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Quantum Theory and the Electronic Structure of Atoms Chapter 7 Properties of Waves Wavelength ( l ) 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 Properties of Waves

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slide2

Properties of Waves

Wavelength (l) 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

slide3

Properties of Waves

Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).

The speed (u) of the wave = l x n

7.1

slide4

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 108 m/s

All electromagnetic radiation

l x n = c

7.1

slide6

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?

l

n

Radio wave

l x n = c

l = c/n

l = 3.00 x 108 m/s / 6.0 x 104 Hz

l = 5.0 x 103 m

l = 5.0 x 1012 nm

7.1

mystery 1 black body problem solved by planck in 1900
Mystery #1, “Black Body Problem”Solved by Planck in 1900

Energy (light) is emitted or absorbed in discrete units (quantum).

E = h x n

Planck’s constant (h)

h = 6.63 x 10-34 J•s

7.1

slide8

Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905

hn

  • Light has both:
  • wave nature
  • particle nature

KE e-

Photon is a “particle” of light

hn = KE + BE

KE = hn - BE

7.2

slide9

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.

E = h x n

E = h x c / l

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)

E = 1.29 x 10 -15 J

7.2

slide12

( )

En = -RH

1

n2

Bohr’s Model of

the Atom (1913)

THIS CALCULATION HAS BEEN REMOVED

  • e- can only have specific (quantized) energy values
  • light is emitted as e- moves from one energy level to a lower energy level

n (principal quantum number) = 1,2,3,…

RH (Rydberg constant) = 2.18 x 10-18J

7.3

slide13

E = hn

E = hn

7.3

slide14

ni = 3

ni = 3

f

i

ni = 2

nf = 2

DE = RH

i

f

( )

( )

( )

Ef = -RH

Ei = -RH

nf = 1

nf = 1

1

1

1

1

n2

n2

n2

n2

THIS CALCULATION HAS BEEN REMOVED

Ephoton = DE = Ef - Ei

7.3

slide15

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 =

DE = RH

i

f

( )

1

1

n2

n2

THIS CALCULATION HAS BEEN REMOVED

Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)

Ephoton = DE = -1.55 x 10-19 J

Ephoton = h x c / l

l = h x c / Ephoton

l= 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J

l=1280 nm

7.3

slide16

Why is e- energy quantized?

De Broglie (1924) reasoned that e- is both particle and wave.

2pr = nll = h/mu

u = velocity of e-

m = mass of e-

THIS CALCULATION HAS BEEN REMOVED

7.4

slide17

What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?

THIS CALCULATION HAS BEEN REMOVED

l = h/mu

h in J•s

m in kg

u in (m/s)

l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)

l = 1.7 x 10-32 m = 1.7 x 10-23 nm

7.4

slide18

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

Mystery element was named Helium

In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay).

slide19

Chemistry in Action: Laser – The Splendid Light

Laser light is (1) intense, (2) monoenergetic, and (3) coherent

slide20

Chemistry in Action: Electron Microscopy

le = 0.004 nm

STM image of iron atoms

on copper surface

schrodinger wave equation
Schrodinger Wave Equation
  • In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-
  • Wave function (Y) describes:
    • . energy of e- with a given Y
    • . 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

quantum numbers
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

ml(magnetic) ---> designates a particular suborbital

The fourth quantum number is not derived from the wave function

s(spin) ---> spin of the electron (clockwise or counterclockwise: ½ or – ½)

schrodinger wave equation23

n=1

n=2

n=3

Schrodinger Wave Equation

Y = fn(n, l, ml, ms)

principal quantum number n

n = 1, 2, 3, 4, ….

distance of e- from the nucleus

7.6

slide24

Where 90% of the

e- density is found

for the 1s orbital

e- density (1s orbital) falls off rapidly

as distance from nucleus increases

7.6

slide25

Schrodinger Wave Equation

Y = fn(n, l, ml, ms)

angular momentum quantum number l

for a given value of n, l= 0, 1, 2, 3, … n-1

l = 0 s orbital

l = 1 p orbital

l = 2 d orbital

l = 3 f orbital

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

7.6

types of orbitals l
Types of Orbitals (l)

s orbital

p orbital

d orbital

slide27

l = 0 (s orbitals)

l = 1 (p orbitals)

7.6

p orbitals
p Orbitals

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

3py orbital

p orbitals29
p Orbitals
  • The three p orbitals lie 90o apart in space
f orbitals
f Orbitals

For l = 3, f sublevel with 7 orbitals

slide32

Schrodinger Wave Equation

Y = fn(n, l, ml, ms)

magnetic quantum number ml

for a given value of l

ml = -l, …., 0, …. +l

if l = 1 (p orbital), ml = -1, 0, or1

if l = 2 (d orbital), ml = -2, -1, 0, 1, or2

orientation of the orbital in space

7.6

slide33

ml = -1

ml = 0

ml = 1

ml = -2

ml = -1

ml = 0

ml = 1

ml = 2

7.6

slide34

Schrodinger Wave Equation

Y = fn(n, l, ml, ms)

spin quantum number ms

ms = +½or -½

ms = +½

ms = -½

7.6

slide35

Schrodinger Wave Equation

Y = fn(n, l, ml, ms)

Existence (and energy) of electron in atom is described

by its unique wave function Y.

Pauli exclusion principle - no two electrons in an atom

can have the same four quantum numbers.

Each seat is uniquely identified (E, R12, S8)

Each seat can hold only one individual at a time

7.6

slide37

How many electrons can an orbital hold?

Schrodinger Wave Equation

Y = fn(n, l, ml, ms)

Shell – electrons with the same value of n

Subshell – electrons with the same values of nandl

Orbital – electrons with the same values of n, l, andml

If n, l, and mlare fixed, then ms = ½ or - ½

Y = (n, l, ml, ½)

or Y = (n, l, ml, -½)

An orbital can hold 2 electrons

7.6

slide38

How many 2p orbitals are there in an atom?

n=2

n=3

l = 1

l = 2

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

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

2p

3 orbitals

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

3d

5 orbitals which can hold a total of 10 e-

7.6

slide39

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

Energy depends on n and l

7.7

slide40

?

?

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

Li 3 electrons

Be 4 electrons

B 5 electrons

C 6 electrons

B 1s22s22p1

Be 1s22s2

Li 1s22s1

H 1 electron

He 2 electrons

He 1s2

H 1s1

7.7

slide41

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

C 6 electrons

N 7 electrons

O 8 electrons

F 9 electrons

Ne 10 electrons

Ne 1s22s22p6

C 1s22s22p2

N 1s22s22p3

O 1s22s22p4

F 1s22s22p5

7.7

slide42

Order of orbitals (filling) in multi-electron atom

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

7.7

why are d and f orbitals always in lower energy levels
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!

slide44

number of electrons

in the orbital or subshell

principal quantum

number n

angular momentum

quantum number l

1s1

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

1s1

Orbital diagram

H

7.8

slide45

What is the electron configuration of Mg?

What are the possible quantum numbers for the last (outermost) electron in Cl?

Mg 12 electrons

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

1s22s22p63s2

2 + 2 + 6 + 2 = 12 electrons

Abbreviated as [Ne]3s2

[Ne] 1s22s22p6

Cl 17 electrons

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

1s22s22p63s23p5

2 + 2 + 6 + 2 + 5 = 17 electrons

Last electron added to 3p orbital

n = 3

l = 1

ml= -1, 0, or +1

ms = ½ or -½

7.8

slide47

2p

2p

Paramagnetic

Diamagnetic

unpaired electrons

all electrons paired

7.8

exceptions to the aufbau principle
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

d4 and d9

For the purposes of this class, we are going to assume that ALL atoms (or ions) that end in d4 or d9 are exceptions to the rule. This may or may not be true, it just depends on the atom.

exceptions to the aufbau principle50
Exceptions to the Aufbau Principle

d4 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 d5 instead of d4.

For example: Cr would be [Ar] 4s2 3d4, but since this ends exactly with a d4 it is an exception to the rule. Thus, Cr should be [Ar] 4s1 3d5.

Procedure: Find the closest s orbital. Steal one electron from it, and add it to the d.

try these
Try These!
  • Write the shorthand notation for:

Cu

W

Au

[Ar] 4s1 3d10

[Xe] 6s1 4f14 5d5

[Xe] 6s1 4f14 5d10

exceptions to the aufbau principle52
Exceptions to the Aufbau Principle
  • The next most common are f1 and f8
  • The electron goes into the next d orbital
  • Example:
    • La [Xe]6s2 5d1
    • Gd [Xe]6s2 4f7 5d1
keep an eye on those ions
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!)
keep an eye on those ions54
Keep an Eye On Those Ions!
  • Tin

Atom: [Kr] 5s2 4d10 5p2

Sn+4 ion: [Kr] 4d10

Sn+2 ion: [Kr] 5s2 4d10

Note that the electrons came out of the highest energy level, not the highest energy orbital!