The ionization energies of rubidium and silver are 4.18 and 7.57 eV, respectively. Calculate the ionization energies of an H atom with its electron in the same outermost orbitals as in these two atoms and account for the differences in values in these different elements.

Question

Corey Leach

Chemistry

17 July, 05:25

The ionization energies of rubidium and silver are 4.18 and 7.57 eV, respectively. Calculate the ionization energies of an H atom with its electron in the same outermost orbitals as in these two atoms and account for the differences in values in these different elements.

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Eliza Jones · Answer 1

0.544 eV; Ag is a smaller atom.

Explanation:

1. Ionization energy of hydrogen

The outermost electrons in Rb and Ag are in 5s orbitals.

The formula for the energy of an electron in a hydrogen atom is

E = - 13.6/n² eV

For a hydrogen atom in a 5s orbital,

E = - 13.6/5² = - 13.6/25 = - 0.544 eV

The ionization energy would be 0.544 eV.

2. Rb vs Ag

The first electrons to be removed from Rb and Ag are in 5s orbitals.

The atomic radius of Ag is less than that of Rb because, as you go from left to right across the Row, you are adding 10 protons to the nucleus and 10 electrons to the outer shell.

The added electrons do not effectively shield each other from the attraction of the nucleus, so the 5s electron of Ag is closer in.

It takes more energy to remove the electron from silver, so the ionization energy of Ag is greater than that of Rb.