Structure in Chemistry

IC. Ionic Compounds: Solutions for Selected Problems

Problem IC1.1.

a) Ta (tantalum)   is lower in the periodic table than V (vanadium)    

b) Hg (mercury)   is lower in the periodic table than Zn (zinc)   

c)  Si (silicon) is to the left in the periodic table compared to S (sulfur)

d)  W (tungsten) is to the left and lower in the periodic table compared to Cu (copper)

Problem IC1.2.

a) [He]2s22px22py12pz1

b)  [He]2s22px22py22pz2

c)  [Ar]4s2

d)  [Ar]

e)  [Ne]3s22px1

f)  [Ne]

g)  [He]2s22px12py12pz1

h)   [He]2s22px22py22pz2

i)   [Ne]2s22px22py22pz1

j)   [Ne]2s22px22py22pz2

 

Problem IC1.3.

a) 0.076 nm; smaller

b) 0.140 nm; larger

c) 0.167 nm; smaller

d)  0.144; larger

 

Problem IC1.4.

a) 0.154; larger

b) 0.071; smaller

c)  0.128; larger

d)  0.133; smaller

Problem IC2.1.

a) 1:1 K:Cl or KCl

b)  1:3 Fe:Cl or FeCl3

c)  1:6 Mo:Cl or MoCl6

d)  1:4 Zr:Cl or ZrCl4

Problem IC2.2.

a)  Li:O 2:1 or Li2O

b)  Fe:O 2:3 or Fe2O3

c)  Cr:O 1:3 or CrO3

d)  Ti:O 1:2 or TiO2

Problem IC2.3.

a)  Li:N 3:1 or Li3N

b)  Ta:N 1:1 or TaN

c)  W:N 1:2 or WN2

d)  Co:N 3:2 or Co3N2

Problem IC2.4.

a) Li4WO4  because tungstate anion would be 4-

b) Li4V4O12 because the tetravanadate ion would be 4-

c)  Li2Mo4O14 because the tetramolybdate ion would be 2-

d)  Li2Cr(OH)6Mo6O18  because the complex polyoxoanion would be 2-

Problem IC3.1.

a)  KCl would have the lowest melting point.  It would be easier to melt than LiCl because the smaller Li+ ion would more strongly attract the counterion, owing to the smaller distance separating the opposite charges.

b)  NaBr would have a lower melting point than NaF. 

c)  CaO would have a lower melting point than BeO.

d)  KBr would have a lower melting point than LiF.

Problem IC3.2.

 The answer is (a).  The sum of the cation and anion radius is longer; therefore the distance betwwen the two is longer, so the force of attraction will be weaker.

Problem IC3.3.

a) CaCl2

b) Na2O

 c) CaO

Problem IC3.4.

a) A little lower since K+ is a little larger than Na+ (actual= 770°C)

b) A little higher since F- is a little smaller than Cl-. (actual =930°C)

c) A lot higher since the anion charge is 2x higher. (actual =1275°C (sublimes but doesn’t melt))

d) A whole lot higher since cation and anion charge are 2x higher. (actual = 2000+ °C (decomposes above this temp))

Problem IC4.1.

a) 25 waters.

b)  2 units (2 anions and 2 cations).

c)  Half the water might only dissolve half the salt: 1 unit.

d)  Four times the water may dissolve four times the salt:  4 units.

Problem IC4.2.

a) LiFshould be more soluble.

b) KF should be more soluble (but keep reading).

c)  LiF should be more soluble.

Problem IC4.3.

The MgO contain more highly charged ions (Mg2+ and O2-) than LiF (Li+ and F-) and so it is more difficult to separate the ions from their solid state.

Problem IC4.4.

The ions interact with the water via electrostatic interactions, too.  The same distance factor that allows small ions to attract each other more strongly also allows small ions to interact more strongly with the water.

Problem IC4.5.

a)  The lattice energy of an ionic solid is a measure of the strength of bonds in that ionic compound.

b)  Charge of ion (directly related to lattice energy); Radius of each ion (inversely related to lattice energy)

c)  Stronger lattice energy results in a stronger bond. The stronger the bond, the more energy required to separate ions.

d)  Stronger lattice energy results higher mp or bp.

e)  Stronger lattice energy results in less soluble crystal lattice.

Problem IC4.6.

a) NaCl NaBr

b) KF CaF2

c) MgO Na2O

d) KF CsCl

e) RbBr CaCl2

Problem IC5.1.

Problem IC5.2.

Generally, but not always, anions are bigger than cations, so cations can pack efficiently into the holes between the anions.

Problem IC5.3.

i.  a) simple cubic; b) cubic hole; c) coordination number = 8; d)  all occupied; e) 8 x 1/8 Cl and 1 Cs (or vice versa, depending on how you define a unit cell); f)  CsCl

ii.  a) face centered cubic; b) octahedral hole; c) coordination number = 6; d)  all occupied; e) 6 x 1/2 plus 8 x 1/8 = 4 Cl and 1 plus 12 x 1/4 = 4 Na (or vice versa, depending on how you define a unit cell); f)  NaCl

iii.  a) face centered cubic; b) tetrahedral hole; c) coordination number = 4; d)  all occupied; e) 6 x 1/2 plus 8 x 1/8 = 4 Ca and 8 F ; f)  CaF2

 

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

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