MCQ
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
- A$+1$
- B$+4$
- C$+3$
- ✓$+2$
✓
Answer
Correct option: D.
$+2$
d
The +2 onidation state will be the most stable oridetion state.
The +2 onidation state will be the most stable oridetion state.
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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