C HAPTER 7: Q UANTUM T HEORY AND THE E LECTRONIC S TRUCTURE OF A TOMS Chemistry 1411 Joanna Sabey 1

P ROPERTIES OF W AVES Wavelength ( ): the distance between identical points on successive waves. Amplitude: the vertical distance from the midline of a wave to the peak or trough. Frequency ( ): the number of waves that pass through a particular point in 1 second. The speed ( u ) of the wave = x 2

E LECTROMAGNETIC RADIATION Electromagnetic radiation: the emission and transmission of energy in the form of electromagnetic waves. Electromagnetic wave: an electric field component and a magnetic field component. Maxwell (1873) proposed that visible light consists of electromagnetic waves. Speed of light ( c ) in vacuum = 3.00 x 10 8 m/s All electromagnetic radiation: x  c 3

E LECTROMAGNETIC RADIATION The wavelength of the green light from a traffic signal is centered at 522 nm. What is the frequency of this radiation? Rearrange radiation equation:  c / λ Convert nm to meters: 522 nm X (10 -9 m / 1 nm) = 5.22 X m Plug in known values  3.00 x 10 8 m/s) / (5.22 X m)  5.75 X / s = 5.75 X Hz 4

L IGHT S PECTRUM 5

P LANCK ’ S Q UANTUM T HEORY When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. Planck stated that atoms and molecules could emit, or absorb, energy only in discrete quantities, quantum. Quantum: the smallest quantity of energy that can be emitted, or absorbed in the form of electromagnetic radiation. E= hv h = planck’s constant= 6.63 X J x s E = h ( c / λ ) 6

P LANCK ’ S Q UANTUM T HEORY o Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00 × 10 4 nm(infrared region) and b) a photon with a wavelength of 5.00 × 10 −2 nm (X ray region): o E = h ( c / λ ) o Convert wavelength to meters: o 5.00 × 10 4 nm X (10 -9 m / 1 nm) = 5.00 X m o Plug in known values: o E = (6.63 X J x s) (3. 00 x 10 8 m/s / 5.00 X m) o E = 3.98 X J o B) E = h ( c / λ ) o Convert wavelength to meters: o 5.00 × 10 −2 nm X (10 -9 m / 1 nm) = 5.00 X m o Plug in known values: o E = (6.63 X J x s) (3. 00 x 10 8 m/s / 5.00 X m) o E= 3.98 X J 7

T HE PHOTOELECTRIC E FFECT Photoelectric Effect: a phenomenon in which electrons are ejected from the surface of certain metals exposed to light of at least a certain frequency. Photons: particles of light, explained by Einstein. If the light used is a high frequency, the electrons will be knocked loose, they will aqcuire kinetic energy: h = KE + W KE = h - W 8

T HE PHOTOELECTRIC E FFECT o The work function of cesium metal is 3.42 × 10 −19 J. a)Calculate the minimum frequency of light required to release electrons from the metal. B) Calculate the kinetic energy of the ejected electron if light of frequency 1.00 × s −1 is used for irradiating the metal. o A) At the minimum frequency, KE = 0: o hv = W o Plug in known values: o v = (3.42 × 10 −19 J / 6.63 X J x s) o v= 5.16 X Hz o B) KE = h – W o Plug in known values: o KE = [(6.63 X J x s)(1.00 × s −1 )] × 10 −19 J o KE = 3.21 X J 9

B OHR ’ S T HEORY OF THE A TOM 10 Line Emission Spectrum of Hydrogen Atoms

11

B OHR ’ S THEORY OF THE ATOM When a photon emits a quantum energy, the energy is referenced by E n = - R H ( ) The transition of a photons energy is represented by R H = 2.18 X J 12 1 n2n2  E = R H ( ) 1 ni2ni2 1 nf2nf2

E MISSION S PECTRUM 13

E MISSION S PECTRUM What is the wavelength of a photon (in nanometers) emitted during a transition from the n i = 5 state to the n f = 2 state in the hydrogen atom? ΔE = R H ( 1/n i 2 – 1/n f 2 ) ΔE = (2.18X J) ( 1/5 2 – 1/2 2 ) ΔE = (2.18X J) ( -0.21) ΔE = X J 14

D E B ROGLIE ’ S T HEORY De Broglie reasoned that electrons act as both particles and waves. λ= h / mu M must be in kg U must be in m/s 1 J = 1 kg m 2 /s 2 Calculate the wavelength of a 6.0 X kg tennis ball traveling at a speed of 68 m/s. Plug in known values: λ= (6.63 X J x s) / (6.0 X kg X 68 m/s) λ= 1.6 X m 15

Q UANTUM N UMBERS Quantum numbers: used to describe the distribution of electrons in atoms. Principal quatnum number, n Angular momentum quantum number, l Magnetic quantum number, m l Spin quantum number, m s 16

P RINCIPAL Q UANTUM N UMBER n: can have integral values, 1,2,3… corresponds to the energy level of an orbital. 17

A NGULAR M OMENTUM Q UANTUM N UMBER l, tells the shape of the orbital. l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital 18

T HE MAGNETIC Q UANTUM N UMBER m l : describes the orientation of the orbital in space. For a certain value of m l, there are (2 l +1) For l =1, there are 3 m l values, -1, 0, and 1. 19

Q UANTUM N UMBERS 20

A TOMIC O RBITAL 21

A TOMIC O RBITAL 22

A TOMIC O RBITAL List the values of n, ℓ, and m ℓ for orbitals in the 4 d subshell. n is the level given: n= 4 d represents the angular momentum: l = 2 m ℓ is dependent on the l value: The number of m ℓ values possible are 5, that range from – l to l -2, -1, 0, 1, 2 23

O RBITAL E NERGIES In a multi-electron atom, energy levels are dependent on the n and l values. 24

E LECTRON C ONFIGURATIONS 25

E LECTRON C ONFIGURATIONS Electron Configuration: represents how the electrons are distributed among the various atomic orbitals. Hydrogen: 1s 1 Pauli Exclusion Principles: no two electrons in an atom can have the same set of four quantum numbers. 26 1s 1

E LECTRON C ONFIGURATION Paramagnetic: contain net unpaired spins and are attracted to a magnet Diamagnetic: do not contain net unpaired spins and are slightly repelled by a magnet. 27

E LECTRON C ONFIGURATIONS S orbitals can hold 2 electrons P orbitals can hold up to 6 electrons D orbitals can hold up to 10 electrons F orbitals can hold up to 14 electrons Each set has to be full before going into the next orbital. 28

E LECTRON C ONFIGURATIONS What is the maximum number of electrons that can be present in the principal level n=3? The third level can contain what orbitals? S, which has 2 electrons P, which has 6 electrons D, which has 10 electrons Total of maximum 18 electron 29

E LECTRON C ONFIGURATIONS An oxygen atom has a total of eight electrons. Write the electron configuration of Oxygen and the four quantum numbers for each of the eight electrons in ground state. O electron configuration: 1s 2 2s 2 2p 4 30

E LECTRON C ONFIGURATIONS Noble Gas Core: shows in brackets the noble gas element that most nearly precedes the element being considered. K: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 K: [Ar] 4s 1 31

E LECTRON C ONFIGURATIONS o Write the ground-state electron configurations for a)sulfur (S) and b)palladium (Pd), which is diamagnetic. o A) Find sulfur on the periodic table: o S contains 16 electrons, write the configuration out until you get to 16: o1s22s22p63s23p4o1s22s22p63s23p4 o [Ne] 3 s 2 3 p 4 o B) For a diamagnetic compound, the final subshell has all paired spins: o 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 8 o This is not diamagnetic, so you take the two electrons in the s orbital to pair up spins: o 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 4 d 10 o [Kr]4 d 10 32

E LECTRON C ONFIGURATIONS When writing electron configurations for ions, you must consider the electrons added or taken away. Write the electron configurations for the following ions: Ca 2+ Ca: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 Ca 2+ :1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 [Ne]3 s 2 3 p 6 O 2- O:1 s 2 2 s 2 2 p 4 O 2- :1 s 2 2 s 2 2 p 6 O 2- : [He] 2 s 2 2 p 6 33