chem 222 intro to inorganic chemistry summer 2011 ...web.uvic.ca/~asirk/222_s11_ps2_key.pdfaddition...

Chem 222 Intro to Inorganic Chemistry Summer 2011

Problem Set 2 Wed May 25, 2011 1. What is the rough trend in the number of oxidation states available to the elements (excluding

the noble gases) as you move a) to the right and b) down the periodic table? Explain the reason for the trend moving down a group.

The number of oxidation states increases as you move to the right across a period and down a group. This is a very trend with numerous exceptions, especially moving across the period. This trend is not as robust as the solid trends in electronegativity or atomic radii. As you move down a group the ionization energy decreases, so it takes less energy to remove electrons and form higher oxidation states.

2. What are the most common oxidation states available to the alkalis, the alkaline earth metals, the noble gases and the halogens? Why do the heavier halogens have more oxidation states available to them?

The most common (and effectively the only) oxidation states available to the alkalis are 0 and +1, to the alkaline earth metals are 0 and +2. The most common oxidation states for the halogens are 0 and -1, but the heavier halogens are capable of supporting other oxidation states, specifically Cl, Br and I exist in the +1, +3, +5 oxidation states and I can also support +7. The ionization energies for the halogens decrease (dues to greater shielding from the Z of the positive nucleus of the outer shell electrons by the inner shell electrons-see Slater’s rules) going down the group allowing formation of these ions.

3. a) What is the trend in the boiling points of the halogens and noble gases as one moves down the group? Explain.

As one moves down the group the principle quantum number n increases. As this increases, the electrons are less tightly bound to the nucleus. Therefore the polarizability (formation of temporary dipoles due the probability nature of electrons) of the atoms or molecules increases and the London dispersion forces (aka instantaneous dipole: induced dipole) increase resulting in greater atom:atom or molecule:molecule attraction and increased boiling points. As neither the halogens, nor the noble gases have any permanent dipole, the only force to consider is the London dispersion forces.

b) What is the trend in melting point as one moves up the group in alkali metals? Explain. The melting point increases as we move up the group in the alkali metals. The bonding to consider is metallic bonding which is affected by 1) Number of electrons in the bond. 2) Size of the atom and 3) Degree of ionicity. In this case all atoms have the same number of electrons in the bond (1). The atoms gets smaller moving up the group and therefore the bond strength increases. The ionization energy


decreases moving down the group which would weaken the bond, but this is not as significant an effect as the change in size.

4. When halogens that form ionic compounds what is the order in which solubility increases and why?

When the halogens form ionic compounds the solubility increases in the order of I>Br>Cl>F. The governing force is the lattice energy which increases as the atomic radii decrease.

5. Explain why the radius of a cation, atom and anion for a given element range in size while the weight remains constant. Are cations always smaller than anions?

For a given element, the radius will change with anions > atoms > cations due to decreased shielding of the positive nucleus from the other electrons and therefore a higher effective nuclear charge (recall calculations on Cl, Cl- and Cl+ from class. The weight of an electron is 9.10938215×10−28 g , while the weight of a proton is 1.672621637(83)×10−24 g or about 10,000 times heavier. Therefore the removal or addition of an electron does not change the mass of the species in any significant fashion for our purposes.

The cation of a given element is always smaller than the anion for the reasons given above. However this does not always hold true when considering different elements. For example Cs+ (167 pm) is larger than F- (133 pm) . Also, there are other more complex ions ie: N(CH3)4

+ that are larger.

6. Write a balanced equation for the electrolysis of a) aqueous NaCl and b) NaCl in a molten NaCl/CaCl2.

a) 2(Na+ + e- Na) Cl- Cl2 + 2e- Overall: 2Na+ + 2Cl- +energy (electrical) 2Na + Cl2(gas) But note that the system is aqueous therefore the Na reacts directly with the water. 2Na + 2H2O 2Na + 2OH- + H2(gas) Therefore the net reaction is 2H2O + 2Cl- 2OH- (cathodic product) + Cl2 (gaseous anodic product) + H2 (gaseous cathodic

product) There are a few other points to consider. The gaseous products must be kept separate or else they will

react together directly to generate HCl. Also, in actuality, Na is not produced at the cathode in appreciable quantities; instead H2 is produced directly. This ratio depends on the relative concentrations of Na+ and the applied voltage and is beyond the scope of this course.

b) In a molten salt there is no water to consider so the net reaction is: 2(Na+ + e- Na) Cl- Cl2 + 2e- Overall: 2Na+ + 2Cl- + energy (electrical) 2Na + Cl2(gas) The CaCl2 does not react (Ca is even more difficult to reduce to the metal than Na: recall reactions with

water) and serves to decrease the melting point of the molten salt.


7. (RC&O 11.1) Write balanced chemical equations for each of the following reactions: a) sodium metal with water 2Na + 2H2O 2Na + 2-OH + H2(gas)

b) rubidium metal with oxygen (O2) Rb + O2 RbO2 (superoxide:O2

- , note that K and Cs also form the superoxide)

c) solid potassium hydroxide with carbon dioxide KOH + CO2 KHCO3

8. What are differences in reactivity with oxygen of the alkali and alkaline earth metals? It’s easiest to describe the reactivity with the alkaline earth metals first. All alkaline earth metals react

with oxygen to give MO, with the M in the 2+ oxidation state (the only oxidation state known to the alkaline earth other than 0) and the O in the standard 2- oxidation state.

2M + O2 2MO For the alkalis, only Li gives the “normal” oxide Li2O, while Na gives the peroxide and K, Rb and Cs

produce the superoxide. In all cases the metal is in the 1+ oxidation state (the only oxidation state known to the alkali other than 0).

4Li + O2 2Li2O (O2-) Na + O2 Na2O2 (O2

2-) K + O2 KO2 (O2

-) Rb + O2 RbO2 (O2

-) Cs + O2 CsO2 (O2

-)

9. (RC&O 2.17(a)) In the classification of elements into metals and nonmetals why is metallic lustre a poor guide?

Several non-metals have very shiny surfaces (e.g. iodine, silicon), and some compounds also look “metallic” (e.g. fool’s gold/pyrite FeS2). Metals can also lose their luster when oxides or other films form on the surface. The best general indicator of the metallic nature of an element is high three-dimensional electrical conductivity.

10. (H&S 6.1) 10. For a cubic close-packed arrangements of spheres give the (a) coordination numbers, (b) number of interstitial holes, and (c) the unit cell.

Cubic close-packed arrangements of spheres both relies layers of spheres in which each sphere is surrounded by six, in-plane, nearest neighbours. The subsequent vertical layer is identical to the first layer, but is offset so that the spheres sit in every second row of “wells” or “dips” left by the first layer. For cubic close-packed arrangements, the third vertical layer is again offset so that it fills the alternate layers of wells, the ones that persisted from the first layer, Thus the ccp layer repeat in an ABC fashion.

(Here’s the picture to illustrate:)


(a) Any sphere is surrounded by 12 nearest neighbours: 6 in the layer plane, 3 in the layer above, and 3 in the layer below.

(b) Even though the spheres in these arrangements are “close packed”, there are still gaps (interstitital holes), which are smaller than the spheres, left between the layers. The octahedral interstitial holes are defined by 6 spheres arranged in two triangles stacked one above the other and pointing in opposite directions, and tetrahedral interstitial holes, which are smaller, and are defined by four spheres in a (!) tetrahedral array. There are as many octahedral interstitial holes and there are spheres, and there are twice as many tetrahedral holes.

2.(c) The unit cell of the ccp structure is a face-centred cube.

11. (H&S 6.2) State the coordination number of a sphere in each of the following arrangements: (a) ccp: 12 (c) bcc: 8 (d) fcc: 12 (e) simple cubic: 6. Note that as discussed above, the fcc and ccp are the same system.

12. (RC&O 4.20) The atoms in barium metal are arranged in a body-centred cubic unit cell. Calculate the metallic radius of a barium atom if the density of barium is 3.50 g/cm3 .

Two different approaches to calculate this : 1) Density = mass/volume From our discussion of body-centred cubic structures, the volume of the barium unit cell is 12.33r3,

where r is the metallic radius of the barium atom. We also showed that there are two spheres (atoms) per unit cell, so the mass of the unit cell should be 2x(atomic mass of Ba) = 2(137.33 g/mol)/(6.022 x 1023 mol-1) = 45.61 x 10–23g.

Therefore: 3.50g/cm3 = (45.61 x 10-23g)/12.33r3 (12.33r3)(3.50 g/cm3) = 45.61 x 10-23


r3 = 1.06 x 10-23 cm3 = 10.6 x 10-24 cm3

r = (10.6)1/3 x 10–8 cm = (10.6)1/3 x 10–10m = 2.20 Å, 2) From diagonal line through BCC of closest neighbours

both of which correspond well with the metallic radius listed in H&S Appendix 6 of 224 pm for barium.

13. (Brown et al. 11.61) Iridium crystallizes in a face-centred cubic unit cell that has an edge length of 3.833Å. The atom in the centre of the face is in contact with the corner atoms, as shown in the drawing (of a face of the cube). (a) Calculate the atomic (metallic) radius of an iridium atom. (b) Calculate the density of iridium metal.

You will probably want to sketch a face-centred cubic unit cell:

a2 +c2 = d2 and a2 +a2 = c2

therefore a2 +a2 + a2 = d2 or d=√3 a and d=4r

a

c

d

ρ= 3.50 g/cm2 (BCC) aw=137.33 g/mole ρ=n(grams per cell)/a3 (volume of cell)

(2 atoms/cell)(137.33 grams/mol)

6.022*1023 atoms/mol

3.50grams/mol a3=n/ρ = =2.36*10-23 cm3

aBCC = 5.07* 10-8 cm

4rBa =√3aBCC

rBa =√3aBCC

4

rBa=2.19*10-8 cm or 2.19*10-10 m or 219 pm or 2.19 A


The length of the diagonal of a face is r + 2r + r = 4r. The relationship of the edge length (e) to the diagonal length is: (Diagonal length)2 = a2 + a2 = 2a2 Diagonal length = (√2)a = ~(1.44)(3.833)Å = 5.52Å (a) Therefore the rmetal for iridium is 1.38Å, which seems consistent with Ir being two periods below Co,

which has rmetal of 125 pm (H&S Appendix 6). (b) The density of iridium metal: = mass/volume = (4 x atomic mass of iridium)/(edge length)3 = (4 atoms/cell in a FCC x 192.22 g/mol)/(6.022 x 1023 mol-1)(3.833 x 10-8 cm)3

= (768.88 x 101)/(6.022)(3.833)3 = 22.67 g/cm3 (This is pretty close to the value of 22.56 g/cm3 listed on p. 750 in H&S.)

14. (H&S 6.3(a)) Lithium metal undergoes a phase change at 80K (1 bar pressure) for the - to the -form; one form is bcc and the other is a close-packed lattice. Suggest, with reasons, which form is which. What name is given to this type of structural change?

Lithium metal has a body-centred cubic structure at room temperature and 1 bar pressure (as do all the other alkali metals and barium, from the alkaline earth metals), you can rationalize that this is the structure at temperatures above 80K, while the structure at lower temperatures is the close-packed lattice, which is more dense. (What are the packing efficiencies for these two structure types?).

This type of structural change is called polymorphism.

Based on the various , , polymorphs of different metals described in H&S 6.4, it looks as though the lowest temperature structure is the , while the next higher temperature structure is the -polymorph.

15. (a) Explain why each of MgO and CaS contains isoelectronic ions. In MgO, the Mg is in the 2+ oxidation state, which means it has the same electron configuration as the

preceding noble gas, neon. Oxygen is in the 2– oxidation state (the oxide ion O2–), so it has the electron configuration of the next noble gas, which is again neon. Thus the two ions have identical electron configurations, and are called isoelectronic.

Likewise, Ca2+ and S2– both have the same electron configuration as the noble gas argon. (Can you write each of these electron configurations?)

(b) (H&S p.167 self study exercise 2) MgO adopts an NaCl structure. How many Mg2+ and O2– ions are present per unit cell?

You will want to draw a NaCl unit cell to work this out: In this unit cell, the anion occupies the corner sites. There are six anions occupying face sites, in which half the sphere is inside the

unit cell (= 3 anions), and eight anions occupying corner sites, in which 1/8 of the sphere is inside the unit cell (= 1 anion). Total equals 4 O2– in the cell.

There is one cation in the central (octahedral) site, which is entirely within the unit cell (= 1 cation). The remaining twelve cations occupy edge sites, in


which ¼ of the sphere is inside the unit cell (= 3 cations). Thus there are 4 Mg2+ in the cell.

16. (RC&O 5.20) Why, in the study of ionic lattices, is the anion packing often considered to be the framework into which the cations fit?

Because the effective nuclear charge felt by the outermost electrons in anionic species is low relative to that felt by neutral or cationic species (i.e. these electrons experience more shielding from the nuclear charge by the “interior” electrons), anions are typically larger than cations. In many cases the cations are actually small enough, relative to the anions, to “fit” into interstitial holes. (Recall octahedral holes have radii up to about 0.44 times the radius of the packed spheres, and the radii of tetrahedral holes are up to 0.23 times the radius of the packed spheres.)

17. (H&S 6.11 (part)) With reference to the CsCl solid structure type, explain what is meant by (a) coordination number, (b) unit cell, (c) ion sharing between unit cells.

(a) Recall that almost all of the alkali halides MX crystallize in the sodium chloride lattice (coordination number 6), including CsF, but the radius ratios of lower halides of cesium (which has a huge rcation of 170 pm) cause them to take a structure with the higher coordination number of 8. This is easiest to picture by visualizing the lattice as extended arrays of interpenetrating simple cubic structures, such that each ion finds itself at the centre of a cube defined by eight of the counterions. Thus coordination number refers to the number of nearest neighbours for each species in the structure.

(b) The unit cell is the smallest repeating unit in a solid state lattice. For the CsCl structure, the unit cell is a body-centred cube where the corners are one ion and the complete sphere in the centre is the counterion. It can be drawn with either ion in either role.

(c) In the cesium chloride unit cell, each ion placed in the corner sites is actually shared between this unit cell and seven other unit cells around it. (Another way to visualize this is to realize that the centre of that ion serves as a corner for eight different unit cells.) the ion occupying the centre of the unit cell is not shared.

18. (RC&O 5.36) Ammonium chloride crystallizes in the cesium chloride lattice. The cations and anions are in contact across the body diagonal of the unit cell and the edge length is 386 pm. Determine a value for the radius of the ammonium ion. (You can obtain ionic radii for elements from H&S appendix 6.)

The lattice of the ammonium chloride crystal is probably easiest to picture with the ammonium ion occupying an approximately spherical space in the centre of a body-centred cubic structure, for which the corners are chloride ions:

The body diagonal of the unit cell will be rCl- + 2rNH4+ + rCl- or 2(rCl- + rNH4+) The relationship of the body diagonal to the edge length is (see in-class example): diagonal = (3)1/3(edge) = (~1.73)(386 pm) = ~668 pm Therefore: 2(rCl- + rNH4+) = 668 pm (rCl- + rNH4+) = 334 pm rNH4+ = 334 pm – rCl- (from Appendix 6, rCl- = 181 pm) rNH4+ = 153 pm (note: rcovalent for N is listed as 75 pm, for H is 30-37 pm)


19. (RC&O 5.22) Suggest the probable structure for crystalline (a) barium fluoride; (b) potassium bromide; (c) magnesium sulfide.

(a) The cation radius ratio for BaF2 is: r+/r– = (142 pm/133 pm) = ~1.07. since this is >1, the Ba2+ probably has a coordination number of 8 in

the crystal lattice. The 1:2 stoichiometry of the salt implies that each F– has a coordination number of 4. This is implemented in a fluorite structure, which is essentially a face-centred cubic array of big cations in which all of the tetrahedral intersitial holes are filled with the anion. (Note, the face-centred cubic array of cations must be slightly stretched, or puffed out, to make the interstitial holes big enough to hold these anions!).

(b) The cation radius ratio for KBr is 138 pm/196 pm = 0.704, which suggests that the cation (and hence the anion in this 1:1 salt) will have a coordination number of 6. This salt probably has a simple NaCl structure.

(c) The cation radius ratio for MgS is 72 pm/184 pm = 0.391, which suggests that the cation (and hence the anion in this 1:1 salt) will have a coordination number of 4, although it is very close to the cutoff for coordination number 6. MgS might actually have an NaCl structure, but it could also have a “zinc blende” or “ZnS” type of structure. If you look at Figure 6.18 on p168 in H&S, you’ll see that this diamond-like structure gives a 1:1 ratio of four coordinate ions by adopting a face-centred cubic array of one ion, and filling one half of the tetrahedral insterstitial holes with the other ion. The ionic radii of Mg and Zn are 72 pm and 60 pm, respectively.

20. (RC&O 5.38) The unit cell of a particular solid has tungsten atoms at the corners, oxygen atoms in the centres of each cube edge, and a sodium atom in the cube centre. What is the empirical formula of the compound?

There are eight W atoms at the corners, so 8(1/8) = 1 W per unit cell. There are twelve O atoms in the centres of the edges, so 12(1/4) = 3 Na

per unit cell. There is one sodium atom in the cube centre, completely within the unit

cell. Therefore, the empirical formula of this compound is NaWO3


21. (RC&O 5.39) Astatine the lowest (and radioactive) member of the halogen series can form the astatinide ion At–, which has an approximate ionic radius of 225 pm. What lattice type would be expected for each of the alkali metal compounds of the astatinide ion?

alkali ion rcation (pm) r+/r– preferred coordination number

probable structure of Mat (1:1 salt)

Li+ 76 0.338 four ZnS (zinc blende or wurtzite)

Na+ 102 0.453 six NaCl

K+ 138 0.613 six NaCl

Rb+ 149 0.662 six NaCl

Cs+ 170 0.756 eight CsCl

Definitions/Concepts: 2nd ionization energy, crystal lattice, metallic radius, covalent radius, ionic radius, coordination number, radius ratio, interstices (or interstitial holes), cubic arrangement, close packing.

chem 222 intro to inorganic chemistry summer 2011 ...web.uvic.ca/~asirk/222_s11_ps2_key.pdfaddition...

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