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Quantum Theory and the Electronic Structure of Atoms Chapter 7.

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Presentation on theme: "Quantum Theory and the Electronic Structure of Atoms Chapter 7."— Presentation transcript:

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2 Quantum Theory and the Electronic Structure of Atoms Chapter 7

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

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

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

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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

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

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

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

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 Planck’s constant (h) h = 6.626 x 10 -34 J s 7.1

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

13 Relationship Between Energy and Frequency E = h h=Planck’s Constant h=6.626 x 10 -34 J s This constant was determined experimentally

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

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

16 7.3 Line Emission Spectrum of Hydrogen Atoms

17 7.3

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 Bohr’s 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

19 E = h 7.3

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

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

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=Planck’s Constant m=mass

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

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

25 Chemistry in Action: Electron Microscopy STM image of iron atoms on copper surface e = 0.004 nm

26 Heisenberg’s 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

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 Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi- electron systems. 7.5

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

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: ½ or – ½)

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

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

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

33 Types of Orbitals ( l ) s orbital p orbital d orbital

34 l = 0 (s orbitals) l = 1 (p orbitals) 7.6

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

36 p Orbitals The three p orbitals lie 90 o apart in space The three p orbitals lie 90 o apart in space

37 l = 2 (d orbitals) 7.6

38 f Orbitals For l = 3, f sublevel with 7 orbitals

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

40 m l = -1m l = 0m l = 1 m l = -2m l = -1m l = 0m l = 1m l = 2 7.6

41  = fn(n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Schrodinger Wave Equation m s = -½m s = +½ 7.6

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

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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 = ½ or - ½  = (n, l, m l, ½ ) or  = (n, l, m l, - ½ ) An orbital can hold 2 electrons 7.6

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

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

47 “Fill up” electrons in lowest energy orbitals (Aufbau principle) 7.7

48 The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). 7.7

49 Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7

50 Why are d and f orbitals always in lower energy levels? d and f orbitals require LARGE amounts of energy It’s 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!

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

52 Outermost subshell being filled with electrons 7.8

53 Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8

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55 Exceptions to the Aufbau Principle Remember d and f orbitals require LARGE amounts of energy If we can’t 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.

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

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

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

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!)

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!


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