Question

The energy needed to remove the first electron from an atom is the first ionization energy. The energy needed to remove the second electron from an atom is the second ionization energy. The definition of third and fourth ionization energy is similar. which of the atoms below would you expect to have the largest 3rd ionization energy?

Please include explanation.

a) Na

b) P

c) Mg

d) Al

e) Si

Answer 1

Electronic configuration of Na = 12226² 20635' Electronic configuration of p= 1322122p6 35+3p3 Electronic configuration of My = 1822332p6392 Electronic configuration of Al = 15²24² 2p 6 34² 3p' Electronic configuration of si = 182122p63323p? Naw for third inpation of all atoms, we remove two election and then Let's us start from . look for removal of third election, how it will be easy on to removal the election (3rd). On that Basis ionization energy value will be After removal of two electorion , electronic configuration wereld Electronic configuration of Ma² = 18 ²2x²2ps Electronic configuration of pr. 1222322p63823p! Electronic configuration of my 2t - 18²2s²2p6 Electronic configuration of Al 24 = 18221²2pl 25' Electronic configuration of sitt = 18223²2p6 38² Now for removal of third electron from Nazt is not rasy as electron is in pi orbited and also it has not attained any noble gas configuration o f any atom attained Noble gas configuration then removal of nent electron will be hard so in all above cases only Might and si2+ attain noble you configuration. But removal of electron from wat will be hard as ap orbital is near to nuclear, so eleetons are tightly held while in care size the elections are in 34 orbital, so third ionization for Na 2t > siet. But for Mght as it hal attained noble Scanned with calgas configuration and also last electrons are in ap orbital

& it will be very hard to remove third electon from مواطنين مع مهل سنت کی مع ہر اللہ نے و در ح ال تو Scanned with CamScanner