Ionization period trend

[luludaly2B]

In one of the review worksheets, the question asked to rank Be, B, C, N, O and F in order from least ionized to most ionized. The correct answer was B<Be<C<O<N<F. Can someone explain why this is correct. I thought that the general trend over the periodic table is that elements have higher ionization as you move to the right. Is that incorrect? Thanks!

[Jeffrey Hablewitz 2I]

You are right that ionization energy generally increases from left to right. However, there are two inconsistencies in the ionization energy of period 2 elements: boron and oxygen. Boron has a lower ionization energy than beryllium because boron has one unpaired electron in the 2p subshell, which is more easily removed than an electron from beryllium's full 2s subshell. Oxygen has a lower ionization energy than nitrogen because oxygen has one paired electron in the 2p subshell, which is more easily removed than an electron from nitrogen's half-full 2p subshell.
As for the ordering of the answer, the direction of the inequality signs suggests that the elements were listed from lowest to highest ionization energy. Hope this helps!

[Chanel Mao 3D]

Hi!
You are correct in that the general trend shows higher ionization energies for elements towards the right. Elements like Be/B and N/O are exceptions to this trend because of their electron configurations and their overall stability after we take away an electron. For example, the electron configuration of N is 1s^2 2s^2 2p^3, which is relatively stable (because of the half filled 2p shell). If we take away an electron, it will require more energy because N is going from a relatively stable state to a less stable state. Hope this helps!

[605721741]

It is harder to take away electrons from electronegative atoms like oxygen or fluorine. This is why there is a trend that ionization energy increases as we move to the right of the periodic table.