The first four Ionization Energies (IE) of an element are given, which of the following elements is it most likely to be?
!!IE_1 = 0.8; IE_2 = 1.7; IE_3 = 14.8; IE_4 = 21!!
In the given element, the Ionization Energy abdrubtly increases after removing two electrons, hence we need to find and element which achieves stability after the removal of 2 electrons.
He: Since Helium has only two electrons, the given data cannot be for helium.
Be: The electronic configuration of Be is !!1s\^2 2s\^2!!. Therefore, after removing two electrons we are left with a fully filled s-orbital. Since s-orbital is closer to nucleus, it will take more energy to remove these electrons. Thus, the given data can be for Berelium.
O: The electronic configuration of O is !!1s\^2 2s\^2 2p\^4!!. Therefore, after removing two electrons, we are left with 2 electrons in p-orbital, which is not a stable configuration, hence we should not expect any jumps in ionization energy here.
Mg: The electronic configuration of Mg is !!1s\^2 2s\^2 2p\^6 3s\^2!!. Therefore, after removing 2 electrons, we are left with a fully filled p-orbital. Since s-orbital is farther away from nucleus compared to s, it shouldn't be very difficult to remove electrons from this orbital. Therefore we should not expect such a high jump in ionization energy.
Therefore, the given element, most likely, is Berelium.