Question

The first four ionisation energies of an element (in $kJ.{mol}^{-1}$) are approximately $738$, $1450$, $7.7\times {10}^3$ and $1.1\times {10}^4$. To which periodic group number does this element belong?

Answer 1

The third ionization energy is very high. Hence, the element belongs to the second group of the periodic table. It is alkaline earth metal with outer electronic configuration $n{s}^2$.

When two electrons are removed from it, the noble gas configuration is achieved. To remove the third electron, this noble gas configuration must be broken which requires very high energy.