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CH. 8 CHEMICAL BONDING

CH. 8 CHEMICAL BONDING. Chemical bonding types Lewis Dot structures Energy of bonds. Ionic properties Formal Charge Trends. Equations. more metallic: ------> & to more nonmetallic. Why bonds formed? recall, toward min E, lowest PE. Properties of (atoms) substs determined by

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CH. 8 CHEMICAL BONDING

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  1. CH. 8 CHEMICAL BONDING Chemical bonding types Lewis Dot structures Energy of bonds Ionic properties Formal Charge Trends

  2. Equations

  3. more metallic: ------> & to more nonmetallic Why bonds formed? recall, toward min E, lowest PE Properties of (atoms) substs determined by * type of bond ( e- config) * strength of bond (p+/n0 -- e- attraction) Metal vs Nonmetal * based on location on p.table RECALL:

  4. List in decreasing lattace E -- MgF2 CaF2 ZrO2 -- ZrO2 > MgF2> CaF2 Zr+4: O-2 Ca+2 : F-1 Mg+2 : F-1 +2 : +2 Ca > Mg size Lattice Energies pg.293 table 8.2

  5. 3 ways of bonding to occur -- metal + non ionic bond; e- transfer; lrg EN diff non + non small diff gain/lose e- metal + metal 1A 2A w/ 7A, top 6A ions formed; forms ionic solid; ion ratio: emperical formula metallic bond; e- pooling; large atoms, few outer e- well shielded; low IE, lose e- ezly but not gain covalent bond; e- share small EN diff.; high IE attract high EA e- pair localized bond of specific length & strength; actual # of atoms: molecular formula Share e-, delocalized; e- flow around cation core, move freely

  6. .Al. . .. .O. .. .Sr . . .P . .. .Na . .C . . LEWIS DOT STRUCTURE element symbol: reps. nucleus & filled inner core e- DOTS: reps. # valence e- metal: # e- transfer to form cation nonmetal: # e- gain to form anion; or share to form covalent bond Octet Rule: forming bonds by gain, lose, or share e- to fill outer-level in easiest way to min. E form noble gas of e-

  7. Mg 3s P Mg 3s 3px 3py 3pz 3s P Mg 3s3px 3py 3pz 3s IONIC BONDING MODEL metal (lose) + nonmetal (gain) # e- lost = # e- gain show Mg+2 & P-3 formation

  8. IE1 = 738 kJ pg.259 table7.2 IE2 = 1451 kJ EA1, 2, 3 = kJ S = [IE1,2 + EA1,2,3] kJ all steps 3[ Mg (g) ----> Mg+2 (g) + 2 e-] 2[ P (g) + 3 e- ----> P-3 (g) ] 3 Mg + 2 P ----> 3 Mg +2+2P-3 e- transfer is endo, absorbs ionic solid formation is exo, great release think, dist/speed E is required to form ions be converted to gaseous atoms but DHf0 Mg3P2(s) = - kJ/mol kJ released/mol Mg3P2(s)formed

  9. So, exo steps larger than endo parts Mg (g) ----> Mg+2 (g) + 2 e-DH0= - kJ P (g) + 3 e- ----> P-3 (g) DH0= + kJ the strong attraction of ions to form solid;exo Lattice E, DHlattice enthalpy change of 1 mol ionic solid to gaseous ions indicated strength of attraction influences: melting pt, hardness, solubility, etc

  10. Calculate the DHlattice of KF(s) from the following K(s)  K(g) DH1 = 90 kJ K(g)  K+(g) + e-DH2 = 419 kJ F2(g)  2 F(g) DH3 = 159 kJ F(g) + e- F-(g) DH4 = -336 kJ K+(g) + F-(g)  KF(s) DH5 = -821 kJ K(s) + 1/2 F2(g)  KF(s) DHf = -569 kJ = (-569) – [90 + 419 + .5(159) + -336 + -821] = -1138 kJ

  11. Trends in Lattice E Ionic Size: down col incr radius; dist bet +/- incr so electro-E less Ionic Charge: consider the electro- eqn; ions w/ > charges require more DH0lattice Mg+2 2188 kJ Li+1 520 kJ O-2 737 kJ F-1 -328 kJ add 1st e-: exo additonal e-: endo

  12. ELECTRONEGATIVITY (EN) --- POLARITY d+d- polar covalent bonds use to show partial “+” charge on less electroneg. atom use to show partial “-” charge on more electroneg. atom EN: bonded atom to attract shared e- pair T9.2 pg 349 EN values for the elements largest value F: 4.0 lowest Cs, Fr: 0.7 EN inverse to size again, think of dist EN inverse to size again, think of dist bet 2 atoms: the more EN element, assign neg charge

  13. I N C R E A S E I N C R E A S E I N C R E A S E I N CREASE ATOMIC SIZE ELECTRONEGATIVITY pg.299 Fig.8.7

  14. H F H F Nonpolar: equal sharing of e- pair in bond occurs when electronegativity value is less than 0.5 all diatomics are nonpolar Polar: unequal sharing e- pair in bond occurs when electronegativity diff. between elements is greater than 0.5 Determine electronegativity value, show δ partial charges, and overall polarity in the molecule, HF 2.1 4.0 4.0 - 2.1 = 1.9 polar covalent

  15. F . H . Electron rich end POLAR COVALENT

  16. EN values between atoms H---Cl Mg---O Ca---P N---Br 2.1 3.0 1.2 3.5 1.0 2.1 3.0 2.8 0.9 2.3 1.1 0.2 Polar Co Ionic Polar Co Nonpolar EN Scale 0.0 0.4 0.5 1.7 2.0 Nonpolar Cov Polar Covalent Ionic Larger the EN diff, more ionic character Let’s take Cl with element in Per 3. What trend can you predict in, terms of: bond length, strength, type, E & EN, metallic. mp bond type: Ionic - NP - PC bond length: decr bond strength: incr bond E: incr EN: incr metallic: decr melting pt: decr

  17. BOND ENERGY (BE): bond enthalpy or bond strength E to break the bond in 1 mol gaseous molecules T8.4 pg 316 DH0break;endo BE > 0 DH0form;exo BE < 0 the strong attraction of ions to form solid;exo CORRELATIONS bond order - bond length - bond E *multiply bonds decr bond length, single bonds weaker *shorter bond length atoms held tighter *higher order is shorter length is higher E

  18. Rxn in 2 steps heat absorbed to beak reactant bonds, +DH° heat released to form pdt bonds, -DH° DH°rxn = DH°broken + DH°formed or DH°rxn = SBEbroken - SBEform Bond E values in terms of absorb, + Covalent bonds are strong, localized bonds BE are in terms of “+” values most are gases,, liquids, low melting solids If such strong bonds, why low mp/bp??? Strong bonds bet atoms w/i molecule, but What can be said of fuels & bond lengths??? more weak bonds is more release of E weaker bonds bet molecules

  19. FORMAL CHARGE f.c. =# val. e- - (# unshare e- + 0.5 # share e-) What does the atom own??? * all unbonded e- pairs * 0.5 of bonding e- Have 2+ possible arrangements, which structure imprt?? * smaller f.c. to large * f.c. not side-by-side * more -f.c. on more -EN atom

  20. .. O :O: :O: .. 6 - [4 + .5(4)] 6 - (4 + 2) 6 - 6 = 0 Oa: 6 val e- 4 unbonded 4 bonded Let’s look at O3 +1 -1 0 b ag 6 - [2 + .5(6)] 6 - (2 + 3) 6 - 5 = +1 Ob: 6 val e- 2 unbonded 6 bonded 6 - [6 + .5(2)] 6 - (6 + 1) 6 - 7 = -1 Og: 6 val e- 6 unbonded 2 bonded

  21. bonds brokebond E(kJ/mol)bonds formbond E(kJ/mol) 3 C-H 3(413) = 1239 1 C-O 358 1 O-H 463 1 C=O 1072 total ______ kJ Calculate DHo from Bond E-values for the following reaction CH3OH + CO ----> CH3COOH 3(413) = 1239 348 799 358 463 ________ kJ 3 C-H 1 C-C 1 C=O 1 C-O 1 O-H DH°rxn = __________________ = _____ kJ or DH°rxn = ________________ = _____ kJ

  22. Na Na Na Na Na Na Na Na Na METALLIC BONDING, simplified “Sea” of electrons delocalized val e- a form of covalent sharing also form gaseous diatomic (Li2) delocalized: no ions bonded together w/ e- pair, so, val e- are “shared” w/ many ions e- e- e- e- e-

  23. Properties of Metals solids; mod-high mp; high bp; mallable; ductile; conduct heat/elec properties influenced by ion structure & mobile val e- Why relative low mp? No actual bond to break trend: 2A alkaline mp >> 1A alkali 2 val e- to, greater attraction to overcome Then why ext. high bp? Require high E to break ion & val e- apart As conductors elec: mobile e- flow thur metal bet terminals heat: deloc e- disperse heat quicker

  24. #1. An analogous Born-Haber cycle has been describedfor LiF. Solve for step 4, the electron affinity of F. 1) K(s)  K(g) DH1 = 90 kJ 2) K(g)  K+(g) + e-DH2 = 419 kJ 3) 1/2 F2(g)  F(g) DH3 = (1/2) (159) kJ 4) F(g) + e- F-(g) DH4 = ? = EA 5) K+ + F-(g)  KF(s) DH5 = -821 kJ 6) K(s) + 1/2 F2(g)  KF(s) DHf = -569 kJ DH4 = EA = DHf - (DH1 + DH2 + DH3 + DH5) EA = ________________________________ EA =_________ kJ #2. Calculate Hlatt for NaCl 1) Na(s)  Na(g) DH1 = 109 kJ 2) Cl2(g)  2 Cl(g) DH2 = 243 kJ 3) Na(g)  Na+1(g) + e-DH3 = 496 kJ 4) Cl(g) + e- Cl-(g) DH4 = -349 kJ 5) Na(s) + 1/2 Cl2(g)  NaCl(s) DHf = -411 kJ DHlattice = DHf - (DH1 + DH2 + DH3 + DH4)

  25. #3. 1) Mg(s)  Mg(g) DH1 = 148 kJ 2) 1/2 Cl2(g)  Cl(g) DH2 = 1/2 (243 kJ) 3) Mg(g)  Mg+(g) + e-DH3 = 738 kJ 4) Cl(g) + e- Cl-(g) DH4 = -349 kJ 5) Mg+(g) + Cl-(g)  MgCl(s) DH5 = -783.5 kJ (= (MgCl)) 6) Mg(s) + 1/2 Cl2(g)  MgCl(s) DHf(MgCl) = ? DHf(MgCl) = DH1 + DH2 + DH3 + DH4 + DH5 #4. 1) Si(s)  Si(g) DH1 = 454 kJ 2) Si(g)  Si4+(g) + 4 e-DH2 = 9949 kJ 3) O2(g)  2 O(g) DH3 = 498 kJ 4) 2 O(g) + 4 e- 2 O2-(g) DH4 = 2(737) kJ 5) Si4+(g) + 2 O2-(g)  SiO2(s) DH5 = DHlattice(SiO2) = ? 6) Si(s) + O2(g)  SiO2(s) DHlattice(SiO2) = -910.9 kJ = _______________________________________________ = _______ kJ = _______________________________________________ = __________________ kJ

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