From the following elements: lithium, beryllium, and fluorine, predict which one will have the largest second ionisation energy. Explain your answer.

The element that has the largest second ionization energy is;

Lithium, Li

Explanation:

The given elements and their positions on the periodic table are;

1) Lithium is located in group 1, period 2 position on the periodic table

2) Beryllium is located in group 2, period 2 position on the periodic table

3) Fluorine is located in group 17, and period 2 position on the periodic table

The first ionization energy of the atoms in the periodic table increases across the periods from left to right and also increases from bottom to the top within a group

However  for the second ionization energy, which is the energy required to remove an electron from an atom that has one less electron, we have;

The electron configuration of the given of ions that take part in the reaction for the removal of the second ionization energy are;

Li⁺ = [He} 2s⁺

Therefore, the second ionization energy is the removal of an electron from the stable duplet of the helium atom (the core of the lithium atom) which require about 7297 kJ/mol

For beryllium, we have;

Be⁺ = [He]2S¹⁺

Therefore, the second ionization requires the removal of the remaining one electron in the valence shell, which will then lead to the remaining structure of the stable helium core and therefore, the second ionization energy of beryllium is about 1757 kJ/mol, which is low compared to helium

For fluorine, we have;

F⁺ = [He] 2s² 2p⁴⁺

Therefore, the second ionization of fluorine remains high due to the increased attraction of the positively charged fluorine ion for electrons than electrically neutral fluorine atom that has equal number of protons and electrons such that the second ionization energy of fluorine is higher than the first ionization energy and is about 3357 kJ/mol

Therefore, the element that has the largest second ionization energy is lithium, Li.

5.6⋅10−^19  j

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