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Ionization (ionisation) Energy and Electronic Configuration

Key Concepts

  • Ionization energy, or ionisation energy, is the energy required to remove an electron from a gaseous atom or ion.
    First Ionization energy : energy required to remove an electron from the gaseous atom
    First Ionization for Hydrogen: H(g) → H+(g) + e

    First Ionization for Carbon: C(g) → C+(g) + e

  • Ionization energy is measured in kilojoules per mole (kJ mol-1) or electronvolts per atom (eV).
    (1 kJ/mol = 96.4869 x eV)

  • Ionization energy is given a number of symbols including I and I.E.
    The first ionization energy of an atom would therefore be given the symbol I1 or I.E1

  • The greater the value of the ionization energy, the harder it is to remove an electron from the gaseous species.
    The lower the value of the ionization energy, the easier it is to remove an electron from the gaseous species.

  • Ionization energy depends on :
  1. The charge on the nucleus: ionization energy increases with increasing numbers of protons in the nucleus, that is, with increasing positive nuclear charge.
    (The number of protons in the nucleus of an atom = the atomic number of the element (Z))

  2. The distance of the electron from the nucleus: ionization energy decreases as the distance between the electron and the nucleus increases.

  3. Repulsion between electrons : ionization energy decreases if 2 negatively charged electrons are in close proximity to each other.

Predicting Electronic Configuration Using First Ionization Energy

Using hydrogen as a base with a first ionization energy of 1312kJ/mol, we can predict the first ionization energy for helium to be 2 x 1312 = 1624kJ/mol since helium has twice as many protons in the nucleus as hydrogen.
By comparing the predicted value for helium's first ionization energy with its measured value, we can model what the electronic configuration of an atom of helium is.
Lithium has 3 protons, so its nuclear charge is 3/2 x nuclear charge of helium. We predict that lithium's first ionization energy would be 3/2 x helium's first ionization energy.
By comparing the predicted first ionization energy of lithium with its measured value we can then model the electronic configuration for lithium.
This process can be continued for each element.

M(g) → M+(g) + e
Element Nuclear Charge
(no. protons)
Predicted I1(p)
(kJ/mol)
Measured I1(m)
(kJ/mol)
Comparision Implication Electronic
Configuration
H
1 1312

_______
H(g) → H+(g) + e

He 2 2x1312
= 2624
2372 I1(m) < I1(p)
e easier to remove than expected
Repulsion between electrons in close proximity
↑↓
_______
He(g) → He+(g) + e

Li 3 3/2x2372
=3558
520 I1(m) << I1(p)
e much easier to remove than predicted
Li's e much further from the nucleus than He's e

_______
 
↑↓
_______
Li(g) → Li+(g) + e

Be 4 4/3x520
= 693
899 I1(m) > I1(p)
e slightly harder to remove than predicted
Be's extra nuclear charge has shrunk the distance a little.
↑↓
_______
 
↑↓
_______
Be(g) → Be+(g) + e

B 5 5/4x899
= 1124
800 I1(m) < I1(p)
e easier to remove than predicted
B's e a little further from the nucleus than Be's e

__
↑↓
__
 
↑↓
________
B(g) → B+(g) + e

C 6 6/5x800
= 960
1086 I1(m) > I1(p)
e a little harder to remove than expected
C's e is a little closer to the nucleus than B's e

__

__
↑↓
__
 
↑↓
________
C(g) → C+(g) + e

N 7 7/6x1086
= 1267
1402 I1(m) > I1(p)
e harder to remove than predicted
N's e is a little closer to the nucleus than C's e

__

__

__
↑↓
__
 
↑↓
________
N(g) → N+(g) + e

O 8 8/7x1402
= 1602
1314 I1(m) < I1(p)
e slightly easier to remove than expected
repulsion between a pair of e's is making it easier to remove one of them
↑↓
__

__

__
↑↓
__
 
↑↓
________
O(g) → O+(g) + e

F 9 9/8x1314
= 1478
1680 I1(m) > I1(p)
e slighly harder to remove than expected
F's e a little closer to the nucleus than O's e
↑↓
__
↑↓
__

__
↑↓
__
 
↑↓
________
F(g) → F+(g) + e

Ne 10 10/9x1680
= 1867
2080 I1(m) > I1(p)
e is slightly harder to remove than expected
Ne's e is a little closer to the nucleus than F's e
↑↓
__
↑↓
__
↑↓
__
↑↓
__
 
↑↓
________
Ne(g) → Ne+(g) + e

Na 11 11/10x2080
= 2288
496 I1(m) << I1(p)
e is much easier to remove than predicted
Na's e is much further from the nucleus than Ne's e
               

__
 
↑↓
__
↑↓
__
↑↓
__
↑↓
__
 
↑↓
________
Na(g) → Na+(g) + e

Ionization Energy: Evidence for Energy Levels and Orbitals

Each of the huge decreases in first ionization indicates an electron at much greater distance from the nucleus than expected, for example, the huge decrease in first ionization for lithium and for sodium indicates the electron being removed is much, much further from the nucleus than expected.
These jumps in distances equate to the beginning of new energy levels in the electronic configuration of the atom.
2        

 

_
 

 
↑↓
_
 

_
↑↓
_
 

_

_
↑↓
_
 

_

_

_
↑↓
_
 
↑↓
_

_

_
↑↓
_
 
↑↓
_
↑↓
_

_
↑↓
_
 
↑↓
_
↑↓
_
↑↓
_
↑↓
_
 
1
_
  ↑↓
_
  ↑↓
_
  ↑↓
_
  ↑↓
____
  ↑↓
______
  ↑↓
________
  ↑↓
________
  ↑↓
________
  ↑↓
_________
Energy
Level
H   He   Li   Be   B   C   N   O   F   Ne
simple electronic configuration
1   2   2,1   2,2   2,3   2,4   2,5   2,6   2,7   2,8

Each volume of space capable of holding 2 electrons is an orbital.
Each energy level contains only one s orbital.
Each energy level can contain a maximum of three p orbitals.
2
 
 
 
 

 

_
s
 

 
↑↓
_
s
 

_
p
↑↓
_
s
 

_
p

_
p
↑↓
_
s
 

_
p

_
p

_
p
↑↓
_
s
 
↑↓
_
p

_
p

_
p
↑↓
_
s
 
↑↓
_
p
↑↓
_
p

_
p
↑↓
_
s
 
↑↓
_
p
↑↓
_
p
↑↓
_
p
↑↓
_
s
 
1
_
s
  ↑↓
_
s
  ↑↓
_
s
  ↑↓
_
s
  ↑↓
____
s
  ↑↓
______
s
  ↑↓
________
s
  ↑↓
________
s
  ↑↓
________
s
  ↑↓
_________
s
Energy
Level
H   He   Li   Be   B   C   N   O   F   Ne
subshell electronic configuration
1s1   1s2   1s22s1   1s22s2   1s22s22p1   1s22s22p2   1s22s22p3   1s22s22p4   1s22s22p5   1s22s22p6

The p orbitals are designated as px, py and pz.
In a given energy level, the energy of an s orbital is less than the energy of a p orbital.


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