Quantum Theory and the Electronic Structure of Atoms
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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 l.jpg

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 l.jpg

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 l.jpg

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 l.jpg

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 l.jpg
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 l.jpg

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 l.jpg

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


Slide10 l.jpg

Line Emission Spectrum of Hydrogen Atoms are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.

7.3


Slide11 l.jpg

7.3 are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.


Slide12 l.jpg

( ) are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.

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 l.jpg

E = h are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.n

E = hn

7.3


Slide14 l.jpg

n are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.i = 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 l.jpg

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 l.jpg

Why is e hydrogen atom when its electron drops from the - 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 l.jpg

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 l.jpg

Chemistry in Action: Element from the Sun 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg

Chemistry in Action: Laser – The Splendid Light 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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


Slide20 l.jpg

Chemistry in Action: Electron Microscopy 2.5 g Ping-Pong ball traveling at 15.6 m/s?

le = 0.004 nm

STM image of iron atoms

on copper surface


Schrodinger wave equation l.jpg
Schrodinger Wave Equation 2.5 g Ping-Pong ball traveling at 15.6 m/s?

  • 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 l.jpg
QUANTUM NUMBERS 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg

n=1 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg

Where 90% of the 2.5 g Ping-Pong ball traveling at 15.6 m/s?

e- density is found

for the 1s orbital

e- density (1s orbital) falls off rapidly

as distance from nucleus increases

7.6


Slide25 l.jpg

Schrodinger Wave Equation 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg
Types of Orbitals ( 2.5 g Ping-Pong ball traveling at 15.6 m/s?l)

s orbital

p orbital

d orbital


Slide27 l.jpg

l 2.5 g Ping-Pong ball traveling at 15.6 m/s? = 0 (s orbitals)

l = 1 (p orbitals)

7.6


P orbitals l.jpg
p Orbitals 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg
p Orbitals 2.5 g Ping-Pong ball traveling at 15.6 m/s?

  • The three p orbitals lie 90o apart in space


Slide30 l.jpg

l 2.5 g Ping-Pong ball traveling at 15.6 m/s? = 2 (d orbitals)

7.6


F orbitals l.jpg
f Orbitals 2.5 g Ping-Pong ball traveling at 15.6 m/s?

For l = 3, f sublevel with 7 orbitals


Slide32 l.jpg

Schrodinger Wave Equation 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg

m 2.5 g Ping-Pong ball traveling at 15.6 m/s?l = -1

ml = 0

ml = 1

ml = -2

ml = -1

ml = 0

ml = 1

ml = 2

7.6


Slide34 l.jpg

Schrodinger Wave Equation 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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

spin quantum number ms

ms = +½or -½

ms = +½

ms = -½

7.6


Slide35 l.jpg

Schrodinger Wave Equation 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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


Slide36 l.jpg

7.6 2.5 g Ping-Pong ball traveling at 15.6 m/s?


Slide37 l.jpg

How many electrons can an orbital hold? 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg

How many 2p orbitals are there in an atom? 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg

Orbital Diagrams 2.5 g Ping-Pong ball traveling at 15.6 m/s?

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 l.jpg

? 2.5 g Ping-Pong ball traveling at 15.6 m/s?

?

“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 l.jpg

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 l.jpg

Order of orbitals (filling) in multi-electron atom one with the greatest number of parallel spins (

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 l.jpg
Why are d and f orbitals always in lower energy levels? one with the greatest number of parallel spins (

  • 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 l.jpg

number of electrons one with the greatest number of parallel spins (

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 l.jpg

What is the electron configuration of Mg? one with the greatest number of parallel spins (

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


Slide46 l.jpg

Outermost subshell being filled with electrons one with the greatest number of parallel spins (

7.8


Slide47 l.jpg

2p one with the greatest number of parallel spins (

2p

Paramagnetic

Diamagnetic

unpaired electrons

all electrons paired

7.8


Slide48 l.jpg

7.8 one with the greatest number of parallel spins (


Exceptions to the aufbau principle l.jpg
Exceptions to the Aufbau Principle one with the greatest number of parallel spins (

  • 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 l.jpg
Exceptions to the Aufbau Principle one with the greatest number of parallel spins (

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 l.jpg
Try These! one with the greatest number of parallel spins (

  • Write the shorthand notation for:

    Cu

    W

    Au

[Ar] 4s1 3d10

[Xe] 6s1 4f14 5d5

[Xe] 6s1 4f14 5d10


Exceptions to the aufbau principle52 l.jpg
Exceptions to the Aufbau Principle one with the greatest number of parallel spins (

  • 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 l.jpg
Keep an Eye On Those Ions! one with the greatest number of parallel spins (

  • 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 l.jpg
Keep an Eye On Those Ions! one with the greatest number of parallel spins (

  • 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!


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