# What is the difference between ground state electron configurations and otherwise?

Nov 26, 2016

You can treat the ground-state electron configuration as basically a point of reference that is (at the moment) lowest in energy.

You could go in three directions from the ground-state:

• Ionization by removal of an electron (forming ${\text{B}}^{+}$)
• Ionization by addition of an electron (forming ${\text{B}}^{-}$)
• Promotion of an electron to a higher energy level (forming $\text{B"^"*}$)

Let's take boron as a normal example. Boron's ground-state configuration is $\textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} 2 {p}^{1}}$.

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(color(white)(uarr darr))" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\stackrel{}{\underbrace{\text{ "" "" "" "" "" "" "" }}}$
$\text{ "" "" } 2 p$

$\underline{\uparrow \downarrow}$
$2 s$

$\underline{\uparrow \downarrow}$
$1 s$

(1) Removal of an electron

If an electron is removed, it corresponds to an ionization energy; energy is required to remove an electron from boron, so ${\text{IE}}_{1}$ is negative.

The new electron configuration, for ${\text{B}}^{+}$, is $\textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} \textcolor{red}{2 {p}^{0}}}$ (the $2 {p}^{0}$ is to emphasize the absence of that electron).

$\underline{\cancel{\textcolor{red}{\uparrow}} \textcolor{w h i t e}{\downarrow}} \text{ " ul(color(white)(uarr darr))" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\stackrel{}{\underbrace{\text{ "" "" "" "" "" "" "" }}}$
$\text{ "" "" } 2 p$

$\underline{\uparrow \downarrow}$
$2 s$

$\underline{\uparrow \downarrow}$
$1 s$

If an electron is added, it corresponds to an electron affinity. Electron affinity basically describes what happens to the energy of the atom when you add an electron to it.

That is, if ${\text{EA}}_{1} > 0$, then you destabilize the atom (which is why electron affinities for noble gases are positive). For boron, it is $- \text{27.0 kJ/mol}$, so boron is slightly stabilized when it gains one electron.

The new electron configuration, of ${\text{B}}^{-}$, is then $\textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} \textcolor{red}{2 {p}^{2}}}$.

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(color(blue)(uarr) color(white)(darr))" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\stackrel{}{\underbrace{\text{ "" "" "" "" "" "" "" }}}$
$\text{ "" "" } 2 p$

$\underline{\uparrow \downarrow}$
$2 s$

$\underline{\uparrow \downarrow}$
$1 s$

(3) Promotion of electron to higher energy level

You can do this by shooting, say, the right wavelength of laser at a sample of boron, and some of the sample will get excited, giving the electron exactly the right energy to get promoted to a higher energy level.

A valid new energy level for boron is the $3 s$ orbital:

$\underline{\textcolor{b l u e}{\uparrow} \textcolor{w h i t e}{\downarrow}}$
$3 s$

$\underline{\cancel{\textcolor{red}{\uparrow}} \textcolor{w h i t e}{\downarrow}} \text{ " ul(color(white)(uarr darr))" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\stackrel{}{\underbrace{\text{ "" "" "" "" "" "" "" }}}$
$\text{ "" "" } 2 p$

$\underline{\uparrow \downarrow}$
$2 s$

$\underline{\uparrow \downarrow}$
$1 s$

That would be known as an electronic excitation. That changes the electron configuration to the one for $\text{B"^"*}$:

$\textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} 2 {p}^{0} \textcolor{red}{3 {s}^{1}}}$

The $2 p$ subshell becomes empty because the only electron in it was excited up to the $3 s$ orbital.

(This is an unstable state, so soon after it forms, the electron will fall back down to the original $2 p$ orbital and we'll see the ground state again, $1 {s}^{2} 2 {s}^{2} 2 {p}^{1}$.)