# Electronic Structure of Atoms Chapter 6. Light is a Wave Electromagnetic wave Wavelength ( ), m Frequency ( ), Hz or s -1 Travels at c (3.00 X 10 8 m/s)

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Electronic Structure of Atoms Chapter 6

Light is a Wave Electromagnetic wave Wavelength ( ), m Frequency ( ), Hz or s -1 Travels at c (3.00 X 10 8 m/s) in a vacuum

Wavelength and frequency are inversely proportional c =

How many complete waves are shown above? What is the wavelength of light shown above?

Electromagnetic Spectrum

Blu-Ray = 405 nanometers (blue light) DVD = 650 nanometers (red light)

Visible light = 4 X 10 -7 m to 7X 10 -7 m (400 to 700 nm)

Waves: Ex 1 Calculate the wavelength of a 60 Hz EM wave

Waves: Ex 2 Calculate the wavelength of a 93.3 MHz FM radio station

Waves: Ex 3 Calculate the frequency of 500 nm blue light.

Planck’s Quantum Hypothesis: Light is a Particle Photon –light particle Photons emitted in “packets” (whole numbers) Light is both a wave and a particle E = h  (for one photon) h = 6.63 X 10 -34 J s (Planck’s constant)

Photons: Ex 1 Calculate the energy of a photon of wavelength 600 nm. 600 nm 1 X 10 -9 m = 6 X10 -7 m 1 nm

Photons: Ex 2 Calculate the energy of a photon of wavelength 450 nm (blue light). Also, calculate the energy per mole. Ans: (4.42 X 10 -19 J, 266 kJ/mol)

Photons: Ex 3 Calculate the energy of laser light with a frequency of 4.69 X 10 14 s -1. Also, calculate the energy per mole. Ans: (3.11 X 10 -19 J, 187 kJ/mol)

Photons: Ex 3a If the laser emits 1.3 X 10 -2 J per pulse, how many photons (quanta) are released during this pulse? Ans: 4.18 X 10 16 photons

Photoelectric Effect (Einstein) When light shines on a metal, electrons are emitted Can detect a current from the electrons Used in light meter, scanners, digital cameras

Three Key Points 1.Below a certain frequency, no electrons are emitted 2.Greater intensity light produces more electrons 3.Greater Frequency light produces no more electrons, but they come off with greater speed

Low FrequencyHigh Frequency Not enough energy to eject electron Can eject electron Energy of photon is greater than W (Work function)

2.More intensity – More photons – More electrons ejected with same KE

3.Greater Frequency – No more electrons ejected – Electrons ejected with greater speed (KE)

Photon/Matter Interactions 1.Electron excitation (photon disappears) 2.Ionization/photoelectric effect (photon disappears) 3.Scattering by nucleus or electron 4.Pair production (photon disappears)

Electron Excitation Photon is absorbed (disappears) Electron jumps to an excited state Ionization/Photoelectric Effect Photon is absorbed (disappears) Electron is propelled out of the atom

Scattering Photon collides with a nucleus or electron Photon loses some energy Speed does not change, but the wavelength increases Pair Production Photon closely approaches a nucleus Photon disappears An electron and positron are created.

Principle of Complimentarity Any experiment can only observe light’s wave or particle properties, not both Different “faces” that light shows

Line Spectra Discharge tube – Low density gas (acts like isolated atoms) – high voltage Light emitted only at certain (discrete) wavelengths

Hydrogen Helium Solar absorption spectrum

Electrons orbit in ground state (without radiating energy) Jumps to excited state by absorbing a photon Returns to ground state by emitted a photon

h = E e - E g

4. Ways to make something glow Bohr Model Photon AbsorptionCollision -Glow in the dark-Heat -Electricity -Chemical Reaction

Photon AbsorptionCollision

Bohr’s Equation Works only for H and other 1- electron atoms (He +, Li 2+, Be 3+, etc…)

E n = -R H n 2 E n = Energy of an orbital R H = Rydberg constant (2.18 X 10 -18 J) n = Principal quantum number of orbital

Bohr’s Equation: Ex 1 Calculate the energy of the first three orbitals of hydrogen E n = - 2.18 X 10 -18 J n 2 E 1 = - 2.18 X 10 -18 J 1 2 E 1 = - 2.18 X 10 -18 J

E 2 = - 2.18 X 10 -18 J 2 2 E 2 = - 5.45 X 10 -19 J E 3 = - 2.18 X 10 -18 J 3 2 E 3 = - 2.42 X 10 -19 J

Bohr’s Equation: Ex 2 What wavelength of light is emitted if a hydrogen electron drops from the n=2 to the n=1 orbit? E 1 = - 2.18 X 10 -18 J E 2 = - 5.45 X 10 -19 J  E = (- 2.18 X 10 -18 J - - 5.45 X 10 -19 J)  E = -1.64 X 10 -18 J  E = h = 2.48 X 10 15 s -1

c =  c/  = (3 X 10 8 m/s)(2.48 X 10 15 s -1 )  = 122 nm

 E = h c =  E = hc/ = hc  E = (6.626 X 10 -34 J s)(3.0 X 10 8 m/s) (1.63 X 10 -18 J) = 1.22 X 10 -7 m = 122 nm (UV)

Bohr’s Equation: Ex 3 Calculate the wavelength of light emitted if a hydrogen electron drops from the n=6 to the n=2 orbit? (ANS: 410 nm (violet))

Bohr’s Equation: Ex 4 Calculate the wavelength of light that must be absorbed to exite a hydrogen electron from the n=1 to the n=3 orbit? (ANS: 103 nm (UV))

Bohr Model: Other Atoms E n = (Z 2 )(- 2.18 X 10 -18 J) n 2 Z= Atomic number of the element (H=1, He=2, etc)

The Batcave Intruder Alert Laser uses a beam that has a frequency of 3.53 X 10 14 Hz. a)Calculate the wavelength in nanometers. (850) b)Calculate the energy per photon. (2.34X10 -19 J) c)Calculate the energy per mole of photons. (1.41 X 10 5 J/mol) d)Calculate the number of photons in a 50.0 mJ pulse of this laser. (2.13 X 10 17 photons) e)Calculate the moles of photons in that pulse. (3.54 X 10 -7 moles) f)Identify the range of the electromagnetic spectrum of this laser.

Wave Nature of Matter Louis DeBroglie All matter has wave and particle properties “matter waves” = h mv momentum Everything has a wavelength

Diffraction pattern of electrons scattered off aluminum foil Wavelike properties only matter for small objects Electrons, protons, light, etc…

DeBroglie Wavelength: Ex 1 Calculate the wavelength of a baseball of mass 0.20 kg moving at 40.25 m/s (90 mph) = h mv = (6.626 X 10 -34 J s) (0.20 kg)(40.25 m/s)  = 8.2 X 10 -35 m

DeBroglie Wavelength: Ex 2 Calculate the wavelength of an electron (9.109 X 10 -31 kg) moving at 2.2 X 10 6 m/s = h mv = (6.626 X 10 -34 J s) (9.11 X 10 -31 kg)(2.2 X 10 6 m/s)  = 3.3 X 10 -10 m or 0.33 nm

DeBroglie Wavelength: Ex 3 What velocity must a neutron (1.67 X 10 -27 kg) move to have a wavelength of 500 pm (1 pm = 1X10 -12 m)? 500 pm1X10 -12 m = 5.0 X 10 -10 m 1 pm = h mv v = h=(6.626 X 10 -34 J s)= 794 m/s m  (1.67 X 10 -27 kg)(5.0 X 10 -10 m/s)

1.Electron is both wave and particle 2.Probabilistic view – Where does the electron “hang out” rather than where the electron “is Quantum Mechanical Model

Electron as a particle Heisenberg Uncertainty Principle – can never know both the position and velocity of an electron at the same time Results a. Electron moves randomly (not like a planet) b. Electron cloud – 90% probability c. Important for transistors/chips Quantum Mechanical Model

nucleus Random electron cloud

Electron as Wave Schrodinger Wave Equation (1926) – treats electron solely as a wave Quantum Mechanical Model

Result One Explains the forbidden zone (waves do not match) Quantum Mechanical Model

Forbidden Zone

Result Two Orbital are not circular a.Orbital – region of space where there is a significant chance of finding an electron b.Does not move like a planet Quantum Mechanical Model

Three Major Discoveries Light is both a wave and a particle Electron Orbitals are Quantized The electron (and all matter) is both a wave and a particle Planck Einstein BohrDe Broglie (Later Schrodinger/ Heisenberg)

1. First (principal) QN (n)– how far the electron is from the nucleus (larger the number, farther away) – Level or shell n = 2 n = 1 Quantum Numbers

2. Second (azimuthal) QN (l) – the shape of the orbital Value of l0123 Letter usedspdf

Quantum Mechanical Model

3. Third (magnetic) QN (m l )– the suborbital Orbital # suborbitals Total e- s02 p3 (p x,p y,p z )6 d5 10 f714 Link to Periodic table

In a magnetic field, you can “see” the three p suborbitals in a line spectrum

4. Fourth (spin) QN (m s )– spin of the electron Pauli Exclusion Principle – No two electrons in an atom can have the same four quantum numbers – Two electrons in the same suborbital (ex: p y ) must have opposite spins – Can have values of +1/2 or -1/2

Shows the two electrons, spin +½ and spin -½

nPossible Values of l Possible Values of m l Subshell name 1001s 20101 0 -1, 0, +1 2s 2p 3012012 0 -1, 0, +1 -2, -1, 0, +1, +2 3s 3p 3d 401230123 0 -1, 0, +1 -2, -1, 0, +1, +2 -3, -2, -1, 0, +1, +2, +3 4s 4p 4d 4f

Quantum Numbers: Ex 1 What is the designation for a subshell n=5 and l =1? How many suborbitals are in this subshell? What are the values of m l for each of the orbitals?

Quantum Numbers: Ex 1 What is the designation for a subshell n=5 and l =1? 5p How many orbitals are in this subshell? 3 What are the values of ml for each of the orbitals? -1, 0, +1

Quantum Numbers: Ex 2 Which of the following sets of quantum numbers are not allowed in the hydrogen atom? A. n=2, l=0, m l =0, m s =1/2 B. n=1, l=0, m l =0, m s = -1/2 C. n=3, l=1, m l = 2, m s =1/2 D. n=4, l=2, m l = -2, m s =1/2

Quantum Numbers: Ex 2 Which of the following sets of quantum numbers are not allowed in the hydrogen atom? A. n=2, l=0, m l =0, m s =1/2 B. n=1, l=0, m l =0, m s = -1/2 C. n=3, l=1, m l = 2, m s =1/2 D. n=4, l=2, m l = -2, m s =1/2

Quantum Numbers: Ex 3 Which of the following sets of quantum numbers are not allowed in the carbon atom? A. n=2, l=0, m l =0, m s =1/2 B. n=1, l=1, m l =0, m s = -1/2 C. n=2, l=1, m l = -1, m s =1/2 D. n=4, l=2, m l = -3, m s =1/2

Quantum Numbers: Ex 3 Which of the following sets of quantum numbers are not allowed in the carbon atom? A. n=2, l=0, m l =0, m s =1/2 B. n=1, l=1, m l =0, m s = -1/2 C. n=2, l=1, m l = -1, m s =1/2 D. n=4, l=2, m l = -3, m s =1/2

1.Electron Configuration – shorthand notation to tell you the locations of all the electrons in an atom or ion 2.Notation 2p 3 OrbitShape# e- Electron Configurations

Electron Configuration

Electron Configurations

Why is “d” one less? Energy is slightly above the next s orbital Complete the orbital filling diagram for manganese

1.Easy Examples HHeOFeS 2.Write e- configuration LiNSrPSe VFArMgKr Electron Configurations

3.Which element is represented by the following electron configurations? 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 7 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 1s 2 2s 2 2p 6 3s 2 3p 3 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 2 3p 2 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 Electron Configurations

1.Rule – Use the noble gas in the previous row 2.Examples Ne and P Ru Kr You try: Br Ar S Ca I Xe Noble Gas (Condensed) Shortcut

Electron Configuration Exceptions Mostly with transition metal elements There is a special stability to filled and half-filled orbitals ElementActual configurationInstead of Cr[Ar]4s 1 3d 5 [Ar]4s 2 3d 4 Mo[Kr]5s 1 4d 5 [Kr]5s 2 4d 4 Cu[Ar]4s 1 3d 10 [Ar]4s 2 3d 9 Ag[Kr]5s 1 4d 10 [Kr]5s 2 4d 9

pee - configuration Sr Sr + Sr 2+ Al 2+ Al 3+ Ions

pee - configuration S S 1- S 2- Br 1- Ba Ba 2+ B 3+ Ions

Transition Metal Ions Lose their “s” electrons first. Fe[Ar]4s 2 3d 6 Fe 1+ Fe 2+ Fe 3+ Cu Cu 2+ Ru Ru 3+

Transition Metal Ions Zn 2+ Co 1+ Co 2+ V 3+ Sc 2+ Sc 3+

4.a) Increaseb) Decrease c) line spectrum(forbidden zone) 14. a) Gamma < (d)yellow < (e)red < (b)93.1 radio < (c)680 AM 16. a) 3.00 X 10 17 Hz b) 3.94 X 10 -3 m c) (a) is X-Ray and (b) is microwave d) 7.64 X 10 -6 m 18. UV has a higher frequency and shorter than IR. UV produces more energy

22.a) 5.09 X 10 14 Hzb) 20.3 kJ c) 3.37 X 10 -19 Jd) Na + 24. AM: 6.69 X 10 -28 JFM: 6.51 X 10 -26 J 26.1.56 X 10 -18 J/photon, 127 nm 28. a) microwaveb) 6.4 X 10 -11 J/hr 34.a) Absorbedb) Emittedc) Absorbed 36.a) 9.7 X10 -8 m, emittedb) 434 nm, emitted c) 1.06 X 10 -6 m, absorbed 38. a) n=1 larger  Eb) 121 nm, 103 nm, 97.2 nm 40. a)2.626X10 -6 m (IR) b) 6  4 transition

42. 3.97 X 10 -10 m (3.97 A) 44. 7.75 X 10 -11 m (0.775 A) 50. a) n = 3 (l=2,1,0) (9 m l values) b) n = 5 (l = 4,3,2,1,0) (25 m l values) 52. a) 2,1,1 2,1,0 2,1,-1 b) 5,2,2 5,2,1 5,2,0 5,2,-1 5,2,-2 54. 2pnot allowed 1s4d not allowed5f 3d not allowed

Solution to 41c 6.941 g X 1kg=0.006941kg/mol 1mol 1000g 0.006941 kg X 1mol= 1.15 X 10 -26 kg/atom 1mol 6.022X10 23 atoms = h= 2.3 X 10 -13 m (1.15 X 10 -26 kg)(2.5 X 10 5 m/s)

66 CPNe Config[He]2s 2 sp 2 [Ne]3s 2 3p 3 [He]2s 2 2p 6 Core2102 Valence458 Unpaired230

68. a) [Ar]4s 2 3d 10 4p 1 (1 unpaired) b) [Ar]4s 2 (0 unpaired) c) [Ar]4s 2 3d 3 (3 unpaired) d) [Kr]5s 2 4d 10 5p 5 (1 unpaired) e) [Kr]5s 2 4d 1 (1 unpaired) f) [Xe]6s 1 4f 14 5d 9 (2 unpaired) g) [Xe]6s 2 4f 14 5d 1 (1 unpaired) 70. a) [Ar]3d 10 b) [Xe]4f 14 5d 8 c) [Ar]3d 3 d) [Ne]3s 2 3p 6 72.a) 7Ab) 4Bc) 3A d) Sm and Pm (f-block)

74.a) [He]2s 2 2p 3 b) [Ar]4s 2 3d 10 4p 4 c) [Kr]5s 2 4d 7 76.a) Ba Ca K Nab) Au (shortest) Na(longest) c) 455 nm, Ba 78 a) 9.37X10 14 s-1b) 374 kJ/mol c) UV-B Shorter, higher energy d)UV-B

100. O 3  O 2 + O  H = 105.2 kJ/mol  H = 1.052 X 10 5 J/mol 1.052 X 10 5 J 1 mol=1.75 X 10 -19 J 1 mol 6.022X10 23 molec.  E = h  = 2.63 X 10 14 Hz  = c/  = 1140 nm

Extra Problem A certain biomolecule requires 598 kJ/mol to break one of its bonds. What wavelength of light (nm)would a single photon need to break this bond? What region of the EM spectrum is this?

A photon of wavelength 550 nm will break a bond in a synthetic dye. Calculate the energy per mole of that bond (kJ/mol) (218 kJ/mol)

a) Calculate the wavelength of light emitted from a 7  2 transition in a hydrogen atom. (398 nm) b) Calculate the speed an electron would need to have the wavelength. (1830 m/s) c) Calculate the wavelength of a proton moving at that same speed. (2.17 X 10 -10 m or 0.217 nm) d) Why is the proton’s wavelength so much smaller? Which has more energy, e - or p + ? m e = 9.109 X 10 -31 kg m p = 1.673 X 10 -27 kg

a) Calculate the wavelength of a proton that has been accelerated to 2 X 10 6 m/s. (1.98X10 -13 m) b) Calculate the frequency. (1.51 X 10 21 Hz) c) Calculate the energy. (1.00 X 10 -12 J/photon) m p = 1.673 X 10 -27 kg

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