Question

The first ionization enthalpies for three elements are 1314, 1680, and 2080 kJ$mo{l}^{-1}$, respectively. The correct sequence of the element is ______.

• A. O,F and Ne
• B. F,O and Ne
• C. Ne,F and O
• D. F,Ne and O

Answer 1

The correct option is A O,F and Ne

Electronic configuration of $O$, $F$, $Ne$ is

$1s^{2}2s^{2}2p^4$, $1s^{2}2s^{2}2p^5$ and $1s^{2}2s^{2}2p^8$ respectively

and its ionization gives monocation ($O^{+}$, $F^{+}$, $N{e}^{+}$) that has configuration as:

$1s^{2}2s^{2}2p^3$, $1s^{2}2s^{2}2p^4$ and $1s^{2}2s^{2}2p^7$

Since $O$ attains stable half-filled configuration after ionization, will be more stable with least ionization enthalpy. $Ne$ having completely filled configuration before ionization, so will have high ionization enthalpy and $F$ will have higher ionization than $O$ but lower than $Ne$.

Therefore order of first ionization enthalpyis:

$O$<$F$<$Ne$.

Option A is correct.