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CHM 112 Final Review Guide

by: Madeline Kaufman

CHM 112 Final Review Guide CHM 112

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Madeline Kaufman
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These notes cover what is on the CHM 112 final exam.
Chemistry 112
Dr Burjor Captain
Study Guide
Energy, internal energy, Enthalpy, Heat Capacity, isothermal, entropy, Gibbs Free Energy, Electrochemistry, rate, Half Life, Catalysis, redox, potential, Electromagnetic spectrum, Planck's equation, deBroglie, Heisenberg uncertainty principle, electron configuration, periodic table, Bonding, dipole, Orbitals, atom hybridization
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This 31 page Study Guide was uploaded by Madeline Kaufman on Monday September 19, 2016. The Study Guide belongs to CHM 112 at University of Miami taught by Dr Burjor Captain in Winter 2016. Since its upload, it has received 20 views. For similar materials see Chemistry 112 in Chemistry at University of Miami.


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Date Created: 09/19/16
CHM 112 FINAL EXAM 1: FREE ENERGY AND THERMODYNAMICS  Internal Energy: ΔΕ o Internal Energy: the sum of the potential energy and kinetic energy of a system o ΔE= q+w  q= heat (added or taken from the system)  w= work (done on or by the system)  Units: 1 cal = 4.148 joules (energy required to change 1g of water by 1 °C o ΔE= E -E finainitial o If ΔE is positive, the system gains energy from the surroundings  Enthalpy: ΔH o Enthalpy: total heat content of a system (amount of heat absorbed or evolved by the system) o An extensive property; must multiply by molecule stoichiometry o +ΔH  Endothermic reaction  Heat is added to the system  Breaking bonds (solidliquid, liquidgas, F 2F) 2 o –ΔH  Exothermic reaction  Heat given off to surroundings  Forming bonds (gasliquid, liquidso2id, 2FF ) o q versus ΔH  q is any heat change, ΔH applies to chemical changes (not applicable to things that remain the same substance)  Energy/Enthalpy Relationship o ΔΕ=ΔH-PΔV  -PΔV= w = pressure x volume change = -nRT  pressure in atm, volume in L  work consequentially in atmL o 101.3 J = 1 atmL  ΔE= internal energy (J)  ΔH= enthalpy (J) o ΔH is heat exchanged  ΔE is heat AND work exchanged  Heat Capacity o Specific Heat Capacity: C: the amount of heat required to change the temperature of 1 gram of a substance by 1°C  Unit: J/g°C o Molar Heat Capacity: the amount of heat required to change the temperature of 1 mole of a substance by 1°C  Unit: J/mol°C o Both intensive properties o q= mCΔT  q: heat (j)  m= mass (g)  C= specific heat (j/g°C)  ΔT= temperature change o To measure specific heat, compare to water  Heat gained by water= -(heat lost by substance)  M C (T -T )=-m C (T -T ) w w f i y y f i y  Tfw=Tfyequal final temperatures)  C w 1.0 cal/g°C o 1 cal= 4.148 joules  Hess’ Law o Hess’ Law applies to all state functions: values that are only dependent on the beginning state and end state, not the path taken to get there o Hess’ Law: if a reaction is carried out in a series of steps, ΔH for that reaction is equal to the sum of the ΔHs of the individual steps o Steps to doing a problem  Balance the equation  Look up enthalpies  Flip and multiply specific intermediate equations to cancel out terms that are not in the final equation  If you multiply an equation, must multiply ΔH f of that reaction  If you flip an equation, must switch sign on ΔHf o ΔH f: enthalpy of formation in standard state  Standard state: the form most stable of a substance at room temperature (25°C), 1 atm, and 1M concentration  ΔH f is the change in enthalpy that accompanies the formation of 1 mole of a substance in their standard state  ΔH f for an element already in its standard state is –  ΔH f of O2(g)Br2(l), 2(l)aphite carbon= 0  ΔH rxn= Σ[n ΔH °fproducts)- n ΔH °(refctants)]  Isothermal Expansion/Compression o Isothermal: process at constant temperature o W= -nRTln(V /V2)=1-nRTln(P /P )1 2 o ΔE of isothermal= 0 o For a reversible process, total work= expansion + compression = 0 o For an irreversible process, w= -PΔV  Calculating q with Phase Change o Cannot use just q=mCΔT, require energy for the phase changes o Example: heat required to convert 1g of ice at -30°C to steam at 120°C  Solidliquidgas  Step 1: use mCΔT for solid region with Tfas freezing point  Step 2: use latent heat of fusion multiplied by mass  Step 3: use mCΔT for gas region: Tfas boiling point, T is freezing point  Step 4: use latent heat of vaporization multiplied by mass  Step 5: use mCΔT for gas region, Tfas final temperature, T is boiling point  Entropy: ΔS o Spontaneous: processes that occur without outside intervention  If a reaction is spontaneous in one direction, it is not spontaneous in the other direction  Can be endothermic or exothermic o Entropy (ΔS): describes the amount of disorder/randomness  Chemical reactions with positive entropy are more likely to be spontaneous (but must also consider enthalpy)  ΔS positive +  Increase randomness  Going from liquid  gas  More moles in product  Increase in temperature  ΔS negative –  Increase in order  Crystal formation  Fewer moles in product  Decrease in temperate  ΔS= S finalSinitial  S= k ln W  Ludwig Boltzman -23  K= Boltzman constant = 1.38 x 10 J/K  W=Ω= number of arrangements/microstates o More microstates, more randomness  ΔS= q/T  Latent heat of vaporization/boiling point= entropy of getting one mole of water to one mole of steam  ΔS= nRln(V /2 )1 nRln(P /P1)2  For reversible isothermal expansion  ΔS universeS system ΔS surroundings  ΔS is also a state function, so we can apply Hess’ Law  ΔS° rxn Σ[n ΔS f(products)- n ΔS f(reactants)]  Where ΔS f= mCln(T /F )i  And phase changes= ΔH fusionlting point or ΔHvaporizationing point  -ΔS implies loss of degrees of freedom. Where n= # of atoms…  Non-linear molecule o Translation: 3 degrees (along x,y,z) o Rotation: 3 degrees (about x,y,z) o Vibration: 3n-6 degrees (stretch/bend)  Linear molecule o Translation: 3 degrees (along x, y, z) o Rotation: 2 degrees o Vibration: 3n-5 degrees  Total degrees of freedom: 3n  Gibb’s Free Energy: ΔG o ΔG=ΔH-TΔS  T= temperature in Kelvin o ΔG system -TΔS universeH systemTΔSsystem  ΔG is proportional to –ΔS universe o If ΔG is NEGATIVE, SPONTANEOUS process  Heating does not necessarily evoke a reaction o Can also apply Hess’ Law  ΔG f= Σ[n ΔG °fproducts)- n ΔG °freactants)] o In summary… ΔH ΔS T ΔG Spontane ity - + All - Forward direction - - Low - Forward - - High + Reverse + + Low + Reverse + + High - Forward + - All + + Reverse o ΔG= ΔG°+RTlnQ pD ¿ ¿ ¿d(pC) c ¿ pB ¿  Q= reaction quotient ¿ ¿b ¿ [D] [C]c ¿ ¿ ¿  R= gas constant = 8.314 j/molK or .008314 kJ/molK  T= temperature in kelvin  ΔG°= standard free energy  Under standard conditions, Q always = 1, ln1=0  ΔGrxnΔG rxn  Under equilibrium conditions,  ΔGrxnalways = 0 and Q=K  ΔG°= -RTlnK −ΔH° ΔS° o lnK= + RT R −ΔH° ( 1 − 1 ) o ln(K2/K1)= R T 2 T 1  Bond Enthalpy o Always need energy to break a bond; ΔH always positive o Use bonds when you cannot find enthalpies of reactants/products o ΔH°= Σ[D(bonds broken)- D(bonds formed)]  D: bond energy: energy required to break 1 mole of the bond  In a sample problem  For broken, sum of bonds broken in reactants and for formed, sum of bonds broken in products (reactants-products; opposite of normal)  Must multiply bond strength (different for doubles and singles) by number of bonds in individual molecule AND by mole number  Laws of Thermodynamics o First Law: total energy of the universe is constant  ΔE systemEsurroundings o Second Law: for any spontaneous process, the entropy of the universe increases  ΔS universe o Third Law: the entropy of a perfect crystal at absolute zero (0K) is zero EXAM 2: CHEMICAL KINETICS AND ELECTOCHEMISTRY  Introductory Chemical Kinetics o Chemical kinetics: the study of rates of chemical reactions o Factors that affect rate  Concentration  Temperature  Catalyst: lowers activation energy  Surface area o Rate= ΔC/ΔT  R= Δ[products]/ΔT= -Δ[reactants]/ΔT  For aA+bB=cC+dD −Δ[A]/a −Δ[B]/b Δ[C]/c Δ[D]/d  R= ΔT = ΔT = ΔT = ΔT  Unit= M/s  Rate Laws o Rate Law: the relationship between reactant concentrations and the reaction rate o Rate=k[Reactant 1] [Reactant 2] … m  K= rate constant  Units: o Zero order: M/s st o 1 order: 1/s o 2ndorder: 1/Ms  n and m= reaction orders  to determine reaction order, choose experimental rate/concentration values where only one reactant concentration changes because then R will only be affected by that reactant. Then compare factor changes of these concentration to the factor change in the rate o Zero order: rate is independent of concentration (no effect) o First order[m=1]: as concentration doubles, rate doubles; rate is directly proportional to concentration o Second order[m=2]: as concentration doubles, rate quadruples; rate is directly proportional to concentration squared o If there are no consistent concentrations rate1 k[A] [B] given, plug into rate2 x n k[A] [B]  m+n= overall reaction order; sum of reactant orders  Stoichiometric coefficients insignificant here  Integrated Rate Laws o Integrated rate laws: expressing reactant concentration as a function of time st  1 order: ln[A]= ln[A o-kt  2ndorder: 1/[A]= 1/[A o+kt  Zero order: [A]=[A o-kt o If something is X% decayed, pretend initial= 100M and solve for 100-X as current concentration o If product changes by a certain concentration, add to initial  If a reactant, subtract from initial  Use ratios of change versus moles to determine changes in concentration of other reactants  Half Lives o Half life: the time required for the amount of reagent to half its initial val1/2(t )  1 order: t1/2ln2/k  Each successive half-life is the same as the preceding  2ndorder: t =1/k[A ] 1/2 o  Each successive half-life is double the previous  Zero order: 1/2[A o/2k  Each successive half-life is half the previous  Pseudo “__” Order Reactions o Simplify more complicated reactions to apply integrated rate laws o 2 reactants are much bigger than a 3 in terms of molarity  The smallest reactant is the limiting reactant; change rate law to k’[small reagent]  K’= the given rate law excluding [small reagent]  Plug into corresponding integrated rate law equation and solve for k’  Plug k’ value back into k’=rate law excluding [small reagent] and solve for k  Temperature and Rate o Activation energy: Ea: minimum energy required for a reaction to occur  Activated complex/Transition stat: the highest energy arrangement of atoms that occurs in the course of a reaction  Reactant+reactantactivated complex product+product o Reaction profile  Ea 1 delta E; exothermic 1 −Ea o k=A RT e  lnk=lnA− Ea RT k2 −Ea 1 1  lnk1 = R (T 2− T1)  A: Arrhenius parameter/pre-exponential factor/frequency factor  A= p x z o P: steric factor; correct orientation −EaZ: collision frequency  RT ; exponential factor/fraction: fraction of e molecules that have enough energy to make it over the activation barrier  Reaction Mechanism o Elementary steps: each step in a reaction mechanism that cannot be broken down into simpler steps that add to the overall reaction  For elementary steps to accurately represent reaction, check that mechanism adds up to entire equation AND that rate law matches up (after not using intermediates) o Reaction intermediate: forms in 1 elementary step and is consumed in another; appear in reaction mechanisms but not in overall equation  Cannot be in rate laws o Molecularity: the number of reactant particles involved in an elementary step  Unimolecular: involves one species (isomerization)  Bimolecular: involves 2 species (2Aproduct)  Termolecular: involves 3 species  *The rate law for an elementary step can be deduced from the balanced equation (not for overall reaction)  The molecularity of an elementary step= overall order of the elementary step  Catalysis o Catalyst: a substance that increases the rate of a chemical reaction and is not consumed by the reaction [lowers activation energy]  Homogenous: catalyst and reactant in the same phase  Heterogenous: catalyst in different phase than reactant o  Higher hump: slower step (requires more energy) o Hydrogenation: breaking 2 bond to add to another compound  High activation energy to break, need catalyst  Common catalyst: Pb, Pt, Rn, Ru  Redox and Balancing o Redox: reaction involving electron exchange  Oxidation: losing electron  Reduction: gaining electron  Oxidizing agent: is reduced to allow oxidation  Reducing agent: is oxidized to allow reduction o Balancing half reactions (acidic)  Assign oxidation states to atoms (H= +1, O= -2, lone atoms=0, ions=charge, sum of molecule charges=0)  Divide into half reactions; 1 oxidation and 1 reduction  Balance all elements other than H and O (by coefficients)  Balance Os by adding H 2  Balance Hs by adding H+ (to side opposite of where you just added H2O double the amount of H 2)  Balance charge by adding electrons; sum of charges on both sides should be equal  Multiply both half reactions so that electrons gained= electrons lost (moles electron equal after multiplication)  Add half reactions, cancel electrons and other species  If in basic solution, then add OH- to both sides, to neutralize H+ to H2O (to side with H+, add same number of OH- and change to H2O then cancel if necessary, and add same amount of OH- to other side o Disproportionation: a redox reaction in which the same substance is oxidized and reduced (aka autoredox reaction)  Comproportionation/symproportionation: the opposite of disproportionation; 2 reactants with different oxidation states combine to reach the same oxidation number in the product  Galvanic Cell o Galvanic/Voltaic Cell: electrochemical cell that produces electrical current from spontaneous chemical reaction o  Anode: electrode where oxidation occurs  Cathode: electrode where reduction occurs  Salt Bridge: used to complete circuit and take care of built up charge  Cations migrate to the cathode to neutralize the accumulation of negative charge after copper cations are reduced  Anions migrate to the anode to neutralize accumulation of positive charge after zinc is oxidized o Voltage: potential difference (energy) between anode and cathode  Units: 1 volt= 1 joule/coulomb o EMF: Electron Motive Force: difference in potential that gives rise to an electric current  Cell EMF/Cell Potential/Ecell: difference in potential between electrodes in a specific voltaic cell (intensive property)  Standard EMF/E°cell: cell potential under standard conditions; 1M concentration (aq) or 1 atm pressure (g)  Standard electrode potential: potential of electrode in each half cell  Sum of standard electrode potentials= standard cell potential (if sum is positive, spontaneous and can occur)  Charge difference, tendency for redox, Ecell  EMF depends on concentrations, conditions, and tendencies  Positive Ecell, spontaneous in the forward direction o Chart of standard electrode potentials at 25° Celsius  Based on SHE: standard hydrogen electrode  All values written as reduction; need to reverse voltage value for oxidation half reactions  Do NOT multiply by coefficients since Ecell is an intensive property  F2is strongest oxidizing agent and weakest reducing agent  E°cell value in table, more reduction potential  Li is strongest reducing agent, wants to be oxidized  E°cell value in the table, reduction potential, oxidation potential o To write a reaction for an anode/cathode  Write half reactions both as reductions, including electrons, and find E°cell reduction from tables  Reverse the half reaction that still yields a total +E°cell  Add half reactions and cancel electrons  Cell Potential Equations o E°cell = E°cathode - E°anode  E°cell = E°final - E°initial o ΔG°= -nFE°cell  n= # of moles of electrons  F= Faraday’s constant: quantity of charge on 1 mole electrons  96485 C/mol  equal to charge on one electron x avogadro’s number  q=nF  E°cell must be positive such that ΔG° is negative for spontaneity  Must use joules o Ecell= E°cell – RT/nF ln Q (Nernst equation)  Common equations at 298 K  Ecell=E°cell- .0592/n log Q  E°cell= .0592/n log K o Solving sample problems  Solve for E°cell using E° values of half reactions (reverse values for oxidation and do not multiply stoichiometrically)  Balance the equations to get n/mole number of electrons  Solve for reaction quotient and plug in  Cannot sum E°values of half reactions if electrons do not completely cancel out (for same  Instead, solve for ΔG° with -nFE° for both, and solve for ΔG° total by adding the two and plugging back into -nFE° for total equation for total moles electrons of complete equation  Standard Line Notation o List anode first what releases the electron o Use single line to indicate phase change (solid  aqueous) o Place oxidized ion on other side (and include H+ ions if possible) o Use double line to separate 2 compartments (now cathode) o Place ion (before reduced), then single line for phase change o End with cathode newly reduced substance o If anode/cathode are not solids… cannot start or end with aqueous  Use inert metal electrode, Pt, as cation and anion  Separate with phase change line, then include both reactants and products of anode oxidation half step (do not include 2 O or electrons)  Double line, then list reactants and products of cathode reduction, then single line then Pt  Separate things in same phase by commas o Include (1M) for aq and (1 atm) for gas o Choose anode based on E° values as to get total positive E°cell  Concentration Cell from 2 Half Reactions o General notes  Both half reactions are the same but a difference in concentration drives the current flow  Standard cell potential would be zero  Electrons flow from the dilute half cell forming Cu+2 ions there ([Cu+2] in the dilute half cell), to the concentrated cell to reduce Cu+2 ions ([Cu+2] in the concentrated half cell)  Dilute concentration at anode, that becomes more concentrated with flow of current o Ecell= E°cell- .0592/n ln ([dilute]/[concentrated])  Units: volts EXAM 3: QUANTUM MECHANICS, PERIODIC PROPERTIES, AND BONDING&MOLECULAR ORBITALS  Electromagnetic Spectrum o From low energy to high energy, long wave length to short wave length, low frequency to high frequency…based on wavelength:  Radiowave; 10 -10-1  Microwave; 10 -10-3  Infrared; 10 -105 -6  Visible light; 10 (100 nm)  Ultraviolet; 10 -10 -10  Xray; 10  Gamma Ray; 10 -102 -14  Photon Energy o Planck’s Photon  Planck: energy is released/absorbed only in chunk of some minimum size  E=hv= hc/λ  V= frequency o v= C/λ o v= speed of light/wave length -34  h= planck’s constant (6.63x10 Js)  E= photon energy o Multiply by Avogadro’s to get mole energy o Einstein’s Photoelectric Effect  Photoelectric effect: radiant energy striking a metal surface emits electrons; however, for each metal there is a minimum frequency of light below which no electrons are emitted  Radiant energy is a stream of tiny packets of energy that behave like a tiny particle of light known as a photon  Deduced each photon must have energy proportional to the frequency of light (E=hv) o Bohr’s Explanation of the Line Spectra  A charged particle (electron) that moves in a circular path should lose energy continuously by emitting electromagnetic radiation  Adopted that energies are quantized, an electron in a permitted orbit has a specific energy and is said to be in an allowed energy state  Electrons may absorb light of certain energies and be excited to orbitals of higher energy. Can also emit light of specific energies and fall to orbitals of lower energy o Energy of light emitted or absorbed; ΔE=R (1/n 2 – 1/n ) H 1 2  RH= Rydberg Constant; 2.18 x 10 J8 (energy to ionize an electron in a H atom)  n1= quantum number of lower energy state  n2= quantum number of higher energy state  wavelength emitted when returning to ground state= wavelength absorbed to return o the more quantum states an electron falls, the shorter the wavelength o Explains line spectra of hydrogen series o De Broglie’s Equation of Dual Nature  Dual nature (particle-wave) is not restricted to light; all moving particles have an associated wave  λ= h/mv  m: mass of particle in kg  v: velocity of particle in m/s  Wave properties of large objects are so small they are generally not considered o Heisenberg’s Uncertainty Principle  (Δpx)(Δr)>h/4π  Uncertainty in momentum and in position  It is impossible to know the exact position and momentum of an electron  Orbital Quantum Numbers o Schrodinger Wave Equation: reduces to 3 equations that lead to first 3 quantum numbers n, l, and ml, that describe orbitals as functions  Hψ=Εψ Ψ: wave function H: Hamiltonian operator E: actual energy of electron o Principle Quantum Number: n  Describes energy level of orbital  1, 2, or 3 o Angular Momentum Number: l  Describes shape of orbital  Possible ls from 0 to (n-1)  0 1 2 3 4 5  s p d f g h o Magnetic Quantum Number: ml  Describes orientation of orbital  Possible mls; from –l to +l o Spin Quantum Number: ms  Describes spin direction  Either +1/2 or -1/2 o Orbital possibilities 2  Possible # of orbitals in a principle level= n  Possible # of orbitals in a sublevel= 2l+1 Each orbital can hold 2 electrons  Electron Configurations o Electron configurations: show particular orbitals that are occupied  Electrons occupy lowest energy orbitals available o Orbital diagram: arrows symbolize electron and its direction represents electron spin  All electrons have the same amount of spin, as spin up (+1/2) and spin down (-1/2)  Pauli exclusion principle: no 2 electrons in an atom can have the same 4 quantum numbers; each orbital can only have 2 electrons, each with opposite spins  Aufbau principle: indicates the pattern of orbital filling (from lowest energy to highest) in an atom   Hund’s rule: (of max multiplicity) ground state is attained when electrons fill degenerate identical orbitals first singly with parallel spins, as to maximize spin  Paramagnetism: unpaired electrons are attracted by a magnetic field o The degree of effect is directly related to the number of unpaired electrons  Diamagnetic: all paired electrons, not attracted by magnetic field almost repelled o Configuration Shortcuts and Tricky things  For atoms with a lot of electrons, use noble gas preceding it  [Ne] re1resents full n=2 principle level, start with 3s after  [Ar] represents full n=3 principle level, start with 4s after  [Kr] represents full n=4 principle level, start with 5s after  Use atomic numbers for electron counts  Period 4 and 5 transition metals; NO d-subshells with 4 or 9 in exponent  Must take 1 electron from preceding s subshell to make exponent 5 or 10  For positive ions, write out normal configuration for traditional element, then cancel electrons from closest s orbital  For negative ions, just add one electron to count and write normal configuration  Periodic Table Patterns o Core electrons: in complete principle energy level  Valence electrons: important in chemical bonding For main groups, valence electrons are those in outermost principle energy level o Group #= # of valence electrons For transition elements, also include outermost D electrons o  4 blocks corresponding to filling in quantum sublevels Row # of main group= highest principle quantum number of that element o Atomic Size: distance to outermost electron  Increases down a group (with more electrons)  Decreases across a period (with more protons in nucleus attracting) o Ionization Energy: energy required to remove an electron  Increases across a period (getting closer to stable octet and do not want to lose an electron)  Decreases down a group (bigger electron cloud, electrons not pulled as tightly down nucleus)  Half-filled principle state causes unusual stability and higher ionization energy  Electronegativity: the ability for an electron to attract electrons; follows same pattern o Electron Affinity: energy change associated with gaining an electron  Usually negative because an atom releases energy when gaining an electron  Generally becomes less negative moving down a column (less energy to receive an electron in higher orbitals in group 1A)  Electron affinity increases across a row Electron affinity more positive for half filled principle state because has special stability  Effective Nuclear Charge o Effective Nuclear Charge: Z°/Zeff: electrons can interfere with nuclear attraction o Zeff= Z-σ  Z: actual nuclear charge/atomic number  σ: shielding constant: charge shielded/screened by other electrons o Calculating the shielding constant  Group subshells in order of principle quantum number  (1s) (2s,2p) (3s,3p)(3d)(4s,4p)(4d)(4f)  Electrons to the right of the electron in question contribute zero to the shielding of the nucleus  All electrons in an ns,np group (same quantum number, minus the one electron itself) shield by 0.35 each  All electrons in an n-1 orbital shield by 0.85 each  All electrons in an n-2 orbital (or lower) shield by 1.00 each  If electron is in a d or f shell:  All electrons in the same nd or nf group (don’t include actual electron in count) shield by 0.35 each  All electrons to the left of the d or f shield by 1.00 each  Radial probability functions o Describe electron density at different distances from the nucleus o # of nodes= n-ι-1  Zero probability of finding an electron at the node o Each consecutive hump is taller and broader; greater probability of finding an electron further with each increased level o  Bond Types o Ionic: if the difference in electronegativity is >2, between a metal and a nonmetal (electron transfer) o Pure Covalent: if the electronegativity is the same (electrons shared equally) (between identical elements) o Polar Covalent: if the electronegativities are similar but not the same (electrons unequally shared)  Quantizing Dipole Moments o μ=Qr  μ: dipole moment: created by separating 2 particles of equal but opposite charge  Measured in debye (=3.34x10-3cm)  Q: electron charge magnitude in units C  To find in terms of electron, compare to 1e- = 1.6x10 c ratio  r: distance between 2 equal and opposite charges (in m)  A°= x 10 m10 o % ionic character: measured dipole moment/dipole moment if electrons completely transferred x 100  = measured μ of x-y/ measured μ of x -y x 100  if % ionic character > 50%, it is ionic (large dipole moment)  measure dipole moment of complete transfer by plugging in traditional electron charge and given distance  Born-Haber Cycle o Born-Haber Cycle: a hypothetical series of steps that represents the formation of an ionic compound from its constituent elements/ a thermochemical cycle analyzing factors contributing to ionic compound stability  Lattice Energy: ΔHlatticeergy required to separate one mole of an ionic compound into its constituent gaseous ions (coming together highly exothermic)  Lattice energy is directly proportional to the charge of the ions and inversely proportional to the distance between the ions o Lattice energies are more negative/exothermic with increasing magnitude of ionic charge, and become less negative with increasing ionic radius 1 q1q2 o 4πEo r  The sum of various steps is used to determine ΔHf°  Enthalpies of sublimation to get from solidgaseous reactive state (extensive)  Energies of 1 ionization then 2dionizationion formation from losing electrons (extensive)  Bond enthalpies breaking X2to 2X (extensive)  Electron affinities for accepting electrons to get anions (extensive)  Finally, lattice energy (-ve) for forming from constituent ions  Molecular Geometry Elect Bon Lon Electr Image Molecul Bond ron d e on ar angle grou gro pair geome geometr s ps ups s try y 2 2 0 Linear Linear 180° 3 3 0 Trigonal Trigonal 120° planar planar 3 2 1 Trigonal Bent <120 planar ° 4 4 0 Tetrahe Tetrahedr 109.5 dral al ° 4 3 1 Tetrahe Trigonal <109. dral pyramida 5° l 4 2 2 Tetrahe Bent/Ang <109. dral ular/ 5° Vshaped 5 5 0 Trigonal Trigonal 120°( bipyra bipyrami eq) midal dal 90°(a xial) 5 4 1 Trigonal See saw <120 bipyra °(eq) midal <90°( ax) 5 3 2 Trigonal T-shaped <90 bipyra midal 5 2 3 Trigonal Linear 180° bipyra midal 6 6 0 Octahe Octahedr 90° dral al 6 5 1 Octahe Square <90° dral pyramid al 6 4 2 Octahe Square 90° dral planar o Other notes  Tetrahedral: 4 faces  Deltahedral: triangular face (trigonal bipyramidal and octahedral)  Hypervalent: compounds that can have more than an octet of electrons Access to d orbitals- period 3 P, S, Se, Xe  Compounds satisfied by less than an octet Hydrogen only needs 2 electrons Boron only needs 6 electrons  Polar: if assymetric (v-shaped, trigonal pyramidal)  Radicals: odd electron; stick on central atom  Resonance structures: different insignificant placement of double bond  Put random atom in equatorial when necessary  Applying formal charge to molecular geometry Formal charge: # of valence electrons- # of nonbonding electrons-1/2(bonding electrons) You want the smallest formal charge possible (less accumulation of electrons) Put more charge on the more electronegative atom Sum of formal charges must equal charge of the molecule  Orbital Surfaces o Include charges! o S orbital  o P orbitals  o D orbitals   dx2-y2lobes along x and y axis  dxy dxz dyz lobes along planes (in between axis)  lobes across from eachother are of same charge  dz2 positive lobes, ring is negative o F orbitals  F z3   Negative rings of electrons  Along z axis  Fxyz   Cube with orbitals pointing to 8 corners  Gerade: symmetric with respect to center of inversion (consider charges); s and d orbitals  Ungerade: anti-symmetric with respect to center of inversion; p and f orbitals  Hybridization of Atomic Orbitals o Hybridization: standard atomic orbitals combine to form new orbitals called hybrid orbitals that correspond to the distribution of electrons in chemically bonded atoms  The number of standard atomic orbitals added together = the number of possible hybrid orbitals formed  The particular type of hybridization that occurs is one that yields the lowest overall energy for the molecule  Number of electrons according to exponents in hybrid orbitals= steric number of corresponding molecular geometry o Types 3  Four sp orbitals; from one 2s and three 2p orbitals; for tetrahedrals     Three sp orbitals; from one s and 2 p orbitals; for trigonal planar   Sigma bond: σ: overlapping, head on interaction (same axis)  Pi bond: π: side by side interaction (different axis) o For double and triple bonds o Check pi bonds to see if molecule is planar or not; if 2 pi bonds, repel each other and no longer planar  Two sp orbitals; from one s and one p orbital; for linear     dsp    orbital from 1 s orbital, 3 p orbitals, 1 d orbital; for  trigonal bipyramidal   d2sp3 orbital from octahedral  o d orbitals (transition metals) can form sigma bonds, pi bonds, AND delta bonds δ;  Face on bonding between 2 nodes; between 4 lobes of 2 d orbitals  Molecular Orbital Theory o Linear Combination of Atomic Orbitals  Bonding orbital: σ: orbital is lower in energy than either of the 2 1s orbitals  Constructive interference; orbitals in same phase, causing high electron density in internuclear region  Anti-bonding orbital: σ*: indicated by star  Destructive interference; summing one orbital and another in opposite (negative) phase; creates a node in internuclear region  Destabilizes  Bond order: # of electrons in bonding molecular orbitals- # e s in antibonding molecular orbitals/2  If positive, more electrons in bonding orbitals, lower energy than in isolated atomic orbitals, chemical bonds will form  The higher the bond order, the stronger the bond, the shorter the bond length  If negative or zero, indicates a bond will not form between atoms  Representation   Number of total molecular orbitals formed will equal the number of atomic orbitals in the set  Fill lowest energy first  o Period 2 Homonuclear Diatomic Molecules   2pz and 2py are the same just 90° rotation  σ-π Crossover  2s and 2 p are very close, orbital mixing can occur between orbitals of similar energy and symmetry  for 2 ,2C , and2N ; large 2s-2p interaction, 2p σ bonding (not anti) is higher energy than 2p π bonding (switch for orbital filling) o Period 2 Heteronuclear Diatomic Molecules  The more electronegative the atom, the lower the energy of the atomic orbitals  When 2 atomic orbitals are not identical (from different elements), the lower energy orbital makes a greater contribution to the bonding molecular orbital  The higher energy atomic orbital makes a greater contribution to the anti-bonding molecular orbital  In HF, Fluorine is so electronegative that all atomic orbitals are lower in energy than hydrogen atomic orbitals   Fluorine’s 2s orbital is so low in energy compared to hydrogen’s 1s, does not contribute to molecular orbitals  Molecular orbitals of HF approximated by linear combination of fluorine 2px orbital and hydrogen 1s orbital o Other 2p orbitals remain localized on fluorine and are nonbonding orbitals o Homo: highest occupied molecular orbital o Lumo: lowest unoccupied molecular orbital


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