MCQ
The correct order of $IE_2$ is
- A$Ne > F > O > N$
- B$O > F > Ne > N$
- ✓$Ne > O > F > N$
- D$O > Ne > F > N$
✓
Answer
Correct option: C.
$Ne > O > F > N$
c
Second $IE$ required to remove $an ^{-}$for a $1+$ cation in the gaseous state,
Second $IE$ required to remove $an ^{-}$for a $1+$ cation in the gaseous state,
$A ^{+}( g ) \rightarrow A ^{2+}+ e ^{-}$
$IE _2$ 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 $IE.$
$\rightarrow F ^{+}$is having less $IE _2$ 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: $Ne \,>\, O \,>\, F\, >\, N$
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