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$+3$
- ✓$+2$
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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$Si_2H_6(g)\,\,+\,\,H_2(g)\,\,\to \,\, 2SiH_4(g),\,\,\Delta H\,\,=\,\,-\,11.7\,\,kJ/mol$
$SiH_4(g)\,\,\to \,\,SiH_2(g)\,\,+\,\,H_2(g),\,\, \Delta H\,\,=\,\,+239.7\,\,kJ/mol$
$\Delta H_f^o\,\,Si_2H_6(g)\,\,=\,\,80.3\,\,kJ/mol$
......$kJ/mol$