Chemistry of the Nonmetals

Transcription

1 22 Chemistry of the Nonmetals Visualizing Concepts 22.2 Acid-base (Brønsted) Charges on species from left to right in the reaction: 0, 0, 1+, 1 NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH (aq) 22.5 Analyze. Given: space-filling models of molecules containing nitrogen and oxygen atoms. Find: molecular formulas and Lewis structures. Plan. Nitrogen atoms are blue, and oxygen atoms are red. Count the number of spheres of each color to determine the molecular formula. From each molecular formula, count the valence electrons (N = 5, O = 6) and draw a correct Lewis structure. Resonance structures are likely. Solve. N 2 O 5 40 valence electrons, 20 e pairs Many other resonance structures are possible. Those with double bonds to the central oxygen (like the right-hand structure above) do not minimize formal charge and are less significant in the net bonding model. N 2 O 4 34 e, 17 e pairs Other equivalent resonance structures with different arrangement of the double bonds are possible. NO 2 17 e, 8.5 e pairs We place the odd electron on N because of electronegativity arguments. N 2 O 3 28 e, 14 e pairs 635

2 (e) NO 11 e, 5.5 e pairs We place the odd electron on N because of electronegativity arguments. (f) N 2 O 16 e, 8 e pairs The right-most structure above does not minimize formal charge and makes smaller contribution to the net bonding model Analyze/Plan. Evaluate the graph, describe the trend in data, recall the general trend for each of the properties listed, and use details of the data to discriminate between possibilities. Solve. The general trend is an increase in value moving from left to right across the period, with a small discontinuity at S. Considering just this overall feature, both first ionization energy and electronegativity increase moving from left to right, so these are possibilities. Atomic radius decreases, and can be eliminated. Since Si is a solid and Cl and Ar are gases at room temperature, melting points must decrease across the row; melting point can be eliminated. According to data in Tables 22.2, 22.5, 22.7, and 22.8, (e) X-X single bond enthalpies show no consistent trend. Furthermore, there is no known Ar-Ar single bond, so no value for this property can be known; (e) can be eliminated. Now let s examine trends in first ionization energy and electronegativity more closely. From electronegativity values in Chapter 8, we see a continuous increase with no discontinuity at S, and no value for Ar. Values in first ionization energy from Chapter 7 do match the pattern in the figure. The slightly lower value of I 1 for S is due to a decrease in repulsion by removing an electron from a fully occupied orbital. In summary, only first ionization energy fits the property depicted in the graph. Periodic Trends and Chemical Reactions Analyze/Plan. Use the color-coded periodic chart on the front-inside cover of the text to classify the given elements. Solve. Metals: Sr, Mn, (e) Rh; nonmetals: P, Se, (f) Kr; metalloids: none Analyze/Plan. Follow the logic in Sample Exercise Solve. O Br Ba O (e) Co Analyze/Plan. Use the position of the specified elements on the periodic chart, periodic trends, and the arguments in Sample Exercise 22.1 to explain the observations. Solve. 636

3 Nitrogen is too small to accommodate five fluorine atoms about it. The P and As atoms are larger. Furthermore, P and As have available 3d and 4d orbitals, respectively, to form hybrid orbitals that can accommodate more than an octet of electrons about the central atom. Si does not readily form π bonds, which would be necessary to satisfy the octet rule for both atoms in SiO. A reducing agent is a substance that readily loses electrons. As has a lower electronegativity than N; that is, it more readily gives up electrons to an acceptor and is more easily oxidized Analyze/Plan. Follow the logic in Sample Exercise Solve. Mg 3 N 2 (s) + 6H 2 O(l) 2NH 3 (g) + 3Mg(OH) 2 (s) Because H 2 O(l) is a reactant, the state of NH 3 in the products could be expressed as NH 3 (aq). 2C 3 H 7 OH(l) + 9O 2 (g) 6CO 2 (g) + 8H 2 O(l) AlP(s) + 3H 2 O(l) PH 3 (g) + Al(OH) 3 (s) (e) Na 2 S(s) + 2HCl(aq) H 2 S(g) + 2NaCl(aq) Hydrogen, the Noble Gases, and the Halogens Analyze/Plan. Use information on the isotopes of hydrogen in Section 22.2 to list their symbols, names, and relative abundances. Solve. a) b) The order of abundance is proteum > deuterium > tritium Analyze/Plan. Consider the electron configuration of hydrogen and the Group 1A elements. Solve. Like other elements in group 1A, hydrogen has only one valence electron and its most common oxidation number is Analyze/Plan. Use information on the descriptive chemistry of hydrogen in Section 22.2 to formulate the required equations. Steam is H 2 O(g). Solve. Mg(s) + 2H + (aq) Mg 2 + (aq) + H 2 (g) 637

4 22.25 Analyze/Plan. Use information on the descriptive chemistry of hydrogen given in Section 22.2 to complete and balance the equations. Solve. NaH(s) + H 2 O(l) NaOH(aq) + H 2 (g) Fe(s) + H 2 SO 4 (aq) Fe (aq) + H 2 (g) + SO 4 (aq) H 2 (g) + Br 2 (g) 2HBr(g) 2Na(l) + H 2 (g) 2NaH(s) (e) Analyze/Plan. If the element bound to H is a nonmetal, the hydride is molecular. If H is bound to a metal with integer stoichiometry, the hydride is ionic; with noninteger stoichiometry, the hydride is metallic. Solve. ionic (metal hydride) molecular (nonmetal hydride) metallic (nonstoichiometric transition metal hydride) Vehicle fuels produce energy via combustion reactions. The reaction H 2 (g) + 1/2 O 2 (g) H 2 O(g) is very exothermic, producing 242 kj per mole of H 2 burned. The only product of combustion is H 2 O, a nonpollutant (but like CO 2, a greenhouse gas) Analyze/Plan. Consider the periodic properties of Xe and Ar. Solve. Xenon is larger, and can more readily accommodate an expanded octet. More important is the lower ionization energy of xenon; because the valence electrons are a greater average distance from the nucleus, they are more readily promoted to a state in which the Xe atom can form bonds with fluorine Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve. ClO 3, +5 HI, 1 ICl 3 ; I, +3; Cl, 1 NaOCl, +1 (e) HClO 4, +7 (f) XeF 4, +4; F, Analyze/Plan. Review the nomenclature rules and ion names in Section 2.8. Solve. iron(iii) chlorate chlorous acid xenon hexafluoride bromine pentafluoride (e) xenon oxide tetrafluoride (f) iodic acid Analyze/Plan. Consider intermolecular forces and periodic properties, including oxidizing power, of the listed substances. Solve. Van der Waals intermolecular attractive forces increase with increasing numbers of electrons in the atoms. F 2 reacts with water: F 2 (g) + H 2 O(l) 2HF(aq) + 1/2 O 2 (g). That is, fluorine is too strong an oxidizing agent to exist in water. 638

5 HF has extensive hydrogen bonding. Oxidizing power is related to electronegativity. Electronegativity decreases in the order given If the lifetime of perchlorate anion, ClO 4, in soils and water is decades, the ion must be extremely stable (unreactive) in aqueous solutions and aerobic environments; it is not easily oxidized by O 2. Although chlorine is in a very high oxidation state in ClO 4, it is not readily reduced, because the ion has a stable, symmetric structure that protects it against reactions. The anion is a symmetrical tetrahedron with several plausible Lewis structures when expanded octets about Cl are considered. (minimum formal charges) More structures with alternate locations of the single and double bonds can be drawn. Resonance stabilization could certainly contribute to the ion s high stability. Oxygen and the Group 6A Elements Analyze/Plan. Consider the industrial uses of oxygen and ozone given in Section Solve. As an oxidizing agent in steel-making; to bleach pulp and paper; in oxyacetylene torches; in medicine to assist in breathing Synthesis of pharmaceuticals, lubricants, and other organic compounds where bonds are cleaved; in water treatment Analyze/Plan. Use information on the descriptive chemistry of oxygen given in Section 22.5 to complete and balance the equations. Solve. PbS(s) + 4O 3 (g) PbSO 4 (s) + 4O 2 (g) 2ZnS(s) + 3O 2 (g) 2ZnO(s) + 2SO 2 (g) (e) 2K 2 O 2 (s) + 2CO 2 (g) 2K 2 CO 3 (s) + O 2 (g) 639

6 22.45 Analyze/Plan. Oxides of metals are bases, oxides of nonmetals are acids, oxides that act as both acids and bases are amphoteric and oxides that act as neither acids nor bases are neutral. Solve. acidic (oxide of a nonmetal) acidic (oxide of a nonmetal) amphoteric basic (oxide of a metal) Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve. H 2 SeO 3, +4 KHSO 3, +4 H 2 Te, 2 CS 2, 2 (e) CaSO 4, +6 Oxygen (a group 6A element) is in the 2 oxidation state in compounds,, and (e) Analyze/Plan. The half-reaction for oxidation in all these cases is: H 2 S(aq) S(s) + 2H + + 2e (The product could be written as S 8 (s), but this is not necessary. In fact it is not necessarily the case that S 8 would be formed, rather than some other allotropic form of the element.) Combine this half-reaction with the given reductions to write complete equations. The reduction in happens only in acid solution. The reactants in are acids, so the medium is acidic. Solve. 2Fe 3 + (aq) + H 2 S(aq) 2Fe 2 + (aq) + S(s) + 2H + (aq) Br 2 (l) + H 2 S(aq) 2Br (aq) + S(s) + 2H + (aq) 2MnO 4 (aq) + 6H + (aq) + 5H 2 S(aq) 2Mn 2 + (aq) + 5S(s) + 8H 2 O(l) 2NO 3 (aq) + H 2 S(aq) + 2H + (aq) 2NO 2 (aq) + S(s) + 2H 2 O(l) Analyze/Plan. For each substance, count valence electrons, draw the correct Lewis structure, and apply the rules of VSEPR to decide electron domain geometry and geometric structure. Solve Analyze/Plan. Use information on the descriptive chemistry of sulfur given in Section 22.6 to complete and balance the equations. Solve. SO 2 (s) + H 2 O(l) H 2 SO 3 (aq) H + (aq) + HSO 3 (aq) ZnS(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 S(g) 8SO 3 2 (aq) + S 8 (s) 8S 2 O 3 2 (aq) 640

7 SO 3 (aq) + H 2 SO 4 (l) H 2 S 2 O 7 (l) Nitrogen and the Group 5A Elements Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve. NaNO 2, +3 NH 3, 3 N 2 O, +1 NaCN, 3 (e) HNO 3, +5 (f) NO 2, Analyze/Plan. For each substance, count valence electrons, draw the correct Lewis structure, and apply the rules of VSEPR to decide electron domain geometry and geometric structure. Solve. The molecule is bent around the central oxygen and nitrogen atoms; the four atoms need not lie in a plane. The right-most form does not minimize formal charges and is less important in the actual bonding model. The geometry is tetrahedral around the left nitrogen, trigonal pyramidal around the right. (three equivalent resonance forms) The ion is trigonal planar Analyze/Plan. Use information on the descriptive chemistry of nitrogen given in Section 22.7 to complete and balance the equations. Solve. Mg 3 N 2 (s) + 6H 2 O(l) 2NH 3 (g) + 3Mg(OH) 2 (s) Because H 2 O(l) is a reactant, the state of NH 3 in the products could be expressed as NH 3 (aq). (e) 2NO(g) + O 2 (g) 2NO 2 (g) N 2 O 5 (g) + H 2 O(l) 2H + (aq) + 2NO 3 (aq) NH 3 (aq) + H + (aq) NH + 4 (aq) N 2 H 4 (l) + O 2 (g) N 2 (g) + 2H 2 O(g) Analyze/Plan. Follow the method for writing balanced half-reactions given in Section 20.2 and Sample Exercises 20.2 and 20.3 Solve. HNO 2 (aq) + H 2 O(l) NO 3 (aq) + 3H + (aq) + 2e, N 2 (g) + H 2 O(l) N 2 O(g) + 2H + (aq) + 2e, 641

8 22.63 Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve. H 3 PO 3, +3 H 4 P 2 O 7, +5 SbCl 3, +3 Mg 3 As 2, 3 (e) P 2 O 5, Analyze/Plan. Consider the structures of the compounds of interest when explaining the observations. Solve. Phosphorus is a larger atom and can more easily accommodate five surrounding atoms and an expanded octet of electrons than nitrogen can. Also, P has energetically available 3d orbitals which participate in the bonding, but nitrogen does not. Only one of the three hydrogens in H 3 PO 2 is bonded to oxygen. The other two are bonded directly to phosphorus and are not easily ionized because the bond is not very polar. PH 3 is a weaker base than H 2 O (PH 4 + is a stronger acid than H 3 O + ). Any attempt to add H + to PH 3 in the presence of H 2 O merely causes protonation of H 2 O. White phosphorus consists of P 4 molecules, with bond angles of 60. Each P atom has four VSEPR pairs of electrons, so the predicted electron pair geometry is tetrahedral and the preferred bond angle is 109. Because of the severely strained bond angles in P 4 molecules, white phosphorus is highly reactive Analyze/Plan. Use information on the descriptive chemistry of phosphorus given in Section 22.8 to complete and balance the equations. Solve. PBr 3 (l) + 3H 2 O(l) H 3 PO 3 (aq) + 3HBr(aq) 4PBr 3 (g) + 6H 2 (g) P 4 (g) + 12HBr(g) Carbon, the Other Group 4A Elements, and Boron Analyze/Plan. Review the nomenclature rules and ion names in Section 2.8. Solve. HCN Ni(CO) 4 Ba(HCO 3 ) 2 CaC Analyze/Plan. Use information on the descriptive chemistry of carbon given in Section 22.9 to complete and balance the equations. Solve. (e) BaC 2 (s) + 2H 2 O(l) Ba 2 + (aq) + 2OH (aq) + C 2 H 2 (g) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) CS 2 (g) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) Ca(CN) 2 (s) + 2HBr(aq) CaBr 2 (aq) + 2HCN(aq) Analyze/Plan. Use information on the descriptive chemistry of carbon given in Section 22.9 to complete and balance the equations. Solve. 642

9 NaHCO 3 (s) + H + (aq) CO 2 (g) + H 2 O(l) + Na + (aq) 2BaCO 3 (s) + O 2 (g) + 2SO 2 (g) 2BaSO 4 (s) + 2CO 2 (g) Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the logic in Sample Exercise 4.8. Solve. H 3 BO 3, +3 SiBr 4, +4 PbCl 2, +2 Na 2 B 4 O 7 10H 2 O, +3 (e) B 2 O 3, Analyze/Plan. Consider periodic trends within a family, particularly metallic character, as well as descriptive chemistry in Sections 22.9 and Solve. Sn; see Table The filling of the 4f subshell at the beginning of the sixth row of the periodic table increases Z and Z eff for later elements. This causes the ionization energy of Pb to be greater than that of Sn. Carbon, Si, and Ge; these are the nonmetal and metalloids in group 4A. They form compounds ranging from XH 4 ( 4) to XO 2 (+4). The metals Sn and Pb are not found in negative oxidation states. Silicon; silicates are the main component of sand Analyze/Plan. Consider the structural chemistry of silicates discussed in Section and shown in Figures Solve. Tetrahedral Metasilicic acid will probably adopt the single-strand silicate chain structure shown in Figure The empirical formula shows 3 O and 2 H atoms per Si atom. The chain has the same Si to O ratio as metasilicic acid. Furthermore, in the chain structure, there are two terminal (not bridging) O atoms on each Si. These can accommodate the 2 H atoms associated with each Si atom of the acid. The sheet structure does not fulfill these requirements Diborane (Figure and below) has bridging H atoms linking the two B atoms. The structure of ethane shown below has the C atoms bound directly, with no bridging atoms. B 2 H 6 is an electron deficient molecule. It has 12 valence electrons, while C 2 H 6 has 14 valence electrons. The 6 valence electron pairs in B 2 H 6 are all involved in sigma bonding, so the only way to satisfy the octet rule at B is to have the bridging H atoms shown in Figure A hydride ion, H, has two electrons while an H atom has one. The term hydridic indicates that the H atoms in B 2 H 6 have more than the usual amount of electron density for a covalently bound H atom. 643

10 Additional Exercises BrO 3 (aq) + XeF 2 (aq) + H 2 O(l) Xe(g) + 2HF(aq) + BrO 4 (aq) H 2 SO 4 H 2 O SO 3 2HClO 3 H 2 O Cl 2 O 5 2HNO 2 H 2 O N 2 O 3 H 2 CO 3 H 2 O CO 2 (e) 2H 3 PO 4 3H 2 O P 2 O PO 4, + 5; NO 3, + 5 The Lewis structure for NO 4 3 would be: The formal charge on N is +1 and on each O atom is 1. The four electronegative oxygen atoms withdraw electron density, leaving the nitrogen deficient. Since N can form a maximum of four bonds, it cannot form a π bond with one or more of 3 the O atoms to regain electron density, as the P atom in PO 4 does. Also, the short distance would lead to a tight tetrahedron of O atoms subject to steric repulsion Ge(l) + 2Cl 2 (g) GeCl 4 (l) GeCl 4 (l) + 2H 2 O(l) GeO 2 (s) + 4HCl(g) GeO 2 (s) + 2H 2 (g) Ge(s) + 2H 2 O(l) Assume that the reactions occur in acidic solution. The half-reaction for reduction of H 2 O 2 is in all cases H 2 O 2 (aq) + 2H + (aq) + 2e 2H 2 O(aq). N 2 H 4 (aq) + 2H 2 O 2 (aq) N 2 (g) + 4H 2 O(l) 2 SO 2 (g) + H 2 O 2 (aq) SO 4 (aq) + 2H + (aq) NO 2 (aq) + H 2 O 2 (aq) NO 3 (aq) + H 2 O(l) H 2 S(g) + H 2 O 2 (aq) S(s) + 2H 2 O(l) (e) 644

11 Integrative Exercises First calculate the molar solubility of Cl 2 in water. Assuming that x is small compared with : x 3 = (0.1384)( ) = ; x = = M We can correct the denominator using this value, to get a better estimate of x: One more round of approximation gives x = = M. This is the equilibrium concentration of HClO SO 2 (g) + 2H 2 S(aq) 3S(s) + 2H 2 O(g) or, if we assume S 8 is the product, 8SO 2 (g) + 16H 2 S(aq) 3S 8 (s) + 16H 2 O(g). Assume that all S in the coal becomes SO 2 upon combustion, so that 1 mol S (coal) = 1 mol SO 2. = = mol H 2 S This is about 210 lb S per ton of coal combusted. (However, two-thirds of this comes from the H 2 S, which presumably at some point was also obtained from coal.) MnSi: more than one element, so not metallic; high melting, so not molecular; insoluble in water, so not ionic; therefore covalent network. MnSi(s) + HF(aq) SiH 4 (g) + MnF 4 (s) 645

12 Reduction of Mn(IV) to Mn(II) is unlikely, because F reducing agent. for F 2 (g) + 2 e 2F (aq) = 2.87 V is an extremely weak about 1.23 Å; 1.34 Å or less. Since consecutive bonds require sp hybrid orbitals on C (as in allene, C 3 H 4 ), we might expect the orbital overlap requirements of this bonding arrangement to require smaller than usual distances. The product has the formula C 3 H 4 O valence e, 14 e pr Three possibilities are shown above. The group on the lower structure is uncommon and less likely than the two symmetrical structures. 646