This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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
\'
41tr21g2 I!“
(I‘M t“ are.“ I'm" (awMath
P‘ '3‘: P70“.
is relM Put . 1‘ Distance from the nucleus a‘hm [5. Bst
5w LI? PctAh: 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 halffilled 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 ...
View
Full Document
 Fall '08
 AKSYONOV
 Chemistry

Click to edit the document details