The successive ionization energies of boron are shown in the table below : Ionization 1 st 2 nd 3 rd 4 th 5 th Ionization energy/ KJ mol - 1 799 2420 3660 25000 32800 Why is there a large increase between the third and fourth ionization energies?

The atomic number of Boron is 5 . The symbol used for Boron is B . The electronic configuration of Boran is 1 s 2 2 s 2 2 p 1 . The first ionization energy of Borane. i.e; B → B + + e - = 799 kJmol - 1 and the electronic configuration becomes 1 s 2 2 s 2 2 p 0 . The second ionization energy of Boron. i.e; B + → B 2 + + e - = 2420 kJmol - 1 and the electronic configuration becomes 1 s 2 2 s 1 2 p 0 . The third ionization energy of Boron. i.e; B 2 + → B 3 + + e - = 3660 kJmol - 1 and the electronic configuration becomes 1 s 2 2 s 0 2 p 0 . The fourth ionization energy of Boron. i.e; B 3 + → B 4 + + e - = 25000 kJmol - 1 and the electronic configuration becomes 1 s 1 2 s 0 2 p 0 . The fifth ionization energy of Boron. i.e; B 4 + → B 5 + + e - = 32800 kJmol - 1 and the electronic configuration becomes 1 s 0 . The large energy gap is seen between third and fourth ionization energy because of the follwoing four factors listed below: within the same shell the penetration effect decreases in the order : s > p > d > f . 1 s 2 orbital is much more stable than 2 s 1 orbital. The nuclear charge becomes higher,so the ionization energy increases.

The successive ionization energies of boron are shown in the table below : Ionization 1 st 2 nd 3 rd 4 th 5 th Ionization energy/ KJ mol - 1 799 2420 3660 25000 32800 Explain how these figures confirm that the electronic structure of Boron is 2 , 3 .

The electronic configuration of Boron is 1 s 2 2 s 2 2 p 1 . So from the electronic configuration, it is clear that the removal of electrons from 2 p 1 requires very less amount of energy, so the first ionization energy will be very less, but after the removal of 2 s 2 electrons, a sharp pick is seen in the ionization energy because of the high nuclear charge or the internuclear force of attraction between the electron and nucleus is very high, so the removal of an electron is very difficult So a great amount of energy is needed to remove electrons from an atom. From the data given it can be concluded that there are three electrons that must be present in the outermost orbit of an atom, and thus it belongs to the group 13 and the electronic configuration would become 2 , 3 .