Ionization energy calculations

In summary, the ionization energy of an electron is not equal to its energy level because of the presence of electron-electron repulsions in multi-electron atoms. Chemists use various methods, such as Zeff and quantum defects, to account for this effect. The ionization energy is equal to the energy required to ionize an atom, which takes into consideration the change in energy of the remaining electrons in the cation.
  • #1
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Why isn't the ionization energy of an electron equal to it's energy level such that:
E(electron)= -13.6(Z^2/n^2) = IP for that electron
But instead it is equal to the energy difference in energy between the atom and its ionized cation:
IP = E(A)-E(A+)
 
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  • #2
fsci said:
Why isn't the ionization energy of an electron equal to it's energy level such that:
E(electron)= -13.6(Z^2/n^2) = IP for that electron
But instead it is equal to the energy difference in energy between the atom and its ionized cation:
IP = E(A)-E(A+)

For a hydrogen atom, this would be true (ignoring for the moment, things like the Lamb shift, etc.)

The problem with multi-electron atoms is that the "truth" is that the picture of electrons sitting in discrete energy levels given by the equation that you have above is not right. The primary problem is that electron-electron repulsions are present and are not insignificant. Imagine that you have a two-electron atom. How many electron-electron repulsions do you have to consider? You could apportion this interaction energy (50:50) to the two electrons to calculate an effective energy level for each electron. But when you ionize one electron, what happens to this interaction? The "energy level" of the remaining electron changes, too, no?

Chemists play all sorts of games to take into account the effect of electron-electron interactions. You will see things like Zeff (an effective nuclear charge) discussed. In other contexts, you will see fudge factors on "n" called a "quantum defect" -- where have I seen that before... no matter.

The ionization energy is, by definition, equal lto the energy required to ionize the atom, which is the Delta E for:

A ----> A+ + e-
 
  • #3
Quantum Defect said:
For a hydrogen atom, this would be true (ignoring for the moment, things like the Lamb shift, etc.)

The problem with multi-electron atoms is that the "truth" is that the picture of electrons sitting in discrete energy levels given by the equation that you have above is not right. The primary problem is that electron-electron repulsions are present and are not insignificant. Imagine that you have a two-electron atom. How many electron-electron repulsions do you have to consider? You could apportion this interaction energy (50:50) to the two electrons to calculate an effective energy level for each electron. But when you ionize one electron, what happens to this interaction? The "energy level" of the remaining electron changes, too, no?

Chemists play all sorts of games to take into account the effect of electron-electron interactions. You will see things like Zeff (an effective nuclear charge) discussed. In other contexts, you will see fudge factors on "n" called a "quantum defect" -- where have I seen that before... no matter.

The ionization energy is, by definition, equal lto the energy required to ionize the atom, which is the Delta E for:

A ----> A+ + e-

Ahhhhhhh, that makes more sense now, thanks a ton!
So if I were to calculate E(A+) for the cation using the new Zeff values for the remaining electrons I am basically accounting for the decrease in electron-electron repulsion due to one less electron? And using IP= -13.6eV (Zeff^2/n^2) for the ionized electron does not make sense because it assumes there is no change in energy in the cation even though electron-electron repulsion has decreased?
 
  • #4
fsci said:
Ahhhhhhh, that makes more sense now, thanks a ton!
So if I were to calculate E(A+) for the cation using the new Zeff values for the remaining electrons I am basically accounting for the decrease in electron-electron repulsion due to one less electron? And using IP= -13.6eV (Zeff^2/n^2) for the ionized electron does not make sense because it assumes there is no change in energy in the cation even though electron-electron repulsion has decreased?

I think that sometimes people calculate a Zeff from the IP, but remember that all of this is a fudge to take into account electron-electron repulsions. It is useful, to some extent, to compare the "shielding" provided by electrons -- Inorganic Chemistry textbooks sometimes talk about ways to estimate what Zeff is based upon the electron configuration. These are all pretty crude approximations.
 

1. What is ionization energy?

Ionization energy is the amount of energy required to remove an electron from an atom or molecule, resulting in the formation of a positively charged ion.

2. How is ionization energy calculated?

Ionization energy is calculated by measuring the amount of energy needed to remove one mole of electrons from one mole of gaseous atoms or molecules. This is typically measured in units of kilojoules per mole (kJ/mol).

3. What factors affect ionization energy?

The main factors that affect ionization energy are the size of the atom or molecule, the number of protons in the nucleus, and the electron configuration. Generally, smaller atoms with fewer electrons and a high nuclear charge will have a higher ionization energy.

4. Why does ionization energy increase as you move across a period on the periodic table?

As you move across a period on the periodic table, the number of protons and electrons increases, resulting in a stronger attraction between the positively charged nucleus and the negatively charged electrons. This makes it more difficult to remove an electron and therefore increases the ionization energy.

5. How does ionization energy relate to an atom's reactivity?

Generally, atoms with lower ionization energy are more reactive because they are more likely to lose electrons and form positively charged ions. This is why elements in the alkali metal group have low ionization energy and are highly reactive.

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