Electronic Configuration
• Electrons are placed in orbitals based on the
energy of the shell (n ; principal quantum
number)
• Each subshell contains only two electrons
• Arrows indicating spin show the difference of
each electron’s energy
• This process is the Aufbau Principal
Ground state electron configurations
• Example: Li
• atomic number = 3
• nucleus has 3 protons
• neutral atom has 3 electrons
• 2 electrons in 1s orbital, 1 electron in 2s orbital
2s
1s
1
Different ways to show electron configuration
Box notation
Energy level diagram
↑↓
1s
2s
↑
2s
1s
Spectroscopic notation
Li 1s2 2s1
Read this “one s two”
not “one s squared”
Write the superscript 1.
Don’t leave it blank
How degenerate orbitals are filled
• Definition of degenerate orbitals
• have the same energy
• are almost always in the same subshell
• Electrons prefer to be in different orbitals with
the same spin (Hund’s rule of maximum multiplicity)
• Example: electrons in a p-subshell
↑
↑↓ ↑
↑
↑
↑
↑↓ ↑↓ ↑
↑
↑
↑
↑↓ ↑↓ ↑↓
2
Electron configuration of O
• Atomic number of O = 8 so neutral atom has 8 e–
Electron configuration of O
↑↓
1s
O 1s2
config. has 2 e–
• Atomic number of O = 8 so neutral atom has 8 e–
• First 2 e– go into 1s orbital. 1s subshell is now full. 6 e– left.
3
Electron configuration of O
↑↓
1s
↑↓
2s
O 1s22s2
config. has 4 e–
• Atomic number of O = 8 so neutral atom has 8 e–
• First 2 e– go into 1s orbital. 1s subshell is now full. 6 e– left.
• Next 2 e– go into 2s orbital. 2s subshell is now full. 4 e– left.
Electron configuration of O
↑↓
1s
↑↓
2s
↑↓ ↑ ↑
2p
O 1s22s22p4
•
•
•
•
config. has 8 e–
Atomic number of O = 8 so neutral atom has 8 e–
First 2 e– go into 1s orbital. 1s subshell is now full. 6 e– left.
Next 2 e– go into 2s orbital. 2s subshell is now full.
Last 4 e– go into 2p subshell.
4
Electron configuration of Co
• Atomic number of Co = 27 so neutral atom has 27 e–
Electron configuration of Co
↑↓
1s
Co 1s2
config. has 2 e–
• Atomic number of Co = 27 so neutral atom has 27 e–
• First 2 e– go into 1s orbital. 1s subshell is now full. 25 e– left.
5
Electron configuration of Co
↑↓ ↑↓
1s 2s
Co 1s22s2
config. has 4 e–
• Atomic number of Co = 27 so neutral atom has 27 e–
• First 2 e– go into 1s orbital. 1s subshell is now full. 25 e– left.
• Next 2 e– go into 2s orbital. 2s subshell is now full. 23 e– left.
Electron configuration of Co
↑↓ ↑↓
1s 2s
Co 1s22s2
config. has 4 e–
• Atomic number of Co = 27 so neutral atom has 27 e–
• First 2 e– go into 1s orbital. 1s subshell is now full. 25 e– left.
• Next 2 e– go into 2s orbital. 2s subshell is now full. 23 e– left.
6
Electron configuration of Co
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
1s 2s
2p
Co 1s22s22p6
•
•
•
•
config. has 10 e–
Atomic number of Co = 27 so neutral atom has 27 e–
First 2 e– go into 1s orbital. 1s subshell is now full. 25 e– left.
Next 2 e– go into 2s orbital. 2s subshell is now full. 23 e– left.
Next 6 e– go into 2p subshell. 2p subshell is now full. 17 e– left.
Electron configuration of Co
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
1s 2s
2p
3s
Co 1s22s22p63s2
•
•
•
•
•
config. has 12 e–
Atomic number of Co = 27 so neutral atom has 27 e–
First 2 e– go into 1s orbital. 1s subshell is now full.
Next 2 e– go into 2s orbital. 2s subshell is now full.
Next 6 e– go into 2p subshell. 2p subshell is now full.
Next 2 e– go into 3s orbital. 3s subshell is now full.
25 e– left.
23 e– left.
17 e– left.
15 e– left.
7
Electron configuration of Co
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
1s 2s
2p
3s
3p
Co 1s22s22p63s23p6
•
•
•
•
•
•
config. has 18 e–
Atomic number of Co = 27 so neutral atom has 27 e–
First 2 e– go into 1s orbital. 1s subshell is now full.
Next 2 e– go into 2s orbital. 2s subshell is now full.
Next 6 e– go into 2p subshell. 2p subshell is now full.
Next 2 e– go into 3s orbital. 3s subshell is now full.
Next 6 e– go into 3p subshell. 3p subshell is now full.
25 e– left.
23 e– left.
17 e– left.
15 e– left.
9 e– left.
Electron configuration of Co
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
1s 2s
2p
3s
3p
4s
Co 1s22s22p63s23p64s2
•
•
•
•
•
•
•
config. has 20 e–
Atomic number of Co = 27 so neutral atom has 27 e–
First 2 e– go into 1s orbital. 1s subshell is now full.
Next 2 e– go into 2s orbital. 2s subshell is now full.
Next 6 e– go into 2p subshell. 2p subshell is now full.
Next 2 e– go into 3s orbital. 3s subshell is now full.
Next 6 e– go into 3p subshell. 3p subshell is now full.
Next 2 e– go into 4s subshell. 4s subshell is now full.
25 e– left.
23 e– left.
17 e– left.
15 e– left.
9 e– left.
7 e– left.
8
Electron configuration of Co
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑
1s 2s
2p
3s
3p
4s
3d
Co 1s22s22p63s23p64s23d7
•
•
•
•
•
•
•
•
↑
config. has 27 e–
Atomic number of Co = 27 so neutral atom has 27 e–
First 2 e– go into 1s orbital. 1s subshell is now full.
Next 2 e– go into 2s orbital. 2s subshell is now full.
Next 6 e– go into 2p subshell. 2p subshell is now full.
Next 2 e– go into 3s orbital. 3s subshell is now full.
Next 6 e– go into 3p subshell. 3p subshell is now full.
Next 2 e– go into 4s subshell. 4s subshell is now full.
Last 7 e– go into 4d subshell.
25 e– left.
23 e– left.
17 e– left.
15 e– left.
9 e– left.
7 e– left.
Simplifying electron configurations
• Build on the atom’s noble gas core
• He 1s2
O 1s22s2sp4
O [He]2s2
↑↓
1s
↑↓
2s
↑↓ ↑ ↑
2p
• Ar 1s22s22p63s23p6
Co 1s22s22p63s23p64s23d7
Co [Ar]4s23d7
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑
1s 2s
2p
3s
3p
4s
3d
↑
9
The periodic table trick
The last subshell in the electron configuration is one of these
(row #) s
(row # – 1) d
(row #) p
(row # – 2) f
Electron configuration of Sn
• Locate Sn on the periodic table
10
Electron configuration of Sn
Sn [Kr]
• The noble gas core is Kr
Electron configuration of Sn
Sn [Kr]5s2
• The noble gas core is Kr
• From Kr, go 2 spaces across the s-block in the 5th row → 5s2
11
Electron configuration of Sn
Sn [Kr]5s24d10
• The noble gas core is Kr
• From Kr, go 2 spaces across the s-block in the 5th row → 5s2
• Then go 10 spaces across the d-block on the 5th row → 4d10
• note: n = row – 1 = 5 – 1 = 4
Electron configuration of Sn—done
Sn [Kr]5s24d105p2
• The noble gas core is Kr
• From Kr, go 2 spaces across the s-block in the 5th row → 5s2
• Then go 10 spaces across the d-block on the 5th row → 4d10
• note: n = row – 1 = 5 – 1 = 4
• Finally go 2 spaces into the p-block on the 5th row → 5p2
12
Practice
• Refer to a periodic table and write the electron
configurations of these atoms using
spectroscopic notation. Also show box notation
for the orbitals not in the noble gas core.
• Zn [Ar]4s23d10
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
• I [Kr]5s24d105p5
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑
5s
4d
5p
The f-block is inserted into to the d-block
13
Find the electron configuration of Au
• Locate Au on the periodic table
Find the electron configuration of Au
• Au [Xe]
• The noble gas core is Xe
14
Find the electron configuration of Au
• Au [Xe]6s2
• The noble gas core is Xe
• From Xe, go 2 spaces across the s-block in the 6th row → 6s2
Find the electron configuration of Au
• Au [Xe]6s24f14
• The noble gas core is Xe
• From Xe, go 2 spaces across the s-block in the 6th row → 6s2
• Then detour to go 14 spaces across the f-block → 4f14
• note: for the f-block, n = row – 2 = 6 – 2 = 4
15
Find the electron configuration of Au
• Au [Xe]6s24f145d9
• The noble gas core is Xe
• From Xe, go 2 spaces across the s-block in the 6th row → 6s2
• Then detour to go 14 spaces across the f-block → 4f14
• note: for the f-block, n = row – 2 = 6 – 2 = 4
• Finally go 9 spaces into the d-block on the 6th row → 5d9
• note: for the d-block, n = row – 1 = 6 – 1 = 5
The periodic table
16
Electron configurations
• Without using a periodic table find the groundstate electron configuration of an atom with
• 6 electrons
• 12 electrons
• 24 electrons
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
6f
7s
7p
7d
7f
Electron configurations
• Without using a periodic table find the groundstate electron configuration of an atom with
• 6 electrons
1s
1s22s22p2
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
6f
7s
7p
7d
7f
• 12 electrons
1s22s22p63s2
• 24 electrons
1s22s22p63s23p64s23d4
17
The periodic table trick
(row #)
(row #) – 1
(row #)
(row #) – 2
Using the periodic table trick
• Atomic number = 6
C [He]2s22p2
• Atomic number = 12
Mg [Ne]3s2
• Atomic number = 24
Cr [Ar]4s23d4
18
Magnetic susceptibility
• C [He]2s22p2
paramagnetic
↑↓
2s
• Mg [Ne]3s2
↑
diamagnetic
↑↓
3s
• Cr [Ar]4s13d5
↑
2p
3p
paramagnetic
↑
4s
↑
↑
↑ ↑
3d
↑
19