3.1.7-1. Na and K.

3.1.7-2. (c) and (d).

3.1.7-4. True, but only for compounds of alkali metals. As simple substances (metal atoms, not ions), the alkali metals have the oxidation state of 0.

3.1.7-5. False. Lithium is the lightest metal, but sodium and potassium are also less dense than H₂O.

3.1.7-6. 6 Na + N₂ = 2 Na₃N. Oxidant because it accepts electrons from Na.

3.1.7-7. While it is true that all alkali metals react with O₂, only Li is converted exclusively to the oxide (Li₂O) when combined with oxygen.

3.1.7-8. All of them are strong bases, but not equally strong.

3.1.7-10. The reaction of Li produced more hydrogen. As the atomic weight of Li is smaller than the atomic weight of Na, 1 g of Li contains more atoms/moles than 1 g of Na. Therefore, the reaction of Li will produce a larger number of molecules/moles of H₂, which translates to a higher volume of the gas.

3.1.7-11. (b) and (e).

3.1.7-12. False. They are all white, Cs salts included. The color of cesium metal is yellow, but Cs salts are white.

3.1.7-13. (b).

3.1.7-14. (c).

3.2.5-1. False. The ability of the alkaline earth metals to donate valence electrons increases in the order: Be < Mg < Ca < Sr < Ba.

3.2.5-2. True. The order of basicity is: Ba(OH)₂ < Mg(OH)₂ < Ca(OH)₂ < Sr(OH)₂ < Ba(OH)₂.

3.2.5-3. (b).

3.2.5-4. The reaction of MgO with water to give Mg(OH)₂ is very sluggish. In contrast, the reaction of CaO with H₂O is fast and highly exothermic. Therefore, CaO cannot survive in a water-containing environment, whereas MgO can.

3.2.5-6. (c).

3.2.5-7. Insoluble CaCO₃ reacts with CO₂ and water to give soluble Ca(HCO₃)₂, see Figures 3-14 and 3-16.

3.2.5-8. False. The info provided relates to temporary, not permanent water hardness.

3.2.5-9. As follows from the equation, MCO₃ + 2 HCl = MCl₂ + H₂O + CO₂↑, 1 mol of MCO₃ produces 1 mol of CO₂. The volume of the CO₂ produced was 22.4 L = 1 mol. Therefore, 100 g of the carbonate salt = 1 mol and, consequently, the molecular weight of the alkaline earth metal carbonate used in the experiment was 100 a.m.u. The molecular weight of the CO₃^2- anion is 60. Therefore, the atomic mass of the metal equals 100 - 60 = 40. This is the atomic mass of Ca, so the compound that was reacted with HCl was CaCO₃.

3.3.11-2. False. See 3.3.2.

3.3.11-3. False. Elemental halogens are too reactive to exist in nature. The only exception (and a notable one!) is antozonite, see subsection 3.3.1.

3.3.11-4. Electrolysis of an aqueous solution of NaCl produces NaOH and Cl₂. This is the way chlorine gas and NaOH are made in industry. Passing Cl₂ through NaOH in water gives bleach, a solution of NaClO.

3.3.11-5.

(a) 2 KBr + Cl₂ = 2 KCl + Br₂

(b) NaF + Cl₂ = no reaction (F is more electronegative than Cl)

(c) 2 Al + 3 Br₂ = 2 AlBr₃

(d) 2 NaI + Br₂ = 2 NaBr + I₂

(e) CuO + 2 HBr = CuBr₂ + H₂O

(f) 2 NaOH + Cl₂ = NaClO + NaCl + H₂O

(g) BaCl₂ + 2 AgNO₃ = Ba(NO₃)₂ + 2 AgCl↓

(h) Ca + I₂ = CaI₂

(i) 2 Fe + 3 Cl₂ = 2 FeCl₃

(j) Fe + 2 HCl = FeCl₂ + H₂

3.3.11-6. Molarity refers to the number of moles of a solute in 1 liter of solution. The given solution contains 0.84 mg = 0.00084 g of NaF in 1 liter. Dividing the amount of NaF in grams by the molecular weight of NaF (42) gives 0.00002 = 2 x 10^-5 mol. This is the molar concentration, 0.00002M or 2 x 10^-5 mol/L.

3.3.11-7. This is a somewhat tricky one. There is no discrepancy. The reaction of Berthollet's salt (KClO₃) with I₂ does occur, is well known, and makes perfect sense. A redox reaction involving any change in the oxidation states of two different halogens can occur if the more electronegative halogen is reduced (gains electrons) and the less electronegative halogen is oxidized (loses electrons). The reaction of Cl₂ with KI occurs because the more electronegative Cl atoms (oxidation state 0) accept electrons from the less electronegative iodine (oxidation state -1). The reverse process is impossible. In the reaction of I₂ with KClO₃, the less electronegative iodine loses electrons (is oxidized; oxidation state 0 changes to +5) and the more electronegative chlorine gains electrons (is reduced; oxidation state +5 changes to 0). The statement that a more reactive halogen displaces a less reactive halogen from its compounds is just a cliché.

3.3.11-8. False. Aqueous solutions of HCl, HBr, and HI are strong acids, but HF is a weak acid. The order of acidity is: HF < HCl < HBr < HI, see subsection 3.3.5.

3.3.11-9. False. The lack of precipitation indicates that the water did not contain chloride, bromide, or iodide anions because otherwise insoluble AgCl, AgBr, or AgI would precipitate. In contrast, AgF is soluble in water and would not precipitate.

3.3.11-10. False. Acids decompose NaClO, the active ingredient of bleach, to toxic and corrosive chlorine gas.

3.3.11-11. (b) and (e).

3.3.11-12. hypochlorous acid – hypochlorite; chlorous acid – chlorite; chloric acid – chlorate; perchloric acid – perchlorate.

3.4.4-1. 1s² 2s² 2p⁴ 3s² 3p⁴. A sulfur atom needs 2 electrons to attain the stable octet. This is realized in the molecule of H₂S due to the formation of two shared electron pairs between an S atom and two H atoms: H:S:H or H-S-H. Being in possession of 8 valence electrons, the S atom in H₂S cannot accept more, which means that it cannot act as an oxidant.

3.4.4-2. Pyrite, FeS₂, is the most common mineral of sulfur that is used to make sulfuric acid. In FeS₂, the iron and sulfur have the oxidation states of +2 and -1, respectively. As this is not obvious at all, do not be upset at yourself if you could not answer this question correctly. The actual 3-dimensional structure of pyrite is too complex to be studied in our introductory course. The fragment of the structure presented below shows that each S atom in pyrite is bonded to one Fe atom and to another S atom.

The sulfur-sulfur bridge between the Fe atoms in pyrite is similar to the O-O bridge in H₂O₂ or metal peroxides. You are expected to remember that the oxidation state of the oxygen atoms of the O-O bridge is -1 and understand why (Volume 2)].

3.4.4-3. (a).

3.4.4-4.

(a) S + O₂ = SO₂

(b) 2 SO₂ + O₂ = 2 SO₃ (in the presence of a vanadium catalyst)

(c) 2 H₂S + SO₂ = 3 S + 2 H₂O

(d) I₂ + SO₂ + 2 H₂O = 2 HI + H₂SO₄

(e) FeS + 2 HCl = FeCl₂ + H₂S↑

(f) MnO₂ + H₂SO₃ = MnSO₄ + H₂O

(g) SO₃ + H₂O = H₂SO₄

(h) SO₂ + 2 NaOH = Na₂SO₃ + H₂O

(i) SO₃ + Ca(OH)₂ = CaSO₄ + H₂O

(j) H₂S + 2 KOH = K₂S + 2 H₂O

Redox reactions: (a), (b), (c), (d), and (f).

3.4.4-7. Because the reaction of SO₃ with water is irreversible, producing odorless H₂SO₄, whereas that of SO₂ is reversible (Figure 3-35). Sulfurous acid is in equilibrium with water and SO₂, hence the smell of SO₂. In contrast, there is no SO₃ in H₂SO₄ or its solutions. See 3.4.2.

3.4.4-8. Mass percent concentration refers to the gram quantity of a solute in 100 g of solution. Oleum is a solution of SO₃ in H₂SO₄. Therefore, 100 g of 30% oleum contains 30 g SO₃ whose molecular weight is 80. This amount of SO₃ in mol = 30/80 = 0.375 mol. The stoichiometry (SO₃ + H₂O = H₂SO₄) indicates that each mol of SO₃ reacts with one mol of H₂O. Therefore, 0.375 mol of water should be used. Given the molecular weight of H₂O (18), the amount of water to be mixed with the oleum is 18 x 0.375 = 6.75 g = 6.75 mL. Important: Similarly to the proper way of H₂SO₄ dilution, it is always oleum that should be carefully added to water, not the other way around.

3.4.4-9. True.

3.4.4-10. (b). Since K₂SO₃ is formed by a strong base (KOH) and a weak acid (H₂SO₃), its solution in water should be alkaline (pH >7).

3.4.4-11. Both are correct. We should keep in mind that the reaction of SO₂ with water to give H₂SO₃ is reversible. Therefore, a solution of SO₂ in water contains both SO₂ and H₂SO₃. Both are reductants since the sulfur atom in both has the same oxidation state of +4. Consequently, the bottom equation in Figure 3-36, Fe₂(SO₄)₃ + H₂SO₃ + H₂O = 2 FeSO₄ + 2 H₂SO₄, may also be rewritten in its different yet equally correct form: Fe₂(SO₄)₃ + SO₂ + 2 H₂O = 2 FeSO₄ + 2 H₂SO₄.