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What’s coming up??? • • • • • • • • • • • • • • • Oct 25 Oct 27 Oct 29 Nov 1 Nov 3,5 Nov 8,10,12 Nov 15 Nov 17 Nov 19 Nov 22 Nov 24 Nov 26 Nov 29 Dec 1 Dec 2 The atmosphere, part 1 Ch. 8 Midterm … No lecture The atmosphere, part 2 Light, blackbodies, Bohr Postulates of QM, p-in-a-box Ch. 8 Ch. 9 Ch. 9 Hydrogen atom Ch. 9 Multi-electron atoms Periodic properties Periodic properties Valence-bond; Lewis structures Hybrid orbitals; VSEPR VSEPR MO theory MO theory Review for exam Ch.10 Ch. 10 Ch. 10 Ch. 11 Ch. 11, 12 Ch. 12 Ch. 12 Ch. 12 PARTICLE IN A BOX (x) A sin kx ENERGY d 2 E 2 8 m dx 2 h 2 (0) = 0 BOUNDARY CONDITION (L) = 0 0 x L PARTICLE IN A BOX 2 2 2 2 ENERGY n=3 h 3 E3 2 8 mL 2 2 n=2 n =1 h 2 E2 2 8 mL 2 h E1 2 8 mL 3 hn En 2 8 mL 2 1 QUANTIZED PARTICLE IN A BOX The application of the BOUNDARY CONDITIONS Gives a series of QUANTIZED ENERGY LEVELS ONLY CERTAIN ENERGIES ALLOWED! DETERMINED BY THE NUMBER n 2 2 hn En 2 8 mL n 1, 2, 3....... n is a QUANTUM NUMBER THE WAVEFUNCTIONS n(x) 2 n sin x L L ENERGY n=3 3 2 n=2 n =1 1 A NODE CHANGES SIGN THE NUMBER OF NODES IS GIVEN BY N-1 ….. INCREASING NUMBER OF NODES AS THE ENERGY INCREASES What does Mean????? The answer lies in WAVE-PARTICLE DUALITY Electrons have both wavelike and particle like properties. Because of the wavelike character of electron we CANNOT say that an electron WILL be found at certain point in an atom! THE HEISENBERG UNCERTAINTY PRINCIPLE He postulated that if... Dx is the uncertainty in the particle’s position Dp is the uncertainty in the particle’s momentum h DxDp 4 For a particle like an electron, we cannot know both the position and velocity to any meaningful precision simultaneously. PARTICLE IN A BOX FOR ONE DIMENSION SCHRODINGER EQUATION d n ( x ) 2 En n ( x ) 2 8 m dx 2 h 2 2 2 hn En 2 8 mL 1-D REQUIRES ONE QUANTUM NUMBER! THREE DIMENSIONS SCHRODINGER EQUATION 2 h { 2 2 2 d d d + + n ( x , y , z ) Enx ny nz nx ny nz ( x , y , z ) 2 2 2 2 8 m dx dy dz } 2 2 nx 2 ny 2 nz + h + Enxnynz 8m Lx Ly Lz ( ) 3-D REQUIRES THREE QUANTUM NUMBERS!!! 2 2 nx 2 ny 2 nz + h + Enxnynz 8m Lx Ly Lz Note that when Lx=Ly=Lz E2,3,4 = E 3,2,4 = E4,2,3 These levels are said to be degenerate – this means they are different wavefunctions, but have the same energy z HYDROGEN ATOM y Electron at (x,y,z) Proton at (0,0,0) x The hydrogen atom is composed of a proton (+1) and an electron (-1). If the proton is located at the origin, the electron is at (x, y, z) • We want to obtain the energy of the hydrogen atom system. We will do this the same way as we got it for the particle-in-a-box: by performing the “energy operation” on the wavefunction which describes the H atom system. H E • Remember that this equation is called the Schrodinger wave equation (SWE) • H = KE operator + PE operator H= (h2 / 82m) 2 +V • H E (h2 / { Where 82m) 2 2 + V } E = {d2/dx2 + d2/dy2 +d2/dz2} • Now there is a difference from the particlein-a-box problem: here there is a potential energy involved …. • Coulombic attraction between the proton and electron • V = Ze2 / r Since the potential energy depends on the separation between the proton and the electron, it is more convenient to think about the problem using a different co-ordinate system: (x,y,z) (r,q,j) z y Electron at (r, q, j) Proton at (0,0,0) x After the transformation we still have the Schrodinger equation 2 2 2 { (h / 8 m) + V } E 2 Where now has terms in {d2/dr2 ; d2/dq2 ; d2/dj2} and V = Ze2 / r • The result of solving the Schrodinger equation this way is that we can split the hydrogen wavefunction into two: (x,y,z) (r,q,j) = R(r) x Y(q,j) Depends on r only Depends on angular variables • The solutions have the same features we have seen already: – Energy is quantized • En = R Z2 / n2 = 2.178 x 10-18 Z2 / n2 J [ n = 1,2,3 …] – Wavefunctions have shapes which depend on the quantum numbers – There are (n-1) nodes in the wavefunctions • Because we have 3 spatial dimensions, we end up with 3 quantum numbers: n, l, ml • n = 1,2,3, …; l = 0,1,2 … (n1); ml = l, l+1, …0…l1, l • n is the principal quantum number – gives energy and level • l is the orbital angular momentum quantum number – it gives the shape of the wavefunction • ml is the magnetic quantum number – it distinguishes the various degenerate wavefunctions with the same n and l n l ml 1 0 (s) 0 2 0 (s) 1 (p) 0 -1, 0, 1 3 0 (s) 1 (p) 2 (d) 0 -1, 0, 1 -2, -1, 0, 1, 2 •En = R Z2 / n2 = 2.178 x 10-18 Z2 / n2 J [ n = 1,2,3 …] … degenerate The result (after a lot of math!) Node at s = 2!! Probability Distribution for the 1s wavefunction: 3/2 1 1 -r/a0 100 e a0 Maximum probability at nucleus A more interesting way to look at things is by using the radial probability distribution, which gives probabilities of finding the electron within an annulus at distance r (think of onion skins) max. away from nucleus 90% boundary: Inside this lies 90% of the probability nodes