# Assigment2_answer - 5.5 5.9 5.44 5.61 5?6 8.6 8.2...

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Unformatted text preview: 5.5 5.9 5.44 5.61 5.?6 8.6 8.2 ta}- (‘3) (a) (M (a) No. This distance traveled to the top of a mountain depends on the path taken by the hiker. Distance is a path function, not a state function. Yes. Change in elevation depends only on the location of the base camp and the heightof themotlntain, notonthe path tJothe top. Changeinelevation is a state function, not a path function. w = —PAV. Because AV for the process is (—), the sign ofw is {+}. AB = q + w. At constant pressure, nH = q. If the reaction is endothermic, the signs of AH and q are {+}. From (a), the sign of w is (+}, so the sign of AB is {+). The internalenergy ofthesystnemincreases duringthechange. (This situation is described by the diagram [ii] in Exercise 5.4.} The sign of AH is positive, so the reaction is endothermic. 11H = —214.4 k] 13,; =AH‘.’ CatOHhtsMH‘: eHzth— 2 AH? Hzoﬂt—AH: (Islets) — 1272 k] = —936.2 k} + 226.7”! k] — 2[—235.33 k1] —;1H‘; CaCﬂs] AH: for CaCz-{s} = —60.6 k] (a) tb} (C) (151} (a) (b) I-INOz, 18 valence e‘, 9 e‘ pairs N02: 18 valence e‘, 9 e‘ pairs H-O-N=O [:6—N=E5 |‘ The formal charge on N is zero, in both species. NOf is expected to exhibit resonance; the double bond can be drawn to either oxygen atom. An alternate resonance structure for HNO: can be drawn, but it has nonzero formal charges on the oxygen atoms. This structure is less liker than the one shown above. Assuming that the structure shown above is the main contributor to the structure ofI-DJO:,theN=ObondlengthinI—INOZ willbeshorterthantheN—Dlengthsin NOf. Because there are two equivalent resonance structures for N02; the N—D lengths are approximately an average of N—O single and double bond lengths. These are longer than the full N=O double bond inI-INIDZ. NaCL 788 Elmo]; ICF, 808 kjfmol Given that crystal structure and ionic charges are the same for the two compounds,ﬂ1ediﬂererwemlatﬁceenergyisbecauseofﬂ|edifferencemion separation (d). Lattice energyr is inverser proportional to ion separation (d), so we expect the compound with the smaller lattice energy, NaCl, to have the larger ion separation. That is, the Na—Cl distance should be longer than the K—F distance. Na—Cl, 1.16 A + 1.6725 = 2.33 A K—F, 1.52 .5. + 1.19 A = 2.71 .5. This estimate of the relative ion separations agrees with the estimate from lattice energies. Ionic radii indicate that the Na—Cl distance is longer than the K—F distance. 8.54 8.74 E 9.4 (a) (b) (b) (a) (b) (a) lib} (C) 18 e", 9 e' pairs :6: 1- :0: 1- | || .- —C H—C— t _.. —O ‘ on H Yes, resonance structures are required to describe the structure. The Lewis structure of C02 {16 e", 8 e' pairs) is DZCZO In C02, the (2—0 bonds are full double bonds with two shared pairs of electrons. In HCOz', the two resonance structures indicate that the (3—0 bonds have partial, but not full, double bond character. The (2—D bond lengths in foal-mate will be longer thanthose in C02. -219 1d,! mol AH“ = AH: Cszigt—AH; emits—AH: Hats) = 4.34.63 — 52.30 — (l = —136.9‘B k] :NEN—O: H :N—NEO: H NZNZD t) +1 -1 -2 +1 +1 '1 +1 D In the leftmost structure, the more electronegative O atom has the negative formal charge, so this structure is likely to be most important. No single resonance structure rationalizes both observed bond lengths. 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 theNatomssharelessthanthreeelectronpairsarecontlibutorstothetrue structure. That the N—O bond length is slightly shorter than a typical N=O 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 although formalcharge canbeused to predict whichresonance formwillbe more important to the observed structure, the inﬂuence of minor contributors on the true structure cannot be ignored. 4e‘ domains Themoleculehasanonzero dipole moment, becausetheC—HandC—Fbond dipoles do not cancel each other. (ii) 9.26 9.56 9.66 bent {b}, linear {I}, octahedral (oh), seesaw [ss), square pyramidal {5p}, square planar (spl), tetrahedral {td}, trigonal bipyramidal (tbp), trigonal planar (tr), trigonal pyramidal (t9). T—shaped m EIECtI‘O‘D' Molecule Valence LeWis domain MOIECular orion electrons structure Geometry geometry (a) A5}?3 26 : Ff—AB—F : its 11d tp : t": F/ {\F H C + H + (b) CH3+ 6 [ I-II H] /é tr tr H \H Elem- MOIECUIE VEIEIIOE' Lewis domain MOIECIJIBI' orion electrons structure geometry geometry :F: F (c) 311:3 23 ’Istr_1:; "‘~1tr_1: tbp T l: " "’4: (d) (1103‘ 26 9—0—9: (1] td 1;. A): var/5‘0 F (e) 21:12:122 22 :T:_+)I<‘e. :1: "‘-)te_. ﬂap 1 3’ t-‘ ' (I) Bro; 20 l:o_sr_o:]‘ ﬁr _ td b o/po 1"llwlore than one resonance structure is possible. All equivalent resonance structures predict the same n‘Lolecular geometry. 1.1 (a) The Lewis structure depicts an anion with a 1— charge. The chemical formula of the given structure is C3H301. This grouping of atoms has 27 valence electrons, whereas the structure shown has 14 electron pairs or 28 electrons. This means that the structure is an anion with a 1— charge. spl Yes, there is one other resonance structure. 'I'hethree CandtwoDatounseaachhaveapn orbital. Therearesixelectronsintheit system ofthemolecule. [falltheCandOatoms are sp2 hybridized, there are seven bonding electron pairs and four nonbonding electron pairs in the 1:: system. This leaves three electron pairs or six electrons in the 1: system. 9.88 (a) (b) (d) 4ﬂe‘, 20e‘ pairs 5 e" domains trigocnal—bipyrainidal electron—domain geometry The greater the electronegativity of the terminal atom, the larger the negative charge centered on the atom, the smaller the effective size of the P—X bonding electron domain. A P—F bond will produce a smaller (and shorter} electron donLaintl'IanaP—Cl bond. The molecular geometry (shape) is also trigonal bipyramidal, because all five electron domains are bonding domains. Because we predicted the P—F electron domain to be smaller, the larger P—Cl bonding domain will occupy the equatorial plane of the molecule, minimizing the number of 90“ P—Cl to P—F repulsions. This is thesameargumentthatplaces a“larger”nonbonding domainintheeqiiatou'ial position of a molecule like 5114. The molecular geometry is distorted from a perfect trigonal bipyr'an‘lid because not all electron domains acre alilce. The 90" P—Cl to P—F repulsions will be greater than the 90“ P—F to P—F repulsions, so the F{axial)—P—Cl angles will be greater than 90“. The equatorial P—P—F angles may distort slightly to ”make room” for the axial F atoms that are ”pushed away" from the equatorial C1 atom ...
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