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Quantum Theory and the
Electronic Structure of Atoms
Chapter 7
Quantum Theory
Electromagnetic radiation
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.
Frequency (n) is the
number of waves that
pass through a
particular point in 1
second (Hz = 1 cycle/s).
Speed of light (c) in vacuum
= 3.00 x 108 m/s
All electromagnetic radiation
lxn=c
What is the frequency of light with wavelength of 500nm?
lxn=c
n = c/l
n = 3.00 x 108 m/s / 500 x 10 -9m
n = 6.00 x 10 14 s-1
Low wavelength --- high energy
Energy of light:
E = h n= h x c / l
Planck’s constant (h), h = 6.63 x 10-34 J•s
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=hxn
E=hxc/l
= 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
Bohr’s Model of the Atom (1913)
light is emitted as e- moves
from one energy level to a
lower energy level
En = -RH (
1
n2
)
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
Transition between
energy states
ni = 3
ni = 3
ni = 2
nf = 2
Ephoton = DE = Ef - Ei
1
Ef = -RH ( 2
nf
1
Ei = -RH ( 2
ni
1
DE = RH( 2
ni
)
)
1
n2f
)
=hxn
nnf f==11
if ni > nf, then emission occurs
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(
1
n2i
1
n2f
)
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
Quantum Mechanical Description of the distribution of
electrons in the atom:
• 4 quantum numbers:
n (principal quantum number)
l (angular momentum quantum number)
ml (magnetic quantum number)
ms (spin quantum number)
Allowed values:
• n = 1,2,3 …,n
• l = 0,1,2,…,(n-1)
• ml = (-l, …, 0. …+l)
• ms = +1/2, -1/2
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
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
l=0
l=1
l=2
l=3
s orbital
p orbital
d orbital
f orbital
Shape of the “volume” of space that the e- occupies
l = 0 (s orbitals)
l = 1 (p orbitals)
l = 2 (d orbitals)
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, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space
ml = -1
ml = -2
ml = 0
ml = -1
ml = 0
ml = 1
ml = 1
ml = 2
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
Two possible spinning
motions of an electron
result in two different
magnetic fields.
ms = +½
ms = -½
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
Shell – electrons with the same value of n
Subshell – electrons with the same values of n and l
Orbital – electrons with the same values of n, l, and ml
Which of the following sets of quantum numbers in an atom is
acceptable?
(a)(1, 0, +½, +½)
(b) (3, 2, -2, -½)
(c) (3, 3, 0, -½).
(d) (4, 2, +3, +½)
(e) (3, 0, +1, +½)
Allowed values:
• n = 1,2,3 …,n
• l = 0,1,2,…,(n-1)
• ml = (-l, …, 0. …+l)
• ms = +1/2, -1/2
(b)
Energy diagram of orbitals in a multi-electron atom
Energy depends on n and l
n=3 l = 2
n=3 l = 0
n=2 l = 0
n=3 l = 1
n=2 l = 1
n=1 l = 0
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
Orbital diagram
H
1s1
Electron Configurations
Periods 1, 2, and 3
Three rules:
1.Electrons fill orbitals starting with lowest n and moving
upwards (Aufbau principle: Fill up electrons in lowest energy orbitals )
2. No more than two electrons can be placed in each orbital.
No two electrons can fill one orbital with the same spin
(Pauli exclusion principle: no two electrons in an atom can have the
same four quantum numbers.)
3. For degenerate orbitals, electrons fill each orbital singly
before any orbital gets a second electron (Hund’s rule: The
most stable arrangement of electrons in subshells is the one with the greatest
number of parallel spins )
Period 4 and Beyond
the d orbitals begin to fill
Which of the following violates the Pauli Exclusion Principle?
(a)↑ ↑__ ↑↑
(b) ↑ ↑↓ ↑_
(c) ↑ ↑↓ ↓_
(d) ↑↓_ __ ↑ _ ↑_ ↑_
(e) ↑_ ↑__ ↑ _ ↓_ ↑↓_
Pauli exclusion principle: no two electrons in an atom
can have the same four quantum numbers. No two electrons can
fill one orbital with the same spin. )
(a)
Which of the following violates the Hund’s rule?
(a)↑ ↑__ ↑↑
(b) ↑ ↑↓ ↑_
(c) ↑ ↑↓ ↓_
(d) ↑↓_ __ ↑ _ ↑_ ↑_
(e) ↑_ ↑__ ↑ _ ↓_ ↑↓_
Hund’s rule: The most stable arrangement of electrons in
subshells is the one with the greatest number of parallel spins.
For degenerate orbitals, electrons fill each orbital singly before
any orbital gets a second electron.
(d)
“Fill up” electrons in lowest energy orbitals (Aufbau principle)
??
B 5 electrons B 1s22s22p1
Be 4 electrons Be 1s22s2
Li 3 electrons Li 1s22s1
He 2 electrons He 1s2
1
H 1 electron H 1s
The most stable arrangement of electrons in
subshells is the one with the greatest number of
parallel spins (Hund’s rule).
Ne 10 electrons Ne 1s22s22p6
F 9 electrons F 1s22s22p5
O 8 electrons
N 7 electrons
O 1s22s22p4
N 1s22s22p3
C 6 electrons
C 1s22s22p2
Core electrons = electrons in inner shells; Valence
electrons = electrons in outer shell
General rules for assigning electrons to atomic orbitals
1. Each shell or principle level of quantum number n
contains n subshells. E.g. if n=2, there are two subshells
(two values of l) of angular momentum quantum number.
2. Each subshell of quantum number l contains (2l+1)
orbitals. E.g. if l=1, there are 3 p orbitals.
3. No more than two electrons can be placed in each orbital.
4. A quick way to determine the maximum number of
electrons that an atom can have in principle level n is to
use the formula of 2n2.
What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2
[Ne] 1s22s22p6
core electrons
valence electrons
What are the possible quantum numbers for the
last (outermost) electron in Cl?
Cl 17 electrons
1s22s22p63s23p5
1s < 2s < 2p < 3s < 3p < 4s
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
Paramagnetic
unpaired electrons
2p
(attracted by a magnetic field)
Diamagnetic
all electrons paired
2p
(not drawn into a magnetic field)
How many unpaired electrons are present in Al,N,Si,S?
Unpaired electrons
Al: 1s22s22p63s23p1
N: 1s22s22p3
Si: 1s22s22p63s23p2
S: 1s22s22p63s23p4
1
3
2
2
Electron Configurations and the Periodic Table
Shorthand way of writing electron configurations:
Write the core electrons corresponding to the filled
Noble gas in square brackets.
Write the valence electrons explicitly.
Example, P: 1s22s22p6 3s23p3
but Ne is 1s22s22p6
Therefore, P: [Ne] 3s23p3
The reason for this irregularity is that a slightly stability is
associated with the half-filled (d5) and completely filled (d10)
subshells.
4f
5f
ns2np6
ns2np5
ns2np4
ns2np3
ns2np2
ns2np1
d10
d5
d1
ns2
ns1
Ground State Electron Configurations of the Elements