Principles of Chemistry 1
Principles of Chemistry 1 CHEM 1211
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Basic Concepts of Chemical Bonding Visualizing Concepts 81 AnalyzePlan Count the number of electrons in the Lewis symbol This corresponds to the lA igroup number of the family Solve a Group 14 or AA b Group 2 or 2A c Group 15 or 5A These are the appropriate groups in the s and p blocks where Lewis symbols are most useful a No Oppositer charged ions combine via Coulombic attraction to form ionic compounds A 1 and A2 have like charges and repel each other b A121 has the largest lattice energy Lattice energy increases as ionic charge increases and decreases as interiionic distance increases Since the magnitude of all charges is 1 this factor doesn t vary among possible ion combinations A1 and Z 1 have the smallest radii which leads to the smallest interiionic distance and the largest lattice energy c A2 Z 2 has the smallest lattice energy because it has the largest interiion distance AnalyzePlan Count the valence electrons in the orbital diagram take ion charge into account and find the element with this orbital electron count on the periodic chart Write the complete electron configuration for the ion Solve a This ion has seven 3d electrons Transition metals or diblock elements have valence electrons in diorbitals Transition metal ions rst lose electrons from the 4s orbital then from 3d if required by the charge This 2 ion has lost two electrons from 4s none from 3d The transition metal with seven 3d7electrons is cobalt Co b The electron con guration of Co is Ar4s23d7 The configuration of Co is Ar3d7 AnalyzePlan This question is a quotreversequot Lewis stIucture Count the valence electrons shown in the Lewis stIucture For each atom assume zero formal charge and determine the number of valence electrons an unbound atom has Name the element Solve A 1 shared e pair 1 valence electron 3 unshared pairs 7 valence electrons F E 2 shared pairs 2 valence electrons 2 unshared pairs 6 valence electrons O D 4 shared pairs 4 valence electrons C Q 3 shared pairs 3 valence electrons 1 unshared pair 5 valence electrons N 8 Chemical Bonding Solutions to Exercises 85 86 X Z 1 shared pair 1 valence electron no unshared pairs H same as X H Check Count the valence electrons in the Lewis structure Does the number correspond to the molecular formula CH ZONF 12 e pair in the Lewis structure CH ZONF 4 2 6 5 7 24 equot 12 e pair The molecular formula we derived matches the Lewis structure AnalyzePlan Since there are no unshared pairs in the molecule we use single bonds to H to complete the octet of each C atom For the same pair of bonded atoms the greater the bond order the shorter and stronger the bond 61 b C a b d Solve Moving from left to right along the molecule the first C needs two H atoms the second needs one the third needs none and the fourth needs one The complete molecule is H H 1 2 3 H CC CEC H In order of increasing bond length 3 lt 1 lt 2 In order of increasing bond enthalpy strength 2 lt 1 lt 3 The central atom Xe is a member of Group 8A currently known as noble gases Prior to 1960 this group was known as inert gases because there were no known compounds of group 8A elements Noble gas elements have 8 valence electrons so they satisfy the octet rule without forming chemical bonds Since forming compounds wasn t necessary for these elements it was assumed that no compounds of group 8A elements existed There are a total of three resonance structures for the given Lewis structure of XeO 3 each with the single bond in a different Xe O bonding domain The given Lewis structure does not satisfy the octet rule because the central Xe atom has more than 8 12 actually electrons Below are four possible Lewis structures for XeOg with formal charges shown Several other resonance structures with one or two double bonds can be drawn Oz l Oz l 1Q Xe Q 1 0 1 3 2 Oz0 O0 0 1 O 0 1 0 It neither minimizes formal charge nor obeys the octet rule so it is not the best Lewis The Lewis structure given in this problem is on the lower left structure for XEOg According to Section 87 the best single Lewis structure is usually the one that obeys the octet rule This structure is shown on the upper left above The structure that minimizes formal charge is on the lower right The quotbestquot view of bonding in XeO 3 is a composite of all correct Lewis structures not any single Lewis structure 181 8 Chemical Bonding Solutions to Exercises L ewis Symb 015 e 87 a Valence electrons are those that take part ln chemlcal bondlng those ln th quot the t e atom lll re oolegas conflguratlon of the atom although lt ls sometlmes only the outer shell electrons b N He 2s22p3 A nitrogen atom hass valence electrons Valence electrons c 1s2 2s22p 3s2 3p2 The atom 51 has 4 valence electrons Ne valence electrons 88 a Atoms w ll garn lose or share electrons to achreve the nearest noblegas electron conflguratlon Except for H anol He thls corresponds to elght electrons in the valence shell thus the term octet rule o s NelsszspA A sulfur atom has slx valence electrons so lt must galn two electrons to achleve an octet c 1s22s22p3 He2522p3 The atom N has flve Wlence electrons anol must galn three electrons to achleve an octet 89 P 1s22s22p63s23p3 A as electron ls a Wlence electron a 2s or 15 electron ls a non valence electron The as Wlence electron ls rnvolveol ln chemrcal oonolrng whlle the 2s r 1s nonrvalence electron ls not 810 Sc 1522522p53523p54523dl Ar4523d Scandlum has three 3 Wlence electrons The e ce electrons are avarlaole for chemrcal oonolrng whlle the core electrons olo not partrcrpate rn bonding 811 5 Al o 812 a K o Ionic Bonding 813 rclr r 814 7 my 815 a A1173 o KZS c yzo3 d Mg3N2 816 3 B30 o RbI c ers d Mglar2 817 a Srz Kr noblegas conflguratlon o le Arlaol2 c Sez Ar4523d 4p5 Kr noblegas conflguratlon 8 Chemical Bonding Solutions to Exercises d Ni Ar3d3 e Br Ar4s23d1 4p6 Kr noblegas configuration f Mn Ar3d4 a Zn Ar3d10 b c Se Ar4s23d1 4p1 This is a very unlikely ion A more stable and commonly found ion would be Sc Sc Ar nobleigas con guration d Ru Kr4d6 e Tl lte6s24f1 5d10 f Au Xe4f145d1 Te Kr5s24d1 5p6 Xe nobleigas con guration a lattice energy is the energy required to totally separate one mole of solid ionic compound into its gaseous ions b The magnitude of the lattice energy depends on the magnitudes of the charges of the two ions their radii and the arrangement of ions in the lattice The main factor is the charges because the radii of ions do not vary over a wide range a NaF 910 kImol MgO 3795 kImol The two factors that affect lattice energies are charge and ionic radii The NarF and MgO separations are similar Na is larger than Mg but F is smaller than 0 The charges on Mg and O are twice those of Na and F so according to Equation 8A the lattice energy of MgO is approximately four times that of NaF b MgCl 2 2326 kImol SrCl 2 2127 k mol The two factors that affect lattice energies are charge and ionic radii The ionic charges are the same in the two compounds The ionic radius of Mg is smaller than that of Sr so the Mng1 distance is slightly smaller than the SrrCl distance Since lattice energy is inversely proportional to the ion separation the lattice energy of MgCl 2 is slightly larger than that of SrCl 2 KF 808 kImol CaO 3414 kImol ScN 7547 kImol The sizes of the ions vary as follows Sc lt Ca lt K and F lt O lt N Therefore the interiionic distances are similar According to Coulomb s law for compounds with similar ionic separations the lattice energies should be related as the product of the charges of the ions The lattice energies above are approximately related as 11 22 33 or 149 Slight variations are due to the small differences in ionic separations a According to Equation 8A electrostatic attraction increases with increasing charges of the ions and decreases with increasing radius of the ions Thus lattice energy i increases as the charges of the ions increase and ii decreases as the sizes of the ions increase 8 Chemical Bonding b Since Solutions to Exercises RbBr lt NaBr lt LiCl lt MgO This order is con rmed by the lattice energies given in Table 82 MgO has the highest lattice energy because the ions have 2 and 27 charges The other compounds have cations with 1 charges and anions with 17 charges They are placed in order of decreasing ionic separation Rb and Br have the largest radii Na is smaller than Rb Li is smaller than Na and Cl is smaller than Br the ionic charges are the same in the two compounds the KrBr and CyCl separations must be approximately equal Since the radii are related as Cs gt K and Br gt Cl the difference between Cs and K must be approximately equal to the difference between Br and Cl This is somewhat surprising since K and Cs are two rows apart and Cl and Br are only one row apart a b C In MgO the magnitude of the charges on both ions is 2 in MgClz the magnitudes of the charges are 2 and 1 Also the Cl ion is larger than the 02 ion so the charge separation is greater in MgCl 2 Thus the lattice energy of MgO is greater because the product of the ionic charges is greater and the ion separation is smaller The ions have 1 and 17 charges in all three compounds In NaCl the cationic and anionic radii are smaller than in the other two compounds so it has the largest lattice energy In RbBr and CsBr the anion is the same but the Cs cation is larger so CsBr has the smaller lattice energy In BaO the magnitude of the charges of both ions is 2 in KF the magnitudes are 1 Charge considerations alone predict that BaO will have the higher lattice energy The distance effect is less clear 02 and F are isoelectronic so F with the larger Z has a slightly smaller radius Ba is two rows lower on the periodic chart than K but it has a greater positive charge so the radii are probably similar In any case the ionic separations in the two compounds are not very different and the charge effect dominates Equation 84 predicts that as the oppositely charged ions approach each other the energy of interaction will be large and negative This more than compensates for the energy required to form Ca and 02 from the neutral atoms see Figure 84 for the formation of NaCl ms 4 CaltggtBr2lt1gt 423mg my a Cawg 1e Ca g gt Ca2g 1e 2Brg 2e gt 2Br g exothermic Ca2g 2Br g gt CaBr2 s exothermic RbC1s gt Rbg Cl g AH lattice energy 7 By analogy to NaCl Figure 84 the lattice energy is AH lat 7AH RbClsAH RbgAHCIgI1RbECI 474305 KJ 858 kJ 1217 kJ 403 kJ 349kJ 692 kJ This value is smaller than that for NaCI 788 kl because Rb has a larger ionic radius than Na This means that the value of d in the denominator of Equation 84 is larger for RbCl and the potential energy of the electrostatic attraction is smaller 184 8 Chemical Bonding Solutions to Exercises 828 By analogy to Figure 84 AH lat AH CaCI 2 AH Cag2AH Cg1Ca2Ca2EC 7958 kJ 1793 kJ 21217 kJ 590 kJ 1145 kJ 2 349 M 2256 kJ From Table 82 the lattice energy of NaCl 788 k mol is considerably less than that of CaF 2 The 2 charge of Ca2 leads to much greater electrostatic attractions and a higher lattice energy Covalent Bonding Electronegativity and Bond Polarity 829 830 831 832 833 834 835 a A covalent bond is the bond formed when two atoms share one or more pairs of electrons b Any simple compound whose component atoms are nonmetals such as H 2 SO 2 and CCl 4 are molecular and have covalent bonds between atoms c Covalent because it is a gas even below room temperature K and Ar K is an active metal with one valence electron It is most likely to achieve an octet by losing this single electron and to participate in ionic bonding Ar has a stable octet of valence electrons it is not likely to form chemical bonds of any type AnalyzePlan Follow the logic in Sample Exercise 83 Solve E 3971 x quot Lquot l 139 2 I Hi el 2 H 91 Check Each pair of shared electrons in SiCl4 is shown as a line each atom is surrounded by an octet of electrons F m 39I 13 gt P 31L a 5Q b A double bond is required because there are not enough electrons to satisfy the octet rule with single bonds and unshared pairs c The greater the number of shared electron pairs between two atoms the shorter the distance between the atoms If 02 has a double bond the 0 10 distance will be shorter than the 0 0 single bond distance 35 CISI The C S bonds in CS 2 are double bonds so the C 8 distances will be shorter than a C S single bond distance a Electronegativity is the ability of an atom in a molecule a bonded atom to attract electrons to itself b The range of electronegativities on the Pauling scale is 07 40 185 8 Chemical Bonding Solutions to Exercises c Fluorine F is the most electronegative element d Cesium Cs is the least electronegative element that is not radioactive a The electronegativity of the elements increases going from left to right across a row of the periodic chart b c Generally the trends in electronegativity are the same as those in ionization Electronegativity decreases going down a family of the periodic chart energy and opposite those in electron af nity That is the more positive the ionization energy and the more negative the electron affinity ignoring a few exceptions the greater the electronegativity of an element Plum Electronegativity increases going up and to the right in the periodic table a Br Check The b C c P d Be values in F39 g 8 quot Electronegativity increases going up and to the right in the periodic table a O b Al c C1 d F The bonds in a c and d are polar because the atoms involved differ in electronegativity The more electronegative element in each polar bond is a F C 0 d1 The more different the electronegativity values of the two elements the more polar the bond a 07F lt CrF lt BerF This order is clear from the periodic trend b SrBr lt CrP lt 0C1 Refer to the electronegativity values in Figure 86 to confirm the order of bond polarity The 3 pairs of elements all have the same positional relationship on the periodic chart The more electronegative element is one row above and one column to the left of the less electronegative element This leads us to conclude that AEN is similar for the 3 bonds which is confirmed by values in Figure 86 The most polar bond OrCl involves the most electronegative element 0 Generally the largest electronegativity differences tend to e between row 2 and row 3 elements The 2 bonds in this exercise involving elements in row 2 and row 3 do have slightly greater AEN than the SrBr bond between elements in rows 3 and 4 c CrS lt N70 lt BrF You might predict that N70 is least polar since the elements are adjacent on the table However the big decrease going from the second row to the third means that the electronegativity of S is not only less than that of 0 but essentially the same as that of C C75 is the least polar AnalyzePlan Q is the charge at either end of the dipole Q u r From Table 83 the values for HF are u 182 D and r 092 A Change Ato m and use the de nition of the Debye and the charge of an electron to calculate the charge in units of e Salve 8 Chemical Bonding Solutions to Exercises 842 843 844 334 x 10 30 Com 1e x 19 041e 160 x 10 C Q 0 182D 14 o x x r 092A 1 x 10 10 m 1D Check The calculated charge on H and F is 041 6 This can be thought of as the amount of charge transferred from H to F This value is consistent with our idea that HF is a polar covalent molecule the bonding electron pair is unequally shared but not totally transferred from H to F a The more electronegative element Br will have a stronger attraction for the shared electrons and adopt a partial negative charge b Q is the charge at either end of the dipole 334 x 10 30 Com 16 X 19 160 x 10 c 01014 0101 e H 121D 14 Q 0 X X r 249A 1 x 10 m 1D The charges on I and Br are 0101 e AnalyzePlan Generally compounds formed by a metal and a nonmetal are described as ionic while compounds formed from two or more nonmetals are covalent Solve a MnO 2 ionic b P 2 S 3 covalent c CoO ionic d copperl sulfide ionic e chlorine trifluoride covalent vanadiumV uoride ionic Generally compounds formed by a metal and a nonmetal are described as ionic while compounds formed from two or more nonmetals are covalent a MnF 3 ionic b CrO 3 ionic c AsBr5 covalent As is a metalloid so the bondin model is not obvious The 8 structure of the compound a symmetrical trigonal bipyramid indicates that it is probably covalent d sulfur tetrafluoride covalent e molybdenumIV chloride ionic f scandiumlll chloride ionic Lewis Structures Resonance Structures 845 Analyze Counting the correct number of valence electrons is the foundation of every Lewis structure PlanSolve a Count valence electrons 4 4 x 1 8 equot 4 equot pairs Follow the procedure in Sample Exercise 86 b Valence electrons 4 6 10 equot 5 equot pairs ICOZ 187 8 Chemical Bonding Solutions to Exercises 846 C d e Valence electrons 6 2 x 7 20 equot 10 e pairs i Place the S atom in the middle and connect each F atom with a single bond this requires 2 e pairs ii Complete the octets of the F atoms with nonbonded pairs of electrons this requires an additional 6 equot pairs iii The remaining 2 e pairs complete the octet of the central S atom 32 valence equot 16 e pairs Choose the Lewis structure that obeys the octet rule Section 87 Follow Sample Exercise 88 20 valence equot 10 e pairs l 14 valence equot 7 equot pairs H N H H Check In each molecule bonding e pairs are shown as lines and each atom is surrounded by an octet of electrons duet for a C 12 equot 6 e pairs b 14 valence equot 7 e pairs H c H ij ij H IL 50 valence equot 25 e pairs d 26 valence equot 13 e pairs F IF quot 39 39 a s a F C C JE I I I a Choose the Lewis structure that obeys the octet rule Section 87 26 valence e 13 e pairs f 10 equot 5 e pairs 9 H CEC H Choose the Lewis structure that obeys they octet rule Section 87 188 8 Chemical Bonding Solutions to Exercises 847 848 849 a Formal charge is the charge on each atom in a molecule assuming all atoms have the same electronegativity b Formal charges are not actual charges They assume perfect covalency one extreme for the possible electron distribution in a molecule c The other extreme is represented by oxidation numbers which assume that the more electronegative element holds all electrons in a bond The true electron distribution is some composite of the two extremes a 26 equot 13 equot pairs 2132 The octet rule is satisfied for all atoms in the structure b F is more electronegative than P Assuming F atoms hold all shared electrons the oxidation number of each F is 1 The oxidation number of P is 3 c Assuming perfect sharing the formal charges on all F and P atoms are 0 d The oxidation number on P is 3 the formal charge is 0 These represent extremes in the possible electron distribution not the best picture By virtue of their greater electronegativity the F atoms carry a partial negative charge and the P atom a partial positive charge AnalyzePlan Draw the correct Lewis structure count valence electrons in each atom total valence electrons and electron pairs in the molecule or ion connect bonded atoms with a line place the remaining equot pairs as needed in nonbonded pairs or multiple bonds so that each atom is surrounded by an octet or duet for Calculate formal charges assign electrons to individual atoms nonbonding equot 12 bonding e39 formal charge valence electrons assigned electrons Assign oxidation numbers assuming that the more electronegative element holds all electrons in a bond Solve Formal charges are shown near the atoms oxidation numbers ox are listed below the structures a 10 equot 5 equot pairs b 32 valence equot 16 equot pairs 1 I NEoz 6 0 1 0 ci P ci 0 ox N 3 0 2 39 1 quot 29312 0 ox F 5 Cl 1 O 2 189 8 Chemical Bonding Solutions to Exercises 850 851 C 32 valence equot 16 equot pairs d 26 valence equot 13 equot pairs 2 1 16 amp391 6 H 262 o I 6 0 9 13Q Cl Q 1 1 Tl ox c1 5 H 1 0 2 1 ox 2 Cl 7 0 2 Check Each atom is surrounded by an octet or duet and the sum of the formal charges and oxidation numbers is the charge on the particle Formal charges are given near the atoms oxidation numbers are listed below the structures a 18 equot 9 equot pairs b 24 equot 12 equot pairs o o o 0 o o 9 5 9 1zo soo 1 1 0 39 I 39 9 ox 2 8 4 0 2 1 ox 2 8 6 0 2 C 26 equot 13 equot pairs d 32 e39 16 equot pairs 1 1 2 262 2 12c 39s39 C 2 1 I 1 3C S O 1 O O O O 1 292 1 ox 2 8 4 0 2 ox 2 8 6 0 2 a Plan Count valence electrons draw all possible correct Lewis structures taking note of alternate placements for multiple bonds Solve 18 equot 9 equot pairs ea 9i H ia wr Check The octet rule is satisfied b Plan Isoelectronic species have the same number of valence electrons and the same electron configuration Solve A single 0 atom has 6 valence electrons so the neutral ozone molecule 03 is isoelectronic with NO 2 39 H Check The octet rule is satisfied 190 8 Chemical Bonding Solutions to Exercises 852 853 854 855 C Since each N O bond has partial double bond character the N O bond length in NO 2 39 should be shorter than in species with formal N O single bonds a 16 equot 8 e pairs C C lt gt OEN lt gt id NEG b More than one correct Lewis structure can be drawn so resonance structures are needed to accurately describe the structure c NO2 has 16 valence electrons Consider other triatomic molecules involving secondrow nonmetallic elements 032 or C34quot are not common or stable CO 2 is common and matches the description as does N 3 azide ion PlanSolve The Lewis structures are as follows 5 equot pairs 8 equot pairs 6c o CEO 12 e pairs 0 2 6 2 O 2 C C C o o 0 o o o o o o The more pairs of electrons shared by two atoms the shorter the bond between the atoms The average number of electron pairs shared by C and O in the three species is 3 for CO 2 for CO 2 and 133 for CO 3239 This is also the order of increasing bond length CO lt CO2 lt C032quot The Lewis structures are as follows 5 e pairs 9 e pairs NEOI i391 N 39o o o o39 12 e pairs 0 6 O N N N o o o o 20 o o o of o The average number of electron pairs in the N O bond is 30 for NO 15 for N02quot and 133 for N031 The more electron pairs shared between two atoms the shorter the bond Thus the N O bond lengths vary in the order NO lt NO 2 39 lt NO 3 a Two equally valid Lewis structures can be drawn for benzene 191 8 Chemical Bonding Solutions to Exercises 856 b a b C H H H H Each structure consists of alternating single and double C C bonds a particular bond is single in one structure and double in the other The concept of resonance dictates that the true description of bonding is some hybrid or blend of the two Lewis structures The most obvious blend of these two resonance structures is a molecule with six equivalent C C bonds each with some but not total double bond character If the molecule has six equivalent C C bonds the lengths of these bonds should be equal The resonance model described in a has six equivalent C C bonds each with some double bond character That is more than one pair but less than two pairs of electrons is involved in each C C bond This model predicts a uniform C C bond length that is shorter than a single bond but longer than a double bond H H H H H H H H H H H H The resonance model of this molecule has bonds that are neither single nor double but somewhere in between This results in bond lengths that are intermediate between C C single and CC double bond lengths GO Exceptions to the Octet Rule 857 858 a b The octet rule states that atoms will gain lose or share electrons until they are surrounded by eight valence electrons The octet rule applies to the individual ions in an ionic compound That is the cation has lost electrons to achieve an octet and the anion has gained electrons to achieve an octet For example in MgClZ Mg loses 2 e to become Mg2 with the electron configuration of Ne Each Cl atom gains one electron to form Cl with the electron configuration of Ar Carbon in group 14 needs to form four single bonds to achieve an octet as in CH4 Nitrogen in group 15 needs to form three as in NH 3 If G group number and n the number of single bonds G n 18 is a general relationship for the representative non metals Check 0 as in H20 G 16 n 2 bonds 18 192 8 Chemical Bonding Solutions to Exercises 859 860 861 The most common exceptions to the octet rule are molecules with more than eight electrons around one or more atoms usually the central atom Examples SF 6 PF 5 In the third period atoms have the space and available orbitals to accommodate extra electrons Since atomic radius increases going down a family elements in the third period and beyond are less subject to destabilization from additional electronelectron repulsions Also the third shell contains d orbitals that are relatively close in energy to 35 and 3p orbitals the ones that accommodate the octet and provide an allowed energy state for the extra electrons a 26 equot 13 e pairs 2 o T o 292 Other resonance structures with one two or three double bonds can be drawn While a structure with three double bonds minimizes formal charges all structures with double bonds violate the octet rule Theoretical calculations show that the single best Lewis structure is the one that doesn t violate the octet rule Such a structure is shown above b 6 equot 3 e pairs H Al H H 6 electrons around Al impossible to satisfy octet rule with only 6 valence electrons c 16 equot 8 e pairs NEN lxl lt gtll NEN lt gtllNll 3 resonance structures all obey octet rule d 20 equot 10 e pairs 51 Cl C H O O l H Obeys octet rule e 40 equot 20 e pairs Does not obey octet rule 10 e around central Sb 193 8 Chemical Bonding Solutions to Exercises 862 863 864 a b a b C 16 equot 8 e pairs oc o lt gt CEO lt gt OEC Three resonance structures all obey the octet rule The left structure minimizes formal charge 26 equot 13 e pairs 292 Other resonance structures with one two or three double bonds can be drawn While a structure with three double bonds minimizes formal charges all structures with double bonds Violate the octet rule Theoretical calculations show that the single best Lewis structure is the one that doesn t Violate the octet rule Such a structure is shown above 6 equot 3 e pairs d 32 equot 16 e pairs H B H H F H F B F 6 electrons around B 39 39 I 22 equot 11 e pair Obeys octet rule Violates the octet rule 10 e around central Xe 16 equot 8 e pairs 1 Be This structure Violates the octet rule Be has only 4 e around it 1Be lt gt ci BeEc1lt gt CIEBe Q The formal charges on each of the atoms in the four resonance structures are Eil Be 39c391 c391Be a cj BeE c1 CIEBe 121 0 0 0 1 2 1 0 2 2 2 2 0 Formal charges are minimized on the structure that Violates the octet rule this form is probably most important Note that this is a different conclusion than for molecules that have resonance structures with expanded octets that minimize formal charge 19 equot 95 e pairs odd electron molecule d g dlt gt d j dH 36 51 63 194 8 Chemical Bonding Solutions to Exercises b C None of the structures satisfies the octet rule In each structure one atom has only 7 e around it If a molecule has an odd number of electrons in the valence shell no Lewis structure can satisfy the octet rule H H 1 2 1 1 1 0 0 1 1 Formal charge arguments predict that the two resonance structures with the odd electron on O are most important This contradicts electronegativity arguments which would predict that the less electronegative atom Cl would be more likely to have fewer than 8 e around it Bond Enthalpies 865 866 Analyze Given structural formulas Find enthalpy of reaction Plan Count the number and kinds of bonds that are broken and formed by the reaction Use bond enthalpies from Table 84 and Equation 812 to calculate the overall enthalpy of reaction AH 61 b C a b C Solve AH 2DOH DOO 4DCH DCC 2DO H 2DO C 4DC H DC C AH DO O DCC 2DO C DC C 146 614 2358 348 304 k AH 5DC H DC E N DCC 5DC H DC E N 2DC C DCC 2DC C 614 2348 82 k AH 6DN Cl 3DCl Cl DN E N 6200 3242 941 467 k AH 3DC Br DC H DCl Cl 3DC Br c c1 DHCl DCH DCl Cl DC C1 DHCl AH 413 242 328 431 104 k AH 4DC H 2DC S 2DS H DCC DHBr 4DSH DCC 2DC Br 4DCH 2DCS DH Br 2DS H 2DC Br AH 2259 366 2339 2276 346 k AH 4DN H DN N DCl Cl 4DNH 2DNCl DNN DCl Cl 2DNCl AH 163 242 2200 5 k 195 8 Chemical Bonding Solutions to Exercises 867 868 Plan Draw structural formulas so bonds can be Visualized Then use Table 84 and Equation 812 Solve a 2 H c H 00 gt 2 H c o H AH 8DC H DOO 6DCH 2DCO 2DOH 2DCH DOO 2DCO 2DO H 2413 495 2358 2463 321 k b H H BrBr gt 2 HBr AH DHH DBr Br 2DH Br 436 193 2366 103 k c 2 HOOH gt 2 HOH o 0 AH 4Do H 2DOO 4Do H DOO AH 2DOO DOO 2146 495 203 k Plan Draw structural formulas so bonds can be Visualized Solve a HH H H AH 12DC H 3DcC 12DC H 6DCC 3DcC 6DCC 3614 6348 246 k C1 C1 b I I H Si H 3 Cl Cl H Cl AH 3DSi H DSi C1 3Dc1 c1 4DSiCl 3DH c1 Cl Si Cl 3 H Cl 3DSiH 3Dc1 c1 3DSi C1 3DH c1 3323 3242 3464 3431 990 k C Plan Use bond enthalpies to calculate AH for the reaction with 88g as a product Then 8stltggt 98H2ltggtsgltggt AHO 588 gt 585 AHl f0r 588 8H28g gt 8H2g88s Aern AH AH 58g 196 8 Chemical Bonding Solutions to Exercises 869 870 871 AH 16DSH 8HH 8S S 16339 8436 8266 192 k Aern AH AH 88g 192 k 1023 k 2943 294 k Plan Draw structural formulas so bonds can be visualized Then use Table 84 and Equation 812 Solve a b a b NEN 3 H H gt 2 H ii H 139 AH DN E N 3DH H 6N H 941 k 3436 kJ 6391 k 97 k 2 mol NH 3 exothermic Plan Use Eq 531 to calculate AH m from AH values AHXr1 2 Zn AH products Zn AH reactants AH NH3g 4619 k Solve AHEX 2AHE NH3ltggt 3 AH H2ltggt AHE N2ltggt AHXn 2 4619 30 0 9233 kJ2 mol NH3 The AH calculated from bond enthalpies is slightly more exothermic more negative than that obtained using AH values CC H H AH 4DC H DCC DHH 6DC H DC C DCC DHH 2DC H DC C AH 614 436 2413 348 124 k AHO 2 AH C2H68 AH C2H48 AH H28 8468 5230 O 13698 k The values of AH for the reaction differ because the bond enthalpies used in part a are average values that can differ from one compound to another For example the exact enthalpy of a C H bond in C 2H 4 is probably not equal to the enthalpy of a C H bond in C 2H 6 Thus reaction enthalpies calculated from average bond enthalpies are estimates On the other hand standard enthalpies of formation are measured quantities and should lead to accurate reaction enthalpies The advantage of average bond enthalpies is that they can be used for reactions Where no measured enthalpies of formation are available The average Ti Cl bond enthalpy is just the average of the four values listed 430 kJmol 197 8 Chemical Bonding Solutions to Exercises 872 a b F i C 2 F F gt F C F F AH 2DFF 4DCF 2155 4485 1630 k F ii CEO 3 F F gt F C F F O F F AH CEO 3DF F 4DCF 2DDF 1072 3155 4485 2190 783 k F iii oco 4 F F gt F C F 2 F O F l AH 2DCO 4DFF 4DCF 4DOF 2799 4155 4485 4190 482 k Reaction is most exothermic The more oxygen atoms bound to carbon the less exothermic the reaction in this series Additional Exercises 873 874 875 Six nonradioactive elements in the periodic table have Lewis symbols with single dots Yes they are in the same family assuming H is placed with the alkali metals as it is on the inside cover of the text This is because the Lewis symbol represents the number of valence electrons of an element and all elements in the same family have the same number of valence electrons By definition of a family all elements with the same Lewis symbol must be in the same family a b a Lattice energy is proportional to Q1 Q2 d For each of these compounds Q1 Q2 is the same The anion Hquot is present in each compound but the ionic radius of the cation increases going from Be to Ba Thus the value of d the cationanion separation increases and the ratio Q1Q2 d decreases This is reflected in the decrease in lattice energy going from BeH 2 to BaH 2 Again Q1 Q 2 for ZnH2 is the same as that for the other compounds in the series and the anion is Hquot The lattice energy of ZnH2 2870 k is closest to that of MgH 2 2791 k The ionic radius of Zn2 is similar to that of Mg Lattice Compound Energy kJ Lattice Compound Energy kJ NaCl 788156k LiCl 834155k1 106 k NaBr 732 104 k LIB 779 Na 1 682 Li I 730 198 8 Chemical Bonding Solutions to Exercises The difference in lattice energy between LiCl and Lil is 104 k The difference between NaCl and NaI is 106 k the difference between NaCl and NaBr is 56 k or 53 of the difference between NaCl and NaI Applying this relationship to the Li salts 053104 k 55 k difference between LiCl and LiBr The approximate lattice energy of LiBr is 834 55 k 779 k b Lattice Lattice Compound Energy kJ Compound Energy kJ NaCl 788 56 k CsCl 657 30 k 106 k NaBr 732 57 k CsBr 627 Na I 682 Cs I 600 By analogy to the Na salts the difference between lattice energies of CsCl and CsBr should be approximately 53 of the difference between CsCl and CsI The lattice energy of CsBr is approximately 627 k C Lattice Lattice Compound Energy kJ Compound Energy kJ MgO 3795 381k MgCl2 23261131k 578 k CaO 3414 199 k c6c12 2195 SrO 3217 SrC12 2127 By analogy to the oxides the difference between the lattice energies of MgC12 and CaC12 should be approximately 66 of the difference between MgC12 and SrClZ That is 066199 k 131 k The lattice energy of CaC12 is approximately 2326 131 k 2195 k 899 x 109 Jom 4160 gtlt 10 19C2 X 876 E 2 10 c 099140 x 10 m 3852 x 10 18 385 x 10 18 J On a molar basis 3852 x 10 186O22 x 1023 2319 x 106 J 2320 k Note that its absolute value is less than the lattice energy 3414 k mol The difference represents the added energy of putting all the Ca2OZ ion pairs together in a three dimensional array similar to the one in Figure 83 877 EQ1Q2d llt899gtlt109mcoul2 899 x 109161 X 160 x 10 19C2 c2 O97196 x 10 10 m 785 x 10 19 The sign of E is negative because one of the interacting ions is an anion this is an attractive interaction On a molar basis 7855 x 10 19 x 6022 x 1023 473 x 105 J 473 k a NaBr E 78547 x 10 19 899 109 160 10 19c 2 b RbBrE X 2 J m x X 910 671 x 10 c 147196 x 10 m On a molar basis 404 x 105 J 404 k 899 x 109 Jom 2 x 160 x 10 19 cf X c2 113184 x 10 10 m On a molar basis 187 x 106 J 187 x 103 k c Sr282E 310 x 1048 199 8 Chemical Bonding Solutions to Exercises 878 879 880 881 a b C A polar molecule has a measurable dipole moment its centers of positive and negative charge do not coincide A nonpolar molecule has a zero net dipole moment its centers of positive and negative charge do coincide Yes If X and Y have different electronegativities they have different attractions for the electrons in the molecule The electron density around the more electronegative atom will be greater producing a charge separation or dipole in the molecule u Qr The dipole moment u is the product of the magnitude of the separated charges Q and the distance between them r Molecule b H 2 S and ion c NO 2 39 contain polar bonds The atoms that form the bonds H8 and N O have different electronegativity values a b C d a 1 130 The most polar bond will be formed by the two elements with the greatest difference in electronegativity Since electronegativity increases moving right and up on the periodic chart the possibilities are 13 0 and TeO These two bonds are likely to have similar electronegativitiy differences 3 columns apart vs 3 rows apart Values from Figure 86 confirm the similarity and show that 130 is slightly more polar TeI Both are in the fifth row of the periodic chart and have the two largest covalent radii among this group of elements TeI2 Te needs to participate in two covalent bonds to satisfy the octet rule and each I atom needs to participate in one bond so by forming a Tel 2 molecule the octet rule can be satisfied for all three atoms 512 IT III B203 Although this is probably not a purely ionic compound it can be understood in terms of gaining and losing electrons to achieve a noblegas configuration If each B atom were to lose 3 e and each 0 atom were to gain 2 equot charge balance and the octet rule would be satisfied P20 3 Each P atom needs to share 3 e and each 0 atom 2 e to achieve an octet Although the correct number of electrons seem to be available a correct Lewis structure is difficult to imagine In fact phosphorus III oxide exists as P406 rather than P 2 O 3 Chapter 22 12 3 15 30 valence equot 15 e pairs H ii H H N H C C C C on I e MO H H Structures with H bound to N and nonbonded electron pairs on C can be drawn but the structures above minimize formal charges on the atoms The resonance structures indicate that triazine will have six equal C N bond lengths intermediate between C N single and C N double bond lengths See 200 8 Chemical Bonding Solutions to Exercises 882 883 884 Solutions 855 and 856 From Table 85 an average C N length is 143 A a CN length is 138 A The average of these two lengths is 1405 A The C N bond length in triazine should be in the range 140 141 Use the method detailed in Section 85 A Closer Look to estimate partial charges from electronegativity values From Figure 86 the electronegativity of Br is 28 and of Cl is 30 Br has 28 30 28 048 of the charge of the bonding e pair Cl has 30 30 28 052 of the charge of the bonding e pair This amounts to 052 x 26 1046 on Cl or 0046 more than a neutral Cl atom This implies a 004 charge on Cl and 004 charge on Br From Figure 75 the covalent radius of Br is 114 A and of Cl is 099 A The Br Cl separation is 213 A 1 1 19 1 1 10 1D 60X0 Cgtlt213Agtlt XEJ m r004e x x M Q 6 A 334 x 10 30cm 041 D Clearly this method is approximate The estimated dipole moment of 041 D is within 28 of the measured value of 057 D 13 39 has a Lewis structure with an expanded octet of electrons around the central 1 II quot12 I F cannot accommodate an expanded octet because it is too small and has no available d orbitals in its valence shell Formal charge FC valence equot nonbonding equot 1 2 bonding equot a 18 equot 9 e pairs FC for the central 0 6 2 12 6 1 b 48 equot 24 e pairs FCforP5 0 12 12 1 The three nonbonded pairs on each F have been omitted c 17 equot 8 e pairs 1 odd e CN IQ The odd electron is probably on N because it is less electronegative than 0 Assuming the odd electron is on N FC for N 5 1 1 2 6 1 If the odd electron is on 0 FC for N 5 2 12 6 0 201 8 Chemical Bonding Solutions to Exercises 885 886 887 d a b C a b NEN C lt N NEO lt gt iiNo 28 e39 14 e39 pairs e 32 e39 16 e39 pairs ii 6 3913 C1 H FCforI74126O Cl FCfor c17 O128 3 14e39 7 e39 pairs 32 e39 16 e39 pairs xi 91 Q C1 O PC on C1761220 FC onCl 7 O 128 3 The oxidation number of Cl is 1 in C10quot and 7 in C10 4 39 The definition of formal charge assumes that all bonding pairs of electrons are equally shared by the two bonded atoms that all bonds are purely covalent The definition of oxidation number assumes that the more electronegative element in the bond gets all of the bonding electrons that the bonds are purely ionic These two definitions represent the two extremes of how electron density is distributed between bonded atoms In C10quot and C104quot Cl is the less electronegative element so the oxidation numbers have a higher positive value than the formal charges The true description of the electron density distribution is somewhere between the extremes indicated by formal charge and oxidation number 0 1 1 2 1 1 1 1 0 In the leftmost structure the more electronegative O atom has the negative formal charge so this structure is likely to be most important In general the more shared pairs of electrons between two atoms the shorter the bond and vice versa That the N N bond length in N20 is slightly longer than the typical NEN indicates that the middle and right resonance structures where the N atoms share less than three electron pairs are contributors to the true structure That the NO bond length is slightly shorter than a typical NO indicates that the middle structure where N and 0 share more than two electron pairs does contribute to the true structure This physical data indicates that while formal charge can be used to predict which resonance form will be more important to the observed structure the in uence of minor contributors on the true structure cannot be ignored AH 8DCH DCC 6DCH DHH 2DCH DCC DHH 2413 348 436 42 k 202 8 Chemical Bonding Solutions to Exercises AH 8DCH 12 DCC DC C 6DCH 2DOH 2DC H 12 DCC DCC 2DOH 2413 12 495 348 2463 200 k The fundamental difference in the two reactions is the formation of 1 mol of H H bonds versus the formation of 2 mol of 0 H bonds The latter is much more exothermic so the reaction involving oxygen is more exothermic 888 a AH 5DCH DC C DC O DOH 6DCH 2DC O DC C DOH DCH DC O 348 k 463 k 413 k 358 k AH 40 k ethanol has the lower enthalpy b AH 4DCH DC C 2DCO 4DCH DC C DCC 2DCO DCC 2358 kJ 799 k AH 83 k acetaldehyde has the lower enthalpy c AH 8DCH 4DC C DCC 8DC H 2DC C 2DCC 2DCC DCC 2348 kJ 614 k AH 82 k cyclopentene has the lower enthalpy 01 AH 3DCH DC N DC E N 3DC H DC C DC E N DC N DC C 293 k 348 k AH 55 k acetonitrile has the lower enthalpy 1r 1r 1r 889 a 4 H c c C H gt 6 NEN 12 oco I I I 10 H o Hoo N N N o O o O o O nitroglycerine AH 20DC H 8DCC 12DCO 24DON 12DNO 6DN a N 24DCO 20DHO DCC AH 20413 8348 12358 24201 12607 6941 24799 20463 495 7129 k 203 8 Chemical Bonding Solutions to Exercises b 890 a b 1 mol C3H5N309 2271 g C3H5N309 7129 k x 4 mol C3H5N309 100 g C3H5N3Og x 785 kJg C3H5N309 4C7H5N306s gt 6N2g 7C02g 1OHZOg 21Cs C3H6N606 126303684e 42e pairs 42 e pairs 24 shared e pairs 18 unshared lone e pairs Use unshared pairs to complete octets on terminal 0 atoms 15 unshared pairs and ring N atoms 3 unshared pairs No CN bonds in the 6membered ring are possible because all C octets are complete with 4 bonds to other atoms NN are possible as shown below There are 8 possibilities involving some combination of N N and NN groups 1 with O NN 3 with 1 NN 3 with 2NN 1 with 3NN A resonance structure with 1 NN is shown below Each terminal CN O group has two possible placements for the NO This generates 8 structures with O NN groups and 3 O N O groups 4 with 1 NN and 2 CN O 2 with 2 NN and 1 CN O and 1 with 3 NN and no CN O This sums to a total of 15 resonance structures that I can Visualize Can you find others C3H6N606s gt3COg 3N2g 3H20g The molecule contains NO NN C H C N N O and N N bonds According to Table 84 N N bonds have the smallest bond enthalpy and are weakest Calculate the enthalpy of decomposition for the resonance structure drawn in part a AH 3DNO 3DN O 3DN N 6DN C 6DCH 3DCEO 3DNEN 6DOH 204 8 Chemical Bonding Solutions to Exercises 891 892 893 3607 3201 3163 6293 6413 31072 3941 6463 1668 kJmolC3H6N606 1 1c H N o 16681lt 50 gC3H6N6O6 x m 6 6 6 6 I 23755238 k While exchanging NO and N O bonds has no effect on the enthalpy calculation structures with NN and 2 N O do have different enthalpy of decomposition For the resonance structure with 3 NN and 6 N O bonds instead of 3 N N 3 N O and 3 NO AH 2121 k mol The actual enthalpy of decomposition is probably somewhere between 1668 and 2121 k mol The enthalpy charge for the decomposition of 50 g RDX is then in the range 38 48 k a AH 2DAA 4DAA b For the reaction to be exothermic AH DAA lt 2DAA c If the reaction is exothermic the second bond between atoms in AA must be weaker than the first bond The first bond is AA If AA lt 2A A then the second bond has a smaller bond dissociation enthalpy than the first 1000 Q 800 gt E GS 3 5 600 CI LTJ quotC5 CI O m 400 200 10 11 12 13 14 15 16 Bond Length When comparing the same pair of bonded atoms C N vs CN vs CEN the shorter the bond the greater the bond energy but the two quantities are not necessarily directly proportional The plot clearly shows that there are no simple length strength correlations for single bonds alone double bonds alone triple bonds alone or among different pairs of bonded atoms all C C bonds vs all C N bonds etc a S N z 177 A sum of the bonding atomic radii from Figure 76 b S O z 175 A the sum of the bonding atomic radii from Figure 76 Alternatively half of the S S distance in 88 102 plus half of the 0 0 distance from Table 85 074 is 176 A c Owing to the resonance structures for SO 2 we assume that the S O bond in 80 2 is intermediate between a double and single bond so the distance of 143 A should be significantly shorter than an S O single bond distance 175 A 205 8 Chemical Bonding Solutions to Exercises d 54 equot 27 equot pair s s 39s39 s O S I I lt 39I39 I S 282 St S S S OSO S The observed S O bond distance 148 A is similar to that in 80 2 143 A which can be described by resonance structures showing both single and double S O bonds Thus 880 must have resonance structures with both single and double S O bonds The structure with the SO bond has 5 e pairs about this 8 atom To the extent that this resonance form contributes to the true structure the 8 atom bound to O has more than an octet of electrons around it Integrative Exercises 894 a Ti2 Ar3d2 Ca Ar4sz Yes The two valence electrons in Ti2 and Ca are in different principle quantum levels and different subshells b According to the Aufbau Principle valence electrons will occupy the lowest energy empty orbital Thus in Ca the 45 is lower in energy than the 3d while in Ti the 3d is lower in energy than the 45 c Since there is only one 45 orbital the two valence electrons in Ca are paired There are five degenerate 3d orbitals so the two valence electrons in Ti2 are unpaired Ca has no unpaired electrons Ti2 has two 895 a Srs gtSrg AH Srg AHsflb Srs Srg gt Srg 1 e I1 Sr Srg gt Sr2g 1 e I2 Sr C12g gt 2Clg 24H Cg DC 2 2Clg 2 equot gt 2Cl39g 2E1 Cl Src12 s gt Srs c12 g AH Src12 Src12 s gt Sr2g 2Cl39g m b AH SrC12s AH Srg 11 Sr 12 Sr 24H Clg 2ECl AHlatt Srcl2 AH SrC12s 1644 k 549 k 1064 k 21217 k 2 349 k 2127 k 804 k 896 The pathway to the formation of K 2 O can be written 2Ks gt 2Kg 2AH Kg 2Kg gt 2K g 2 e 2 11 K 12 02 g gt Og AH 09 0g 1 equot gt O39 E1O O g 1 e gt 021 E2O 2Kltggt02 ltggt 4K20ltsgt 4H1a1lt20ltsgt 2Ks 12 02g gt 1905 AH KZOs 206 8 Chemical Bonding Solutions to Exercises 8 97 AH K20S2AH K921KAH 09 E1OE2O AHla1K2OS E20AH K2OSAH1a1 K20S2AH K921KAH 09E 0 E 2 O 73632 k 2238 k 7 28999 k 7 2419 k 7 2475 k 7 7141 k a b c d e a b C d 750 k Assume 100 g A 877 g ln11482 0764H1011n07640384m 2 123 g S3207 0384 H101S 03840384 1 B 782 g ln11482 0681 mol In 0681068 01 218 g S3207 0680 H101 S 06800680 1 C 705 g ln11482 0614H1011n06140614 1 295 g S3207 0920 H101S 09200614 15 A InzSBInSCIn2S3 A Inl B Inll C Inlll Inl Kr5524d10 11111 Kr5514d10 InIII 1ltr4c110 None of these is a noble7gas con guration The ionic radius of In in compound C will be smallest Removing successive electrons from an atom reduces electron repulsion increases the effective nuclear charge experienced by the valence electrons and decreases the ionic radius The higher the charge on a cation the smaller the radius Lattice energy is directly related to the charge on the ions and inversely related to the interionic distance Only the charge and size of the In varies in the three compounds Inl in compound A has the smallest charge and the largest ionic radius so compound A has the smallest lattice energy and the lowest melting point Inlll in compound C has the greatest charge and the smallest ionic radius so compound C has the largest lattice energy and highest melting point Even though Cl has the greater more negative electron af nity F has a much larger ionization energy so the electronegativity of F is greater F kIE7EA k1681 7 7328 k2009 C1 kIE7EA k1251 7 7349 k1600 Electronegativiy is the ability of an atom in a molecule to attract electrons to itself It can be thought of as the ability to hold its own electrons as measured by ionization energy and the capacity to attract the electrons of other atoms as measured by electron affinity Thus both pr0perties are relevant to the concept of electronegativity EN kIE 7 EA For F 40 k2009 k 402009 20 X 103 Cl EN 20 X 10 3 1600 32 0 EN 20 X 103 13147 7141 29 207 8 Chemical Bonding Solutions to Exercises 899 8100 a b C b C 01 These values do not follow the trend on Figure 86 The Pauling scale on the figure shows 0 to be second only to F in electronegativity more electronegative than Cl The simple definition EN lltIE EA that employs thermochemical properties of isolated gas phase atoms does not take into account the complex bonding environment of molecules Assume 100 g 1 l 1452 g c x 1209 mol c12091209 1 12011 g C 1 mol 183 g H x 1816 mol H 1816 1209 15 1008 g H 1 mol 6430 gCl gtlt 1814 mol Cl 1814 1209 15 35453 g Cl 1 mol 1935 g O x 1209 molO12091209 10 159994 g 0 Multiplying by 2 to obtain an integer ratio the empirical formula is C 2 H 3 Cl 3 O 2 The empirical formula weight is 2120 310 3355 216 1655 The empirical formula is the molecular formula 44 equot 22 e pairs Elflz EIS H ZCZI C C H 1C H Assume 100 g 1 mol 6204 g Ba x 04518 mol Ba 04518 04518 2 10 13733 g Ba 1 mol 3796 g N x 2710 mol N 2710 04518 2 60 14007 g N The empirical formula is BaN6 Ba has an ionic charge of 2 so there must be two 1 azide ions to balance the charge The formula of each azide ion is N 3 16 equot 8 e pairs NNNl lt gt tNEN Nl lt gt1i NEN 1 1 1 0 1 2 2 1 0 The left structure minimizes formal charges and is probably the main contributor The two N N bond lengths will be equal The two minor contributors would individually cause unequal N N distances but collectively they contribute equally to the lengthening and shortening of each bond The N N distance will be approximately 124 A the average NN distance 208 8 Chemical Bonding Solutions to Exercises 8101 a C2H210e 5e pair N210e 5e pair H C C H IN I b N2 is an extremely stable unreactive compound Under appropriate conditions it can be either oxidized Section 227 or reduced Sections 147 and 151 C 2H2 is a reactive gas used in combination with 02 for welding and as starting material for organic synthesis Section 254 c 2N2g502ggt2N205g 2C2H2g502ggt4CO2g2H2Og d AHOM N 224H N205924Hi N295AH 029 21130 7 20 7 50 260 k AHZX1130kJmot N2 AH xn C2H 2 44H OO2924H H 20924H 02H 295AH 029 47393 k1 2724182 k 7 22267 k 7 50 725110 k AH C2H 2 712565 kJmot C2H 2 The oxidation of C 2H2 is highly exothermic which means that the energy state of the combined products is much lower than that of the reactants The reaction is downhill in an energy sense and occurs readily The oxidation of N 2 is mildly endothermic energy of products higher than reactants and the reaction does not readily occur This is in agreement with the general reactivities from part b Referring to bond enthalpies in Table 84 when the CrH bonds are taken into account even more energy is required for bond breaking in the oxidation of C 2H 2 than in the oxidation of N 2 The difference seems to be in the enthalpies of formation of the products CO 2g and H2Og have extremely exothermic AH values which cause the oxidation of C 2 H 2 to be energetically favorable N 2 O 5g has an endothermic AH value which causes the oxidation of N 2 to be energetically unfavorable 8102 a Assurne 100 g of compound 696g s X 1 mol 5 7 217 mol 5 3207 g 1 IN 304 g N X Inquot 7 217 mol N 1401g S and N are present in a 11 mol ratio so the empirical formula is SN The empirical formula weight is 46 MMFW 184346 4 The molecular formula is SANA b 44 e 22 e pairs Because of its small radius N is unlikely to have an expanded octet Begin with alternating S and N atoms in the ring Try to satisfy the octet rule with single bonds and lone pairs At least two double bonds somewhere in the ring are required 8 Chemical Bonding Solutions to Exercises C d sii s ii s ii oo oo oo 9 N S N N N S l IL l l Q A l L G N 151 5 11 8 11 8 9 These structures carry formal charges on S and N atoms as shown Other possibilities include 39s39ii 39s39 ii quot IN IS IN S 1 1 H 1 ll 39 N N S These structures have zero formal charges on all atoms and are likely to contribute to the true structure Note that the S atoms that are shown with two double bonds are not necessarily linear because 8 has an expanded octet Other resonance structures with four double bonds are e li oo is l l 3 lTl T S N 5 N e H N e N 5 In either resonance structure the two extra electron pairs can be placed on any pair of S atoms in ring leading to a total of 10 resonance structures The sulfur atoms alternately carry formal charges of 1 and 1 Without further structural information it is not possible to eliminate any of the above structures Clearly the S 4 N 4 molecule stretches the limits of the Lewis model of chemical bonding Each resonance structure has 8 total bonds and more than 8 but less than 16 bonding e pairs so an average bond will be intermediate between a S N single and double bond We estimate an average S N single bond length to be 177 A sum of bonding atomic radii from Figure 76 We do not have a direct value for a S N double bond length Comparing double and single bond lengths for CC 134 A 154 A NN 124 A 147 A and 0 0 121 A 148 A bonds from Table 85 we see that on average a double bond is approximately 023 A shorter than a single bond Applying this difference to the S N single bond length we estimate the S N double bond length as 154 A Finally the intermediate S N bond length in S4N4 should be between these two values approximately 160 165 A The measured bond length is 162 A S4N4 gt 45g 4Ng AH 4AH Sg 4AH Ng AH S4N4 AH 42228 k1 44727 k1 480 k 2302 k This energy 2302 k represents the dissociation of 8 SN bonds in the molecule the average dissociation energy of one SN bond in S4N4 is then 2302 k 8 bonds 2878 k 210 8 Chemical Bonding Solutions to Exercises 8103 a b C d 8 8104 a b C 8105 a b C d Yes In the stIucture shown in the exercise each P atom needs 1 unshared pair to complete its octet This is con rmed by noting that only 6 of the 10 valence e pairs are bonding pairs There are six P7P bonds in P 4 The atomization reaction is P 4g gt 4Pg AH aom 4AH P9AH 319 43164 k 7 589 k 12067 k Since there are six P7P bonds in P4 the bond dissociation enthalpy for an P7P bond 130113 120676 201 k From Table 8A DN7N 163 k Our calculated value for DP7P is 201 k so the PrP bond is stronger than the NrN bond HF8 Hg Fg DH7F 567 k Hg Hg 1 e 1H 1312 k Fg 1 e a Fig EF 732814 HF8 Hg Fg AH 1551 k AH DH7C1 1H ECl AH 431 k 1312 k 7349 k 1394 k AH DH7Br 1H EBr AH 366 k 1312 k 7325 k 1353 k C eHeg 6Hg 6Cg 4H 64H Hg 64H Cg 74H 06H 69 AH 621794 k 60184 k 7 829 k 5535 k C 6 H 6 8 6CH8 C5H5g 6Hg 6Cg AH 5535 k 6Hg 6Cg 6CHg 7 6DC7H 7 6413 k CEHE g 6CHg 3057 k 3057 k is the energy required to break the six CrC bonds in C 6H6g The average bond dissociation energy for one carbonicarbon bond in C 6 H 6 g is 3057 k 6 C 7 C bonds The value of 5095 k is between the average value for a CrC single bond 348 k and a CC double bond 614 k It is somewhat greater than the average of these two values indicating that the carbonicarbon bond in benzene is a bit stronger 5095 k than we might expect 8 Chemical Bonding 8106 a b C d Solutions to Exercises Br2lgt2Brg AH 72AH Brg21118kJ2236kJ 13141 gt Cg 4Clg AH AH Cg4AH 0g7AH 00 4a 7184 k 41217k 7 71393 k 13445 61 kJ H2020 2Hltggt 20g 2Hltggt 20g 20Hltggt H2020 20Hg Dlto foxl 2AH Hltggt 2AHF 09 AH H 2020 gt72D07Hg 221794 k 22475 k 7 71878 k 7 2463 k 193 k The data are listed below bond D gas kJmnl D liquid kJmnl BrrBr 193 2236 CrCl 328 3361 070 146 1 92 7 Breaking bonds in the liquid requires more energy than breaking bonds in the gas phase For simple molecules bond dissociation from the liquid phase can be thought of in two steps molecule 1 gt molecule g molecule g gt atoms g The rst step is evaporation or vaporization of the liquid and the second is bond dissociation in the gas phase Average bond enthalpy in the liquid phase is then the sum of the enthalpy of vaporization for the molecule and the gas phase bond dissociation enthalpies divided by the number of bonds dissociated This is greater than the gas phase bond dissociation enthalpy owing to the contIibution from the enthalpy of vaporization 212 Atoms Molecules and Ions Visualizing Concepts 21 a The path of the charged particle bends because it is repelled by the negatively charged plate and attracted to the positively charged plate b Like charges repel and opposite charges attract so the sign of the electrical charge on the particle is negative c The greater the magnitude of the charges the greater the electrostatic repulsion or attraction As the charge on the plates is increased the bending will increase d As the mass of the particle increases and speed stays the same linear momentum mv of the particle increases and bending decreases See A Closer Look The Mass Spectrometer In general metals occupy the left side of the chart and nonmetals the right side metals red and green nonmetals blue and yellow alkaline earth metal red noble gas yellow Since the number of electrons negatively charged particles does not equal the number of protons positively charged particles the particle is an ion The charge on the ion is 2 Atomic number number of protons 16 The element is S sulfur Mass number protons neutrons 32 32 27 16 S In a solid particles are close together and their relative positions are fixed In a liquid particles are close but moving relative to each other In a gas particles are far apart and moving All ionic compounds are solids because of the strong forces among charged particles Molecular compounds can exist in any state solid liquid or gas Since the molecules in ii are far apart ii must be a molecular compound The particles in i are near each other and exist in a regular ordered arrangement so i is likely to be an ionic compound Formula IF 5 Name iodine penta uoride Since the compound is composed of elements that are all nonmetals it is molecular 17 2 Atoms Molecules and Ions Solutions to Exercises 26 Cations red spheres have positive charges anions blue spheres have negative charges There are twice as many anions as cations so the formula has the general form CA2 Only CaNOS2 calcium nitrate is consistent with the diagram Atomic Theory and the Discovery of Atomic Structure 27 Postulate 4 of the atomic theory is the law of constant composition It states that the relative number and kinds of atoms in a compound are constant regardless of the source Therefore 10 g of pure water should always contain the same relative amounts of hydrogen and oxygen no matter Where or how the sample is obtained 28 a 6500 g compound 0384 g hydrogen 6116 g sulfur b Conservation ofmnss c According to postulate 3 of the atomic theory atoms are neither created nor destroyed during a chemical reaction If 0384 g of H are recovered from a compound that contains only H and S the remaining mass must be sulfur 176 5711 0 0 g oxygen 70 g 0571105711 10 3082 g nitrogen l g N 3520 1142 O goxygen 7g 114205711 20 3082 g nitrogen lg N 7040 2284 0 g oxygen 7g 228405711 40 3082 g nitrogen lg N 8800 g oxygen 7 2855 g 0 7 285505711 50 3082 g nitrogen lg N b These masses of oxygen per one gram nitrogen are in the ratio of 1245 and thus obey the law of multiple proportions Multiple proportions arise because atoms are the indivisible entities combining as stated in Dalton s theory Since atoms are indivisible they must combine in ratios of small Whole numbers 356 fl 2 10 a 1 0749 g fluorinel g iodine 475 g iodine 343 fl 2 0449 g fluorinel g iodine 764 g iodine 986 fl 3E 105 g fluorinel g iodine 941 g iodine b To look for integer relationships among these values divide each one by the smallest If the quotients aren t all integers multiply by a common factor to obtain all integers 18 2 Atoms Molecules and Ions Solutions to Exercises 1 07490449 167 167 X 3 5 2 04490449 100 100 X 3 3 3 1050449 234 234 X 3 7 The ratio of g uorine to g iodine in the three compounds is 537 These are in the ratio of small whole numbers and therefore obey the law of multiple proportions This integer ratio indicates that the combining uorine units atoms are indivisible entities Evidence that cathode rays were negatively charged particles was 1 that electric and magnetic elds de ected the rays in the same way they would de ect negatively charged particles and 2 that a metal plate exposed to cathode rays acquired a negative charge Since the unknown particle is de ected in the opposite direction from that of a negatively charged beta 3 particle it is attracted to the plate and repelled by the plate The unknown particle is positively charged The magnitude of the de ection is less than that of the 3 particle or electron so the unknown particle has greater mass than the electron The unknown is a positively charged particle of greater mass than the electron a b a b C If the positive plate were lower than the negative plate the oil drops coated with negatively charged electrons would be attracted to the positively charged plate and would descend much more quickly The more times a measurement is repeated the better the chance of detecting and compensating for experimental errors That is if a quantity is measured five times and four measurements agree but one does not the measurement that disagrees is probably the result of an error Also the four measurements that agree can be averaged to compensate for small random uctuations Millikan wanted to demonstrate the validity of his result via its reproducibility The droplets carry different total charges because there may be 1 2 3 or more electrons on the droplet The electronic charge is likely to be the lowest common factor in all the observed charges Assuming this is so we calculate the apparent electronic charge from each drop as follows A 160gtlt10 1160gtlt10 C B 315 X 10 2 158 X 10 C C 481 X 10 3 160 X 10 C D 631 X 10 4 158 X 10 C The reported value is the average of these four values Since each calculated charge has three significant gures the average will also have three signi cant gures 160 X 10 C 158 X 10 C 160 X 10 C 158 X 10 C 4 159 X 10 C 19 2 Atoms Molecules and Ions Solutions to Exercises Modern View of Atomic Structure Atomic Weights 215 a b C b C b C b C d b 1 X 10 10 m 1nrn 19Agtlt x 1A 1gtlt10 9m 019nrn 1 X 10 10 m X 1pm 19Agtlt 12 1A 1gtlt10 m 19 gtlt102 or190 pm 1A100pm Aligned Kr atoms have diameters touching d 2r 219 38 1A X lKr atom 17m X l 26 X 106 Kratoms 1000 mm 1 X 1010 m 38A 10 mm gtlt 1x10 10 m X 100cm V43m3 r19Agtlt 19gtlt10 sqn V 43119 X 10 83 cm3 29 X 10 23 cm3 28 108 rd2ramp gtlt 1 A814A l gtlt 10 cm 28 10 8 w 1m 14 gtlt10 10m 2 100m Aligned Sn atoms have diameters touching d 28 X 10 8 cm 28 X 10 10 m 1 x 10 6 m 1Sn atom gtlt 102l X 104 Sn atoms 1pm 28 X 10 m 60 pm gtlt V 43 1 r3 r 14 X 10 10 m V 43114 gtlt 10 1quot3 m3 1149 gtlt 10 29 11 gtlt 10 29 m3 proton neutron electron proton 1 neutron 0 electron 1 The neutron is most massive the electron least massive The neutron and proton have very similar masses The nucleus has most of the mass but occupies very little of the volume of an atom True The number of electrons in an atom is equal to the number of protons in the atom True Atomic number is the number of protons in the nucleus of an atom Muss number is the total number of nuclear particles protons plus neutrons in an atom The mass number can vary Without changing the identity of the atom but the atomic number of every atom of a given element is the same 20 2 Atoms Molecules and Ions 220 225 226 Solutions to Exercises a an are isoto es 0 t e same e ement ecause t e ave i entica fgx digx p fh 1 b hyh d 1 atomic numbers b These are isotopes of the element sulfur S atomic number 16 p protons n neutrons e electrons a 40Ar has 18 p 22 n 18 e b 65Zn has 30 p 35 n 30 e c 70Ga has 31 p 39 n 31 e d 80Br has 35 p 45 n 35 e e 184W has 74 p 110 n 74 e f 246Am has 95 p 148 n 95 e a 32P has 15 p 17 n b 51Cr has 24 p 27 n c 60Co has 27 p 33 n d 99Tc has 43 p 56 n e 131I has 53 p 78 n f 20111 has 81 p 120 n Symbol 52Cr 55Mn sz Han 207Pb Protons 24 25 48 86 82 Neutrons 28 30 64 136 125 Electrons 24 25 48 86 82 Mass no 52 55 112 222 207 Symbol 121Sb 103Rh 88Sr 127Te 239Pu Protons 51 45 38 52 94 Neutrons 70 58 50 75 145 Electrons 51 45 38 52 94 Mass No 121 103 88 127 239 a 1798613t b 2 Kr C 3 AS C 3 Mg Since the two nuclides are atoms of the same element by definition they have the same number of protons 54 They differ in mass number and mass because they have different numbers of neutrons 129Xe has 75 neutrons and 130Xe has 76 neutrons a 162 C b Atomic weights are really average atomic masses the sum of the mass of each naturally occurring isotope of an element times its fractional abundance Each B atom will have the mass of one of the naturally occurring isotopes While the atomic weight is an average value The naturally occurring isotopes of B their atomic masses and relative abundances are 1013 10012937 199 11B 11009305 801 21 2 Atoms Molecules and Ions Solutions to Exercises 228 a 12 amu b The atomic weight of carbon reported on the frontinside cover of the text is the abundanceweighted average of the atomic masses of the two naturally occurring isotopes of carbon 12C and 13C The mass of a 12C atom is exactly 12 amu but the atomic weight of 12011 takes into account the presence of some 13C atoms in every natural sample of the element Atomic weight average atomic mass E fractional abundance gtlt mass of isotope Atomic weight 06917629296 03083649278 635456 6355 amu Atomic weight average atomic mass E fractional abundance gtlt mass of isotope Atomic weight 001420397302 024120597444 022120697587 052420797663 20722 207 amu The result has 0 decimal places and 3 sig gs because the fourth term in the sum has 3 sig figs and 0 decimal places a Compare Figures 24 and 213 referring to Solution 212 In Thomson s cathode ray experiments and in mass spectrometry a stream of charged particles is passed through a magnetic eld The charged particles are de ected by the magnetic field according to their mass and charge For a constant magnetic field strength and speed of the particles the lighter particles experience a greater de ection b The xaxis label independent variable is atomic weight and the yaxis label dependent variable is signal intensity c Uncharged particles are not de ected in a magnetic eld The effect of the magnetic eld on moving charged particles is the basis of their separation by mass a The purpose of the magnet in the mass spectrometer is to change the path of the moving ions The magnitude of the de ection is inversely related to mass which is the basis of the discrimination by mass b The atomic weight of CI 355 is an average atomic mass It is the average of the masses of two naturally occurring isotopes weighted by their abundances c The single peak at mass 31 in the mass spectrum of phosphorus indicates that the sample contains a single isotope of P and the mass of this isotope is 31 amu a Average atomic mass 078992398504 010002498584 011012598259 2431 amu 22 2 Atoms Molecules and Ions Solutions to Exercises b 234 a b C 79 signal intensity 01 10 11 24 25 26 Atomic Weight amu The relative intensities of the peaks in the mass spectrum are the same as the relative abundances of the isotopes The abundances and peak heights are in the ratio 24Mg 25Mg 26Mg as 78 10 11 Three peaks 1H 1H 1H 2H 2H 2H 1H 1H 2100783 201566 amu 1H 2H 100783 201410 302193 amu 2H 2H 2201410 402820 amu The mass ratios are 1 149923 199845 or 1 15 2 1H 1H is largest because there is the greatest chance that two atoms of the more abundant isotope will combine 2H 2H is the smallest because there is the least chance that two atoms of the less abundant isotope will combine The Periodic Table Molecules and Ions 235 a e 236 a d s 237 a C e 238 Cr metal b He nonmetal c P nonmetal d Zn metal Mg metal f Br nonmetal g As metalloid sodium metal b titanium metal c gallium metal uranium metal e palladium metal f selenium nonmetal krypton nonmetal K alkali metals metal b I halogens nonmetal Mg alkaline earth metals metal d Ar noble gases nonmetal S chalcogens nonmetal C carbon nonmetal Si silicon metalloid Ge germanium metalloid Sn tin metal Pb lead metal 23 2 Atoms Molecules and Ions Solutions to Exercises 239 240 241 242 243 244 245 246 An empirical formula shows the simplest ratio of the different atoms in a molecule A moleculurformulu shows the exact number and kinds of atoms in a molecule A structural formula shows how these atoms are arranged Compounds with the same empirical but different molecular formulas differ by the integer number of empirical formula units in the respective molecules Thus they can have very different molecular structure size and mass resulting in very different physical properties a AIBrg b C4145 c C2H40 d P205 e C3H2Cl BNH2 A molecular formula contains all atoms in a molecule An empirical formula shows the simplest ratio of atoms in a molecule or elements in a compound a molecular formula C 6 H 6 empirical formula CH b molecular formula SiCl 4 empirical formula SiCl 4 14 is the simplest ratio c molecular B 2 H 6 empirical BH 3 d molecular C 6 H 1 20 6 empirical CH 2 O a 6 b 6 c 12 a 4 b 6 C 9 H H H H a C2H6O H clj o clj H b C2H6O H clj clj o H l l l l H cCH4O H clj o H dPF3 F P F A F H H H H a C2H5Br H clj clj Br b C2H7N H clj N clj H l l l l l H c CH2C12 H clj c1 d NHzOH H N o 391 l l 24 2 Atoms Molecules and Ions Solutions to Exercises 247 Symbol Cow Sez 1920s ZOUHg Protons 27 34 76 80 Neutrons 32 46 116 120 Electrons 24 36 74 78 Net Charge 3 2 2 2 248 Symbol 75As3 Ni 1271 197Au3 Protons 33 28 53 79 Neutrons 42 31 74 118 Electrons 36 26 54 76 Net Charge 3 2 1 3 249 a Mg b A1 c K d 82 e F 250 a Sr b Sc or Sc c PS d I39 e Sez 251 a GaFS galliumII uoride b LilI lithium hydride c AHS aluminum iodide d K28 potassium sul de 252 a AgI b AgZS c AgF 253 a CaBr2 b KZCO3 c A1C2H3023 d NH4ZSO4 e Mg 3 PO 4 2 254 a CuZS b FeZO3 c ngCO3 d Ca3AsO42 e NH4 2 CO 3 255 Molecular all elements are nonmetals a B2 H6 b CH3 OH f NOCl g NF3 Ionic formed by a cation and an anion usually contains a metal cation c LiNOS d Sc203 e CsBr h AgZSO4 256 Molecular all elements are nonmetals a PFS c SCI h N2 04 Ionic formed from ions usually contains a metal cation b NaI d CaNO3 2 e FeCl3 f LaP g CoCO3 Naming Inorganic Compounds Organic Molecules 257 a CIOZ b C1 c CIOS d CIO4 e C10 258 a selenate b selenide c hydrogen selenide biselenide 25 2 Atoms Molecules and Ions Solutions to Exercises 259 260 261 262 263 264 265 266 267 268 269 d a C e g i a C e g i a e a e a d a d a d a d a d a d a b potassium chromate hydrogen selenite biselenite magnesium oxide b aluminum chloride lithium phosphate d barium perchlorate copperll nitrate cupric nitrate f ironll hydroxide ferrous hydroxide calcium acetate h chromiumlll carbonate chromic carbonate ammonium sulfate lithium oxide b sodium hypochlorite strontium cyanide d chromiumlll hydroxide chromic hydroxide ironIII carbonate ferric carbonate cobaltll nitrate cobaltous nitrate ammonium sulfite h sodium dihydrogen phosphate potassium permanganate silver dichromate AlOH 3 b K 2 SO 4 c Cu 2 0 d ZnNO 3 2 HgBr 2 Fe 2 CO 3 3 g NaBrO Na 3 PO 4 b CoNO 3 2 c BaBrO 3 2 d CuClO 4 2 MgHCO32 CrC2H3O23 g K2Cr2O7 bromic acid b hydrobromic acid c phosphoric acid HClO e H10 3 H 2 SO 3 HBr b H 2 S c HNO 2 carbonic acid e chloric acid acetic acid sulfur hexa uoride b iodine penta uoride c xenon trioxide N204 e HCN f P486 dinitrogen monoxide b nitrogen monoxide c nitrogen dioxide dinitrogen pentoxide e dinitrogen tetroxide ZnCO3ZnOCO2 b HFSiO2SiF4H2O c SO2H2OH2SO3 PH 3 e HClO 4 Cd CdClO 4 2 VBr 3 NaHCO 3 b CaClO 2 c HCN MgOH 2 e SnF CdS H 2 SO 4 H 2 S A hydrocarbon is a compound composed of the elements hydrogen and carbon only HHHH H C C C C H HHHH molecular C4H10 empirical C2H5 2 Atoms Molecules and Ions Solutions to Exercises 270 ane b Hexane has 6 carbons in its chain H H H H H H H C C C C C C H H H H H H H molecular C6H14 empirical C3H7 271 a Functional groups are groups of specific atoms that are constant from one molecule to the next For example the alcohol functional group is an OH Whenever a molecule is called an alcohol it contains the OH group b OH C H H H H H C C C C OH H H H H 272 a They both have two carbon atoms in their molecular backbone or chain In 1propanol one of the H atoms on an outer terminal C atom has been replaced by an OH group Additional Exercises 273 a Based on data accumulated in the late eighteenth century on how substances react with one another Dalton postulated the atomic theory Dalton s theory is based on the indivisible atom as the smallest unit of an element that can combine with other elements By determining the effects of electric and magnetic fields on cathode rays Thomson measured the masstocharge ratio of the electron He also proposed the plum pudding model of the atom in which most of the space in an atom is occupied by a diffuse positive charge in which the tiny negatively charged electrons are imbedded By observing the rate of fall of oil drops in and out of an electric field Millikan measured the charge of an electron After observing the scattering of alpha particles at large angles when the particles struck gold foil Rutherford postulated the nuclear atom In Rutherford s atom most of the mass of the atom is concentrated in a small dense region called the nucleus and the tiny negatively charged electrons are moving through empty 27 2 Atoms Molecules and Ions Solutions to Exercises space around the nucleus Radioactivity is the spontaneous emission of radiation from a substance Becquerel s discovery showed that atoms could decay or degrade implying that they are not indivisible However it wasn t until Rutherford and others characterized the nature of radioactive emissions especially the particle nature of on and 3 rays that the full signi cance of the discovery was apparent a b C a b C Most of the volume of an atom is empty space in which electrons move Most alpha particles passed through this space The path of the massive alpha particle would not be signi cantly altered by interaction with a puny electron Most of the mass of an atom is contained in a very small dense area called the nucleus The few alpha particles that hit the massive positively charged gold nuclei were strongly repelled and essentially de ected back in the direction they came from The Be nuclei have a much smaller volume and positive charge than the Au nuclei the charge repulsion between the alpha particles and the Be nuclei will be less and there will be fewer direct hits because the Be nuclei have an even smaller volume than the Au nuclei Fewer alpha particles will be scattered in general and fewer will be strongly back scattered Droplet D would fall most slowly It carries the most negative charge so it would be most strongly attracted to the upper plate and most strongly repelled by the lower plate These electrosta c forces would provide the greatest opposition to gravity Calculate the lowest common factor A 384 gtlt 10 8 288 gtlt 10 8 133 133 gtlt 3 4 B 480 gtlt 10 8 288 gtlt 10 8 167 167 gtlt 3 5 C 288 gtlt 10 8 288 gtlt 10 8 100 100 gtlt 3 3 D 864 gtlt 10 8 288 gtlt 10 8 300 300 gtlt 3 9 The total charge on the drops is in the ratio of 4539 Divide the total charge on each drop by the appropriate integer and average the four values to get the charge of an electron in warmombs A 384 X 10 8 4 960 X 10 9 wa B 480 X 10 8 5 960 X 10 9 wa C 288 X 10 8 3 960 X 10 9 wa D 864 X 10 8 9 960 X 10 9 wa The charge on an electron is 960 X 10 9 wa The number of electrons on each drop are the integers calculated in part b A has 4 e B has 5 e C has3 e39 and D has 9 e 28 2 Atoms Molecules and Ions Solutions to Exercises C a b C d e a b X 716 600 X 107 waC 1e 160 gtlt 10 C 3He has 2 protons 1 neutron and 2 electrons 3H has 1 proton 2 neutrons and 1 electron 3He 216726231 gtlt 10 g 16749286 gtlt 10 g 291093897 gtlt 10 28 g 5021996 gtlt 10 g 3H 16726231 gtlt 10 g 216749286 gtlt 10 g 91093897 gtlt 10 28 g 5023391 gtlt 10 g Tritium 3H is more massive The masses of the two particles differ by 00014 X 10 24 g Each particle loses 1 electron to form the 1 ion so the difference in the masses of the ions is still 14 X 10 quot A mass spectrometer would need precision to 1 X 10 27 g to differentiate 3He and 3H 2 protons and 2 neutrons the nuclear strong force The charge of an on particle is twice the magnitude of the charge of an electron with the opposite sign That is 2 16022 X 10 C 32044 X 10 19 C 32044 X 10 19 c 4 66448 X 1024 g 48224 X 10 gC l amu 66448 X 1024 X g 166054 X 10 24 g 40016 amu The sum of the particle masses in an on particle is 210073 amu and 210087 amu 40320 amu The actual particle mass 40016 amu is less than the sum of the masses of the components The difference is the nuclear binding energy the energy released when protons and neutrons combine to form a nucleus Mass and energy are interchangeable according to the Einstein relationship E mcz Calculate the mass of a single gold atom then divide the mass of the cube by the mass of the gold atom 1 1970 amu 823 32713 X 1022 3271 x 1022 ggold atom gold atom 6022 X 10 amu 193 1 1d t g X go 3930 590 X 1022 Auatorns in the cube cube 3271 X 10 g The shape of atoms is spherical spheres cannot be arranged into a cube so that there is no empty space The question is how much empty space is there We can calculate the two limiting cases no empty space and maximum empty space The true diameter will be somewhere in this range 29 2 Atoms Molecules and Ions Solutions to Exercises C a b V No empty space volume cubenumber of atoms volume of one atom V 431 r3 r 31 V41339 d 2r 10 cm3 590 X 1022 Au atoms 17x10 23cm3 vol of cube 10 X 10 X 10 1695 X 10 23 r 1r 1695 X 10 23 cm3413 34 X 10 8 cm d 2r 68 X 10 8 cm Maximum empty space assume atoms are arranged in rows in all three directions so they are touching across their diameters That is each atom occupies the volume of a cube with the atomic diameter as the length of the side of the cube The number of atoms along one edge of the gold cube is then 590 X 1022 3 3893 X 107 389 X 107 atoms10 cm The diameter ofa single atom is 10 cm389 X 107 atoms 2569 X 10 8 26 X 10 8 cm The diameter ofa gold atom is between 26 X 10 8 cm and 68 X 10 8 cm 26 68 A Some atomic arrangement must be assumed since none is speci ed The solid state is characterized by an orderly arrangement of particles so it isn t surprising that atomic arrangement is required to calculate the density of a solid A more detailed discussion of solidstate structure and density appears in Chapter 11 In arrangement A the number of atoms in 1 cm2 is just the square of the number that t linearly in 1 cm 1 atom 1 X 1010 A 1 a gtlt 495A imam X 107 20 X 107 atomscm 100cm 10an lm 10 cm2 202 X 1072 4081 X 1014 41 X 1014 atomscm2 In arrangement B the atoms in the horizontal rows are touching along their diameters as in arrangement A The number of Rb atoms in a 10 cm row is then 20 X 107 Rb atoms Relative to arrangement A the vertical rows are offset by 12 of an atom Atoms in a column are no longer touching along their vertical diameter We must calculate the vertical distance occupied by a row of atoms which is now less than the diameter of one Rb atom Consider the triangle shown below This is an isosceles triangle equal side lengths equal interior angles with a sidelength of 2d and an angle of 60 Drop a bisector to the uppermost angle so that it bisects the opposite side 30 2 Atoms Molecules and Ions Solutions to Exercises 281 C a b 300 600 9 I The result is a right triangle with two known side lengths The length of the unknown side the angle bisector is 2h two times the vertical distance occupied by a row of atoms Solve for h the height of one row of atoms 2h2 d2 2d2 4h2 4d2 d2 3d2 h2 3d24 h 3c12412 3495 13324 2 42868 429 A The number of rows of atoms in 1 cm is then 1row 1 X 1010A 1m x o x 2333 x 107 23 x 107 42868 A 1 m 100 cm 10 cm x The number of atoms in a 10 cm2 square area is then 2020 x 107 atoms 1 row x 2333 x 107 rows4713 x 1014 47 x 1014 Note that we have ignored the loss of 12 atom at the end of each horizontal row Out of 20 x 107 atoms per row one atom is not significant The ratio of atoms in arrangement B to arrangement A is then 4713 x 1014 atoms 4081 x 1014 1555 1221 Clearly arrangement B results in less empty space per unit area or volume If extended to three dimensions arrangement B would lead to a greater density for Rb metal diameter of nucleus 1 x 10394 A diameter of atom 1 A V437c r3r d2rn 05 x 10394Ara 05A volume of nucleus 43 7 05 x 103943 A3 volume of atom 43 7 053 A3 volume of nucleus 43 7 05 x 10 43 A3 3 O3 1 x 10 12 437c05 A volume fraction of nucleus 2 volume of atom diameter of atom 5 A r a 25 A 437c05 x 10 43 A3 3 O 3 8 x 10 15 437c25 A volume fraction of nucleus 2 Depending on the radius of the atom the volume fraction of the nucleus is between 1 x 103912 and 8 x 1039 that is between 1 part in 1012 and 8 parts in 1015 mass of proton 10073 amu 31 2 Atoms Molecules and Ions Solutions to Exercises a 19 10073 amu gtlt 166054 X 10 24 gamu 16727 X 10 24 g 50 X 10 14cm diameter 10 X 10 15 m radius 050 X 10 15 m gtlt 103 m Assuming a proton is a sphere V 43 1 r3 16727 10 24 g 7 X g 332x1015gcm3 densit 77 y cm3 437I50gtlt10 143q39n 10 no 10 All isotopes are atoms of the same element oxygen with the same atomic number Z 8 8 protons in the nucleus and 8 electrons Elements with similar electron arrangements have similar chemical properties Section 25 Since the 3 isotopes all have 8 electrons we expect their electron arrangements to be the same and their chemical properties to be very similar perhaps identical Each has a different number of neutrons 8 9 or 10 a different mass number A 16 17 or 18 and thus a different atomic mass FkQ1Qzd2k90gtlt109NmzC2d053gtlt10 1quotm Q electron 16 X 10 19 C Q proton Q electron 16 X 10 19 C 1 a b a b a 90 X 109 Nm2 X 716 X 1019 c X 16 X 1019 c C 8202 X 108 82 X 108 N 053 X 10 2 m2 The 68926 amu isotope has a mass number of 69 with 31 protons 38 neutrons and the symbol nga The 70925 amu isotope has a mass number of 71 31 protons 40 neutrons and symbol giGa All Ga atoms have 31 protons The average mass ofa Ga atom is 6972 amu Letx abundance of the lighter isotope 1 x abundance of the heavier isotope Then x68926 1 x70925 6972 x 06028 0603 69Ga 603 71Ga 397 There are 24 known isotopes of Ni from 51Ni to 7 lNi The ve most abundant isotopes are 58Ni 57935346 amu 68077 60Ni 59930788 amu 26223 62Ni 61928346 amu 3634 61Ni 60931058 amu 1140 64Ni 63927968 amu 0926 Data from Handbook of Chemistry and Physics 74th Ed Data may differ slightly in other editions A Brz molecule could consist of two atoms of the same isotope or one atom of each of the two different isotopes This second possibility is twice as likely as the rst Therefore the second peak twice as large as peaks 1 and 3 represents a Br2 molecule containing different isotopes The mass numbers of the two isotopes are 32 2 Atoms Molecules and Ions Solutions to Exercises determined from the masses of the two smaller peaks Since 157836 m 158 the rst peak represents a 79Br 79Br molecule Peak 3 161832 m 162 represents a 81Br Br molecule Peak 2 then contains one atom of each isotope 79Br Br with an approximate mass of 160 amu b The mass of the lighter isotope is 157836 amu2 atoms or 78918 amuatom For the heavier one 161832 amu2 atoms 80916 amuatom c The relative size of the three peaks in the mass spectrum of Br2 indicates their relative abundance The average mass of a Br 2 molecule is 02569157836 04999159834 02431161832 15979 amu Each product has four signi cant gures and two decimal places so the answer has two decimal places 15979 amu avg Br2 molecule 1 Br2 molecule d 79895 amu 2 Br atoms e Let x the abundance of 79Br 1 x abundance of 81Br From b the masses of the two isotopes are 78918 amu and 80916 amu respectively From d the mass of an average Br atom is 79895 amu x78918 1 x80916 79895 x 05110 79Br 5110 81Br 4890 a Five signi cant gures 1H is a bare proton with mass 10073 amu 1H is a hydrogen atom with 1 proton and 1 electron The mass of the electron is 5486 X 10 4 or 00005486 amu Thus the mass of the electron is signi cant in the fourth decimal place or fth signi cant figure in the mass of 1H b Mass of 1H 10073 amu proton 00005486 amu electron 10078 amu We have not rounded up to 10079 since 49 lt 50 in the final sum mass of e 5486 X 10 4 am Mass of electron X 100 u X 100 005444 mass of 1H 10078 amu copper Cu 113 coinage metals transition metal tin Sn 4A zinc Zn 213 phosphorus P 5A lead Pb 4A a an alkali metal K b an alkaline earth metal Ca c a noble gas Ar d a halogen Br e a metalloid Ge f a nonmetal in 1A H g a metal that forms a 3 ion Al h a nonmetal that forms a 2 ion 0 33 2 Atoms Molecules and Ions Solutions to Exercises 290 291 292 293 294 295 296 297 298 299 an element that resembles Al Ga Sg has 106 protons 160 neutrons and 106 electrons Sg is in Group 63 or 6 and immediately below tungsten W We expect the chemical properties of Sg to most closely resemble those of W chlorine gas C12 ii b propane C9ng V c nitrate ion 21 sulfur trioxide 503 iii e methylchloride CH3Cl iv nickelll oxide 2 b manganeselV oxide 4 chromiumlll oxide 3 d molybdeniumVI oxide 6 103 b 104 c 10 d HIO e HIO4orH5lO erbromate ion b selenite ion P Aso43 d HTeo4 sodium chloride b sodium bicarbonate or sodium hydrogen carbonate sodium hypochlorite d sodium hydroxide ammonium carbonate f calcium sulfate potassium nitrate b sodium carbonate c calcium oxide d hydrochloric acid e magnesium sulfate f magnesium hydroxide a CaS CaHS 2 b HBr HBrO c AlN AlNO 2 3 d FeO Fe 2 O 3 e NH 3NH 4 f K2503 KHSO3 g ngClz HgCl2 h HClO3 HClO 4 Formula Name Density Melting Boiling gmL Point C Point C a PF 3 phosphorus 3907 1515 1015 trifluoride b SiCl 4 silicon 1483 70 5757 tetrachloride c C 2 H 6 0 ethanol 07893 1173 785 C 2 H 5 OH In an alkane all C atoms have 4 single bonds so each C in the partial structure needs 2 more bonds All alkanes are hydrocarbons so 2 H atoms will bind to each C atom in the ring 2 Atoms Molecules and Ions Solutions to Exercises 2100 b The molecular formula of cyclohexane is C6H12 the molecular formula of nhexane is C6H14 see Solution 264d Cyclohexane can be thought of as nhexane in which the two outer terminal C atoms are joined to each other In order to form this C C bond each outer C atom must lose 1 H atom The number of C atoms is unchanged and each C atom still has 4 single bonds The resulting molecular formula is C 6 H 144 C 6 H 12 c On the structure in part a replace 1 H atom with an OH group H H H c o HT H C 0 0 H c HH H H H Elements are arranged in the periodic table by increasing atomic number and so that elements with similar chemical and physical properties form a vertical column or group By its position in the periodic chart we know whether an element is a metal nonmetal or metalloid and the common charge of its ion Members of a group have the same common ionic charge and combine in similar ways with other elements 35
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