Question

The correct order of second ionisation energy is:

• A. Ne $>$ F $>$ O $>$ N
• B. O $>$ F $>$ Ne $>$ N
• C. Ne $>$ O $>$ F $>$ N
• D. O $>$ Ne $>$ F $>$ N

Answer 1

Answer: (C)

Ne $>$ O $>$ F $>$ N

Second $IE$ required to remove an $e^{-}$ for a $1+$ cation in the gaseous state,

$A^{+}\left(g\right)⟶A^{2+}+e^{-}$

$I{E}_2$ is affected by the size, effective nuclear charge and electronic configuration. So, on moving from $N$ to $Ne$ size of ion decreases, thereby increasing the $IE$.

$⟶F^{+}$ is having less $I{E}_2$ than $O^{+}$ due to the interelectronics repulsion due to its' high electronegativity, it requires comparatively less energy to remove $e^{-}$ from its' outer shell.

Thus, the correct order is: $Ne>O>F>N$

Hence, the correct option is $\text{C}$