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The Final Exam

The Final Exam. December 13 (Monday) 9:00 – 12:00 Cumulative (covers everything!!) Worth 50% of total mark Multiple choice. The Final Exam. From my portion, you are responsible for: Chapter 8 … material from my lecture notes Chapter 9 … everything Chapter 10 … everything

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The Final Exam

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  1. The Final Exam • December 13 (Monday) • 9:00 – 12:00 • Cumulative (covers everything!!) • Worth 50% of total mark • Multiple choice

  2. The Final Exam • From my portion, you are responsible for: • Chapter 8 … material from my lecture notes • Chapter 9 … everything • Chapter 10 … everything • Chapter 11 … everything • Chapter 12 … everythingexcept 12.7

  3. The Final Exam • You will need to remember • Relationship between photon energy and frequency / wavelength • De Broglie AND Heisenberg relationships • Equations for energies of a particle-in-a-box AND of the hydrogen atom • VSEPR shapes AND hybribizations which give them

  4. My office hours next week • Wednesday Dec 8: 10-12AND2-4 • Friday Dec 10: 10-12 AND 2-4

  5. Planck Postulated “Energy can only be transferred in discrete quantities.” n is the frequency of the energy h is Planck’s constant, 6.626 x 10-34 J s. Energy is not continuous Energy is quantized “forced to have only certain discrete values” Planck…….

  6. THE PHOTOELECTRIC EFFECT KE of electron light electron metal n0 Frequency of light (n) When n <n0, no electrons are ejected at any light intensity. When n >n0, the number of electrons is proportional to the light intensity. KE of the ejected electrons depends only on the light’s frequency This lead Einstein to use Planck’s idea of quanta

  7. EINSTEIN POSTULATED “Electromagnetic radiation can be viewed as a stream of particle-like units called photons.” The energy of the photon depends upon the frequency Energy of a Photon:

  8. Diffraction of an electron beam…. We can relate these spacings to the electron wavelength De Broglie:

  9. SPECTROSCOPY EMISSION Sample heated. Many excited states populated

  10. n = Ion ... 8 } n = 4 Excited states n = 3 n = 2 Energy n = 1 Ground state SPECTROSCOPY EMISSION Sample heated. Many excited states populated The spectrum…..

  11. We will describe atoms and molecules using wavefunctions, which we will give symbols … like this: Y • These wavefunctions contain all the information about the item we are trying to understand • Since they are waves, they will have wave properties: amplitude, frequency, wavelength, phase, etc.

  12. We obtain the energy by performing the “energy operation” on the wavefunction – the result is a constant (the energy) times the wavefunction HY = EY • This equation is called the Schrodinger wave equation (SWE) • Let’s see how this might work

  13. + V 2 2 -(h2 / 8p2m) • So H = KE operator + PE operator H = • HY = EY Y = EY { + V } -(h2 / 8p2m)

  14. PARTICLE IN A BOX Y (x) = A sin kx ENERGY Y(0) = 0 Y(L) = 0 L 0 x BOUNDARY CONDITION

  15. PARTICLE IN A BOX The SCHRODINGER WAVE EQUATION for a for a 1-D particle with no potential acting on it!

  16. THE HEISENBERG UNCERTAINTY PRINCIPLE Dx is the uncertainty in the particle’s position Dp is the uncertainty in the particle’s momentum For a particle like an electron, we cannot know both the position and velocity to any meaningful precision simultaneously.

  17. What doesY 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!

  18. p n (x) Y = A sin x n L PARTICLE IN A BOX The probability of finding the particle on a segment of the x-axis of length dx surrounding the point x is…. If we sum all of these infinitesimal probabilities, the total must be equal to one, since there must be some finite probability of finding the particle somewhere..

  19. L 0 ENERGY OF A PARTICLE IN A BOX ENERGY INTEGRAL NUMBER OF HALF-WAVELENGTHS n = 3 n = 2 n =1

  20. PROBABILITIES Y2 ENERGY n = 3 ZERO NO CHANCE OF FINDING ELECTRON n = 2 n =1 AND….

  21. z y Electron at (r, q, j) 2 2 Proton at (0,0,0) x After the transformation we still have the Schrodinger equation { Y = EY + V } -(h2 / 8p2m) Where now has terms in {d2/dr2 ; d2/dq2 ; d2/dj2} and V = -Ze2 / r

  22. The result of solving the Schrodinger equation this way is that we can split the hydrogen wavefunction into two: Y(x,y,z)  Y(r,q,j) = R(r)xY(q,j) Depends on angular variables Depends on r only

  23. The solutions have the same features we have seen already: • Energy is quantized • En = -RZ2 / 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

  24. Because we have 3 spatial dimensions, we end up with 3 quantum numbers: n, l, ml • n = 1,2,3, …; l = 0,1,2 … (n-1); ml = -l, -l+1, …0…l-1, 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

  25. n l ml 1 0 (s) 0 2 0 (s) 0 1 (p) -1, 0, 1 3 0 (s) 0 1 (p) -1, 0, 1 2 (d) -2, -1, 0, 1, 2

  26. The result (after a lot of math!)

  27. 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

  28. The Radial Probability Distribution for the 3s, 3p, and 3d Orbitals

  29. Another quantum number! Electrons are influenced by a magnetic field as though they were spinning charges. They are not really, but we think of them as having “spin up” or “spin down” levels. These are labeled by the 4th quantum number: ms, which can take 2 values.

  30. 4d 5s 4p 3d 4s 3p E 3s 2p 2s 1s THE MULTI-ELECTRON ATOM ENERGY LEVEL DIAGRAM Remember the energies are < 0

  31. THE PAULI PRINCIPLE No two electrons in the same atom can have the same set of four quantum numbers (n, l, ml , ms). An orbital is described by three quantum numbers, Then each electron in a given orbital must have a different ms HOW MANY ELECTRONS IN AN ORBITAL? each orbital may contain a maximum of two electrons, and they must have opposite spins.

  32. THE ELECTRON CONFIGURATIONS FOR NITROGEN TO NEON Nitrogen Oxygen Fluorine Neon 1s 2s 2p N: 1s22s22p3 1s 2s 2p O: 1s22s22p4 1s 2s 2p F: 1s22s22p5 1s 2s 2p Ne: 1s22s22p6

  33. SCREENING AND PENETRATION Is 3p THE 1s close to the nucleus PENETRATES WELL SEES A CHARGE OF Z=2 THE 3p DOES NOT SCREEN THE NUCLEUS EFFECTIVE NUCLEAR CHARGE…...

  34. THE 3s orbital penetrates better than 3p orbital The 3p orbital penetrates better than 3d orbital Zeff(s) > Zeff(p) > Zeff(d) 3s 3p 3d

  35. TRANSITION METALS the metals that fill the d orbitals in their valence shell. WHEN n=3 HUND’S RULE OBEYED FOR ALL EXCEPT Cr and Cu Cr: [Ar] 4s23d4 EXPECTED Cr: [Ar] 4s13d5 …. The d-shell is ½ filled this way; all spin up OBSERVED FOR COPPER

  36. 1 18 1 2 3 4 5 6 7 1s 1s 2 13 14 15 16 17 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p La 6s 5d 6p Ac 6d 7s 4f 5f Zeff INCREASES RADIUS DECREASES DOWN GROUP...

  37. 1 18 1 2 3 4 5 6 7 1s 1s 2 13 14 15 16 17 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p La 6s 5d 6p Ac 6d 7s 4f 5f RADIUS INCREASES Zeff DECREASES

  38. TRENDS IN EA ELECTRON AFFINITY ELECTRON AFFINITY MORE NEGATIVE MORE NEGATIVE ELECTRON AFFINITY MORE NEGATIVE

  39. TRENDS IN FIRST IE First Ionization energies decrease down the group Zeff DECREASES IONIZATION ENERGY IONIZATION ENERGY Zeff INCREASES UP THE GROUP Electrons closer to nucleus more tightly held

  40. TRENDS IN FIRST IE Zeff INCREASES IONIZATION ENERGY IONIZATION ENERGY Greater effective nuclear charge across period Poor shielding by electrons added

  41. VALENCE SHELL ELECTRON PAIR REPULSION: VSEPR Based on the idea that all electron pairs repel each other. The bonding and lone pairs push apart as far as possible…….. This means that atoms bound to a central atom are as far apart as possible……. we can find the molecular shape! Lets see how it works…...

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