Definition (First Ionization Energy):
The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.
\( M(g) \rightarrow M^+(g) + e^- \)
Key point: State symbols must be (g).
Factors Affecting Ionization Energy
- Nuclear Charge (Z): More protons = stronger attraction = Higher IE.
- Distance / Shielding: Further from nucleus + more inner shells blocking = Weaker attraction = Lower IE.
Trend 1: Down a Group (Decreases)
As you go down a group (e.g. Li → Na → K):
- Atomic Radius Increases: Valence electrons are further away.
- Shielding Increases: More inner shells repel the outer electron.
- Result: Attraction to the nucleus is weaker, so less energy is needed to remove an electron.
Trend 2: Across a Period (General Increase)
As you go across a period (e.g. Li → Ne):
- Nuclear Charge Increases: More protons added.
- Shielding is Constant: Electrons are added to the same main energy level.
- Result: Attraction increases, radius decreases, difficult to remove electron → Higher IE.
Deep Think Concept
Critical Exceptions (The Evidence for Sublevels)
There are two dips in the general upward trend across Period 3 (and 2). These prove sub-shells exist.
Mg → Al (Dip)
Group 2 to 13
Mg: \([Ne] 3s^2\)
Al: \([Ne] 3s^2 3p^1\)
The 3p electron in Al is higher in energy and further from the nucleus than the 3s. It is easier to remove.
P → S (Dip)
Group 15 to 16
P: \(3p^3\) (All singly occupied)
S: \(3p^4\) (One pair)
In Sulfur, the 4th electron must pair up. Electron-electron repulsion within the same orbital makes it easier to remove.
Student Practice Set
1.
Which equation represents the First Ionization Energy of Sodium?
2.
Why does Oxygen have a lower First IE than Nitrogen?