Electronic configuration of period 3 element from its successive ionization energies Given below are the first five successive ionization energies $(\mathrm{IE})$ in $\pu{kJ mol-1}$ of an element in period 3.
  
  $$
\begin{array}{ccccc}
\hline
\mathrm{IE_1} &amp; \mathrm{IE_2} &amp; \mathrm{IE_3} &amp; \mathrm{IE_4} &amp; \mathrm{IE_5} \\
\hline
801 &amp; 2427 &amp; 3659 &amp; 25022 &amp; 32822 \\
\hline
\end{array}
$$
  
  Deduce the electronic configuration of the valence energy level of the element.
  Explain how you came to this answer.


I need clarification on this question. I have explained into using my own knowledge, therefore correct me if am wrong.

I believe that the element in period three would be Al, since there is a big jump from $\pu{3659 kJ/mol}$ to $\pu{25022 kJ/mol}$. For the first three electron $\mathrm{(3s)^2(3p)^1}$, they are all in the same energy level which is the third shell and that takes $\pu{801 kJ/mol} + \pu{2427 kJ/mol} + \pu{3659 kJ/mol}$ to remove them, after that we notice a big jump after the third ionization energy indicating the its entering a new level of energy which is the second energy level.

Please advice me if its correct or not.
Yes, your reasoning is logical and makes sense, but the question must have a typo because those IE values are the exact match for boron B, an element from period 2 right above aluminium Al (data from [1, p. 10-204], converted from $\pu{eV}$ to $\pu{kJ mol-1}$):

\begin{array}{lccccc}
\hline
\text{Element} &amp; \mathrm{IE_1} &amp; \mathrm{IE_2} &amp; \mathrm{IE_3} &amp; \mathrm{IE_4} &amp; \mathrm{IE_5} \\
\hline
\ce{B} &amp; 801 &amp; 2427 &amp; 3659 &amp; 25022 &amp; 32822 \\
\ce{Al} &amp; 578 &amp; 1816 &amp; 2745 &amp; 11580 &amp; 14842 \\
\ce{Ga} &amp; 580 &amp; 1980 &amp; 2965 &amp; 6102 &amp; 8299 \\
\ce{In} &amp; 559 &amp; 1821 &amp; 2704 &amp; 5210 &amp; \\
\ce{Tl} &amp; 589 &amp; 1971 &amp; 2878 &amp; &amp; \\
\hline
\end{array}



Another hint that it's not aluminium is the absolute value of the jump in energy between $\mathrm{IE_3}$ and $\mathrm{IE_4}$, which is too high for latter group 13 elements.
Probably the author meant group 13, not period 3, but that's just a speculation at this point.

References


Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.


Appendix

Source code for the plot, pgfplots:

\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
    \begin{axis}[
        title={Successive ionization energies of group 13 elements},
        xmin=1, xmax=5,
        ymin=0, ymax=35000,
        xlabel = $i$,
        ylabel = {$\mathrm{IE}_i$/\si{\kilo\joule\per\mole}},
        minor tick num = 1,
        mark = *,
        grid,
        legend cell align = {left},
        legend pos = north west,
    ]

    \addplot coordinates {
        (1,801)
        (2,2427)
        (3,3659)
        (4,25022)
        (5,32822)
    };
    \addlegendentry{Boron}

    \addplot coordinates {
        (1,578)
        (2,1816)
        (3,2745)
        (4,11580)
        (5,14842)
    };
    \addlegendentry{Aluminium}

    \addplot coordinates {
        (1,580)
        (2,1980)
        (3,2965)
        (4,6102)
        (5,8299)
    };
    \addlegendentry{Gallium}

    \addplot coordinates {
        (1,559)
        (2,1821)
        (3,2704)
        (4,5210)
    };
    \addlegendentry{Indium}

    \addplot coordinates {
        (1,589)
        (2,1971)
        (3,2878)
    };
    \addlegendentry{Thallium}

    \end{axis}
\end{tikzpicture}

\end{document}


Converted to PNG with the following command:

magick convert -density 120 G13_ionization_energies.pdf -quality 100 G13_ionization_energies.png