Descriptive Inorganic, Coordination, and Solid

State Chemistry

2nd Edition

Glen E. Rodgers

Student Manual

1

Chapter 1 Objectives The student should be able to

• explain what the realm of inorganic chemistry encompasses • appreciate when chemistry first became recognized as a separate academic discipline • appreciate when and how inorganic chemistry became established as a subdiscipline of

chemistry • appreciate the role and some of the accomplishments of inorganic chemists in the latter half

of the nineteenth century • explain the role of the quantum revolution in changing the nature of inorganic chemistry • appreciate some of the major accomplishments of inorganic chemists in the first half of the

twentieth century • appreciate some of the major accomplishments of inorganic chemists in the second half of

the twentieth century

2

Chapter 2 An Introduction to Coordination Chemistry

The sections and subsections in this chapter are listed below. 2.1 The Historical Perspective 2.2 The History of Coordination Compounds Early Compounds The Blomstrand-Jørgensen Chain Theory The Werner Coordination Theory 2.3 The Modern View of Coordination Compounds 2.4 An Introduction to the Nomenclature of Coordination Compounds

Chapter 2 Objectives

You should be able to

• define some important terms used in coordination chemistry • give a few examples of coordination compounds encountered in earlier courses • put coordination chemistry into the historical context of the conceptual development of

atomic structure, the periodic table, and chemical bonding • relate how the formulas and properties of early coordination compounds were but

incompletely rationalized by the Blomstrand-Jørgensen chain theory • explain how Werner's coordination theory, with its concept of primary and secondary

valences, more completely rationalized the properties of early coordination compounds • draw structural formulas for coordination compounds using both the Blomstrand-

Jørgensen chain theory and the Werner coordination theory • explain how Werner established that the secondary valence of cobalt(III) is directed to the

corners of an octahedron • work with a variety of coordination compounds involving monodentate, multidentate,

bridging, and ambidentate ligands • name coordination compounds involving a variety of metals, ligands, and counterions

Solutions to Odd-Numbered Problems 2.1. Dalton's atomic theory had three essential components. First, he proposed that all elements were composed of tiny, indivisible particles called atoms. All the atoms of a given element were the same in every way while those of different elements were different in properties such as mass and size. Second, compounds were composed of atoms of more than one element in small, whole number ratios, 1:1, 2:1, etc. Third, a chemical reaction involved the shuffling of atoms from one compound to another. The idea of hooks embedded into atoms could have been a way to account for the number of other atoms with which a given atom could combine. That is, an atom of oxygen, which often combines with two other atoms of a second element in such compounds as water, H2O, might have two embedded hooks while hydrogen atoms might have only one. 2.3. In carbon monoxide, the valence of carbon is one while in carbon dioxide it is two. Therefore, these two compounds are not consistent with the concept of fixed valence. In practice, it is found that a given element may have several characteristic valences. 2.5. The quantum mechanical atom pictures an electron in an atom to occupy only certain, allowed energy levels. When this atom is excited, that is, when energy is put into it, the electron can be moved from one allowed level to a higher one. Once the atom moves away from the energy source, the electron will return to its lower "ground state" energy level. The energy lost, E, according to Planck's equation, E = hv, would correspond to a specific frequency or wavelength of light emitted. As various electrons are excited and then return to their lower levels, a set of energies (and therefore a set of wavelengths), characteristic of the particular atom, would be emitted. This characteristic set of wavelengths is known as an emission spectrum. 2.7. The disappointment came when the Blomstrand-Jørgensen chain theory could not predict the conductivity or the number of ions that would exist in aqueous solution or rationalize the correct number of isomers of certain compounds. Specifically, IrCl3(NH3)3 was predicted by the chain theory to have two precipitatable chlorides and a nonzero conductivity. In fact this compound did not form ions in solution and therefore its aqueous solution was a nonconductor and had no precipitatable chlorides. Furthermore, it was not clear why this compound could not have a number of different isomers rather than only the two that could be prepared. 2.9. (a) CrCl3·6H2O + 3AgNO3(aq) → 3AgCl(s) + Cr(NO3)3(aq) + 6H2O (b) The conductivity should decrease due to fewer and lower charged ions in solutions. For example, compound (1) would produce a +3 cation and three -1 anions while compound (3) would produce a +1 cation and one -1 anion. H2O (c) i) Blomstrand-Jørgensen ii) Werner

Cr

OH2Cl

Cl

OH2H2O

H2O

Cl

H2O H2O-Cl / Co-H2O-H2O-H2O-H2O-Cl \ Cl

3

iii) Modern [Cr(H2O)5Cl]Cl2 2.11. One would suspect that Jørgensen's idea of putting the more reactive chlorides at the end of a chain carried over to Werner's idea of putting them outside the coordination sphere. 2.13. There would be three possible isomers.

Cl Cl ClCl

H3N

NH3

NH3

Cl

NH3 NH3

NH3NH3

H3N

NH3

Cl

NH3NH3

NH3

2.15. tricarbonyltrichlorochromium(III)

Pt

Cl

Cl

ClCO

OC

OC

1,2,3

Pt

Cl

CO

ClCl

OC

OC

1,2,4

Pt

Cl

CO

ClCO

OC

Cl

1,3,5

2.17. (a) The chloride ligands satisfy both primary and secondary valences while the waters satisfy only secondary.

Cr

Cl

OH2

ClH2O

H2O

Cl

(b) Three trigonal prismatic geometrical isomers

4

Cl Cl ClCl

H2O

OH2

OH2

Cl

Cl

OH2OH2

H2O

ClOH2

ClCl

OH2OH2

Two octahedral isomers

H2O OH2

Cl Cl

OH2

Cl Cl

H2OOH2

H2OCl

Cr Cr

Cl 2.19. Only ammonia and nitrite are possible ligands as they are the only ones to have lone pairs that can be donated to a metal ion. Formula No. of ionic No. of geometric nitrites isomers [Co(NH3)6](NO2)3 3 1 [Co(NH3)5(NO2)](NO2)2 2 1 [Co(NH3)4(NO2)2]NO2 1 2 [Co(NH3)3(NO2)3] 0 2 K[Co(NH3)2(NO2)4] 0 2 K2[Co(NH3)(NO2)5] 0 1 K3[Co(NO2)6] 0 1 2.21 (a) Formulas [Pt(NH3)4](SCN)2 (b), (i) tetraammineplatinum(II) thiocyanage [Pt(NH3)3SCN]SCN (contains cation of highest charge) [Pt(NH3)2(SCN)2] NH4[Pt(NH3)(SCN)3]

5

(NH4)2[Pt(SCN)4] (b), (ii) ammonium tetrathiocyanatoplatinate(II) (contains anion of highest charge) (c)

Pt

SCN

NH3

SCN

H3N

neutral compound: two isomers, cis and trans 2.23 (a)

NH2

NH2

NH2

NH2

CoH2C

H2CCH2

CH2

NH2

NH2

NH2Co

H2C

H2C

H2N

CH2

CH2Cl

Cl

Cl

Cl

+ +

(b) There would have been four structural isomers.

NH2

NH2

Cl

NH2

Cl

ClCl

NH2H2C

H2CCH2

CH2

NH2

NH2

H2C

H2CH2N

CH2

CH2

NH2

H2N

NH2Cl

Cl

H2NCH2

CH2

CH2

CH2

Cl

NH2

NH2Cl

H2N

H2NCH2

CH2

H2C

CH2

NH2

6

(c) There are two possible isomers.

Co

NH2

Cl

ClNH2

H2N

H2NCo

Cl

ClNH2H2N

H2N NH2CH2

CH2

H2C

H2CH2C

H2CH2C

H2C

2.25. (a) hexaammineplatinum(IV) chloride (b) (acetylacetonato)tetrakis(triphenylphosphine)nickel(II) nitrate (c) ammonium trioxalatoferrate(II) (d) tricarbonyldinitrosyltungsten(0) 2.27 (a) tris(ethylenediamine)iron(III) hexachloroiridate(III) (b) amine(methylamine)silver(I) dichlorodinitritoplatinate(II) (c) dichlorobis(ethylenediamine)vanadium(III) hexacyanoferrate(II) 2.29. (a) bis(ethylenediamine)cobalt(III)-μ-thiocyanato-μ-isothiocyanato- bis(acetylacetonato)chromium(III) nitrate (b) triaquacopper(I)-μ-diacetatotriaquacopper(I) (c) diaquatris(thiocyanato)chromium(III)-μ-hydroxopentaamminecobalt(III) sulfate 2.31. (a) [Ag(CH3NH2)2]C2H3O2 (b) Ba3[CoBr2(C2O4)2]2 (c) NiCO[P(C6H5)3]3 2.33 (a) [Co(NH3)6][CuCl5] (b) [Pt(C5H5N)4][PtCl4] (c) [Pd(NH3)2(PPh3)2][Au(C2O4)2]2

2.35.

(a)

[(NH3)5Cr

O

Cr(NH3)5]Cl5 [(NH3)2(en)Cr

O

O O

Co(NH3)4]Br6

(b)

OH

7

Chapter 3 Structures of Coordination Compounds The sections and subsections of this chapter include 3.1 Stereoisomers 3.2 Octahedral Coordination Spheres Compounds with Monodentate Ligands Compounds with Chelating Ligands 3.3 Square Planar Coordination Spheres 3.4 Tetrahedral Coordination Spheres 3.5 Other Coordination Spheres 3.6 Structural Isomers

Chapter 3 Objectives You should be able to

• distinguish among the various types of stereoisomers and structural isomers • define, describe, test for, and be familiar with the nomenclature of chirality • determine the number and types of and name stereoisomers commonly encountered in octahedral

coordination compounds • describe the general approach to resolving isomers of ionic coordination compounds • demonstrate how Werner forever laid to rest the idea that chirality was a property associated only

with carbon • determine the number and types of and name stereoisomers commonly encountered in square

planar and tetrahedral coordination compounds • appreciate and give examples of coordination compounds with coordination numbers other than

four or six • define and give an example of fluxional five-coordinate compounds • recognize and predict the incidence of coordination and ionization structural isomers • describe, name, and predict the incidence of linkage isomers that occur with the common

ambidentate ligands

8

Solutions to Odd-Numbered Problems 3.1. Ethanol and dimethylether are isomers because they have the same number and types of atoms but different properties. Furthermore, they are structural isomers because they have different numbers and types of chemical bonds. For example, ethanol has one C-O bond while dimethylether has two. 3.3. The two chain theory formulations would be considered to be stereoisomers because they have the same numbers and types of chemical bonds (three each of Co-N, NH3-NH3, and NH3-Cl) but differ in the spatial arrangements of those bonds. 3.5. A light beam of a given wavelength can be polarized such that only one plane of the electric or magnetic field is allowed. A molecule is optically active if it is capable of rotating that plane to the right or to the left when the light beam is passed through the substance. *3.7. For each case (MA4BB2 and MA3B3B ), these are geometric isomers because they have the same numbers and types of chemical bonds but the different spatial arrangements of these bonds result in different geometries. In the MA4BB2 set the 1,6 case does not contain an internal mirror plane and is chiral. The 1,5 isomer is actually the nonsuperimposable mirror image of the 1,6 isomer. Similarly, in the MA3B3B the 1,2,4 isomer is chiral. The 1,2,5 isomer is its nonsuperimposable mirror image. 3.9. These two mirror images are nonsuperimposable. No matter how one is rotated in space, it cannot be made equivalent to the other. Therefore, this molecule is chiral. The same conclusion is arrived at by considering the fact that this molecule does not contain any internal mirror planes. Cl

C

I

FBr

Cl

C

BrI

F

3.11. (a) If the mirror image of a molecule cannot be rotated in space so as to be equivalent to the original molecule, it is said to have a nonsuperimposable mirror image. Such a molecule is chiral. (b) If a molecule does not possess an internal mirror plane (a plane that passes through the molecule such that every atom in the molecule can be reflected through the plane into another equivalent atom) it is chiral. [There are a few exceptions to this rule but they are beyond the scope of this text.] 3.13. Only AlClBrI has an internal mirror plane, the one that the entire molecule sits in. Therefore, this molecule is not chiral whereas the corresponding phosphorus compound is. 3.15. The structural formulas are as follows: (For (b) and (c), the structure of only the cations are shown.)

9

(a)

Cl

OC

PPh3

PPh3

IrH3N

H3NCr

HN3

HN3

(b)

Cl

NO2

CoONO py

py ONO

NH3

NH3

(c)

+ +

3.17. Neither of these compounds is chiral because each contains at least one internal mirror plane.

Co

Cl

Cl

Cl

CC

O

OC

OC

O

O

O

OO O

HCC

CH3C

H3CO

AsPh3

Cl

AsPh3

Re

(a)

(b)

K3

3.19 (a) [Cr(C2O4)3] chiral due to propeller shape with no internal mirror planes

10 (b) Not chiral due to several internal mirror planes including the plane of the molecule.

Pt

Cl

ClNH2

NH2

CH2

CH2 *3.21. Prior to Werner's synthesis and resolution of coordination compounds containing chelating agents, the evidence for his coordination theory had been "negative." That is to say, his theory often predicted only two isomers for a given MA4BB2 or MA3B3B complex (where A and B are monodentate ligands) if the geometry of the coordination sphere was octahedral. Planar hexagonal and trigonal prismatic geometries, on the other hand, yielded three isomers in each case. His and other research groups indeed could synthesize only two isomers of these compounds. But what if the correct geometry was hexagonal or trigonal prismatic and the third isomer was just particularly difficult to synthesize? Werner had only "negative" evidence for his octahedral coordination theory. It was the absence of a given isomer that Werner claimed supported his ideas. However, octahedral coordination spheres for a complex of general formula M(AA)2BB

.25.

2 (where AA = a chelating bidentate ligand) should have one chiral and one nonchiral form. Werner and his group were able to confirm the existence of this number of isomers and therefore provide "positive" evidence for his theory and the octahedral configuration of his secondary valence. *3.23. Structures (iii), (vi) and (vii) are chiral as they lack an internal mirror plane.

aa

a

a

a

a

a

a

aa

aa

aa

A

A

A

AA

A

A

AA

A

AA

A

A

B

B

B

B

B

B

B

B

B

B

B

B

B B

(i) (ii) (iii)

(iv) (v) (vi) (vii)

3

11

12

NH2

NH2

NH2

NH2

CoH2C

H2CCH2

CH2

NO2

NO2

trans

NH2

NH2

NH2Co

H2C

H2C

H2N

NO2

R/S-cis

CH2

CH2

NO2

1+

H2C

1+

sym

m

3.27. No, the structure would then have a plane of symmetry and would no longer be chiral. The plane of metry would contain the C-C bond of the methyl-substituted ethylenediamine, pass through the

platinum cation, and then bisect the C-C bond of the phenyl (C6H5) - substituted ethylenediamine.

3.29. (a) potassium chloro(nitrilotriacetato)thiocyanatocobaltate(III)

(b) Since the four binding cites of the NTA must be adjacent to each other, the two monodentate ligands (SCN- and Cl-) also must always be cis to each other. However, either the chloride or the thiocyanate could be trans to the nitrogen of the NTA as shown below. In either case, the complex anion contains an internal

irror plane and is therefore not chiral.

Co

ON

O

C

O

CH2CH2

CO

O

C

CH2

Cl

O

SCN

O

Co

O

Cl

N

O

C

O

CH2CH2

CO

O

C

CH2

2 K2

SCN

*3.31. Both of the square planar structures possess a plane of symmetry (the plane of the molecule), but the

K

one tetrahedral structure does not. Therefore the square planar versions are nonchiral while the tetrahedral is chiral.

3.33. Since the nitrate ligand is capable only of O-bonding to a metal center, both the bis(ethylene-iamine)dinitratocobalt(III) chloride and the bis(ethylenediamine)dinitritocobalt(III) chloride possess four o3+-N and two Co3+-O interactions. Therefore it follows that both of these compounds should be the same olor while the bis(ethylenediamine)dinitrocobalt(III) chloride, in which the complex cation contains six

3+

irs available to bind with a metal center while the nitrogen has only one. Assuming the four electron airs a - omplexes should possess a nonlinear M-S-C interaction. Given the linear arrangement of the triple bond nd the one lone pair about the nitrogen atom, on the other hand (or is it on the other tooth?) N-bonded iocyanate complexes should possess a linear M-N-C interaction. These two cases are shown in structures ii) and (iv), respectively.

3.37. S- bonded

NH2

O

NH2

O

NiCH2

CH2C

C

O O

NH2

O

Ni CH2C

O

O

NH2

C

H2C

O

Ni

H2NCH2

C

OO

H2N

H2CC

O

O

*dCcCo -N interactions, should be a different color. *3.35. The Lewis and VSEPR structures of thiocyanate are shown in (i) and (ii) below. The sulfur has three lone pap round the sulfur (three lone pairs and one bonding pair) are tetrahedrally dispersed, S-bonded SCNcath(i

13

S C N_

(i)

S C N

(ii)

M M<109ο

(iii) (iv)

180o

S C N S C N M

*

thiocyanate will take up more space around a metal atom or ion because the -SCN is able to rotate about the M-S bond and sweep out a fairly large cone-shaped volume. N-bonded thiocyanate, on the other hand (tooth?) rotates about the M-N bond and sweeps out a much smaller, needle-like cylindrical volume. In the [Co(NH3)5NCS]2+ cation, the five ammonia ligands sweep out cone-like volumes and take up a fair amounof volume. They do not leave much volume available for the thiocyanate so it is more stable in the isothiocyanate "needle" form that does not require as much volume. In the [Co(CN)

t

tle volume and so the S-bonded SCN , that takes up more space, is favored.

.39. [Cu(NH3)3Br][Pt(NH3)Br3], triamminebromocopper(II) amminetribromoplatinate(II) [Pt(NH3)3Br][Cu(NH3)Br3], triamminebromoplatinum(II) amminetribromocuprate(II) [Pt(NH3)4][CuBr4], tetraammineplatinum(II) tetrabromocuprate(II)

3.41. Since all four positions in a tetrahedral are adjacent to each other, there would only one possible eometric isomer for [M(dmen)2]. This isomer would chiral because it would lack an internal mirror plane.

.43. (a) [Pt(NH3)4Cl2]Br2: An ionization isomer would be [Pt(NH3)4Br2]Cl2, dibromotetraammine- platinum(IV) chloride. (b) [Cu(NH3)4][PtCl4]: A coordination isomer would be [Cu(NH3)3Cl][Pt(NH3)Cl3], triamminechlorocopper(II) amminetrichloroplatinate(II).

3.45. At least one of each type of isomer except optical is possible as shown below.

[Pd(NH3)4][Pd(NO2)4] tetraamminepalladium(II) tetranitropalladate(II) ↑ coordination isomer

optical isomer ↑ tetrathiocyanat0cobaltate(III) neaquacobalt(III)

ocyanatocobaltate(III) [Co(acac)2][Co(NH3)(H2O)(NCS)4] bis(acetylacetonato)cobalt(III)

trans-ammineaquatetraisothiocyanatocobaltate(III))

5SCN]3- anion, the linearCN- ligands take up very lit -

3 [Cu(NH3)4][PtBr4], tetraamminecopper(II) tetrabromoplatinate(II) *g 3 * [Pd(NH3)2(NO2)2] cis-diamminedinitropalladium(II) ↓ geometric isomer [Pd(NH3)2(NO2)2] trans-diamminedinitropalladium(II) ↓ linkage isomer [Pd(NH3)2(ONO)2] trans-diamminedinitritoopalladium(II) *3.47. [Co(acac)2(NH3)(H2O)][Co(SCN)4] S-cis-bis(acetylacetonato)ammineaquacobalt(III) [Co(acac)2(NH3)(H2O)][Co(SCN)4] R-cis-bis(acetylacetonato)ammi coordination isomer ↓ tetrathiocyanatocobaltate(III) [Co(acac)2][Co(NH3)(H2O)(SCN)4] bis(acetylacetonato)cobalt(III) geometric isomer ↓ cis-ammineaquatetrathi linkage isomer ↓ trans-ammineaquatetrathiocyanatocobaltate(III) [Co(acac)2][Co(NH3)(H2O)(NCS)4] bis(acetylacetonato)cobalt(III)

14

15

Chapter 4 Bonding Theories for Coordination Compounds

The sections and subsections in this chapter are listed below. 4.1 Early Bonding Theories The Lewis Acid-Base Definition Crystal Field, Valence Bond, and Molecular Orbital Theories 4.2 Crystal Field Theory Shapes of 3d Orbitals Octahedral Fields Tetragonally-Distorted Octahedral and Square Planar Fields Tetrahedral Fields 4.3 Consequences and Applications of Crystal Field Splitting Crystal Field Splitting Energies versus Pairing Energies Crystal Field Stabilization Energies Factors Affecting the Magnitude of the Crystal Field Splitting Energies Magnetic Properties Absorption Spectroscopy and the Colors of Coordination Compounds

Chapter 4 Objectives You should be able to

• explain how Lewis acid-base theory applies to coordination compounds • define and use the effective atomic number rule for coordination compounds • briefly outline how the valence bond, molecular orbital, and crystal field theories generally are

applied to coordination compounds • qualitatively sketch out the shapes of the five independent 3d orbitals • explain how the two dependent 3dz²-y² and 3dz²-x² orbitals are related to the 3dz² orbital • explain in detail how the five independent 3d orbitals split in an octahedral field • explain in detail the changes in the relative energies of the 3d orbitals when the z-axis ligands of

an octahedral field are gradually withdrawn to produce a tetragonally-elongated and ultimately a square planar crystal field

• explain in detail how the five independent 3d orbitals split in a tetrahedral field • describe the strong- and weak-field, low- and high-spin cases as applied to coordination

compounds • calculate the crystal field stabilization energy (CFSE) in terms of the crystal field splitting (Δ) and

pairing (P) energies for any given dn case in an octahedral, tetrahedral, or square planar crystal field

• explain how factors such as the type of field, the size and charge of the metal ion, and the size and charge of the ligand should affect the magnitude of the crystal field splitting energy assuming a completely ionic M-L interaction

• rationalize the degree to which the spectrochemical series can and cannot be explained by the crystal field theory

• explain how admitting a certain degree of covalent character to the M-L interaction can be used to better rationalize the spectrochemical series

16

• sketch out how molar susceptibility is measured and how its value can be related to the magnetic moment, μ, and to the number of unpaired electrons in a coordination compound

• rationalize why so many coordination compounds are highly colored and how these colors can, in some cases, be simply related to the size of crystal field splitting energy

Solutions to Odd-Numbered Problems 4.1 Hydrogen ions: (1) Arrhenius acid -- treats acids as substances that produce H+ in solution (2) Brønsted-Lowry acid -- capable of transferring a proton (H+) (3) Lewis acid -- H+ is an electron pair acceptor Hydroxide ions: (1) Arrhenius base -- treats bases as substances that produce OH- in solution (2) Brønsted-Lowry base -- capable of accepting a proton (H+) (3) Lewis base -- OH- is an electron pair donor 4.3 The oxygen molecule has a number of lone pairs of electrons associated with it. One of these lone pairs can be donated to the central iron ion (Fe2+) of the hemoglobin. The O2, then, has served as an electron pair donor and is a Lewis base. The Fe2+ is an electron pair acceptor and is a Lewis acid. 4.5. (a) Lewis base = hydroxide ion; Lewis acid = Zn2+

(b) Lewis base = water; Lewis acid = SO2 (Sulfurous acid is often better described as SO2·H2O but also as OS(OH)2, a form that would necessitate a rearrangement after the initial Lewis acid- base reaction.) (c) Lewis bases = two ammonias; Lewis acid = Ag+

4.7. Metal electrons Ligand electrons Total Electrons Result (a) Cu2+ 27 4 NH3 8 35 does not follow EAN (b) Ag+ 46 2 NH3 4 50 does not follow (c) Fe2+ 24 6 CN- 12 36 follows EAN (d) Mo0 42 6 CO 12 54 follows EAN (e) Fe3+ 23 3C2O4

2- 12 35 does not follow EAN 4.9. Coulomb's law says that the potential energy (PE) is equal to Q1Q2/r. If r is infinite, then the PE is zero. But we also recognize that the two electrons have the same charge and, therefore, the net energy will be positive for finite values of r. When the two electrons are placed side by side, the distance r between the two particles decreases and the value of the potential energy (= Q1Q2/r) becomes more and more positive. 4.11. (nodes = dashed lines)

17

*4.13.

4.15. The six independent orbitals are classified as dependent because any one of them can be written as a linear combination of the others. If one writes a linear combination of the 3dz²- y²_and 3dz²- x²_orbitals to make the 3d z²_orbital, the resulting five orbitals are no longer dependent but independent. *4.17. (a) 3dx2-y2 z2-x2 3d3d z2-x2

barycenter ΔO

1/2ΔO

1/2ΔO

3dxy 3dxz 3dyz

E

(b) The 3dxz orbital points in between the x and z axes and therefore avoids, to some extent, the ligands located on these axes. On the other hand the 3dz²- x² orbital points directly along the x and z axes and therefore points directly at the ligands. Given these relative positions, it is not surprising that a d electron prefers not to occupy the 3dz²- x²_orbital. That is, the 3d z²- x²_orbital is of higher energy. *4.19. Given that the 3dz²_ orbital is a linear combination of the 3dz²-y²_and 3dz²-x²_orbitals which are identical in shape and relative orientation (that is, they all point directly at the ligands of the octahedral field) to the 3dx²-y²_orbital, all of these orbitals (including the 3dz² and 3dx² - y²) should be degenerate in such a crystal field. *4.21. (a) The 3dz

2_ orbital will point directly at the two ligands along the z axis of the linear field. The

3dyz and 3dxz ligands point in between the z axis and either the y or the x axis and therefore are destabilized somewhat. The 3dxy and 3dx² - y²_orbitals, being fully in the xy plane do not point at the z axis ligands at all are the most stabilized.

18

3dz2

3dyz 3dxzx

3dxy 3dx2-y2

0.74Δ -x

0.91Δ

barycenter

(b) To calculate the position of the barycenter, one assigns the position roughly between the first and second levels and uses the relationship that

Estabilization = Edestabilization

2(0.74Δ - x) = 2x + 1(0.91Δ + x)

1.48Δ - 2x = 3x + 0.91Δ

0.57Δ = 5x

x = 0.11Δ

19

4.23. By analogy to other fields, the crystal field splitting energy of the 3p

int at any ligand) would be stabilized by 2/3 Δsp.

.25. (a) Since a cubic field is two tetrahedral fields superimposed

hedral field.

f the cube and therefore is closer to the ligands than the 3dx² - y²_

x² - y² (and the similar 3d z²_) orbital is stabilized.

3px 3py

1/3Δsp

2/3Δsp

3pz

orbitals in a square planar field can be labeled as Δsp. In order for the energy of stabilization to be equal to the energy of destabilization, the 3px and 3py orbitals (which point directly at the ligands in the xy plane) would have to be destabilized by 1/3 Δsp while the 3pz orbital (which does not po 4 3dxy 3dupon each other (as shown in Figure 4.7), its crystal field splitting diagram is qualitatively the same as that for a tetra(b) As shown in Figure 4.8, the 3dxy orbital points towards four of the edges o

yz 3dxz

Δ c

3dz2 3dx2-y2orbital that points towards the centers of four of the faces of the cube. It follows that the 3dxy and similar orbitals are destabilized while the 3d

*4.27. (a) Both the 3dx²-y² and 3dz² orbitals point directly at ligands and therefore are destabilized. (The dx²-y² points at four ligands and is more destabilized than the 3dz² which points at only one.) The 3dxy, dyz, and 3dxz orbitals point in between the ligands and are stabilized. (The 3dyz and 3dxz orbitals have

)

2[(0.37Δ + (0.17Δ - x)] + (0.17Δ - x) = x + (0.83Δ + x)

1.25Δ - 3x = 2x + 0.83Δ

x = 0.084Δ

Therefore, the barycenter sits abo tween the 3dz_ and 3dxy orbitals.

*4.29.

.31. Tetrahedral complexes are alm is about half the magnitude of Δo nd almost always less than the pairing energy, P.

33components along the z-axis, where there is only one ligand, and therefore are more stable than the 3dxyorbital.)

3dx2-y2

0.83

0.17

0.37

3dyz3dxz

3dz2

x

0.17Δ − x

3dxy

0.83 + x

0.37 + (0.17 - x)Δ

Δ

Δ

Δ

Δ

Δ

(b

Estabilization = Edestabilization

0.42Δ = 5x

ut midway be

2 Δ L

1/3Δ LPyPx

Δ L

Pz/3

4 ost always high spin because Δta

20

*4.33. Only d8 would have high- and low-spin ossibilities. d7 would involve the sum of the p-most energy split and the middle-level split

.35. Cr3+ 3; Co2+ 7; Pd4+

.37. d3 CFSE = 3(2/5 Δo) = 6/5 Δo

2/5 Δ =

d6 low-spin: CFSE = 12/5 Δo - 2P d is so high in the spectrochemical series)

(b) [Ru(NH3) ] Ru d low-spin: CFSE = 2 Δo - 2P on s nd Ru3+ is large and

P

CFSE = 6(0.40Δo) - 2(0.60Δo) CFSE = 4(0.428Δo + 0.086Δo) + 2(0.428Δo) -1(0.228Δo) - 1(0.228Δo + Δo)

1.2 Δo = 1.456 Δo

ptoand so does not meet the condition set in the statement of the problem. d8 high-spin d8 low-spin 4 6; Pt2+ 8; Cu2+ 9 4 d8 CFSE = 6( o) - 2(3/5 Δo) 6/5 Δo 4.39. (a) [Fe(CN)6]4- Fe2+

(low-spin chosen because the cyanide ligan3+ 3+ 5 6

(low-spin because amm ia ligand i moderately high in the series a highly charged leading to a large crystal field splitting energy) (c) [Co(NH3)6]3+ Co3+ d6 low-spin: CFSE = 12/5 Δo - 2 (low-spin because ammonia ligand is moderately high in the series and Co3+ is highly charged leading to a moderately high crystal field splitting energy) *4.41.

21

0.60ΔO

0.40ΔO

al

ΔO

0.23

0.43

0.086ΔO

d8 square planar

ΔO

ΔO

8 octahed dr =

Therefore, the square planar configuration is of lower energy and is favored. The stronger the field

o), the m s.

FSE = 4(0.63Δ) - 2(0.11Δ) - 1(0.91Δ + 0.11Δ)

= 1.28Δ

.45. In a tetragonal compression, the ligands located along the z axis movement results in more steric hindrance about the metal and causes the ligands located in the xy plane move away from the metal center.

l series. This set of conditions would probably result in a fairly rge crystal field splitting energy and a strong-field low-spin state.

) The larger Rh allows for the fluoride ligands to approach more closely to the d orbitals without

draws the ligands in more closely and thereby splits the d orbitals

sphines are able to accept π-electron density from filled metal d orbitals and ereby increase both the partial positive and partial negative charges on the metal and ligands,

ts of ds

l series. Sterically, PF3 is much smaller and compact than the bulky PPh3. Therefore, the F3 will get closer to the metal and more significantly influence the relative energies of the d-orbitals.

on density

(Δ ore favorable the square planar configuration become 4.43. C

0.91Δ

0.11Δ

0.63Δ

4 ove toward the metal ion. This mto 4.47. In [Ru(NH3)6]2+, the Ru2+ is a moderately charged, large metal cation and the ammonia ligand is moderately high in the spectrochemicala 4.49. Δ(Rh3+) > Δ(Co3+) > Δ(Co2+) (a) (b)

3+(asteric hindrance. (b) The more highly charged Co3+

more than the Co2+. 4.51. Phosphines have d orbitals with which to form π-interactions with a transition metal whereas ammines do not. Phothrespectively. The resulting enhanced electrostatic M-L interaction increases the split between the semetal d orbitals in phosphine complexes. Phosphines, therefore, are generally stronger-field liganthan ammines. *4.53. Trifluorophosphine would be the stronger ligand, that is, it would be higher in the spectrochemicaPElectronically, the high electronegativity of the fluorine atoms in PF3 will help draw electrtoward the ligand. This extra electron-withdrawing power will make the π-back bonding stronger in the M-PF3 bond than in the M-PPh3 bond. The empty 3d orbital of the phosphorus is capable of accepting electron density back from a filled t2g type orbital. The presence of the three electron-withdrawing fluorine atoms enhances this effect. As a result the M-PF3 bond will be more polar and the splitting of the d orbitals more enhanced.

22

4.55. d6 octahedral: strong-field: t2g

6, n = 0, μS = 0 weak-field: t2g

4eg2, n = 4, μS = )24)(4( + = 4.90 BM

4 d6 tetrahedral: strong-field: e t22, n = 2, μS = )22)(4(3 3

+ = 2.83 BM 0 BM

spin state (BM) electrons config.

n(CN)6]4- 1.8 1 Mn2+ d5 t2g5 low-spin

)6]n(N S) n2+ d high-spin

3+ 4

weak-field: e t2 , n = 4, μS = 4.9 4.57. μexptl # of unpaired Mn+ dm elect. [M[Mn(CN 3- 3.2 2 Mn3+ d4 t2g

4 low-spin 5[M C 6] 6.1 5 M t4-

2g3eg

2 [Mn(acac)3] 5.0 4 Mn3+ d4 t2g

3eg1 high-spin

23

4.59. [CoI ]6 2g3- Co d6 high-spin t 4eg

2 n = μ = )6)(4(S = 4.90 BM

.61. For the [Ru ]4- ion, μ = 2.84 (χ T)½ -2 ½ = 4.93 BM which is consistent t and the re no

npaired electrons. Four unpaired electrons are consistent with a d6 ion in a weak field while zero

same litting abilities of the ligands involved, these

omplexes should absorb at a variety of wavelengths and therefore be of various colors.

only the square lanar splitting could support the presence of a diamagnetic compound.

4 F6 M = 2.84[(1.01 x 10 )(298)]with 4 unpaired electrons. For the [Ru(PR ) ]3 6

2+ ion, χ = 0 indicates tha μ = 0 refore there aMuunpaired electrons correspond to a d6 ion in a strong field. 4.63. Yes, it would be surprising if all the compounds involving these complex ions were to be thecolor. Given the spectrum (pun intended) of crystal field spc 4.65. (a) [Ni(H2O)2(en)2] ─ heat → [Ni(en)2] + 2H2O (b) The [Ni(en)2] could be either tetrahedral or square planar in nature. As shown below, p

tetrahedral -- can onlybe paramagnetic

square planar can be paramagneticor diamagnetic

*4.67. When the two water ligands are driven off by heating, the nickel changes from an octahedral to a

uare planar environment. We know that [Ni(deen)2]Br2 must be square planar because the alternative oordination sphere, tetrahedral, cannot yield a diamagnetic species for the product. One way to

e

4.69. The lectrons. The [Ni(CN)4]2- must be a strong-field, square planar complex in order to be diamagnetic.

I

sqcrationalize the color change in this reaction is shown below. When the waters are driven off, there is more room for the somewhat bulky deen bidentates to get closer to the nickel therefore increasing the magnitude of the splitting between the uppermost two d-orbitals. As shown, this could result in thobserved change of color. Again, we must be wary of such an oversimplification but, in any case, it islogical that the color should change in a fairly radical manner.

[Ni(deen)2]Br2 (orange)[Ni(H2O)2(deen)2]Br2 (blue)

R O Y G B I V R O Y G B I V

absorbed atransmitted transmitted

increase in crystalfield splitting

bsorbed * octahedral [Ni(H2O)6]2+ has a t2g

6eg2 configuration corresponding to two unpaired

e

H2O

OH2

OH2

Ni

NC

NC

CN

CN

Ni

H2O

H2O

2-H2O 2+

t must absorb at considerably higher frequencies (most likely into the ultraviolet) than the hexaaqua complex and therefore transmits or reflects all the visible frequencies and appears colorless.

24

eg

t2g

25

4.71. To picture the bonding in coordination compounds as strictly ionic is, at first thought, rather nrealistic. However, this theory does a surprisingly good job at accounting for the existence and high ability of these compounds as well as their magnetic characteristics and colors. The experimental vidence for the theory, as covered in this chapter, comes from two major types of studies: magnetic

onic

ustemeasurements and absorption spectroscopy. Magnetic measurements generate values for the molar magnetic susceptibilities of a series of coordination compounds. Molar susceptibilities, in turn, generatemolecular-level magnetic moments that can be related to the number of unpaired electrons and electrconfigurations of a complex. These results are explained quite adequately by the crystal field theory. Absorption spectroscopy gives information on the frequencies of light absorbed by coordination compounds. These frequencies vary in ways (depending on the size and charge of the metal and ligands) that can be explained in many cases by the ionic crystal field theory.

26

Chapter 5 Rates and Mechanisms of Reactions of Coordination Compounds

The sections and subsections in this chapter are listed below. 5.1 A Brief Survey of Reaction Types 5.2 Labile and Inert Coordination Compounds 5.3 Substitution Reactions of Octahedral Complexes Possible Mechanisms Experimental Complications Evidence for Dissociative Mechanisms Explanation of Inert versus Labile Complexes 5.4 Redox or Electron Transfer Reactions Outer-Sphere Mechanisms Inner-Sphere Mechanisms 5.5 Substitution Reactions in Square Planar Complexes: The Kinetic Trans Effect

27

Chapter 5 Objectives You should be able to

• recognize the various types of reactions that coordination compounds undergo substitution (including hydrolysis and anation), dissociation, addition, electron transfer (including oxidative-addition and reductive-elimination), reactions of coordinated ligands

• write expressions for overall and stepwise equilibrium constants for a given substitution reaction

• distinguish between and correctly use the kinetic terms labile and inert as well as the thermodynamic terms stable and unstable

• write dissociative (D), associative (A), and interchange (I) mechanisms for substitution reactions

• explain, when determining the rate law of a given reaction, how the concentration of a reactant may be masked by certain experimental conditions

• cite evidence from rates of exchange of water molecules as well as anation and aquation reactions, that substitution reactions of octahedral complexes most often follow a dissociative mechanism

• discuss the contributions of metal size and charge, steric hindrance among ligands, overall charge on the complex, and M-L covalent overlap to the energy of activation of the bond-breaking, rate-determining step of a substitution reaction

• discuss the contribution of the change of crystal field stabilization energy to the energy of activation of the bond-breaking, rate-determining step of a substitution reaction

• summarize why complexes of the first row transition metal ions, with the exception of Cr3+ and Co3+, are generally labile while those of most second and third row transition metal ions are inert.

• describe outer-sphere electron transfer mechanisms and how their rates are a function of the relative M-L distances in the reactants

• describe inner-sphere electron transfer mechanisms and how their rates are a function of the polarizability of the bridging ligand

• describe the kinetic trans effect and rationalize it in terms of the polarizability of the trans-directing ligand

• be able to use the kinetic trans effect to outline the synthesis of various square planar complexes

28

Solutions to Odd-Numbered Problems 5.1. K1K2K3K4 = [{Cu(NH3)(H2O)3}2+] x _[{Cu(NH3)2(H2O)2}2+]__ x [{Cu(H2O)4}2+][NH3] [{Cu(NH3)(H2O)3}2+][NH3] ___[{Cu(NH3)3(H2O)}2+]__ x _____[{Cu(NH3)4}2+]____ [{Cu(NH3)2(H2O)2}2+][NH3] [{Cu(NH3)3(H2O)}2+][NH3] = ___[{Cu(NH3)4}2+]___ = β4 [{Cu(H2O)4}2+][NH3]4 5.3. K1 = _[Cr(CO)5PPh3]_ [Cr(CO)6][PPh3] K2 = __[Cr(CO)4(PPh3)2]__ β3 = K1K2K3 = [Cr(CO)3(PPh3)3] [Cr(CO)5PPh3][PPh3] [Cr(CO)6][PPh3]3

K3 = ___[Cr(CO)3(PPh3)3]__ [Cr(CO)4(PPh3)2][PPh3] 5.5. K1 = [{Cr(H2O)4(C2O4

2-)+}]_ [{Cr(H2O)6}3+][C2O4

2-] K2 = ___[{Cr(H2O)2(C2O4)2}-]___ β3 = K1K2K3 = ____[{Cr(C2O4)3}3-]___ [{Cr(H2O)4(C2O4)}+][C2O4

2-] [{Cr(H2O)6}3+][C2O42-]3

K3 = ______[{Cr(C2O4)3}3-]_____ [{Cr(H2O)2(C2O4)2}-][C2O4

2-] 5.7. Kb = [NH4

+][OH-], the value of [H2O] is incorporated into the value of Kb. [NH3] 5.9. (a) redox (b) redox, dissociation (c) reaction of a coordinated ligand (d) substitution (e) addition 5.11. The O17 label would end up in the water molecule that is given as a product of the reaction. The Cr3+-O bonds in the hexaaquachromium(III) ion are not broken in this reaction of a coordinated ligand. Instead, the OH- reactant abstracts a proton (H+) from one of the waters leaving an OH- ligand in its place in the coordination sphere. *5.13. Strong-field low-spin d6 cobalt(III) complexes are inert to substitution because they lose crystal field stabilization energy upon breaking a metal-ligand bond to form a five-coordinate transition state. Accordingly, it is often easier (kinetically) to prepare cobalt(III) compounds by oxidizing Co(II)

29

complexes that are not inert. 5.15. (a) [Co(NH3)6]2+ Co2+ labile (b) [Co(NH3)5NO2]2+ Co3+, d6, moderately strong-field inert (c) [CoI6]4- Co2+, d6, weak field labile (d) [Fe(H2O)5(NCS)]2+ Fe3+ labile (e) Ni(en)3]2+ Ni2+ labile (f) [IrCl6]3- Ir3+ inert 5.17. [Ni(CN)4]2- Ni2+ d8 square planar, room for attack of incoming ligand; assuming an associative mechanism – most likely to be labile [Mn(CN)6]3- Mn3+ d4 in between, high charge density but only a moderate loss of CFSE – an in between case [Cr(CN)6]3- Cr3+ d3 high charge density; significant loss of CFSE – most

likely to be inert *5.19. The situation we are faced with is as follows: k1 (1) ML5X ─→ ML5 + X (fast) k-1 k2 (2) ML5 + Y ─→ ML5Y (slow) The rate-determining step is (2), so therefore, Overall rate = Rate of step (2) = k2[ML5][Y] Now the fast step (1) will establish an equilibrium "behind" the rate-determining second step such that k1[ML5X] = k-1[ML5][X] Solving this last equilibrium expression for [ML5] yields [ML5] = k1[ML5X]/k-1[X] Substituting this expression into the one for the overall rate yields Overall rate = k2k1[ML5X] [Y] k-1[X] *5.21. Since the first step is now considered to be faster than the second, the concentration of ML5XY would now be able to build up to a significant extent. Given that the second step is now much slower than the first, not much of the ML5XY would be used up in the rate-determining step. It follows that we expect the concentration of ML5XY to be relatively high.

30

5.23. In the absence of experimental complications, an associative mechanism should yield a rate law first order in both the reactant complex and the entering ligand while the dissociative mechanism would yield a rate law first order only in the reactant complex. In addition, an A mechanism might be proven by the isolation of the seven-coordinate intermediate while the D could be verified by the isolation of the five-coordinate intermediate. 5.25. Dissociative or interchange-dissociative mechanisms for the substitution reactions in octahedral coordination compounds are supported by the observed rates of (a) water exchange, (b) anation, and (c) aquation. The rates of exchange of water molecules in a hydrated metal ion are observed to be faster for metals of lower charge and/or larger size. Specifically, as the charge density decreases, the rate of exchange is found to be faster. This observation is consistent with a dissociative (or interchange-dissociative) mechanism because decreasing the charge density weakens the bond between the metal and the ligand-to-be-replaced and therefore leads to a faster rate-determining dissociative step. The fact that the rates of anation reactions are independent of the incoming anion is also consistent with the rate-determining step being the splitting off of a coordinated ligand followed by the addition of the incoming anion. Finally, the fact that the rates of aquation reactions are found to be dependent on the identity of the ligand to be replaced is also consistent with a dissociative mechanism. Data from all three of these types of reactions indicate that the strength of the original metal-ligand bond (broken in the rate-determining step of dissociative mechanism) determines the rate at which this ligand is replaced. 5.27. These data are generally consistent with a dissociative mechanism. Log k does not vary appreciably with the identity of the entering ligand, L. (The variations that do occur might very well be due to the ligand L "lurking" outside the coordination sphere of the reactant as in an Id mechanism.) The rate-determining step would be [Cr(NH3)5H2O]3+ ─ slow → [Cr(NH3)5]3+ + H2O Since the rate-determining step does not involve the L ligand, the rates are not dependent on [L]. 5.29. If the rate constant is 1.0 x 10-4 s-1 then log k is -4.0, a value of the same order of magnitude as those for the anation reactions of Problem 5.26. This similarity implies that again the rate-determining step is the breaking of the Co3+-OH2 bond and this reaction occurs via a dissociative mechanism. *5.31. If the rate constant is 3 x 10-6 M-1s-1 then log k is -5.5, a value that differs significantly from most of the other data given in Problem 5.30. This data seems to lead to the conclusion that these reactions occur by an associative rather than a dissociative mechanism. If the splitting of a Cr3+-OH2 were the rate-determining step, than the rates of all these reactions should all be of the same order of magnitude. 5.33. The table shows that these aquation reactions depend quite strongly on the identify of the L-ligand. This result is consistent with a rate-determining step in which the M-L bond is broken to form a five-coordinate transition state. 5.35. The change in the CFSE on going from the octahedral reactant to the five-coordinate intermediate directly affects the kinetics of substitution reactions. If CFSE is gained by forming the five-coordinate intermediate, the energy of activation (Ea) will be reduced by that amount and the reaction will be faster than expected in the absence of crystal field effects. If, on the other hand, CFSE is lost upon forming the

31

intermediate, the Ea will be increased by that amount and the reaction will be slower. 5.37. For a d6 octahedral case, CFSE = 6(0.40ΔO) - 2P = 2.40 - 2P For the square pyramidal case, CFSE = 4(0.457ΔO) + 2(0.086ΔO) = 2.00 - 2P ΔCFSE = -0.40ΔO 5.39. For a d8 octahedral case, CFSE = 6(0.40Δo) - 2(0.60Δo) = 1.20Δo For a d8 trigonal bipyramidal case, CFSE = 4(0.272Δo) + 3(0.082Δo) - 0.707Δo = 0.627Δo(a) There is a substantial loss (1.20Δo - 0.627Δo = - 0.57Δo) of CFSE upon going to the trigonal bipyramidal intermediate transition state, therefore we expect this reaction to be slower than it would be in the absence of crystal field effects. On the other hand, these ions only have a +2 charge and therefore would probably not be classified as inert. (b) The loss in CFSE for the square pyramidal state (see Table 5.4) is only 0.20Δo. Since loss of CFSE adds directly to the energy of activation, the Ea will most likely be smaller for the mechanism involving the square pyramidal transition state. This mechanism will therefore be favored rather than the one involving the trigonal bipyramidal intermediate. 5.41. The Fe3+ and V3+, while possessing a 3+ charge do not lose appreciable amounts of crystal field stabilization energy upon forming a square-pyramidal five-coordinate transition state. The Fe3+ cation is a d5 case that loses only 0.09ΔO or zero CFSE (strong-field and weak-field cases, respectively), while the V3+ cation is a d2 case that gains 0.11ΔO of CFSE. In order for a first row transition metal to be inert it appears that it must be 3+ charged and lose 0.20ΔO or more of CFSE. *5.43. (a) There would be a loss of 0.20Δo (see Table 5.4) when an octahedral d8 complex goes to a five-coordinate intermediate. This loss combined with a 3+ charge would make these complexes inert to substitution. (b) Cu3+ compounds contain d8 metals that are more likely to form square planar compounds rather than octahedral ones. *5.45. The greater steric hindrance provided by the bulkier N,N-diethylethylenediamine in complex (b) should help to weaken the Ni2+- OH2 bond that breaks in the rate determining step. 5.47. Given the lack of a bridging ligand, an inner-sphere mechanism for this cross reaction is not possible. The rate of this reaction, which presumably then occurs by an outer-sphere mechanism, is rather slow because the electron that the Co(III) accepts from the Ru(II) must be placed in an eg orbital. This orbital points directly at the ligands and therefore, because of electron-electron repulsions, significantly increases the metal-ligand distance. The result is that bond distances in the hexaammine cobalt complexes must be quite different. The energy necessary to rearrange the Con+-NH3 bonds so that electron transfer can take place contributes to the energy of activation and is a major factor in causing a low rate constant for the reaction. 5.49. A second-order rate law for a reaction involving two different reactants would be of the general form, rate = k[A][B]. Putting units with the rate and concentrations and then solving for the rate constant yields the following result: k (M-1s-1) = rate (M/s) [A][B] M2

5.51. [Ru(en)3]2+ + [Ru(en*)3]3+ ─→ [Ru(en)3]3+ + [Ru(en*)3]2+, where * = a radioactive label Lacking a potential bridging ligand in the reactant complexes, this self-exchange reaction proceeds by an outer-sphere mechanism. The reaction should be very rapid because little adjustment of the Ru-N distances need be made before the electron can be transferred. 5.53. (a) [IrCl6]2- + [W(CN)8]4- ─→ [IrCl6]3- + [W(CN)8]3-: Due to the inert nature of these highly charged iridium and tungsten ions and the fact that no ligand exchange seems to occur, this is most likely an outer-sphere mechanism. (b) [Co(NH3)5CN]2+ + [Cr(H2O)6]2+ ─→ [Cr(H2O)5NC]2+ + [Co(NH3)5(H2O)]2+: With the transfer of the bridging cyanide ligand and the fact that it reverses its denticity from cyano to isocyano, an inner-sphere mechanism seems likely. (c) [*Cr(H2O)6]2+ + [Cr(H2O)5F]2+ ─→ [*Cr(H2O)5F]2+ + [Cr(H2O)6]2+: Again, a bridging ligand seems to be transferred and an inner-sphere mechanism seems likely. 5.55. The mechanism that produces the S-thiocyanato would be much the same as the above written for Problem 5.54 except that the thiocyanate would bridge from the cobalt to the chromium ions using two lone pairs on the sulfur atom of the ligand. 5.57. (a) The most straightforward associative mechanism for the substitution of square planar complexes would have a rate-determining step in which the entering ligand attacks the square planar reactant complex to form a five-coordinate intermediate. In this case, the rate would be first-order in the entering ligand and in the reactant complex and therefore second-order overall. There is, however, a second possibility. Since most all these reactions are carried out in aqueous solution, the rate-determining step could be the attack of a water molecule to form a five-coordinate intermediate. The resulting aqua complex could then react with the incoming ligand to form the final product in a faster, non-rate-determining step. In this case the rate law would be first order in the reactant complex and first order in water. However, as in other cases we have discussed, given the large and nearly constant concentration of water in aqueous solutions, this latter rate law might be pseudo-first order in only the concentration of the reactant complex. (b) The rate of these associative mechanisms would decrease with the increasing steric bulk of the nonreacting ligands and the bulk of the entering ligand. It would increase with the charge of the entering ligand and the overall charge carried by the reactant complex. 5.59. Since the chloride is higher in the trans-directing series than ammonia, the second ammine is placed preferentially in the cis position. (The charges on the various complexes are omitted for clarity.)

Pt

Cl Cl

Cl Cl

NH3Pt

Cl NH3

Cl Cl

NH3Pt

Cl NH3

Cl NH3

5.61. Organic sulfides are large, polarizable ligands. They should be good trans-directors. The ability of various ligands to be good trans directors is explained, in part, by the polarization theory attributed to A. Grinberg.

32

Chapter 6 Applications of Coordination Compounds

The sections and subsections in this chapter are listed below. 6.1 Applications of Monodentate Complexes 6.2 Two Keys to the Stability of Transition Metal Complexes Hard and Soft Acids and Bases The Chelate Effect 6.3 Applications of Multidentate Complexes 6.4 Chelating Agents as Detergent Builders 6.5 Bioinorganic Applications of Coordination Chemistry Oxygen Transport Therapeutic Chelating Agents for Heavy Metals Platinum Antitumor Agents

Chapter 6 Objectives You should be able to

• cite examples of transition metal complexes involving monodentate ligands in qualitative analysis, dyes, silver and gold ore processing, nickel purification, and black/white photography

• define, characterize, rationalize, and use the concept of hard and soft acids and bases as applied to the stability of metal-ligand interactions

• define, explain, and give examples of the chelate effect • cite and explain applications of multidentate complexes drawn from complexometric

quantitative analytical methods • explain why EDTA is used to remove hard water deposits from hot water boilers and heaters

and often added to foods and other consumer products • explain the function of detergent builders and the advantages and disadvantages of using

phosphates, nitrilotriacetic acid, and carbonates to carry out that function over the years • explain the role of hemoglobin and its oxygen complexes in the process of respiration • explain how carbon monoxide and cyanide poisoning work • explain how and why EDTA, pencillamine, and British anti-lewisite function as therapeutic

chelating agents for heavy metals • briefly cite some of the history and symptoms of lead and mercury poisoning • explain how cisplatin and its derivatives serve as antitumor agents

33

Solutions to Odd-Numbered Problems 6.1. Diamminesilver(I), [Ag(NH3)2]+, contains Ag(I), a d10 metal. With no vacancies in d orbitals, coordination compounds containing this cation cannot readily absorb visible light and so remain colorless. On the other hand, tetraamminecopper(II), [Cu(NH3)4]2+, contains Cu(II), a d9 metal that does have vacancies in its d orbitals and therefore readily absorbs visible light. *6.3. Cyanide acts as a halide. In fact, ions such as cyanide (CN-), thiocyanate (SCN-), and others (see Chapter 18 for more details) are often referred to as pseudohalides because of their similarities to these ions. Accordingly, cyanide can often be quantitatively analyzed in the same way as the halide ions. One method of doing this is to add a slight excess of a solution of silver nitrate, AgNO3, to the cyanide until silver cyanide, AgCN(s), is precipitated, isolated, and then weighed quantitatively. The Liebig method actually refers to the method in which a standardized solution of silver nitrate is used to titrate a solution of cyanide until the first appearance of a precipitate of silver dicyanoargentate(I), Ag[Ag(CN)2]. This reaction becomes the basis of a titrimetric determination of cyanide. 2CN-(aq) + 2Ag+ ─→ Ag[Ag(CN)2](s) 6.5. Since hard acids are small, often highly-charged metal ions (Lewis acids) with high charge densities and low polarizabilities whereas hard bases are small, highly electronegative ligands (Lewis bases), the hard-hard M-L interaction is one between fairly highly charged species. It follows that the bonding between such species would tend to be predominantly ionic. On the other hand, soft acids are large, highly polarizable metals of low (even zero) oxidation states and soft bases are large, polarizable ligands of low charge, it follows that the soft-soft M-L bond between such species would not be highly ionic. 6.7. S-thiocyanato complexes involve thiocyanate, SCN-, bound through its relatively soft sulfur atom. N-thiocyanato complexes are bound through the relatively hard nitrogen atom. Mercury(II) is a soft acid whereas cobalt(III) is hard. Given the rules of HSABs, the Hg-S and Co-N bonds are favored and not vice versa. 6.9. The diacetatotetraaquairon(II) ion involves two acetate ions acting as monodentate ligands. The malonate ion has two acetate moities bound together through a common -CH2- group. Both of the acetate moities can act as Lewis bases and make malonate a bidentate ligand. The difference between the equilibrium constants, then, is an example of the chelate effect. The malonate complex involves a chelating, multidentate ligand that is more stable than the equivalent compound involving monodentate ligands. These two situations are shown in the following two equations. (The acetates could also form a trans complex.)

34

.11. The diacetatotetraaquairon(II) ion involves two acetate ions acting as monodentate ligands. The xalate ion has two carboxylate (COO-) moieties bound together through a carbon-carbon bond. Both of e carboxylate moities can act as Lewis bases and make oxalate a bidentate ligand. The difference

etween the equilibrium constants, then, is an example of the chelate effect. The oxalate complex volves a chelating, multidentate ligand that is more stable than the equivalent compound involving onodentate ligands. These two situations are shown in the following two equations.

H2O

H2O OH2

OH2

Fe

H2O

H2O

2+H2O

H2O O

O

Fe

H2O

H2O

C

O

CH3

CCH3

O

H2O

H2OO

O

Fe

H2O

H2O

C

O

CH2

C

O

2 CH3COO-

-OOCCH2COO-

6othbinm

35

H2O

H2O OH2

OH2

Co

H2O

H2O

2+H2O

H2O O

O

Co

H2O

H2O

C

O

CH3

CCH3

O

H2O O

H2O

H2O

2 CH3COO-

-OOCCOO-

C

O

H2O OCo C

O

6.13. Acetone is a monodentate ligand whereas acetylacetonate, [CH3COCHCOCH3]-, made up of twacetone [or acetyl, CH

o e

acac he equivalent

compounds involving two monodentate ligands.

3CO] moities connected together by a common -CH- group, is bidentate. Thdifference between the stability of acetone and acetylacetonate, then, is due to the chelate effect. The complexes involve chelating, bidentate ligands that form more stable compounds than t

6.15. Using Equations (6.9) and (6.10) as shown below, we can calculate the equilibrium constant for the substitution of three ethylenediamine ligands for six ammines in the Ni(II) coordination sphere. [Ni(NH3)6]2+(aq) ─→ Ni2+(aq) + 6NH3(aq) K = 1/β = 1/(4.0 x 108) Ni2+(aq) + 3en(aq) ─→ [Ni(en)3]2+(aq)________________ β = 2.0 x 1018

[Ni(NH3)6]2+(aq) + 3en(aq) ─→ [Ni(en)3]2+(aq) + 6NH3(aq) K = 2.0 x 1018 = 5.0 x 109

4.0 x 108

6.17. The dmgH ligand has a molecular formula of C4H7O2N2 and a molecular weight of 115.11 g/mol. Since one mole of Ni yields one mole of Ni(dmgH)2, we can calculate the mass of nickel in the ore as follows: 0.7815g Ni(dmgH)2 x 1 mol Ni(dmgH)

2 x ___1 mol Ni____ x 58.69 g Ni = 0.1588g Ni 288.91g Ni(dmgH)2 1 mol Ni(dmgH)2 1mol Ni % Ni = 0.1588g Ni x 100 = 45.94% 0.3456g ore 6.19. One mole of EDTA complexes one mole of Ca2+, therefore we can calculate the grams and percentage of calcium in the sample as follows. .02394L x 0.04672 mol EDTA 0

36

x __1 mol Ca__ x 40.08 g Ca = 0.04483g Ca L 1 mol EDTA 1 mol Ca

% Ca = _0.04483g Ca__

x 100 = 22.41%

0.2000 g sample

6.21. One mole of trien complexes one mole of Cu2+, therefore we can calculate the grams of copper in e 10.00mL aliquot (or portion) of the original solution.

.02275L x 0.01000 mol trien

*th 0 x _1 mol Cu_ x 63.54g Cu = 0.01446g Cu

L 1 mol trien 1 mole Cu

there are 0.01446g Cu2+ in the 10.00mL aliquot, there must be 0.1446g in the full 100mL volumetric ask. Therefore we can calculate the percentage of copper in the sample as follows:

Cu = _0.1446g Cu__

Iffl % x 100 = 72.10%

0.2005g sample 6.23. ___8.7g P__ x 1 mol P x 1mol Na5P3O10 x 367.85g Na5P3O10 = 34g Na5P3O10 = 34% 100g sample 30.97g P 3 mol P 1 mol Na5P3O10 100g sample

6.25. Dear Econ Major: Phosphates have been an important component of synthetic detergents because they let the

ap-like part of a detergent do its job unhindered by hard water. In other words, phosphates render the

readily etergent manufacturer. (See footnote at the end of this chapter.)

ecause the ab or some wavelengths of visible ght an either llow o through or flect them to our eyes so that we perceive something to be colored. Sometimes this process is

a wavelength in the range of 750-610 nm. What causes light to be absorbed? Usually this is ttributed to various energy levels in an atom or molecule. In the hemoglobin of the blood there are a

ng the d tals f e i n d appears red. When

e blood is oxygenated, an oxygen molecule (O2) also exerts a slight influence on the split among the d oglobin" absorbs at a slightly different range of wavelengths so

a brighter shade of red. (See footnote at the end of this chapter.)

590-570 yellow

blue 450-370 violet

. Hemes are made up of four nitrogen oms (incorporated into a impressive array of organic heterocyclic rings) that are bound to a central iron

al part of the ytochromes (c, a, a3, etc.) that play a large role in cellular respiration, the process whereby O2 is

anide, CN-, is that it binds or coordinates to the central iron tially shuts down cellular respiration. For this reason, of course, it is

portant to minimize patient exposure to cyanide. (See footnote at the end of this chapter.)

s Lewis ase sites would most likely go unused.

sohard water ions (calcium, magnesium, and iron) incapable of interfering with the action of the soap-like molecules in detergents. Phosphates have also been important because they are inexpensive andavailable to the d *6.27. Dear Biology Major: You may recall from your mind-bending introductory chemistry class that things are colored b y s b li d a ther wavelengthsreexpressed by a color wheel like the one shown on the next page. This wheel says that if an object, for example, absorbs blue-green wavelengths it will therefore reflect or transmit red light to our eyes. Red light has anumber of iron ions surrounded by a so-called heme group. This heme group causes a split amoorbi o th ro resulting in the absorption of blue-green light so that our bloothorbitals of the iron. Now the "oxyhemthat the blood now appears Wavelength Range (nm) Color 750-610 red 610-590 orange 570-500 green 500-450 6.29. Dear Doctor: You may recall a group of molecules called the hemesatcation. For example, hemes are an essential part of hemoglobin that is, as you know, responsible for transporting oxygen molecules (O2) from the lungs to the cells. Hemes are also a vitcconverted to water. The problem with cycation of cytochromes and essenim 6.31. This is potentially an octadentate ligand. It would literally surround a Mn ion. Some of itb

37

N

CH2 CH2

N

CH2 CH2

N

CH2COO--OOCH2C

: CH2COO-CH2COO- :

5-

:-OOCH2C

6.33. Mercury is a soft metal and will prefer to interact with soft groups like the sulfur-containing mercaptans. 6.35. Both histidine and cysteine are tridentates that could readily coordinate with cobalt and remove i This sequestering effect would be the basis of its use for cobalt detoxification. 6.37. Of serine and cysteine, the latter with its soft mercapto (-SH) group, would be the better antidote against the soft heavy metals.

t.

6.39. Dear English Major: You may recall from your AP chem course that cis m

re both from the Latin.) Cisplatin then has two groups hlorine atoms as it turns out) on the same side of a platinum atom. When cisplatin interacts with fast-

e that the two chlorides are replaced with adjacent from that long ago AP experience) of DNA. In this capable of reproducing faithfully and the growth of the

he trans analog of cisplatin is used, it cannot interact ificantly interfere with the replication of the tumor cells.

wing footnote.)

ootnote: ear readers: I know you can do better than I did in writing letters to various academic majors and

ld love to see the various short paragraphs that you all come up with. You'll be mous I m

* eans "on the same side of" while trans means "over" or "across." (These terms a(cgrowing tumor tissue, many researchers believsections ("nitrogen-bases" if you recall this much form (attached to a platinum atom) the DNA is intumor cells is retarded or even reversed. When twith adjacent sections of DNA and cannot signThe tumor continues to grow unchecked. (See follo F(Dprofessionals. I woufa ; ight put your paragraph in the next version of this Solution Manual! (If there is a next version.) Submit your entries to the author at Department of Chemistry, Allegheny College, Meadville, PA 16335.

38

39

Chapter 7 Solid State Structures

The sections and subsections in this chapter are listed below. 7.1 Types of Crystals Ionic Crystals Metallic Crystals Covalent Network Crystals Atomic/Molecular Crystals 7.2 A-type Crystal Lattices A-type Lattices 7.3 ABn-type Crystal Lattices Cubic, Octahedral, and Tetrahedral Holes Radius Ratios Ionic Radii AB Structures AB2 Structures 7.4 Structures Involving Polyatomic Molecules and Ions 7.5 Defect Structures 7.6 Spinel Structures: Connecting Crystal Field Effects with Solid State Structures

40

Chapter 7 Objectives

You should be able to

• distinguish among and characterize the various types of crystals: ionic, metallic, covalent, and atomic/molecular

• identify and characterize common A-type structures by the coordination number of a given sphere, the number of spheres per unit cell, the fraction of space occupied, and the density expression

• identify and appreciate the significance of the Bravais lattices • characterize atomic, van der Waals, and metallic radii • calculate the density of a given metallic crystal knowing its structure and relevant atomic

weight and radii data • identify and determine the relative sizes of cubic, octahedral, and tetrahedral holes in various

A-type structures • use radius ratios to predict the type of hole occupied by the smaller species in an ABn structure • rationalize why the radius ratio represents the lower limit to the range in which a given type of

hole will be occupied • discuss the origin and variation of Shannon-Prewitt ionic radii • identify and characterize common AB structures (rock salt, zinc blende, wurtzite, and cesium

chloride) by the structure and coordination numbers of both their anions and cations • calculate the density of a given compound knowing the type of AB structure it assumes • identify and characterize common AB2 structures (fluorite, cadmium iodide, rutile, and anti-

fluorite) by the structure and coordination numbers of both their anions and cations • identify and discuss ABn structures involving polyatomic molecules and ions • identify and characterize common defect structures such as Schottky and Frenkel defects and

edge dislocations • characterize normal and inverse spinel structures and provide a rationale for the structure of a

given compound based on calculations of crystal field stabilization energies

Solutions to Odd-Numbered Problems

7.1. Metallic Crystals – a lattice of cations held together by a sea of free electrons, for example, metallic copper. Covalent Crystals – atoms or groups of atoms in a lattice held together by an interlocking network of covalent bonds, for example, diamond. Atomic/Molecular Crystals – a lattice of atoms or molecules held together by intermolecular forces (Van der Waals, dipole-dipole, hydrogen bonds), for example, argon, water. *7.3. C covalent network interlocking covalent bonds (diamond) CCl2F2 molecular London dispersion forces and dipole-dipole forces CaCO3 ionic ionic bonds among Ca2+ and CO3

2- ions NH4F ionic ionic bonds among NH4

+ and F- ions *7.5. Ca metallic metallic forces CaF2 ionic ionic bonds among Ca2+ and F- ions CF4 molecular London dispersion forces 7.7. The free, non-hydrogen-bonded water molecules fill some of the large, hexagonal holes in the ice structure resulting in a greater mass in approximately the same volume, that is, a higher density. 7.9.

= in body of eight unit cells (7 dotted, 1 solid)

= at corner of solid unit cell

7.11. As shown below, there is 1 (= 8 x 1/8) circle and 1 (= 2 x 1/2) "x" in the unit cell for a total of two particles per unit cell.

X

X

7.13. A corner 12 x 1/6 = 2 Since there are three A and three B atoms in a unit face 2 x 1/2 = 1 cell, the empirical formula would be AB. B interior 3 x 1 = 3

41

7.15. This formula corresponds to the fraction of space occupied in a body-centered cubic unit cell composed of spheres of diameter d. There are two spheres per unit cell [ 1 + 8 (1/8)] and each sphere has a volume of 4/3πr3 or, substituting d/2 for r, 4/3 π(d/2)3. The volume of the unit is the cell edge (2d/ 3 , see Figure 7.9) cubed. 7.17. Gold is face-centered cubic.

42

(a) 4r = l 2

1 = 4.070A r = (l 2 ) / 4 = (4.070 2 ) / 4 r = 1.44Å (b) (4 atoms) x 196.67 g/mol __ density = 6.02 x 1023 atoms/mol = 1.31 x 10-21 grams = 19.3 g/cm3

[2(1.44 x 10-8cm) ]3 6.76 x 10-23 cm32 The actual density is 19.3 g/cm3, therefore the percent error is zero. 7.19. Europium is body-centered cubic. (2 atoms) x 151.96 g/mol_____ density = 5.26 g/cm3 = 6.02 x 1023 atoms/mol ) l3 cm3

l = 4.58 x 10-8 cm or 4.58Å 2d = 4r = l 3 , therefore, r = (4.58 3 )/4 = 1.98Å 7.21. Krypton is cubic close-packed or face-centered cubic. (a) 4 atoms per unit cell (b) (4 atoms) x 83.80 g/mol_____ density = 3.5 g/cm3 = (6.02 x 1023 atoms/mol ) l3 cm3

l = 5.4 x 10-8 cm or 5.4Å (c) 2d = 4r = l 2 , therefore r = l 2 /4 = 5.4 2 /4 = 1.9Å 7.23. Magnesium is hexagonal close-packed. Mg-Mg distance = 2rMg, therefore rMg = 1.602Å (6 atoms) 24.3 g/mol_____ density = (6.02 x 1023 atoms/mol ) = 2.42 x 10-22 grams = 1.74 g/cm3

24 2 (1.602 x 10-8cm)3 1.39 x 10-22 cm3

7.25. In a simple cubic lattice, the eight spheres

43

are in contact along the cell edge. The diagram at right shows a cross section through the face diagonals of the top and bottom faces. From the Pythagorean theorem,

2rsphere+2rhole

cell edge =2rsphere

face diagonal =2rsphere 2

(2rsphere + 2rhole)2 = (2rsphere)2 + (2rsphere 2 )2

2rsphere + 2rhole = (2 3 ) rsphere

( 3 - 1) rsphere = rhole

rhole/rsphere = 0.732 7.27. (a) Fluorite [see Figure 7.23(a)] The M atoms are in a cubic close-packed unit cell and therefore there are 4 per unit cell. The X atoms are in all the tetrahedral holes and therefore there are 8 per unit cell. It follows that there are 4 MX2 formula units per cell. (b) Rutile [see Figure 7.23(c)] The M atoms are located at the cell corners [8(1/8)] and in the center (1) and therefore there are two M atoms per cell. There are 4 X atoms in the top and bottom faces of the cell giving 2 [ = 4(1/2)] atoms. The other 2 X atoms are fully within the cell giving a total of 4 X atoms/cell. It follows that there are 2 MX2 formula units per unit cell. *7.29. All radii in Ångstrom units for coordination number = 6 (a) r(Be2+) = 0.59 = 0.47 predicts c.n. = 6 (b) r(Be2+) = 0.59 = 0.35 predicts c.n. = 4 r(O2-) 1.26 r(S2-) 1.70 incorrect, BeO assumes correct, BeS assumes wurtzite structure zinc blende structure (c) r(Mg2+) = 0.86 = 0.68 predicts c.n. = 6 (d) r(Mg2+) = 0.86 = 0.51 predicts c.n. = 6 r(O2-) 1.26 r(S2-) 1.70 correct, MgO assumes correct, MgS assumes rock salt structure rock salt structure (e) r(Ag+) = 1.29 = 0.77 predicts c.n. = 8 (f) r(Ag+) = 1.29 = 0.71 predicts c.n. = 6 r(Cl-) 1.67 r(Br-) 1.82 incorrect, AgCl assumes correct, AgBr assumes

rock salt structure rock salt structure (g) r(Ag+) = 1.29 = 0.63 predicts c.n. = 6 (h) r(Tl+) = 1.64 = 0.98 predicts c.n. = 8 r(I-) 2.06 r(Cl-) 1.67 incorrect, AgI assumes correct, TlCl assumes wurtzite structure CsCl structure 7.31. Cs+Au-

Assume that dCs-Au = r(Cs+) + r(Au-) = 3.69Å r(Cs+) = 1.81 = 0.96; predicts c.n. = 8; r(Au-) = 3.69 - r(Cs+) = 3.69 - 1.81 = 1.88Å 1.88 Therefore, a CsCl structure is reasonable. 7.33. Zinc blende S2- ions are in a face-centered cubic or cubic close-packed (ABCABC) structure. Zn2+ ions occupy half of the tetrahedral holes. 7.35. (a) If all the cubic holes of fluorite were occupied by calcium ions, the stoichiometry would be 1 to 1, i.e., CaF. (b) If all the octahedral holes in cadmium iodide were occupied by cadmium ions, the stoichiometry would also be 1 to 1, i.e., CdI. 7.37. Start with a calculation of the radius ratio: r(F-) = 1.19 = 0.66 r(Cs+) 1.81 (a) This result predicts a rock salt structure for CsF in which the Cs+ cations form a face-centered cubic lattice and the F- anions occupy all the octahedral holes. (b) There would be 4 Cs+ cations [8(1/8) + 6(1/2)] and 4 F- anions [12(1/4) + 1] in the lattice. The cations and anions are in contact along the cell edge, therefore l = 2r(Cs+) + 2r(F-) = 2(1.81Å) + 2(1.19Å) = 6.00Å. Density (4 atoms) x 132.9 g/mol + (4 atoms) x 19.00 g/mol____ g/cm3 = __________(6.02 x 1023 atoms/mol) (6.02 x 1023 atoms/mol) (6.00 x 10-8)3 cm3

= 4.67 g/cm3

7.39. In the NaCl (rock salt) structure, the anions are in a close-packed arrangement whereas in CsCl they are not. Accordingly, the CsCl should be more compressible under conditions of high pressure and could give way to the close-packed NaCl structure. *7.41. KCl, according to Table 7.8, has the rock salt structure. Since the cations and anions can be assumed to be contact along a cell edge (but note that the face-centered cubic anions cannot be assumed to be contact along the face diagonal – see the figure in the solution to Problem 7.38), it follows that l,

44

the cell edge, is just equal to 2r(K+) + 2r(Cl-). l = 2(1.52) + 2(1.67) = 6.38Å Volume of unit cell = l3 = (6.38Å)3 = 260Å3 (or 2.60 x 10-22 cm3) 4(39.1) + 4(35.45) grams Density = 6.02 x 1023 = 1.91 g/cm3

2.60 x 10-22 cm3

The actual density is listed as 1.984 g/cm3. 7.43. CaF2 assumes a fluorite structure (see Table 7.10). When one pictures a Ca2+ cation in the middle of the unit cell, the body diagonal of the unit cell is just 4r(Ca2+) + 4r(F-). body diagonal = l 3 = 4(1.26) + 4(1.19) = 9.80Å (Note that the radius for the Ca2+ is the c.n. = 8 value.) Therefore, l = 5.65Å or 5.65 x 10-8 cm 4(40.08) + 8(19.00) grams Density = 6.02 x 1023 = 2.88 g/cm3

(5.65 x 10-8 cm)3

The actual density is listed as 3.180 g/cm3. *7.45. We cannot assume that the oganium(II) cations are in contact along the cell edge, However, the cations and anions can be assumed to be in contact along the body diagonal. Therefore, body diagonal = l 3 = 2(1.26Å) + 2(1.5Å) = 5.5Å and l = 3.2Å. 1(298) + 1(16.0) grams Density = 6.02 x 1023 = 16 g/cm3

(3.2 x 10-8 cm)3

7.47. r(Rb+) = 1.66 = 0.912 It follows that a reasonable structure would be CsCl since the r(Br-) 1.82 radius ratio predicts a coordination number of eight for the cation. The ions would be in contact along the body diagonal of the unit cell such that l 3 = 2r(Rb+) + 2r(Br-) = 2(1.66) + 2(1.82) = 6.96Å; therefore, l = 4.02Å 1(85.47) + 1(79.90) grams Density = 6.02 x 1023 = 4.23 g/cm3

(4.02 x 10-8)3 cm3

45

46

The actual density is given as 3.35 g/cm3. These values are significantly different (percent error = 26%). If the c.n. = 8 value of 1.75Å for the Rb+ cation is used, the calculated density comes out to be 3.92 g/cm3, only 17% off from the actual density. If instead of a CsCl structure, RbBr assumes a rock salt structure (as, in fact, Table 7.8 indicates), then the ions would be in contact along the edge of the face-centered unit cell. Therefore, l = 2r(Rb+) + 2r(Br-) = 2(1.66) + 2(1.82) = 6.96Å and the density would be 4(85.47) + 4(79.90) grams Density = 6.02 x 1023 = 3.26 g/cm3

(6.96 x 10-8)3 cm3

This latter value is much closer to the experimental value (percent error = 3%). 7.49. r(NH4

+) = 1.37 = 0.665 r(NH4+) = 1.37 = 0.820 r(F-) = 1.19 = 0.869

r(I-) 2.06 r(Cl-) 1.67 r(NH4+) 1.37

indicates c.n. = 6; indicates c.n. = 8; indicates c.n. = 8; correct, NH4I correct, NH4Cl incorrect, NH4F has a rock salt has a CsCl structure has a wurtzite structure structure 7.51. Assuming the Na+ and SbF6

- ions to be in contact along the cell edge, we can set up the following expression for the density of NaSbF6. (FW of SbF6

- = 235.75 g/mole) 4(22.99) + 4(235.75) grams Density = 6.02 x 1023 = 4.37 g/cm3

[2(1.16 x 10-8) + 2r(SbF6-)]3 cm3

Solving for r(SbF6

-) yields a value of 2.50 x 10-8 cm or 2.50Å 7.53. Transition metal oxides are more frequently non-stoichiometric because these elements usually have a variety of oxidation states, more so than non-transition metal oxides. With more oxidation states, the charge of missing ions can be made up by oxidizing some of the remaining ions to a higher stable oxidation state. 7.55. The oxide ions are face-centered cubic, therefore there are [8(1/8) + 6(1/2) =] 4 per unit cell. There are 8 tetrahedral holes in a fcc unit cell (1 for each corner of the cube). If 1/8 of these are occupied by the AII cations, there is 1 AII per unit cell. There are 1 + 12(1/4) = 4 octahedral holes in a fcc unit cell. If one half of these are occupied by BIII cations, there are 2 BIII's per unit cell. It follows that the stoichiometry of such a unit cell and therefore the entire compound is AIIBB2

IIIO4. 7.57. Mn3O4 is made up of Mn(II) and Mn(III), that is, it is MnIII

2MnIIO4. In a normal spinel structure, the Mn(III) ions would occupy half the octahedral holes in the face-centered cubic array of oxide ions while the Mn(II) ions would occupy one eighth of the tetrahedral holes. In an inverse spinel the Mn(II) would exchange places with one half of the Mn(III) cations. Manganese has an electronic configuration

of [Ar]4s23d5, therefore Mn(II) is [Ar]3d5 and Mn(III) is [Ar]3d4. In a high-spin crystal field, the CFSE of a d5 ion is zero for both tetrahedral and octahedral cases. (See Figure 7.27(b) for these calculations.) The calculations of the CFSE for a d4 ion in both octahedral and tetrahedral fields is shown below. We see that the Mn(III) d4 ion clearly prefers an octahedral environment. Therefore, since the AIII ions occupy the octahedral holes in a normal spinel, this is the structure that the compound is expected to assume.

3/5Δ t

2/5Δ t

e

t2

CFSE = 2(3/5Δ t) -2(2/5Δ t) = 2/5Δ t = 1/5ΔO

tetrahedral field

Mn(III) d4

3/5ΔO

2/5ΔO

CFSE = 3(2/5ΔO) -1(3/5ΔO) = 3/5ΔO

octrahedral field

47

48

Chapter 8 Solid State Energetics

The sections and subsections in this chapter are listed below. 8.1 Lattice Energy: A Theoretical Evaluation 8.2 Lattice Energy: Thermodynamic Cycles Electron Affinities Heats of Formation for Unknown Compounds Thermochemical Radii 8.3 Lattices Energies and Ionic Radii: Connecting Crystal Field Effects with Solid State Energetics

Chapter 8 Objectives You should be able to

• define and discuss both the electrostatic and short-range repulsive forces that contribute to the overall lattice energy of an ionic compound

• derive and discuss the components of the Born-Landé equation for the lattice energy of an ionic compound

• discuss the role of charge density as a factor in determining the magnitude of the lattice energy of a compound

• discuss the role of Kapustinskii equation in estimating the lattice energy of an ionic compound • write a Born-Haber thermodynamic cycle that can be used to derive a value of the lattice

energy of an ionic compound • discuss the conditions under which significant covalent contributions to lattice energy would

be expected • use thermodynamic cycles and the Born-Landé or Kapustinskii equations to estimate values of

electron affinity, heats of formation for unknown compounds, and thermochemical radii • account qualitatively and quantitatively for the role of crystal field effects in determining the

cationic radii and lattice energies of transition metal salts

Solutions to Odd-Numbered Problems

49

8.1. (a) r(Rb+) = 1.66 = 0.91 r(Br-) 1.82 This radius ratio is consistent with a coordination number of eight.Therefore, a reasonable

* unit cell would be that of cesium chlorid

that is, a simple cubic array of bromide ions with rubidium cations in the center of each cell. A sketch of such a cell is found at right. (

e,

l of RbBr is rock salt, but CsCl is a reasonable structure

o = 1389

*The actual unit cel

Br Br

Br

BrBr

Br Br

Rb

nevertheless for this compound.)

o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n11 = 1389

1.82) (1.66)763.1)(1)(1(

+−+

⎟⎠⎞

⎜⎝⎛ −

1011(b) U = - 633 kJ/mol

.3. Th e a a roaches to this problem. (1) Since we do not know the crystal

- 632 kJ/mol = 1202

8 er re two reason ble appstructure, we can simply use the Kapustinskii Equation and solve for r(Fr+).

or)1)(1)(2( −+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

or345.01

Solving for ro yields a value of 3.42Å; 3.42Å = r(Fr+) + r(Cl-) = r(Fr+) + 1.67

) A second approach: since all the other alkali halides assume the rock salt structure, it would be for

o = 1389

Therefore, r(Fr+) = 1.75Å (2logical to assume that FrCl does also. In that case we can use the Born-Landé Equation with n = 14Fr+, which has a [Rn] shell (the value of 14 was obtained by extrapolating from the data given in Table 8.2), n = 9 for Cl- which has an [Ar] shell. The average value of n is 11.5.

or)748.1)(1)(1( −+

⎟⎠⎞

⎜⎝⎛ −

5.1111U = - 632 kJ/mol; Solving for ro yields a value of 3.51Å

r(Fr+) + r(Cl-) = r(Fr+) + 1.67 = 3.51Å; therefore, r(Fr+) = 1.84Å *8.5. (a) r(Bk4+) = 0.97 = 0.77 which is consistent with a coordination number of 8 and

o = 1389

r(O2-) 1.26 a fluorite lattice.

o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n11 = 1389

1.26) (0.97)519.2)(2)(4(

+−+

⎟⎠⎞

⎜⎝⎛ −

5.1011U = 111,500 kJ/mol

he value of n is obtained by averaging n = 7 for O2-, which has a [Ne] shell and a value of n = 14 for T

Bk4+, which has a [Rn]4f7 electronic configuration. n = 14 is obtained by extrapolation of the data in Table 8.2.

(b) For a CdI2 structure, M = 2.191 instead of 2.519 for fluorite.

ow the lattice energy comes out to be – 9880 kJ/mol, about a 13% difference.

Kap = 1202

N

1.26 0.97)2)(4)(3(

+−+

⎟⎠⎞

⎜⎝⎛ −

23.2345.01(c) U = - 10,900 kJ/mol

he Kapustinskii value lies in between the above two values. Without knowing for sure what the crystal

.7. The lattice energy of MgO should be approximately four times that of NaF because the former is a

8.9. Assumptions to be made for the estimation of the lattice energy of Xe+F-:

(1) radius of Xe (isoelectronic with I but with one more proton;

(metallic

.90Å (a little smaller than I)

Tstructure of BkO2 is, the UKap seems to be a reasonable approximation of the lattice energy. 8+2/-2 lattice while the latter is +1/-1. This difference will make MgO harder, as well as higher melting and boiling. *

+ one more electron and proton than I-): rI ) = 1.96Å r(I-) = 2.06Å Use r(Xe+) = 1 (2) r(F-) = 1.19 = 0.63 Since this is consistent with c.n. = 6, we can

(3) n(Xe ) = 12, n(Ne) = 7, therefore, average n = 9.5

o = 1389

r(Xe+) 1.90 reasonably assume a rock salt structure.

+

o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n11 = 1389

1.90) (1.19)748.1)(1)(1(

+−+

⎟⎠⎞

⎜⎝⎛ −

5.911Born-Landé: U = - 703 kJ/mol

Kapustinskii: UKap = 1202o

-

r Z Z+ν

⎟⎟⎠

⎞⎜⎜⎝

⎛−

or345.01 = 1202

3.09)1)(1)(2( −+

⎟⎠⎞

⎜⎝⎛ −

09.3345.01 = - 691 kJ/mol

8.11. U = NAZ+Z-Mr-1 + NBr-n

dU

*

+

50

= - NAZ Z-M - n NB = 0 when r = ro

Solving for B yields B =

dr r2 rn+1

n r

rMZAZ 1n

o2o

- ++

⋅−

Substituting this expression for B in the original expression for U yields the following:

U = NAZ+Z-Mr -1 + Nr – n n

r r

MZAZ 1no

2o

- ++

⋅⋅

└────────────┘ - NAZ+Z-Mro

-1 (1/n)

U = o

-

r MZNAZ+

⎟⎠⎞

⎜⎝⎛ −

n11

ber Cycle:

= ΔH°f - ΔHsubl - IE1 - IE2 - 2ΔH(g) - 2EA .0) - 2 (-349.0) = - 2289.9 kJ/mol

orn-Landé Equation:

r(Ba

8.13. Born-Ha

Ba(s) + Cl2(g)

ΔHsubl

ΔHof

BaCl2(s)

2 Cl(g)2 EA

2 Cl-(g) + Ba2+(g)

Ba(g)IE1 + IE2

U2ΔH(g)

U = -858.6 - (175.6) - (502.7) - (965.0) - 2(243 B

2+

51

) = 1.49 = 0.89 Indicates c.n. = 8, so a fluorite structure is reasonable.

Ba has a [Xe] configuration, therefore n = 12

o = 1389

r(Cl-) 1.67

2+ Cl- has an [Ar] configuration, so n = 9

o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n11 = 1389

1.67) (1.49)519.2)(1)(2(

+−+

⎟⎠⎞

⎜⎝⎛ −

5.1011U = - 2000 kJ/mol

here is 12.5% difference between these two results.

T

8.15.

ΔH°f = ΔHsubl + IE + ΔH(g) + EA + U

Uo = 1389

Δ

52

Cu(s) + 1/2 F2(g)

ΔHsubl

Hof

CuF(s)

X(g)EA

F-(g) + Cu+(g)

Cu(g) IE

UΔH(g)

o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n1 1 Cu+, n = 9; F-, n = 7

= 1389 1.19) (0.91

)638.1)(1)(1(+

−+⎟⎠⎞

⎜⎝⎛ −

811 = - 948.0 kJ/mol

ΔH°f = 338.3 + 745.3 + 79.0 + (- 328.0) + (- 948.0) = -113.4 kJ/mol

.17.

ΔH°f Cl-Cl F-F F-Cl

½ DF-F = ΔH°f - ½ DCl-Cl + DCl-F

= - 56.1 - ½(243.4) + 246.4 = 68.6 kJ/mol

DF-F = 137.2 kJ/mol

.19. We want to calculate a value for the ΔH° for the following equation:

CsCl (ordinary form) → CsCl (rock salt form)

hen we calculate the ΔH°f for these two compounds, all the factors will be the same except the lattice

ΔH = ΔH°f (rock salt form) - ΔH°f (ordinary form)

8

Δ 1/2 Cl2(g) + 1/2 F2(g)Ho

fClF(g)

1/2 DCl-Cl1/2 DF-F

Cl(g) + F(g)-DCl-F

= ½D + ½D - D *8 Wenergies. In fact, the ΔH of the above process can be calculated as follows:

= U(rock salt form) - U (ordinary form) The lattice energies will differ only because the Madelung constants differ. Knowing this, we can

o (rock salt form) = 1389

calculate the Born-Landé lattice energies for each form and proceed to calculate ΔH.

o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n11 = 1389

1.67) (1.81)748.1)(1)(1(

+−+

⎟⎠⎞

⎜⎝⎛ −

5.1011U = - 631 kJ/mol

o (ordinary form) = 1389 o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n11 = 1389

1.67) (1.81)763.1)(1)(1(

+−+

⎟⎠⎞

⎜⎝⎛ −

5.1011U = - 637 kJ/mol

ΔH = - 631 kJ/mol - (- 637 kJ/mol) = 6 kJ/mol

.21. (a)

) From part (b) of Problem 8.1, we know that Uo = - 633 kJ/mol

EA = ΔH°f - ΔHsub - IE - ΔH(g) - U

EA = -394.6 - 80.9 - 403.1 - 111.9 - (- 633) = - 358 kJ/mol

his result is somewhat different from the value given in Table 8.3 (- 324.7 kJ/mol) because RbBr

8.23. Uo = 1389

Δ

53

8 (b B-L

Tactually assumes the rock salt (NaCl) structure not the CsCl structure used in Problem 8.1.

*o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n11 = 1389

1.70) (0.74)641.1)(2)(2(

+−+

⎟⎠⎞

⎜⎝⎛ −

911 = - 3320 kJ/mol

[The value of r(Zn2+) is for coordination number 4.]

)

Rb(s) + 1/2 Br2(l)

ΔHsubl

Hof

RbBr(s)

Br(g)EA

Br-(g) + Rb+(g)

Rb(g) IE

UΔH(g)

Δ(b Zn(s) + S(s)

ΔHsubl

Hof

ZnS(s)

S(g) S2-(g) + Zn2+(g)

IE1 + IE2

UΔH(g)

EA1 EA2

Zn(g)

ΔH°f = ΔHsubl + IE1 + IE2 + ΔH(g) + EA1 + EA2 + U

- 3320) = 280 kJ/mol

.25. 2CaCl(s) ─→ CaCl2(s) + Ca(s)

ΔH = ΔH°f [CaCl2(s)] + ΔH°f [Ca(s)] - 2ΔH°f [CaCl(s)]

es, we would predict that this reaction would probably occur. It has a negative heat of reaction and

.27.

= IE + ΔH(g) + EA + U

Kap = 1202

EA2 = ΔH°f - ΔHsubl - IE1 - IE2 - ΔH(g) - EA1 - U EA2 = - 192.6 - 130.8 - 906.4 - 1733 - 278.8 - (- 200.4) - ( (Note: there are only two significant figures in this value.) 8 ΔH = -792 + 0 - 2(- 130) = - 532 kJ/mol Ysince there are two moles of solids on either side of the equation, the ΔS is probably not appreciably negative. These factors would make ΔG negative and the reaction (under standard state conditions) spontaneous from left to right as written. Δ

54

8 Ne(g) + 1/2 Cl2(g)Ho

fNe+Cl-(s)

Cl(g)EA

Cl-(g) + Ne+(g)

IE

UΔH(g)

ΔH°f

o

-

r Z Z+ν

⎟⎟⎠

⎞⎜⎜⎝

⎛−

or345.01 = 1202

3.07)1)(1)(2( −+

⎟⎠⎞

⎜⎝⎛ −

07.3345.01 U = - 695 kJ/mol

able 7.2 gives a value of 1.54Å for the Van der Waals radius of neon. Removing one electron from a

ΔH°f = 2080 + 121.7 + (- 349.0) + (- 695) = 1158 kJ/mol

eCl would not form because the ionization energy of neon is too large. Not enough energy is released

8.29. The equation representing the reaction in which CaCl3 might be produced from its constituent

Ca(s) + 3/2 Cl2(g) ─→ CaCl3(s)

Tfilled shell should not decrease the size much at all. Let's arbitrarily use a value of 1.40Å for the radius of Ne+. This combined with the ionic radius of Cl- of 1.67Å gives a value for ro of 3.07Å. Nduring the formation of the NeCl(s) lattice to make up for the energy needed to ionize an electron from the stable noble gas electronic configuration of neon. *elements in their standard states would be as follows:

The entropy change accompanying the above reaction would be negative since one and a half moles of

.31. The effective radius (or so-called thermochemical radius) of a polyatomic ion can be measured by

,

ed

.33. (a)

ΔH°f [NaOH(s)] = ΔH°f [OH-(g)] + ΔHsubl + IE + U

J/mol

(b) UKap = 1202

gaseous reactants is going to zero moles of gaseous products. This negative ΔSf°, combined with the +1600 kJ/mol calculated for the ΔH°f of this compound, would always yield a positive value for ΔG°f. Given these values of ΔH°f and ΔSf°, the entropy (which will always be negative) could never be used toforce the formation of CaCl3. 8taking a series of salts containing the ion for which the all the relevant thermochemical data are known except the lattice energies. Using Born-Haber cycles, the lattice energies can then be determined. Nextassuming the crystal structures of these salts are known, we can use the Born-Landé equation to estimate the radius of the polyatomic ion in each salt. (If the crystal structures are not known, we can use the Kapustinskii equation to estimate to estimate the radius in each case.) These radii can then be averagto yield an estimate of the effective radius of the polyatomic ion.

55

8 U = ΔH°f [NaOH(s)] - ΔH°f [OH-(g)] - ΔHsubl - IE U = - 425.6 - (- 143.5) - 107.3 - 495.9 = - 885.3 k

o

-

r Z Z+ν

⎟⎟⎠

⎞⎜⎜⎝

⎛−

or345.01 = 1202

or)1)(1)(2( −+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

or345.01 = - 885.3

Solving for ro yields values of 2.31 and 0.41Å, the latter value being physically impossible.

ro = r(Na ) + r(OH ) = 1.16 + r(OH ) = 2.31Å

Solving for r(OH ) yields a value of 1.15Å

his result is within 14% of the value given in Problem 7.50. This latter value is obtained by averaging

+ - -

- Tthe results of a number of thermochemical calculations involving the hydroxide anion.

Na(s) + 1/ 2 O2(g) + 1/2 H2(g) ΔH f[NaOH(s)]NaOH(s)

OH-(g) + Na+(g)

U

ΔHof[OH-(g)]

Na(g)

ΔΗsubl

IE

o

8.35.

ΔH°f [NaBH4(s)] = ΔH°f [BH4-(g)] + ΔHsubl + IE + U

4 kJ/mol

ΔH°f [K2PtCl6(s)] = ΔH°f [PtCl62-(g)] + 2(ΔHsubl) + 2(IE) + U 226 kJ/mol

8.39. UKap = 1202

56

Na(s) + B(s) + 2 H2(g)ΔHo [NaBH (s)]f 4 NaBH4(s)

Na(g)

UΔHo

f[BH4-(g)]

BH4-(g) + Na+(g)

ΔHsubl

IE

ΔH°f [BH4

-(g)] = ΔH°f [NaBH4(s)] - ΔHsubl - IE - U ΔH°f [BH4

-(g)] = - 183.3 - 107.3 - 495.9 - (- 703) = - 8 8.37. Δ

2 K(s) + Pt(s) + 3 Cl2(g)Ho [K PtCl (s)]f 2 6

2 K(g)

U

PtCl62-(g) + 2 K+(g)

2ΔHsubl

2 x IE

K2PtCl6(s)

ΔHof[PtCl6

2-(g)]

ΔH°f [K2PtCl6(s)] = -774 + 2(89.2) + 2(418.9) + (-1468) = - 1

= 1202 r

)2)(2)(2(

o

−+⎟⎟⎠

⎞⎜⎜⎝

⎛−

r345.01oo

-

r Z Z+ν

⎟⎟⎠

⎞⎜⎜⎝

⎛−

or345.01 = - 2911 kJ/mol

olving for ro yields values of 2.91 and 0.39Å, the latter value being physically impossible.

8.41. density =

SIf ro = r(Ca2+) + r(C2

2-) = 1.14 + r(C22-) = 2.91Å, then r(C2

2-) = 1.77Å ( )

( )3

23

4

4

221002.6

44

−+

−+

+

+

ClONa

ClONa

rrx

AWAW

=

[ ]

( ) ([ ]) 388

23

1080.41032.21002.6

45.99(4)99.22(4

−− +

+

xxx = 2.25 g/cm3

8.43.

he high-spin Fe2+ has two electrons occupying the eg orbitals that point directly at the octahedrally-

8.45.

egeg

t2g

t2g

low-spin Fe2+ (d6) high-spin Fe2+ (d6)

Tplaced ligands, therefore pushing these ligands back. The result is that the high-spin iron(II) is effectively larger than the low-spin. *

3400

3600

3800

4000

4200

d0 d1 d2 d3 d4 d5 d6 d7 d8 d9 d10

50

actual latticeenergies

gradual increase expecteddue to decrease in radii ofM2+ ions

Ca2+ Ti2+ V2+ Mn2+ Fe2+ Co2+ Ni2+ Zn2+

Number of d electrons

240

57

58

Chapter 9 Building a Network of Ideas to Make Sense of the Periodic Table

The sections and subsections of this chapter are listed below. 9.1 The Periodic Law Effective Nuclear Charge Atomic Radii Ionization Energy Electron Affinity Electronegativity 9.2 The Uniqueness Principle The Small Size of the First Elements The Increased Likelihood of π-bonding in the First Elements The Lack of Availability of d orbitals in the First Elements 9.3 The Diagonal Effect 9.4 The Inert Pair Effect 9.5 Metal, Nonmetal, and Metalloid Regions 9.6 Conclusions

Chapter 9 Objectives You should be able to

• discuss the contributions of Dmitrii Mendeleev in formulating the periodic table of the elements • calculate and rationalize values of effective nuclear charges • discuss the central role of effective nuclear charge in determining the major periodic trends in

atomic radii, ionization energy, electron affinity, and electronegativities • define the uniqueness principle and discuss the three main reasons behind it • define the diagonal effect and discuss the three main factors behind it • define the inert pair effect and discuss the two main reasons behind it • locate the metal/nonmetal line on the periodic table and discuss why it is positioned the way it is • summarize the first five components of the interconnected network of ideas for making sense of

the periodic table

59

Solutions to Odd-Numbered Problems 9.1. The properties of eka-silicon (germanium) can be interpolated from those of silicon and tin (above and below it) and gallium and arsenic (to its left and right). Some reasonable predicted and actual values are given below. Actual for Ge Predicted vertically Predicted horizontally atomic weight (g/mol) 72.59 73.4 72.3 density (g/cm3) 5.32 4.8 5.8 melting point (ºC) 937 823 421 boiling point (ºC) 2830 2312 --- electronegativity 1.8 1.8 1.8 ionization energy (kJ/mol) 760 748 763 electron affinity (kJ/mol) -118 -120 -56 atomic radius (Å) 1.39 --- --- 9.3. Main-group or representative elements are defined in this chapter as those in which the ns and np orbitals are partially filled. Under this definition, the noble gases, with their filled shells, would not be so defined. On the other hand, if the noble gases are not representative elements, how should we classify them? The only alternative would be to designate a separate term for them, a designation that is probably not worth making. 9.5. According to Webster's New Universal Unabridged Dictionary, 2nd Edition, Dorset & Baber, 1979, the prefix chalco- is the combining form of copper or brass while the suffix -gen indicates "something that produces." Therefore the word chalcogen means something that produces copper or brass. Sulfides and oxides are the main components of copper ores and therefore do produce copper. This is the origin of the group name for the Group 6A elements. 9.7. The periodic law as applied to the table of the elements refers to the fact that elemental properties are periodic, that is, they occur at regular intervals. For example, the properties of lithium resemble those of sodium that occurs eight elements later in a list of the elements ordered by atomic number. After another eight elements, potassium has properties also resembling those of lithium and sodium. 9.9. If the atomic volume of rubidium was missing from the plot of atomic volumes given in Problem 9.8, one could interpolate its value from those of potassium and cesium. Roughly, potassium has a volume of 45 while cesium is about 70. The average of these is about 58, which is very close to the actual atomic volume of rubidium. 9.11. Orbital diagrams: Phosphorus [Ne] ↑ ↓ ↑ ↑ ↑__ 3s 3p Copper [Ar] ↑ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ 4s 3d Arsenic [Ar] ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↑ ↑__ 4s 3d 4p

60

Thallium [Xe] ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑__ ___ ___ 6s 4f 5d 6p 9.13. Uuo, element-116, is a chalcogen. Starting with Rn, its abbreviated orbital diagram follows: [Rn] ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ _ ↑ _ 7s 5f 6d 7p Its most prevalent oxidation state is most likely +4. 9.15. Effective nuclear charge, Zeff = Z - σ (a) calcium, Zeff = 20 - 18 = 2 (b) silicon, Zeff = 14 - 10 = 4 (c) gallium, Zeff = 31 - 28 = 3, where the 28 electrons are the [Ar]3d10 pseudonoble gas

configuration. (d) argon, Zeff = 18 - 10 = 8 *9.17. Slater's rules are listed in Table 9.1. (a) Ne has the electronic configuration 1s22s22p6. An electron occupying the 3s orbital will be shielded by a factor of 0.85 by the eight electrons in the 2s and 2p orbitals and by a factor of 1.00 by the two electrons in the 1s orbital. Therefore, Zeff = 10 - 8(0.85) - 2(1.00) = 1.20 (b) On the other hand, a 2p electron being ionized from neon will be shielded by a factor of only 0.35 by the other seven electrons in the 2s,2p group and by a factor of 0.85 by the two inner core 1s electrons. Therefore, Zeff = 10 - 7(0.35) - 2(0.85) = 5.85 Again we see that the effective nuclear charge attracting an incoming electron to a noble gas such as neon is very small (but not zero) while the Zeff holding a valence electron against ionization is very large (but not 8 as calculated in Problem 9.16). 9.19. Effective nuclear charge is the nuclear charge experienced by a valence electron. From lithium to neon, the number of shielding electrons remains the same (2), while the actual nuclear charge increases from +3 to +10. therefore, the effective nuclear charge (= Z- σ) increases from +1 to +8. *9.21. For the purposes of calculating the effective nuclear charge using Slater's rules, the electronic configuration of calcium is written as follows: 1s22s22p63s23p64s2. For a 4s electron of calcium, Zeff = 20 - 1(0.35) - 8(0.85) - 10(1.00) = 2.85 For a 3p electron of calcium, Zeff = 20 - 7(0.35) - 8(0.85) - 2(1.00) = 8.75 9.23. The electronic configurations and Zeff's of Al, Al+, Al2+, and Al3+ are as follows: Al 1s22s22p63s23p1 Zeff = 13 - 2(0.35) - 8(0.85) - 2(1.00) = 3.50 Al+ 1s22s22p63s2 Zeff = 13 - 1(0.35) - 8(0.85) - 2(1.00) = 3.85

61

Al2+ 1s22s22p63s1 Zeff = 13 - 8(0.85) - 2(1.00) = 4.20 Al3+ 1s22s22p6 Zeff = 13 - 7(0.35) - 2(0.85) = 8.85 Based on these results, we expect that the ionization energies will gradually increase for Al, Al+, and Al2+ and then be much larger for Al3+. *9.25. As can be seen by close inspection of the plot, the 3s, 3p, and 4s orbitals all have small but significant probabilities of placing electrons very close to the nucleus. That is, these orbitals penetrate through the inner-core electrons to the nucleus just a little better than does a 3d orbital. (nf orbitals, not shown in the figure, have an even smaller probability of placing electrons close to the nucleus.) Therefore, Slater's rules give a smaller contribution to the shielding constant for orbitals inside an (ns,np) group than they do for orbitals inside an (nd) or (nf) group. 9.27. Ionization energy: the amount of energy required to completely remove an electron from a neutral, gaseous atom (or ion). Horizontally, ionization energies increase because Zeffs increase. Vertically, ionization energies decrease down a group because the electrons to be removed are farther and farther away from the same Zeff (as calculated assuming σ is the number of inner shell electrons). *9.29. P [Ne]3s23p3 The ionization energy is less for sulfur because the fourth 3p elec- S [Ne]3s23p4 tron is paired with one of the others. Therefore, this 3p electron, being repelled by its partner electron, is easier to remove than might normally be expected on the basis of trends in effective nuclear charge. 9.31. To see the trends in electron affinities, it helps to have the electron configurations to refer to. Al [Ne]3s23p1

Si [Ne]3s23p2

P [Ne]3s23p3

As the effective nuclear charge increases from Al to Si to P, we expect the electron affinities to increase in magnitude. However, at phosphorus the incoming electron would have to be paired with an electron already occupying a 3p orbital. The extra electron-electron repulsion involved in this interaction makes the electron affinity of phosphorus of lower magnitude than expected. 9.33. C [He]2s22p2 The electron affinity of nitrogen is less than that of carbon be- N [He]2s22p3 cause the electron being added to nitrogen must be paired with an electron already occupying the 2p orbital. The extra electron-electron repulsion involved in this interaction makes the electron affinity of nitrogen of lower magnitude than expected. 9.35. The polarizing power, in this case of a cation, is the ability to induce a dipole moment in a neighboring anion (or atom). The greater the cationic charge, the greater the pull it has on the electron cloud of the anion and the more easily a dipole moment can be induced in that anion. Also, the closer the cation can get to the anion, the greater its ability to induce a dipole moment. The center of the positive charge of a smaller cation can get closer to a neighboring anion than that of a larger cation, so decreasing the radius of cation increases its polarizing power. 9.37. Beryllium chloride should be more covalent than magnesium chloride because the charge density of beryllium (Z+/r+ = 4.9, see Figure 9.14 for radii) is greater than that of magnesium (Z+/r+ = 2.8) and therefore the polarizing power of the former cation is greater. The Be2+ distorts or polarizes the electron cloud of the Cl- increasing the likelihood of orbital overlap or covalent character in BeCl2.

62

Noting how similar the values of charge density are for the beryllium and aluminum cations, one would expect that BeCl2 might be very similar to AlCl3, that is, that they would have similar degrees of covalent character. 9.39. The planar molecular layers of graphite are held together by pi bonds formed by the parallel overlap of 2p orbitals. Because silicon atoms are larger than carbon atoms, this pπ-pπ overlap is much less effective. Silicon is characterized by strong σ bonds but weak pπ bonds. 9.41. Nitrogen atoms can form strong π bonds because they are so small. Phosphorus, on the other hand is significantly larger and therefore pπ-pπ overlap is much less effective. Phosphorus is characterized by strong σ bonds but weak pπ bonds. 9.43. Nitrogen only forms compounds such as nitrogen trichloride whereas phosphorus, arsenic, and antimony can form both tri- and pentachlorides because nitrogen does not have readily available d orbitals and therefore cannot form compounds in which the central atom exhibits an expanded octet. P, As, and Sb do have available nd orbitals which can participate in the bonding in the compounds of these heavier elements. Specifically, P, As, and Sb can form dsp3 hybrid orbitals and readily form the pentachlorides. 9.45. Oxygen only forms compounds such as OF2 but sulfur, selenium, and tellurium form the hexafluorides because the oxygen does not have readily available d orbitals and therefore cannot form compounds in which the central atom expands its octet. S, Se, and Te do have available nd orbitals, which can participate in the bonding in the compounds of these heavier elements. Specifically, S, Se, and Te can form d2sp3 hybrid orbitals used in the bonding of these hexafluorides. 9.47. Carbon only forms compounds such as CCl4 but tin forms polyatomic anions such as SnCl62-

because the carbon does not have readily available d orbitals and therefore cannot expand its octet. Tin has available 5d orbitals that can participate in the bonding in SnCl62-. Tin forms SnCl2 because of the inert pair effect. The cost in energy required to ionize the two 5s2 electrons in tin is sometimes not repaid by forming the two additional Sn-Cl bonds. 9.49. The inert pair effect refers to the fact that the ns2 electrons going down a group (3A, 4A, 5A principally) become more stable and less readily ionized. This leads to an increased stability of the n-2 oxidation state where n equals the group number. The ns2 electrons are less readily ionized because ionization energies increase somewhat after an nd10 or nf14 subshell is filled. (The d10 or f14 electrons do not effectively shield succeeding electrons from the nucleus very well and therefore the electrons outside these filled subshells "feel" a greater than expected effective nuclear charge.) At the same time that IE is slightly increasing or not decreasing as much, the M-X bond energies are decreasing going down the groups because the bonds get longer and weaker. The combination of higher than expected IEs and lower bond energies means that the cost in energy required to persuade the ns2 electrons to participate in bonds is not fully repaid when bonds are formed. Therefore the elements lower in these groups often do not form the extra two bonds as their lighter congeners do. 9.51. When moving from mercury to indium to germanium to phosphorus to oxygen, one crosses the metal-nonmetal line. Going from bottom left to top right in the periodic table means that the elements move from being metals (Hg, In) to metalloids (Ge) to nonmetals (P, O).

63

Chapter 10 Hydrogen and Hydrides

The sections and subsections of this chapter are listed below. 10.1 The Origin of the Elements (and of Us!) 10.2 The Discovery, Preparation, and Uses of Hydrogen 10.3 Isotopes of Hydrogen 10.4 Radioactive Processes involving Hydrogen Alpha and Beta Decay, Nuclear Fission, and Deuterium Tritium 10.5 Hydrides and the Network Covalent Hydrides Ionic Hydrides Metallic Hydrides 10.6 The Role of Hydrogen in Various Alternative Energy Sources Hydrogen Economy Nuclear Fusion

64

Chapter 10 Objectives You should be able to

• explain the nature of the "big bang" theory • give a brief general account of the origin of the elements • represent the nucleus of an element properly • explain the source of energy in nuclear reactions • explain the difference between fusion and fission • relate how and by whom hydrogen was discovered • explain how hydrogen gas is commonly generated in the laboratory • describe the common industrial preparations and uses of hydrogen • describe the dangers and uses of the reaction between hydrogen and oxygen • describe the similarities and differences among the three isotopes of hydrogen • describe the isolation and uses of "heavy water," D2O • describe, represent and explain the deuteration of various compounds • describe and give examples of α, ß-, and ß+ decay • explain the concept of half life as it applies to radioactive isotopes • explain and represent fission • describe the conditions for the occurrence of a nuclear fission chain reaction • relate the first example of nuclear transmutation • describe how tritium is produced in nature and by humankind • discuss the placement of hydrogen among the groups of the representative elements • describe the three major oxidation states of hydrogen • describe and give examples of covalent hydrides • describe and give examples of ionic or "saline" hydrides and their uses • describe and give examples of metallic hydrides and their uses • discuss the nature of nonstoichiometric interstitial hydrides • describe what is meant by the "hydrogen economy" • describe the potential of nuclear fusion as a future energy source

Solutions to Odd-Numbered Problems 10.1. “Ylem” is the word Aristotle used to describe the substance out of which all the universe was created. Today we tend to use a word such as “singularity” to describe the nature of the universe before the big bang or what is referred to in the text as the “expansion of ylem”. The big bang is the accepted scientific theory for how the universe came into being. For reasons not known, the ylem or singularity, at a definite point in time, exploded to create the very fabric of space and, eventually, the universe as we know it. The first elements to appear were hydrogen and helium. Gradually, as these swirling clouds began to cool and coalesce together due to gravitational forces, the hydrogen started to sequentially fuse to produce elements as heavy as iron. So the first stars started to “burn” hydrogen and lighted up the blackness of space. 10.3. 11

5B 5 protons, 6 neutrons 17

8O 8 protons, 9 neutrons 60

27Co 27 protons, 33 neutrons 239

94Pu 94 protons, 145 neutrons 10.5. (a) 12

6C + 11H ─→ 137N

(b) 137N ─→ 13

6C + 0+1e (positron) (c) 13

6C + 11H ─→ 147N

(d) 147N + 11H ─→ 15

8O (e) 15

8O ─→ 157N + 0+1e (positron)

(f) 157N + 11H ─→ 12

6C + 42He (alpha particle) 10.7. (a) 11H + 11H ─→ 21H + 0+1e Δm = Σm(reactants) - Σm(products) = 2(1.007825) - (2.0140 + 5.488 x 10-4) = 0.0011 u

E = (0.0011 u) ⎟⎟⎠

⎞⎜⎜⎝

⎛u

kJ/mol 10 x 8.984 10

= 9.88 x 107 kJ/mol

(b) proton-proton cycle: 4 11H ─→ 42He + 2 0+1e Δm = Σm(reactants) - Σm(products) = 4(1.007825) - [4.0015 + 2(5.488 x 10-4)] = 0.0287 u

E = (0.0287 u) ⎟⎟⎠

⎞⎜⎜⎝

⎛u

kJ/mol 10 x 8.984 10

= 2.58 x 109 kJ/mol

*10.9. While Robert Boyle was, according to Asimov, the first chemist to collect a gas and demonstrate that the volume of a gas is inversely related to its pressure, no mention is made about his work with hydrogen. Even though others (including Boyle) had worked with hydrogen before him, Henry Cavendish is credited with the discovery of this element because, to quote Asimov, he "was the first to

65

investigate its properties systematically." *10.11. ΔH = ΔH°f[CO(g)] + ΔH°f[H2(g)] - ΔH°f[C(graphite)] - ΔH°f[H2O(g)] ΔH = (-110.5) + 0 - 0 - (-241.8) = 131.3 kJ/mol Since this is an endothermic reaction, heat is absorbed and is a reactant. Therefore when the temperature is raised to higher values, this reaction, according to Le Châtelier's principle, shifts to the right. Recall from your previous study of this principle that when the pressure of a reaction at equilibrium is decreased, it will shift in the direction producing the larger number of moles of gases. This reaction goes from one mole of gaseous reactants to two moles of gaseous products, therefore lowering the pressure will shift the reaction to the right and increase its yield. (Standard heats of formation vary somewhat from source to source, so your values may not be exactly the same as those given here.) 10.13. MO(s) + H2(g) ─→ M(s) + H2O(l) M is reduced from 2+ to 0; H is oxidized from 0 to 1+ Ag2O(s) + H2(g) ─→ 2Ag(s) + H2O(l) Bi2O3(s) + 3H2(g) ─→ 2Bi(s) + 3H2O(l) 10.15. The atomic weight or mass of an element is the weighted average of a collection of its isotopes. In this case, AWH = 0.99985(1.007825) + 0.00015(2.0140) = 1.00798u 10.17. Temperature is a measure of mean kinetic energy, which is given by ½mv². The heavier D2O molecules move slower than the lighter H2Os and therefore need a higher temperature to achieve sufficient velocity to escape from the surface of the solid or the liquid, that is, to melt or boil. 10.19. The electrolysis of water first produces H2 and O2 gases. Although naturally occurring water also contains small amounts of D2O and HDO, these are electrolyzed to D2 (and O2) slower because the heavier D+ cation travels more slowly to the electrodes. The result is that the concentration of D2O increases in a sample of water as it is being electrolyzed. As D2O has a greater density than H2O, the overall density of the water sample increases during the process of electrolysis. *10.21. First we can calculate the grams of D2O per L of ordinary water.

⎟⎟⎠

⎞⎜⎜⎝

⎛ watergal 400

OD cup 1 2⎟⎟⎠

⎞⎜⎜⎝

⎛OD cup 1

OD L 0.236

2

2⎟⎟⎠

⎞⎜⎜⎝

⎛OD L

OD mL 1000

2

2⎟⎟⎠

⎞⎜⎜⎝

⎛OD mLOD g 1.10

2

2 ⎟⎠⎞

⎜⎝⎛

water L 3.78 watergal 1 = 0.172

waterLOD g 2

Next we can calculate the liters of water in the oceans.

(3.2 x 108 km3) 3

kmm 1000⎟⎠⎞

⎜⎝⎛

3

mcm 100

⎟⎠⎞

⎜⎝⎛ ⎟

⎠⎞

⎜⎝⎛

cm 1mL 1

3 ⎟⎠⎞

⎜⎝⎛

mL 1000L 1 = 3.2 x 1020 L

Finally we are able to calculate the grams of D2O in the oceans of the world.

(3.2 x 1020 L) ⎟⎠⎞

⎜⎝⎛

waterLOD g 0.172 2 = 5.5 x 1019 g D2O (whew!)

66

10.23. Assume 1 mole of each substance:

H2 ⎟⎠⎞

⎜⎝⎛

molg2x1.0080

⎟⎟⎠

⎞⎜⎜⎝

⎛L22.4

mol 1 = 0.0900 g/L D2 ⎟⎠⎞

⎜⎝⎛

molg2x2.0140

⎟⎟⎠

⎞⎜⎜⎝

⎛L22.4

mol 1 = 0.180 g/L

Assuming ideal behavior, the calculated densities are very close to the actual. 10.25. Only the hydrogen atoms bound directly to oxygen atoms will be replaced with deuterium when mixed with heavy water, D2O. This is because the polar O-D bonds of deuterium oxide will interact with the polar O-H bonds of the glucose but not with the essentially nonpolar C-H bonds. 10.27. The allies support an undercover mission to destroy a heavy water plant under Nazi control in Norway. The allies were concerned that the heavy water could be used as a moderator in a possible Nazi effort to produce an atomic bomb. 10.29. beta-minus decay: 40

19K ─ β-→ 0-1e + 4020Ca

beta-plus decay: 4019K ─ β-→ 0+1e + 40

18Ar 10.31. 238

92U + 10n ─→ [23992U] ─ β-→ 0-1e + 239

93Np 10.33. Lise Meitner was an Austrian-Swedish physicist who, in 1906, was the first woman to receive a Ph.D. in physics at the University of Vienna. She was a pioneer in the field of radioactivity for more than 30 years, most of them working with Otto Hahn in Germany. Immediately after the first world war, they announced their discovery of protactinium, the 91st element. When Hitler came to power, Meitner, a Jew, was forced to leave Germany and took a position in Sweden. Hahn wrote to her that he had some somewhat surprising evidence that the bombardment of uranium with neutrons apparently yielded the element barium. Meitner, working with her cousin, Otto Frisch, wrote a paper in which they calculated that this experimental result was due to the fission of uranium to produce both barium and krypton. Element 109 was named Meitnerium in honor of this pioneering woman of nuclear physics. 10.35. 235

92U + 10n ─→ 9637Rb + 138

55Cs + 210n

10.37. 232

90Th + 10n ─→ 23390

Th ─ β-→ 23391Pa + 0-1e ─ β-→ 233

92U + 0-1e

10.39. 113

48Cd + 10n ─→ 11448Cd + γ

105B + 10n ─→ 42He + 73Li

10.41. 11H + 14

7N ─ cosmic rays→ 148O + 10n

67

10.43.

Osp3

sp3

H

H O

HH

H

+

H3O+with its very large charge density andpolarizing power, the proton severelydistorts the sp3 orbital of water forcingoverlap and covalent character

H+

*10.45. Using the Born-Haber cycle shown at right, ΔH

X-(g)

1/2 D

1/2 X2(g) + e-(g)

X(g)EA

we see that ΔH = ½D + EA For X = H ΔH = ½D + EA = ½(436.4) + (-77) = 141 kJ/mol For X = F ΔH = ½D + EA = ½(150.6) + (-333) = -258 kJ/mol In the process of forming a salt such as NaF, we see that the formation of F-(g) is an exothermic process and therefore will make a negative (and therefore favorable) contribution to the ΔH°f of such alkali metal fluorides. On the other hand if a saline hydride such as NaH is formed, the formation of the corresponding H-(g) is endothermic and does not make a favorable contribution to the ΔH°f of these hydrides. If such hydrides are to form, they must involve highly electropositive metals that have low ionization energies. Such metals will take little energy to ionize and will still have negative lattice energies and therefore have negative standard heats of formation.

10.47. Uo = 1389 o

-

r MZZ+

⎟⎠⎞

⎜⎝⎛ −

n11 = 1389

3.05)748.1)(1)(1( −+

⎟⎠⎞

⎜⎝⎛ −

711 = - 682 kJ/mol

where ro = r(K+) + r(H-) = 1.52 + 1.53 = 3.05Å; where n = (9 + 5)/2 = 7 (values from Tables 7.4 and 7.6) (values from Table 8.2) There is a 3.5% difference between the values obtained from the Born-Haber cycle and the Born-Landé equation (with the Born-Haber cycle taken as the most accurate value).

10.49. UKap = 1202 o

-

r Z Z+ν

⎟⎟⎠

⎞⎜⎜⎝

⎛−

or345.01 = 1202

2.44)1)(1)(2( −+

⎟⎠⎞

⎜⎝⎛ −

44.2345.01 = - 846 kJ/mole

where ro = r(Cu+) + r(H-) = 0.91 + 1.53 = 2.44Å (Note: there is a very small difference, 0.2 on the Pauling scale, between the electronegativities of copper and hydrogen. This result is indicative of a high degree of covalent character in the Cu-H bond.

68

The high degree of covalent character causes the actual (Born-Haber) lattice energy to be much higher in magni-tude than that derived from the Kapustinskii equation. See the text for further discussion of these ideas.)

10.51. UKap = 1202 o

-

r Z Z+ν

⎟⎟⎠

⎞⎜⎜⎝

⎛−

or345.01 = 1202

2.67)1)(2)(3( −+

⎟⎠⎞

⎜⎝⎛ −

67.2345.01 = - 2350 kJ/mole

where ro = r(Ca2+) + r(H-) = 1.14 + 1.53 = 2.67Å 10.53. 5CaH2(s) + Ta2O5(s) ─→ 5CaO(s) + 2Ta(s) + 10H2(g) *10.55. Adding these three equations together yields the net reaction H2O(g) ─→ H2(g) + ½O2(g) The heats of reaction for the three reactions can be calculated as follows: (1) ΔH = ΔH°f [H2(g)] + 6ΔH°f[HCl(g)] + ΔH°f[Fe3O4(s)] - 3ΔH°f[FeCl2(s)] - 4ΔH°f[H2O(g)] = 0 + 6(-92.3) + (-1118.4) - 3(-341.8) - 4(-241.8) = 320.4 kJ/mol (2) ΔH = ½ΔH°f [O2(g)] + 3ΔH°f[H2O(g)] + 3ΔH°f[FeCl3(s)] - ΔH°f[Fe3O4(s)] - 3/2ΔH°f[Cl2(g)] - 6ΔH°f[HCl(g)] = ½(0) + 3(-241.8) + 3(-399.5) - (-1118.4) - 3/2(0) - 6(-92.3) = -251.7 kJ/mol (3) ΔH = 3/2ΔH°f [Cl2(g)] + 3ΔH°f[FeCl2(s)] - 3ΔH°f[FeCl3(s)] = 3/2(0) + 3(-341.8) - 3(-399.5) = 173.1 kJ/mol The sum of these heats is ΔH1 + ΔH2 + ΔH3 = 241.8 kJ/mol The heat of the reaction H2O(g) ─→ H2(g) + ½O2(g) can also be calculated as follows: ΔH = ½ΔH°f [O2(g)] + ΔH°f[H2(g)] - ΔH°f[H2O(g)] ΔH = ½(0) + (0) - (-241.8) = 241.8 kJ/mol (Standard heats of formation vary somewhat from source to source, so your values may not be exactly the same as those given here.) 10.57. High temperatures ensure high atomic or nuclear velocities and kinetic energies. This kinetic energy, at the moment of impact of a collision, is largely converted to the potential energy needed to get over the energy of activation of the reaction. High temperatures also increase the number of atomic or nuclear collisions which will serve to increase the rate of these reactions. One significant difference between chemical and nuclear reactions is the magnitude of the energy of activation. The very large repulsions between nuclei make Eas for nuclear reactions very large indeed.

69

70

Chapter 11 Oxygen, Aqueous Solutions, and the Acid Base Character of

Oxides and Hydroxides

The sections and subsections of this chapter are listed below. 11.1 Oxygen Discovery Occurrence, Preparation, Properties, and Uses 11.2 Water and Aqueous Solutions The Structure of the Water Molecule Ice and Liquid Water Solubility of Substances in Water The Self-Ionization of Water 11.3 The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution Oxides: Survey and Periodic Trends in Oxides in Aqueous Solution (Acidic and Basic Anhydrides) The E-O-H Unit in Aqueous Solution An Addition to the Network 11.4 The Relative Strengths of Oxo- and Hydroacids in Aqueous Solution Oxoacids Nomenclature of Oxoacids and corresponding salts (Optional) Hydroacids 11.5 Ozone 11.6 The Greenhouse Effect and Global Warming

71

Chapter 11 Objectives

You should be able to

• relate how and by whom oxygen was discovered • explain why Priestley was able to isolate many more gases than scientists before him • describe the influence of the phlogiston theory of combustion on the discovery of oxygen • explain why Priestley could be called the "father of the modern soft drink industry" • explain the general relationship between carbon dioxide and oxygen in the biosphere • explain and represent how oxygen gas is generally prepared industrially and in the laboratory • describe some of the most important uses of oxygen • describe the structure and polarity of an individual water molecule • describe the utility of the FONCl rule of hydrogen bonding • relate the nature and properties of ice and snowflakes to the structure of solid water • describe and represent the "flickering cluster" model of water structure • describe, explain, and represent the solubility of ionic and covalent substances in liquid water • describe and represent the self-ionization of water • describe the utility and limitations of the concept of the hydronium ion in water and as a product

of acids in aqueous solution • show and explain the distribution of ionic, covalent polymeric, and discrete molecular oxides in

the periodic table • explain why metal oxides are basic anhydrides • explain why nonmetal oxides are acidic anhydrides • describe and give an example of an amphoteric anhydride and oxide • explain why substances containing an M-O-H unit are basic in aqueous solution: give several

examples • explain why substances containing an NM-O-H unit are acidic in aqueous solution; give several

examples • explain the effect on the strength of an oxoacid of changing the central atom; give several

examples • explain the effect on the strength of an oxoacid of changing the number of nonhydroxyl oxygens

attached to the central atom; give several examples • name a variety of oxoacids and their salts • explain the variation of acid strength in hydroacids within a period and a group of the periodic

table; give several examples • describe and represent ozone • describe how ozone is formed and some of its beneficial uses • describe and represent how ozone is formed and destroyed in the stratosphere in the absence of

external agents • describe the manner in which various greenhouse gases potentially cause radiative forcing and

global warming • list some methods whereby the amount of greenhouse gases released to the atmosphere could be

reduced

Solutions to Odd-Numbered Problems 11.1. HgO(s) ─heat→ Hg(l) + O(g) 11.3. Of the major components of air (N2, O2, H2O, Ar), only O2 will react with NO to produce NO2. Air is approximately 20 percent oxygen while "dephlogisticated air" is, of course, 100 percent oxygen. therefore, "dephlogisticated air" would have consumed four or five (depending on his experimental accuracy) times as much NO as "common air." NO(g) + ½O2(g) ─→ NO2(g) 11.5. 15

8O ─ β+→ 0+1e + 157N

*11.7. AW = 0.99763(15.994915) + 0.00037(16.999134) + 0.00200(17.999160) = 15.999u This result is the same as listed on the periodic table (at least to 3 places to the right of the decimal point). 11.9. C5H12(l) + 8O2(g ) ─→ 5CO2(g) + 6H2O(l) 11.11.

(a) NH3

N H

H

HH

H

HN

107.5o

(b) SO3

(c) CO2 (d) OF2

(e) NO2-

(f) CS2

O

C

SO

O O S

O

O O SO

O

O O O FF

O N O

-

O N O

-C SS

OS

O

O

120o

O C O

180o

μnet= 0

FO

F o<109.5

ON

O o<120

S C S

180o

11.13. A lone pair of electrons is confined by (or is under the influence of) only one nucleus whereas a bonding pair is confined by two nuclei. The lone pair is therefore freer to spread out and occupy more space.

72

73

*11.15. If every zinc and sulfur atom of the wurtzite structure (Figure 7.22d) were replaced by an oxygen of a water molecule (with hydrogen atoms bridging among oxygens), the wurtzite structure would become the ice structure (Figure 11.3). 11.17. Of CH4, CHF3, CH3OH, glucose, and CH3COOH, only the last three have hydrogen atoms bonded to an electronegative FONCl atom, in this case, oxygen. Only the hydrogen atoms covalently bonded to the oxygen atoms could participate in hydrogen bonds. In CHF3 the hydrogen atom is not bound to the fluorine atom but rather to the nonFONCl (how about that for a word?) carbon atom. *11.19. Dear English Major: Ice floats because its structure contains large hexagonally shaped holes. Thus it contains much open space and is light. (A poem could be written about what it would be like to be in one of those open spaces!) When ice melts, water molecules break off the ice structure and wander into some of the holes. Therefore, dear EM, liquid water contains less open space and is heavier than ice. The lighter ice rises to float upon the heavier liquid. (Dear readers: I know you can do better than I did. I would love to see the various short paragraphs that you all come up with. You'll be famous; I might put your paragraph in the next version of this Solutions Manual! (If there is next version.) Submit your entries to the author at Department of Chemistry, Allegheny College, Meadville, PA 16335. 11.21. The boiling points of the Group 7A hydrides (HF, HCl, HBr, HI) should, in the absence of hydrogen-bond effects, gradually increase because the molecular weight (and therefore the number of electrons) increases in that direction. The more electrons there are, the greater the chance of instantaneous dipole-induced dipole (London) forces. The greater these London forces the higher temperature is needed to vaporize the substance. 11.23. Carbon dioxide, while non-polar, has individual bond dipole moments that can interact with the polar water. Therefore CO2 is reasonably water-soluble. Sulfur dioxide is a polar molecule and interacts even more with the polar water. Ammonia is a polar molecule that interacts with water molecules by strong (for intermolecular forces) hydrogen bonds. Ammonia is also roughly the same shape as a water molecule (with only an additional hydrogen atom) that fits nicely into the existing structure of liquid water. Therefore, ammonia is very soluble in water. *11.25. O2 is nonpolar and does not appreciably dissolve in polar water. The water molecules prefer-entially hydrogen-bond to each other and exclude the dioxygen molecules. (Although, of course, given the abundance of aquatic life, some of these molecules must dissolve in water.) NO, by virtue of the fact that the electronegativity of oxygen is slightly greater than that of nitrogen, is slightly polar and therefore more soluble in water. NO2 is a much more polar molecule possessing a significant permanent dipole moment. Therefore, it is significantly more soluble in liquid water. The structure of the resulting solution is shown below.

N

O O

H H

O O

HH

H H

O O

H

H OH

O

H

H

O

H

H

H

O

H H

H

O

H

free water

11.27. Glucose has 5 -OH groups and also an oxygen atom in the ring. These 6 polar groupings can interact with water molecules and therefore make glucose soluble. A linear hydrocarbon, on the other hand, is nonpolar. The water molecules will therefore prefer to hydrogen bond among themselves and exclude (and therefore not dissolve) this solute.

74

*11.29. 11.31. H2O + H2O H3O+(aq) + OH-(aq)

Co3+

O

O

O

O

O

H H

H

H

HH

HH

H

H

O

H

H

N

O

H

O O

H H

O

O O

H

H

H

OH

H

OH

H

1-

Kw = [H3O+(aq)][OH-(aq)] = x2 = 1.0 x 10-14

x = [OH-(aq)] = [H3O+(aq)] = 1.0 x 10-7

pH = - log[H3O+(aq)] = 7.00

11.33. In aqueous solution, the equilibrium among hydronium ions, hydroxide ions, and water molecules must always be satisfied. That is, Kw = [H3O+(aq)][OH-(aq)] must always be equal to 1.0 x 10-14. Therefore, if [H3O+(aq)] is raised, then it follows that [OH-(aq)] must be lowered such that Kw is still equal to 1.0 x 10-14. 11.35. Strong acids a high co

provide ncentration

ons,

ations)

1.37. Oxides: K2O, Ga2O3, As2O3 or As2O5, SeO2 or SeO3

1.39. The existence of S3 (if analogous to O3)

n llel ov

11.41. N2O3 + H2O ─→ 2HNO2 ─→ 2H (aq) + NO2 (aq)

O

H

HH

O

O

O

H

H

H

HH

H H

OH

= H11O5+

+

of H3O+, hydronium iin solution. These species are hydrated by other H2O groups as shown at right. If these hydrates are stableenough (and of highenough concentrthey may be isolated from such solutions. 1 basic amphoteric acidic but close acidic (along M/NM to borderline border) 1

S S S S S Sdepends on the strength of S=S double bonds. Sulfur atoms are considerably larger than oxygeatoms (re: uniqueness principle) and therefore para erlap among their p orbitals (so-called pπ-pπ overlap) is much less effective. Therefore, we are unlikely to find an S3 analog of ozone.

+ -*

75

O

N

O

N

O

H

O

H

δ−

δ+

δ

−

δ+ O

N

O

H

H

O

N

O

11.43. (One resonance structure of SO2 is used here for clarity.)

1.45.

1.47. A polar X-H bond (in which X is a highly electronegative atom) would readily interact with a olar heavy water molecule, D2O. In the course of this interaction, the H of the X-H bond could be xchanged with a deuterium atom. However, a non-polar C-H bond of a hydrocarbon is not polar enough be attracted to the a polar heavy water molecule and therefore cannot be deuterated.

1.49. Sulfuric acid the central sulfur atom. The SO2(OH)2 representation better portrays this situation.

1.51. (a) NO(OH) acidic (b) Be(OH)2 amphoteric (c) Ti(OH)4 basic

(c) H3PO3 > H3AsO3 due to the higher electronegativity of the phosphorus

1.55. (a) HClO2 HBrO2 > HBrO due to additional nonhydroxyl oxygen atom HClO2 > HBrO2 due to the more electronegative central atom

H2SeO3 due to the more electronegative central atom lO4 >> H2SO3 due to the more electronegative central atom and sO4 > H3AsO3 due to the extra nonhydroxyl oxygen atom

H3PO4 > H3AsO4 due to the more electronegative central atom

76

O

S

O

O

H

H

δ−

δ+

δ−

δ+S

O OH

O

H2OS

O OH

O_

+ H3O+

H

1 1peto 1 has two non-hydroxyl oxygen atoms as well as two hydroxyl groups (-OH) bound to 1 (d) Si(OH)4 acidic 11.53. (a) HClO3 > HIO3 due to the higher electronegativity of the chlorine (b) H3AsO4 > H3SbO4 due to the higher electronegativity of the arsenic (d) HSO4

- > HSeO4- due to the higher electronegativity of the sulfur

1 (b) HClO4 H2SO3 > HC (c) H3PO4 H3A 11.57. (a) HIO2 (b) HBrO4 (c) H2SO5

δ−

Sr2+ O

O

H Hδ+

Sr2+ + 2OH-2-

11.59. (a) HSeO3-

) Perchloric acid would b s to withdraw lectron density from its O-H bond. In addition, the electronegativity of chlorine is greater than that of dine also helping to make the O-H bond of HClO4 more polar and more susceptible to attack by water olecules. ) The acid anhydride of perchloric acid is dichlorine heptoxide: Cl2O7 + H2O ─→ 2 HClO4. The

cid anhydride of iodous acid is diiodine trioxide: I2O3 + H2O ─→ 2 HIO2.

1.63. HCl is a stronger hydroacid than H2S because while the H-Cl and H-S bonds are about the same ngth and strength, the H-Cl bond is the more polar and the more susceptible to attack by polar water olecules.

gth

de as the primary reenhouse gas? Have we actually started to slow the increase of the output of any of these gases to the

11.69. What articles did you find? The Kyoto protocol was a tentative agreement among industrialized

vidence that steps taken have actually reduced the level of greenhouse gases that the world releases to

(b) BrO2-

(c) H2AsO3-

(d) IO4-

*11.61.

77

O

O OO

O

O

Cl I

H H

H

perchloric acid iodous acid

δ−

δ−

δ+

δ+

δ−

δ+

(b e the stronger because it has two additional oxygen atomeiom(ca 1lem *11.65. Electronegativity differences are not the only effect in determining acid strengths. The lenand strength of H-X bonds are also very important. For example, H-F bonds are more polar than H-Cl bonds but the latter are significantly longer and weaker making HCl a stronger hydroacid than HF. 11.67. What articles did you find? Can anyone say that the greenhouse effect has actually started to lead to global warming? Have any other gases started to challenge carbon dioxigatmosphere? Has this action led to any type of economic slowdown in the United States or other developed countries? *countries to try to reduce the level of greenhouse gases released to the atmosphere. Is it still viable? What countries have actually ratified the protocol? What countries have not? What are the primary issues that the national governments cite as standing in the way of their ratification? Is there any ethe atmosphere?

78

Chapter 12 Group 1A: The Alkali Metals

The sections and subsections of this chapter are listed below. 12.1 Discovery and Isolation of the Elements 12.2 Fundamental Properties and the Network Hydrides, Oxides, Hydroxides, and Halides Application of the Uniqueness Principle and Diagonal Effect 12.3 Reduction Potentials and the Network 12.4 Peroxides and Superoxides Peroxides Superoxides 12.5 Reactions and Compounds of Practical Importance 12.6 Selected Topic in Depth: Metal-Ammonia Solutions

Chapter 12 Objectives

You should be able to

• briefly relate how and by whom the alkali metals were discovered • rationalize the trends in radii, ionization energies, electron affinities, and electronegativities of

the group • describe the general nature of the hydrides, oxides, hydroxides, and halides of the group • describe and represent the uniqueness of lithium in the group • describe and represent the diagonal relationship between lithium and magnesium • define and use the concepts of oxidation, reduction, oxidizing agent, reducing agent, and

standard reduction potential • describe the role of the standard hydrogen electrode in defining standard reduction potentials • describe and represent the relationship among standard reduction potential, change in free

energy, and the spontaneity of a reaction • predict the spontaneity of an oxidation-reduction reaction by analyzing its net reduction

potential • explain the trends in the reduction potentials of the alkali metals • explain why the lighter alkali metals are less reactive than the heavier • describe and represent the general preparation, structure, and uses of hydrogen peroxide • describe the occurrence and uses of superoxides • describe the application of various alkali metals as fertilizers, treatment for bipolar disorders,

reducing agents, battery components, and agents for chronometric procedures • rationalize the properties of alkali/alkaline earth metal liquid ammonia solutions in terms of

the solvated electron • describe and give several examples of electrides and alkalides

Solutions to Odd-Numbered Problems 12.1. Cavendish report of his discovery of hydrogen, 1766 Priestley published discovery of dephlogisticated air (oxygen), 1774 Lavoisier characterization of dephlogisticated air as oxygen, late 1770s Davy discovery of Group 1A and 2A elements, 1807-1808 Arfwedson discovery of lithium, 1817 Bunsen discovery (with Kirchhoff) of cesium and rubidium by spectroscopy, Mendeleev formulation of periodic table, 1869 *12.3. According to my dictionary, a voltaic pile, named after Alessandro Volta, is a device that produces current electricity, as distinguished from static electricity. So a voltaic pile is a name for one of the first types of battery. 12.5. Melting point (or freezing point) depression is a so-called colligative property, dependent only upon the number of solute particles present in a solution. Adding some CaCl2 provides other particles (ions, in this case) that make it more difficult to freeze out the "solvent," in this case sodium chloride. In order to freeze out the sodium chloride, a lower temperature must be achieved. Turning this argument around means that the melting point of NaCl is also lowered. 12.7. 223

87Fr —α→ 42He + 21985At

*12.9. In each of the plots (shown on the next page) the rate of increase (of radii) or decrease (of the other properties as plotted) slows down at rubidium. Rubidium is the first alkali metal after the filling of the first period of transition metals. The electrons in the filled 3d10 sub-shell do not shield the succeeding electrons as well from the effective nuclear charge. Rubidium therefore is a little smaller and has higher values of ionization energy and electronegativity, and absolute value of electron affinity than otherwise might be expected.

79

3.0

2.5

2.0

1.5

Li Na K Rb Cs Fr

atomic radii (A)

Li Na K Rb Cs Fr

ionzation energy (kJ/mol)

500

450

400

350

an oxidizing agent.

2.19. 2 [ Li(s) ─→ Li+(aq) + e- ] Ε° = +3.05v; ΔG° = -2(96.5kJ/v)(3.05v) = -589 kJ _____1 [ 2H

80

12.11. K2O(s) + H2O ─→ 2 KOH(aq) 12.13. Na+H-(s) + H2O ─→ Na+(aq) + OH-(aq) + H2(g) *12.15. Sodium nitride would be of greater ionic character than lithium nitride. Lithium has a greater charge density and polarizes a relatively large anion like N3- to produce a significant degree of covalent character. 12.17. LEORA stands for "loses electrons oxidized reducing agent." That is, if a substance loses electrons its oxidation state increases and we say it is oxidized. The electrons it loses go to another substance, which is in turn reduced. The substance that is oxidized then is serving as a reducing agent. GEROA stands for "gains electrons reduced oxidizing agent." In this case, the substance gains electrons, which causes a decrease in its oxidation state, and we say it is reduced. The electrons that it gains must come from another substance, which in turn is oxidized. The substance that is reduced then is serving as

1_ +(aq) + 2e- ─→ H2(g) ]____ Ε° = 0.00v; ΔG° = 0.00 kJ2 Li(s) + 2H+(aq) ─→ 2 Li+(aq) + H2(g) ΔG° = -589 kJ

E° = - ___-589kJ__ = 3.05V i = reducing agent; H+ = oxidizing agent (2)(96.5kJ/V)

12.21. Li+(aq) + e- ─→ Li(s) E° = -3.05V; ΔG° = -(1)(96.5kJ/v)(-3.05v) = 294 kJ Na+(aq) + e- ─→ Na(s) E° = -2.71V; ΔG° = -(1)(96.5kJ/v)(-2.71v) = 261 kJ

ased on these standard reduction potentials, it will be more difficult to electrolyze LiCl to lithium metal an it will be to electrolyze NaCl to sodium metal (under standard state conditions).

12.23. The heats of hydration of the gaseous alkali metal cations would be negative. The formation of n-dipole forces among the cations and water molecules would be an exothermic process. Because the

harge densities of the cations decrease going down the group, the heats of hydration would become less would be farther and farther away from the centers of charge

L * Bth *iocnegative in that direction. Water molecules

Li Na K Rb Cs Fr

elecron affinity (kJ/mol)

-30

0

0

-60

0

Li Na K Rb Cs Fr

electronegativity (Pauling)

0.4

0.6

0.8

1.0

1.2-7

-5

-4

81

and the interactions would be become weaker going down the group.

2.25. Hydrogen peroxide should be very soluble in water due to the formation of hydrogen bonds

ed on the ideas we have discussed under the uniqueness principle, the O-O bond distance in 2O2 is considerably shorter than the corresponding S-S bond distance in H2S2. This longer bond

1among the O-H groups and lone pairs of H2O2 and the water molecules. 12.27. BasHdistance lets the lone pairs on the two sulfur atoms rotate by each other freely so that there is no significant hindered rotation in dihydrogen disulfide. *12.29. 2 [ H2O2(aq) + 2H+(aq) + 2e- ─→ 2H2O(l) ] E° = +1.77V 1 [ 2H2O(l) ─→ O2(g) + 4H+(aq) + 4e- ] E° = - 1.23V 2H2O2(aq) ─→ O2(g) + 2H2O(l) E° = +0.54V 12.31. Half reactions under acid conditions:

Co ° = - 1.82V 2 [ Co2+(aq) ─→ 3+(aq) + e- ] E 1 [2H+(aq) + H2O2(aq) + 2e- ─→ 2H2O(l) ] E° = +1.77V 2Co2+(aq) + 2H+(aq) + H2O2(aq) ─→ 2Co3+(aq) + 2H2O(l) E° = - 0.05V

Under standard state conditions, H2O2 would not oxidize Co (aq) to Co3+(aq).

dium l, we need

Al (aq) + Na(s) ─→ Al(s) + Na (aq)

2+

12.33. In order to decide whether so metal could be used to reduce Al3+ to aluminum meta

to calculate the E° of the following equation:

3+ + 1 [Al3+(aq) + 3e- ─→ Al(s) ] E° = - 1.66V 3 [______Na(s) ─→ Na+(aq) + e-_]_______ E° = +2.71V Al3+(aq) + 3 Na(s) ─→ 3 Na+(aq) + Al(s) E° = +1.05V

3+ Yes, sodium could be used to reduce Al (aq) to Al(s).

the oxygen atoms in the molecule reduced from –1 to the –2 found in water.

*12.35. (a) When H2O2 acts as an oxidizing agent, the oxidation state of is (b) 1 [H2O2(aq) + 2H+(aq) + 2e- ─→ 2H2O(l) ] 1_[ 2I-(aq) ─→ I2(s) + 2e- ]__________________ H2O2(aq) + 2H+(aq) + 2I-(aq) ─→ 2H2O(l) + I2(s) 12.37. As shown below, the standard reduction potentials increase in the order oxygen, dichromate,

s t t, und te conditions, hydrogen is orst. All four, however,

Cr2O7 (aq) +1.33V nO4

-(

permanganate, and then hydrogen peroxide. This mean ha er standard staperoxide is the best oxidizing agent of these four while oxygen actually the ware excellent oxidizing agents. O2(g) +1.23V

2- M aq) +1.51V

H2O2(aq) +1.77V

-12.39. The oxidation state of each oxygen in O2 is -½.

2.41. When a person inhales oxygen from and exhales carbon dioxide and water vapor into breathing

net reac as does supplying carbon dioxide. The more CO2 pplied and the more O2(g) removed by the user, the more the equipment responds to his or her needs.

2.43. 2 [ Na + e ─→ Na(s) ] E° = -2.71V Note that a significant amount

1equipment based on potassium superoxide, the reaction is put into motion. Removing oxygen shifts the

tion, shown in Equation (12.21c), to the rightsu

+ -1 1 [___2Cl- ─→ Cl2(g) + 2e-______] E .36V° = -1 of energy is needed to make

° = -4 ction go. ° = -(2)(96.5kJ/V)(-4.07V) = 785 kJ

2.45. (a) TiCl4(s) + 4Na(s) ─→ Ti(s) + 4NaCl(s)

ucture from which argon gas cannot escape, the gas starts to ithin even ally be crushed in the

Knowing the half-life of e volcanic event. When

arious remains are found in such volcanic rock, they too can be linked to the time when the volcanic

elow, is not quite as large as 18-crown-6 and therefore might not as readily cat

in the through it.

2Na+ + 2Cl- ─→ 2Na(s) + Cl2(g) E .07V this rea ΔG1 (b) ZrCl4(s) + 4Na(s) ─→ Zr(s) + 4NaCl(s) 12.47. Potassium-argon dating is based on the beta plus decay of potassium 40 to argon 40 gas. When volcanic magma hardens to form a crystal straccumulate w the crystal. If the rock is left undisturbed, it can tulaboratory to release the argon gas that then can be monitored quantitatively. the beta plus decay, the amount of Ar 40 measured can be directly linked to thvrock formed. 12.49. 135

55Cs ─β-→ 0-1e + 13556Ba

12.51. 12-crown-4, shown baccommodate the large cesium ion.

82

CH2CH2 CH2

12.53. 2,2,2-crypt is just the right size to bind to the sodium cation. The larger Cs+ does not fitcage of this molecule while the smaller Li+ slips right

CH2CH2

CH2

CH

CH2

CH

O O

2

CH2

2

OO

CH2 CH2

83

Chapter 13 Group 2A: The Alkaline Earth Metals

The sections and subsections of this chapter are listed below. 13.1 Discovery and Isolation of the Elements 13.2 Fundamental Properties and the Network Hydrides, Oxides, Hydroxides, Halides Uniqueness of beryllium and diagonal relationship to aluminum 13.3 Reactions and Compounds of Practical Importance Beryllium Disease Radiochemical Uses Metallurgical Uses Fireworks and X-rays Hard water 13.4 Selected Topic in Depth: The Commercial Uses of Calcium Compounds CaCO3 (limestone) CaO (quicklime) and Ca(OH)2 (slaked lime)

84

Chapter 13 Objectives You should be able to

• explain the origin of the term "alkaline earth" • briefly relate how and by whom the alkaline earth metals were discovered • describe the process by which the Curies isolated radium • rationalize the trends in radii, ionization energies, electron affinities, and electronegativities of

the group • rationalize, on the basis of standard reduction potentials, why the alkaline earths are good

reducing agents • describe the general nature of the hydrides, oxides, hydroxides, and halides of the group • list and briefly rationalize the ways in which beryllium is different from its heavier congeners • list and briefly rationalize the ways in which beryllium is similar to aluminum • briefly describe the role of the alkaline earths in beryllium disease, neutron production, nuclear

fission products, metallurgy, fireworks, x-ray technology, and hard water • describe how limestone and marble structures are damaged by acid rain • briefly describe and represent the Solvay process • briefly describe and represent the use of calcium carbonate in smokestack scrubbers • briefly list some other uses of calcium carbonate • briefly describe the use of calcium oxide (quicklime) in the steel and glass industries and in

acetylene production • briefly describe the use of calcium hydroxide (slaked lime) as a mortar, for water treatment, in

the paper and pulp industries, and for bleaches • list several uses of calcium sulfate

Solutions to Odd-Numbered Problems 13.1. Limelight: a brilliant light created by the oxidation of lime and formerly used in theaters to throw an intense beam of light upon a particular part of the stage. (Limelight is actually the result of heating calcium oxide, CaO, to incandescence with a very hot hydrogen/oxygen or acetylene/oxygen torch. [See J. Chem. Educ., 64(4), 306 (1987) for further information.] 13.3. 3SrO(s) + 2Al(s) ─→ 3Sr(s) + Al2O3(s) 13.5. The Greek "barys" meaning "heavy" is the source of the name of baryta, the oxide of barium that has a high density. It would seem likely that the same Greek name was used to designate baryons, the heavier subnuclear particle compared to hadrons. *13.7. First we need to find out what weight 0.01% of seawater is. This is how much magnesium will have to be removed.

(1.5 x 109 km3) 3

kmm 1000⎟⎠⎞

⎜⎝⎛

3

mcm 100

⎟⎠⎞

⎜⎝⎛ ⎟

⎠⎞

⎜⎝⎛

seawater cm 1seawater g 1.025

3 ⎟⎟⎠

⎞⎜⎜⎝

⎛seawater g 100

Mg g 0.01 = 1.5 x 1020 g Mg

Now we can calculate how long it would take to consume the above weight of magnesium.

1.5 x 1020 g Mg ⎟⎟⎠

⎞⎜⎜⎝

⎛g 454

1b 1⎟⎠⎞

⎜⎝⎛

lb 2000 ton1

⎟⎟⎠

⎞⎜⎜⎝

⎛tons10 x 100

yr6 = 1.7 x 106 yrs, i.e., between 1 and 2 million

years 13.9. BeF2(s) + Mg(s) ─→ MgF2(s) + Be(s) ΔG° = ΔG°f [MgF2(s)] + ΔG°f [Be(s)] - ΔG°f [BeF2(s)] - ΔG°f [Mg(s)] ΔG° = -1070.2 + 0 - (-979.4) - 0 = -90.8 kJ/mol 13.11. 226

88Ra ─β-→ 0-1e + 22689Ac

13.13. 226

88Ra ─α→ 42He + 22286Rn

*13.15. FrCl(s) + Ag+(aq) ─→ AgCl(s) + Fr+(aq) 0.00476 g 0.00263 g

(0.00263 g AgCl) ⎟⎟⎠

⎞⎜⎜⎝

⎛AgCl g 143.32

AgCl mol 1⎟⎟⎠

⎞⎜⎜⎝

⎛AgCl mol 1FrCl mol 1

= 1.84 x 10-5 mol FrCl

Formula weight of FrCl = __0.00476 g__ = 259 g/mol 1.84 x 10-5 mol Atomic Weight of Fr = FW of FrCl - AW of Cl = 259 - 35.453 = 224 g/mol

85

13.17. No, barium has a very low ionization energy and therefore should readily be ionized when heated. 13.19. We need to see if calcium will reduce water. The half reactions will be as shown below. 1 [Ca(s) ─→ Ca2+(aq) +2e- ] E° = +2.87v; ΔG° = -2(96.5kJ/v)(+2.87) = -554 kJ 1 [2H2O + 2e- ─→ 2OH-(aq) + H2(g) ]______ E° = - 0.83v; ΔG° = -2(96.5kJ/v)(- 0.83) =+160 kJ Ca(s) + 2 H2O ─→ Ca2+(aq) +2OH-(aq) + H2(g) ΔG° = -394 kJ E° = - _(-394kJ) = 2.04V 2(96.5kJ/V) The reaction in which calcium reduces water is thermodynamically spontaneous under standard state conditions. *13.21.

ΔHhyd

Mn+(aq) + ne-

Mn+(g)

ΔHred

Sum of n IEsM(g)

M(s)

ΔHatom

In general, ΔHred = -(ΔHatom + Sum of n IEs + ΔHhyd) ΔHred[Li] = -[159 + 520 + (-519)] = - 160 kJ/mol E° = -3.05V ΔHred[Be] = -[324 + 899 + 1757 + (-2494)] = -486 kJ/mol E° = -1.85V As mentioned in the start of Section 13.2, the standard reduction potential for beryllium is not the most negative in its group as is lithium within the alkali metals. The reason for this difference, it was cited in the third paragraph [starting with "Presumably (here we go again)..."] of Section 13.2, is "because the energy required to ionize the beryllium to the +2 state is not fully compensated for by the energy released when the Be2+ is hydrated." Here we have an opportunity to test the above hypothesis. Note that it takes 2136 kJ [= (899+1757) - 520] more to ionize the two electrons of beryllium than the one of lithium. Further note that only 1975kJ [= 2494 - 964] more energy is released when Be2+ is hydrated than when Li+ is. So the above statement comparing the ionization energies with the energies of hydration has been shown to be true. *13.23.

3.25. Picture calcium hydroxide as containing two Ca-O-H units as shown below.

Ca2+ H

O

H2OCa2+(aq) + H2(g) + 2OH-(aq)

δ−

δ+

H H 2 H

1

86

δ+ δ- δ+ δ- δ+

H---O---Ca---O---H

he most polar bonds in this structure are those between the calcium and the oxygen atoms. The partial gen (as opposed to

etwee hyd n in a larger font. When this compound is placed in water, the lecu polar sites within the molecule, that is, the Ca-O bonds. The

alcium-oxygen bonds will be broken producing Ca2+ and OH- ions in solution. It follows that this

y of e Be cation so its separate existence as a free ion is ighly doubtful. That is, the Be is "really a formality."

more soluble in non-polar organic solvents than otherwise might be expected. This last prediction is, in arbon tetrachloride and ether. Being electron deficient,

luminum chloride will act as an electron-pair acceptor, that is, as a Lewis acid.

3.33. 13153I ─β-→ 0-1e + 131

54 Xe hile Sr 90 would replace calcium in bone structure and cause radiation damage, I 131 would go to the yroid where it would be used to make thyroxine.

3.35. ZrCl4 + 2Mg(s) ─Δ→ Zr(s) + 2MgCl2

Tcharges caused by the larger difference in electronegativity between calcium and oxyb n rogen and oxygen) are showpolar water mo les will attack the moreccompound will be basic and should be written as Ca(OH)2, calcium hydroxide, rather than H2CaO2, which might be referred to as "calcic acid." 13.27. According to my dictionary, formality is defined as "something done merely for form's sake." Referring to a separate Be2+ ion is something we do for form's sake. That is, we write this ion in thisform because its similar to that used to write other metallic cations. However, Be(II) compounds are highly covalent due to the high charge densit th 2+

2+h 13.29. Be(OH)2 + 2OH-(aq) ─→ Be(OH)4

2-

87

*13.31. As we know from the diagonal effect, the aluminum +3 cation will be much like the beryllium +2 cation. Both ions have large charge densities and will form more covalent compounds than we might otherwise expect. Therefore, we might predict that AlCl3 will not be as soluble in water as we would expect. (It turns out, however, that AlCl3 is quite water-soluble.) We also would predict that it will be

: O

H O Be O H

OH

:

: : : :

: :

Be

OO

O

O

H

H

H

All O-Be-O bond angles would be 109.5o;all H-O-Be bond angles would be somewhatless

H2-

H

han 109.5o

Be

OO

O

O

H

H

H

All Be-O bonds would be sp3-sp3;all O-H bonds would be sp3-s

H

t

fact, quite true as this compound is soluble in ca 1Wth 1

88

Na CO 2 Na+(aq) + CO32-(aq)

HfCl4 + 2Mg(s) ─Δ→ Hf(s) + 2MgCl2

3.37. CaCO3(s) + 2HCl(aq) ─→ H2O + CO2(g) + CaCl2(aq) 13.39.

3.41. CaCO3(s) + 2HCl(aq) ─→ CO2(g) + H2O(l) + CaCl2(aq)

13.45.

of water, 2x moles of Na+ and x moles of CO32-are

roduced. The solubility product, Ksp, will be as follows.

(aq)] [CO3 (aq)] = (2x) (x) = 4x3

x can be determined from the solubility: x =

1*

2 3

C O

O

H

O C

H O

CO2 + H2O + CO32-

δO O

δ+

−

δ+

δ−

1 13.43. NaHCO3(aq) + HCl(aq) ─→ H2O + CO2(g) + NaCl(aq) *

When x moles of sodium carbonate dissolve in 1Lp

+ 2 2- 2 Ksp = [Na

⎟⎠⎞

⎜⎝⎛

332

cm 100CONa g 7.1

⎟⎟⎠

⎞⎜⎜⎝

⎛g 106.0

mol⎟⎟⎠

⎞⎜⎜⎝

⎛mLcm3

⎟⎠⎞

⎜⎝⎛

LmL 1000 = 0.67M

Ksp = 4(0.67)3 = 1.2 13.47. Equations (13.22) are not redox. The oxidation states of the various elements do not change in

of these reactions. the course

89

Chapter 14 The Group 3A Elements

The sections and subsections of this chapter are listed below. 14.1 Discovery and Isolation of the Elements Boron Aluminum Gallium Indium and Thallium 14.2 Fundamental Properties and the Network Hydrides, Oxides, Hydroxides, and Halides 14.3 Structural Aspects of Boron Chemistry Allotropes Borides Borates 14.4 Aluminum, Gallium, Indium, and Thallium Aluminum Metal and Alloys Alum Alumina Gallium, Indium, and Thallium Compounds 14.5 Selected Topic in Depth: Electron-Deficient Compounds

Chapter 14 Objectives

You should be able to:

• briefly state how and by whom the Group 3A elements were discovered • rationalize the trends in radii, ionization energies, electron affinities, and electronegativities of

the group with particular emphasis on the anomalies in these properties shown by the heavier elements

• describe the general nature of the group oxides, hydroxides, and halides • list and briefly rationalize the ways in which boron is different from its heavier congeners • list and briefly rationalize how the chemistry of the heavier Group 3A congeners is organized

by the inert pair effect • describe the relationship between the general chemical nature of the borides and their

applications • describe the general molecular nature and a few of the applications of ortho-, meta-, and

perborates • list and briefly rationalize several applications of aluminum and its alloys • list and briefly rationalize several applications of alums and alumina • describe the safety and health hazards that accompanied the discovery of the borohydrides • describe the nature of three-center, two-electron bonds • draw and rationalize semi-topological diagrams of the simple boranes • represent and rationalize the preparations and major reactions of the boranes • describe and represent several electron-deficient compounds other than the boranes

90

Solutions to Odd-Numbered Problems 14.1. (a) Gay-Lussac and Thénard's preparation of boron: 6K(s) + B2O3(s) ─Δ → 2B(s) + 3K2O(s) (b) Wöhler's preparation of aluminum: AlCl3(s) + 3K(s) ─Δ → Al(s) + 3KCl(s) 14.3. Turkish samples: AW = 0.2030(10.01294) + (0.7970)(11.00931) = 10.807u California samples: AW = (0.1910)(10.01294) + (0.8090)(11.00931) = 10.819u 14.5. 204

81Tl ─β-→ 0-1e + 20482Pb

14.7. (a) Periodic law: (i) atomic radii increase whereas ionization energies, electronegativities, and (the magnitudes of) electron affinities generally decrease going down the group; (ii) all the elements have three valence electrons. (b) Uniqueness principle: (i) the B3+ cation does not exist as a separate entity – if it did, it would be very small and have a correspondingly high charge density. Because its polarizing power would be high, the +3 oxidation state is only found in covalent compounds. (ii) The electron affinity of boron is less than that of aluminum. This anomaly is attributed to the fact that boron is so small making it more difficult to add an electron to this atom than expected on the basis of the usual trends in size and effective nuclear charge. 14.9. Acid-base character: (i) B2O3 is an acidic anhydride whereas Tl2O is a basic anhydride. (ii) Aluminum hydroxide, Al(OH)3, is amphoteric acting as either an acid or a base depending on the circumstances. 14.11. We would expect the atomic radii to increase down the group because the electrons occupy larger and larger orbitals. The actual values show that the expected trend is observed. Thallium, however, does not have a very much larger radius than indium because its 6p electron experiences a slightly greater effective nuclear charge then expected due to the relative inability of the filled d and f subshells to shield it from the nuclear charge. 14.13. If boron trichloride were to be an ionic compound and therefore be written as B3+Cl-3, the highly polarizing (charge density = 12.0, a very large value) B3+ ion (that exists only in a formal sense) would severely distort the chloride ion resulting in a large degree of covalent character. In the final analysis, BCl3 is best thought of as a covalent compound. 14.15. Boron oxide reacts with water to produce boric acid that is characterized by 3 B-O-H units in which the O-H bonds are (just slightly) more polar than the B-O bonds. Although one might suspect that polar water molecules would attack at least one of these O-H bonds and split it to produce H+ and H2BO3

- ions, this turns out not to be the case. Instead B(OH)3 is an acid in the sense explained in the solution to Problem 14.14. Indium oxide, on the other hand, does react with water to produce an E-O-H unit in which the E-O bond is clearly more polar and more susceptible to attack by water molecules. Therefore, the In-O-H unit yields hydroxide ions in aqueous solution and indium oxide is considered to be basic. 14.17. 2Al(s) + 3Cl2(g) ─→ 2AlCl3(s) Al(OH)3(aq) + 3HCl(aq) ─→ AlCl3⋅ 3H2O(s)

14.19. In BF4-, the boron is sp3 hybridized and there is no chance for pπ interactions. In BF3, however,

there is an opportunity for the filled p orbitals of the fluorine to donate

B

F

FF

F

sp3-p

sp3-psp3-p

p-sp3

-

F Bp-sp2

2p 2psp2-p

sp2-pF

F

2p

2p

electron density to the unfilled, unhybridized p orbital of the boron. Such a pπ-pπ interaction makes the B-F bond stronger and s horter. 14.21. This diagram is shown in Figure 14.6. The carbons and boron are sp3-hybridized while the fluorines use either sp3 hybrids or just native p orbitals. The hydrogens, of course, use 1s orbitals. 14.23. Thallium is a metal and therefore does not tend to form the covalent bonds that characterize the M2X6 structures. In addition, as per the inert pair effect, thallium usually forms compounds of general formula TlX. 14.25. In order to determine the ΔG° of Tl3+(aq) + Tl(s) going to Tl+(aq), we need to use the various half equations and accompanying standard reduction potentials that are available to us. 1 [Tl3+(aq) + 2e- ─→ Tl+(aq) ] E° = +1.25V; ΔG° = -2(96.5)(1.25) = - 241 kJ 2 [Tl(s) ─→ Tl+(aq) + e- ] E° = +0.33V; ΔG° = -2(96.5)(0.33) = - 64 kJ Tl3+(aq) + 2 Tl(s) ─→ 3Tl+(aq) ΔG° = - 305kJ Yes, this reaction would be spontaneous. Tl+(aq) is more stable (under standard state conditions) than is Tl3+(aq). 14.27. (a) Tl(s) + 2H+(aq) ─→ Tl+(aq) + H2(g) 2 [Tl(s) ─→ Tl+(aq) + e- ] E° = +0.33V; ΔG° = -(2)(96.5)(0.33) = -64 kJ 1 [2H+(aq) + 2e- ─→ H2(g)] E° = 0.00V; ΔG° = 0.00 2Tl(s) + 2H+(aq) ─→ 2Tl+(aq) + H2(g) E° = 0.33V; ΔG° = -64 kJ (b) 2Tl(s) + 6H+(aq) ─→ 2Tl3+(aq) + 3H2(g) 2 [Tl(s) ─→ Tl3+(aq) + 3e-] E° = -0.72V; ΔG° =-(6)(96.5)(-0.72) = 417 kJ 3 [ 2H+(aq) + 2e- ─→ H2(g)] E° = 0.00V; ΔG° = 0.00 2Tl(s) + 6H+(aq) ─→ 2Tl3+(aq) + 3H2(g) E° = -0.72V; ΔG° = 417 kJ Note that the reaction to Tl+(aq) is spontaneous (E° = 0.33v) but that to Tl3+(aq) is not (E° = -0.72V). 14.29. There are 4 B12 [ = 8(1/8) +6(1/2)] in the unit cell. (Note that the diagram, for purposes of clarity, does not show the B12 units in the back faces or the back left corner of the cube.) There are also 4 Sc atoms [ = 12(1/4) + 1] in the unit cell. (Again, the diagram omits the M's found in back edges and in the

91

center of the cube.) 14.31.

B

O O

O

B

B

O

O

3-

B

O B

O

O

O

2-O The hybridization of the boron atoms is sp2. All O-B-O bond angles are 120°. B-O-B bond angles are somewhat less than 109.5°. 14.33. The diagrams for these two anions are found in Figure 14.13 (d) and (e). In perborate, the O-B-O angles are all 109.5° while all the others are somewhat less than 109.5°. Both the oxygen and boron atoms are sp3 hybridized. In borax, the boron atoms at the top and the bottom of the figure have O-B-O angles of 109.5° and are sp3 hybridized. The borons on the left and right of the figure have O-B-O angles of 120° and are sp2 hybridized. All the oxygen atoms are sp3 hybridized. The H-O-B bond angles are somewhat less 109.5°. 14.35. Pure aluminum metal is not very strong structurally so it cannot be used for such things as aircraft frames or automobile engines. A copper alloy is used for aircraft while a Cu/Si alloy is used for automobile engines. *14.37. Al(s) ─→ Al3+(aq) + 3e- E° = +1.66V ΔG° = -(3)(96.5kJ/v)(1.66v) = -481 kJ Fe(s) ─→ Fe2+(aq) + 2e- E° = +0.44V ΔG° = -(2)(96.5kJ/v)(0.44v) = -85 kJ From the E°'s and ΔG°'s, we see that under standard state conditions, it is easier to oxidize aluminum than it is iron. 14.39. When AlCl3 is dissolved in water, the hydrated Al(H2O)6

3+ cation is formed. When four equivalents of OH- are added, each one removes a hydrogen ion from a water attached to the Al3+ ion to produce [Al(OH)4 (H2O)2]-. No Al-O bonds are broken during this process.

92

14.41. Given the structure of Al2Cl6 and the multiple

M

Cl

O

M

HO OH

HOH

OH

lone pairs on a hydroxide ion, the structure given at right might be a reasonable one for Al2(OH)5Cl⋅ 2H2O. The two waters of hydration might be hydrogen-bonded to the OH groups. *14.43. 2Al(s) + Fe2O3(s) ─→ Al2O3(s) + 2Fe(s) ΔH° = ΔHf°[Al2O3(s)] + 2ΔHf°[Fe(s)] - 2ΔHf°[Al(s)] - ΔHf°[Fe2O3(s)] = -1675.7 + 2(0) - 2(0) - (-822.2) = -853.5 kJ/mol 2Al(s) + Cr2O3(s) ─→ Al2O3(s) + 2Cr(s) ΔH° = ΔHf°[Al2O3(s)] + 2ΔHf°[Cr(s)] - 2ΔHf°[Al(s)] - ΔHf°[Cr2O3(s)] = -1675.7 + 2(0) - 2(0) - (-1128.4) = -547.3 kJ/mol 14.45. B6H10 -- consult Figure 14.15(e) for structure

B

B

B

BB

B

H HH

H H

HH

H H

H

Available electrons: Distribution of electrons: 6B x 3e- = 18e- 6 B-H = 12e-

10H x 1e- = 10e- 5 B-H-B = 10e-

28e- 2 BBB = 4e-

1 B-B =_ 2e-

28e-

HH

H

H

B

H HB B

+14.47. B3H6+ cation

Available electrons: Distribution of electrons: 3B x 3e- = 9e- 3 B-H = 6e- Each boron is 6H x 1e- = 6e- 3 B-H-B = 6e- sp3 hybridized. (+) = -1e- 1 BBB = _2e-

14e- 14e-

14.49. In going from B6H6

2- to B5H9 to B4H10 a boron atom is removed from a closo octahedron to form a nido square pyramid and then another is removed to produce the arachno structure. 14.51. One of the easiest ways to make diborane in the laboratory is to react boron trifluoride with sodium hydride. 8BF3 + 6NaH ─→ 6NaBF4 + B2H6 This is quite different from the method that Stock used which was to react magnesium boride with hydrochloric acid to produce a mixture of various boranes. After laboriously separating the highly

93

reactive boranes from each other, B4H10 was gently heated to decompose it to diborane. *14.53.

HH

H

H

Be

H

B BH

H H

HH

H

H

Be

H HB B

H

H

94

*14.55. B5H8-

Available electrons: Distribution of electrons: 5B x 3e- = 15e- 5 B-H = 10e- 8H x 1e- = 8e- 3 B-H-B = 6e-

(-) 1e- 3 B-B = 6e-

24e- 1 BBB = 2e-

24e- 20e-

H

HH H

B B

B

BBH

HH

H

_

There would be two additional resonance structures. 14.57. The carbon and boron atoms use sp3 hybrids. The inteto 95

rior H-B-H bond angles are probably close o as in diborane. The exterior H (or CH3)-B-H angles are probably greater than 109.5o.

H3C CH CHH3H

HH H

BB

3H

HH

BB

H3C

CH3

H

HH

H

BB

CH3

14.59.

general, the bonding in Al2(CH3)6 is similar to that in diborane with aluminum atoms replacing borons

H3C CH3

CH3C CH3

AlAl

H3C

H3AlAl

C

H

H H

Inand methyl groups replacing hydrogen atoms. All the carbon and aluminum atoms are sp3 hybridized. The aluminum-carbon-aluminum bond is a 3c-2e bond with the carbon and one of the aluminums providing the electrons (as shown by shading in the figure).

95

96

Chapter 15 The Group 4A Elements

The sections and subsections of this chapter are listed below. 15.1 Discovery and Isolation of the Elements Carbon, Tin, and Lead Silicon Germanium 15.2 Fundamental Properties and the Network Hydrides Oxides and Hydroxides Halides

15.3 An Eighth Component of the Interconnected Network: pπ-dπ Bonding involving Elements of the Second and Third Periods

15.4 Reactions and Compounds of Practical Importance Diamond, Graphite and the Graphenes Tin Disease Radiochemical Uses Carbon Compounds Lead Compounds and Toxicology 15.5 Silicates, Silica, and Aluminosilicates Silicates and Silica Aluminosilicates 15.6 Selected Topics in Depth: Semiconductors and Glass Semiconductors Glass

97

Chapter 15 Objectives You should be able to:

• briefly describe the form in which some of the Group 4A elements were known to the ancients • briefly relate how and by whom the remaining Group 4A elements were discovered • outline the general manner in which the periodic law, uniqueness principle, inert pair effect, and

the acid-base character of oxides are applicable to this group • explain why catenation is much more prevalent in the carbon hydrides than it is in those of the

heavier congeners, particularly silicon • compare and contrast the incidence of pπ-pπ bonding in carbon compounds with pπ-dπ bonding

in those of silicon • compare and contrast the bonding in trimethyl- and trisilylamines • describe why pπ-dπ bonding becomes important in Group 4A chemistry and will become more so

going farther right in the third period • describe the general nature of the group oxides, hydroxides, and halides • compare and contrast the major allotropes of carbon • describe the discovery and the modes of preparation of the fullerenes and nanotubes • give some examples of elements that have been incorporated into fullerenes of various sizes • list three types of reactions that the fullerenes undergo and give an example of each • describe the structures, properties, and applications of the nanotubes • briefly describe the role of the Group 4A elements in tin disease, radiochemical chronometry, and

tracing methods • describe several applications of carbon oxides; ionic, covalent, and interstitial carbides; carbon

disulfide; and cyanides • discuss some major applications of lead compounds including the storage battery • discuss the general toxicology of lead compounds • list, represent, and discuss the variety of silicates including the relationship of submicroscopic

structures to macroscopic properties • describe the structure and uses of aluminosilicates • describe the nature of intrinsic, n-type, and p-type Group 4A semiconductors • represent and describe the preparation of semiconductor-grade silicon • describe the general nature of common, quartz, laboratory-grade, crystal, colored, photochromic,

and fiber glasses

Solutions to Odd-Numbered Problems 15.1. SnO2(cassiterite) + 2C(s) ─→ Sn(l) + 2CO(g) 15.3. SiO2(s) + 4HF(aq) ─→ SiF4(g) + 4H2O(l) 15.5. Amorphous (re: Webster's New Universal Unabridged Dictionary, Deluxe 2nd Edition, Simon & Schuster, NY, 1983) means having no regular structure, noncrystalline; formless, characterless, unorganized, vague. This definition implies that Berzelius isolated an impure, poorly characterized form of silicon. 15.7. Going down the group, the ionic radii increase fairly regularly except for lead. Here's another example in which the electrons in filled d and f subshells do not shield valence electrons from the nuclear charge as effectively as electrons in other subshells do. 15.9. The ionization energies, like the ionic radii and electronegativities, behave fairly regularly going down the group. For the most part, it gets progressively easier to ionize the valence electrons of the Group 4A elements. As usual, the exception comes in the heavier elements, particularly lead in this case. Once again, we see that the valence electrons must experience a greater than expected effective nuclear charge due to the relative inability of electrons in filled d and f subshells to shield succeeding electrons from the nuclear charge. 15.11. LiH, while not as saline as the other hydrides in its group, is still of significant ionic character. BeH2 is a polymeric, electron-deficient hydride showing significant covalent character while B2H6 is the most famous borane also with largely (but atypical) covalent character. With methane, CH4, we reach an almost fully covalent hydride characterized by good ole normal two-center, two-electron bonds. 15.13. Given the elaborate element-to-element bonds in the allotropes of boron, it is most certainly appropriate to include boron as an element that shows catenation. 15.15. Structures for tetrasilane and pentasilane

(a) H Si Si Si Si H

H H H H

H H H H

H Si Si

H H H

H SiH

HSi

H H

H

98

(b) H Si Si Si Si H

H H H H

H H H H

H Si Si

H H H

HSi

H

HSi

H H

H

Si

H

H

H Si Si

HSi

H

H SiH

Si

H H

H

H

H

H H

15.17. The alcohol can react more readily with silane because the oxygen of the alcohol can donate one of its lone pairs into one of the empty d orbitals of the silicon atom and readily form a low- energy transition state. Since no low-energy d orbitals exist in the carbon of methane, this mode of attack is not open and methane reacts less readily. 15.19.

H Ge Ge Ge Ge Ge Ge H

H H H H H H

H H H H H H

H Ge Ge Ge Ge Ge

H H H H H

H H H H

H

GeH H

H Ge Ge Ge Ge

H H H

H H HGeH H

GeH H

H

H

H

H Ge Ge Ge Ge

H H

H H HGeH H

GeH H

H

H

H

H

H Ge Ge Ge Ge Ge

H H H H H

H H H

H

GeH HH

H

H

99

100

15.21. Ionic character of Li, Be, B, C oxides

Li2O BeO B2O3 CO2 ───→ direction of increasing covalent character v. ionic intermediate predominantly covalent

basic amphoteric weakly acidic acid-base character anhydride anhydrides anhydrides helps characterize oxides

5.23. The double bonds found in the C3O2 suboxide are made possible by pπ-pπ bonds between the arbon and oxygen atoms. Due to ineffective overlap, the larger elements beyond the second period do ot form stable pπ-pπ bonds. Therefore a silicon analog of C3O2 does not exist.

5.25. The direct combination of elemental germanium and tin with chlorine yields the tetrachlorides. ichlorides.

(a) GeCl4 + Ge ─→ 2GeCl2

5.27. The carbon halides are not electron deficient as are most of the Group 3A halides. Therefore, the oct of ele ntral

.

t of

36V; ΔG° = -(2)(96.5kJ/v)(-1.36v) = 262kJ →

1cn 1Other measures must be taken to produce the d Ge + 2Cl2 ─→ GeCl4 (b) Sn + HCl(g) ─→ SnCl2 + H2 Sn + 2Cl2 ─→ SnCl4 1carbon halides need not form bridged structures in order to achieve an et ctrons around the ceatom 15.29. PbCl2 would be more soluble as it is the more ionic. The charge density of Pb2+ is less than thaPb4+ giving the latter a greater polarizing power and an increased ability to form more covalent compounds. 15.31. 1 [2Cl-(aq) ─→ Cl2(g) + 2e- ] E° = - 1. 1 [PbO2(s) + 4H+(aq) + 2e- ─_______________ Pb2+(aq) + 2H2O(l) ] E° = +1.46V; ΔG° = -(2)(96.5kJ/v)(1.46v) = -282kJ Cl

q) + 2H2O E° = +0.10V; ΔG° = -20kJ

The phosphorus atom is sp hybridized. Each of these four ybrid orbitals is used to form a single bond with an oxygen atom.

ms a second bond with the phosphorus atom. This 3d orbital of phosphorus and a 2p

f oxygen as shown at right. (The signs are the phases of the orbitals.)

2 -(aq) + PbO2(s) + 4H+(aq) ─→ Cl2(g) + Pb2+(a This oxidation is possible under standard state conditions. 15.33. 3

hThree of these oxygen atoms are, in turn, singly bonded to a hydrogen atom. The fourth oxygen atom does not have a hydrogen bonded to it but rather forsecond bond is formed using aorbital o

P O+_

3d 2p

+ _

+

_

15.35. The chlorine atom is sp3 hybridized. Each of these four

101

hybrid orbitals is used to form a single bond with an oxygen atom.

tom. The other three oxygen atoms do not have hydrogens rine atom.

his second bond is formed using a 3d orbital of chlorine and a 2p

trical co hite is possible because of the mobility of

.

5.39. The fullerenes can be described as graphite sheets that are formed into spheres. They still have pπ t nslate into d calized pi electron

presented by MitsuhoYoshida, the largest smaller?

ly, 2001)

NTs are exceptionally strong materials while still being very light weight. Any application here this combination of properties is important may well employ SWNTs. Such applications would

;

eight, they may be used a new generation of transmission power lines.

he 55.8ºF. Gray tin has a diamond structure but

e single bonds holding it together are both longer and weaker and therefore it tends to crumble quite

in disease has been en (in addition to the

arassment by defending Russian troops and their campaign of devastating the land as they retreated) the in

y C

++_

Cl O

3d 2p

+ __

One of these oxygen atoms is, in turn, singly bonded to a hydrogen abonded to them but rather form a second bond with the chloTorbital of oxygen as shown at right. (The signs are the phases of theorbitals.) 15.37. Charcoal is composed of very small pieces of graphite. The elec nductivity of graprelies upon keeping its horizontal planes intact. (The conductivitypπ electrons delocalized above and below a plane of carbon atoms in graphite.) If these planes are broken into very small bits, the substance loses its ability to conduct electricity 1electrons delocalized above and below the plane, a situation that ra s loedensity both inside and outside the C60 sphere, for example. 15.41. On the website entitled “Fullerene Structure Library,” fullerene listed is C100 and the smallest is C20. Did you find any larger orhttp://shachi.cochem2.tutkie.tut.ac.jp/Fuller/fsl/fsl.html (accessed Ju 15.43. SWwinclude aircraft, space craft, automobile and truck parts (SWNTs in a new generation of NASA Space Shuttles); fibers, ropes and cables (a cosmic funiculaire is only the most spectacular of such applications)bridge supports allowing longer spans then present day materials; and, of course, sports equipment! SWNTs can also be excellent electrical conductors. Combined with their lightwin 15.45. Tin disease is characterized by metallic (or white or β-Sn) changing to gray (or α-Sn) tin when tmetal is exposed to prolonged periods of temperatures belowtheasily. When metallic tin materials change to gray tin, they lose their metallic qualities. Tin organ pipes in unheated European cathedrals, churches and priories are a good example where tdevastating. Tin disease may have changed the course of European history whhtin buttons on the uniforms of Napoleon’s army crumbled away as he drove his troops toward Moscow the unusually cold winter of 1812. 15.47. One reason is that rocks do not take in carbon. Therefore there is no steady state concentration or ratio of C 14 to C 12 in rocks. The second reason is that rocks are generally too old to be measured b14 dating. The half-life of C 14 is only about 5730 years, not a very long time compared to the age of rocks, which is often in the millions of years. 15.49. Now we can see that radium will be found with uranium of which it is a radioactive daughter and not necessarily with its congener barium.

*15.51. Each source of lead is the product of a different combination of three radioactive series. The leimpurities derived from the combustion of coal will have a different set of isotopes than that set from a smelter. A set of isotopes is, in essence, the signature of that particular source of lead. The set of lead isotopes used in a smelter in Missouri is very likely to have a different signature than that of the lead in a Californian smelter. 15.53. “Supra” was a sweetener po

ad

used

pular in Roman times. It was prepared by boiling down sour wines in ad pans. The widespread use of this Roman sugar may have been responsible for spreading a low level

. n parallels of supra. Improperly glazed pottery that releases

ad into various liquids, particularly those with a mildly acidic character is a direct parallel. Children

lPO4 is SiO2

uartz) with half of the silicon atoms replaced by aluminum and the other half replaced by phosphorus. It stands to reason then that aluminum phosphate forms in quartzlike structures.

leof lead poisoning throughout the Roman ruling class. The early signs of lead poisoning include anemia, fatigue, headache, irritability, and psychological problems. Greater neurological damage can result from higher levels of lead. Leaders with such symptoms would show less sound judgment than healthy onesLead in paint and pottery could both be moderlewho eat lead-containing paint chips is also quite similar, particularly because these chips are said to have asweet taste. 15.55.

102

15.57. The equation representing the overall reaction occurring in a lead storage battery shows sulfuric acid as a reactant. If the battery had been used extensively and needed charging, a significant amount of H2SO4 would have been consumed and the density of the electrolyte solution would be low. *15.59.

Cl C C

OO

o

Op - p sp2 - sp2

l C>120

C

Cl Cl<120o Cl Clsp3- sp2 sp2 - sp3

*15.61. The AlP pair is isoelectronic with two silicon atoms. If we reformulate AlPO4 as AlO2PO2 we seethat AlPO4 is isoelectronic with two SiO2 units. Another way to look at this situation is that A(q

103

*15.63. Dear Aunt Emily,

So, as I was saying, aluminosilicates are made up of aluminum, silicon, and oxygen atoms. They rm structures with atomic- or molecule-sized holes in them. These structures also contain more

lectrons than protons and therefore are negatively charged molecules, or what my professor calls polyatomic anions.” I know you’ve heard the term “hard water” but many people think this means it is mehow literally harder than regular water. However, hard water is just an expression meaning regular ater that contains significant amounts of calcium, magnesium, and iron ions (atoms that have lost one or ore electrons to form positively charged “cations”). If not neutralized in some way, these hard water

rbonate is also a water ftener. Easily dissolved in water, it forms shiny crystalline precipitates with hard water ions and so

revents them from interfering with soaps and detergents. (The precipitates containing the hard water ns simply get thrown out with the wash water.) Some detergent manufacturers use only “suspended

ed in the sense that it is uniformly scattered throughout the product) to soften ater whereas others have been starting to use both sodium carbonate and aluminosilicates. It’s difficult

le-hard

milarly,

e

foe“sowmions can interfere with the cleansing action of soaps and detergents. (They also form a gummy solid with hand and bath soaps – this is the stuff that “bathtub rings” are made up of.) Anyway, when the detergent is added to hard water, the calcium, magnesium, or irons ions are attracted into the holes of the negatively-charged aluminosilicates and we say the water is “softer”. Sodium casopiosodium carbonate” (suspendwto know for sure, but perhaps using the combination of the two ingredients makes for an even better detergent because there are then two ways to neutralize the effect of the hard water ions. This doubbarreled action would be particularly important in geographic areas where the water is particularly and the consumer does not have a water softener installed in the house. Anyway, I hope you enjoyed andunderstood my explanation of how aluminosilicates and sodium carbonate work in detergents. Thanks forasking about how things are going in my inorganic chemistry course. 15.65. Gallium arsenide, GaAs, is isoelectronic with germanium and therefore would also serve as an intrinsic semiconductor. We can prepare n-type semiconductors from GaAs by doping it with atoms suchas germanium or selenium that carry one more electron than gallium or arsenic, respectively. Sip-type semiconductors are prepared by doping GaAs with atoms such as zinc or germanium that carry onless electron than gallium or arsenic, respectively. 15.67. When sodium carbonate or limestone (CaCO3) is heated (as they would be in the preparation ofglass) they liberate carbon dioxide gas and sodium oxide and calcium oxide, respectively.

104

Chapter 16 Group 5A: The Pnicogens

The sections and subsections of this chapter are listed below. 16.1 Discovery and Isolation of the Elements Antimony and Arsenic Phosphorus Bismuth Nitrogen 16.2 Fundamental Properties and the Network Uniqueness Principle pπ-dπ Bonding involving Elements of the Second and Third Periods Other Network Components Hydrides Oxides and Hydroxides Halides 16.3 A Survey of Nitrogen Oxidation States Nitrogen (-3) Compounds: Nitrides and Ammonia Nitrogen (-2), Hydrazine, N2H4 Nitrogen (-1), Hydroxylamine, NH2OH Nitrogen (+1), Nitrous Oxide, N2O Nitrogen (+2), Nitric Oxide, NO Nitrogen (+3), Dinitrogen Trioxide, N2O3, and Nitrous Acid, HNO2 Nitrogen (+4), Nitrogen Dioxide, NO2 Nitrogen (+5), Dinitrogen Pentoxide, N2O5, and Nitric Acid, HNO3 16.4 Reactions and Compounds of Practical Importance Nitrogen Fixation Nitrates and Nitrites Nitrogen Air Bags Matches and Phossy Jaw Phosphates 16.5 Selected Topic in Depth: Photochemical Smog

105

Chapter 16 Objectives You should be able to

• briefly describe the relationship between some of the Group 5A elements and the ancient practice of alchemy

• briefly relate how and by whom the Group 5A elements were discovered • outline the general manner in which the periodic law, uniqueness principle, inert pair effect, and

the acid/base character of oxides are applicable to this group • explain how the phosphazenes summarize the relative π-bonding abilities of nitrogen and

phosphorus • compare and contrast the hydrides of the nitrogen, phosphorus and arsenic • discuss the role of pπ-dπ bonding in phosphorus chemistry • represent the structural similarities among P4, P4O6, and P4O10 • discuss and represent the structures of phosphoric acid, ortho-phosphate, and the various

condensed phosphates • discuss and represent the structures of phosphorous acid, ortho-phosphites and the various

condensed phosphites • discuss the structures, preparations, and properties of the phosphorus halides • describe and draw the structure of borazine • briefly describe the oxidation state, structure, properties, history, preparations and reactions of

1) ammonia 2) hydrazine 3) nitrous oxide 4) nitric oxide 5) nitrous acid and nitrites 6) nitrogen dioxide 7) nitric acid and nitrates

• briefly discuss three methods of nitrogen fixation • explain why nitrogen compounds are often components of high explosives; give three examples • briefly explain how automobile air bags work • briefly relate the history and hazards of matchmaking • briefly explain the role of phosphates in fertilizers, food processing, soft drink formulations, and

dental hygiene • compare and contrast the general causes and characteristics of London and photochemical smog • describe the daily production and abatement of both the primary and secondary pollutants that

constitute photochemical smog

Solutions to Odd-Numbered Problems 16.1. The name pnicogen means "choking producer." Obviously if we breathed in just nitrogen and not "common air" we would choke. If phosphorus is burned in air, it consumes the oxygen and makes the air a "choker." The same could be said of Priestley's test for the goodness of air, reacting it with nitric oxide. The result would be an "air" that would serve as a choker. Arsenic, the famous poison, and its compounds could also be loosely viewed as chokers as they cut off life. 16.3. As2S3(s) + 3CaCO3(s) + 6O2(g) ─Δ → As2O3(s) + 3CaSO4(s) + 3CO2(g) 2As2O3(s) + 3C(s) ─→ 4As(s) + 3CO2(g) *16.5 The mouse consumed a good portion of the oxygen (dephlogisticated air) in the air by respiration. Burning a candle and phosphorus would consume any left over oxygen. A strong alkali (potassium or sodium hydroxide) would react with the carbon dioxide originally in the air plus that generated by the mouse and the burning candle. Some of these processes are represented by the following equations: C(s) + O2(g) ─→ CO2(g) P4(s) + O2(g) ─→ P4O6(s) (oxygen is the limiting reagent) 2NaOH(s) + CO2(g) ─→ Na2CO3(s) + H2O(l)

106

16.7. The N-N bond is weaker

N NH

H

H

H<109.5o

C CH

H

H

H

<109.5oH H

due to the close proximity of the lone pairs on the two nitrogen atoms. 16.9. The nitrogen of N(SiH3)3 has a filled 2p orbital that can be used to overlap the empty 3d orbitals of the silicon atom. The resulting pπ-dπ bond of this planar molecule is shown in Figure 15.4. The filled 3p orbital of the phosphorus atom of P(SiH3)3 is not large enough to form an effective pπ-dπ bond with the silicon atoms. While the 3d orbital of the phosphorus might be able to stretch to form a dπ-dπ bond with the silicon, neither of these 3d orbitals has any electrons in them so no π bond can form. 16.11. There are two possible resonance structures for the isocyanate ion. (I) is favored as it places a -1 formal charge on oxygen, the most electronegative atom. Analogous phosphorus compounds are not possible as this element is not capable of pπ-pπ bonding.

C NO

(-) -C NO

(-) -O C N O C N

180oI II

sp3 - sp sp - sp

px - pxpy - py

16.13. The arsenic atom is sp3 hybridized using its 4s and p orbitals. Each hybrid orbital overlaps with oxygen rbitals (sp3 for the OH oxygens and sp2 for the non- ydroxyl oxygen) to form a single bond. The shorter s-O bond is a double bond where the second pπ-dπ

d ed 2p orbital of the oxygen.

beconia.

4): than arsenic and therefore

e P would draw more electron density from the O-H bond then would the As. The O-H bond of H3PO4 and H3PO4 would be the stronger acid.

(b) (NH4)H2PO4

ydrogen atom bound to an oxygen tom is acidic. The P-H bonds are nonpolar and not scepti water.

6.25.

107

4ohAbond is formed using the empty 4d orbital of arsenic ana fill 16.15. Of NH3 and PH3, phosphine is the stronger acid ause the P-H

O

2p orbital

4d orbital

sp3 hybridized

pπ-dπbond As

HO

HO

OH

bonds are longer and weaker than the N-H bonds of amm 16.17. Phosphoric acid (H3PO4) versus arsenic acid (H3AsOPhosphorus is more electronegative thwould be more susceptible to attack by water molecules 16.19. (a) Na2HPO4 (c) K2H2P2O7 (d) CaHPO4 *16.21. H4P2O6, three acidic hydrogens

O

16.23. Only the one h P

O

H

H

Hasu ble to attack by 1

16.27. (a) H3SbO3 (b) CuHAsO3 (c) NH4H2AsO4 16.29. NH3(g) + F2(g) ─Cu→ NF3(g) + 3HF(g)

O As H

O

H

O

H

As

O

O

H

a

OH H

O

rsenic acid arsenious acid

16.31. E2O3(s) + 6HF(g) ─→ 2EF3(s) + 3H2O → 2EX3(s)

where E = As or Sb; X = Cl, Br, or I

16.33. Borazine is shown in Figure 16.7 while the trimeric phophazene is shown in Figure 16.3(a). he similarities include (a) alternating X (= B or P) and N in a 6-membered ring, (b) both have elocalized X-N bonds, (c) nitrogen is sp2 hybridized in both compounds, and (d) all the π electrons in

differences include (a) two chlorine atoms are 6 but there is one each on each of the three boron and nitrogen

toms in

2E + X2(g,l,s) ─ *Tdboth structures come from the 3 nitrogen atoms. Thebonded to each phosphorus in P3N3Cla BB bridized in its compound but the boron is sp2 hybridized

the borazine derivative, and (c) the phosphorus uses its d orbitals for π bonding while the boron uses p

6.37. (NH4)2Cr2O7(s) ─→ N2(g) + 4H2O(g) + Cr2O3(s)

he nitrogen is oxidized from -3 to 0; the chromium is reduced from +6 to +3.

16.39. (a) 1 [ N2H5+(aq) ─→ N2(g) + 5H+(aq) + 4e- ] E° = +0.23V

_________ 4 [ Fe

3N3Cl6, (b) the phosphorus is sp hy3

inorbitals. 16.35. (dashed lines = hydrogen bonds)

N

H

H

N

H

N

HH

H

H

H

N

H

H

H

H

H

N

H

H

H

H N H

1 -3 +1 +6 -2 0 +1 -2 +3 -2 T *

3+ - 2+_

108

(aq) + e ─→ Fe (aq) ]_________ E° = +0.77VN2H5

+(aq) + 4Fe3+(aq) ─→ N2(g) + 5H+(aq) + 4Fe2+(aq) E° = +1.00V

Yes, hydrazine could accomplish this reduction under standard state conditions.

) 5 [ N2H5+(aq) ─→ N2(g) + 5H+(aq) + 4e- ] E° = +0.23V

_____4 [MnO

(b

109

_ 4-(aq) + 8H+ - 2+(aq) + 5e ─→ Mn (aq) + 4H2O(l) ]_______ E° = +1.51V

5N2H5+(aq) + 4MnO4 + 7H (aq) ─→ 5N-(aq) n2+(aq) + 16H2O(l) E° = +1.74V

ally) reaction.

6.41. As shown on the right, there

. 1 [ 2e + N2O + 2H ─→1 [ N

+2(g) + 4M

This would be a most spontaneous reaction (thermodynamic 1are several possible sites within the hydroxylamine molecule where hydrogen bonding could take place.

NH

O H O

H

H

H

N

O H

H

H

NH *16.43 - + N2 + H2O ] E° = +1.77V

+ 2O + H2O ─→ 2NO + 2H + 2e- ] E° = - 1.59V 2N2O ─→ N2 + 2NO E° = + 0.18V

ince E° is greater than zero, ΔG° will be less than zero and the reaction should be spontaneous under andard state conditions.

6.45. N2(g) + O2(g) ─→ 2NO(g), ΔS° = 24.7 J/mol-K his entropy change favors

Sst 1T the increased production of NO at high temperatures because ΔG° = ΔH° - ΔS°. A positive ΔS° leads to a negative contribution to ΔG°, which in turn makes the reaction more ontaneous left to right.

16.47. Internet search for NO: What did you find? NO news is definitely still good news. In addition those uses given in this text, the following quote is from the “Nitric Oxide Homepage,” http://www.apnet.com/no/>, (accessed August 2001): “From diabetes to hypertension, cancer to drug

, sunburn to p ondition where nitric

e. On ce ly (w n the discovered as a product

most of these published within the last five years.”

Tsp *to<addiction, stroke to intestinal motility, memory and learning disorders to septic shockanorexia, male impotence to tuberculosis, there is probably no athological coxide does not play an important rol ly re nt ithi last 13 years)of enzymatic synthesis in mammals, there are more than 32,000 scientific papers dealing with this remarkable molecule with

110

16.49.

*16.51.

The expected bond angles decrease as the nitrogen atom first adds a single nonbonding electron and then a full lone pair. 16.53. N2O5 + H2O ─→ 2HNO3This is not a redox equation; the oxidation state of nitrogen is +5 in both N2O5 and HNO3. 16.55. Within nitrogen chemistry there are both excellent oxidizing agents (for example, HNO3) and excellent reducing agents (for example, hydrazine). Therefore, it is an overgeneralization to claim that all the good oxidizing agents are on the right side of the table and all the good reducing agents on the left. *16.57. P2O5 + 2HNO3 + 2H2O ─→ N2O5 + 2H3PO4 iron *16.59. N2(g) + 3H2(g) ──────→ 2NH3(g) ΔH° = -92.6 kJ/mol high T&P High pressure is justified because it would favor the side (in this case the product side) with the fewer moles of gases. Higher temperature would be justified thermodynamically only if this were an endothermic reaction, which it is not. The high temperature must be necessary in order to increase the rate of the reaction to an acceptable level. The iron is a catalyst that increases the rate but does not affect the thermodynamic spontaneity of the reaction.

N NO O N N

O<120oO

N N

OO

sp2 - sp2 sp2 - sp2

sp2 - sp2

2p-2p2p-2p

NO2+ NO2 NO2

-

O N O+

O N O

O N O

O N O

O N O

-

-

NO O

180o

N

O ON

>120o

N

O ON

<120o

olecules.

6.63. Ammoniumith its strong N≡N bond), dioxygen (with its strong O=O bond), and water (with its relatively strong O-

single bonds). The production of these strong bonds makes this reaction highly exothermic.

6.65. The double bonds of the phosphorus sulfides cannot be of the pπ-pπ type because this overlap is ot effective in the third period elements. These bonds cannot be of the dπ-dπ type either because the d rbitals are empty in both of these elements. The bonds must be of the pπ-dπ type with each sulfur roviding two electrons in its 4p orbital to overlap with the empty d orbitals of the phosphorus atoms. ote that the phosphorus atoms are sp3 hybridized and cannot provide the p orbitals.

6.67. KHC4H4O6 + NaHCO3 ─→ CO2(g) + H2O + KNaC4H4O6(s)

16.69. PO3F2- is isoelectronic with the orthophosphate ion, PO43-. One would speculate, then, that the

n would be tetrahedral in structure.

6.71. Hydroxyl radicals serve as “reaction intermediates” in the production of photochemical smog. hey combine with the lower hydrocarbons to produce various methyl, ethyl, etc. radicals that, in turn,

16.61. The oxidation state of the nitrogen atom in hydroxylamineis –1. This compound will be very soluble in water because the lone pairs of both the oxygen and nitrogen atoms as well as the hydrogen atoms themselves are capable of hydrogen bonding with water N

O

H

H

Hm 1 nitrate, NH4NO3, as shown in Equation (16.52), can explode to produce dinitrogen (wH 1nopN 1 *io 1Tcan readily keep the oxidant-producing photochemical reactions going that facilitate the production of smog.

111

112

Chapter 17 Sulfur, Selenium, Tellurium, and Polonium

The sections and subsections of this chapter are listed below. 17.1 Discovery and Isolation of the Elements Sulfur Tellurium and Selenium (Earth and Moon) Polonium 17.2 Fundamental Properties and the Network Hydrides Oxides and Oxoacids Halides 17.3 Allotropes and Compounds Involving Element-Element Bonds Allotropes Polycations and Anions Catenated Halides and Hydrides Catenated Oxoacids and Corresponding Salts 17.4 The Sulfur Nitrides 17.5 Reactions and Compounds of Practical Importance Sodium-Sulfur Batteries Photoelectric Uses of Selenium and Tellurium Sulfuric Acid 17.6 Selected Topic in Depth: Acid Rain

113

Chapter 17 Objectives You should be able to

• briefly relate how and by whom sulfur, selenium, tellurium, and polonium were discovered • outline the general manner in which the periodic law, uniqueness principle, inert pair effect,

acid-base character of oxides, standard reduction potentials, and pπ-dπ bonding involving elements of the second and third periods are applicable to the Group 6A elements

• summarize the preparations and properties of the non-catenated hydrides of sulfur and its heavier congeners

• summarize the preparations and properties of the oxides and non-catenated oxoacids of sulfur and its heavier congeners

• summarize the preparations and properties of the non-catenated halides of sulfur and its heavier congeners

• discuss the major allotropes of sulfur in the solid, liquid, and gaseous phases • give some examples and discuss the properties and structures of some catenated polycations

and anions of sulfur and its heavier congeners • give some examples and discuss the properties, preparations, and structures of the catenated

halides and hydrides of sulfur and its heavier congeners • give some examples and discuss the properties, preparations, and structures of catenated

sulfur-containing oxoacids and their corresponding salts • give some examples and discuss the properties, preparations, and structures of some

representative sulfur nitrides • describe the nature of sodium-sulfide batteries • describe the role of selenium and tellurium in the Xerox process and in II-VI semiconductors • describe the industrial preparation and some of the uses of sulfuric acid • describe the causes, consequences, and possible control of acid rain

Solutions to Odd-Numbered Problems 17.1. Since gunpowder was introduced in the thirteenth century and matches soon after the discovery of phosphorus in late seventeenth century by Brandt, these inventors most likely did not regard sulfur as an element. Only in the first decade of the nineteenth century was sulfur so recognized. 17.3. 210

84Po ─α→ 42He + 20682Pb

17.5. The Zeff is essentially constant but atomic sizes increase down the group. Therefore, the atomic radii increase, electronegativites decrease (electrons are attracted to positions farther from the Zeff) and ionization energies decrease (electrons to be removed are also farther from the Zeff). 17.7. H2O H2S H2Se H2Te H2Po bp (°C) 100 -61 -41 -2 --- H2O has the highest boiling point due to the strong hydrogen bonds that characterize its inter-molecular forces. Oxygen is significantly more electronegative than its heavier congeners due to its ability to attract bonding electrons to positions much closer to its effective nuclear charge. Therefore O-H bonds are much more polar than bonds between H and S, Se, and Te. It follows that compounds characterized by O-H bonds are capable of forming the stronger hydrogen bonds and are higher boiling. *17.9. Oxidizing agent: oxygen (reduced from 0 to -2) Reducing agent: silver (oxidized from 0 to +1) Tarnished silver objects (and therefore silver sulfide) are black 17.11. The acid strength should increase in the order H2S to H2Se to H2Te. The H-X bond strength decreases down the group (because the bond length increases) and therefore the acids are readily dissociated into H+ and HX- ions. 17.13. SO2 ·H2O(aq) + H2O HSO3

-(aq) + H3O+(aq) *17.15. SO3

2-(aq) + H2O HSO3-(aq) + OH-(aq)

Kb = Kw = __[H+][OH-]__ = [HSO3

-][OH-] = 1.00 x 10-14 = 1.6 x 10-7

Ka2 [H+][SO32-] [SO3

2-] 6.3 x 10-8

[HSO3-]

17.17. SO3

2-(aq) + Cl2(g) ─→ SO42-(aq) + Cl-(aq)

Balance by half equations: SO3

2- + H2O ─→ SO42- + 2H+ + 2e- ] 1

Cl2 + 2e- ─→ 2Cl- ] 1 SO3

2- + Cl2 + H2O ─→ SO42- + 2H+ + 2Cl-

Given that H+ is a product, high acid concentrations would shift the position of equilibrium of this

114

reaction back toward the left. This is an example of Le Châtelier's Principle. 17.19.

O O

O OS

OS S

O O O

O

S

OO

O

S

O O

S

O O

O

O

Structures with terminal S=O bonds using pπ-dπ bonds are also possible here. 17.21. Sulfuric acid has one more nonhydroxyl-group oxygen atom withdrawing electron density from the O-H bonds. This makes these bonds more polar and therefore more susceptible to attack by polar water molecules. It follows that H2SO4 is the stronger acid. 17.23. 2e- + S(s) + 2H+(aq) ─→ H2S(g) E° = 0.14V; ΔG° = -(2)(96.5kJ/V)(0.14V) = -27 kJ SO4

2-(aq) + 8H+(aq) + 6e- ─→ S(s) + 4H2O(l) E° = 0.37V; ΔG° = -(6)(96.5kJ/V)(0.37V) =-214kJ8e- + 10H+(aq) + SO4

2-(aq) ─→ H2S(g) + 4H2O(l) ΔG° = -241 kJ E° = - __ -241kJ__ = 0.31V (8)(96.5kJ/V) 17.25. (a) Sn(s) ─→ Sn2+(aq) + 2e- E° = +0.14V 2e- + 2H+(aq) ─→ H2(g) E° = 0.00V SO4

2-(aq) ─→ SO42-(aq)___________________________________

Sn(s) + 2H+(aq) + SO42-(aq) ─→ Sn2+(aq) + H2(g) + SO4

2-(aq) E° = +0.14V (b) Sn(s) ─→ Sn2+(aq) + 2e- E° = +0.14V Sn2+(aq) ─→ Sn4+(aq) + 2e- E° = -0.13V 2 [SO4

2-(aq) + 4H+(aq) + 2e- ─→ SO2(g) + 2H2O(l) ] E° = +0.20V Sn(s) + 2SO4

2-(aq) + 8H+(aq) ─→ 2SO2(g) + 2H2O(l) + Sn4+(aq) E° = +0.21V *17.27. Fluorine is isoelectronic with OH so fluorosulfuric acid is analogous with sulfuric acid but with an OH replaced by a F.

O S O

F

O

H

Fluorine is more electronegative than oxygen so even more electron density is removed from the one remaining O-H bond than in sulfuric acid. That bond is therefore more polar and more susceptible to attack by polar water molecules.

115

17.29. Sulfamide

O S ON

N

HH

HH

SN

NO

O

H

H

H

H

17.31. The intermolecular forces in Te(OH)6, telluric acid, should be hydrogen bonds among adjacent O-H groups. *17.33. 2[ 4H2O(l) + Mn2+(aq) ─→ MnO4

-(aq) + 8H+(aq) + 5e- ] E° = -1.51V ______ 5[ 2e- + S2O8

2-(aq) ─→ 2SO42-(aq) ] E° = 2.01V

8H2O(l) + 2Mn2+(aq) +5S2O82-(aq) ─→ 2MnO4

-(aq) +16H+(aq) +10SO42-(aq) E° = +0.50V

ΔG° = -nFE° = -(10)(96.5 kJ/V)(+0.50V) = -482 kJ/mole 1[ 7H2O(l) + 2Cr3+(aq) ─→ Cr2O7

2-(aq) + 14H+(aq) + 6e- ] E° = -1.33V ______ 3[ 2e- + S2O8

2-(aq) ─→ 2SO42-(aq) ] E° = 2.01V

7H2O(l) + 2Cr3+(aq) + 3S2O82-(aq) ─→ Cr2O7

2-(aq) + 14H+(aq) + 6SO42- E°= +0.68V

ΔG° = -(6)(96.5 kJ/V)(+0.68V) = -394 kJ/mol 17.35.

ll the terminal S-O bonds could also be shown as double bonds involving pπ-dπ bonds. For simplicity, ey all are shown as single bonds.

7.37 FeCl3 ⋅ 6H2O(s) + 6SOCl2(l) ─→ FeCl3(s) + 6SO2(g) + 12HCl(g) Thionyl chloride is an excellent dehydrating agent.

17.39. (a) The single bond energy of sulfur is greater than that of oxygen because, although the S-S ond distances are longer and therefore, one might suspect, weaker than the O-O bonds, the repulsions mong the lone pairs of electrons on the oxygen atoms in a O-O bond make it a weaker bond than xpected. (b) The double bond energy of O=O is greater than that of S=S because the pi bond between

O

SO

S

O O

O <109.5o

<109.5o

O S O

O

S O

O

O O

H H

OO

H H

O

SO

S

O O

O

OO

H Hsp3

sp3 sp3

Ath 1 *baeoxygen atoms is of the stronger (more effective overlap) pπ-pπ type while that between sulfur atoms is of

116

the dπ-pπ type. 17.41. Cyclohexasulfur, S6 S S S

117

boiling

he 6-membered ring most likely exists in "chair"

6 - 4 = 32e

) single bonds 9 x 2 = 18e

Tand/or "boat" forms much like cyclohexane. 17.43. Total electrons available in Te6

4+

6 x -

SS

S

Te

Te

Te

Te

Te

Electrons in cation as shown at right:

- Te12) lone pairs 6 x 2 = 12e-

30e-

The structure cannot be accounted for using all single ight very well be elocalized in the triangular faces giving some dπ-pπ character to these bonds and therefore making them

17.45. q) S2O32-

t forms "H2SO3" (sulfurous acid) which immediately breaks apart su ioxide is driven off and the above equilibrium shifts to the left decreasing u and increasing the amount of solid elemental sulfur present.

bonds. The extra two electrons mdslightly stronger and shorter. * S(s) + SO3

2-(a (aq) The addition of H+ to the SO3

2- reactanto SO2(g) and H2O. The lfur dthe concentration of thios lfate 17.47. Pb2+(aq) + 2e - ─→ Pb(s) ]1 2H2O + S2O4

2-(aq) ─→ 2SO32-(aq) + 4H+ -(aq) + 2e ]1

Pb (aq) + 2H2O + S2O4 ─→ Pb(s) + 2SO32-(aq) + 4H+(aq)

Ag+(aq) + 1e- ─→ Ag(s) ]2

2+ 2-(aq) 2H2O + S2O4

2-(aq) ─→ 2SO32-(aq) + 4H+(aq) + 2e- ]1

2Ag+(aq) + 2H2O + S2O42-(aq) ─→ 2Ag(s) + 2SO3

2-(aq) + 4H+(aq) *17.49. S4N4 4S = 24e-

4N = 20e-

44e

The VSEPR structure is the same as Figure 17.8(a).

S SN

S

NS

N N N

S S S-

N N N S

17.51. S4N4 (4x6) + (4x5) = 44e- two transannular S-S bonds needed

(8x6) - 2e- = 46e- one transannular S-S bonds needed S8 (8x6) = 48e- no transannular S-S bonds needed

or S4(NH)4, there are (4x6) + (4x5) + (4x1) = 48e-, therefore we would predict that this cycloocta ring he hydrogen atoms would be

ngle-bonded to the nitrogens.

17.53. For O, EN = NO, nitrogen oxide, a stable, discrete odd-electron species best presented as shown at right. This molecule is characterized by strong pπ-pπ

S,

ulfur uses d orbitals

miconductor of the type covered in Chapter 15 (p. 401). It is not rprising to expect that AlP is a III-V intrinsic semiconductor isoelectronic with Si (or 2Si).

NH + H2 NH4)2SO4(aq) + 2H2O he wat cou d be ammonium sulfate.

7.61. B th w t an dry sc n the principle that adding calcium oxide (or calcium carbonate r plant effluents serves to convert sulfur dioxide to

o calcium sulfate, CaSO4. In wet scrubbers, an aqueous lution of the calcium oxide or carbonate is sprayed into the smokestack and the result is a difficult-to-

less water is used so that the excess is driven

S82+

Fwith alternating S and N atoms should contain no transannular S-S bonds. Tsi *rebonds typical of the second period elements (re: the uniqueness principle). For

N OEN =SN, the polymeric, fibrous material that acts as a one-dimensional metal. The S-N bonds of this polymer are characterized by dπ-pπ, S-N bonds. Againthe nitrogen employs p orbitals to form π bonds while the sto accept the electron density from the nitrogen. *17.55. Silicon is an intrinsic sesu 17.57. roasting 2MS + 3O2 ────→ 2SO2(g) + 2MO V2O5 SO2(g) + O2 ────→ 2SO3(g) (as air) SO3(g) + H2SO4(aq) ─→ H2S2O7(aq) H2S2O7(aq) + H2O(l) ─→ 2H2SO4(aq) 17.59. 2 4OH(aq) SO4(aq) ─→ (T er l continuously evaporated off to leave solid 1 o e d rubbers work othat loses carbon dioxide to produce the oxide) to powecalcium sulfite, CaSO3, and sulfur trioxide tsomanage slurry of the sulfites and sulfates. In dry scrubbers, off and the result is a dry mixture of the sulfites and sulfates.

118

*17.63.

O S

O>109.5o <109.5o

OO

H

S

O

HO0

0

0 00<109.5o

<109.5o

. The Lewis structure gives formal charges of zero for all atoms. (Resonance structures with single S-O bonds would also be important for this molecule.)

7.65. The burning of foss jor greenhouse gas. If we ere to continue to strongly rely on combustion of coal to produce our energies, enormous amounts of is gas might accumulate in the atmosphere resulting in significant global warming.

1 il fuels would still produce carbon dioxide, a mawth

119

120

Chapter 18 Group 7A: The Halogens

The sections and subsections of this chapter are listed below. 18.1 Discovery and Isolation of the Elements Chlorine Iodine Bromine Fluorine Astatine 18.2 Fundamental Properties and the Network Hydrides Halides Oxides 18.3 Oxoacids and Their Salts Hypohalous Acids, HOX, and Hypohalites, OX- Halous acids, HOXO and Halites, XO2

-

Halic acids, HOXO2 and Halates, XO3-

Perhalic acids, HOXO3 and Perhalates, XO4-

18.4 Neutral and Ionic Interhalogens 18.5 Reactions and Compounds of Practical Importance Fluoridation Chlorination Bleaches Bromides 18.6 Selected Topic in Depth: Chlorofluorocarbons (CFCs) – A Threat to the Ozone Layer

121

Chapter 18 Objectives You should be able to

• briefly relate how and by whom the halogens were discovered • outline the general manner in which the periodic law, uniqueness principle, acid-base

character of oxides, and standard reduction potentials are applicable to the group • explain why fluorine is such an extraordinarily powerful oxidizing agent and how this property

made it so difficult to isolate • discuss the preparations and properties of the hydrogen halides • summarize and give representative examples of the various ways to prepare halides of the

representative elements • list and discuss the major applications of the halogen oxides • discuss the preparations, properties, and applications of

1) the hypohalous acids and the hypohalites 2) the halous acids and the halites 3) the halic acids and the halates 4) the perhalic acids and the perhalates

• discuss the preparation, structures, and major reactions of the interhalogens • discuss the history of and the chemistry central to systemic and topical fluoridation • briefly explain the role of the halogens in the chlorination of water supplies, the nature of

bleaches, and the applications of the bromides • discuss the history of and chemistry central to identifying the role of chlorofluorocarbons and

halons as threats to the ozone layer

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Solutions to Odd-Numbered Problems *18.1. (a) Priestley’s isolation of hydrogen chloride gas: NaCl + HX ─→ HCl(g) + NaX, where HX is a strong acid. He used mercury because HCl gas is soluble in water. (b) Scheele’s isolation of chlorine gas: 4NaCl(aq) + 2H2SO4(aq) + MnO2(s) ─→ 2Na2SO4(aq) + MnCl2(aq) + 2H2O(l) + Cl2(g). MnO2 is the oxidizing agent and NaCl is the reducing agent. 18.3. 2NaI(s) + 2H2SO4(aq) ─→ I2(s) + SO2(g) + 2H2O + Na2SO4(aq) An analogous attempt to make chlorine gas, Cl2(g), would produce only HCl because sulfuric acid is not a strong enough oxidizing agent to oxidize chlorides to chlorine. 18.5. Cl AW = 35.45u Br Predicted AW of Br = 35.45 + 126.9 = 81.2u I AW = 126.9u 2 The actual AW of bromine is 79.9u. There is only a 1.6% error between the predicted and actual values. 18.7. Gay-Lussac ─→ Fremy ─→ Moissan ─→ Stock 18.9. halogen mp covalent ionization (°C) radius (Å) energy (kJ/mol) F -220 0.72 1680 Cl -101 0.99 1251 Br -7.3 1.14 1143 I 114 1.33 1009 At 302 ----- 916 These are very regular periodic trends. The melting points increase as the number of electrons (and atomic weight) increase making for stronger intermolecular forces. The covalent radii increase as the sizes of the individual atoms regularly increase down the group. The ionization energies regularly decrease as the electrons to be ionized are farther and farther away from an approximately constant effective nuclear charge. 18.11. (a) 2F2(g) + CH4(g) ─→ 4HF(g) + CF4(g) (b) 3F2(g) + 2NH3(g) ─→ 6HF(g) + 2NF3(g) 18.13. H2(g) + F2(g) ─→ 2HF(g) The weakness of the F-F bond in F2 combined with the high strength of the H-F bond in HF means that a great amount of energy is released in the above reaction. In other words, much more energy is released when the bonds of the products are formed than is needed to break the bonds of the reactants. The above situation makes for an extremely exothermic reaction. 18.15. Fluorine is thought to have an anomalously low electron affinity because of its very small size. When an eighth electron is added to the valence shell of fluorine to produce a neon inert gas electronic configuration, that incoming electron is severely repelled by the seven existing electrons occupying the n = 2 shell. It is therefore more difficult to add the eighth electron than expected and the amount of energy

released is also less than expected. *18.17.

1/2 F2(g) + e- ΔHored F-(aq)

F(g)EA

F-(g)

ΔH(g) ΔHhyd

ΔHored = ΔH(g) + ΕΑ + ΔHhyd

The above reaction is very thermodynamically spontaneous because (a) ΔH(g), corresponding to breaking of the weak F-F bond of F2, is so low and (b) the ΔHhyd of F-(g) is so large and negative. The latter is due to the very high charge density of F- leading to a very strong and exothermic interaction with the polar water solvent. 18.19. NaF(aq) + H2SO4(aq) ─→ 2HF(aq) + Na2SO4(aq) NaCl(aq) + H2SO4(aq) ─→ 2HCl(aq) + Na2SO4(aq) 2NaBr(aq) + 2H2SO4(aq) ─→ Br2(l) + SO2(g) + 2H2O(l) + Na2SO4(aq) 2NaI(aq) + 2H2SO4(aq) ─→ I2(s) + SO2(g) + 2H2O(l) + Na2SO4(aq) Sulfuric acid is a strong enough oxidizing agent to convert bromides to bromine and iodides to iodine but it is not strong enough to produce fluorine and chlorine from the fluorides and chlorides, respectively. 18.21. 2[ 2I-(aq) ─→ I2(s) + 2e- ] E° = - 0.54V 1[O2(g) + 4H+(aq) + 4e- ─→ 2H2O(l) ] E° = +1.23V 4HI(aq) + O2(g) ─→ 2I2(s) + 2H2O(l) E° = +0.69V 2[ 2Br-(aq) ─→ Br2(l) + 2e- ] E° = - 1.07V 1[O2(g) + 4H+(aq) + 4e- ─→ 2H2O(l) ] E° = +1.23V 4HBr(aq) + O2(g) ─→ 2Br2(l) + 2H2O(l) E° = +0.16V Both of these reactions are spontaneous under standard state conditions. The hydrohalic acids are not immediately converted to the halogens because the concentration or partial pressure of O2(g) is not great enough in aqueous solutions. 18.23. Note that the H-F- - -H

123

F

F

H

FH

F

H

F

HH

F

H

hydrogen bonds must be linear.

*18.25. 2 [ NO3

-(aq) + 4H+(aq) + 3e- ─→ NO(g) + 2H2O ] E° = 0.96V 2 [ 2Cl-(aq) ─→ Cl2(g) + 2e- ] E° = -2.87V2NO3

-(aq) + 8H+(aq) + 6Cl-(aq) ─→ 2NO(g) + 3Cl2(g) + 4H2O E° = -1.91V No, HNO3 could not be used to produce chlorine from chloride. 18.27. Al2O3(s) + 6HF(g) ─→ 2AlF3(s) + 3H2O(l) 2Al(s) + 3Cl2(g) ─→ 2AlCl3(s) 18.29. (a) 2Co3O4(s) + 6ClF3(g) ─→ 6CoF3(s) + 3Cl2(g) + 4O2(g) (b) 2B2O3(s) + 4BrF3(l) ─→ 4BF3(g) + 2Br2(l) + 3O2(g) (c) 3SiO2(s) + 4BrF3(l) ─→ 3SiF4(g) + 2Br2(l) + 3O2(g) 18.31. (a) PCl3(s) + 3H2O ─→ H3PO3(aq) + 3HCl(aq) (b) PCl5(s) + 4H2O ─→ H3PO4(aq) + 5HCl(aq) *18.33. NaSCN(aq) + AgNO3(aq) ─→ AgSCN(s) + NaNO3(aq) 18.35. Cl2O C 2

Cl O Cl O Cl O

O

Cl Cl<109.5o

Cl

O O<109.5o

lO 18.37. Oxidizing agent: I2O5 Reducing agent: CO *18.39. Equation (18.30) is a redox reaction with X2 being both the oxidizing and the reducing agent. That is, X2 disproportionates to HOX and X-. 18.41. Ba(BrO3)2(aq) + H2SO4(aq) ─→ BaSO4(s) + 2HBrO3(aq) Because barium sulfate is a fine white precipitate, this is sometimes called a "milkshake" reaction. *18.43. HIO3 + 5HI ─→ 3I2 + 3H2O Oxidizing agent: HIO3 Reducing agent: HI 18.45. 83

34SeO42- ─β-→ 83

35BrO4- + 0-1e

124

18.47. 1 [ BrO3

-(aq) + H2O(l) ─→ BrO4-(aq) + 2H+(aq) + 2e- ] E° = -1.74V

1 [XeF2(aq) + 2H+(aq) + 2e- ─→ Xe(g) + 2HF(aq) ] E° = +2.64VXeF2(aq) + H2O(l)(aq) + BrO3

-(aq) ─→ Xe(g) + 2HF(aq) + BrO4-(aq) E° = +0.90V

The above reaction is thermodynamically feasible. 18.49. 2NH4ClO4(s) ─Δ→ N≡N(g) + Cl2(g) + 2 O=O(g) + 4H2O(l) This reaction is very exothermic because large amounts of energy are released when the very strong N≡N and O=O bonds are formed. 18.51. The bromine atom is sp3 hybridized. Three of these four

Br O

4d 2p

+ _

++

_

_hybrid orbitals are used to form a single bond with an oxygen atom. The remaining hybrid orbital contains a lone pair of electrons. One of the three oxygen atoms is singly bonded to a hydrogen atom. The other two oxygen atoms do not have hydrogens bonded to them but rather form a second bond with the bromine atom. This second bond is formed using a 4d orbital of bromine and a 2p orbital of oxygen as shown at right. (The signs are the phases of the orbitals.) 18.53. Br2(s) + F2(g) ─→ 2BrF(g) Br2(s) + 3F2(g) ─→ 2BrF3(l) BrF3(l) + F2(g) ─→ BrF5(l) 18.55. (a) Se(s) + 4ClF(g) ─→ SeF4(s) + 2Cl2(g) (b) 6NiO(s) + 4ClF3(g) ─→ 6NiF2(s) + 2Cl2(g) + 3O2(g) (c) 3SiO2(s) + 4BrF3(l) ─→ 3SiF4(s) + 2Br2(l) + 3O2(g) (d) P4(s) + 20ICl(s) ─→ 4PCl5(s) + 10I2(s) 18.57. IF3 + 2H2O ─→ 3HF + HIO2 IF5 + 3H2O ─→ 5HF + HIO3 IF7 + 4H2O ─→ 7HF + HIO4 18.59.

125

(a)

IF

F F

F

+

I

F

F

F

F

(b) II I I

I

I

<109.5o

<90o

<120o

+

126

18.61. Fluorine-20 can be incorporated in small amounts into teeth and bones in the form of fluoro-apatite, Ca10(PO4)6

20F2. For a given geographical area with a given amount of fluoride in the water supply, a hominid should have a predictable amount of fluoride in his or her bones or teeth. Knowing the isotopic abundance of fluorine-20, one could calculate the expected levels of F-20 in the teeth or bones at death. After death, the amount of F-20 would decrease. Measurement of the existing level of F-20 (by counting beta minus particles), combined with knowledge of the half life of F-20 would yield a determination of the age of the sample. 18.63. Fluoridation is a good idea because there is significant evidence that it reduces the number of dental caries in children without subjecting them to undue health risks. It is not a good idea, some would argue, because it forces a medication on an entire population area whether they (a) want it or not and (b) in the case of adults, they can significantly benefit from it or not. *18.65. Dear Aunt Emily, I enjoyed our last talk about aluminosilicates, sodium carbonate, and water softeners in general. (See the answer to Problem 15.63). Have you had a chance lately to inspect some more detergent box labels to see if aluminosilicates are catching on with other manufacturers? Now, in my class, we are discussing fluoride additives in toothpastes. Water fluoridation (an example of “systemic fluoridation”), I’m sure you know, continues to be a controversial topic in some communities, but these toothpaste fluoride additives (an example of “topical fluoridation”) don’t seem to be a problem. That is probably the case because people can choose these topical products but are forced to drink fluoridated water if that is what his or her community has decided to do. Colgate’s MFP is based on phosphate, another polyatomic anion like the silicates and aluminosilicates we discussed last. Phosphate has a formula of PO4

3-; it has a phosphorus atom in the middle and four oxygen atoms around it. MFP stands for monofluorophosphate and has a fluorine atom in place of one of the oxygen atoms. When MFP is placed in water, it releases some fluoride ions that can protect teeth, particularly the developing teeth in young children, against tooth decay. Evidently, when fluoride becomes part of a child’s tooth enamel, her teeth are less susceptible to attack by the acids that result when saliva breaks down sugars in the mouth. Crest’s fluoristat is simply sodium fluoride (very similar to ordinary table salt with fluoride in place of chloride) that dissolves in water when the toothpaste is used. Again, this fluoride ion can be taken in and made a part of the tooth enamel. You also asked about fluoristan, the additive that Crest used before fluoristat. The –stan suffix comes from the Latin stannum for relating to or containing tin, so fluoristan was Crest’s word for stannous fluoride. The tin was incompatible with newer forms of tooth polishing compounds in toothpastes so stannous fluoride had to be replaced. There were also some worries about the toxic effects of tin so perhaps that was another reason to go from –stan to –stat stat. (My chemistry professor really enjoys puns like that!) Anyway, thanks for continuing to be interested in my studies. It’s fun to explain these things to you. 18.67. Cl2(g) + 2NaOH(aq) ─→ NaOCl(aq) + NaCl(aq) + H2O 2Cl2(g) + 2Ca(OH)2(aq) ─→ Ca(OCl)2(aq) + CaCl2(aq) + 2H2O *18.69. Dear Aunt Emily, I enjoyed our earlier conversation about water softeners (see Problem 15.63) and our more recent talk about various fluoride toothpaste additives (see Problem 18.66). I hear that you are interested in talking about threats to the ozone layer as represented in Figure 18.6. You may have wondered if we have any concrete evidence – a “smoking gun” so to speak – that chlorine compounds really are a threat

127

to ozone. In the mid-70s, back when you and Uncle Tim had those wonderful suntans I have seen in pictures – suntans due to the uv component of sunlight – chlorofluorocarbons (CFCs) were implicated to pose a threat to the stratospheric ozone that shields us from receiving too much solar uv radiation. (Too much of this powerful radiation is dangerous because it can cause skin cancer.) The theory proposed at that time was that CFCs, very inactive in the troposphere (the part of the atmosphere where we live), are eventually swept up into the stratosphere where the chlorine atoms are broken off by uv radiation and start to destroy ozone. The theory proposed in the 1970s said that chlorine oxide, ClO, was an integral part of that reaction system and the compound that actually reacted directly with the ozone, O3. (At that time no one knew for sure if this was true even though CFCs were quickly removed from products such as aerosol spray cans.) Figure 18.6, which plots data gathered in the 1980s as aircraft flew towards the South Pole, shows the intimate connection between ClO and O3. Notice that the concentration of ozone (represented in the upper light line of the graph) always decreases when the concentration of ClO (the lower darker line) increases. These two lines are very nearly mirror images of each other(!) and this plot was solid proof that ClO concentrations directly affect ozone concentrations – a chemical smoking gun if there ever was one! Again, it was fun to “talk chemistry” with you, Aunt Emily. 18.71. The Antarctic gets colder than the Arctic and the Antarctic polar vortex is more stable. This means that during the Antarctic winter, inert chlorine reservoirs can be readily set up from which active ozone-destroying chlorine and chlorine compounds can be released in the polar Spring.

128

Chapter 19 Group 8A: The Noble Gases

The sections and subsections of this chapter are listed below. 19.1 Discovery and Isolation of the Elements Argon Helium Krypton, Neon, and Xenon Radon 19.2 Fundamental Properties and the Network 19.3 Compounds of Noble Gases History Fluorides Structures Other Compounds 19.4 Physical Properties and Elements of Practical Importance 19.5 Selected Topic in Depth: Radon as a Carcinogen

Chapter 19 Objectives You should be able to

• briefly relate how and by whom the noble gases were discovered • outline the general manner in which the periodic law is applicable to this group • briefly relate how Bartlett came to prepare the first compound of xenon • discuss the preparations, properties and structures of some representative fluorides and

oxides of xenon • give several examples of compounds of the noble gases other than xenon • list a few representative uses of the noble gases • relate and discuss the discovery and mechanism of radon as a major environmental health

problem

Solutions to the Odd-Numbered Problems 19.1. P4(s) + 5O2(g) ─→ P4O10(s) *19.3. NaOH(s) + CO2(g) ─→ NaHCO3(s) 19.5. Assuming air to be 20% oxygen (O2) and 80% nitrogen (N2), its average molecular weight is given by the following equation: Average MW = 0.20(32) + 0.80(28) = 29u

*19.7. ⎟⎠⎞

⎜⎝⎛

LAr 1.78g

⎟⎠⎞

⎜⎝⎛

mol22.4L = 39.9 g/mol

19.9 The ionization energies decrease going down the group because the electron to be ionized becomes farther and farther away from an approximately constant effective nuclear charge. *19.11. The effective nuclear charges felt by the valence electrons are indeed very large but the Zeff felt by an incoming electron (that would occupy an ns orbital outside of the closed shell of valence electrons) is very small. Therefore, these atoms do not attract free electrons. 19.13. The only intermolecular forces operating between xenon and water molecules in xenon hydrates are London dispersion forces. 19.15. 235UF6(g), being slightly lighter, diffuses faster than 238UF6(g).

6

2386

235

UFof diffusion of rateUFof diffusion of rate =

6(19.0) (235)6(19.0) (238)

++ =

349352 = 1.004

19.17. Radon would be more reactive toward such a strong oxidizing and fluorinating agent as PtF6 because it has a lower ionization energy than does xenon. 19.19. 4KI(s) + XeF4(g) ─→ 4KF(s) + Xe(g) + 2I2(s) *19.21. Oxidizing agent: XeF2 Reducing agent: H2O

129

*19.23.

PtF

FF

F

FF

Pt

F

F

F

-

F

F

19.25. (b), (f), and (h) are consistent with the VSEPRT.

(b)

Xe

F

F

F

FXe F F

F

F90o

(f)

Xe

O

O

O

OXe

90oO

O

(h)

Xe

OF

FXe F F

F

F90oF

F

O90o

4-O

O

O

O

O

O

130

(d) There would be nine pairs of electrons around the large xenon atom in the XeF82- anion.

Consistent with VSEPR theory, eight pairs would logically be dispersed in a square antiprism. The ninth pair is theorized to occupy the spherical 5s orbital of xenon and not affect the geometry of the other eight. 19.27. The F-Xe-F may be slightly less than 90º but not much because the xenon atom is

131

very large and there is plenty of room for the lone pair to spread out without unduly influencing the Xe-F bonds Xe

FF

FXe F F

F

F~90oF

F~90o+ F

19.29. XeO3F2

OO

FXe

F

OXe

F

F

O

O

<120o

<90o

O

19.31. 129

53ICl2- ─β-→ 0-1e + 12954XeCl2

Xe Cl Xe

Cl

Cl

180oCl

19.33. Oxidizing agent: O3

3

19.35. Xe=F double bonds are not likely because they require a +1 formal charge on the fluorine

9.37. 3[ XeO6 (aq) + 12H (aq) + 8e ─→ Xe(g) + 6H2O ]

Reducing agent: XeO *atom. Xe=O double bonds are not forbidden because of formal charges but would involve dπ-pπ bonding between a 6d orbital of xenon and a 2p orbital of the oxygen atom. The disparity in size would make this 6d-2p orbital overlap not particularly effective.

4- + -1 4[ 7H2O + Mn2+(aq) ─→ MnO4

-(aq) + 8H+ + 5e- ] XeO6

4-(aq) + 2H2O + 2Cr3+(aq) ─→ 3Xe(g) + 4Cr2O72-(aq) + 20H+(aq)

132

*19.39. 23892U ─α→ 42He + 234

90Th (from clevite)

21819.41. 84 Po ─α→ 42He + 21482Pb

9.43. Smokers are at a greater risk to the hazards of radon than nonsmokers because the sited

1radioactive products (especially Po-218) of radon decay adhere to the smoke particles depoin smokers’ lungs.