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]2s²2p_x²2p_y¹2p_z¹
2. [He]2s²2p_x²2p_y²2p_z²
3. [Ar]4s²
4. [Ar]
5. [Ne]3s²2p_x¹
6. [Ne]
7. [He]2s²2p_x¹2p_y¹2p_z¹
8. [He]2s²2p_x²2p_y²2p_z²
9. [Ne]2s²2p_x²2p_y²2p_z¹
10. [Ne]2s²2p_x²2p_y²2p_z²

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 O^2-) 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

Contributors

Chris P Schaller, Ph.D., (College of Saint Benedict / Saint John's University)