Question
The first, second and third ionisation energies $(E_1 ,E_2$ & $E_3 )$ for an element are $7\,eV, 12.5\,eV$ and $42.5\,eV$ respectively. The most stable oxidation state of the element will be
It can be seen that $E_{3}$ is very high at 42.5 ev in to $E_{1}(7 eV )$ and $E_{2}$ comparision to $(12.5 eV )$ which implies the element reached a stable electronic configuration at $2^{n d}$ ionization and hence, $E_{3}$ is very high.
Hence, D is the correct anower
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| सूची $I$ (यौगिक) | सूची $II$ ($Pk_a$ मान) |
| $A$.ethanol | $I$. $10.0$ |
| $B$. phenol | $II$. $15.9$ |
| $C$. $m$-nitrophenol | $III$. $7.1$ |
| $D$. $p$-nitrophenol | $IV$. $8.3$ |
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