Content check — Chemistry: Atoms First, Chapter 18: Representative Metals, Metalloids, and Nonmetals

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 18: Representative Metals, Metalloids, and Nonmetals

58 problems · 27 figures

How do alkali metals differ from alkaline earth metals in atomic structure and general properties?

Answer

The alkali metals all have a single s electron in their outermost shell. In contrast, the alkaline earth metals have a completed s subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period.

18-3 §18.1 source ↗

Predict the formulas for the nine compounds that may form when each species in column 1 of the table reacts with each species in column 2.

1	2
Na	I
Sr	Se
Al	O

Answer

\begin{array}{l} Na + I_{2} ⟶ 2NaI \\ 2Na + Se ⟶ {Na}_{2} Se \\ 2Na + O_{2} ⟶ {Na}_{2} O_{2} \end{array}

\begin{array}{l} Sr + I_{2} ⟶ {SrI}_{2} \\ Sr + Se ⟶ SrSe \\ 2Sr + O_{2} ⟶ 2SrO \end{array}

\begin{array}{l} 2Al + {3I}_{2} ⟶ {2AlI}_{3} \\ 2Al + 3Se ⟶ {Al}_{2} {Se}_{3} \\ 4Al + {3O}_{2} ⟶ {2Al}_{2} O_{3} \end{array}

18-5 §18.1 source ↗

Sodium chloride and strontium chloride are both white solids. How could you distinguish one from the other?

Answer

The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test 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{35.7 g}{100 mL}

compared with

\frac{53.8 g}{100 mL}

for SrCl₂. Heating to 100 °C provides an easy test, since the solubility of NaCl is

\frac{39.12 g}{100 mL},

but that of SrCl₂ is

\frac{100.8 g}{100 mL} .

Density determination on a solid is sometimes difficult, but there is enough difference (2.165 g/mL NaCl and 3.052 g/mL SrCl₂) that this method would be viable and perhaps the easiest and least expensive test to perform.

18-7 §18.1 source ↗

Write a balanced equation for the reaction of elemental strontium with each of the following:

(a) oxygen

(b) hydrogen bromide

(d) phosphorus

(e) water

Answer

(a)

2Sr (s) +

O_{2} (g) ⟶

2SrO (s);

(b)

Sr (s) +

2HBr (g) ⟶

{SrBr}_{2} (s) +

H_{2} (g);

(c)

Sr (s) +

H_{2} (g) ⟶

{SrH}_{2} (s);

(d)

6Sr (s) +

P_{4} (s) ⟶

{2Sr}_{3} P_{2} (s);

(e)

Sr (s) +

{2H}_{2} O (l) ⟶

Sr {(OH)}_{2} (a q) +

H_{2} (g)

18-9 §18.1 source ↗

What is the mass of fish, in kilograms, that one would have to consume to obtain a fatal dose of mercury, if the fish contains 30 parts per million of mercury by weight? (Assume that all the mercury from the fish ends up as mercury(II) chloride in the body and that a fatal dose is 0.20 g of HgCl₂.) How many pounds of fish is this?

Answer

11 lb

18-11 §18.1 source ↗

Does metallic tin react with HCl?

Answer

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

18-13 §18.1 source ↗

Compare the nature of the bonds in PbCl₂ to that of the bonds in PbCl₄.

Answer

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

18-15 §18.2 source ↗

Write an equation for the reduction of cesium chloride by elemental calcium at high temperature.

Answer

2CsCl (l) + Ca (g) \overset{\begin{array}{l} countercurrent \\ fractionating \\ tower \end{array}}{\to} 2Cs (g) + {CaCl}_{2} (l)

18-17 §18.2 source ↗

Give balanced equations for the overall reaction in the electrolysis of molten lithium chloride and for the reactions occurring at the electrodes. You may wish to review the chapter on electrochemistry for relevant examples.

Answer

Cathode (reduction):

{2Li}^{+} +

{2e}^{-} ⟶

2Li (l);

Anode (oxidation):

{2Cl}^{-} ⟶

{Cl}_{2} (g) +

{2e}^{−};

Overall reaction:

{2Li}^{+} +

{2Cl}^{-} ⟶

2Li (l) +

{Cl}_{2} (g)

18-19 §18.2 source ↗

What mass, in grams, of hydrogen gas forms during the complete reaction of 10.01 g of calcium with water?

Answer

0.5035 g H₂

18-21 §18.2 source ↗

Magnesium is an active metal; it burns in the form of powder, ribbons, and filaments to provide flashes of brilliant light. Why is it possible to use magnesium in construction?

Answer

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

18-23 §18.2 source ↗

Describe the production of metallic aluminum by electrolytic reduction.

Answer

Extract from ore:

AlO (OH) (s) +

NaOH (a q) +

H_{2} O (l) ⟶

Na [Al {(OH)}_{4}] (a q)

Recover:

2Na [Al {(OH)}_{4}] (s) +

H_{2} {SO}_{4} (a q) ⟶

2Al {(OH)}_{3} (s) +

{Na}_{2} {SO}_{4} (a q) +

{2H}_{2} O (l)

Sinter:

2Al {(OH)}_{3} (s) ⟶

{Al}_{2} O_{3} (s) +

{3H}_{2} O (g)

Dissolve in Na₃AlF₆(l) and electrolyze:

{Al}^{3+} +

{3e}^{-} ⟶

Al (s)

18-25 §18.2 source ↗

A chemist dissolves a 1.497-g sample of a type of metal (an alloy of Sn, Pb, Sb, and Cu) in nitric acid, and metastannic acid, H₂SnO₃, is precipitated. She heats the precipitate to drive off the water, which leaves 0.4909 g of tin(IV) oxide. What was the percentage of tin in the original sample?

Answer

25.83%

18-27 §18.2 source ↗

What mass of magnesium forms when 100,000 A is passed through a MgCl₂ melt for 1.00 h if the yield of magnesium is 85% of the theoretical yield?

Answer

39 kg

18-29 §18.3 source ↗

Write a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding.

(a) H₃BPH₃

(b) ${BF}_{4}^{-}$

(d) B(CH₃)₃

(e) B(OH)₃

Answer

(a) H₃BPH₃:

This Lewis structure is composed of a boron atom single bonded to a phosphorus atom. Each of these atoms is single bonded to three hydrogen atoms.

(b) ${BF}_{4}^{−} :$

This Lewis structure is composed of a boron atom single bonded to four fluorine atoms, each of which has three lone pairs of electrons. The structure is surrounded by brackets, and a negative sign appears as a superscript outside the brackets.

(d) B(CH₃)₃:

This Lewis structure is composed of a boron atom that is single bonded to three carbon atoms, each of which is single bonded to three hydrogen atoms.

(e) B(OH)₃:

This Lewis structure is composed of a boron atom that is single bonded to three oxygen atoms, each of which has two lone pairs of electrons. Each oxygen atom is single bonded to a hydrogen atom.

18-31 §18.3 source ↗

Using only the periodic table, write the complete electron configuration for silicon, including any empty orbitals in the valence shell. You may wish to review the chapter on electronic structure.

Answer

1s²2s²2p⁶3s²3p²3d⁰.

18-33 §18.3 source ↗

Describe the hybridization of silicon and the molecular structure of the following molecules and ions:

(a) (CH₃)₃SiH

(b) ${SiO}_{4}^{4−}$

(d) Si(OH)₄

(e) ${SiF}_{6}^{2−}$

Answer

(a) (CH₃)₃SiH: sp³ bonding about Si; the structure is tetrahedral; (b)

{SiO}_{4}^{4−} :

sp³ bonding about Si; the structure is tetrahedral; (c) Si₂H₆: sp³ bonding about each Si; the structure is linear along the Si-Si bond; (d) Si(OH)₄: sp³ bonding about Si; the structure is tetrahedral; (e)

{SiF}_{6}^{2−} :

sp³d² bonding about Si; the structure is octahedral

18-35 §18.3 source ↗

Classify each of the following molecules as polar or nonpolar. You may wish to review the chapter on chemical bonding.

(a) SiH₄

(b) Si₂H₆

(d) SiF₄

(e) SiCl₂F₂

Answer

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

18-37 §18.3 source ↗

Name each of the following compounds:

(a) TeO₂

(b) Sb₂S₃

(d) SiH₄

(e) GeH₄

Answer

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

18-39 §18.3 source ↗

Why is boron limited to a maximum coordination number of four in its compounds?

Answer

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

18-41 §18.3 source ↗

From the data given in Appendix G, determine the standard enthalpy change and the standard free energy change for each of the following reactions:

(a) ${BF}_{3} (g) +$ ${3H}_{2} O (l) ⟶$ $B {(OH)}_{3} (s) +$ $3HF (g)$

(b) ${BCl}_{3} (g) +$ ${3H}_{2} O (l) ⟶$ $B {(OH)}_{3} (s) +$ $3HCl (g)$

Answer

(a) ΔH° = 87 kJ; ΔG° = 44 kJ; (b) ΔH° = −109.9 kJ; ΔG° = −154.7 kJ; (c) ΔH° = −510 kJ; ΔG° = −601.5 kJ

18-43 §18.3 source ↗

Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it?

Answer

A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond.

18-45 §18.4 source ↗

Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable P₄ molecules instead of P₂ molecules?

Answer

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

18-47 §18.4 source ↗

Determine the oxidation number of each element in each of the following compounds:

(a) HCN

(b) OF₂

Answer

(a) H = 1+, C = 2+, and N = 3−; (b) O = 2+ and F = 1−; (c) As = 3+ and Cl = 1−

18-49 §18.4 source ↗

Arrange the following in order of increasing electronegativity: F; Cl; O; and S.

Answer

S < Cl < O < F

18-51 §18.5 source ↗

Why does hydrogen not exhibit an oxidation state of 1− when bonded to nonmetals?

Answer

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.

18-53 §18.5 source ↗

In drawing Lewis structures, we learn that a hydrogen atom forms only one bond in a covalent compound. Why?

Answer

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

18-55 §18.5 source ↗

What mass of hydrogen gas results from the reaction of 8.5 g of KH with water?

KH +

H_{2} O ⟶

KOH +

H_{2}

Answer

0.43 g H₂

18-57 §18.6 source ↗

Complete and balance the following chemical equations:

(a) hardening of plaster containing slaked lime
$Ca {(OH)}_{2} +$ ${CO}_{2} ⟶$

(b) removal of sulfur dioxide from the flue gas of power plants
$CaO + {SO}_{2} ⟶$

(c) the reaction of baking powder that produces carbon dioxide gas and causes bread to rise
${NaHCO}_{3} +$ ${NaH}_{2} {PO}_{4} ⟶$

Answer

(a)

Ca {(OH)}_{2} (a q) +

{CO}_{2} (g) ⟶

{CaCO}_{3} (s) +

H_{2} O (l);

(b)

CaO (s) +

{SO}_{2} (g) ⟶

{CaSO}_{3} (s);

(c)

{2NaHCO}_{3} (s) +

{NaH}_{2} {PO}_{4} (a q) ⟶

{Na}_{3} {PO}_{4} (a q) +

{2CO}_{2} (g) +

{2H}_{2} O (l)

18-59 §18.7 source ↗

Write the Lewis structures for each of the following:

(a) NH²⁻

(b) N₂F₄

(d) NF₃

(e) $N_{3}^{-}$

Answer

(a) NH²⁻:

(b) N₂F₄:

This Lewis structure shows two nitrogen atoms, each with one lone pair of electrons, single bonded to one another and each single bonded to two fluorine atoms. Each fluorine atom has three lone pairs of electrons.

(d) NF₃:

(e) $N_{3}^{−} :$

18-61 §18.7 source ↗

Explain how ammonia can function both as a Brønsted base and as a Lewis base.

Answer

Ammonia acts as a Brønsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate.
Brønsted base:

{NH}_{3} +

H_{3} O^{+} ⟶

{NH}_{4}^{+} +

H_{2} O

Lewis base:

{2NH}_{3} +

{Ag}^{+} ⟶

{{[H}_{3} N - Ag - {NH}_{3}]}^{+}

18-63 §18.7 source ↗

For each of the following, draw the Lewis structure, predict the ONO bond angle, and give the hybridization of the nitrogen. You may wish to review the chapters on chemical bonding and advanced theories of covalent bonding for relevant examples.

(a) NO₂

(b) ${NO}_{2}^{-}$

Answer

(a) NO₂:

Nitrogen is sp² hybridized. The molecule has a bent geometry with an ONO bond angle of approximately 120°.
(b) ${NO}_{2}^{−} :$

Nitrogen is sp² hybridized. The molecule has a bent geometry with an ONO bond angle slightly less than 120°.
(c) ${NO}_{2}^{+} :$

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

18-65 §18.7 source ↗

Although PF₅ and AsF₅ are stable, nitrogen does not form NF₅ molecules. Explain this difference among members of the same group.

Answer

Nitrogen cannot form a NF₅ molecule because it does not have d orbitals to bond with the additional two fluorine atoms.

18-67 §18.8 source ↗

Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.

(a) PH₃

(b) ${PH}_{4}^{+}$

(d) ${PO}_{4}^{3−}$

(e) PF₅

Answer

(a)

(b)

(c)

(d)

This Lewis structure shows a phosphorus atom single bonded to four oxygen atoms, each with three lone pairs of electrons. The structure is surrounded by brackets and has a superscript 3 negative sign outside the brackets.

(e)

This Lewis structure shows a phosphorus atom single bonded to five fluorine atoms, each with three lone pairs of electrons.

18-69 §18.8 source ↗

Complete and balance each of the following chemical equations. (In some cases, there may be more than one correct answer.)

(a) $P_{4} + Al ⟶$

(b) $P_{4} + Na ⟶$

(d) $P_{4} + {Cl}_{2} ⟶$

(e) $P_{4} + O_{2} ⟶$

(f) $P_{4} O_{6} + O_{2} ⟶$

Answer

(a)

P_{4} (s) +

4Al (s) ⟶

4AlP (s);

(b)

P_{4} (s) +

12Na (s) ⟶

{4Na}_{3} P (s);

(c)

P_{4} (s) +

{10F}_{2} (g) ⟶

{4PF}_{5} (l);

(d)

P_{4} (s) +

{6Cl}_{2} (g) ⟶

{4PCl}_{3} (l)

P_{4} (s) +

{10Cl}_{2} (g) ⟶

{4PCl}_{5} (l);

(e)

P_{4} (s) +

{3O}_{2} (g) ⟶

P_{4} O_{6} (s)

P_{4} (s) +

{5O}_{2} (g) ⟶

P_{4} O_{10} (s);

(f)

P_{4} O_{6} (s) +

{2O}_{2} (g) ⟶

P_{4} O_{10} (s)

18-71 §18.8 source ↗

What volume of 0.200 M NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of PCl₃ is an excess of water? Note that when H₃PO₃ is titrated under these conditions, only one proton of the acid molecule reacts.

Answer

291 mL

18-73 §18.8 source ↗

How many tons of Ca₃(PO₄)₂ are necessary to prepare 5.0 tons of phosphorus if the yield is 90%?

Answer

28 tons

18-75 §18.8 source ↗

Draw the Lewis structures and describe the geometry for the following:

(a) ${PF}_{4}^{+}$

(b) PF₅

(d) POF₃

Answer

(a)

This Lewis structure shows a phosphorus atom single bonded to four fluorine atoms, each with three lone pairs of electrons. The structure is surrounded by brackets and has a superscript positive sign outside the brackets. The label, “Tetrahedral,” is written under the structure.

(b)

This Lewis structure shows a phosphorus atom single bonded to five fluorine atoms, each with three lone pairs of electrons. The label, “Trigonal bipyramidal,” is written under the structure.

(c)

A Lewis structure shows a phosphorus atom single bonded to six fluorine atoms, each with three lone pairs of electrons. The structure is surrounded by brackets and has a superscript negative sign outside the brackets. The label, “Octahedral,” is written under the structure.

(d)

This Lewis structure shows a phosphorus atom single bonded to three fluorine atoms, each with three lone pairs of electrons. The phosphorus atom is also double bonded to an oxygen atom with two lone pairs of electrons. The label, “Tetrahedral,” is written under the structure.

18-77 §18.8 source ↗

Assign an oxidation state to phosphorus in each of the following:

(a) NaH₂PO₃

(b) PF₅

(d) K₃PO₄

(e) Na₃P

(f) Na₄P₂O₇

Answer

(a) P = 3+; (b) P = 5+; (c) P = 3+; (d) P = 5+; (e) P = 3−; (f) P = 5+

18-79 §18.9 source ↗

Predict the product of burning francium in air.

Answer

FrO₂

18-81 §18.9 source ↗

Write balanced chemical equations for the following reactions:

(a) zinc metal heated in a stream of oxygen gas

(b) zinc carbonate heated until loss of mass stops

(d) zinc added to a solution of hydrobromic acid

Answer

(a)

2Zn (s) +

O_{2} (g) ⟶

2ZnO (s);

(b)

{ZnCO}_{3} (s) ⟶

ZnO (s) +

{CO}_{2} (g);

(c)

{ZnCO}_{3} (s) +

2 {CH}_{3} COOH (a q) ⟶

Zn {({CH}_{3} COO)}_{2} (a q) +

{CO}_{2} (g) +

H_{2} O (l);

(d)

Zn (s) +

2HBr (a q) ⟶

{ZnBr}_{2} (a q) +

H_{2} (g)

18-83 §18.9 source ↗

Illustrate the amphoteric nature of aluminum hydroxide by citing suitable equations.

Answer

Al {(OH)}_{3} (s) +

3 H^{+} (a q) ⟶

{Al}^{3+} +

3 H_{2} O (l);

Al {(OH)}_{3} (s) +

{OH}^{-} ⟶

{[Al {(OH)}_{4}]}^{−} (a q)

18-85 §18.9 source ↗

Write balanced chemical equations for the following reactions:

(a) sodium oxide added to water

(b) cesium carbonate added to an excess of an aqueous solution of HF

(d) a solution of sodium carbonate added to solution of barium nitrate

(e) titanium metal produced from the reaction of titanium tetrachloride with elemental sodium

Answer

(a)

{Na}_{2} O (s) +

H_{2} O (l) ⟶

2NaOH (a q);

(b)

{Cs}_{2} {CO}_{3} (s) +

2HF (a q) ⟶

2CsF (a q) +

{CO}_{2} (g) +

H_{2} O (l);

(c)

{Al}_{2} O_{3} (s) +

6 {HClO}_{4} (a q) ⟶

2Al {({ClO}_{4})}_{3} (a q) +

3 H_{2} O (l);

(d)

{Na}_{2} {CO}_{3} (a q) +

Ba {({NO}_{3})}_{2} (a q) ⟶

2 {NaNO}_{3} (a q) +

{BaCO}_{3} (s);

(e)

{TiCl}_{4} (l) +

4Na (s) ⟶

Ti (s) +

4NaCl (s)

18-87 §18.9 source ↗

Which is the stronger acid, HClO₄ or HBrO₄? Why?

Answer

HClO₄ is the stronger acid because, in a series of oxyacids with similar 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 bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid.

18-89 §18.9 source ↗

Which is the stronger acid, H₂SO₄ or H₂SeO₄? Why? You may wish to review the chapter on acid-base equilibria.

Answer

As H₂SO₄ and H₂SeO₄ are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, H₂SO₄ is the stronger acid.

18-91 §18.10 source ↗

Give the hybridization and oxidation state for sulfur in SO₂, in SO₃, and in H₂SO₄.

Answer

SO₂, sp² 4+; SO₃, sp², 6+; H₂SO₄, sp³, 6+

18-93 §18.10 source ↗

Determine the oxidation state of sulfur in SF₆, SO₂F₂, and KHS.

Answer

SF₆: S = 6+; SO₂F₂: S = 6+; KHS: S = 2−

18-95 §18.10 source ↗

Oxygen forms double bonds in O₂, but sulfur forms single bonds in S₈. Why?

Answer

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

18-97 §18.10 source ↗

Write two balanced chemical equations in which sulfuric acid acts as an oxidizing agent.

Answer

There are many possible answers including:

Cu (s) +

2 H_{2} {SO}_{4} (l) ⟶

{CuSO}_{4} (a q) +

{SO}_{2} (g) +

2 H_{2} O (l)

and

C (s) +

2 H_{2} {SO}_{4} (l) ⟶

{CO}_{2} (g) +

2 {SO}_{2} (g) +

2 H_{2} O (l)

18-99 §18.10 source ↗

How many grams of Epsom salts (MgSO₄⋅7H₂O) will form from 5.0 kg of magnesium?

Answer

5.1

\times

10⁴ g

18-101 §18.11 source ↗

Why is SnCl₄ not classified as a salt?

Answer

SnCl₄ is not a salt because it is covalently bonded. A salt must have ionic bonds.

18-103 §18.11 source ↗

Which is the stronger acid, HClO₃ or HBrO₃? Why?

Answer

In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. HClO₃ is stronger than HBrO₃; Cl is more electronegative than Br.

18-105 §18.11 source ↗

Predict the molecular geometries and draw Lewis structures for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.

(a) IF₅

(b) $I_{3}^{-}$

(d) SeF₄

(e) ClF₃

Answer

(a)

This Lewis structure shows an iodine atom with one lone pair of electrons single bonded to five fluorine atoms, each of which has three lone pairs of electrons. The image is labeled, “Square pyramidal.”

(b)

(c)

This Lewis structure shows a phosphorus atom single bonded to five chlorine atoms, each of which has three lone pairs of electrons. The image is labeled, “Trigonal bipyramidal.”

(d)

This Lewis structure shows a selenium atom with one lone pair of electrons single bonded to four fluorine atoms, each of which has three lone pairs of electrons. The image is labeled “Seesaw.”

(e)

This Lewis structure shows a chlorine atom with two lone pairs of electrons single bonded to three fluorine atoms, each of which has three lone pairs of electrons. The image is labeled, “T-shaped.”

18-107 §18.11 source ↗

Name each of the following compounds:

(a) BrF₃

(b) NaBrO₃

(d) NaClO₄

(e) KClO

Answer

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

18-109 §18.11 source ↗

What is the oxidation state of the halogen in each of the following?

(a) H₅IO₆

(b) ${IO}_{4}^{-}$

(d) ICl₃

(e) F₂

Answer

(a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1−; (e) F: 0

18-111 §18.12 source ↗

Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding.

(a) XeF₂

(b) XeF₄

(d) XeO₄

(e) XeOF₄

Answer

(a) sp³d hybridized; (b) sp³d² hybridized; (c) sp³ hybridized; (d) sp³ hybridized; (e) sp³d² hybridized;

18-113 §18.12 source ↗

Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry.

(a) XeF₂

(b) XeF₄

(d) XeO₄

(e) XeOF₄

Answer

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

18-115 §18.12 source ↗

A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 M sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon.

Answer

The empirical formula is XeF₆, and the balanced reactions are:

\begin{array}{l} Xe (g) + {3F}_{2} (g) \overset{Δ}{\to} {XeF}_{6} (s) \\ {XeF}_{6} (s) + {3H}_{2} (g) ⟶ 6HF (g) + Xe (g) \end{array}

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Problems and answers are adapted from Chemistry: Atoms First 2e by Paul Flowers, Edward J. Neth, William R. Robinson, Klaus Theopold and Richard Langley, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/chemistry-atoms-first-2e/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.