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Chemistry 111 Week 9 Lecture Notes

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Chemistry 111 Week 9 Lecture Notes 111/40551

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- chapter 7: atomic spectra, bohr's model of the hydrogen atom, Beer's law, de Broglie equation, Heisenberg's uncertainty, quantum mechanical model, quantum numbers and atomic orbitals, shapes of o...
General Chemistry I
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This 10 page Class Notes was uploaded by Notetaker on Friday November 6, 2015. The Class Notes belongs to 111/40551 at University of St. Thomas taught by Uzcategui-White in Summer 2015. Since its upload, it has received 14 views.


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Date Created: 11/06/15
Chemistry  Week  9  Lecture  Notes:   7.10   An  X  Ray  has  a  a  wavelength  of  1.3  Angstroms.  Calculate  the  energy  (in  J)  of  one  photon  of  this   radiation.   Y  =  ?                      c=  y  x  v  à  c/y   E=  hv   E=  h  x  c/  Y  à  6.626  x  10  ^  -­‐34  J  s  x  3.00  x  10^  8  m/s  /  1.3  A  x  10  ^-­‐10  m/  1  A  =  1.5  x  10^-­‐15  J/   photon   7.13   A  gamma  ray  emitted  by  Cobalt  60  has  an  energy  of  1.33  MeV  (million  electron  volts,  1  eV=  1.60   x  10^-­‐19  J).  What  is  the  frequency  (in  Hz)  and  the  wavelength  (in  m)  of  the  ray?   E=  1.33  MeV   V=  ?                                                                               Y  (wavelength)  =  ?   E=  hv  à  v  =  E/  h  -­‐à  v  =  1.33  MeV  (10^6  eV/  1  MeV)  (1.602  x  10^-­‐19  J/  1  eV)/  6.626  x  10^-­‐34  J  s   V=  3.22  x  10^20  s^-­‐1  (Hz)   Y  =  c/v  -­‐à  3.00  x  10^8  m/s/  3.22  x  10^20  1/s  =  9.32  x  10  ^-­‐13  m   Section  7.2:  Atomic  Spectra-­‐   o   Particle/  atom  (energy)  à  “excited”  (move  electrons  to  higher  energy  level)  à  give   away  extra  energy  à  atomic  spectra  (only  see  colored  lines)   o   Atomic  spectra:  line  spectrum  of  different  wavelengths   o   Line  emission  spectra:  series  of  lines  with  black  in  between  (Balmer  series)   o   Equation  =  Rydberg  Equation:  (predict  different  wavelengths,  n2  >  n1)   1  /  Y  =  R  (1/  n1^2  –  1/n2^2)   R=  Rydberg  constant=  1.096776  x  10^7  m^-­‐1   Bohr’s  Model  of  the  Hydrogen  Atom:   o   H  atom  only  has  certain  energy  levels=  stationary  states   o   Higher  the  energy  level,  farther  from  the  nucleus   o   H  atom  in  first  orbit=  lowest  energy=  ground  state   o   In  stationary  state-­‐  does  not  radiate  energy   o   Atoms  changes  states  by  emitting/  absorbing  photons   o   Electrons  fall  from  high  energy  à  lower  =  emit  light   o   Light  emitted  by  excited  atoms=  specific  lines  at  certain  wavelengths   o   Get  excited  by  giving  energy  through  heat  or  electricity   •   Only  true  for  Hydrogen  atoms   •   Equation:     E=  -­‐2.18  x  10^-­‐18  J  (z^2/n^2)      z  =  1   Change  in  E  =  -­‐2.18  x  10^  -­‐18  J  (1/  n^2  final  –  1/  n^2  intial)   •   Conditions:   a)   Emitting  energy:  electrons  move  closer  to  the  nucleus  change  in  E  <  0   b)   Absorbing  energy:  electrons  move  away  from  the  nucleus  change  in  E  >  0   What  do  we  use  excited  electrons  for?   1.   Visible  light   2.   Identify  atoms  (using  wavelength)   3.   Quantify  color  solutions  (energy  absorbing/emitting  light)   Beer’s  law:   A  =  E  b  c   A=  absorption  of  light,  E  =  constant  (in  problem),  b=  1  cm  (constant),  c  =  concentration   7.19   Use  the  Rydberg  Equation  to  find  Y  (in  nm)  of  photon  emitted  when  an  H  atom  undergoes  a   transition  from  n=  5  to  n=2   Y=  ?     1/Y  =  1.096776  x  10^7  m^-­‐1  x  (1/  2^2  –  1/  5^2)  =  2303229.6  m^-­‐1   Y=  1/  2303229.6  1/m  x  1  x  10^-­‐9  m/  1  nm)  =  434.17  nm   7.21     Calculate  the  energy  difference  for  1  mol  of  H  atoms  (using  #19)   E=  -­‐2.18  x  10^-­‐18  J  (1/  2^2  –  1/5^2)=  -­‐4.58  x  10^-­‐19  J/  photon   -­‐4.58x  10^-­‐19  J/  photon  x  6.022  x  10^23  photon/  1  mol  =  -­‐2.76  x  10^5  J/  mol   7.23     Arrange  the  H  atom  electron  transitions  in  order  of  increasing  frequency.   a)   N=  2  to  n=  4   b)   N=  2  to  n=1   c)   N=2  to  n=5     d)   N=  4  to  n=3     •   The  bigger  the  difference=  the  more  energy=  higher  frequency   •   Wavelength  =  opposite   Answer:  D  <  A  <  C  <  B   7.26   An  electron  in  the  n=  5  level  of  an  H  atom  emits  a  photon  of  wavelength  1281  nm.  To  what   energy  level  does  it  move?   N=  5   N=  ?   Y  =  1281  nm  à  1281  nm  x  1  m/  1  x  10^-­‐9  nm  =  1.281  x  10^-­‐6  m   Emits  =  moves  down;  n2  =  5   1/  (1.281  x  10^-­‐6  m)  (1.099776  x  10^7  1/m)  =  .07118=  1/  n1^2  –  1/25   1/n1^2  =  .1118   n1^2=  1/  .1118  à  8.994   n1=  3   Section  7.4:  The  wave-­‐  particle  duality  of  matter  and  energy   Louis  de  Broglie:   -­‐   Matter  is  wave-­‐  like   -­‐   Energy  à  wave  à  matter  of  particle   Y=  h/  m  x  U   Y=  wavelength  (J  x  s),  m=  mass  (in  kg),  U=  velocity  of  particle  (m/s)   -­‐   Wavelength  is  inversely  proportional  to  mass   -­‐   We  do  have  a  wavelength  in  our  movement  à  hard  to  see   -­‐   Mass  increases,  speed  increases,  wavelength  decreases   o   Momentum:     Y=  h/P  (P=  momentum,  P=  m  x  U)                            de  Broglie/  Einstein  Equation   -­‐   Shorter  wavelengths  =  higher  energy=  higher  momentum   -­‐   Shorter  wavelengths  =  more  dangerous   Heisenberg’s  Uncertainty:   o   “It  is  impossible  to  know  simultaneously  the  position  and  momentum  (mass  x  speed)  of   a  particle”.   Change  in  X  x  mass  x  change  in  U  >  h/  4π   U  =  uncertainty   X=  position   7.31     A  232  lb  fullback  runs  40  yards  at  19.8  +-­  .1  mi/h   a)   what  is  the  de  Broglie  wavelength  (in  m)?   b)   What  is  the  uncertainty  in  his  position?   a)   Y=  h/  m  x  U   M=  232  lb  x  1  kg/  2.205  lbs  =  105  kg   U=  19.8  mi/  h  x  1  km/  .620  mi  x  1  hr/  3600  s  x  1000  m/  1  km  =  8.87  m/s   Y  =  6.626  x  10^  -­34  J/s/  105  kg  x  8.87  m/s  =  7.10  x  10^  -­37  m   b)   X  x  m  x  U  >  h/  4π   U  =  .1  mi/  h  x  1km/  .620  mi  x  1000  m/  1  km  x  1  hr/  3600  s  =  5  x  10^2  m/s   X  =  6.626  x  10^-­34  J  /  105  kg  x  5  x  10^2  m/s  x  4 π =  1  x  10^-­35  m   The  Quantum-­  Mechanical  Model  of  the  Atom:     o   The  matter  wave  of  the  electron  occupies  the  space  near  the  nucleus  and  is   continuously  influenced  by  it.   o   The  Schrodinger  wave  equation  allows  us  to  solve  for  the  energy  states   associated  with  a  particular  atomical  orbit   HW  =  EW   W^2  =  probability  of  finding  electrons  (electron  density)   Orbitals  à  location  of  an  electron  (regions  of  space  within  an  electron)   Quantum  Numbers  and  Atomic  Orbitals:   o   Quantum  number  is  used  to  identity  energy  states  and  orbitals   o   Orbital  specified  by  3  quantum  numbers:   1)   Principal  quantum  number  (n):  positive  integers   -­‐   Indicates  the  relative  size  of  the  orbital  and  its  distance  from  the  nucleus   -­‐   Electron  shell  or  energy  level   2)   Angular  Momentum  Quantum  Number  (l):  integer  #  that  goes  from  0  to  n-­1   -­‐   Value  l  indicates  the  shape  of  the  orbital   -­‐   Subshell  or  sublevel   L  =  0  à  ‘s’  sublevel   L=  1  à  ‘p’  sublevel   L=  2  à  ‘d’  sublevel   L=  3  à  ‘f’  sublevel   3)   Magnetic  Quantum  Number  (ml):  value  of  –l,  0,  +l   -­‐   Indicates  the  spatial  orientation  of  the  orbitals  (#  of  orbitals)   N=  1   L=  0   Ml  =  0   N=  2   L=  0,1     Ml  =  0,  -­1,  0,  +1  (4  total  orbitals)   Shape  of  Orbitals   o   ‘s’  orbital  =  spherical   o   node  region  =  0  probability  of  finding  electrons   o   3s  >  2s  >  1s  (in  size)   o   angular  number  (l)   -­‐   l=  0  à  spherical   -­‐   ‘p’  sublevel  (dumbbell)  à  two  regions  of  high  electron  density   -­‐   1  node   -­‐   ml  =  -­1,0,+1   o   l=  0  à  s  sublevel   o   l=  1à  p  sublevel     o   l=  2  à  d  sublevel   o   l=3  à  f  sublevel   o   d  sublevel:     -­‐   l=  2  à  m1=  -­2,  -­1,  0,  1  ,  2  (s  orbitals/regions  of  high  electron  density)   -­‐   4  clover  leaf,  1  donut  shaped   -­‐   both  have  2  nodes     o   n=  level  of  energy  (size  of  orbital)     o   bigger  n  value  =  bigger  orbital  ex:  4p  >  3p  >  2p  >1p   7.45   Give  the  sublevel  designation,  the  ml  values  and  the  #  of  orbitals   a)   n=  2,  l=0     sub:  2s  ml=  0  orb  #  =  1   b)   n=3  l=2   sub:  3d  ml=  -­2,  -­1,  0,  1,  2  orb=  5   c)   n=5  l=1   sub:  5p  ml=  -­1,  0,  1  orb=  3   7.49   Are  the  following  combinations  allowed?  If  not,  fix  them.   a)   N=1,  l=  0,  ml=0     correct   b)   N=2,  l=2,  ml=  +1   Wrong;;     N=2,  l=0,  ml=0   N=2,  l=1,  ml=-­1   c)   N=7,  l=1,  ml=  +2   Wrong;;   Ml=  -­1  or  0  or  +1   d)   N=3,  l=1,  ml=  -­2   Wrong;;   Ml=  -­1  or  0  or  +1   Or   N=3,  l=2,  ml=  +2   Chapter  8:  Electron  Configuration  and  Chemical  Periodicity:   Spin  quantum  number:     o   Symbol=  ms   o   Values=  +1/2,  -­1/2   o   Property:  direction  of  spin   Pauli  Exclusion  Principle:   o   no  two  electrons  in  the  same  atom  can  have  the  same  4  quantum  numbers     o   an  atomic  orbital  can  hold  a  maximum  of  2  electrons  and  must  have  opposing   spin  values   Energy  Level  Splitting:   o   electrostatic  interactions  between  electrons  with  atoms  with  more  than  one   electron  causes  the  energy  levels  to  split  into  sublevels.     o   The  electrostatic  effects  include:   -­‐   Effect  of  nuclear  charge  (Z)   -­‐   Effect  of  electron  repulsion  and  shielding     -­‐   Effect  of  orbital  shape  on  orbital  energy  (penetration)   1.   Nuclear  charge:   -­‐   Tendency  of  nucleus  to  attract  electrons   -­‐   The  closer  the  electrons  are  to  the  nucleus,  the  more  stable  they  are   -­‐   The  higher  the  nuclear  charge  à  lower  energy/  energy  level   2.   Electron  Repulsion/  shielding:   -­‐   Shielding  effect:  preventing  attraction  to  nucleus   -­‐   Attraction  stronger  when  no  other  electrons  are  present   -­‐   Shielding  by  inner  electrons  greatly  lowers  the  full  nuclear  charge  (Z)  to  an   effective  nuclear  charge  (Zeff)                


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