Question

If for an element (X), the values of successive ionization energies ${{I}_{1}}$, ${{I}_{2}}$, ${{I}_{3}}$, ${{I}_{4}}$, and ${{I}_{5}}$ are 800, 2427, 3658, 25024, 32824 kJ $mo{{l}^{-1}}$ respectively, then the number of valence electron present are?

Answer 1

Hint: Ionization energies are calculated for an isolated gaseous atom, when electrons are removed from its valence shell and convert it into a cation. Ionization energies are also called ionization enthalpies. when consecutive electrons are removed, then successive ionization energies are observed.

Complete answer:
We have given the successive ionization energies, ${{I}_{1}}$,${{I}_{2}}$,${{I}_{3}}$,${{I}_{4}}$, and ${{I}_{5}}$ for an element (X), which means after removal of first five electrons from the valence shell, the ionization energies are 800, 2427, 3658, 25024, 32824 kJ $mo{{l}^{-1}}$ respectively. We have to calculate the number of valence electrons in the element (X).
The ionization energies, to unpair a paired electron are high; also the ionization energy in removing electrons from the inner shell that has a fully filled configuration is much higher.
As we can see that ${{I}_{1}}$ is lower than ${{I}_{2}}$ and ${{I}_{3}}$, this means that after ${{I}_{1}}$, the remaining electrons are paired. But we can see an effective increase in the ionization energies after removal of 3 electrons, ${{I}_{1}}$,${{I}_{2}}$ and ${{I}_{3}}$, and then for ${{I}_{4}}$, and ${{I}_{5}}$the difference is not much. So, this means that ${{I}_{4}}$, and ${{I}_{5}}$ have been removed from a different shell, but ${{I}_{1}}$,${{I}_{2}}$ and ${{I}_{3}}$ are removed from the same shell, which is the valence shell.
Hence, the number of electrons in the valence shell of element (X) is 3.

Note: The ionization energies of ${{I}_{4}}$, and ${{I}_{5}}$are higher as, the electrons in them are removed from the stable noble gas configuration. So, this also proves the presence of 3 electrons in the valence shell.