# What is the order of the second ionization energies for elements in the second row of the Periodic Table?

Jun 25, 2017

WARNING! Long answer! The order is $\text{Be < C < B < N < F < O < Ne < Li}$.

#### Explanation:

The second ionization energy (${\text{IE}}_{2}$) is the energy required to remove an electron from a 1+ cation in the gaseous state.

$\text{X"^"+""(g)" → "X"^"2-""(g)" + "e"^"-}$

Just like the first ionization energy, ${\text{IE}}_{2}$ is affected by size, effective nuclear charge, and electron configuration.

We would expect second ionization energies to increase from left to right as the ionic size decreases.

Here's a table listing the electron configurations of the ions involved.

$\boldsymbol{\text{Group"color(white)(m)"Element"color(white)(m)"Config"color(white)(m)"Ion"color(white)(m)"Config}}$
$\textcolor{w h i t e}{m l l} 1 \textcolor{w h i t e}{m m m m l l} \text{Li"color(white)(mmml)"2s"color(white)(mmml)"Li"^"+"color(white)(mm)"1s"^2}$
$\textcolor{w h i t e}{m l l} 2 \textcolor{w h i t e}{m m m m l l} \text{Be"color(white)(mmm)"2s"^2color(white)(mmm)"Be"^"+"color(white)(mll)"2s}$
$\textcolor{w h i t e}{m} 13 \textcolor{w h i t e}{m m m m m} {\text{B"color(white)(mmml)"2s"^2 "2p"color(white)(mll)"B"^"+"color(white)(mml)"2s}}^{2}$

$\textcolor{w h i t e}{m} 14 \textcolor{w h i t e}{m m m m m} \text{C"color(white)(mmml)"2s"^2 "2p"^2color(white)(ml)"C"^"+"color(white)(mml)"2s"^2 } 2 p$
$\textcolor{w h i t e}{m} 15 \textcolor{w h i t e}{m m m m m} {\text{N"color(white)(mmml)"2s"^2 "2p"^3color(white)(ml)"N"^"+"color(white)(mml)"2s"^2 "2p}}^{2}$
$\textcolor{w h i t e}{m} 16 \textcolor{w h i t e}{m m m m m} {\text{O"color(white)(mmml)"2s"^2 "2p"^4color(white)(ml)"O"^"+"color(white)(mml)"2s"^2 "2p}}^{3}$

$\textcolor{w h i t e}{m} 17 \textcolor{w h i t e}{m m m m m} {\text{F"color(white)(mmmll)"2s"^2 "2p"^5color(white)(ml)"F"^"+"color(white)(mml)"2s"^2 "2p}}^{4}$
$\textcolor{w h i t e}{m} 18 \textcolor{w h i t e}{m m m m m} {\text{Ne"color(white)(mmm)"2s"^2 "2p"^6color(white)(ml)"Ne"^"+"color(white)(mll)"2s"^2 "2p}}^{5}$

And here is a plot of the ionization energies.

We notice three things:

1. $\text{Li}$ has the highest ${\text{IE}}_{2}$, because to remove the second electron we must break the stable ${\text{1s}}^{2}$ noble gas shell.

2. $\text{B}$ has a greater ${\text{IE}}_{2}$ than $\text{C}$. This is probably due to the extra stability of the ${\text{s}}^{2}$ subshell in the $\text{B"^"+}$ ion.

3. $\text{O}$ has a greater ${\text{IE}}_{2}$ than $\text{F}$. The $\text{F"^"+}$ ion has a ${\text{p}}^{4}$ configuration in which electronic repulsions raise the energy and decrease the ${\text{IE}}_{2}$.

Thus, the order of second ionization energies is

$\text{Be < C < B < N < F < O < Ne < Li}$.