Reading Assignments: Chapter 7 in R. Chang, Chemistry, 9th Ed., McGraw-Hill, 2006. or previous editions.

Quantum Theory and the Electronic Structure of Atoms

Or related topics in other textbooks. Consultation outside lecture room: Office Hours: Tuesday & Thursday 10 AM -12 noon, Wednesday 1-4 PM

Chapter 7

@Room 313-3 or by appointment Chemistry for Engineers, SCS126

Chemistry for Engineers, SCS126

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 =

Chemistry for Engineers, SCS126

7.1

Planck’s constant (h) h = 6.63 x 10-34 J•s

Chemistry for Engineers, SCS126

7.1

A photon has a frequency of 6.0 x 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region?

When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.

λ

λxν=c

E=hxν

ν

Chemistry for Engineers, SCS126

7.1

Chemistry for Engineers, SCS126

7.2

Chemistry for Engineers, SCS126

7.3

E = hν

Bohr’s Model of the Atom (1913) 1. e- can only have specific (quantized) energy values 2. light is emitted as emoves from one energy level to a lower energy level

E = hν

n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J Chemistry for Engineers, SCS126

7.3

ni = 3

Ephoton = ∆E = Ef - Ei

ni = 3

Ef = -RH ( ni = 2

Ei = -RH (

nf = 2

1 n2f 1 n2i

) )

nnf f==11

Chemistry for Engineers, SCS126

7.3

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. Ephoton = ∆E = RH( Ephoton = 2.18 x

1 n2i

10-18

1 n2f

Chemistry for Engineers, SCS126

7.3

Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the eWave function (Ψ) describes:

)

1.

J x (1/25 - 1/9)

2.

Ephoton = ∆E = -1.55 x 10-19 J

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

Ephoton = h x c / λ λ = h x c / Ephoton λ = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J Chemistry for Engineers, SCS126

7.3

Chemistry for Engineers, SCS126

7.5

Schrodinger Wave Equation

Schrodinger Wave Equation

Ψ = fn(n, l, ml, ms)

Ψ = fn(n, l, ml, ms)

principal quantum number

angular momentum quantum number for a given value of n, l = 0, 1, 2, 3, … n-1

n = 1, 2, 3, 4, ….

n=1

n=2

n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2

n=3

Chemistry for Engineers, SCS126

l=0 l=1 l=2 l=3

7.6

l = 0 (s orbitals)

s orbital p orbital d orbital f orbital

Chemistry for Engineers, SCS126

7.6

Chemistry for Engineers, SCS126

7.6

l = 2 (d orbitals)

l = 1 (p orbitals)

Chemistry for Engineers, SCS126

7.6

Schrodinger Wave Equation Ψ = fn(n, l, ml, ms) magnetic quantum number ml = -1

for a given value of l ml = -l, …., 0, …. +l

ml = 0

ml = 1

if l = 1 (p orbital), ml = -1, 0, or 1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2

ml = -2 Chemistry for Engineers, SCS126

ml = -1

ml = 0

ml = 1

7.6

Schrodinger Wave Equation

ml = 2

Chemistry for Engineers, SCS126

7.6

Schrodinger Wave Equation Ψ = fn(n, l, ml, ms)

Ψ = fn(n, l, ml, ms)

Existence (and energy) of electron in atom is described by its unique wave function Ψ.

spin quantum number

Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers.

ms = +½ or -½

ms = +½

ms = -½

Chemistry for Engineers, SCS126

Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 7.6

Chemistry for Engineers, SCS126

7.6

Schrodinger Wave Equation Ψ = fn(n, l, ml, ms) – electrons with the same value of n – electrons with the same values of n and l – electrons with the same values of n, l, and ml How many electrons can an orbital hold? If n, l, and ml are fixed, then ms = ½ or - ½

Ψ = (n, l, ml, ½) or Ψ = (n, l, ml, -½) Chemistry for Engineers, SCS126

7.6

Chemistry for Engineers, SCS126

Energy of orbitals in a How many 2p orbitals are there in an atom?

7.6

electron atom

Energy only depends on principal quantum number n

n=2 If l = 1, then ml = -1, 0, or +1 2p n=3

l=1 n=2

How many electrons can be placed in the 3d subshell? n=3

If l = 2, then ml = -2, -1, 0, +1, or +2

3d l=2

n=1 Chemistry for Engineers, SCS126

7.6

Chemistry for Engineers, SCS126

7.7

Energy of orbitals in a

-electron atom

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

Energy depends on n and l

n=3 l = 2 n=3 l = 0 n=2 l = 0

n=3 l = 1

?? Be Li B5 C 3 64electrons electrons 22s 222s 22p 12 1 BBe 2s Li1s1s 1s

n=2 l = 1

H He12electron electrons n=1 l = 0

He 1s12 H 1s Chemistry for Engineers, SCS126

7.7

Chemistry for Engineers, SCS126

7.9

Order of orbitals (filling) in multi-electron atom

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

Ne97 C N O F 6 810 electrons electrons electrons 22s 222p 22p 5 246 3 C N Ne O F 1s 1s222s

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s Chemistry for Engineers, SCS126

7.7

Chemistry for Engineers, SCS126

7.7

What is the electron configuration of Mg?

Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom.

Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s

number of electrons in the orbital or subshell

1s1 principal quantum number n

angular momentum quantum number l What are the possible quantum numbers for the last (outermost) electron in Cl?

Orbital diagram

Cl 17 electrons

1s < 2s < 2p < 3s < 3p < 4s

H 1s1 Chemistry for Engineers, SCS126

7.8

n=3

l=1

ms =for ½ or -½ ml = -1, 0, or +1 Chemistry Engineers, SCS126

Nuclear Chemistry unpaired electrons 2p

Chapter 23

all electrons paired 2p Chemistry for Engineers, SCS126

7.8

Chemistry for Engineers, SCS126

7.8

Atomic number (Z) = number of protons in nucleus

Reading Assignments:

Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons

Chapter 23 in R. Chang, Chemistry, 9th Ed., McGraw-Hill, 2006. or previous editions.

Mass Number Atomic Number

A ZX

Element Symbol

Or related topics in other textbooks. Consultation outside lecture room: Office Hours: Tuesday & Thursday 10 AM -12 noon, Wednesday 1-4 PM

proton 1H 1p 1 or 1

neutron 1n 0

electron 0 0 -1e or -1β

positron 0 0 +1e or +1 β

α particle 4He or 42α 2

A

1

1

0

0

4

Z

1

0

-1

+1

2

@Room 313-3 or by appointment Chemistry for Engineers, SCS126 23.1

Chemistry for Engineers, SCS126

Balancing Nuclear Equations

212Po

decays by alpha emission. Write the balanced nuclear equation for the decay of 212Po.

1. Conserve mass number (A). The sum of protons plus neutrons in the products must equal the sum of protons plus neutrons in the reactants. 235 92 U

+ 10n

138 55 Cs

+

96 37 Rb

alpha particle - 42He or 42α 212Po 84

4He 2

+ AZX

+ 2 10n

235 + 1 = 138 + 96 + 2x1

2. Conserve atomic number (Z) or nuclear charge.

212 = 4 + A

A = 208

84 = 2 + Z

Z = 82

The sum of nuclear charges in the products must equal the sum of nuclear charges in the reactants. 235 92 U

+ 10n

138 55 Cs

+

96 37 Rb

+ 2 10n

92 + 0 = 55 + 37 + 2x0Chemistry for Engineers, SCS126

23.1

Chemistry for Engineers, SCS126 23.1

Nuclear Stability and Radioactive Decay Beta decay +-10β + ν

14C 6

14N 7

40 19K

40Ca 20

Decrease # of neutrons by 1

+ -10β + ν 1n 0

Increase # of protons by 1 1p 1

+ -10β + ν

Positron decay 11C 6

11B 5

38 19K

38Ar 18

++10β + ν

Increase # of neutrons by 1

++10β + ν

Decrease # of protons by 1

1p 1

+ν

37 18 Ar

+ -10e

37Cl 17

55Fe 26

+ -10e

55Mn 25 1p 1

Increase # of neutrons by 1

+ν

Decrease # of protons by 1

+ -10e

1n 0

+ 01n

14C 6

14C 6 14N 7

+ 11H

+ -10β + ν

t½ = 5730 years

Uranium-238 Dating 238U 92

+ν

23.2

Radiocarbon Dating 14N 7

Electron capture decay

++10β + ν

ν and ν have A = 0 and ZChemistry = 0 for Engineers, SCS126

Chemistry for Engineers, SCS126 23.1

Nuclear Stability and Radioactive Decay

1n 0

206Pb 82

+ 8 24α + 6-10β

t½ = 4.51 x 109 years

Alpha decay 212Po 84

4He 2

+

208Pb 82

Decrease # of neutrons by 2 Decrease # of protons by 2

Spontaneous fission 252Cf 98

1 2125 49 In + 2 0n Chemistry for Engineers, SCS126 23.2

Chemistry for Engineers, SCS126 23.3

Nuclear Fission

Nuclear Fission Nuclear chain reaction is a self-sustaining sequence of nuclear fission reactions. The minimum mass of fissionable material required to generate a self-sustaining nuclear chain reaction is the critical mass.

235U 92

+ 01n

90Sr 38

1 + 143 54Xe + 3 0n + Energy

Non-critical

Energy = [mass 235U + mass n – (mass 90Sr + mass 143Xe + 3 x mass n )] x c2

Energy = 3.3 x 10-11J per 235U

Critical

= 2.0 x 1013 J per mole 235U Combustion of 1 ton of coal = 5 x 107 J Chemistry for Engineers, SCS126 23.5

Chemistry for Engineers, SCS126 23.5

Schematic Diagram of a Nuclear Reactor

Nuclear Fusion Fusion Reaction 2 2 3 1 1 H + 1H 1 H + 1H 2 1H

+ 13H

4 2 He

+ 01n

6Li 3

+ 12H

2 42He

Energy Released 6.3 x 10-13 J 2.8 x 10-12 J 3.6 x 10-12 J

Tokamak magnetic plasma confinement

Chemistry for Engineers, SCS126 23.5

Chemistry for Engineers, SCS126 23.6