04CHM114_0216

04CHM114_0216 - Exm TH t Fe 3 m Reman ou wEJ 93¢ my CHEM...

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Unformatted text preview: Exm TH t: Fe: 3 m, Reman- ou wEJ: 93¢. my- CHEM 114. Spring 2004 Monday FebEh 2004 m Lecture 12 auhbfim CHAPTER 3 : DEVELOPING A MODEL FOR THE ATOM 3. Mo. (CONTINUED) LAST LECTURE: (A) Developed de Broglie's concept of electrons (1) as MATTER waves (2) when trapped by positive charge (nucleus), as waves localized in 3D space. THE ATOM AS A MUSICAL INSTRUMENT THE EQUIVALENT OF TIGHTENlN G THE STRING? Electrostatic tension Coulomb's law: E = qlqzlr : q1 =charge 011 electron, q2 = charge at nucleus H vs He" (each has one 6-, but He+ has two protons: therefore higher tension) ' DRUMS vs STRINGS: ........ .. WHY WE DIG DRUMS 2D VS ID COMPLEXITY AND RICHNESS The ATOM IS A 3D RESONATOR: LIKE A MINIATURE, extemely miniature, BOOMBOX STRINGS 1D Icharacteristic no. (11) DRUMS 2D 2 ” "s (n, 1) ATOMS 3D 3 " "s (n,l,m) ,H ,n.-- .3”... - -. A .-L :rnyrknhg.$ -r'|r?t_lb_,,.. , .u .-...- ...- _-I_.3 . x SCHRODINGER AND THE FINAL ANSWER TO THE RIDDLE OF THE H ATOM : the Schrodinger wave equation. Eur = H1]! simple enough? it an Meat is in the H ( Hamiltonian operator) III is a “wave function” which describe the "amplitude" of the electron in a “stationary” state (one that doesn’t eliminate itself by destructive interference). These stable solutions of the wave equation give allowed v w. .J :i'nfig‘x":"'- . n 0 H energIes of electron Ol‘bltalS , (the word which replaces the “orbits” of the Bohr model). Electron "distributions" obtained from : é17:13 (wave function)2 vs r (from nucleus) e.g. for the ground state in which n = 1, we obtain V I5 3 1| SW z 1. FIE. a. g an? (P;rok.al:'l g) r f \' 41tr21|g2 I!“ (I‘M t“ are.“ I'm" (aw-Math P‘ '3‘: P70“. is rel-M Put . 1‘ Distance from the nucleus a-‘hm [5. Bst 5w LI? Pct-Ah: Different possible STATES of hydrogen ARE DISTINGUISHED BY THEIR WAVE FUNCTIONS qr (pronounced psi) Solutions of Schrodinger's equation for n > 1 describe the energies and positions of the electron wave for different excited are harmonics of the fundamental ground state of this system. Different possible harmonics, called orbitals, are identified by sets of 3 quantum numbers (n, 1, In) related by rules: l.®can never be zero or neg., but all other values are possible [email protected]’can never be equal to n, but is always giro or Qositive Q9 n. t 3 3 @c be equal to, but never greater than, a, and can be pos. or neg. as well as h 1: 7‘ ‘ -.-.. O | s ‘ ‘5 zero. Typical question: "which of the following sets of quantum numbers is impossible" M :- _ .- "- ‘ i, o. I ‘ 1 h ORBITAL SHAPES: w?- and 4msz and angular parts. 0 Orbitals with Q_.N.s n,0,0 are radially symmetrical. THEY ARE CALLED "s" ORBITALS. THUS 13, 35, 45 239.0 23 55.": u: uwr'p' pm} Number of nodes! is equal to the principal quantum number for s orbitals .- s k” ‘u-“ o . £ - o C P {am “Rafi-‘3‘) 0 Orbitals with l > 0 have special shapes: For 1 - 1 (11 must be 2 or more) there are three orbitals (m = 1,0, -1). they look like dumbells (2 "lobes") at right angles to each other.... I'CALLED "p" ORBITALS. Thus 2p, 3p, etc «a (113 is impossible typical question : "Which o _ e following orbitals designations must be wrong? -, -, 1p, - -) 0 For 1 = 2 (11 must be 3 or‘ higher) they are more complicated, e.g. 4 lobes. They are called "(1" orbitals. There are five of them (m = 2,1,0,-1,-2). 3d orbitals are responsible for the color or capper minerals, and ruby. QUESTION: 1. Malachite is a copper mineral typical of Arizona, which has deep green color. The color is due to electrons in orbitals of type (x, y, s, f, d) 2. Which type of orbital typically has electron density distributed in four different lobes? Summary: for each value of the principal quantum number there can be 1 s orbital 3 p orbitals 5 d orbitals 7 f orbitals with the restriction that the total number of orbitals for a given shell = n2 Thus 11 = l ls only: total 1 11:2 lZsancl3 2p total4 n=3 13s,33pand5 3d total9 0 TOTAL niumber of orbitals in a MAIN SHELLadefined by principal quantum number n (wave mechanical equivalent of the simple Bohr’s orbit) is n2 QUESTION: Which of the following is the total possible number of orbitals, all starting with the same quantum number, when the principal quantum number (n) is 4? ( -, -, -, -, l6) '\ Ca Nwl' ‘MUW' 0 n,l combination defines SUBSHELL, e. g. 3d or 4p. 0 SPIN QUANTUM NUMBER ELECTRONS BEHAVE LIKE THEY ARE SPINNING. THEREFORE HAVE MAGNETIC PROPERTIES. the spin may be either up or down corresponding to spin Q,N. (1115) = +l/2 or - 1/2. Thus in total, four quantum numbers are needed to specify the state of a given electron in an orbital around the proton nucleus. Orbital degeneracy ' In the hydrogen atom - with no magnetic fields acting on it - all orbitals with a common value of n, have the same energy. e. g n = 5 there are 25 different orbitals...put the electron in to any one of them and it will have the same energy. We say the orbitals are "degenerate". With magnetic fields - or in any atom containing more than one e“, the degeneracy is "lifted". The sharp lines are split. (That’s why the lines of Bohr’s atom split into groups of lines when the field turned on, to Bohr’s dismay). OK, NOW WE UNDERSTAND HYDROGEN, AND THE TRUE ORIGIN OF THE BALMER SERIES OF THE HYDROGEN EMISSION SPECTRUM. TIME TO GO FURTHER. LET'S LOOK AT HIGHER ELEMENTS. THE MULTIELECTRON ATOM AND THE PERIODIC TABLE. Here the idea is to preserve the energy level scheme worked out for the hydrogen atom (assume the same scheme applies to atoms with more than one proton at the center and an equal number of electrons to neutralise the positive charge). AUFBGO PR urn! . The we find out where the electrons are by adding the total elctron count according to rules, as follows: 1. 2; go into the lowest energy levels available f _2_;_The individual orbitals can only hold two e— maximum (Pauli exclusion pn'nciple: no two e- can have the same four m) 3 m“m *‘I. I 1" ‘hr 3. When orbitals are de nerate, the e- occupy them singly till all are half-filled 1m;- rule multiElicity“ UGH! I. O. K. let's go! 1. Electronic formula \3 2. He @‘S 3. Li (alkali metal) shielding of nuclear charge Moseley Text page p. 78 4. Be 5. B 6. C 7. N paramagnetism 8.0 9.F 10 ...
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04CHM114_0216 - Exm TH t Fe 3 m Reman ou wEJ 93¢ my CHEM...

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