Inorganic Ch. 17

Lakukan tugas rumah & ujian kamu dengan baik sekarang menggunakan Quizwiz!

The thiocyanate ion, SCN-, also behaves as a pseudo-halide ion. (a) Write the formula of the parent pseudo-halogen (b) Deduce an insoluble compound of the thiocyanate ion

(a) (CN)2; (b) AgCN, or Pb(CN)2, or Hg2(CN)2.

Give one example of how the cyanide ion resembles (a) the fluoride ion; (b) the chloride ion; (c) the iodide ion

(a) Hydrocyanic acid is a weak acid like hydrofluoric acid. (b) Silver cyanide reacts with ammonia to give the soluble diamminesilver ion, as does the silver chloride; or, cyanogen reacts with water to give the cyanide and cyanate ions in a parallel reaction to that of chlorine; or, cyanide forms complexes such as [Cu(CN)4]2- which parallel their chloride counterparts. (c) Cyanide ion is oxidized by copper(II) ion to give cyanogen, just as iodide ion is oxidized to iodine.

Write balanced chemical equations for the following chemical reactions: (a) lead metal with excess dichlorine (b) magnesium metal with dilute hydrochloric acid (c) the hypochlorite ion with sulfur dioxide gas (d) mild heating of potassium chlorate (e) solid iodine monobromide with water (f) phosphorus and iodine monochloride

(a) Pb(s) + 2 Cl2(g) --> PbCl4(l) (b) Mg(s) + 2 HCl(aq) --> MgCl2(aq) + H2(g) (c) ClO (aq) + SO2(g) + H2O(l) --> SO42 (aq) + Cl (aq) + 2 H+(aq) (d) 4 KClO3(l) --> KCl(s) + 3 KClO4(s) (e) IBr(s) + H2O(l) --> HBr(aq) + HIO(aq) (f) P(s) + 3 ICl(l) PCl3(l) + 3/2 I2(s)

Describe the uses of (a) sodium hypochlorite; (b) chlorine dioxide; (c) ammonium perchlorate; (d) iodine monochloride

(a) Sodium hypochlorite is used for bleaching wood pulp and textiles and as a household bleach and disinfectant. (b) Chlorine dioxide is used to bleach flour and wood pulp. (c) Ammonium perchlorate is used as the oxidizer in solid-fuel rockets. (d) Iodine monochloride is used as a reagent to test for degree of unsaturation in oils and fats.

Suggest how you would prepare (a) selenium tetrachloride, SeCl4, from selenium; and (b) diselenium dichloride, Se2Cl2, from selenium

(a) To form the higher oxidation state compound of a nonmetal, excess dichlorine should be used: Se(s) + 2 Cl2(g) --> SeCl4(s) (b) To form the lower oxidation state of a nonmetal, an excess of that element should be used: 2 Se(s) + Cl2(g) --> Se2Cl2(s)

Suggest how you would prepare (a) chromium(III) chloride, CrCl3, from chromium metal; and (b) chromium(II) chloride, CrCl2, from chromium metal

(a) To form the higher oxidation state of a metal, dichlorine should be used: 2 Cr(s) + 3 Cl2(g) --> 2 CrCl3(s) (b) To form the lower oxidation state of a metal, iodine monochloride should be used: Cr(s) + 2 ICl(l) --> CrCl2(s) + I2(s)

Write balanced chemical equations for the following chemical reactions: (a) uranium(IV) oxide with hydrogen fluoride (b) calcium fluoride with concentrated sulfuric acid (c) liquid sulfur tetrachloride with water (d) aqueous difluorine and hot sodium hydroxide solution (e) diiodine with difluorine in a 1:5 mole ratio (f) bromine trichloride and water

(a) UO2(s) + 4 HF(g) --> UF4(s) + 2 H2O(l) (b) CaF2(s) + H2SO4(l) --> 2 HF(g) + CaSO4(s) (c) SCl4(l) + 2 H2O(l) --> SO2(g) + 4 HCl(g) (d) 3 Cl2(aq) + 6 NaOH(aq) --> NaClO3(aq) + 5 NaCl(s) + 3 H2O(l) (e) I2(s) + 5 F2(g) --> 2 IF5(s) (f) BrCl3(l) + 2 H2O(l) --> 3 HCl(aq) + HBrO2(aq)

Dichlorine heptaoxide, Cl2O7, is a colorless, oily liquid. (a) Calculate the oxidation state of chlorine in the compound (b) Draw the probably structure of the compound (c) Write a balanced chemical equation for the reaction of this compound with water (d) Write the formula of an analogous compound of a metallic element. Give your reasoning. (e) Write the formula of two probably isoelectronic and isostructural ions

(a) Using the calculation method 2[Cl] + 7[O] = 0, [Cl] = +7. (b) The bond angle about the oxygen will be about 109.5º. Each chlorine-oxygen bond is shown as a single bond, but there will certainly be some double bond character. (c) Cl2O7(l) + H2O(l) 2 HClO4(l) (or (aq)) (d) Mn2O7 (using the (n) and (n+10) group similarity)

What are the shapes of the following species? (a) BrF2+ (b) BrF3 (c) BrF4-

(a) V-shaped; (b) T-shaped; (c) square planar

Write balanced chemical equations corresponding to each transformation in the element reaction flowcharts for fluorine, chlorine, and iodine

.

Predict some physical and chemical properties of astatine as an element.

As iodine can almost be classed as a semimetal (for example, it is known in the +1 oxidation state), we would predict astatine to start to show some metallic properties; thus the diatomic element might be a significant electrical conductor. Like the other halogens, astatine should have a common oxidation state of −1 and form an insoluble compound with silver ion: Ag+(aq) + At (aq) --> AgAt(s) The compound should be insoluble in concentrated ammonia solution. All of the other halogens should displace its anion. For example, iodine should react as follows: I2(aq) + 2 At (aq) --> 2 I (aq) + At2(s) Astatine should form interhalogen compounds, such as AtF and AtI, in which astatine has a positive polarity. In fact, being so near the metal/nonmetal border, astatine will probably have a significant cation chemistry, forming, perhaps, At+ and At3+.

Why would you expect the hydrogen difluoride ion to form a solid compound with potassium ion?

As low-charge-density cations stabilize low-charge anions, large, lowcharge-density potassium ion is likely to form a stable solid compound with the hydrogen difluoride ion.

Use the principle of formal charge to determine the average bond order in the phosphate ion and the perchlorate ion. Use these two results to suggest the average bond order in the sulfate ion.

As the preferred formal charge structure for the phosphate ion contains one double bond and that of perchlorate contains three, then sulfate should contain two. This would give an average S-O bond order of 1(1/2)

Draw the electron-dot structure of the dichlorine dioxide, ClOOCl, molecule. Deduce the oxidation state of the chlorine atoms and of the oxygen atoms

Chlorine oxidation state = +1, oxygen = -1

Explain why the cyanide ion is often considered a pseudo-halogen

Cyanide ion is considered a pseudohalogen for the following reasons: cyanide ion is the conjugate base of a weak acid, hydrocyanic acid, which is analogous to the fluoride ion forming hydrofluoric acid; it can be oxidized to (CN)2, which is analogous to an elemental halogen; the anion forms white precipitates with silver ion, lead(II) ion, and mercury(I) ion, like chloride ion; the silver thiocyanate reacts with dilute ammonia solution, like silver chloride; it forms pseudointerhalogen compounds, such as BrCN; and it forms complexes with metal ions that have formulas analogous to those of the chloride complexes.

Which of dichlorine oxide and dichlorine heptaoxide is likely to be the more acidic oxide? Give your reasoning

Dichlorine heptaoxide. It is the oxide in the higher oxidation state (with more oxygen atoms) that will be acidic

Why cannot difluorine be produced electrolytically from an aqueous solution of sodium fluoride by a similar process to that used to produce dichlorine from sodium chloride solution?

Difluorine cannot be produced electrolytically from aqueous solution because the potential needed for the oxidation of water is less than that needed for the oxidation of the fluoride ion. Thus dioxygen would be produced instead: 2 H2O(l) --> O2(g) + 4 H+(aq) + 4 e

Draw a probable structure of the dichlorine trioxide molecule. Will it be completely linear or bent? If bent, suggest an approximate bond angle.

Each chlorine atom will be approximately tetrahedrally coordinated, one chlorine with one lone pair and the other chlorine with two. Thus the bond angles will be approximately 109½º. Though the electron-dot structure depicts singly bonded oxygens, there is certainly multiple-bond character

Summarize the unique features of fluorine chemistry

Fluorine has a very weak fluorine-fluorine bond; it is usually limited to one or two covalent bonds; its compounds with metals are often ionic when those of the comparable chlorides are covalent; it has an extremely high electronegativity and forms the strongest hydrogen bonds known; it tends to stabilize high oxidation states; the solubility of its metal compounds for a particular metal is often quite different than those of the other halides.

Explain why, as the solution becomes more concentrated, hydrofluoric acid ionizes to a lesser extent at first, then to a greater extent at high concentrations

In concentrated hydrofluoric acid, fluoride ion reacts with un-ionized hydrofluoric acid to give the HF2 ion, resulting in greater ionization: F (aq) + HF(aq) --> HF2(aq)

Suggest a reason why hydrofluoric acid is a weak acid, whereas the binary acids of the other halogens are all strong acids

In the ionization of a hydrohalic acid, HX(aq) + H2O(l) --> H3O+(aq) + X (aq) the H X bond must be broken. The H F bond is particularly strong, and as a result the equilibrium will tend to lie to the left and hydrofluoric acid will behave as a weak acid.

Explain why iron(III) iodide is not a stable compound

Iron(III) iodide will not be stable because iodide ion is a reducing agent, hence it will reduce iron(III) to iron(II): 2I --> I2 + 2 e Fe3+ + e --> Fe2+

Diiodine reacts with an excess of dichlorine to form a compound of formula IClx. One mole of IClx reacts with an excess of iodide ion to produce chlorine gas and 2 moles of diiodine. What is the empirical formula of IClx?

Ix+ + 3I + 3e- --> 2I2(g) Thus x must be 3, and the formula must be ICl3.

Fluorine forms only one oxide, F2O. Daw the electron-dot structure of the compound and determine the oxidation state of fluorine and of oxygen in the compound. Explain why the oxidation state of oxygen is unusual. Suggest why one would not expect an other fluorine oxides. Use as comparisons the compounds Cl2O and Cl2O7.

Oxygen, being the second most electronegative element, usually has a negative oxidation state. It is only with fluorine that oxygen has a positive oxidation state. Fluorine would not form other oxides such as one analogous to Cl2O7. In Cl2O7, chlorine has an oxidation state of +7. Fluorine cannot form compounds in which it has positive oxidation states.

Phosphorus forms halide and pseudo-halide compounds of the from PX3. Write the formula for the compound with cyanide.

P(CN)3

Construct an electron-dot formula for the triiodide ion. Thus, deduce the shape of the ion

The I3 ion will be linear as a result of the three equatorial lone pairs.

The melting point of ammonium hydrogen difluoride, (NH4)+(HF2)-, is only 26 degrees C. This is much lower than what one would expect for an ionic lattice. Suggest what might be happening.

The ammonium hydrogen fluoride may be decomposing and dissolving in the hydrogen fluoride produced. (NH4)+(HF2)- (s) --> NH4+ + F-

Construct an approximate molecular-orbital-energy-level diagram to depict the bonding in chlorine monofluoride

The bond order will be (3 2) = 1.

Iodine forms an oxide, diiodine pentoxide, that resembles dinitrogen pentoxide in structure. Construct the electron-dot structure of the iodine compound and contrast the bonding with that in N2O5. Explain why the bonding differs. What is the oxidation state of each iodine atom, of the bridging oxygen atom, of each terminal oxygen atom?

The bonding is different because iodine has two more valence electrons. Thus there is a lone pair on each iodine atom. As a result, each iodine has a tetrahedral environment. The terminal iodine-oxygen bonds are shown as single bonds, though they probably have multiple bond character. Oxidation states: iodine = +5; oxygen (bridging and terminal) = -2.

Another compound of chlorine and oxygen, Cl2O4, is more accurately represented as chlorine perchlorate, ClOClO3. Draw the electron-dot structure of this compound and determine the oxidation number of each chlorine in the compound

The central chlorine atom has a +7 oxidation number and the end chlorine has a +1 oxidation number. Both oxidation numbers are common for chlorine.

Describe the tests used to identify each of the halide ions

The common test for distinguishing chloride, bromide, and iodide ions involves addition of silver ion to give precipitates: Ag+(aq) + X (aq) --> AgX(s) Silver chloride is white and reacts with dilute anmmonia solution: AgCl(s) + 2 NH3(aq) --> [Ag(NH3)2]+(aq) + Cl (aq) Silver bromide reacts with concentrated ammonia in a similar manner;while silver iodide does not react even with concentrated ammonia.

The concentration of hydrogen sulfide in a gas supply can be measured by passing a measured volume of gas over solid diiodine pentoxide. They hydrogen sulfide reacts with the diiodine pentoxide to give sulfur dioxide, diiodine, and water. The diiodine can then be titrated with thiosulfate ion and the hydrogen sulfide concentration calculated. Write chemical equations corresponding to the two reactions.

The half-equations for the first reaction, H2S(g) + 2 H2O(l) --> SO2(g) + 6 H+(aq) + 6 e I2O5(s) + 10 H+(aq) + 10 e --> I2(s) + 5 H2O(l) give an overall equation of 10 H2S(g) + 6 I2O5(s) --> 10 SO2(g) + 6 I2(s) + 10 H2O(l) For the second reaction: I2(s) + 2 S2O32 (aq) --> 2 I (aq) + S4O62 (aq)

Explain why, of the tetraphosphonium halides, [PH4]I is the most stable toward decomposition

The large low-charge iodide anion will stabilize the large low-charge cation.

Carbon tetrachloride has a melting point of -23C; carbon tetrabromide, 92C, and carbon tetraiodide 171C. Provide an explanation for this trend. Estimate the melting point of carbon tetrafluoride.

The only intermolecular forces for these nonpolar molecules are dispersion forces. These depend upon the number of electrons in a molecule. On that basis, by graph or calculation, the melting point of carbon tetrafluoride should be about 90°C—in fact, it is much lower, at -187°C.

Why is ammonium perchlorate an explosive hazard while sodium perchlorate is much less hazardous? Use a balanced equation to illustrate you argument and identify which elements undergo a change in oxidation state.

The perchlorate ion is a strong oxidizing agent, but it needs to be mixed with an oxidizable compound or element in order to have explosive properties. In ammonium perchlorate, the easily oxidizable ammonium ion is an integral part of the compound. Thus no additional component is required to cause a vigorous redox reaction. 2 NH4ClO4(s) --> N2(g) + Cl2(g) + 2 O2(g) + 4 H2O(g) Nitrogen is oxidized from 3 to 0, chlorine is reduced from +7 to 0, and oxygen is oxidized from 2 to 0.

Suggest an explanation for why difluorine is so reactive toward other nonmetals

The reaction with nonmetals is strongly enthalpy-driven because weak fluorine-fluorine bonds are broken and strong bonds are formed between fluorine and other nonmetals.

For chlorine, the only two naturally occurring isotopes are chlorine-35 and chlorine-37, while for bromine, they are bromine 79 and bromine-81. Suggest why chlorine-36 and bromine-80 are not stable isotopes

These would both be odd-proton odd-neutron nuclei. There are only four stable examples of these isotopes in the whole periodic table.

Deduce an electron-dot structure of the cyanogen (CN)2, molecule and draw the molecular shape. Experimental measurements of the carbon-carbon bond length show it to be shorter than the simple bonding model predicts. Suggest an explanation.

This would have the linear structure N--- C- C ---N. It is probable that the pi bond system extends across the whole molecule, increasing the bond order across the C C link while slightly weakening the C N bond.

The highest fluoride of sulfur is sulfur hexafluoride. Suggest why sulfur hexaiodide does not exist.

Three arguments can be used: first, that the sulfur atom is too small to accomodate six iodine atoms around it; second, that iodine is reducing, thus such a high oxidation state cannot be stabilized; and third, that the sulfur-iodine bond energy is not sufficient to provide an exothermic balance to the decrease in entropy that would result from consuming six moles of gas per mole of compound formed.

Thallium forms an iodide, TlI3. Suggest the actual formula of the compounmd given the following information: Tl3+ + 2e- --> Tl+ E = 1.25V I3- + 2e- --> 3I2 E = 0.55 V Why would this formula be expected?

Tl+(I3)-. Iodide is a reducing agent and it would be expected to reduce thallium(III) to thallium(I).

Why is hydrochloric acid used as a common laboratory acid in preference to nitric acid?

Unlike nitric acid, hydrochloric acid is not oxidizing.

Tetramethylammonium fluoride, (CH3)4NF, reacts with iodine heptafluoride, IF7, to give an electrically conducting solution. Write a chemical equation for the reaction

[(CH3)4N]+F- + IF7 --> [(CH3)4N]+ + IF8-


Set pelajaran terkait

Chapter 29: Management of Patients with Complications from Heart Disease

View Set

Chapter 15 Nursing Care during Labor and Birth

View Set

220-901 & 902 CompTIA A+ Certification Exam Prep

View Set

Energy & Weight Balance Nutrition Ch 10

View Set

Exam 3 Terms menstrual cycle to parturition

View Set

SOC 101 ch 9 Constructing Gender and Sexuality

View Set