Electronic configuration of Aluminum in excited state?

2 Answers

Answer:

There could be three different configuration.

Explanation:

Aluminium can release three electrons (don't be confused; electrons are released one by one) at most, so we call it trivalent.

Whenever Aluminium releases a electron, It needs a minimum energy supply, which is called ionisation potential.

When it releases its first electron from its valence shell, the ionistation potential or ionisation enthalpy needed is called first ionisation potential.

Like this, we can define second and third ionisation potentials.

So, every time it releases elctrons, it absorbs more energy and becomes excited.

So, Electronic Configuration after becoming a univalent ion (#Al^+#):
#1s^2 2s^2 2p^6 3s^2# [#3p# shell is empty now.]

Electronic Configuration after becoming a bivalent ion (#Al^(2+)#):
#1s^2 2s^2 2p^6 3s^1#

Electronic Configuration after becoming a trivalent ion (#Al^(3+)#):

#1s^2 2s^2 2p^6# [Now #3s# is empty]

The Most Excited State here is the third one.

Jun 3, 2017

#[Ne]3s^2 color(red)(3p^0) 4s^1#

for the first excited state, #3p -> 4s# transition, changing the valence electron configuration from #3s^2 3p^1# to #3s^2 4s^1#. This is represented as (ignoring the #3s#):

#ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))#
#underbrace(" "" "" "" "" "" "" "" "" "" "" "" "" "" ")#
#" "" "" "" "" "" "3d#

#ul(uarr color(white)(darr))#
#4s#
#" "#

#ul(cancel(uarr) color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))#
#underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "3p#


The normal electron configuration of aluminum is #[Ne]3s^2 3p^1#. Atomic excitations follow the following selection rules:

  • The change in total angular momentum, #DeltaL#, must be #bb(pm1)#, where #L = |sum_k m_(l,k)|#, #k# indicates the #k#th electron, and #m_l# is the magnetic quantum number of the orbital.
  • There must be no change in total spin, #S = |sum_i m_(s,i)|#, where #k# indicates the #k#th electron and #m_s# is the spin quantum number of the electron.

For the total spin #S# to not change, i.e. #DeltaS = 0#, our only option for aluminum is to transition from a half-filled orbital to an empty orbital.

Initially, since we have only one #3p# electron, there is no sum and we have:

#S_i = |m_(s,1)|#

#= 1/2#

for:

#ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))#
#underbrace(" "" "" "" "" "" "" "" "" "" "" "" "" "" ")#
#" "" "" "" "" "" "3d#

#ul(color(white)(uarr darr))#
#4s#
#" "#

#ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))#
#underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "3p#

Now, since for a transition upwards, only #DeltaL = pm1# is allowed, we should consider that right now, for the one #3p# electron,

#L_i = m_(l,1) = |-1| = 1#.

The nearby orbitals in energy are #3d# and #4s#. For aluminum, the #4s# orbital is closer in energy to the #3p# than the #3d# is.

That means the #3p# electron can transition upwards into the #4s# for the first excited state (which is what we want). This gives:

#DeltaL = L_f - L_i = 0 - 1 = -1#,

which is allowed. Therefore, the first excited state electron configuration is:

#color(blue)([Ne]3s^2 color(red)(3p^0) 4s^1)#

due to a #3p -> 4s# transition, with #DeltaL = 0 - 1 = -1#, and #DeltaS = 1/2 - 1/2 = 0#. We have ended up with:

#ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))#
#underbrace(" "" "" "" "" "" "" "" "" "" "" "" "" "" ")#
#" "" "" "" "" "" "3d#

#ul(uarr color(white)(darr))#
#4s#
#" "#

#ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))#
#underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "3p#