Chapter 18 Chemical Bonds Section 1 Review Answers

1 .

The alkali metals all have a unmarried southward electron in their outermost trounce. In dissimilarity, the alkaline earth metals accept a completed southward subshell in their outermost beat out. In full general, the alkali metals react faster and are more reactive than the corresponding alkali metal globe metals in the aforementioned menses.

3 .

$\begin{array}{l}\text{Na}+{\text{I}}_{\mathrm{two}}⟶\text{2NaI}\\ \text{2Na}+\text{Se}⟶{\text{Na}}_{\mathrm{ii}}\text{Se}\\ \text{2Na}+{\text{O}}_2⟶{\text{Na}}_2{\text{O}}_{\mathrm{ii}}\end{array}$
$\begin{array}{l}\text{Sr}+{\text{I}}_2⟶{\text{SrI}}_2\\ \text{Sr}+\text{Se}⟶\text{SrSe}\\ \text{2Sr}+{\text{O}}_2⟶\text{2SrO}\end{array}$
$\begin{array}{l}\text{2Al}+{\text{3I}}_2⟶{\text{2AlI}}_3\\ \text{2Al}+\text{3Se}⟶{\text{Al}}_2{\text{Se}}_{\mathrm{iii}}\\ \text{4Al}+{\text{3O}}_2⟶{\text{2Al}}_2{\text{O}}_3\end{array}$

five .

The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame exam that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At 20 °C, NaCl dissolves to the extent of $\frac{\text{35.vii g}}{\text{100 mL}}$ compared with $\frac{\text{53.8 g}}{\text{100 mL}}$ for SrCl₂. Heating to 100 °C provides an easy test, since the solubility of NaCl is $\frac{\text{39.12 g}}{\text{100 mL}},$ but that of SrCl_ii is $\frac{\text{100.8 g}}{\text{100 mL}}.$ Density determination on a solid is sometimes difficult, but at that place is enough difference (2.165 g/mL NaCl and 3.052 g/mL SrCl₂) that this method would exist viable and perhaps the easiest and least expensive test to perform.

7 .

(a) $\text{2Sr}(s)+{\text{O}}_2(\mathrm{grand})⟶\text{2SrO}(s);$ (b) $\text{Sr}(s)+\text{2HBr}(\mathrm{1000})⟶{\text{SrBr}}_2(\mathrm{south})+{\text{H}}_{\mathrm{ii}}(\mathrm{thou});$ (c) $\text{Sr}(s)+{\text{H}}_{\mathrm{two}}(\mathrm{thou})⟶{\text{SrH}}_{\mathrm{two}}(s);$ (d) $\text{6Sr}(\mathrm{due\; south})+{\text{P}}_4(s)⟶{\text{2Sr}}_3{\text{P}}_2(\mathrm{south});$ (due east) $\text{Sr}(s)+{\text{2H}}_2\text{O}(l)⟶\text{Sr}{(\text{OH})}_2(aq)+{\text{H}}_2(g)$

11 .

Yeah, tin reacts with hydrochloric acid to produce hydrogen gas.

13 .

In PbCl₂, the bonding is ionic, as indicated by its melting point of 501 °C. In PbCl₄, the bonding is covalent, equally evidenced by it existence an unstable liquid at room temperature.

fifteen .

$\text{2CsCl}(l)+\text{Ca}(\mathrm{yard})\stackrel{\begin{array}{l}\text{countercurrent}\\ \text{fractionating}\\ \text{tower}\end{array}}{\to }\text{2Cs}(g)+{\text{CaCl}}_{\mathrm{two}}(\mathrm{50})$

17 .

Cathode (reduction): ${\text{2Li}}^{+}+{\text{2e}}^{-}⟶\text{2Li}(l)\text{;}$ Anode (oxidation): ${\text{2Cl}}^{-}⟶{\text{Cl}}_2(g)+{\text{2e}}^{\text{−}};$ Overall reaction: ${\text{2Li}}^{+}+{\text{2Cl}}^{-}⟶\text{2Li}(l)+{\text{Cl}}_2(\mathrm{chiliad})$

21 .

Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come up in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Simply if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid called-for.

23 .

Extract from ore: $\text{AlO}(\text{OH})(\mathrm{southward})+\text{NaOH}(aq)+{\text{H}}_2\text{O}(\mathrm{fifty})⟶\text{Na}\left[\text{Al}{(\text{OH})}_4\right](aq)$
Recover: $\text{2Na}\left[\text{Al}{(\text{OH})}_4\right](s)+{\text{H}}_2{\text{SO}}_4(aq)⟶\text{2Al}{(\text{OH})}_{\mathrm{iii}}(s)+{\text{Na}}_2{\text{And then}}_{\mathrm{four}}(aq)+{\text{2H}}_2\text{O}(l)$
Sinter: $\text{2Al}{(\text{OH})}_{\mathrm{iii}}(s)⟶{\text{Al}}_{\mathrm{two}}{\text{O}}_3(s)+{\text{3H}}_{\mathrm{ii}}\text{O}(k)$
Dissolve in Na_threeAlF₆(50) and electrolyze: ${\text{Al}}^{\mathrm{three+}}+{\text{3e}}^{-}⟶\text{Al}(s)$

29 .

(a) H₃BPH_iii:

(b) ${\text{BF}}_4{}^{\text{−}}:$

(c) BBr₃:

(d) B(CH_iii)₃:

(e) B(OH)_three:

33 .

(a) (CH₃)_iiiSiH: sp ³ bonding about Si; the structure is tetrahedral; (b) ${\text{SiO}}_{\mathrm{four}}{}^{\text{iv−}}:$ sp ⁱⁱⁱ bonding virtually Si; the construction is tetrahedral; (c) Si₂H₆: sp ³ bonding about each Si; the structure is linear along the Si-Si bond; (d) Si(OH)₄: sp ³ bonding about Si; the structure is tetrahedral; (due east) ${\text{SiF}}_6{}^{\text{2−}}:$ sp ^three d ^two bonding nearly Si; the structure is octahedral

35 .

(a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar

37 .

(a) tellurium dioxide or tellurium(4) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride

39 .

Boron has only s and p orbitals bachelor, which tin can accommodate a maximum of iv electron pairs. Different silicon, no d orbitals are available in boron.

41 .

(a) ΔH° = 87 kJ; ΔK° = 44 kJ; (b) ΔH° = −109.9 kJ; ΔG° = −154.seven kJ; (c) ΔH° = −510 kJ; ΔG° = −601.5 kJ

43 .

A mild solution of hydrofluoric acid would dissolve the silicate and would non harm the diamond.

45 .

In the N₂ molecule, the nitrogen atoms have an σ bail and two π bonds holding the two atoms together. The presence of 3 stiff bonds makes N₂ a very stable molecule. Phosphorus is a third-period chemical element, and every bit such, does not form π bonds efficiently; therefore, it must fulfill its bonding requirement by forming 3 σ bonds.

47 .

(a) H = i+, C = two+, and N = 3−; (b) O = ii+ and F = 1−; (c) Equally = iii+ and Cl = i−

51 .

The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself.

53 .

Hydrogen has but one orbital with which to bail to other atoms. Consequently, only one two-electron bond can form.

57 .

(a) $\text{Ca}{(\text{OH})}_2(aq)+{\text{CO}}_2(g)⟶{\text{CaCO}}_3(\mathrm{south})+{\text{H}}_{\mathrm{ii}}\text{O}(l);$ (b) $\text{CaO}(s)+{\text{So}}_2(g)⟶{\text{CaSO}}_3(\mathrm{south});$
(c) ${\text{2NaHCO}}_{\mathrm{three}}(\mathrm{due\; south})+{\text{NaH}}_{\mathrm{ii}}{\text{PO}}_4(aq)⟶{\text{Na}}_3{\text{PO}}_4(aq)+{\text{2CO}}_{\mathrm{two}}(\mathrm{one\; thousand})+{\text{2H}}_{\mathrm{two}}\text{O}(l)$

59 .

(a) NH²⁻:

(b) North₂F₄:

(c) ${\text{NH}}_{\mathrm{ii}}{}^{\text{−}}:$

(d) NF₃:

(east) ${\text{N}}_3{}^{\text{−}}:$

61 .

Ammonia acts as a Brønsted base because information technology readily accepts protons and as a Lewis base in that it has an electron pair to donate.
Brønsted base: ${\text{NH}}_3+{\text{H}}_3{\text{O}}^{+}⟶{\text{NH}}_4{}^{+}+{\text{H}}_{\mathrm{two}}\text{O}$
Lewis base: ${\text{2NH}}_{\mathrm{iii}}+{\text{Ag}}^{+}⟶{{\text{[H}}_{\mathrm{three}}\text{North}-\text{Ag}-{\text{NH}}_3]}^{+}$

63 .

(a) NO₂:

Nitrogen is sp ² hybridized. The molecule has a bent geometry with an ONO bond angle of approximately 120°.
(b) ${\text{NO}}_2{}^{\text{−}}:$

Nitrogen is sp ² hybridized. The molecule has a bent geometry with an ONO bail angle slightly less than 120°.
(c) ${\text{NO}}_2{}^{\text{+}}:$

Nitrogen is sp hybridized. The molecule has a linear geometry with an ONO bond angle of 180°.

65 .

Nitrogen cannot grade a NF₅ molecule because it does not have d orbitals to bail with the boosted two fluorine atoms.

69 .

(a) ${\text{P}}_4(s)+\text{4Al}(s)⟶\text{4AlP}(s);$ (b) ${\text{P}}_4(s)+\text{12Na}(\mathrm{due\; south})⟶{\text{4Na}}_{\mathrm{three}}\text{P}(s);$ (c) ${\text{P}}_{\mathrm{four}}(s)+{\text{10F}}_2(\mathrm{1000})⟶{\text{4PF}}_5(\mathrm{50});$ (d) ${\text{P}}_4(s)+{\text{6Cl}}_2(\mathrm{yard})⟶{\text{4PCl}}_3(l)$ or ${\text{P}}_4(\mathrm{south})+{\text{10Cl}}_2(\mathrm{one\; thousand})⟶{\text{4PCl}}_5(\mathrm{50});$ (e) ${\text{P}}_4(s)+{\text{3O}}_{\mathrm{two}}(g)⟶{\text{P}}_{\mathrm{iv}}{\text{O}}_6(s)$ or ${\text{P}}_4(s)+{\text{5O}}_2(g)⟶{\text{P}}_{\mathrm{iv}}{\text{O}}_{\mathrm{ten}}(\mathrm{southward});$ (f) ${\text{P}}_4{\text{O}}_{\mathrm{half\; dozen}}(s)+{\text{2O}}_2(\mathrm{thou})⟶{\text{P}}_{\mathrm{iv}}{\text{O}}_{10}(\mathrm{due\; south})$

77 .

(a) P = 3+; (b) P = five+; (c) P = 3+; (d) P = 5+; (east) P = 3−; (f) P = v+

81 .

(a) $\text{2Zn}(s)+{\text{O}}_2(k)⟶\text{2ZnO}(s);$ (b) ${\text{ZnCO}}_{\mathrm{iii}}(\mathrm{southward})⟶\text{ZnO}(s)+{\text{CO}}_2(g);$ (c) ${\text{ZnCO}}_3(\mathrm{due\; south})+2{\text{CH}}_3\text{COOH}(aq)⟶\text{Zn}{({\text{CH}}_{\mathrm{three}}\text{COO})}_2(aq)+{\text{CO}}_{\mathrm{ii}}(g)+{\text{H}}_2\text{O}(l);$ (d) $\text{Zn}(s)+\text{2HBr}(aq)⟶{\text{ZnBr}}_{\mathrm{ii}}(aq)+{\text{H}}_2(g)$

83 .

$\text{Al}{(\text{OH})}_3(s)+\mathrm{three}{\text{H}}^{\text{+}}(aq)⟶{\text{Al}}^{\mathrm{3+}}+3{\text{H}}_2\text{O}(\mathrm{50});$ $\text{Al}{(\text{OH})}_3(s)+{\text{OH}}^{-}⟶{\left[\text{Al}{(\text{OH})}_4\right]}^{\text{−}}(aq)$

85 .

(a) ${\text{Na}}_2\text{O}(s)+{\text{H}}_{\mathrm{ii}}\text{O}(l)⟶\text{2NaOH}(aq);$ (b) ${\text{Cs}}_2{\text{CO}}_{\mathrm{iii}}(s)+\text{2HF}(aq)⟶\text{2CsF}(aq)+{\text{CO}}_2(g)+{\text{H}}_{\mathrm{ii}}\text{O}(l);$ (c) ${\text{Al}}_2{\text{O}}_{\mathrm{three}}(s)+\mathrm{half\; dozen}{\text{HClO}}_4(aq)⟶\text{2Al}{({\text{ClO}}_{\mathrm{four}})}_{\mathrm{three}}(aq)+\mathrm{three}{\text{H}}_2\text{O}(\mathrm{fifty});$ (d) ${\text{Na}}_{\mathrm{two}}{\text{CO}}_3(aq)+\text{Ba}{({\text{NO}}_3)}_{\mathrm{ii}}(aq)⟶2{\text{NaNO}}_3(aq)+{\text{BaCO}}_{\mathrm{iii}}(\mathrm{due\; south});$ (e) ${\text{TiCl}}_4(l)+\text{4Na}(s)⟶\text{Ti}(s)+\text{4NaCl}(s)$

87 .

HClO_iv is the stronger acid because, in a series of oxyacids with like formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bail, making the hydrogen more easily released. The weaker this bond, the stronger the acrid.

89 .

As H₂Then_iv and H_iiSeO_four are both oxyacids and their central atoms both take the aforementioned oxidation number, the acrid strength depends on the relative electronegativity of the central atom. Equally sulfur is more electronegative than selenium, H₂SO_four is the stronger acid.

91 .

Then₂, sp ² 4+; And so₃, sp ², 6+; H_twoSo_four, sp ⁱⁱⁱ, half-dozen+

93 .

SF_vi: S = 6+; And then₂F₂: S = half-dozen+; KHS: Southward = 2−

95 .

Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen.

97 .

At that place are many possible answers including: $\text{Cu}(\mathrm{southward})+2{\text{H}}_2{\text{And so}}_4(l)⟶{\text{CuSO}}_4(aq)+{\text{And so}}_2(g)+\mathrm{two}{\text{H}}_2\text{O}(\mathrm{50})$ and $\text{C}(s)+\mathrm{two}{\text{H}}_2{\text{And so}}_4(\mathrm{50})⟶{\text{CO}}_2(g)+\mathrm{ii}{\text{Then}}_2(g)+2{\text{H}}_2\text{O}(l)$

101 .

SnCl₄ is non a salt because it is covalently bonded. A common salt must accept ionic bonds.

103 .

In oxyacids with similar formulas, the acrid strength increases every bit the electronegativity of the central cantlet increases. HClO₃ is stronger than HBrO₃; Cl is more electronegative than Br.

107 .

(a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite

109 .

(a) I: seven+; (b) I: vii+; (c) Cl: 4+; (d) I: iii+; Cl: 1−; (e) F: 0

111 .

(a) sp ³ d hybridized; (b) sp ^three d ^two hybridized; (c) sp ³ hybridized; (d) sp ³ hybridized; (e) sp ³ d ⁱⁱ hybridized;

113 .

(a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (due east) polar

115 .

The empirical formula is XeF_six, and the balanced reactions are: $\begin{array}{}\\ \\ \text{Xe}(g)+{\text{3F}}_{\mathrm{two}}(g)\stackrel{\text{Δ}}{\to }{\text{XeF}}_6(\mathrm{southward})\\ {\text{XeF}}_{\mathrm{six}}(s)+{\text{3H}}_{\mathrm{two}}(\mathrm{1000})⟶\text{6HF}(k)+\text{Xe}(m)\end{array}$

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Source: https://openstax.org/books/chemistry-2e/pages/chapter-18

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