# Why is the first ionization energy of boron LESS than that of beryllium? Boron has a higher atomic number.

May 13, 2016

Consider electronic configurations of $B e$ versus $B$. Should a $p$ electron be easier to remove?
$B e , Z = 4 , 1 {s}^{2} 2 {s}^{2.}$
$B , Z = 5 , 1 {s}^{2} 2 {s}^{2} 2 {p}^{1.}$
A priori , should it be easier to remove a $p -$orbital electron, which by definition has zero electron density at the nucleus? An $s -$orbital electron, even those of the $2 s$ shell has some probability density at or near the nucleus, and this probably accounts for the differential ionization energies.