Skip to main content
Chemistry LibreTexts

IC6. Ionic Compounds: Solutions for Selected Problems

Problem IC1.1.
  1. Ta (tantalum) is lower in the periodic table than V (vanadium)    
  2. Hg (mercury) is lower in the periodic table than Zn (zinc)   
  3. Si (silicon) is to the left in the periodic table compared to S (sulfur)
  4. W (tungsten) is to the left and lower in the periodic table compared to Cu (copper)
Problem IC1.2.
  1. [He]2s22px22py12pz1
  2. [He]2s22px22py22pz2
  3. [Ar]4s2
  4. [Ar]
  5. [Ne]3s22px1
  6. [Ne]
  7. [He]2s22px12py12pz1
  8. [He]2s22px22py22pz2
  9. [Ne]2s22px22py22pz1
  10. [Ne]2s22px22py22pz2
Problem IC1.3.
  1. 0.076 nm; smaller
  2. 0.140 nm; larger
  3. 0.167 nm; smaller
  4. 0.144; larger
Problem IC1.4.
  1. 0.154; larger
  2. 0.071; smaller
  3. 0.128; larger
  4. 0.133; smaller
Problem IC2.1.
  1. 1:1 K:Cl or KCl
  2. 1:3 Fe:Cl or FeCl3
  3. 1:6 Mo:Cl or MoCl6
  4. 1:4 Zr:Cl or ZrCl4
Problem IC2.2.
  1. Li:O 2:1 or Li2O
  2. Fe:O 2:3 or Fe2O3
  3. Cr:O 1:3 or CrO3
  4. Ti:O 1:2 or TiO2
Problem IC2.3.
  1. Li:N 3:1 or Li3N
  2. Ta:N 1:1 or TaN
  3. W:N 1:2 or WN2
  4. Co:N 3:2 or Co3N2
Problem IC2.4.
  1. Li2WO4  because tungstate anion would be 2-
  2. Li4V4O12 because the tetravanadate ion would be 4-
  3. Li2Mo4O14 because the tetramolybdate ion would be 2-
  4. Li2Cr(OH)6Mo6O18  because the complex polyoxoanion would be 2-
Problem IC3.1.
  1. 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.
  2. NaBr would have a lower melting point than NaF. 
  3. CaO would have a lower melting point than BeO.
  4. KBr would have a lower melting point than LiF.
Problem IC4.1.
  1. 25 waters.
  2. 2 units (2 anions and 2 cations).
  3. Half the water might only dissolve half the salt: 1 unit.
  4. Four times the water may dissolve four times the salt:  4 units.
Problem IC4.2.
  1. LiF should be more soluble.
  2. KF should be more soluble (but keep reading).
  3. 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 IC5.1.

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

Problem IC5.2.
  1. 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
  2. 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
  3. 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