1 / 18

Ch 2 Applications of Quantum Mechanics

Ch 2 Applications of Quantum Mechanics. Atomic Wave Functions Solving a 3D problem: positively charged nucleus and one negative electron Use methods similar to Particle in a Box to find E and Y For a 1D problem, we generated one quantum number = n

kristy
Download Presentation

Ch 2 Applications of Quantum Mechanics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Ch 2 Applications of Quantum Mechanics • Atomic Wave Functions • Solving a 3D problem: positively charged nucleus and one negative electron • Use methods similar to Particle in a Box to find E and Y • For a 1D problem, we generated one quantum number = n • For a 3D problem, we will generate 3 quantum numbers= n, l, ml • Later, we will add the 4th quantum number to describe e spin (ms) • Quantum Numbers • n = principle quantum number = responsible for Energy of electron • l = orbital angular momentum = responsible for shape of orbital • ml = magnetic angular momentum = responsible for orbital position in space • ms = spin angular momentum = describes orientation of e- magnetic moment • When no magnetic field is present, all ml values have the same energy and both ms values have the same energy • Together, n, l, and ml define one atomic orbital

  2. Conversions: x = r sinq cosf y = r sinq sinf z = r cosq • Spherical Coordinates • Cartesian Coordinates: x, y, z define a point • Spherical Coordinates: r, q, f define a point • r = distance from nucleus for the electron • q = angle from the z-axis (from 0 to p) • f = angle from the x-axis (from 0 to 2p) Spherical Volumes: 3 sides = rdq, r sinq df, and dr V = product = r2 sinq dq df dr Volume of shell between r and r + dr

  3. In Spherical Coordinates, Y is the product of the angular factors • Radial factor describes e- density at different distances from nucleus • Angular factor describes shape of orbital and orientation in space • Y(r,q,f) = R(r)Q(q)F(f) = R(r)Y(qf) [Y combines angular factors] • The Radial Function • R(r) is determined by n, l • Bohr Radius = ao =52.9 pm = r at Y2 maximum probability for a H 1s orbital • Used as a unit of distance for r in quantum mechanics (r = 2ao, etc…) • Radial Probability Function = 4pr2R2 • Describes the probability of finding e- at a given distance over all angles • Plots of R(r) and 4pr2R2 use r scale with ao units • Electron Density falls off rapidly as r increases • For 1s, probability = 0 by the time r = 5ao • For 3d, max prob is at r = 9ao ; prob = 0 at r = 20ao • All orbitals have prob = 0 at the nucleus 4pr2R2 = 0 at r = 0 • Maxima: combination of rapid increase of 4pr2 with r and the rapid decrease of R2 with r • Shape and distance of e- from nucleus determine reactivity (valence)

  4. Radial Probability Functions

  5. The Angular Functions • q(q) and F(f) show how the probability changes at the same distance, but different angles = shape/orientation of the atomic orbitals • Angular factors are determined by l and ml • Table 2-2 • Center = shape due to q portion only • Far right = shape due to q and f = 3d orbital shape • Shaded lobes = where Y is negative • Y2 = probability is same for +/-, but useful for bonding

  6. Nodal Surfaces = surfaces where Y2 = 0 (Y changes sign) • Appear naturally from Y mathematical forms • 2s orbital: Y changes sign at r = 2ao giving a nodal sphere • Y2 = prob = 0 for finding the electron here • Y(r,q,f) = R(r)Y(q,f) and Y2 = 0 • Either R(r) = 0 or Y(q,f) = 0 • Determines Nodal Surfaces by finding these conditions • Radial Nodes = Spherical Nodes R(r) = 0 • Gives Layered appearance of orbitals • R(r) changes sign • 1s, 2p, 3d have no radial nodes • Number of radial nodes increases with n • Number of radial nodes = n – l -1

  7. Angular Nodes: Y = 0 = planar or conical • Easiest to see in Cartesian Coordinates (x, y, z) • Can find where Y changes from +/- • Total number of nodes = n - 1 • How can e- have probability on both sides of a nodal plane? • Wave properties • Like a node on a violin string; string still exists even when it has nodes • Example 1 • pz • pz because z appears in the Y expression • Y = 0 for angular node: z = 0 = xy plane is a node • Y = + where z > 0, Y is – where < 0 • 2pz has no spherical nodes

  8. Example 2 • dx2-y2 • Y = 0 when x = y and x = -y • Nodal planes contain z-axis and make 45o angle with x and y axes • Y = + when x2 > y2 and Y = - when x2 < y2 • No spherical nodes • Exercises 2-1 and 2-2 • Linear Combinations • i appears in p and d wave functions • Fortunately, any linear combination of solutions to the Schrodinger Equation is also a solution to the Schrodinger Equation • Simplify p by taking sum and difference of ml = +1 and ml = -1 • Normalize by constants • Now these are real functions (YY* = Y2)

  9. The Aufbau Principle = the Build-Up Principle • Multi-electron atoms have limitations • Any combination of Q#’s works for a 1-electron atom • Electrons will have to interact in multi-electron systems • Rules • Electrons are placed in orbitals to give the lowest energy • Lowest values of n and l filled first • Values of ml and ms don’t effect energy • Pauli Exclusion Principle = every e- has a unique set of quantum numbers • At least on Q# must be different: 2 e- in same orbital have ms = +/- ½ • Not derived from Schrödinger Equation; Experimental Observation • Hund’s Rule = always maximum spin if you have degenerate orbitals • 2 e- in same orbital higher energy than 1 e- each in degenerate orbitals • Electrostatic repulsion explains this (Coulombic Repulsion E = Pc) • Multiplicity = # of unpaired e- + 1 (n + 1)

  10. Exchange Energy = Pe = quantum mechanical result depending on the possible number of exchanges of 2 e- with same E/spin • 2p2 example __ __ __ __ __ __ __ __ __ __ __ __ • P = total pairing E = Pc + Pe Pc = + and approximately constant Pe = - and approximately constant Favors the unpaired configuration vii. Another example p4 and Exercise 2-3 __ __ __ vs. __ __ __ 1Pc – 3Pe 2Pc – 2Pe Lowest E 1 2 2 1 + 1 Pc, 0 Pe 0 Pc, 0Pe 0 Pc, -1Pe Increasing Energy

  11. D) Shielding • Predicting exact order of e- filling is difficult • Shielding Provides an approach to figuring it out • Each e- acts as a shield for e- farther out • This reduces the attraction to the nucleus, increasing E • Figure 2-10 is the accepted energy ordering • n is most important • l does change order for multi-electron systems • As Z increases, the attraction for e- increases and the Energy of the orbitals decreases irregularly • Table 2-6 gives actual e- configurations • Slater’s Rule: Z* = effective nuclear attraction = Z – S • Grouping: 2s,2p/3s,3p/3d/4s,4p/4d/4f/5s5p • e- in higher groups don’t shield lower groups • For ns/np valence e- • e- in same group contributes 0.35 (1s = 0.3) • e- in n-1 group contributes 0.85 • e- in n-2 group contributes 1.00

  12. For nd/nf valence electrons • e- in same group contributes 0.35 • e- in groups to the left contribute 1.00 • S = sum of all contributions • Examples • Oxygen: (1s2)(2s22p4) Z* = Z – S = 8 – 2(0.85) – 5(0.35) = 4.55 Last e- held 4.55/8.00 = 57% of force expected for 8+ nucleus • Nickel: (1s2)(2s22p6)(3s23p6)(3d8)(4s2) Z* = 28 – 10(1.00) – 16(0.85) – 1(0.35) = 4.05 for 4s electron Ni2+ loses 4s2 electrons, not 3d electrons first (d8 metal) c) Exercises 2-4, 2-5 • Why does it work? (Figure 2-6) • 3s,3p 100% shield 3d because their probability is higher than 3d near nucleus = shielding • 2s,2p only shield 3s,3p 85% because 3s/3p have significant regions of high probability near the nucleus

  13. Why are Cr [Ar]4s13d5 and Cu [Ar]4s13d10 • Traditional Explanation: filled and half-filled subshells are particularly stable • Electron Interaction Model • 2 parallel E levels having only one kind of spin • Separated by Pc amount of Energy • E levels slant downward as Z increases (more attraction) • 3d orbitals slant faster than 4s, which has more complete shielding • Fill from bottom up with only e- of one spin type Ti __ __ __ __ __ 3d __ 4s 4s23d2 __ 4s Fe __ __ __ __ __ 3d __ __ __ __ __ 3d __ 4s 4s23d6 __ 4s Cr __ 4s __ __ __ __ __ 3d 4s13d5 __ 4s Cu __ 4s __ __ __ __ __ 3d __ 4s __ __ __ __ __ 3d 4s13d10

  14. Formation of cations: lowers d energy more than s energy, so transition metals always lose s electrons first III. Periodic Properties • Ionization Energy • An+(g) ----> A(n+1)+(g) + e- DU = Ionization Energy • Trends • Increase in DU across period as Z increases so does the attraction to e- • Breaks in trend • B p-orbital 2s22p1 easier to remove • O 2s22p4 paired e- easier to remove __ __ __ • Similar pattern in other periods

  15. Transition Metals have only small differences • Increased shielding • Increased distance from nucleus • Large decreases at start of new period because new s-orbital much higher E • Noble gases decrease as Z increases because e- are farther from nucleus

  16. Electron Affinity • A-(g) ----> A(g) + e- DU = EA • Endothermic except for noble gases and Alkaline Earths • More precisely, this is the Zeroth Ionization Energy • Similar trends to Ionization Energy • Much smaller Energies involved; easier to lose e- from negative charged ion • Ionic Radii • Gradual decrease across a period as Z increases (greater attraction for electrons) 2) General increase down a Group as the size of the valence shell increases 3) Nonpolar Covalent Radii for neutral atoms are found in Table2-7 4) Ionic Radii from crystal data are found in Table 2-8 • Cations are smaller than neutral atoms • Anions are larger than neutral atoms • Radius decreases as + charge increases

More Related