First ionization energy (special cases)

 

Generally, ionization energy increases from left to right along a period, but there are special cases against this general rule. For example, the ionization energy of of Boron (B) is in fact lower than the one of Be (Beryllium), while the general trend indicates the first ionization energy of B is higher than Be. The reason this exception happens is that Boron (B) has only one electron in the p orbital and p orbital has higher energy than s orbital, so this electron in the p orbital tends to go to a lower energy level or lose in order to make the atom more stable. (The lower the energy, the more stable the atom will be.) Therefore the energy required to remove one electron from the p orbital of Boron (B) is lower than the one required to remove one electron from the s orbital (fully filled) of Beryllium (Be).  The same thing applies to the other periods as well, such as the first ionization energy of Al is lower than that of Mg.

 

Another special case is, in the same period,the atom that has 3 electrons in the p orbital has a higher first ionization energy than the one that has 4 electrons in the p orbital. The thing happens because we all know that the first three electrons in the p orbital are separately placed in P1,P2 and P3 orbitals, according to Hund's Rule. Then the 4th one has to be paired with the electron in the P1 orbital. Since these two electrons have negative charges, they repel each other slightly, making this 4th electron easier to lose, and that is why the the atom that has 3 electrons in the p orbital has a higher first ionization energy than the one that has 4 electrons in the p orbital in a period. Example is, the first ionization energy of oxygen(O) is lower than that of nitrogen (N). This also applies to the other periods.