Increasing order of ionization energy:
Ba < Pb < P
Electronic configurations of given elements
Pb: [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p²
P: [Ne] 3s² 3p³
Ba: [Xe] 6s²
- All the given elements have stable half-filled subshells.
- But phosphorus has the smallest atomic size among them, so it will have the lowest tendency to remove an electron from its valence shell and hence will require the highest amount of ionization energy.
- Barium has a higher atomic size than lead, therefore it has lower ionization energy than lead.
How are ionization energy and atomic size related?
- The force of attraction is stronger when the electron is closer to the nucleus, and as a result, more energy is needed to overcome this attraction and remove the electron.
- Therefore, the ionization energy increases with decreasing radius and vice versa.
Learn more about trends of ionization energy here:
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