Given values (in kJ mol⁻¹):
B = 801, Al = 577, Ga = 579, In = 558, Tl = 589
General Expected Trend
Down a group, ionization enthalpy is expected to decrease due to:
- Increase in atomic size.
- Increase in shielding of valence electrons from the nucleus.
So, a simple trend would be: B > Al > Ga > In > Tl. The actual values deviate from this.
Why Ga > Al?
- Al comes immediately after the s-block elements (Mg), so it has no d-electrons in the inner shells.
- Ga comes after the 3d series. The presence of 10 inner 3d electrons gives poor shielding.
- Poor shielding by 3d electrons increases the effective nuclear charge felt by Ga’s valence electrons.
- Therefore, Ga holds its valence electron slightly more strongly than Al, so IE₁(Ga) > IE₁(Al).
Why Tl > In?
- From Ga to In, normal effects dominate: size increases and shielding increases, so IE₁ decreases (579 → 558 kJ mol⁻¹).
- Tl comes after both 4f and 5d electrons.
- 4f (and also d) electrons shield the valence electrons very poorly, so the effective nuclear charge on the valence shell of Tl becomes relatively high.
- As a result, Tl’s outer electron is held more strongly than in In, so IE₁(Tl) > IE₁(In), giving the rise from 558 to 589 kJ mol⁻¹.
Summary
The deviations B > Al < Ga, and In < Tl are explained by poor shielding of d- and f-electrons, which increases effective nuclear charge and hence ionization enthalpy for Ga and Tl.