let's go through a
practice problem regarding ionization
energy. so the question is place the
following processes in order of
increasing ionization energy, so the
first process is IE1 for thallium or
the first ionization energy for thallium
meaning the energy associated with
removing the first electron from neutral
thallium. the next is IE1 for aluminum
so the first ionization energy for
aluminum, then IE3 for aluminum or the
third ionization energy for aluminum
meaning we've already removed one
electron we've removed the second
electron and now we're removing a third
electron from aluminum, and IE2 for
sodium meaning the second ionization
energy for sodium meaning we've already
removed one electron and now we are
removing the second. so if this seems
confusing go back and check out my
tutorial on the periodic table and
periodic trends where we do talk about
ionization energy and successive
ionization energies and when that all
makes sense give this one a try.
so here
are the transitions we are looking at
and let's find these elements on the
periodic table because this is a
periodic trend that we're looking at. so
here is thallium number 81, here is
aluminum number 13, and here is sodium
number 11. now let's compare these one
pair at a time. first let's look at these
two, the first ionization energy of
thallium and the first ionization energy
of aluminum. so what can we say about
these elements? they are in the same
group and thallium has many more shells
it has three more shells
then aluminum does so it's a much larger
atom. now when we remove an electron from
an atom we are removing the electron
from the outermost shell, we are removing
a valence electron and the farther away
an electron is from the nucleus the
easier it is to remove because the
farther away it is from the nucleus the
less electromagnetic attraction it feels
to the nucleus because that attraction
drops off as the distance increases. so
because thallium has more shells the
outermost electron in thallium is much
farther away than the outermost electron
in aluminum and it is therefore much
easier to remove. so the first ionization
energy for thallium is much smaller than
the first ionization energy for aluminum
because it is easier to remove that
electron, we need less energy to do it. so
we have first ionization energy of
thallium as being less than the first
ionization energy for aluminum. now let's
look at this second pair here, the first
ionization energy of aluminum and the
third ionization energy of aluminum. well
let's be clear what we mean about each
of these. the first ionization energy is
when aluminum is ionized to become
aluminium plus, the second ionization
energy goes from aluminum plus to
aluminum two plus, and the third goes
from aluminum two plus to aluminum
three plus. now for any element each
successive ionization energy is going to
get greater because as an atom becomes
more and more positively charged it's
going to become increasingly difficult
to remove negatively charged particles
away from that atom because it's going
to continue to get more and more
positive and more and more unstable, so
that means that the first ionization
energy
for aluminium is less than the third
ionization energy for aluminum, and that
would be true for any element because
the ionization energies get increasingly
larger as we remove more and more
electrons. lastly let's look at the third
ionization energy for aluminum versus
the second ionization energy
for sodium. once again the third
ionization energy is aluminum 2 plus
going to aluminum 3 plus and that means
we're going from a sodium electron
configuration, right if aluminum loses
two electrons go two elements to the
left we get to sodium it will have 11
electrons, it will have sodium's electron
configuration, and aluminum 3 plus is a
neon electron configuration which is a
noble gas and noble gas electron
configurations are very favorable, so for
aluminum 2 plus to go to aluminum 3 plus
is not that bad because it's going
towards a noble gas electron
configuration. however sodium's second
ionization energy involves going from
sodium plus to sodium two plus, now sodium
plus is the neon electron configuration
that is very favorable
that's why sodium's first ionization
energy is so low, it's very easy for
sodium to lose one electron to attain
neon electron configuration because that
is a noble gas electron configuration.
however the second ionization energy is
going to be very large because that
involves moving away from the very
stable neon electron configuration to
get to fluorine electron configuration.
so we have the third ionization energy
of aluminum which highlights a
relatively favorable process whereby
aluminum 2 plus is becoming aluminum 3
plus and attains noble gas electron
configuration versus a very unfavorable
situation, the second ionization energy
of sodium where it has to lose that
favorable noble gas electron
configuration so definitely the second
ionization energy for sodium is going to
be much larger than the third ionization
energy for aluminum and so that is how
that trend will finish up. and as it
turns out we have this already in the
correct order
we have IE 1 thallium being less than IE
1 aluminum being less than IE 3 aluminum
being less than ie 2 sodium so that is
the order for this trend.
