Transcript Electronic Structure

Sublevels and Orbitals
Effective Nuclear Charge
Inner (core) electrons act to shield outer
(valence) electrons from the positive charge
of the nucleus.
Some orbitals penetrate to the nucleus more
than others, therefore s < p < d < f.
As a result, there are different energy levels
for the different sublevels for any given
principal quantum number.
8-1
energy
Sublevels
2s
2p
1s
2s
2p
2s
1s
H
2p
1s
Li
F
8-2
Aufbau Principle
The Aufbau Principal is used to write the
electron configurations of atoms.
For any element, the number of electrons in
the neutral atom equals the atomic number.
Start filling orbitals, from lowest to highest.
If two or more orbitals exist at the same
energy level, they are degenerate. Do not
pair the electrons until you have to.
8-3
Orbital Filling Order
.
s
p
d
f
1
1s
2
2s
2p
3
3s
3p
3d
4
4s
4p
4d
4f
5
5s
5p
5d
5f
6
6s
6p
6d
6f
7
7s
7p
7d
7f
8-4
Hund’s Rule
When putting electrons into orbitals with the
same energy, place one electron in each
orbital before putting two in any one.
In the orbital diagram, each electron must
have opposite spins.
8-5
energy
Applying The Aufbau Principal
2s
2p
2s
2p
2s
1s
1s
1s
C
O
F
2p
8-6
Sublevels on the Periodic Table
s
p
H
d
Li Be
Na Mg
K Ca Sc Ti
Rb Sr
Y
V
He
B
C
N
O
F
Al
Si
P
S
Cl Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
Cs Ba La Hf Ta W Re Os Ir
Fr Ra Ac
f
Ne
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
8-7
Using the Periodic Table to apply
the Aufbau Principle
Main Group Elements
Add electrons to the ns orbital as you
move through s-block.
Add electrons to the np orbital as you
move through the p-block.
8-8
Using the Periodic Table to apply
the Aufbau Principle
Transition Elements
Add electrons to the (n-1)d orbital as you
move through d-block.
Add electrons to the (n-2)f orbital as you
move through f-block.
8-9
Writing Electron Configurations
Expanded Format
O
Ti
Br
1s22s22p4
1s22s22p63s23p64s23d2
1s22s22p63s23p64s23d104p5
Abbreviated Format
O
Ti
Br
[He]2s22p4
[Ar]4s23d2
[Ar]4s23d104p5
8 - 10
Electron Configurations for Ions
Electron configurations can also be written
for ions.
Start with the ground-state configuration for
the atom.
For cations, remove the number of the
outermost electrons equal to the charge.
For anions, add the number of outermost
electrons equal to the charge.
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Electron Configurations for Anions
Example: Cl- (chloride)
First, write the electron configuration for
chlorine:
Cl
[Ne]3s23p5
Because the charge is 1-, add one electron.
Cl-
[Ne]3s23p6
8 - 12
Electron Configurations for Cations
Example: Ba2+ (barium)
First, write the electron configuration for
barium.
Ba
[Xe]6s2
Because the charge is 2+, remove two
electrons.
Ba2+
[Xe]
8 - 13
Isoelectronic Configurations
Species having the same electron
configurations.
Each of the following has an electron
configuration of 1s22s22p6
O2-
F-
Na+ Mg2+
Ne
Al3+
8 - 14
Electron Dot Diagrams
A simple way to show the valence electrons
present in an atom.
Valence electrons are those electrons found
in the highest numbered principal energy
level (PEL).
Valence electrons are found only in the s and
p sublevels and in most cases are the
electrons responsible for bonding.
8 - 15
Electron Dot Diagrams
The chemical symbol represents the kernel of
the atom.
The kernel of an atom consists of the nucleus
and the core electrons.
Example
s
pz X px
py
Start with s and
proceed cw!
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Electron Dot Diagrams
Examples
.
Na
..
..
. Br
..
.
[Ne]3s1
[Ar]4s23d104p5
Cr
[Ar]4s13d5
8 - 17
Heisenberg Uncertainty Principle
 In order to observe an electron, one would
need to hit it with photons having a very
short wavelength.
 Short wavelength photons would have a
high frequency and a great deal of energy.
 If one were to hit an electron, it would
cause the motion and the speed of the
electron to change.
 Lower energy photons would have a
smaller effect but would not give precise
data.
8 - 18
Quantum Numbers
Principal Quantum Number (n)
 Tells the size of an orbital and
determines its energy.
 n = 1, 2, 3 …
Angular Momentum (l)
 The number of subshells that a principal
level contains. It tells the shape of the
orbitals.
 l = 0 to n - 1
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Quantum Numbers
Magnetic Quantum Number (ml)
 Describes the orientation of the orbital.
 ml = -l to +l (all integers, including zero)
 For example, if l = 3, then ml would have
values of -3, -2, -1, 0, 1, 2, 3
8 - 20
Quantum Numbers
.
n
l
ml
Sublevel
Orbitals
1
2
0
0
1
0
1
2
0
1
2
3
0
0
-1,0,1
0
-1,0,1
-2,-1,0,1,2
0
-1,0,1
-2,-1,0,1,2
-3,-2,-1,0,1,2,3
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
1
1
3
1
3
5
1
3
5
7
3
4
8 - 21
The s Orbital
The s orbital is a sphere. Every PEL has one
s orbital.
8 - 22
The p Orbitals
There are three p orbitals: px, py and pz
8 - 23
Electron Spin
Pauli added one additional quantum number
that would allow only two electrons to be in
an orbital.
Spin quantum number (ms).
 It can have values of +1/2 and -1/2
Pauli Exclusion Principle
 Pauli also proposed that no two
electrons in an atom can have the same
set of four quantum numbers.
8 - 24
Waves
Wavelength (l)
 The distance measured from crest to
crest or from trough to trough (m, cm,
nm).
Amplitude
 The vertical distance from the node to
the height of a wave (m, cm, nm).
8 - 25
Frequency (f)
 The number of cycles or complete
vibrations that pass a point each second
(s-1, vib/s).
8 - 26
Electromagnetic Radiation
A form of energy consisting of perpendicular
electrical and magnetic fields that change, at
the same time and in phase with time.
The SI unit of frequency (f) is the hertz (Hz)
 1 Hz = 1 s-1 = 1/s
Wavelength and frequency are related
c=fλ
c = 3.00 x108 m/s
8 - 27
Energy Problems
Calculate the frequency of a quantum of light, a
photon, with a wavelength of 6.00 x 10-7 m.
λ = 6.00 × 10-7 mc = 3.00 × 108 m/s
C=f×λ
8 m/s
14 /s
3.00
×
10
C
5.00
×
10
=
f =
=
6.00 × 10-7 m
λ
8 - 28
Energy Problems
Calculate the energy of a photon, with a
wavelength of 6.00 x 10-7 m.
λ = 6.00 × 10-7 mc = 3.00 × 108 m/s
h = 6.63 × 10-34 J•s
E=h×f
c=f×λ
f = c/λ
E = h × c/λ
8 - 29
E = h × c/λ
-34 J•s × 3.00 × 108 m/s
6.63
×
10
E=
6.0 × 10-7 m
E = 3.3 × 10-19 J
8 - 30