Can lead become isoelectronic with xenon? How many electrons would it have to lose?

1 Answer
Truong-Son N.
Sep 26, 2017

Theoretically, yes... Practically, no.

The electron configuration of lead (#Z = 82#) is:

#[Xe] 4f^14 5d^10 6s^2 6p^2#

In order to become isoelectronic with xenon, it has to lose all electrons past the xenon core, and consequently have the same number of electrons as xenon.

That's hard... It would have to lose #28# electrons and form #"Pb"^(28+)#. I would argue that that's impossible... You would have to ionize #28# times, and that means you add up the first #28# successive ionization energies.

From NIST, approximate ionization energies can be obtained. (The further you go in increased positive cation charge, the less precisely the #i#th ionization energy is known.)

That adds up to #"10089.7 eV"#... you could ionize #770# hydrogen atoms with that much energy...

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