Slide 1

Slide 2 Quantum Theory and the Electronic Structure of Atoms Chapter 7 Slide 3 Properties of Waves Wavelength ( ) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. 7.1 Slide 4 Properties of Waves Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = x 7.1 Slide 5 Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 10 8 m/s All electromagnetic radiation x c 7.1 Slide 6 Slide 7 Relationship Between Frequency and Wavelength Frequency and Wavelength are inversely related C = Where c = speed of light, is the wavelength, and is the frequency Slide 8 C = = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz = 5.0 x 10 3 m A photon has a frequency of 6.0 x 10 4 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? = 5.0 x 10 12 nm 7.1 No, this is a radio wave Example 1 Slide 9 Example 2 Red light has a wavelength that ranges from 7.0 x 10 -7 m to 6.5 x 10 -7 m. What is the frequency of light found in the red region? 3.0 x 10 8 m/s = (7.0 x 10 -7 m)( ) =4.29 x 10 14 Hz 3.0 x 10 8 m/s = (6.5 x 10 -7 m)( ) =4.62 x 10 14 Hz range =4.29 x 10 14 Hz - 4.62 x 10 14 Hz Slide 10 Example 3 A gamma ray has a frequency of 3.0 x 10 21 Hz. What is the wavelength, in nm, of the ray? 3.0 x 10 8 = (3.0 x 10 21 s -1 ) x 10 -13 m x 10 -4 nm Slide 11 Mystery #1, Black Body Problem Solved by Planck in 1900 Energy had been thought to be emitted continuously at all wavelengths. However, energy (light) is emitted or absorbed in discrete units called quanta. E = h x Plancks constant (h) h = 6.626 x 10 -34 J s 7.1 Slide 12 Discovered that light has both: 1.wave nature 2.particle nature Mystery #2, Photoelectric Effect Solved by Einstein in 1905 Photon is a particle of light h KE e - 7.2 Slide 13 Relationship Between Energy and Frequency E = h h=Plancks Constant h=6.626 x 10 -34 J s This constant was determined experimentally Slide 14 Example 1 What is the energy of a wave that has a frequency of 1.05 x 10 16 Hz? E = h E = (6.626 x 10 -34 J s)(1.05 x 10 16 s -1 ) E = 6.96 x 10 -18 J Slide 15 Example 2 What is the wavelength of a photon of light that has 4.79 x 10 -19 J of energy? What color light is represented? E = h 4.79 x 10 -19 J = (6.626 x 10 -34 J s) 7.23 x 10 14 Hz c = 3.0 x 10 8 m/s = (7.23 x 10 14 s -1 ) 4.1 x 10 -7 m; violet Slide 16 7.3 Line Emission Spectrum of Hydrogen Atoms Slide 17 7.3 Slide 18 1.e - can only have specific (quantized) energy values 2.light is emitted as e - moves from one energy level to a lower energy level Bohrs Model of the Atom (1913) E n = -R H ( ) 1 n2n2 n (principal quantum number) = 1,2,3, R H (Rydberg constant) = 2.18 x 10 -18 J 7.3 Slide 19 E = h 7.3 Slide 20 E photon = E = E f - E i E f = -R H ( ) 1 n2n2 f E i = -R H ( ) 1 n2n2 i i f E = R H ( ) 1 n2n2 1 n2n2 n f = 1 n i = 2 n f = 1 n i = 3 n f = 2 n i = 3 7.3 Slide 21 E photon = 2.18 x 10 -18 J x (1/25 - 1/9) E photon = E = -1.55 x 10 -19 J = 6.63 x 10 -34 (Js) x 3.00 x 10 8 (m/s)/1.55 x 10 -19 J = 1280 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = h x c / = h x c / E photon i f E = R H ( ) 1 n2n2 1 n2n2 E photon = 7.3 Slide 22 De Broglie (1924) reasoned that the e - is both particle and wave (he called this the wave/particle duality). Why is e - energy quantized? 7.4 = h/m h=Plancks Constant m=mass Slide 23 = h/m = 6.63 x 10 -34 J s/ (2.5 x 10 -3 kg x 15.6 m/s) = 1.7 x 10 -32 m = 1.7 x 10 -23 nm What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? m in kgh in J s in (m/s) 7.4 Slide 24 Chemistry in Action: Element from the Sun In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay). Mystery element was named Helium Slide 25 Chemistry in Action: Electron Microscopy STM image of iron atoms on copper surface e = 0.004 nm Slide 26 Heisenbergs Uncertainty Principle Heisenberg stated that it is impossible to know precisely both the velocity and the position of a particle *If you measure the position of a particle it will be disturbed therefore altering the velocity Slide 27 Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e - E = H Wave function ( ) describes: 1. energy of e - with a given 2. probability of finding e - in a volume of space Schrodingers equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi- electron systems. 7.5 Slide 28 Probability of Finding an Electron Max Born: later found it was more useful to determine the probability of finding an electron in a certain location 2 Slide 29 QUANTUM NUMBERS The shape, size, and energy of each orbital is a function of 3 quantum numbers which describe the location of an electron within an atom or ion n (principal) ---> energy level l (orbital) ---> shape of orbital m l (magnetic) ---> designates a particular suborbital The fourth quantum number is not derived from the wave function s(spin) ---> spin of the electron s (spin) ---> spin of the electron (clockwise or counterclockwise: or ) Slide 30 Schrodinger Wave Equation fn(n, l, m l, m s ) principal quantum number n n = 1, 2, 3, 4, . n=1 n=2 n=3 7.6 distance of e - from the nucleus Slide 31 e - density (1s orbital) falls off rapidly as distance from nucleus increases Where 90% of the e - density is found for the 1s orbital 7.6 Slide 32 = fn(n, l, m l, m s ) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the volume of space that the e - occupies l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital Schrodinger Wave Equation 7.6 Slide 33 Types of Orbitals ( l ) s orbital p orbital d orbital Slide 34 l = 0 (s orbitals) l = 1 (p orbitals) 7.6 Slide 35 p Orbitals this is a p sublevel with 3 orbitals These are called x, y, and z this is a p sublevel with 3 orbitals These are called x, y, and z There is a PLANAR NODE thru the nucleus, which is an area of zero probability of finding an electron 3p y orbital Slide 36 p Orbitals The three p orbitals lie 90 o apart in space The three p orbitals lie 90 o apart in space Slide 37 l = 2 (d orbitals) 7.6 Slide 38 f Orbitals For l = 3, f sublevel with 7 orbitals Slide 39 = fn(n, l, m l, m s ) magnetic quantum number m l for a given value of l m l = -l, ., 0, . +l orientation of the orbital in space if l = 1 (p orbital), m l = -1, 0, or 1 if l = 2 (d orbital), m l = -2, -1, 0, 1, or 2 Schrodinger Wave Equation 7.6 Slide 40 m l = -1m l = 0m l = 1 m l = -2m l = -1m l = 0m l = 1m l = 2 7.6 Slide 41 = fn(n, l, m l, m s ) spin quantum number m s m s = + or - Schrodinger Wave Equation m s = -m s = + 7.6 Slide 42 Existence (and energy) of electron in atom is described by its unique wave function . Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation = fn(n, l, m l, m s ) Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 7.6 Slide 43 Slide 44 Schrodinger Wave Equation = fn(n, l, m l, m s ) Shell electrons with the same value of n Subshell electrons with the same values of n and l Orbital electrons with the same values of n, l, and m l How many electrons can an orbital hold? If n, l, and m l are fixed, then m s = or - = (n, l, m l, ) or = (n, l, m l, - ) An orbital can hold 2 electrons 7.6 Slide 45 How many 2p orbitals are there in an atom? 2p n=2 l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals How many electrons can be placed in the 3d subshell? 3d n=3 l = 2 If l = 2, then m l = -2, -1, 0, +1, or +2 5 orbitals which can hold a total of 10 e - 7.6 Slide 46 Orbital Diagrams Energy of orbitals in a multi-electron atom n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2 7.7 Energy depends on n and l Slide 47 Fill up electrons in lowest energy orbitals (Aufbau principle) 7.7 Slide 48 The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hunds rule). 7.7 Slide 49 Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7 Slide 50 Why are d and f orbitals always in lower energy levels? d and f orbitals require LARGE amounts of energy Its better (lower in energy) to skip a sublevel that requires a large amount of energy (d and f orbtials) for one in a higher level but lower energy This is the reason for the diagonal rule! BE SURE TO FOLLOW THE ARROWS IN ORDER! Slide 51 Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s 1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s 1 7.8 Slide 52 Outermost subshell being filled with electrons 7.8 Slide 53 Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8 Slide 54 Slide 55 Exceptions to the Aufbau Principle Remember d and f orbitals require LARGE amounts of energy If we cant fill these sublevels, then the next best thing is to be HALF full (one electron in each orbital in the sublevel) There are many exceptions, but the most common ones are d 4 and d 9 For the purposes of this class, we are going to assume that ALL atoms (or ions) that end in d 4 or d 9 are exceptions to the rule. This may or may not be true, it just depends on the atom. Slide 56 Exceptions to the Aufbau Principle d 4 is one electron short of being HALF full In order to become more stable (require less energy), one of the closest s electrons will actually go into the d, making it d 5 instead of d 4. For example: Cr would be [Ar] 4s 2 3d 4, but since this ends exactly with a d 4 it is an exception to the rule. Thus, Cr should be [Ar] 4s 1 3d 5. Procedure: Find the closest s orbital. Steal one electron from it, and add it to the d. This process is called hydridization Slide 57 Try These! Write the shorthand notation for: Cu W Au [Ar] 4s 1 3d 10 [Xe] 6s 1 4f 14 5d 5 [Xe] 6s 1 4f 14 5d 10 Slide 58 Exceptions to the Aufbau Principle The next most common are f 1 and f 8 The electron goes into the next d orbital Example: La [Xe]6s 2 5d 1 Gd [Xe]6s 2 4f 7 5d 1 Slide 59 Keep an Eye On Those Ions! Electrons are lost or gained like they always are with ions negative ions have gained electrons, positive ions have lost electrons The electrons that are lost or gained should be added/removed from the highest energy level (not the highest orbital in energy!) Slide 60 Keep an Eye On Those Ions! Tin Atom: [Kr] 5s 2 4d 10 5p 2 Sn +4 ion: [Kr] 4d 10 Sn +2 ion: [Kr] 5s 2 4d 10 Note that the electrons came out of the highest energy level, not the highest energy orbital!