The successive 5 ionisation energies of an element are $800,2427,3658,25024$ and $32824 \mathrm{~kJ} / \mathrm{mol}$, respectively. By using the above values predict the group in which the above element is present :

Question

Aicharya · Accepted Answer

The successive 5 ionisation energies of an element are $800,2427,3658,25024$ and $32824 \mathrm{~kJ} / \mathrm{mol}$, respectively.
$\begin{aligned}
& \frac{\mathrm{IE}_2}{\mathrm{IE}_1}=\frac{2427}{800}=3.03 \\
& \frac{\mathrm{IE}}{\mathrm{IE}_2}=\frac{3658}{2427}=1.51 \\
& \frac{\mathrm{IE}}{\mathrm{IE}_3}=\frac{25024}{3658}=6.84
\end{aligned}$
$\frac{\mathrm{IE}_5}{\mathrm{IE}_4}=\frac{32824}{25024}=1.31$
Since ( $\mathrm{IE}_4 / \mathrm{IE}_3$) value is maximum, the element belongs to group 13.

The successive 5 ionisation energies of an element are $800,2427,3658,25024$ and $32824 \mathrm{~kJ} /…

Solution