Electron Configurations

The section on quantum numbers talked about the different types of orbitals (s, p, d, etc.) that electrons are in around a nucleus. Here, we're going talk about filling those orbitals up with electrons. I should note that for a few cases (mainly in the transition metals), there are exceptions. These exceptions won't be covered (for now, anyway)

You know, in a perfect world (perfect for chem students, anyway), you'd take all those little electrons, start with a 1s orbital, and work your way up in numerical order (1s, 2s, 2p, 3s, etc). "But Noooooooooo," as John Belushi once said. What actually happens is once you start putting electrons in those orbitals, they don't keep the order they were in after 3p. This is because once you add electrons to orbitals, their energies change and causes a lot of criss-crossing. The actual order for filling in electrons is:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, etc.

Oh no, MEMORIZATION! No, lucky for us - yes, myself included - there have been a couple of memory aids designed to help keep these things in line. One of them has been right under your nose the whole time!

Memory Aids for Electron Configurations

__Memory aid #1__

This way has a bunch of variations. This is just one of them. First, write out the orbitals in numerical order, starting a new row for every n. Make sure you keep you l's in order (s, p, d, f; you can keep going, but there aren't any known g orbitals with electrons in them)

Starting at 1s and moving diagonally, always right to left (not zig-zaging), we get the actual order of orbitals.

__Memory aid #2__

This is the one I like. I always forgot how to draw the thing above and how to draw the arrows. But with this memory aid the thing has already been draw out. In fact, you've been looking at it since you started the course. It's . . .

*Drum roll*

. . .the *Periodic Table!* You see, the periodic table wasn't draw the way it is
because it looked pretty. It was put together based on the fact that many of the atoms
behave similar in chemical reactions. Even more important here, it's also based on the
atoms' electron configurations. If we drew out the electron configurations for every
element, we would see a pattern. But tell you what, just to keep things moving along (and
to keep my fingers from cramping), let's just do four. Let's look at the a couple of
elements from the first column. Sodium (Na) has 11 electrons. If We start filling in these
in using our known order above (remember s orbitals can't hold any more than two, and p's
no more than six, etc.), we see that Na has this configuration (Note: the superscripts
tell us how many electrons are in each orbital)

1s^{2}, 2s^{2}, 2p^{6}, 3s^{1}

And Potassium (K), with 19 electrons, has:

1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6},
4s^{1}

Now look at the last orbital for each element, both have are ns^{1},
where the row number the element is in is the same number as n.

Let's pick a couple of transition metals (you know, that skinny part in the middle). Vanadium (V), with 23 electrons has this configuration:

1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6},
4s^{2}, 3d^{3}

The one right below it, molybdenum (Mo), has 42 (I'll show you a shorter way to do this in a minute):

1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6},
4s^{2}, 3d^{10}, 4p^{6}, 5s^{2}, 4d^{3}

Again, this column is just like the first one. They both end with
the same number of electrons in the same type of orbital, nd^{3}. This time,
however, the n is one *less* than the row it's in.

So with this pattern you can divide the periodic table into four "blocks." The first block is the "s block," which are the first two columns, the alkalines and alkaline earth metals (say, can't you only fit two electrons in an s orbital?), then the "p block" has the last six columns (say can't you only fit a total of six electrons in the p orbitals?). For both of these blocks, the primary quantum number is the same as the row number. The "d block" is the transition metals (say, there are ten columns of transition metals, and can't . . . oh you get the picture), their primary quantum number is the row number minus 1. Finally, the "f block" is the lanthanide and actinide series at the bottom, their primary quantum number is the row minus two.

Examples

Aluminum (Al)

Since I like the second method, I'll start with it. Aluminum is atom
#13. To get to aluminum from atom #1, H, we have to go through the 1st row, 1s (even
though He is above the p block it's consider to be part of the s), the second row,
orbitals 2s and 2p, and the s block of the third row, 3s. Aluminum is the *first*
atom in the 3rd row *p* block, so what we have is

1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{1}

If you use the first method, it's pretty much the same idea. But, like I said, here you don't have to remember how to draw it, it's right there in the periodic table. Also, since Al is the first atom in the 3rd row p block, the last orbital, 3p, has 1 electron in it. So you really don't have to keep track of how many electrons you've used by method #2 as you would for #1, which is a good thing when you're dealing with atoms like . . .

Iodine (I)

It's the same as with Al only LONGER. Row 1, no iodine, so we have
1s^{2}. Row 2? Row 3? Nope, no iodine, so to this point we have 1s^{2}, 2s^{2},
2p^{6}, 3p^{2}, and 3p^{6}. Going through the forth row we go
through the first two columns (4s^{2}), then we go through the transition metals
for the first time, which is *3*d^{10}. Remember, the d-block is the row
number minus one. The easiest way to remember how many electrons go into d (and this is
also true for s, p, and f) is that there are ten atoms in each row of transition metals.
So, anyway, we go through 4p, 5s, *4*d and, finally, there's iodine: the *fifth*
atom in 5p. So the whole configuration looks like this:

1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6},
4s^{2}, 3d^{10}, 4p^{6}, 5s^{2}, 4d^{10}, 5p^{5}

Rare (Noble) Gas Configurations: A MUCH Shorter Way to Do This Stuff!

I'm happy to say that chemists, like most people, hate doing tedious stuff and constantly look for easier ways to do things. And, believe it or not, sometimes the short ways actually make sense! A common shorthand for electron configurations is called the rare gas (a.k.a. noble gas) configuration. If you look at all the atoms in a given row, you'll see they all have something in common. They all have the exact same configuration up to the last atom (i.e. the last rare gas) in the previous row. So you take that rare gas atom, place it brackets, and this way you only have to write out the electron configurations for whatever row the atom in question is in. So for aluminum, the rare gas configuration is:

[Ne] 3s^{2}, 3p^{1}

And for iodine it becomes much easier:

[Kr] 5s^{2}, 4d^{10}, 5p^{5}

Also, the rare gas notation for a rare gas is just that atom in
brackets, so for argon it's [Ar], *not* [Ne] 2s^{2}, 2p^{6}.

There's one more way to show electron configurations
and that's using a thing called box orbital notation. The order and hints are the same,
but here each set of orbitals are represented by boxes (so this isn't just a clever name,
eh?). Since we only have one s orbital for every n, we use one box. There and *three*
p orbitals, so we show this with three boxes *joined together*, and so on.

To show that there are electrons in these orbitals, we use arrows.
Now before you start drawing arrows all over the place, there are two things to keep in
mind. First, *Do not pair up electrons unless you have to.* For an s orbital we don't
have a choice, but for all the others you have an electron in every orbital (3 in the p's,
5 in the d's, etc.) before you started pairing up arrow. This brings up the second thing
we have to consider, that *paired electrons are shown by two arrows pointing in opposite
directions.* Remember when we talked about quantum numbers,
I only briefly mentioned m_{s} and said that it was equal to +1/2 or -1/2. We can
show that here by placing one arrow up to show one direction and the down to show the
other.

Examples

Magnesium (Mg)

Vanadium (V)

Again, notice that the order of the orbital didn't change, all we
did was add boxes and arrows to further show how the electrons are placed. Now notice that
in magnesium all of our electrons are paired up. This is an example of a *diamagnetic*
atom. When an atom has one or more unpaired electrons, it's *paramagnetic*. I guess
the easiest way to remember this is the di- means two, so if every box is filled with two
electrons it's diamagnetic.

How does electron configurations change when you deal with ions? Well, the number of electrons change, doesn't it? If you have an anion (negative charge), you add electrons. With cations (positive charge), you take them away. This is exactly what we do with our configurations.

Example

K^{+}

The electron configuration for neutral potassium is

1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}

Since we have a +1 charge we must remove an electron from the
neutral configuration. What you need to remember is that you always start removing
electrons *from the highest shell (n)*. Here it's really no problem, our
highest is n=4, so we remove one electron from the 4s to get

1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}

If we want to show this in rare gas configuration, you got to notice that

this is the same configuration as a rare gas, argon (the hint is that we end in "p^{6}").
So the configuration is [Ar].

Ga^{3+}

Let's just start out with the rare gas configuration. With neutral Ga it's

[Ar]4s^{2}3d^{10}4p^{1}

When we take our three electrons away, you *don't take any out of
3d*! Remember, __highest n__. Again, it's 4, but we have a 4p *and *a
4s. You have to take all of the 4's out before you go into the 3's. We have
three electrons in n=4 so we end up with

[Ar]3d^{10}

This, by the way is called a *pseudo-rare gas configuration*.
You get this anytime you get a configuration like "[R.G.] nd " or
"[R.G.] nd^{10} n-1)f^{14}" (R.G. = rare gas). Notice
there are no s^{2} orbitals after the rare gas. A configuration with
"[R.G.] ns^{2}(n-1)d^{10}" may have all orbitals full, but it is
**not** a *pseudo-rare gas*.

P^{3-}

Let's try an anion, in rare gas form. For neutral, P we get

[Ne]3s^{2}3p^{3}

Here we have to add three electrons. Now you may want to put

[Ne]3s^{2}3p^{6}

WRONG!

Don't forget that when you get to a "p^{6}," we go up to
the next rare gas, so it's actually

[Ar]