
States that it is fundamentally impossible
to know precisely both the velocity and
the position of a particle at the same
time
› It is inaccurate to assign fixed paths for
electrons like the circular orbits in Bohr’s
model
› The only quantity that can be known is the
probability for an electron to occupy a
certain region around the nucleus
Similar to Bohr’s model, it limits an
electron’s energy to certain values
 Unlike Bohr’s model, it does not attempt
to describe the electron’s path around
the nucleus

Quantum mechanical model- shows the
allowed energies an electron can have
and how likely it is to find the
electron in various locations

The Schrodinger equation predicts a
three-dimensional region around the
nucleus where there is a high probability
of finding an electron
› This region is called the atomic orbital

Principle quantum number (n): indicates
the relative size and energy of atomic
orbitals
 As n increases, orbital becomes larger, electron may be
further from nucleus, atom’s energy increases
 n specifies the atom’s main/principle energy level

Principle energy levels contain energy
sublevels
› The # of sublevels in the principle energy level
increases as n increases

Different atomic orbitals are denoted by
letters (s, p, d, f)
› Describes the shape of the orbital

Each atomic orbital can hold 2 electrons
› s orbital holds up to 2 electrons
 1 orbital (2 per orbital = 2 total)
› p orbital set holds up to 6 electrons
 3 orbitals (2 x 3 = 6 total)
› d orbital set holds up to 10 electrons
 5 orbitals (2 x 5 = 10 total)
› f orbital set holds up to 14 electrons
 7 orbitals (2 x 7= 14 total)


Show how the electrons are distributed among the
various atomic orbitals and energy levels
Electrons fill orbitals in order of increasing energy
# of electrons in the
orbital or subshell
The element we
are finding the
electron
configuration for.
He
Energy
level
2
1s
Orbital or
subshell
 Electrons
occupy orbitals of the lowest
energy first

Correct
› Boron:
1s

2s
2p
3s
Incorrect because did not fill the lowest
energy orbital first
› Boron:
1s
2s
2p
3s

A maximum of 2 electrons can occupy
an orbital at any one time and those two
electrons must have opposite spins
› An orbital holds two electrons that spin in
opposite directions
Sodium: 1s22s22p63s1
 Correct:

› Na:
1s

2s
2p
3s
2s
2p
3s
Incorrect:
› Na:
1s
When electrons fill orbitals of the same
energy level, they do so in a way that
maximizes the number of particles with
the same spin.
 Single electrons occupy available
orbitals first before doubling up

“Empty Seat Rule”
Nitrogen: 1s22s22p3
 Correct:

› N:

Incorrect:
› N:
› N:

Orbital Notation
O
1s
8e-
2s
2p
• Exponential Notation (electron configurations)
8
O
15.9994
2
2
4
1s 2s 2p
s
1
2
3
4
5
6
7
p
1s
2s
3s
f
2p
d (n-1)
3p
4s
3d
4p
5s
4d
5p
6s
5d
6p
7s
6d
7p
6
(n-2)7
4f
5f
1s
Write the symbol of the noble gas before
the element in brackets
2. Then the rest of the electrons.
1.
Example:
 Aluminum - full configuration.

 1s22s22p63s23p1

Aluminum- short configuration:
› [Ne] 3s23p1
S
16
32.066

Longhand Configuration
S 16e- 1s2 2s2 2p6 3s2 3p4
Core Electrons
Valence Electrons
• Shorthand Configuration
S
16e
2
4
[Ne] 3s 3p

Write longhand electron configurations:

Write shorthand electron configurations:
1.
2.
3.
4.
5.
6.
H
He
Li
N
Cu
Kr
1.
2.
3.
4.
Ca
Rb
Cl
Ag

Electrically charged atoms that have lost or
gained electrons
› Cation: positive ion
 Lose electrons
 Less electrons than protons
 Metals
› Anion: negative ion
 Gain electrons
 More electrons than protons
 Nonmetals

Why do atoms lose or gain electrons?

Atoms tend to gain, lose, or share
electrons until they have eight valence
electrons (full s and p orbitals)
› This fills the valence shell and tends to give
the atom the stability of the inert gasses

Ions form because the atom is trying to
become more stable
› They become isoelectronic with the noble
gases
 they have the same electron configuration

1
2
3
4
5
6
7
Each family has the same number of
valence electrons so will form the same
type of ions
Ion Electron Configuration
› Write the e- configuration for the closest
Noble Gas
 EX: Oxygen ion  O2-  Ne
 Oxygen normally has 8 electrons, but if it
becomes 2- that means it gains two
electrons so you must write an electron
configuration with 10 electrons (8 + 2 = 10)
O2-
10e-
[He] 2s2 2p6

How many protons
and electrons?
› Protons
Equals atomic number
= 34
› Electrons
If Se was neutral, it would
have 34 but gained 2 so
= 36
79
34
Se
-2
Selenium ion
› Aluminum (Al)
 Electron configuration neutral Al:
 How many valence electrons?
 What type of ion will it form?
 Ion symbol: ____
 How many total electrons in ion?
 Electron configuration ion:
› Sulfur (S)
 Electron configuration neutral S:
 How many valence electrons?
 What type of ion will it form?
 Ion symbol: ____
 How many total electrons in ion?
 Electron configuration ion: