Tip:
Highlight text to annotate it
X
Let's look at the electron configuration
of sodium. First we need to look at our periodic table
and find sodium. Sodium has
atomic number 11. Sodium has an atomic number of 11
so we need 11 electrons in our electron configuration.
Using the periodic table set up in blocks
we find that sodium is right here.
So when we start to write the expanded notation
we start with the 1s and put two electrons
in the 1s orbital. So we write 1s2.
We're up to two electrons. We need nine more.
We move down a row.
We fill the 2s orbital so we write 2s2.
We're up to four electrons.
We can fill the 2p orbital
with six electrons so we write 2p6.
We have 10 electron so far,
we need to add one more.
Here is sodium on the periodic table
so we write 3s1.
We can go back and see how many electrons we put in by counting
the superscripts. We have two,
4, 10, 11 electrons.
The superscripts tell us
the number of electrons in each subshell,
and if you count them you'll figure out how many electrons you have.
So in the noble gas configuration
we move from
sodium's place on the periodic table we move
up a row and all the way over to find
the nearest noble gas and for this
its neon. So
neon takes the place of 1s2 2s2 2p6.
We put Neon
in brackets and then whatever's left
we write next to it. So the noble gas configuration
is "neon 3s1". Again to find the noble gas
we look at sodium, move
up a row and all the way over to find
neon. And neon has ten electrons which is represented here
as 1s2 2s2 2p6 so we write [Ne]3s1.
We can also find the
valence shell electrons, and the valence shell is the highest value of n.
And if we look at our expanded notation
the highest value of n is n = 3.
so our third shell is our valence shell.
We count the number of electrons
in that shell, and we only have
one electron in the n = 3 shell.
So we have one valence electron.