The order of the energies would be :
number or shell for $\mathrm{H}$ - like species.
$E_{n^{t h}}=\frac{-13.6}{n^{2}} e V$
$s, p, d$ are subshell of the same orbit
Therefore, all the electrons will have the same energy.
$\therefore E_{1}=E_{2}=E_{3}$
$\therefore$ Option $\mathrm{C}$ is correct.
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Reason : $sp$ hybridised carbon atoms of acetylene are considerably electronegative.
$\begin{array}{*{20}{c}}
{C{H_3} - CH = C - C{H_2} - C{H_3}} \\
{|\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2} - C{H_2} - C{H_3}}
\end{array}$