Question

Question #280f7

Answer 1

We know that ionization potential or the first ionization potential is the energy required to remove the outermost electron from a neutral gaseous atom of an element.

Similarly second and third or successive ionization potentials relate to the removal of the second and third or successive electrons respectively. As removal of each successive electron leaves increasing positive charge on the nucleus these are higher than the first ionization potential. For an element `"M"`

`"M " -> "M"^+ + e^-` (`"I"_1 =` First Ionization Potential)
`"M"^(+" ") -> "M"^(++) + e^-` (`"I"_2 =`Second Ionization Potential)
`"M"^(++) -> "M"^(+++) + e^-` (`"I"_3 =`Third Ionization Potential)

The ionization potential depends on various factors which include

• size of the atom,
• charge on the nucleus,
• screening effect of the inner electrons, and
• type of electron to be removed.

For lithium the atomic number is `3`, therefore its electronic configuration is [`"He"`] `2"s"^1`

Now ionization potential of `"Li"^(++)` ion is same as `"I"_3 =`Third Ionization Potential) of `"Li"`
Notice the striking resemblance of `"Li"^(++)` ion with Hydrogen atom where we have single electron orbiting the nucleus.

We know that for a single electron system Bohr's atomic model is able to predict the binding energy levels quite accurately. As such using the formula for a nucleus with `Z` protons, the energy levels are given by

`E_{n}=-Z^{2}R_ {E } / n^{2}`
where `R_E` is called the Rydberg energy and is `=13.6eV` or `2.178 × 10^(−18) J`
`n` is principal quantum number of the electron.

With `Z= 3 and n= 1`, the 3rd ionization energy of the `"Li"` atom or

Ionization potential of `"Li"^(++)` ion`= 9xx13.6 =122.4 eV`

This calculated value compares well with experimental result of `122.45429eV`.