3.5.9-1. The molecule of N₂ is exceptionally inert and, as a consequence, does not react with other substances in the environment. In contrast, elemental phosphorus (white phosphorus, P₄) is a highly reactive substance that easily combines with oxygen in the atmosphere.

3.5.9-2. The nitrogen-nitrogen bond is a triple bond (N≡N or N:::N), which makes it very strong and energetically demanding to cleave. In the molecule of N₂, each N atom has 8 electrons in the outermost shell, which makes the N atoms "happy" (the octet rule).

3.5.9-3. N₂ + 3 H₂ = 2 NH₃; (c) and (d).

3.5.9-4. Both. The higher the pressure, the better it is for the process. In contrast, the temperature of the NH₃ synthesis is carefully optimized. Although raising the temperature accelerates the reaction, it also shifts the equilibrium N₂ + 3 H₂ ⇄ 2 NH₃ to the left because the reaction is exothermic. See Volume 2, section 2.12.5.

3.5.9-5. (b).

3.5.9-6. (c).

3.5.9-8. (d).

3.5.9-10. False. Although the formation of hydrogen in a reaction of HNO₃ with a metal is seldom observed, some metals do react with dilute HNO₃ to give H₂. For example, H₂ is the main reduction product of the reaction of highly diluted nitric acid with magnesium.

3.5.9-11. The nature of the metal and acid concentration.

3.5.9-12. Cu, Hg, and Ag are the least reactive metals that can reduce HNO₃. On treatment with concentrated HNO₃, these metals produce NO₂. Dilute HNO₃ reacts with these metals to give NO. For the balanced equations, see Figures 3-51, 3-52, and 3-53.

3.5.9-13. (a) NO₂; (b) NO and NO₂; (c) NH₄⁺, N₂, N₂O, and NO. See 3.5.5.

3.5.9-14. Both. Al and Fe do not react with highly concentrated HNO₃, but do react with dilute HNO₃. Containers made of aluminum or steel can therefore be used to store and transport HNO₃, but only of high concentration.

3.5.9-15. (b).

3.5.9-20. Phosphoric acid is a tribasic acid that can form three types of salts depending on how much metal hydroxide is used per certain amount of the acid (see Figure 3-70). 49 g of H₃PO₄ = 0.5 mol; 40 g of NaOH = 1 mol. The 0.5:1 = 1:2 molar ratio of H₃PO₄ to NaOH ensures the formation of Na₂HPO₄: H₃PO₄ + 2 NaOH = Na₂HPO₄ + 2 H₂O.

3.6.6-1. Both carbon and silicon have four valence electrons. To attain the electron configuration of a noble gas by becoming a cation or anion, a carbon or silicon atom would need to either donate 4 electrons or accept 4 electrons, which is too many. Consequently, carbon and silicon atoms prefer to form shared electron pairs.

3.6.6-2. Graphite and diamond.

3.6.6-5. True. See 3.6.2.

3.6.6-6. Watch this video demonstration.

3.6.6-7. False. While CO₂ does react with H₂O to give H₂CO₃, CO does not react with water to give an acid.

3.6.6-8. CO + H₂O = CO₂ + H₂ (in the presence of a catalyst). Water-gas shift reaction is run on a large industrial scale to make hydrogen.

3.6.6-9. (c).

3.6.6-10. Mixing Na₂CO₃ with acetic acid in a 1:1 molar ratio would produce NaHCO₃ but also sodium acetate (Na salt of acetic acid) that are not easy to separate from each other. Instead, a portion of Na₂CO₃ should be treated with excess vinegar to produce CO₂ that should be bubbled, in excess, through a solution of another portion of Na₂CO₃ to give a solution of baking soda:

Na₂CO₃ + H₂O + CO₂ = 2 NaHCO₃

Leaving the resultant solution to evaporate will produce pure solid NaHCO₃.

3.6.6-11. The main ingredients of toothpaste are CaCO₃ and water. Heating toothpaste would first result in evaporation of the water. Further heating at a much higher temperature would give rise to CaO due to the decomposition: CaCO₃ = CaO + CO₂. Mixing the thus produced CaO with water would produce limewater, an aqueous solution of Ca(OH)₂: CaO + H₂O = Ca(OH)₂. If limewater turns cloudy on bubbling a gas through it, the gas contains CO₂: Ca(OH)₂ + CO₂ = CaCO₃↓ + H₂O.

3.6.6-12. Almost true, but not true. Silicon is the second most abundant element in the Earth's crust, next to oxygen.

3.6.6-13. True.

3.6.6-15. (a) SiO₂ and CaO; (b) SiO₂, CaCO₃, and Na₂CO₃.

3.7.4-1. All.

3.7.4-2. Bauxite.

3.7.4-4. (b).

3.7.4-5. The amount of H₂ produced will be the same. Regardless of whether it is an alkaline or acidic solution that is used for the reaction, it is hydrogen in the oxidation state of +1 that is reduced. 1 g of Al metal can reduce a particular quantity of H (+1) regardless of its source.

3.7.4-7. False. Aluminum metal does react with water if the protecting layer of alumina is somehow removed from the surface. See section 3.7.3 and Figure 3-90.

3.7.4-8. If you were able to solve this problem, you are great, congratulations! If you could not solve it, do not be too upset, as this is a somewhat tricky one. I have met PhD chemists who were not able to find a solution to this problem. On mixing solutions of AlCl₃ and NaOH, first insoluble Al(OH)₃ is formed and precipitates out: AlCl₃ + 3 NaOH = Al(OH)₃↓ + 3 NaCl. However, if NaOH is used in excess, the insoluble amphoteric Al(OH)₃ first formed will dissolve due to the reaction producing soluble NaAlO₂ (see Figure 3-93, right). To tell which test tube contains which solution, a small amount (say a couple of drops) of either solution should be carefully added to the other solution without agitation. A white precipitate of Al(OH)₃ will be formed at the top of the liquid within the time of mixing due to the formation of Al(OH)₃. Now, swirl the contents of the tube. If the precipitated Al(OH)₃ dissolves on agitation, the solution that was originally in that test tube is NaOH because Al(OH)₃ dissolves in NaOH (see above). If the precipitate of Al(OH)₃ does not dissolve on thorough mixing, the test tube with the mixed solution originally contained AlCl₃ with which insoluble Al(OH)₃ does not react.

3.8.5-1. Magnetite (Fe₃O₄) and hematite (Fe₂O₃).

3.8.5-2. (c).

3.8.5-3. Fe₃O₄ is a mixed Fe(II)/Fe(III) oxide that can be represented as FeO•Fe₂O₃. One of the three Fe atoms in Fe₃O₄ has an oxidation state of +2 and the other two +3.

3.8.5-5.

(a) Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂
(b) 2 Fe₂O₃ + 3 C = 4 Fe + 3 CO₂
(c) Fe₂O₃ + 3 H₂ = 2 Fe + 3 H₂O
(d) Fe₂O₃ + 2 Al = 2 Fe + Al₂O₃
(e) Cu cannot reduce Fe₂O₃ because copper is less reactive (poorer electron donor) than Fe, see Figure 3-50.

3.8.5-7.

(1) Iron dissolves in HCl: Fe + 2 HCl = FeCl₂ + H₂
(2) On treatment of FeCl₂ with KOH in water, Fe(OH)₂ precipitates out: FeCl₂ + 2 KOH = Fe(OH)₂ + 2 KCl
(3) In the presence of alkali and air, Fe(OH)₂ is oxidized to Fe(OH)₃: 4 Fe(OH)₂ + O₂ + 2 H₂O = 4 Fe(OH)₃
(4) Fe(OH)₃ decomposes on heating: 2 Fe(OH)₃ = Fe₂O₃ + 3 H₂O
(5) Fe(OH)₃ reacts with HNO₃ to give Fe(NO₃)₃: Fe(OH)₃ + 3 HNO₃ = Fe(NO₃)₃ + 3 H₂O
(6) Ferric nitrate decomposes on heating: 4 Fe(NO₃)₃ = 2 Fe₂O₃ + 12 NO₂ + 3 O₂

3.8.5-8. (d).

3.8.5-9. Sample A; see subsection 3.8.4.

3.9.2-1.

(a) HSO₃^- + H₂O - 2 e^- → SO₄^2- + 3 H⁺ (acidic conditions)

(b) 2 NO₃^- + 12 H⁺ + 10 e^- → N₂ + 6 H₂O (acidic conditions)

(c) 2 NO₃^- + 10 H⁺ + 8 e^- → N₂O + 5 H₂O (acidic conditions)

(d) NO₃^- + 2 H⁺ + e^- → NO₂ + H₂O (acidic conditions)

(e) H₂O₂ + 2 OH^- - 2 e^- → O₂ + 2 H₂O (basic conditions)

(f) P + 4 H₂O - 5 e- → PO₄^3- + 8 H⁺ (acidic conditions)

(g) Cl₂ + 12 OH^- - 10 e^- → 2 ClO₃^- + 6 H₂O (basic conditions)

(h) NO₃^- + 6 H₂O + 8 e^- → NH₃ + 9 OH^- (basic conditions)

(i) MnO₄^- + 2 H₂O + 3 e^- → MnO₂ + 4 OH^- (neutral conditions)

(j) H₂ - 2 e^- → 2 H⁺ (acidic conditions)

(k) Cl₂ + 4 OH^- - 2 e^- → 2 ClO^- + 2H₂O (basic conditions)

(l) NO₂^- + H₂O - 2 e^- → NO₃^- + 2 H⁺ (acidic conditions)

(m) NO₂ + 2 OH^- - e^- → NO₃^- + H₂O (basic conditions)

(n) Just kidding. There is no change in the oxidation states of N or H in the formation of NH₄⁺ from NH₃.

(o) H₂S + 4 H₂O - 8 e^- → SO₄^2- + 10 H⁺ (acidic conditions)

(p) Cr₂O₇^2- + 14 H⁺ + 6 e^- → 2 Cr³⁺ + 7 H₂O (acidic conditions)

(q) MnO₄^- + 8 H⁺ + 5 e^- → Mn²⁺ + 4 H₂O (acidic conditions)

(r) I^- + 3 H₂O - 6 e^- → IO₃^- + 6 H⁺ (acidic conditions)

(s) Zn + 4 OH^- - 2 e^- → ZnO₂^2- + 2 H₂O (basic conditions)

(t) 2 Cu²⁺ + 2 OH^- + 2 e^- → Cu₂O + H₂O (basic conditions)

(u) Cu₂O + 2 H⁺ - 2 e- → 2 Cu²⁺ + H₂O (acidic conditions)

(v) ClO₃^- + 2 H⁺ + e^- → ClO₂ + H₂O (acidic conditions)

(w) Mn₂O₇ + 14 H⁺ + 10 e^- → 2 Mn²⁺ + 7 H₂O (acidic conditions)

(x) CrO₄^2- + 4 H₂O + 3 e^- → Cr(OH)₃ + 5 OH^- (basic conditions)

(y) Al + 4 OH^- - 3 e^- → AlO₂^- + 2 H₂O (basic conditions)

(z) FeO + 2 H⁺ - e^- → Fe³⁺ + H₂O (acidic conditions)