Identification of an Element based on Ionization Energies

Question

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!!

a) He
b) Be
c) O
d) Mg

Solution

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.

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