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## explain steps for (using periodic table) for electron configuration regarding any nuetral atoms.

Please show me how to use the periodic table to write electron configuration/ orbital notation.

Please show me everything I need to do.

A link with videos is great for me...

Show me how to do this...

Count  across the periodic table from left to right etc. Show me every thing...

I really want instructions to configure any ions now to.

Thanks...

Carie

Something to also keep in mind when you are doing electron configuration is that the electrons will go into the lower energy levels to partially fill the orbitals before they go to the next energy level.  You can't go to the next energy level until the previous orbital is at least partially full. This doesn't come into play with the above example but will for something like Chromium it does. Looking at the periodic table for Chromium it is Atomic Number 24 and is in the "d block" on the periodic table so it will have some electrons in d orbitals. (see this image from wikipedia for a clear picture of the "blocks" http://en.wikipedia.org/wiki/File:Periodic_Table_2.svg).

The complete configuration for chromium is 1s2 2s2 2p6 3s2 3p6 3d5 4s1.  This might be confusing because on the periodic table, you first go through 4s and then 3d so you would expect it to be: 1s2 2s2 2p6 3s2 3p6 3d4 4s2, but that isn't the case.  Let's look at the orbitals of the 3p and 4s shells to really understand this.

3d has 5 available orbitals and can hold 10 electrons:  ____ ____ ____ ____ ____

4s has 2 available orbitals and can hold 4 electrons: ____ ____

In chromium, there are 6 electrons between the 3d and 4s orbitals so there is no way for every orbital to get its own electron.  So you have to start by filling in the 3d since it is one energy level below 4s.  In order for 3d to be stable, each of these orbitals needs to have at least one electron so the electrons will fill one at a time:

3d: _↑_    _↑_    _↑_   _↑_    __

Since the 4s is higher in energy, until the 3d is stable, one of the electrons that you would think was in the 4s orbital, is actually in the 3d orbital.  We have used 5 of our available electrons so there is only one left.  So the 4s actually looks like this:

4s:  _↑_   ___

This same principle applies to any element in the same column as chromium (the 4th column of the d block) or the same column as copper (the 9th column of the d block).  Copper will end up with 3d10 4s1 for the same reason.

If you are ever asked to do this without a periodic table there is a neat trick to remembering the order.  Write down the orbitals in a triangle like this:

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d

7s 7p

Then you can draw an arrow from the top right like this picture: http://www.m2c3.com/c101/chemistry_in_contextx/ch2/simple_electron_configuration/elecfill.gif

That is the order that the electrons fill the orbitals in most cases, except in those cases like the one I described above where the previous energy level has empty orbitals.  Since an s orbital can hold 2 electrons, a p holds up to 6, and a d up to 10, you can do the electron configuration just from knowing the atomic number.

As for ions, you can also determine the electron configuration of ions without doing the neutral atom first but the process is the same as Liza described.  Looking at the periodic table, if you are asked for the +1 ion of an atom, the electron configuration for this ion will be the same as the neutral electron configuration for the atom directly to the left of it on the periodic table.  For the -1, it will be the same as the neutral electron configuration for the atom to the right on the periodic table.

To use the example of Chlorine, the Cl- electron configuration will be the exact same as neutral Argon (Ar) which is 1s2 2s2 2p6 3s2 3p6.

Last paragraph should read "...neutral argon (Ar)..."

Thanks Dick B., I've corrected the mistake!

I have a very interactive site that I used with my honors class.  It will help you build each atom's configuration consecutively.

http://intro.chem.okstate.edu/WorkshopFolder/Electronconfnew.html

Remember, the periodic table is divided into 4 sections to represent sublevels (s,p,d,f), with the rows representing the energy levels.  The placement of the atom on the table identifies the LAST configuration in its electron arrangement.

For example,

the atom Potassium has the last configuration 4s1, (4th row, s-section, 1st atom)

the atom Scandium has the last configuration 3d1 (3rd row, d-section, 1st atom)

******remember the d-sublevel begins on the 3rd energy level

the atom Boron has the last configuration 2p1 (2nd row, p-section, 1st atom)

******remember the p-sublevel begins on the 2nd energy level

Thus, each atom begins with the same configuration and ends in a different configuration.

For example, the atom Hydrogen 1s1

the atom Helium       1s2

the atom Lithium       1s2, 2s1

the atom Berylium      1s2, 2s2

Each atom is its own building and must always begin with a "1st floor"...........1s1

To follow up on Liza's answer: Once you have the electron configuration for the neutral atom, you add or subtract electrons (number based on charge, so a -1 charge would add an electron, a +2 charge would subtract 2 electrons). Remember that electrons have negative charges, so if you add electrons, the ionic charge becomes more negative; if you remove electrons, the charge becomes more positive. Another hint: ion usually (but not always) tend toward full electron orbitals, so ou tend to end with p6 or possibly s2.

Hey!

Videos are awesome for this, I agree. I'm just not sure if I'd violate some kind of rule (which I'm sure I would) if I posted a bunch of video links. Let's try to do it this way. So, let's pretend that we are writing a configuration for Cl, which is # 17 in the table. To get to Cl, if we start on top, we need to move "through" the first row, which only has s elements (something you need to remember). So we have 1s2 --  1 for the first row, and two for the two elements that there are. Then, going back to the left in the second row, we have 2 s elements and 6 p elements till the end of the row. So, 2s2 2p6. Then, since Cl is in the third row, we need to work through the third row elements until we "bump" into Cl. So, 2 s elements in the third row, and 5 more p elements until we are at Cl (we count Cl as one). So, 3s2 3p5. So, the full thing is 1s2 2s2 2p6 3s2 3s5.Hope this helps!