Transcript Chapter 7: Quantum Theory and the Electronic Structure of Atoms

CHAPTER 7: QUANTUM
THEORY AND THE ELECTRONIC
STRUCTURE OF ATOMS
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Chemistry 1411
Joanna Sabey
PROPERTIES OF WAVES
Wavelength (l): the distance between identical
points on successive waves.
 Amplitude: the vertical distance from the midline
of a wave to the peak or trough.
 Frequency (n): the number of waves that pass
through a particular point in 1 second.
 The speed (u) of the wave = l x n

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ELECTROMAGNETIC 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 108 m/s
 All electromagnetic radiation:
lxn=c

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ELECTROMAGNETIC 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:


Convert nm to meters:


n=c/λ
522 nm X (10-9 m / 1 nm) = 5.22 X 10-7 m
Plug in known values
n = (3.00 x 108 m/s) / (5.22 X 10-7 m)
 n= 5.75 X 1014 / s = 5.75 X 1014 Hz

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LIGHT SPECTRUM
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PLANCK’S QUANTUM THEORY
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 X10-34 J x s
E = h (c / λ)
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PLANCK’S QUANTUM THEORY
o
o
o
Calculate the energy (in joules) of (a) a photon with a
wavelength of 5.00 × 104 nm(infrared region) and b) a
photon with a wavelength of 5.00 × 10−2 nm (X ray
region):
E = h (c / λ)
Convert wavelength to meters:
o
o
5.00 × 104 nm X (10-9 m / 1 nm) = 5.00 X 10-5 m
Plug in known values:
E = (6.63 X10-34 J x s) (3. 00 x 108 m/s / 5.00 X 10-5 m)
o E = 3.98 X 10-21 J
o
o
o
B) E = h (c / λ)
Convert wavelength to meters:
o
o
5.00 × 10−2 nm X (10-9 m / 1 nm) = 5.00 X 10-11 m
Plug in known values:
E = (6.63 X10-34 J x s) (3. 00 x 108 m/s / 5.00 X 10-11 m)
o E= 3.98 X 10-15 J
o
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THE PHOTOELECTRIC EFFECT



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:
hn = KE + W
KE = hn - W
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THE PHOTOELECTRIC EFFECT
o
o
o
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 × 1015 s−1 is used for irradiating the
metal.
A) At the minimum frequency, KE = 0:
hv = W
Plug in known values:
v = (3.42 × 10−19 J / 6.63 X10-34 J x s)
o v= 5.16 X 1014 Hz
o
o
o
B) KE = hn – W
Plug in known values:
KE = [(6.63 X10-34 J x s)(1.00 × 1015 s−1 )] - 3.42 × 10−19 J
o KE = 3.21 X 10-19 J
o
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BOHR’S THEORY OF THE ATOM
Line Emission Spectrum of Hydrogen Atoms
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BOHR’S THEORY OF THE ATOM

When a photon emits a quantum
energy, the energy is referenced by
1
En = -RH( 2 )
n

The transition of a photons energy
is represented by
1
DE = RH( 2
ni

1
nf2
)
RH= 2.18 X 10-18 J
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EMISSION SPECTRUM
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EMISSION SPECTRUM

What is the wavelength of a photon (in
nanometers) emitted during a transition from the
ni = 5 state to the nf = 2 state in the hydrogen
atom?

ΔE = RH( 1/ni2 – 1/nf2)

ΔE = (2.18X10-18 J) ( 1/52 – 1/22)

ΔE = (2.18X10-18 J) ( -0.21)

ΔE = -4.58 X 10-19 J
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DE BROGLIE’S THEORY

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 m2/s2
 Calculate the wavelength of a 6.0 X 10-2 kg tennis
ball traveling at a speed of 68 m/s.
 Plug in known values:

λ= (6.63 X10-34 J x s) / (6.0 X 10-2 kg X 68 m/s)
 λ= 1.6 X 10-34 m

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QUANTUM NUMBERS

Quantum numbers: used to describe the
distribution of electrons in atoms.

Principal quatnum number, n

Angular momentum quantum number, l

Magnetic quantum number, ml

Spin quantum number, ms
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PRINCIPAL QUANTUM NUMBER

n: can have integral values, 1,2,3… corresponds to
the energy level of an orbital.
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ANGULAR MOMENTUM QUANTUM NUMBER

l, tells the shape of the orbital.

l=0
s orbital

l=1
p orbital

l=2
d orbital

l=3
f orbital
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THE MAGNETIC QUANTUM NUMBER

ml : describes the orientation of the orbital in
space.

For a certain value of ml , there are (2l +1)

For l=1, there are 3 ml values, -1, 0, and 1.
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QUANTUM NUMBERS
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ATOMIC ORBITAL
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ATOMIC ORBITAL
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ATOMIC ORBITAL


List the values of n, ℓ, and mℓ for orbitals in the 4d
subshell.
n is the level given:


d represents the angular momentum:


n= 4
l=2
mℓ is dependent on the l value:
The number of mℓ values possible are 5, that range
from –l to l
 -2, -1, 0, 1, 2

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ORBITAL ENERGIES

In a multi-electron atom, energy levels are
dependent on the n and l values.
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ELECTRON CONFIGURATIONS
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ELECTRON CONFIGURATIONS
Electron Configuration: represents how the
electrons are distributed among the various
atomic orbitals.
 Hydrogen:
 1s1

1s1

Pauli Exclusion Principles: no two electrons in an
atom can have the same set of four quantum
numbers.
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ELECTRON CONFIGURATION
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.

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ELECTRON CONFIGURATIONS

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.
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ELECTRON CONFIGURATIONS


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
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ELECTRON CONFIGURATIONS
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: 1s2 2s2 2p4

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ELECTRON CONFIGURATIONS

Noble Gas Core: shows in brackets the noble gas
element that most nearly precedes the element
being considered.
K: 1s2 2s2 2p6 3s2 3p6 4s1
 K: [Ar] 4s1

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ELECTRON CONFIGURATIONS
o
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Write the ground-state electron configurations for
a)sulfur (S) and b)palladium (Pd), which is diamagnetic.
A) Find sulfur on the periodic table:
S contains 16 electrons, write the configuration out until
you get to 16:
1s22s22p63s23p4
o [Ne] 3s23p4
o
o
B) For a diamagnetic compound, the final subshell has
all paired spins:
o
o
1s22s22p63s23p64s23d104p65s24d8
This is not diamagnetic, so you take the two electrons in
the s orbital to pair up spins:
1s22s22p63s23p64s23d104p64d10
o [Kr]4d10
o
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ELECTRON CONFIGURATIONS
When writing electron configurations for ions, you
must consider the electrons added or taken away.
 Write the electron configurations for the following
ions:
 Ca2+

Ca: 1s22s22p63s23p64s2
 Ca2+:1s22s22p63s23p6
 [Ne]3s23p6


O2O:1s22s22p4
 O2-:1s22s22p6
 O2-: [He] 2s22p6

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