Question

Magnesium loses electrons successively to form Mg⁺, Mg²⁺ and Mg³⁺ ions. Which step will have the highest ionisation energy and why?

Answer 1

\[\ce{\underset{\text{Neutral atom}}{Mg_{(g)}} -> Mg^+ + e}\] (I.E₁ = X₁) (Deb: e or e⁻)

\[\ce{\underset{\text{Unipositive cation}}{Mg^+} -> Mg^2+ + e^-}\] (I.E₂ = X₂)

\[\ce{\underset{\text{Dipositive cation}}{Mg^2+} -> Mg^3+ + e^-}\] (I.E₃ = X₃)

1. The third step will have the highest ionization energy. I.E₃ > I.E₂ > I.E₁
2. Because from a neutral gaseous atom, the electron removal is easy and less amount of energy is required. But from a di positive cation, there will be more protons than the electrons and there are more forces of attraction between the nucleus and electron. So the removal of electrons in a di positive cation becomes highly difficult and more energy is required.