The first ionization energy of polonium, Po, is 812 kJ/mol. Is the first ionization energy of Lv expected to be greater than, less than, or equal to that of Po? Justify your answer in terms of Coulomb’s law.

The two atoms (Lv and Po) have comparable effective nuclear charges but the valence electron in Lv would be at the greater distance from the nucleus than those in Po.
By Coulomb's law, the attractive forces between the valence electron and the nucleus decrease by the inverse square of the distance between them.

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pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-09-04T01:08:22+02:00

Following Coulomb's law, the first ionization energy of Lv is expected to be less than that of Po.

According the Coulomb's law, the magnitude of attractive/repulsive force between charged particles is directly proportional to he product of the charges and inversely proportional to the square of the distance between the charges. It is an inverse square law.

Hence;

[tex]F = \frac{Kq1q2}{r^2}[/tex]

The first ionization energy is affected by the size of the nuclear charge as well as the distance of the outermost electron from the nucleus.

As we descend from Po to Lv, the increase in size of the nuclear charge is counteracted by the addition of more shells hence the distance of the outermost electrons increases.

According to Coulomb's law, as the distance between the charges increases, the first ionization energy of Lv decreases.

Hence, first ionization energy of Lv is less than that of Po.

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