Question
The first, second and third Ionisation energies $(E_1 , E_2$ and $E_3)$ for an element are $7\, eV, 12.5\,eV$ nnd $42.5\, eV$ respectively. The most stable oxidation state of the element will be
✓
Answer
d
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→$\begin{matrix}
\begin{matrix}
\,\,\,\,\,C{{H}_{3}} \\
| \\
\end{matrix}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}-C-C{{H}_{2}}C{{H}_{2}}OH \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\end{matrix}$ $\xrightarrow[{{H_2}O,heat}]{{{K_2}C{r_2}{O_7};{H_2}S{O_4}}}A$ $\xrightarrow{{SOC{l_2}}}B\xrightarrow[{(2{\kern 1pt} mole)}]{{{{(C{H_3})}_2}NH}}C$ $\xrightarrow[{2.{\kern 1pt} {H_2}O}]{{\begin{array}{*{20}{c}}
{1.{\kern 1pt} LiAl{H_4}} \\
{diethyl{\kern 1pt} {\kern 1pt} ether}
\end{array}}}D$