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How to Write an Electron Configuration for an - ChemistryforMrL

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How to Write an Electron Configuration for an Atom of Any Element
Being able to write electron configurations of an atom is necessary to ace a chemistry class, but writing them can seem hard, but they are actually easy, if you know the basics.
What you Need: A Periodic Table of Elements
Find out how many electrons the atom has. On the periodic table, the atomic number is the number of protons of the atom, and thus equals the number of electrons in an atom with zero charge. Using the method you learned in the "Element-ary Code" Game, follow down through the table, writing the configuration as you go.
For example: Look at magnesium. It's number 12. So, you have to have an electron configuration that shows 12 electrons (these are the superscripts in the configuration).
Remember, you always start at hydrogen. So, that's in row one.
Write "1"
Look at the block you are in. If it is the alkali metals, alkaline earth metals, or hydrogen and helium, it is "s".
Now you have 1s.
Since on the first row, you went through 2 elements, you write a superscript "2" to indicate you have placed two electrons in 1s.
Now, you have 1s2.
That's the end of Period 1, so now you move to period 2 (remember you are making your way to Magnesium).
You are now in Period 2, so you write that.
Look where you are in the period (alkali/alkaline earth)... write an "s"
Since you went through 2 more elements, you write another superscript 2
You aren't done with Period 2 so keep going across the period. Now you have reached the "p block" these are the columns 3A - 8A (remembering that helium isn't part of it). You changed "blocks" so you must write 2 again. Then, write p.
You are going all the way to the end of the block (still on your way to Magnesium). You pass through 6 elements, so write a superscript "6" to place those 6 electrons.
You finally made it to the row with magnesium! It's in period 3. You are in the "s" block again, and it's two elements to magnesium. So, what do you think you might add? That's right, 3s2
If you add the superscripts, 2+2+6+2, that equals 12... hey, that's magnesium's atomic number and thus the number of electrons it has!
All the other ones we will be doing in class are the same. You just have to remember that the "d block" is one energy level behind. So, when you get to Scandium (#21), that is 3d, not 4d.
Some for you to try. No abbreviated noble gas notation for this exercise!
1. hydrogen
2. carbon
3. nitrogen
4. helium
5. calcium
6. aluminum
7. iron (careful... 3d, not 4d)
8. zinc
9. sodium
10. argon
11. vanadium
12. lithium
13. chlorine
14. silicon
15. cobalt
16. cadmium (Cd)
17. bromine
18. strontium
19. lead
20. fluorine
* The number following the letter is actually a superscript, so don't make that mistake on a test. * To find the atomic number of the atom when it is in electron configuration form, just add up all of the numbers that follow the letters (S, P, D, and F). * You can also figure out the electron configuration of an atom without remembering the order of the orbital sets. Just look at the periodic table. Have you ever wondered why it's shaped like that? Take the top right element (Helium) and shift it all the way to the left. The two left columns are the S orbitals sets. The middle 10 columns are the D orbital sets. and the right 6 columns are the P orbital sets. The two rows on the bottom are the F orbital sets. Now, all you need to do is go horizontally across the periodic table until you arrive at the atom you are calculating. So, if you were looking for scandium, you first go through the S orbital sets, which are hydrogen and helium. Write down 1s2 because you are on the first row and you passed through all of the first S orbital set without arriving at scandium. Now, go to the second row. Go through the 2S orbital, which are lithium and beryllium. Write down 2s2. Now, go through the 2P orbital set and continue through all of them until you arrive at your atom. * Beware, after you go through the 6S orbital set, you go to the 4F orbital set, instead of 5D. Same thing goes for 7S (you go through 5F before going to 6D). * Writing long electron configurations can be avoided by writing them in their noble gas configurations. Simply find the last symbol containing p6 (such as 3p6 or 5p6) and add up all of the numbers following the letters of every symbol before and including the p6 orbital set. Then, using the sum of the numbers, locate the element with the atomic number equal to the sum you just calculated. It should appear at the very right of the periodic table. That's called a noble gas. Now, just remove all of the symbols that you added up and put in the noble gas' symbol in brackets. So, for an antimony atom, the noble gas configuration would be [Kr] 5s2 4d10 5p3. Notice that you don't add up the 5s2 and 4d10 because they are after 4p6, which you do add. * You can also write an element's electron configuration by just writing the valence configuration, which is the last S and P orbital set. So, the valence configuration of an antimony atom would be 5s2 5p3. * When the atom is an ion, it means that the number of protons does not equal the number of electrons. The charge of the atom will them be displayed at the top right (usually) corner of the chemical symbol. So, an antimony atom with charge +2 has an electron configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p1. Notice that the 5p3 changed into a 5p1. Be careful when the configuration of an uncharged atom ends in anything but an S and P orbital set. When you take away electrons, you can only take them away from the valence orbitals (the S and P orbitals). So, if a configuration ends in 4s2 3d7, and the atom gains a charge of +2, then the configuration would change to end with 4s0 3d7. Notice that 3d7 does not change. Instead, the S orbital electrons are lost. * There are circumstances when an electron needs to be "promoted." When an orbital set is one electron away from being half occupied or completely occupied, remove one electron from the nearest S or P orbital set and move it to the orbital set that needs the electron. * Every atom desires to be stable, and the most stable configurations have full S and P (s2 and p6) orbital sets. The noble gases have this configuration, which is why they are non-reactive and are on the right side of the periodic table. So, if a configuration ends in 3p4, it only needs two more electrons to become stable (losing six, including the S orbital set's electrons, takes more energy, so losing four is easier). And if a configuration ends in 4d3, it only needs to lose three electrons to reach a stable state. 
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