Chapter 9: Models of Chemical Bonding

1 Chem 6A Michael J. Sailor, UC San DiegoChapter 9: Models of Chemical BondingChem 6A, Section D Oct 11, 20111 Chem 6A Michael J. Sailor, UC San DiegoAnnouncements: Practice Final is posted on the web: Thurs Nov 10 quiz (#7) will be on Chapter 7 Tues of thanksgiving week is review No office hours Weds Nov 2312 Chem 6A Michael J. Sailor, UC San DiegoQuiz 6 score histogramChem 6A Michael J. Sailor, UC San Diego0 10 20 30 40 50 60 70 80 90 100 F D C- C C+ B- B B+ A- A A+ number of studentsGrades so far(after quizzes 1-6)34 Chem 6A Michael J. Sailor, UC San Diego4 Chem 6A Michael J. Sailor, UC San Diego3 Types of Chemical BondsFig 6A Michael J. Sailor, UC San DiegoBonding in CompoundscovalentionicCovalent bond = neutral atoms held together by sharing a pair of electronsIonic bond = charged atoms (ions) held together by electrostatic forcesAn assembly of atoms held together by covalent bonds is a molecule E=z1z2q24 or1 2 Coulomb s law:chargedistanceNa+Cl-HHOClClwater moleculechlorine moleculeChem 6A Michael J.

2 Sailor, UC San DiegoIonic BondingNa+Cl-The rock salt lattice78 Chem 6A Michael J. Sailor, UC San DiegoLattice enthalpies and ionic Enthalpy vs 1/(Ionic radii)Lattice enthalpy, kJ/mol1/(Na-X) distance, -1 NaFNaClNaBrNaI Chem 6A Michael J. Sailor, UC San DiegoPROBLEM: Lattice Enthalpies(see problem )The thermite reaction (shown below) is highly exothermic, mainly due to the larger energy of the Al2O3 crystal lattice relative to (s) + 2Al(s) Al2O3(s) + 2Fe(s) Using the data in the table below, calculate the lattice enthalpy of Al2O3 and Fe2O3. What is the main reason for the larger lattice energy of Al2O3?910 Chem 6A Michael J. Sailor, UC San DiegoPROBLEM: Lattice EnthalpiesCalculate the lattice enthalpy of Al2O3(s) at 25 C from the following data: Process Enthalpy ( H), kJ/molLattice enthalpy Al2O3(s) ?

3 First ionization energy of Al(g) +578 Second ionization energy of Al(g) +1820 Third ionization energy of Al(g) +2750 Enthalpy of formation of Al(g) +294 Enthalpy of formation of O2(g) 0 Bond energy of O2(g) +498 Electron affinity of O(g) -141 Electron affinity of O-(g) +844 Enthalpy of formation of Al2O3(s) -1676 Chem 6A Michael J. Sailor, UC San DiegoPROBLEM: Lattice EnthalpiesCalculate the lattice enthalpy of Fe2O3(s) at 25 C from the following data: Process Enthalpy ( H), kJ/molLattice enthalpy Fe2O3(s) ?First ionization energy of Fe(g) +759 Second ionization energy of Fe(g) +1561 Third ionization energy of Fe(g) +2957 Enthalpy of formation of Fe(g) +415 Enthalpy of formation of O2(g) 0 Bond energy of O2(g) +498 Electron affinity of O(g) -141 Electron affinity of O-(g) +844 Enthalpy of formation of Fe2O3(s) -8261112 Chem 6A Michael J.

4 Sailor, UC San DiegoPROBLEM: Lattice EnthalpiesRewrite process in terms of Chemical equations: Process Enthalpy ( H), kJ/molLattice enthalpy Fe2O3(s) ?First ionization energy of Fe(g) +759 Second ionization energy of Fe(g) +1561 Third ionization energy of Fe(g) +2957 Enthalpy of formation of Fe(g) +415 Enthalpy of formation of O2(g) 0 Bond energy of O2(g) +498 Electron affinity of O(g) -141 Electron affinity of O-(g) +844 Enthalpy of formation of Fe2O3(s) -8262Fe(s)+ 3/2O2(g) Fe2O3(s) O-(g) + e- O2-(g) O(g) + e- O-(g) O2(g) 2O(g) Fe(s) Fe(g) Fe2+(g) Fe3+(g) + e- Fe+(g) Fe2+(g) + e- Fe(g) Fe+(g) + e- 2Fe3+(g)+ 3O2-(g) Fe2O3(s) Chem 6A Michael J.

5 Sailor, UC San DiegoSOLUTION: Lattice EnthalpiesRearrange equations to add up to 2Fe3+(g)+ 3O2-(g) Fe2O3(s): Process Enthalpy ( H), kJ/mol2Fe(s)+ 3/2O2(g) Fe2O3(s) -826 3O2-(g) 3O-(g) + 3e- 3(-844) 3O-(g) 3O(g) + 3e- 3(+141) 3O(g) 3/2 O2(g) 3/2(-498) 2Fe(g) 2Fe(s) 2(-415) 2Fe3+(g) + 2e- 2Fe2+(g) 2(-2957) 2Fe2+(g) + 2e- 2Fe+(g) 2(-1561) 2Fe+(g) + 2e- 2Fe(g) 2(-759) 2Fe3+(g)+ 3O2-(g) Fe2O3(s) -15,066 ?1314 Chem 6A Michael J. Sailor, UC San DiegoSOLUTION: Lattice EnthalpiesDo the same for 2Al3+(g)+ 3O2-(g) Al2O3(s): Process Enthalpy ( H), kJ/mol2Al(s)+ 3/2O2(g) Al2O3(s) -1676 3O2-(g) 3O-(g) + 3e- 3(-844) 3O-(g) 3O(g) + 3e- 3(+141) 3O(g) 3/2 O2(g) 3/2(-498) 2Al(g) 2Al(s) 2(-294) 2Al3+(g) + 2e- 2Al2+(g) 2(-2750) 2Al2+(g) + 2e- 2Al+(g) 2(-1820) 2Al+(g) + 2e- 2Al(g) 2(-578) 2Al3+(g)+ 3O2-(g) Al2O3(s) -15,416 Chem 6A Michael J.

6 Sailor, UC San DiegoSOLUTION: Lattice EnthalpiesSummary: Process Enthalpy ( H), kJ/mol2Al3+(g)+ 3O2-(g) Al2O3(s) -15,416 2Fe3+(g)+ 3O2-(g) Fe2O3(s) -15,066 Difference:2Al3+(g)+ Fe2O3(s) Al2O3(s) + 2Fe3+(g) -350 kJ/mol Al2O3(s) lattice is more stable than Fe2O3(s) lattice by 350 kJ/mol2Al(s)+ Fe2O3(s) Al2O3(s) + 2Fe(s) -850 kJ/molSo 350/850, or 41% of the energy for this reaction comes from the difference in lattice energies. Why is Al2O3(s) so much more stable?The enthalpy of the thermite reaction is:1516 Chem 6A Michael J. Sailor, UC San Diego350 kJ/mol, or 41% of the energy for the thermite reaction comes from the difference in lattice energies. Why is Al2O3(s) so much more stable than Fe2O3(s)?

7 SOLUTION: Lattice Enthalpies2Al(s)+ Fe2O3(s) Al2O3(s) + 2Fe(s) H = -850 kJ/molIonIonic radius(pm)Fe3+ 64Al3+ 54O2-140Fe3+OAl3+O64 + 140 = 20454 + 140 = 194 Chem 6A Michael J. Sailor, UC San DiegoSOLUTION: Lattice EnthalpiesCalculate the electrostatic energy in one M-O bond: E=z1z2q24 or1 2 Coulomb s law:chargedistanceFe-O: E=3 2 ( 10 19)24 ( 10 12)(204 10 12)= x 10-18 J Al-O: E=3 2 ( 10 19)24 ( 10 12)(194 10 12)= x 10-18 J Difference: Al-O ionic bond is stronger than Fe-O ionic bond by210 kJ/molFe3+OAl3+O64 + 140 = 20454 + 140 = 1941718 Chem 6A Michael J. Sailor, UC San DiegoLattice enthalpies and hardness01800360054007200010002000300040 0050006000 Hardness (Knoop scale)Lattice enthalpy density, kJ/mLdiamondSiCsapphire (Al2O3)hematite (Fe2O3)halite (NaCl)Hematite stone braceletRock salt (halite)Sapphire ringDiamond solitaireChem 6A Michael J.

8 Sailor, UC San Diego19 PROBLEM: Lattice EnthalpiesA related (but simpler) problem:Calculate the enthalpy of formation of AgF(s) from the following data: Process Enthalpy ( H), kJ/mol Lattice enthalpy of AgF(s) -971 First ionization energy of Ag(g) +731 Enthalpy of formation of Ag(g) +284 Enthalpy of formation of F(g) +79 Electron affinity of F(g) +328 a) +451 kJ/mol b) -284 kJ/mol c) -205 kJ/mol d) -246 kJ/mol e) none of the above1920 Chem 6A Michael J. Sailor, UC San Diego20 PROBLEM: Lattice EnthalpiesA related (but simpler) problem:Calculate the enthalpy of formation of AgF(s) from the following data: Process Enthalpy ( H), kJ/mol Ag+(g) + F-(g) AgF(s) -971 Ag(g) Ag+(g) + e- +731 Ag(s) Ag(g) +284 F2(g) F(g) +79 F(g) + e- F-(g) -328 a) +451 kJ/mol b) -284 kJ/mol c) -205 kJ/mol d) -246 kJ/mol e) none of the aboveChem 6A Michael J.

9 Sailor, UC San DiegoLewis Dot StructuresFig used to indicate covalent bonds2122 Chem 6A Michael J. Sailor, UC San DiegoDrawing Lewis Dot Structures Count up all valence electrons Pair up electrons to form bonds or lone pairs Satisfy octet rule (every atom has 8 electrons, either as lone pairs or in shared Bonding pairs)Examples: CH4, O3, NF3 Chem 6A Michael J. Sailor, UC San DiegoPROBLEM: Drawing Lewis structuresDraw the Lewis structure for ozone, electrons for O: 3x6 Total electrons: 18 Total pairs of electrons: 9 OOOS atisfy octet rule with shared electron pairs2324 Chem 6A Michael J. Sailor, UC San DiegoPROBLEM: Drawing Lewis structuresHow many lone pairs of electrons are there in the Lewis structure of NF3?

10 A. 1 b. 3 c. 6 d. 9 e. 10 ANSWER: eNFFFLone pairsTotal electrons: 26 Total pairs of electrons: 13 Total Bonding pairs: 3 Total lone pairs: 10 Chem 6A Michael J. Sailor, UC San DiegoBonding in CompoundscovalentionicCovalent bond = neutral atoms held together by sharing a pair of electronsIonic bond = charged atoms (ions) held together by electrostatic forcesAn assembly of atoms held together by covalent bonds is a molecule E=z1z2q24 or1 2 Coulomb s law:chargedistanceNa+Cl-HHOClClwater moleculechlorine molecule2526 Chem 6A Michael J. Sailor, UC San DiegoCovalent BondsThe bond energy of F2 is 159 kJ/mol. The bond energy of H2 is 432 kJ/mol. What is the energy of the HF bond?