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Given their places on the periodic table I'd assume Aluminium has a higher ionization energy, because it has fewer energy levels, and is on a "righter" row on the periodic table, but in reality it is the opposite. Does anyone here know why?

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  • $\begingroup$ Note that instead of Al vs Ca comparison, it is better to compare Al vs Mg and Mg vs Ca. $\endgroup$
    – Poutnik
    May 21 at 13:11

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Ionization energy is the minimum amount of energy needed to remove most loosely bound electron (valence electron).

The atomic number of Calcium is 20. So it's orbital configuration is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$. This is the arrangement of electrons in the shells. All orbitals are completely filled. So, for removal of an electron from the outermost shell, higher amount of energy is required.

In case of Aluminium, the atomic number is 13. Its orbital configuration is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^1}$. One single electron is present in outermost p-orbital, so it can be easily removed as compared to Calcium.

So, ionization energy of Calcium is higher than the ionization energy of Aluminium.

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The 3p electrons have higher energy than 3s electrons, mainly because of better nucleus charge shielding. See also Slater's rules for electron electrostatic shielding effect.

Therefore, 3p electrons have lower ionization energy.

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  • $\begingroup$ The first ionization of calcium does not break the $3s$ subshell. $\endgroup$ May 21 at 11:40
  • $\begingroup$ @OscarLanzi Sure, it does not, it is 4s. $\endgroup$
    – Poutnik
    May 21 at 11:51

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