Electronic Configuration

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

1s 2s

Different ways to show electron configuration

Read this “one s two”

not “one s squared”

1s 2s

 

1s 2s

Li 1s

2s

Energy level diagram Box notation

Spectroscopic notation

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

  



  

  

  

 

Electron configuration of O

• Atomic number of O = 8 so neutral atom has 8 e^–

Electron configuration of O

O 1s

• 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.



1s

Electron configuration of O

O 1s

2s

• 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.





1s 2s

Electron configuration of O

O 1s

2s

2p

• 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.

  





1s 2s 2p

Electron configuration of Co

• Atomic number of Co = 27 so neutral atom has 27 e^–

Electron configuration of Co

Co 1s

• 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.



1s

Electron configuration of Co

Co 1s

2s

• 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.





1s 2s

Electron configuration of Co

Co 1s

2s

• 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.





1s 2s

Electron configuration of Co

Co 1s

2s

2p

• 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.

  





1s 2s 2p

Electron configuration of Co

Co 1s

2s

2p

3s

• 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.

• Next 2 e^– go into 3s orbital. 3s subshell is now full. 15 e^– left.

  





1s 2s 2p



3s

Electron configuration of Co

Co 1s

2s

2p

3s

3p

• 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.

• Next 2 e^– go into 3s orbital. 3s subshell is now full. 15 e^– left.

• Next 6 e^– go into 3p subshell. 3p subshell is now full. 9 e^– left.

  





1s 2s 2p

  



3s 3p

Electron configuration of Co

Co 1s

2s

2p

3s

3p

4s

• 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.

• Next 2 e^– go into 3s orbital. 3s subshell is now full. 15 e^– left.

• Next 6 e^– go into 3p subshell. 3p subshell is now full. 9 e^– left.

• Next 2 e^– go into 4s subshell. 4s subshell is now full. 7 e^– left.

  





1s 2s 2p

  



3s 3p



4s

Electron configuration of Co

Co 1s

2s

2p

3s

3p

4s

3d

• 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.

• Next 2 e^– go into 3s orbital. 3s subshell is now full. 15 e^– left.

• Next 6 e^– go into 3p subshell. 3p subshell is now full. 9 e^– left.

• Next 2 e^– go into 4s subshell. 4s subshell is now full. 7 e^– left.

• Last 7 e^– go into 4d subshell.

  





1s 2s 2p

  



3s 3p

  



4s 3d

 

Simplifying electron configurations

• Build on the atom’s noble gas core

• He 1s²

O 1s²2s²sp⁴ O [He]2s²

• Ar 1s²2s²2p⁶3s²3p⁶

Co 1s²2s²2p⁶3s²3p⁶4s²3d⁷ Co [Ar]4s²3d⁷

  





1s 2s 2p

  



3s 3p

  



4s 3d

 

  





1s 2s 2p

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

Electron configuration of Sn

Sn [Kr]

• The noble gas core is Kr

Electron configuration of Sn

Sn [Kr]5s

• The noble gas core is Kr

• From Kr, go 2 spaces across the s-block in the 5th row  5s²

Electron configuration of Sn

Sn [Kr]5s

4d

• 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]5s

4d

5p

• 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

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]4s

3d

• I [Kr]5s

4d

5p



4s 3d

  



5s 4d

     5p

    

The f-block is inserted into to the

d-block

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

Find the electron configuration of Au

• Au [Xe]6s

• 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]6s

4f

• 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

Find the electron configuration of Au

• Au [Xe]6s

4f

5d

• 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

Electron configurations

• Without using a periodic table find the ground- state 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 ground- state electron configuration of an atom with

• 6 electrons 1s²2s²2p²

• 12 electrons 1s²2s²2p⁶3s²

• 24 electrons

1s²2s²2p⁶3s²3p⁶4s²3d⁴

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d 6f

7s 7p 7d 7f

The periodic table trick

(row #)

(row #)

(row #) – 1

(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

Magnetic susceptibility

• C [He]2s

2p

paramagnetic

• Mg [Ne]3s

diamagnetic

• Cr [Ar]4s

3d

paramagnetic

 



2s 2p



3s 3p

  



4s 3d

 