Why is the ionization energy of beryllium, #Z=4#, GREATER than the ionization energy of boron, #Z=5#?

1 Answer
Aug 27, 2017

Answer:

Consider the electronic configuration of each element.......

Explanation:

And we consider the ELEMENT not the molecule, inasmuch as this is how ionization enthalpies are defined......

#"Atom(g)"+Deltararr"Atom(g)"^+ + e^(-)#

For #Be# we gots #Z=4#, #1s^(2)2s^(2)#

For #B# we gots #Z=5#, #1s^(2)2s^(2)2p^1#

The ionization of boron involves removal of a #"p-electron"#, which has NO electron density at the nuclear core, and thus should be easier to remove than a #s# electron, which can lie closer to the nuclear core.

For #N# we gots #Z=7#, #1s^(2)2s^(2)2p^3#

For #O# we gots #Z=8#, #1s^(2)2s^(2)2p^4#

For nitrogen, we have a half-filled #"p-shell"# which is energetically stabilized with respect to #"Hund's rule"# of maximum multiplicity. For the oxygen atom, even tho it has #Z=8#, the electronic configuration of the cation is SLIGHTLY stabilized by Hund's rule. None of this treatment is any substitute for reading the relevant section of your text.