Chapter 9

Simple Blue

Chapter 9 Chemical Bonding IBasic Concepts1Lewis Dot SymbolsConsist of the symbols of an element & 1 dot for each valence e in an atom of that element.To keep track of valence e in chemical rxnsElements in same group same # of valence e (same with group number)Lewis Dot Symbols

The Ionic BondElectrostatic force that holds ions together in an ionic cpdBtwn metal & nonmetale from metal transferred to non metalMetal cationNonmetal anionExample (Lewis e dot & e configuration): Na & Cl, Ba & HLattice energy of ionic cpdsEnergy required to completely separate 1 mole of solid ionic cpd to gaseous ionsCannot be measured directlyUse Coulombs Law to calculate if structure & composition of ionic cpd is knownPotential energy (E) btwn 2 ions is indirectly proportional to the product of their charges and inversely proportional to distance of separation btwn themCoulombs LawE is ve; charge for F is veFormation of ionic bond is exothermic energy must be supplied to reverse the processLattice E of LiF is +ve = bonded pair of Li+ & F- ions is more stable than separate Li+ & F-

Born-Haber cycleto indirectly determine lattice energy by assuming formation of ionic cpd has a series of stepsRelates lattice energies of ionic cpds to ionization energies, electron affinities, & other atomic & molecular propertiesBased on Hesss Law

Born-Haber cycle (LiF)Standard enthalpy of formation of LiF is -594.1 kJ/molLi(s) + F2(g) LiF(s)Sum of enthalpy changes for all steps required to separate LiF ion gaseous ions = -594.1 kJ/mol

StepsConvert Li(s) to Li(g)Li(s) Li(g); H1 = 155.2 kJ/mol2. Dissociate F2(g) to 2F(g)F2(g) 2F(g); Ha = 150.6 kJ/mol F2(g) F(g); H2 = 75.3 kJ/mol3. Ionize Li(g)Li(g) Li+(g) + e; H3 = 520 kJ/molSteps4. Add 1 mol e to 1 mol F(g)F(g) + e F-(g); H4 = -328 kJ/mol5. Combine 1 mol Li+(g) and 1 mole F-(g) to form solid LiF(s)Li+(g) + F-(g) LiF(s); H5 = ?

The reverse of step 5 = lattice energy of LiFBorn-Haber cycle

Using Hesss Law to calculate1 Li(s) Li(g)155.22 F2(g) F(g)75.33 Li(g) Li+(g) + e5204 F(g) + e F-(g)-3285 Li+(g) + F-(g) LiF(s)?Li(s) + F2(g) LiF(s)-594.1

Born-Haber cycleThe greater the lattice energy, the more stable the ionic cpdRough correlation btwn lattice energy & mpLarger lattice energy more stable ions more strongly held togetherRequires more heat to melt the solid

Lattice energy & formulas of ionic cpdsLattice energy - a measure of stability of ionic cpds.Ionisation of elements rapidly when e are removed one by oneMg 1st ionisation energy > 2nd ionisation energy Mg2+ has noble gas conf very stable

But stability gained because lattice energy for formation of MgCl2 is higher than 2nd ionisation energy of MgNaClEnergy released to form NaCl too small compared to energy absorbed to form Na2+O2-High lattice energy for formation of Na2O or MgOThe Covalent BondSharing of valence e btwn 2 atomsRepresented by a lineEach e is attracted to nuclei of both atomsLewis structure shared e pairs shown a s lines or pairs of dots btwn 2 atoms & lone pairs pair of dots on individual atomExample (Lewis dot) H2, F2, H2OThe Covalent BondF2, H2O octet rule an atom other than H form bonds until surrounded by 8e.Works for 2nd period elementsType of cov bondsSingle atoms held by 1 e pairMultiple 2 atoms share 2 or more pairs of eExample: C2H4, N2Comparison of properties of cov & ionic cpdsAttractive force in cov cpds force holding atoms togetherForce btwn molecules (intermolecular force)Molecules of a covalent cpds are held weaklyGases, iquid, low mp solidsComparison of properties of cov & ionic cpdsElectrostatic force btwn ionic cpds strongSolid at room T & high mpSoluble in H2O & aqueous soln conducts electricityMolten conduct electricityCov cpds insoluble & nonelectrolytesMolten do not conduct electricity

Electronegativity Polar (covalent) bond = e spend more time being near to 1 atom compared to the otherProperty that helps to identify polar bond electronegativity = ability of an atom to attract more e to itself in a chemical bondElectroneg related to e affinity & ionisation energyElectronegativity Relative concept can only be measured in relative to electronegativity of another atomAtoms of elements with very diff electroneg tend to form ionic bondAtoms of same element that forms a bond purely covalentElectronegativity

Electronegativity Ionic bond electronegativity diff btwn 2 bonding atoms is 2.0 or morePercent ionic character directly related to diff in electroneg (Fig 9.7)Electron affinity isolated atoms attraction for an additional eElectroneg in a chemical bondPractice exerciseWhich of the following bond is covalent, polar covalent and ionic?Bond in CsClBond in H2SNN bond in H2NNH2Electroneg & oxidation numberOxn number conceptRefers to num of chgs an atom would have if e were transferred completely to the other more electroneg atomN in NH2, O in H2O2

Writing Lewis structuresBasic stepsWrite skeletal structure of cpdLeast electroneg centerH, F terminalCount total num of e presentPolyatomic ions, add num of ve chg to the total. (SO42- - 2 more)Cations subtract the num of +ve chgWriting Lewis structuresDraw single covalent bond btwn central atom & surrounding atomsValence shell for H is full with only 2 ee of center & surroundings must be shown as lone pairs if they are not involved in bondingIf center atom has less than 8 e, try add = or bond using lone pairs fr surrounding atoms (CONS)Practice exerciseWrite Lewis structure for CS2, HCOOH, NO2-Formal Charge & Lewis structureFormal chg = electrical chg diff btwn valence e in an isolated atom& the num of e assigned to the atom in Lewis structureMethod:All atomss nonbonding e are assigned to the atomBreak bonds btwn the atom & the other atom & assign of the bonding e to the atomFormal Charge & Lewis structure (O3)O=O-OValence e6 6 6E assigned to atom 6 5 7___________________________Difference (formal chg) 0 +1 -1

So formal chg:

Formal chg & Lewis structureHelpful rulesSum of the chgs must become 0 molecules are electrically neutral spCations sum of formal chgs equal to +ve chgAnions sum of formal chgs equal ve chgFormal chg & Lewis structureFormal chg help keep track of valence eGet qualitative picture of chg distr in molecNot complete transfer of eO3, experimental studies prove that central O has partial +ve chg & end O has partial ve chg.Practice exerciseWrite formal chg for NO-Formal chg & Lewis structureMolecules can hv more than 1 acceptable Lewis structureThe most acceptable one can be deduced using formal chgA Lewis structure with 0 formal chg is more preferableLewis structures with smaller formal chg more acceptableFormal chgs on more electroneg atoms are more acceptable

The concept of resonanceO3 double bond can be put on either side of the moleculeBut bond lengths btwn O atoms equal (= bond shd be shorter than single)So both structures can be used to represent the molecule

The concept of resonanceThe 2 structures are called resonance structuresResonance structure 1 of 2 or more Lewis structures for a molecule that cannot be presented accurately by only one Lewis structActual molecule has own unique structureResonance to address limitations to the bonding modelThe concept of resonanceOther examplesCO32-Benzene (C6H6)

The positions of e can be rearranged in diff resonance structuresExceptions to the octet ruleRecall octet rule applies to period 2 elementsIncomplete octet - # of e surrounding central atom < 8, valence e less than 4eBeH2, BF3

Exceptions to the octet ruleBF3 readily react with NH3

B-N bond diff fr covalent bonds discussed so far both e are provided by NBond = coordinate covalent bond (dative covalent bond)

Odd electron moleculesNO & NO2Odd-e molecules sometimes called radicals Radicals highly reactiveUnpaired e forms cov bond with another unpaired e of another moleculeThe expanded octetMore than 8 e surround central atomAtoms of element in 3 period & beyondExample: SF6

Bond enthalpyMeasure the stability of moleculeEnthalpy change required to break a particular bond in 1 mole of gaseous moleculesTable 9.4Polyatomic molecules average bond enthalpyTriple bond > double > single Use of bond enthalpies in thermochemistryPredict approximate enthalpy of rxn using avg bond enthalpies

H = BE(reactants) - BE(products) = total energy input total energy releasedH = -ve = exo, +ve = endoDiatomic molecules accurately knownPolyatomic gives approximate resultsPractice exerciseFor the rxn; H2(g) + C2H4(g) C2H6(g)Estimate the enthalpy of rxn, using bond enthalpy values in Table 9.4Calculate the enthalpy of reaction, using std enthalpies of formation. (Hf (H2) = 0, Hf (C2H4) = 52.3, Hf (C2H6) = -84.7)Hrxn = nHf (products) - mHf (reactants)End of Chapter 9