1. Hint. Refer to Section 4.1 in Wulfsberg, especially Table 4.1.
 
 

2. Hint. To get the anhydride (oxide), determine the oxidation state of the central atom and write its formula as an oxide. This is equivalent to removing 2 H and 1 O (H₂O) from the acid or base, until you have only an oxide remaining.

Example: Pu(OH)₃. Plutonium in this compound is in the +3 oxidation state, so its basic anhydride is the oxide of Pu(III), Pu₂O₃. [equation: 2 Pu(OH)₃ ---> Pu₂O₃ + 3 H₂O]

Example: H₄XeO₆. Xenon is in the +8 oxidation state, so its acid anhydride is the oxide of Xe(VIII), XeO₄. [equation: H₄XeO₆ ---> XeO₄ + 2 H₂O]

Example. Write balanced equations for the reaction of XeO₃ with H₂O and with Na₂O.

Solution. Xe is in the +6 oxidation state. Table 2.7 (p. 40 in Wulfsberg) shows that the oxo anion formed by Xe(VI) is XeO₄^2-. Thus the reactions are expected to be

XeO₃ + H₂O ---> H₂XeO₄

XeO₃ + Na₂O ---> Na₂XeO₄

Example. Write a balanced equation for the reaction of SiO₂ with MgO.

Solution. This is more complex because there are so many silicate anions. Two possibilities are:

MgO + SiO₂ ---> MgSiO₃ (chain or cyclic silicate)

2 MgO + SiO₂ ---> Mg₂SiO₄ (orthosilicate)

Either equation would be correct unless you are told that there is excess MgO available, in which case only the second equation is correct.

Example (amphoteric oxide). Write balanced equations for the predicted reactions of Sb₂O₃ with HNO₃ and with NaOH.

Solution. Sb₂O₃ is an amphoteric oxide of antimony(III) (Table 4.1). The antimonite anion has the formula SbO₃^3- (Table 2.7). Thus the anticipated reactions are

Sb₂O₃ + 6 HNO₃ ---> 2 Sb(NO₃)₃ + 3 H₂O

Sb₂O₃ + 6 NaOH ---> 2 Na₃SbO₃ + 3 H₂O

6. Hints. All compounds in part (a) have the correct formula to be perovskites. All compounds in part (b) have the correct formulas to be spinels. What is necessary is to figure out whether the cations have the correct radii.

• Perovskites, typical formula ABO₃ where A is a divalent metal ion and B is a tetravalent metal ion (could also have A monovalent and B pentavalent). The prototype is CaTiO₃. The "A" cation should be within 20% of the radius of an oxide ion, as the A cation and the 3 oxide ions together make face-centered cubic packing. The "B" cation should be within 20% of the radius of Ti⁴⁺. Use the ionic radii in Table C of Wulfsberg to determine whether the ionic radii of both A and B fit into these requirements.
• Spinels, typical formula AB₂O₄. The oxide ions are cubic close-packed. The "A" cations are usually +2 metal ions with a radius of 80-110 pm, which occupy 1/8 of the tetrahedral holes in the oxide lattice. The "B" cations are usually +3 metal ions with a radius of 75-90 pm, which occupy half of the octahedral holes.

8. Hints. See Sections 4.7 and 4.8 on silicates and isomorphous substitution. Table 4.8 gives the correlation between silicate formulation and structure. The replacement of silicon by aluminum or other similar-sized +3 ions can lead to aluminosilicates with the same basic structure as the silicate from which it was derived. Thus, K[AlSi₃O₈] and Ba[Al₂Si₂O₈] each have a 1:2 ratio of (Al+Si) to oxygen in the aluminosilicate anions. This means that they are framework silicates, i.e. 3-dimensional networks resembling the SiO₂ 3-dimensional structure but more complex because of the need to incorporate the cations to balance anion charge when Al³⁺ replaces Si⁴⁺.