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GChem 2 2016 Answer Key Hmwk chapter 9 & 10 9.30 Plan: The lattice energy of NaCl is represented by the equation NaCl(s) → Na+(g) + Cl–(g). Use Hess’s law

and arrange the given equations so that they sum up to give the equation for the lattice energy. You will need to reverse the last equation (and change the sign of ∆Hº); you will also need to multiply the second equation (∆Hº) by ½.

Solution: ΔHº Na(s) → Na(g) 109 kJ 1/2Cl

2(g) → Cl(g) 243/2 = 121.5 kJ

Na(g) → Na+(g) + e

–

496 kJ Cl(g) + e

–

→ Cl–

(g) –349 kJ NaCl(s) → Na(s) + 1/2Cl

2 (g) 411 kJ (Reaction is reversed; sign of ΔHº changed.)

NaCl(s) → Na+

(g) + Cl–

(g) 788.5 = 788 kJ ThelatticeenergyforNaClislessthanthatofLiF,whichisexpectedsincelithiumandfluorideionsaresmallerthansodiumandchlorideions,resultinginalargerlatticeenergyforLiF.

9.31 Lattice energy: MgF2 (s) → Mg2+(g) + 2 F–(g) Use Hess’s law: ∆Hº Mg(s) → Mg(g) 148 kJ F2 (g) → 2F(g) 159 kJ

Mg(g) → Mg+(g) + e– 738 kJ Mg+(g) → Mg2+(g) + e– 1450 kJ 2F(g) + 2e– → 2F–(g)

2(–328) = – 656 kJ MgF2 (s) → Mg(s) + F2 (g) 1123 kJ (Reaction is reversed and the sign of ∆Hº changed.)

MgF2 (s) → Mg2+(g) + 2F–(g) 2962 kJ

The lattice energy for MgF2 is greater than that of LiF and NaCl, which is expected since magnesium ions

have twice the charge of lithium and sodium ions. Lattice energy increases with increasing ion charge.

9.32 The lattice energy in an ionic solid is directly proportional to the product of the ion charges and inversely proportional to the sum of the ion radii. The strong interactions between ions cause most ionic materials to be hard. A very large lattice energy implies a very hard material. The lattice energy is predicted to be high for Al2 O3 since the ions involved, Al3+ and O2–, have fairly large charges and are relatively small ions.

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10.1 Plan: To be the central atom in a compound, an atom must be able to simultaneously bond to at least two other atoms. Solution He, F, and H cannot serve as central atoms in a Lewis structure. Helium (1s2) is a noble gas, and as such, it does not need to bond to any other atoms. Hydrogen (1s1) and fluorine (1s22s22p5) only need one electron to complete their valence shells. Thus, they can only bond to one other atom, and they do not have d orbitals available to expand their valence shells. 10.2ResonancemustbepresentanytimethatasingleLewisstructureisinadequateinexplainingoneormoreaspectsofamoleculeorion.ThetwoN–ObondsinNO

2areequivalentinbondlengthandbondenergy;nosingleLewis

structurecanaccountforthis.ThefollowingLewisstructuresmaybedrawnforNO2:

TheaverageofalltheseequivalentN-ObondswithabondlengththatisbetweenN-OandN=O.10.3

Plan:

For an element to obey the octet rule it must be surrounded by eight electrons. To determine

the number of electrons present,

(1) count the individual electrons actually shown adjacent to a particular atom (lone pairs), and (2) add two times the number of bonds to that atom: number of electrons = individual electrons + 2(number of bonds). Solution: (a) 0 + 2(4) = 8; (b) 2 + 2(3) = 8; (c) 0 + 2(5) = 10; (d) 2 + 2(3) = 8; (e) 0 + 2(4) = 8; (f) 2 + 2(3) = 8; (g) 0 + 2(3) = 6; (h) 8 + 2(0) = 8. All the structures obey the octet rule except: c and g. 10.4Foranatomtoexpanditsvalenceshell,itmusthavereadilyavailabledorbitals.Thedorbitalsdonotbecomereadilyavailableuntilthethirdperiodorbelowontheperiodictable.FortheelementsintheproblemF,S,H,Al,Se,andCl,theperiodnumbersare2,3,1,3,4,and3,

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respectively.Alloftheseelements,exceptthoseinthefirsttwoperiods(HandF),canexpandtheirvalenceshells.10.6Totalvalenceelectrons:PH4

+has8;C2F4has36;andSbH3has8.IgnoringH,theatominthelowergroupnumberiscentral:P,C,andSb.Addedproof:HandFarenevercentral.ThetwocentralCatomsmustbeadjacent.Placealltheotheratomsaroundthecentralatom.SplittheFatomssothateachCgetstwo.Connectalltheatomswithsinglebonds.ThisusesalltheelectronsinPH4

+,andgivesPanoctet.TheHatomsneednoadditionalelectrons.TheCatomshavesixelectronseach,butcanachieveanoctetbyformingadoublebond.Splittingthetwenty-fourremainingelectronsinC2F4intotwelvepairsandgivingthreepairstoeachFleaveseachFwithanoctet.ThelasttwoelectronsinSbH3endasalonepairontheSb,andcompleteitsoctet.

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10.27 Determine the total number of valence electrons present. Next, draw a Lewis structure. Finally, use VSEPR or valence bond theory to predict the shape.

10.28 The molecular shape and the electron-group arrangement are the same when there are no lone pairs on the

central atom.

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