount H₂  P V R T  727 mm Hg (1 atm/760 mmHg)





 0.104 L 0.08206 Latm K-1_mol-1_{ 296 K} =4.10 10 3 _{mol H} 2

The electrode reactions are the following.

) ( H aq) ( OH 2 e 2 O(l) H 2 : Cathode e 2 g) ( Cl aq) ( Cl 2 : Anode   ₂   ₂      ₂ g

Thus 2 mol OH–(aq) are produced per mole of H2(g); 8.20 × 10–3 mol OH–(aq) are produced.

M 0.0328 M 10 28 . 3 soln L 250 . 0 OH mol 10 20 . 8 ] OH [  _  3  _{ } 2 _

Then we compute the ion product and compare its value to the value of Ksp for Mg(OH)2.

2 sp 11 4 2 2 2

sp [Mg ][OH ] (0.220)(0.0328) 2.37 10 1.8 10 K for Mg(OH)

Q           

67. (D) (a) We determine the amount of Ca2+ _{associated with each anion in 10}6 _{g of the water.} mol 44 . 1 HCO mol 2 Ca mol 1 HCO g 02 . 61 HCO mol 1 HCO g 176 ) (HCO Ca amount mol 592 . 0 SO mol 1 Ca mol 1 SO g 06 . 96 SO mol 1 SO g 9 . 56 ) (SO Ca amount 3 2 3 3 3 3 2 2 4 2 2 4 2 4 2 4 2 4 2 –                     

Then we determine the total mass of Ca2+, numerically equal to the ppm Ca2+.

      _{   } 2 _ 2 2 2 2 2 _{81.4gCa 81.4ppmCa} Ca mol 1 Ca g 08 . 40 Ca mol ) 44 . 1 592 . 0 ( Ca mass

(b) The reactions for the removal of HCO3–(aq) begin with the formation of hydroxide ion

resulting from dissolving CaO(s). Hydroxide ion reacts with bicarbonate ion to form carbonate ion, which then combines with calcium ion to form the CaCO3(s) precipitate.

CaO(s) H O(l) Ca2 (aq) 2OH (aq)

2   _    s) ( CaCO aq) ( CO aq) ( Ca aq) ( CO O(l) H aq) ( HCO aq) ( OH 3 2 3 2 2 3 2 3            

For each 1.000 × 106 g of water, we need to remove 176 g HCO3–(aq) with the added CaO(s).

CaO g 7 . 48 CaO mol 1 CaO g 08 . 56 OH mol 2 CaO mol 1 HCO mol 1 OH mol 1 HCO g 02 . 61 HCO mol 1 water g 10 000 . 1 HCO g 176 water g 10 602 CaO mass 3 3 3 6 3 3               

(c) Here we determine the total amount of Ca2+ in 602 kg of water, and that added as CaO.

Ca2+₌ 602 kg 1 106 kg    (0.592+1.44) mol Ca

2+_{+ 48.7 g CaO} 1 mol CaO 56.08 g CaO 1 mol Ca2+ 1 mol CaO    = 2.09 mol Ca 2+

The amount of Ca2+ that has precipitated equals the amount of HCO3– in solution, since each

mole of HCO3– is transformed into 1 mole of CO32–, which reacts with and then precipitates

one mole of Ca2+_{. Thus the amount of Ca}2+ _{that has precipitated is (0.602)(1.44) = 0.867 mol}

Ca2+ as CaCO3(s). Then we can determine the concentration of Ca2+ remaining in solution.

M 10 03 . 2 water L 1 water g 10 water g 10 Ca mol 867 . 0 total Ca mol 2.09 ] Ca [ 3 3 6 2 2 2 _    _{  } 

To consider the Ca2+ “removed”, its concentration should be decreased to 0.1% (0.001) of its initial value, or 2.03 × 10–6 M. We use the Ksp expression for CaCO3 to determine the

needed [CO32–] 2 2 9 6 sp 3 9 2 3 6 [Ca ][CO ] 2.8 10 1.46 10 M 2.8 10 [CO ] 0.0014 M 2.03 10 K                

This carbonate ion concentration can readily be achieved by adding solid Na2CO3.

(d) The amount of CO32– needed is that which ends up in the precipitate plus that needed to

attain the 0.0014 M concentration in the 602 kg = 602 L of water.

n CO₃-2= 602 L 0.00203 mol Ca2+ 1 L water  1 mol CO₃ 2-1 mol Ca2+       +0.0014 mol CO3 2-1 L water       = 2.1 mol CO3

This CO32– comes from the added Na2CO3(s).

3 2 2 3 2 3 2 3 3 2 2 3 3 2 2.2 10 gNa CO CO Na mol 1 CO Na g 99 . 105 CO mol 1 CO Na mol 1 CO mol 0 . 2 CO Na mass – 2      

68. (M) (a) In the cell reaction, three moles of electrons are required to reduce each mole of Al 3+.

mass Al 8.00 h 3600 s 1 h  1.00 105 _C 1 s  1 mol e 96,485 C 1 mol Al 3 mol e  26.98 g Al 1 mol Al  2.68  10 5 _{g Al}

The 38% efficiency is not considered in this calculation. All of the electrons produced must pass through the electrolytic cell; they simply require a higher than optimum voltage to do so, leading to resistance heating (which consumes some of the electrical energy).

(b) total energy 4.5 V  8.00 h 3600 s 1 h  1.00 105 C 1 s  1 J 1 VC 1 kJ 1000 J  1.3  10 7 _kJ

mass coal = 1.3 107 _{kJ electricity} 1 kJ heat 0.35 kJ electricity 1 g C 32.8 kJ  1 g coal 0.85 g C 1 metric ton 106_g mass coal = 1.3 metric tons of coal

69. (M) We begin by rewriting Equation 21.25 for the electrolysis of Al2O3(s) with n = 12 e-.

3 C(s) + 2 Al2O3(s)  4 Al(s) + 3 CO2(g)

G = 4(0 kJ/mol) + 3(-394 kJ) – [3(0 kJ/mol) + 2(-1520 kJ/mol)] = +1858 kJ G = 1.858 × 106 _{J =-nFE = -12 mole e}-_{(96,485 C/mol e}-_)E

E = -1.605 V Note: this is just an estimate because G values are at 298 K, whereas the reaction occurs at a temperature that is much higher than 298 K.

If the oxidation of C(s) to CO2(g) did not occur, then the cell reaction would just be the

reverse of the formation reaction of Al2O3 with n = 6 e-.

E = Eo₌Go

nF =

1.520 106_J

6mole-_{ 96485C/mole}-  -2.626 V 

70. (M) It would not be unreasonable to predict that Raoult’s law holds under these circumstances. This is predicated on the assumption, of course, that Pb(NO3)2 is completely

P_water  x_water P_water x_water  Pwater

P_water  97%  0.97

Thus, there are 97 mol H2O in every 100 mol solution. The remaining 3 moles are 1 mol

Pb2+ and 2 mol NO3–. Thus there is 1 mol Pb(NO3)2 for every 97 mol H2O. Compute the

mass of Pb(NO3)2 in 100 g H2O. mass_Pb(NO 3)2= 100.0 g H2O 1 mol H₂O 18.02 g H₂O 1 mol Pb(NO₃)₂ 97 mol H₂O  331.2 g Pb(NO₃)₂

1 mol Pb(NO₃)₂ = 19 g Pb(NO3)2

If we did not assume complete ionization of Pb(NO3)2, we would obtain 3 moles of

un-ionized Pb(NO3)2 in solution for every 97 moles of H2O. Then there would be 57 g

Pb(NO3)2 dissolved in 100 g H2O. A handbook gives the solubility as 56 g Pb(NO3)2/100 g

H2O, indicating only partial dissociation or, more probably, extensive re-association into ion

pairs, triplets, quadruplets, etc.

71. (M) Considerable energy is required to produce the Pb4+ cation—four ionization steps, one for the removal of each electron. It therefore needs quite a large lattice energy to compensate for its energy of production. Both Br– and I– are large anions, and therefore the Pb4+—Br– and Pb4+—I– interionic distances are long. But lattice energy depends on the charge of the cation (which is quite large) multiplied by the charge of the anion (which is reasonably small) divided by the square of the interionic distance (long, as we have said). Thus, we predict a small lattice energy that is insufficient to stabilize the Pb4+ _{cation. We would}

predict, however, that PbF4 and PbCl4 (both with small anions) and PbO2 and PbS2 (both

with small, highly charged anions) would be stable compounds. In addition, notice that

Eo {Pb4+|Pb2+} = 1.5 V (from Table 21-6) is sufficient to oxidize Br– to Br2 and I– to I2, since

Eo {Br2|Br–} = +1.065 V and Eo {I2|I–} = +0.535 V. Thus, even if PbI4 or PbBr4 could be

prepared they would be thermodynamically unstable.

72. (M) Na2B4O7.10 H2O(s) + 6 CaF2(s) + 8 H2SO4(aq)  4 BF3(aq) + 6 CaSO4(aq) + 17 H2O(l) + 2

NaHSO4(aq)

73. (D) (a) 1.00 M NH4Cl is a somewhat acidic solution due to hydrolysis of NH4+(aq); MgCO3

should be most soluble in this solution. The remaining two solutions are buffer solutions, and the 0.100 M NH3–1.00 M NH4Cl buffer is more acidic of the two. In this solution,

MgCO3 has intermediate solubility. MgCO3 is least soluble in the most alkaline solution,

namely, 1.00 M NH3–1.00 M NH4Cl. 2 2 8 3 3 sp 5 4 3 2 2 11 3 3 3 2 2

MgCO (s) Mg (aq) CO (aq) 3.5 10 NH (aq) OH (aq) NH (aq) H O(l) 1/ 1/1.8 10 CO (aq) H O (aq) HCO (aq) H O(l) 1/ 1/ 4.7 10

b K K K                            14 2 3 2 3 4 3 3

2 H O(l) H O (aq) OH (aq) 1.0 10

MgCO (s) NH (aq) Mg HCO ( ) NH (aq)

w K aq                

sp w 8 14 7 5 11 b 2 K K _{3.5 10 1.0 10} K 4.1 10 K K 1.8 10 4.7 10       _{  }         In 1.00 M NH4Cl 2 3 4 3 3

Reaction: MgCO (s) NH (aq) Mg (aq) HCO (aq) NH (aq)

Initial: 1.00 M 0 M 0 M 0 M Changes: x M x M x M x               2 3 3 3 4 3 7 3 3 M Equil: (1.00 )M M M M [Mg ][HCO ][NH ] [NH ] 1.00 1.00

4.1 10 7.4 10 M molar solubility of MgCO in1.0

x x x x x x x x K x x  _                 0 M NH Cl₄

(x 1.00 M, thus the approximation was valid) (b) In 1.00 M NH3–1.00 M NH4Cl

2

3 4 3 3

Reaction: MgCO (s) NH (aq) Mg (aq) HCO (aq) NH (aq)

Initial: 1.00 M 0 M 0 M 1.00 M Changes: M M M M Equil: (1.00 x x x x                 2 2 3 3 4 )M M M (1.00 )M [Mg ][HCO ][NH ] (1.00 ) 1.00 [NH ] 1.00 1.00 x x x x x x x x K x              7 4 3 3 4

4.1 10 6.4 10 M molar solubility of MgCO in 1.00 M NH -1.00 M NH Cl

x_{ }  _{ }  _

(x 1.00 M, thus the approximation was valid)

(c) In 0.100 M NH3–1.00 M NH4Cl

2

3 4 3 3

Reaction: MgCO (s) NH (aq) Mg (aq) HCO (aq) NH (aq) Initial: 1.00 M 0 M 0 M 0.100 M          Changes: M M M M Equil: (1.00 )M M M (0.100 )M x x x x x x x x          2 2 3 3 4 7 3 3 3 4 [Mg ][HCO ][NH ] (0.100 ) 0.100 1.00 1.00 [NH ] 4.1 10 2.0 10 M solubility of MgCO in 0.100 M NH -1.00 M NH Cl 0.100 ( is 2 % of 0.100 M, so x x x x K x x x                  